Curves, surfaces and manifolds

Differential geometry of curves and surfaces, and smooth manifolds.

In this chapter 17

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Curves and surfaces
Smooth manifolds

Curves and surfaces

Theorem

Curves and the Frenet-Serret formulas

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Attaches to a space curve of nonvanishing curvature an orthonormal moving frame $(\mathbf{t}, \mathbf{n}, \mathbf{b})$ , derives the three Frenet-Serret formulas that express its rate of turning through the curvature $\kappa$ and the torsion $\tau$ alone, and states the fundamental theorem that $\kappa$ and $\tau$ determine the curve up to a rigid motion, with the uniqueness half proved here.

Prerequisites: Partial derivatives and the chain rule, Vector algebra and index notation, Line integrals and conservative fields, Existence and uniqueness of solutions

Definition(Arc length, the Frenet trihedron, curvature and torsion)AdaptedAdapted from do Carmo (2016), §1-5, pp. 17–20 and Remark 2, pp. 22–23; Riley, Hobson & Bence (2006), §10.3, Eq. (10.12), p. 341 and pp. 342–343

Let $I \subseteq \mathbb{R}$ be an open interval. A regular parametrized curve is a map $\mathbf{r} : I \to \mathbb{R}^3$ with continuous derivatives of every order and $\mathbf{r}'(u) \ne \mathbf{0}$ for all $u$ .

Its arc length measured from $u_0 \in I$ is $s(u) = \int_{u_0}^{u} \lvert \mathbf{r}'(\tilde u)\rvert\, d\tilde u$ (line integrals). Since $s'(u) = \lvert\mathbf{r}'(u)\rvert > 0$ , the function $s$ has a smooth inverse $u(s)$ , and $\boldsymbol{\gamma}(s) = \mathbf{r}(u(s))$ traces the same set with $\lvert \boldsymbol{\gamma}'(s)\rvert = 1$ . We call $\boldsymbol{\gamma}$ the unit-speed (or arc-length) parametrization, and write $\dot{\phantom{x}} = d/ds$ .
The unit tangent is $\mathbf{t}(s) = \dot{\boldsymbol{\gamma}}(s)$ . Differentiating $\mathbf{t}\cdot\mathbf{t} = 1$ gives $\dot{\mathbf{t}}\cdot\mathbf{t} = 0$ , so $\dot{\mathbf{t}} \perp \mathbf{t}$ . The curvature is $\kappa(s) = \lvert \dot{\mathbf{t}}(s)\rvert = \lvert \ddot{\boldsymbol{\gamma}}(s)\rvert \;\ge\; 0 ,$ and $\rho = 1/\kappa$ is the radius of curvature.
Standing hypothesis. From here on $\kappa(s) > 0$ for every $s \in I$ . (A point with $\ddot{\boldsymbol{\gamma}} = \mathbf{0}$ is what do Carmo calls a singular point of order $1$ ; the frame below is not defined there.) The principal normal is then the unit vector $\mathbf{n}(s)$ fixed by $\dot{\mathbf{t}}(s) = \kappa(s)\,\mathbf{n}(s),$ and the binormal is $\mathbf{b}(s) = \mathbf{t}(s)\times\mathbf{n}(s)$ . The ordered triple $(\mathbf{t}, \mathbf{n}, \mathbf{b})$ is an orthonormal right-handed basis of $\mathbb{R}^3$ at each $s$ , the Frenet trihedron.
The plane through $\boldsymbol{\gamma}(s)$ spanned by $\mathbf{t}$ and $\mathbf{n}$ is the osculating plane; the plane spanned by $\mathbf{n}$ and $\mathbf{b}$ is the normal plane, and the one spanned by $\mathbf{t}$ and $\mathbf{b}$ the rectifying plane.
The torsion $\tau(s)$ is the number defined by $\dot{\mathbf{b}}(s) = -\tau(s)\,\mathbf{n}(s) ,$ which the statement below shows to be well defined. Where $\tau \ne 0$ , $\sigma = 1/\tau$ is the radius of torsion.

Sign convention. We follow Riley, Hobson and Bence, who set $d\hat b/ds = -\tau\hat n$ . do Carmo defines torsion by $b' = \tau n$ , with no minus sign, and warns in the same paragraph that authors differ; his $\tau$ is therefore the negative of ours. Every formula below is written in our convention, and the two places where the difference shows (the third Frenet formula and the sign in the arbitrary-parameter formula for $\tau$ ) are flagged again in the remark.

Theorem(The Frenet-Serret formulas and the fundamental theorem)AdaptedAdapted from do Carmo (2016), §1-5, pp. 19–20, where the binormal and normal derivatives and the Frenet formulas are obtained, and the fundamental theorem stated on p. 20; Riley, Hobson & Bence (2006), §10.3, Eqs. (10.13)–(10.16), pp. 342–343

Let $\boldsymbol{\gamma} : I \to \mathbb{R}^3$ be a unit-speed curve with $\kappa(s) > 0$ throughout, and let $(\mathbf{t}, \mathbf{n}, \mathbf{b})$ be its Frenet trihedron.

(Torsion is well defined.) $\dot{\mathbf{b}}(s)$ is a scalar multiple of $\mathbf{n}(s)$ , so there is exactly one function $\tau$ with $\dot{\mathbf{b}} = -\tau\mathbf{n}$ , and $\tau = -\mathbf{n}\cdot\dot{\mathbf{b}}$ is smooth.
(Frenet-Serret formulas.) $\dot{\mathbf{t}} = \kappa\,\mathbf{n}, \qquad \dot{\mathbf{n}} = -\kappa\,\mathbf{t} + \tau\,\mathbf{b}, \qquad \dot{\mathbf{b}} = -\tau\,\mathbf{n} .$ Equivalently, the matrix of the map $(\mathbf{t},\mathbf{n},\mathbf{b}) \mapsto (\dot{\mathbf{t}},\dot{\mathbf{n}},\dot{\mathbf{b}})$ in the frame itself is skew-symmetric with entries $\kappa$ and $\tau$ only.
(Plane curves.) If $\kappa > 0$ throughout, then $\tau \equiv 0$ on $I$ if and only if $\boldsymbol{\gamma}(I)$ lies in a plane.
(do Carmo's fundamental theorem for curves.) Let $\kappa(s) > 0$ and $\tau(s)$ be smooth on $I$ . Then there is a unit-speed curve $\boldsymbol{\gamma} : I \to \mathbb{R}^3$ whose arc length is $s$ , whose curvature is $\kappa$ and whose torsion is $\tau$ . Any other curve $\bar{\boldsymbol{\gamma}}$ with the same $\kappa$ and $\tau$ differs from it by a rigid motion: there are an orthogonal linear map $R$ of $\mathbb{R}^3$ with $\det R = +1$ and a vector $\mathbf{c}$ with $\bar{\boldsymbol{\gamma}} = R\circ\boldsymbol{\gamma} + \mathbf{c}$ .

do Carmo states part 4 with his own sign for $\tau$ . The statement is unaffected by the convention, because the data $\tau$ ranges over all smooth real-valued functions either way: a curve realizing $(\kappa, \tau)$ in his convention realizes $(\kappa, -\tau)$ in ours.

Idea of the proofOriginal

Each of the three frame vectors has constant length $1$ , so each derivative is orthogonal to the vector itself; that alone forces the matrix of derivatives in the frame to be skew-symmetric, leaving only three independent entries. Two of them are named $\kappa$ and $\tau$ by definition. The third, the $\mathbf{t}$ -component of $\dot{\mathbf{b}}$ , is not a new quantity: differentiating $\mathbf{b}\cdot\mathbf{t} = 0$ and using $\dot{\mathbf{t}} = \kappa\mathbf{n}$ makes it vanish. So the local geometry is controlled by exactly two scalar functions. The uniqueness half of the fundamental theorem then follows by a stability argument: move one curve by a rigid motion until the two frames agree at a single point, and watch the total squared discrepancy $\lvert\mathbf{t}-\bar{\mathbf{t}}\rvert^2 + \lvert\mathbf{n}-\bar{\mathbf{n}}\rvert^2 + \lvert\mathbf{b}-\bar{\mathbf{b}}\rvert^2$ ; the Frenet formulas make its derivative cancel identically, so it stays at its initial value $0$ .

ProofAdaptedAdapted from do Carmo (2016), §1-5, pp. 19–20 and pp. 21–22 (uniqueness)

Part 1. $\mathbf{b}$ is a unit vector, so differentiating $\mathbf{b}\cdot\mathbf{b} = 1$ gives $\dot{\mathbf{b}}\cdot\mathbf{b} = 0$ : the derivative has no $\mathbf{b}$ -component. For the $\mathbf{t}$ -component, differentiate $\mathbf{b} = \mathbf{t}\times\mathbf{n}$ by the product rule for the cross product,

\dot{\mathbf{b}} = \dot{\mathbf{t}}\times\mathbf{n} + \mathbf{t}\times\dot{\mathbf{n}} = \kappa\,(\mathbf{n}\times\mathbf{n}) + \mathbf{t}\times\dot{\mathbf{n}} = \mathbf{t}\times\dot{\mathbf{n}},

where $\dot{\mathbf{t}} = \kappa\mathbf{n}$ was used (this is where the hypothesis $\kappa > 0$ enters, since it is what makes $\mathbf{n}$ and hence $\mathbf{b}$ exist). A cross product with $\mathbf{t}$ is orthogonal to $\mathbf{t}$ , so $\dot{\mathbf{b}}\cdot\mathbf{t} = 0$ . Having no $\mathbf{b}$ - and no $\mathbf{t}$ -component in an orthonormal basis, $\dot{\mathbf{b}}$ is a multiple of $\mathbf{n}$ . Writing that multiple as $-\tau$ defines $\tau$ uniquely, and $\tau = -\mathbf{n}\cdot\dot{\mathbf{b}}$ is smooth because $\mathbf{n}$ and $\mathbf{b}$ are.

Part 2. The first and third formulas are the definitions of $\kappa$ and $\tau$ . For the middle one, use $\mathbf{n} = \mathbf{b}\times\mathbf{t}$ (valid because the frame is right-handed and orthonormal) and differentiate:

\dot{\mathbf{n}} = \dot{\mathbf{b}}\times\mathbf{t} + \mathbf{b}\times\dot{\mathbf{t}} = (-\tau\mathbf{n})\times\mathbf{t} + \mathbf{b}\times(\kappa\mathbf{n}) = \tau\,(\mathbf{t}\times\mathbf{n}) + \kappa\,(\mathbf{b}\times\mathbf{n}) = \tau\,\mathbf{b} - \kappa\,\mathbf{t},

using $\mathbf{t}\times\mathbf{n} = \mathbf{b}$ and $\mathbf{b}\times\mathbf{n} = -\mathbf{t}$ . So no third scalar appears.

Part 3. Suppose $\boldsymbol{\gamma}(I)$ lies in a plane with unit normal $\mathbf{m}$ . Then $(\boldsymbol{\gamma}-\boldsymbol{\gamma}(s_0))\cdot\mathbf{m} \equiv 0$ ; differentiating twice gives $\mathbf{t}\cdot\mathbf{m} = 0$ and $\kappa\,\mathbf{n}\cdot\mathbf{m} = 0$ , so, since $\kappa > 0$ , both $\mathbf{t}$ and $\mathbf{n}$ are orthogonal to $\mathbf{m}$ , whence $\mathbf{b} = \pm\mathbf{m}$ is constant and $\tau = 0$ .

Conversely let $\tau \equiv 0$ . Then $\dot{\mathbf{b}} = \mathbf{0}$ , so $\mathbf{b} \equiv \mathbf{b}_0$ is a fixed unit vector, and

\frac{d}{ds}\bigl[(\boldsymbol{\gamma}(s)-\boldsymbol{\gamma}(s_0))\cdot\mathbf{b}_0\bigr] = \mathbf{t}(s)\cdot\mathbf{b}_0 = 0 .

The bracket vanishes at $s = s_0$ , so it vanishes identically: the curve lies in the plane through $\boldsymbol{\gamma}(s_0)$ normal to $\mathbf{b}_0$ .

Part 4, existence. Fix $s_0 \in I$ . Regard the nine components of $(\mathbf{t},\mathbf{n},\mathbf{b})$ as unknowns in the linear system of ordinary differential equations given by the three Frenet formulas, with the coefficients $\kappa(s)$ and $\tau(s)$ as given data and with an arbitrary right-handed orthonormal triple as initial value at $s_0$ . The coefficients are smooth, so the system has a unique solution on all of $I$ (existence and uniqueness for ODEs; the system is linear, so the solution does not escape in finite time). The skew-symmetry of the coefficient matrix makes the six inner products $\mathbf{t}\cdot\mathbf{t}, \mathbf{t}\cdot\mathbf{n}, \dots$ constant, so the solution stays orthonormal, and $\boldsymbol{\gamma}(s) = \int_{s_0}^{s}\mathbf{t}$ is then a unit-speed curve with the prescribed curvature and torsion. do Carmo states this part and defers its proof to the appendix of his Chapter 4; we record it as his statement.

Part 4, uniqueness. Arc length, curvature and torsion are built from inner and cross products of derivatives, and a rigid motion $\mathbf{x}\mapsto R\mathbf{x}+\mathbf{c}$ with $\det R = +1$ leaves all of these unchanged; so we may move $\bar{\boldsymbol{\gamma}}$ by such a motion and assume $\bar{\boldsymbol{\gamma}}(s_0) = \boldsymbol{\gamma}(s_0)$ and $\bar{\mathbf{t}}(s_0) = \mathbf{t}(s_0)$ , $\bar{\mathbf{n}}(s_0) = \mathbf{n}(s_0)$ , $\bar{\mathbf{b}}(s_0) = \mathbf{b}(s_0)$ (the motion taking one right-handed orthonormal frame to another has positive determinant, which is why no reflection is needed). Put

D(s) = \lvert\mathbf{t}-\bar{\mathbf{t}}\rvert^2 + \lvert\mathbf{n}-\bar{\mathbf{n}}\rvert^2 + \lvert\mathbf{b}-\bar{\mathbf{b}}\rvert^2 .

Both frames obey the Frenet formulas with the same $\kappa$ and $\tau$ , by hypothesis. Hence

\tfrac12 \dot D = (\mathbf{t}-\bar{\mathbf{t}})\cdot(\dot{\mathbf{t}}-\dot{\bar{\mathbf{t}}}) + (\mathbf{n}-\bar{\mathbf{n}})\cdot(\dot{\mathbf{n}}-\dot{\bar{\mathbf{n}}}) + (\mathbf{b}-\bar{\mathbf{b}})\cdot(\dot{\mathbf{b}}-\dot{\bar{\mathbf{b}}}) .

Substituting the three formulas, the first term is $\kappa\,(\mathbf{t}-\bar{\mathbf{t}})\cdot(\mathbf{n}-\bar{\mathbf{n}})$ ; the third is $-\tau\,(\mathbf{b}-\bar{\mathbf{b}})\cdot(\mathbf{n}-\bar{\mathbf{n}})$ ; the second is $-\kappa\,(\mathbf{n}-\bar{\mathbf{n}})\cdot(\mathbf{t}-\bar{\mathbf{t}}) + \tau\,(\mathbf{n}-\bar{\mathbf{n}})\cdot(\mathbf{b}-\bar{\mathbf{b}})$ . The four terms cancel in pairs, so $\dot D \equiv 0$ . Since $D(s_0) = 0$ and $D \ge 0$ , we get $D \equiv 0$ , that is $\mathbf{t} \equiv \bar{\mathbf{t}}$ (and likewise for $\mathbf{n}, \mathbf{b}$ ). Finally $\frac{d}{ds}(\boldsymbol{\gamma}-\bar{\boldsymbol{\gamma}}) = \mathbf{t}-\bar{\mathbf{t}} = \mathbf{0}$ , and the difference vanishes at $s_0$ , so $\boldsymbol{\gamma} \equiv \bar{\boldsymbol{\gamma}}$ . $\square$

Remark(Where each hypothesis is used)Original

$\kappa > 0$ everywhere. This is the hypothesis that does all the work. It is what makes $\mathbf{n} = \dot{\mathbf{t}}/\kappa$ , and with it $\mathbf{b}$ and the osculating plane, exist at all; it is used again in Part 3 to pass from $\kappa\,\mathbf{n}\cdot\mathbf{m}=0$ to $\mathbf{n}\cdot\mathbf{m}=0$ . Without it the frame can jump. do Carmo's Exercise 10 (p. 26) builds a smooth curve, two half-branches of $z = e^{-1/t^2}$ glued in perpendicular planes at $t=0$ , which is regular and has $\kappa(0) = 0$ (his part (b) asks one to show that $k$ is nonzero except at $t = 0$ and $t = \pm\sqrt{2/3}$ , so the vanishing is not confined to the join), yet whose osculating planes have different limits from the two sides; there $\tau$ can be set to $0$ although the curve is not planar, so Part 3 genuinely needs $\kappa>0$ .
Unit speed. Only here do $\lvert\ddot{\boldsymbol{\gamma}}\rvert$ and $\kappa$ coincide, and only here is $\dot{\mathbf{t}}\perp\mathbf{t}$ automatic. For a general parameter one must either reparametrize or use the formulas of the method below; $\lvert\mathbf{r}''(u)\rvert$ is not the curvature, because it also contains the change of speed.
Regularity $\mathbf{r}' \ne \mathbf{0}$ . This is what makes $s(u)$ strictly increasing and so invertible; a curve with a cusp, such as $u \mapsto (u^2, u^3)$ at $u = 0$ , has no unit-speed reparametrization through that point.
Right-handedness. $\mathbf{n} = \mathbf{b}\times\mathbf{t}$ in Part 2 and the "no reflection needed" step in the uniqueness proof both use that $(\mathbf{t},\mathbf{n},\mathbf{b})$ is positively oriented. A reflection of $\mathbb{R}^3$ preserves $\kappa$ but reverses the sign of $\tau$ , which is why the fundamental theorem asks for $\det R = +1$ .
Convention. In do Carmo's sign convention the third formula reads $b' = \tau n$ and the middle one $n' = -kt - \tau b$ ; his arbitrary-parameter formula (his Exercise 12c) carries a minus sign that ours does not. Mixing the two conventions inside one calculation flips the handedness of the answer.

Method(Curvature and torsion from an arbitrary parameter)AdaptedAdapted from do Carmo (2016), §1-5, Exercise 12, pp. 26–27; Riley, Hobson & Bence (2006), §10.3, Eq. (10.12), p. 341

Given $\mathbf{r}(u)$ regular, with $\mathbf{r}', \mathbf{r}'', \mathbf{r}'''$ the derivatives with respect to $u$ :

Speed and arc length. $v(u) = \lvert\mathbf{r}'(u)\rvert$ , and $s = \int_{u_0}^{u} v$ . Never reparametrize unless the integral is elementary; steps 2–4 do not need it.
Curvature. $\kappa = \frac{\lvert \mathbf{r}'\times\mathbf{r}''\rvert}{\lvert\mathbf{r}'\rvert^{3}} .$
Torsion (requires $\mathbf{r}'\times\mathbf{r}'' \ne \mathbf{0}$ , i.e. $\kappa \ne 0$ ): $\tau = \frac{(\mathbf{r}'\times\mathbf{r}'')\cdot\mathbf{r}'''}{\lvert\mathbf{r}'\times\mathbf{r}''\rvert^{2}} .$ In do Carmo's convention the right-hand side carries a minus sign.
Frame. $\mathbf{t} = \mathbf{r}'/v$ , $\mathbf{b} = (\mathbf{r}'\times\mathbf{r}'')/\lvert\mathbf{r}'\times\mathbf{r}''\rvert$ , $\mathbf{n} = \mathbf{b}\times\mathbf{t}$ .
Check. $\kappa$ must come out $\ge 0$ ; a negative value means an arithmetic slip. For a plane curve step 3 must give $\tau = 0$ .

Why steps 2 and 3 are right: write $\mathbf{r}' = v\,\mathbf{t}$ and differentiate with the chain rule $d/du = v\,d/ds$ , giving $\mathbf{r}'' = v'\mathbf{t} + \kappa v^2\mathbf{n}$ and then $\mathbf{r}''' = (v''-\kappa^2v^3)\mathbf{t} + (3\kappa v v' + \kappa' v^2)\mathbf{n} + \kappa\tau v^3\mathbf{b}$ . Hence $\mathbf{r}'\times\mathbf{r}'' = \kappa v^3\,\mathbf{b}$ , which gives step 2 on taking lengths, and $(\mathbf{r}'\times\mathbf{r}'')\cdot\mathbf{r}''' = \kappa^2\tau v^6 = \tau\lvert\mathbf{r}'\times\mathbf{r}''\rvert^2$ , which gives step 3.

Example(A conical spiral)Original

Take the curve that winds up the cone $z^2 = x^2+y^2$ with exponentially growing radius,

\mathbf{r}(u) = \bigl(e^{u}\cos u,\; e^{u}\sin u,\; e^{u}\bigr), \qquad u \in \mathbb{R}.

Step 1: derivatives. Writing $C = \cos u$ , $S = \sin u$ and factoring out $e^u$ each time,

\mathbf{r}' = e^{u}(C-S,\; S+C,\; 1), \quad \mathbf{r}'' = e^{u}(-2S,\; 2C,\; 1), \quad \mathbf{r}''' = e^{u}(-2S-2C,\; 2C-2S,\; 1).

Step 2: speed and arc length. $\lvert\mathbf{r}'\rvert^2 = e^{2u}\bigl[(C-S)^2+(S+C)^2+1\bigr] = e^{2u}(1+1+1) = 3e^{2u}$ because $(C-S)^2+(C+S)^2 = 2C^2+2S^2 = 2$ . So $v(u) = \sqrt{3}\,e^{u}$ , and the arc length from $u=0$ to $u=1$ is

s = \int_0^1 \sqrt{3}\,e^{u}\,du = \sqrt{3}\,(e-1) = 2.97615\ldots

Step 3: the cross product. Expanding the determinant and taking the common factor $e^{2u}$ out of both rows,

\mathbf{r}'\times\mathbf{r}'' = e^{2u}\bigl(\,S-C,\;\; -(C+S),\;\; 2\,\bigr),

because the third entry is $(C-S)(2C) - (S+C)(-2S) = 2C^2 - 2SC + 2S^2 + 2SC = 2$ . Its squared length is

\lvert\mathbf{r}'\times\mathbf{r}''\rvert^2 = e^{4u}\bigl[(S-C)^2 + (C+S)^2 + 4\bigr] = e^{4u}\bigl[(1-2SC)+(1+2SC)+4\bigr] = 6\,e^{4u},

so $\lvert\mathbf{r}'\times\mathbf{r}''\rvert = \sqrt{6}\,e^{2u}$ . As a cross-check, the Lagrange identity $\lvert\mathbf{a}\times\mathbf{b}\rvert^2 = \lvert\mathbf{a}\rvert^2\lvert\mathbf{b}\rvert^2 - (\mathbf{a}\cdot\mathbf{b})^2$ (vector algebra) with $\lvert\mathbf{r}''\rvert^2 = e^{2u}(4S^2+4C^2+1) = 5e^{2u}$ and $\mathbf{r}'\cdot\mathbf{r}'' = e^{2u}\bigl[-2S(C-S)+2C(S+C)+1\bigr] = 3e^{2u}$ gives $3e^{2u}\cdot 5e^{2u} - (3e^{2u})^2 = 6e^{4u}$ , the same answer.

Step 4: curvature. $\lvert\mathbf{r}'\rvert^3 = (\sqrt3\,e^{u})^3 = 3\sqrt{3}\,e^{3u}$ , so

\kappa(u) = \frac{\sqrt{6}\,e^{2u}}{3\sqrt{3}\,e^{3u}} = \frac{\sqrt{2}}{3}\,e^{-u}.

Step 5: torsion. Pairing $\mathbf{r}'\times\mathbf{r}'' = e^{2u}\bigl(S-C,\,-(C+S),\,2\bigr)$ with $\mathbf{r}''' = e^{u}\bigl(-2(S+C),\,2(C-S),\,1\bigr)$ ,

(\mathbf{r}'\times\mathbf{r}'')\cdot\mathbf{r}''' = e^{3u}\bigl[-2(S-C)(S+C) - 2(C+S)(C-S) + 2\bigr] = e^{3u}\bigl[-2(S^2-C^2) - 2(C^2-S^2) + 2\bigr] = 2\,e^{3u},

the two quadratic terms cancelling exactly. Hence

\tau(u) = \frac{2e^{3u}}{6e^{4u}} = \frac{1}{3}\,e^{-u}.

Step 6: the numbers. At $u = 0$ the curve passes through $(1,0,1)$ with

\kappa(0) = \frac{\sqrt2}{3} = 0.4714045\ldots, \qquad \tau(0) = \frac13 = 0.3333\ldots, \qquad \frac{\kappa}{\tau} = \sqrt2 \text{ (constant)}.

The ratio $\kappa/\tau$ being constant says the curve is a generalized helix: its tangent makes a fixed angle with the cone's axis, here $\arccos\bigl(1/\sqrt3\bigr) = 54.7356^\circ$ , since $\mathbf{t}\cdot\hat{\mathbf{z}} = 1/\sqrt3$ . Both $\kappa$ and $\tau$ decay like $e^{-u}$ : the spiral straightens as it opens out, exactly as the self-similarity of the cone demands.

Application(Kinematics, track design and elastic rods)Original

Acceleration. For a particle at $\mathbf{r}(t)$ with speed $v = ds/dt$ , the chain rule and the first Frenet formula give

\mathbf{a} = \frac{d}{dt}\,(v\,\mathbf{t}) = \dot v\,\mathbf{t} + \kappa v^2\,\mathbf{n} = \dot v\,\mathbf{t} + \frac{v^2}{\rho}\,\mathbf{n}.

The tangential part changes the speed, the normal part is the centripetal acceleration, and the binormal part is zero whatever the path: an acceleration can never point out of the osculating plane. This decomposition is how banking angles, $g$ -loads on a roller coaster and the lateral force on a vehicle in a bend are computed.

Transition curves. Because the lateral force is $mv^2\kappa$ , a track joining a straight ( $\kappa=0$ ) directly to a circular arc ( $\kappa = 1/R$ ) subjects a passenger to a jump in force. Railways and highways therefore insert a clothoid, the plane curve with $\kappa$ proportional to $s$ , so that at constant speed the lateral force ramps up linearly and the jerk stays bounded. That such a curve exists and is unique up to a rigid motion is precisely the plane case of the fundamental theorem.

Charged particles. A particle of charge $q$ and mass $m$ moving with speed $v$ in a uniform field $\mathbf{B}$ follows a circular helix, the unique curve with constant $\kappa$ and constant $\tau$ . Write $\alpha$ for the pitch angle between $\mathbf{v}$ and $\mathbf{B}$ and $\omega = qB/m$ for the gyrofrequency. The orbit is $\bigl(a\cos\omega t,\, a\sin\omega t,\, v\cos\alpha\,t\bigr)$ with gyroradius $a = v\sin\alpha/\omega = mv\sin\alpha/(qB)$ , and the formulas of the method give the constants

\kappa = \frac{a\omega^2}{v^2} = \frac{\omega\sin\alpha}{v} = \frac{qB\sin\alpha}{mv}, \qquad \tau = \frac{\omega\cos\alpha}{v} = \frac{qB\cos\alpha}{mv} .

So the radius of curvature is not the gyroradius: $\rho = 1/\kappa = a/\sin^2\alpha$ , the two agreeing only for a purely perpendicular orbit, $\alpha = \pi/2$ . (At $\alpha = 60^\circ$ with $v = \omega = 1$ : $a = 0.8660$ , $1/a = 1.1547$ , but $\kappa = 0.8660$ .) Magnetic-mirror and tokamak confinement arguments are usually phrased in exactly these two constants.

Rods and polymers. In Kirchhoff's theory of a thin elastic rod, the strain of the centreline is measured by $\kappa$ and $\tau$ relative to their unstressed values, and the elastic energy is a quadratic form in them. The same two functions are the standard coarse-grained description of double-stranded DNA, where bending stiffness couples to $\kappa$ and the supercoiling that packs the molecule into a cell couples to $\tau$ .

Definition

Regular surfaces and the first fundamental form

Adapted

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Defines a regular surface in $\mathbb{R}^3$ by charts, proves that the tangent vectors at a point fill a two-dimensional plane spanned by $\mathbf{x}_u$ and $\mathbf{x}_v$ , and introduces the first fundamental form $I_p(w) = \lvert w\rvert^2$ with coefficients $E, F, G$ , from which lengths of curves, angles between them and areas of regions are computed without further reference to the ambient space.

Prerequisites: Surface integrals and flux, Partial derivatives and the chain rule, Implicit and inverse function theorems

Definition(Regular surface, tangent plane, first fundamental form)AdaptedAdapted from do Carmo (2016), §2-2, Def. 1, pp. 54–55, §2-4, pp. 86–87 and p. 89, and §2-5, Def. 1, pp. 94–95; Lee (2013), Ch. 13, pp. 327–329

Regular surface. A set $S \subseteq \mathbb{R}^3$ is a regular surface if for every $p \in S$ there are an open set $V \subseteq \mathbb{R}^3$ containing $p$ , an open set $U \subseteq \mathbb{R}^2$ , and a map $\mathbf{x} : U \to V \cap S$ onto $V \cap S$ such that

$\mathbf{x}$ is smooth: writing $\mathbf{x}(u,v) = (x(u,v), y(u,v), z(u,v))$ , all partial derivatives of all orders of the three components exist and are continuous on $U$ ;
$\mathbf{x}$ is a homeomorphism onto $V \cap S$ , that is, $\mathbf{x}^{-1} : V\cap S \to U$ exists and is continuous;
(regularity) for every $q \in U$ the differential $d\mathbf{x}_q : \mathbb{R}^2 \to \mathbb{R}^3$ is injective.

Such an $\mathbf{x}$ is a parametrization (or local coordinate system) at $p$ , and $V \cap S$ a coordinate neighbourhood. Writing $\mathbf{x}_u = \partial\mathbf{x}/\partial u$ and $\mathbf{x}_v = \partial\mathbf{x}/\partial v$ , the matrix of $d\mathbf{x}_q$ has columns $\mathbf{x}_u(q)$ and $\mathbf{x}_v(q)$ , so condition 3 says exactly that these two vectors are linearly independent, equivalently $\mathbf{x}_u\times\mathbf{x}_v \ne \mathbf{0}$ .

do Carmo requires derivatives of all orders in condition 1, remarking (p. 90) that in practice one rarely needs more than four, and that the blanket $C^\infty$ hypothesis is adopted only to avoid tracking the minimum smoothness case by case. We keep his convention.

Tangent plane. A tangent vector to $S$ at $p$ is a vector $\boldsymbol{\alpha}'(0)$ for some smooth curve $\boldsymbol{\alpha} : (-\varepsilon,\varepsilon) \to S$ with $\boldsymbol{\alpha}(0) = p$ . By the statement below these vectors form a two-dimensional subspace $T_p S \subset \mathbb{R}^3$ , the tangent plane, and $\{\mathbf{x}_u(q), \mathbf{x}_v(q)\}$ , with $q = \mathbf{x}^{-1}(p)$ , is a basis of it, the basis associated with $\mathbf{x}$ . A parametrization also singles out one of the two unit vectors orthogonal to $T_pS$ ,

N = \frac{\mathbf{x}_u\times\mathbf{x}_v}{\lvert\mathbf{x}_u\times\mathbf{x}_v\rvert}.

First fundamental form. Since $T_pS \subset \mathbb{R}^3$ , the Euclidean inner product of $\mathbb{R}^3$ restricts to an inner product $\langle\,\cdot\,,\cdot\,\rangle_p$ on $T_pS$ . The first fundamental form of $S$ at $p$ is the associated quadratic form

I_p : T_pS \to \mathbb{R}, \qquad I_p(w) = \langle w, w\rangle_p = \lvert w\rvert^2 \;\ge\; 0 .

In the basis associated with a parametrization its coefficients are the three functions

E = \langle \mathbf{x}_u, \mathbf{x}_u\rangle, \qquad F = \langle \mathbf{x}_u, \mathbf{x}_v\rangle, \qquad G = \langle \mathbf{x}_v, \mathbf{x}_v\rangle ,

smooth on $U$ , and for $w = a\,\mathbf{x}_u + b\,\mathbf{x}_v$ ,

I_p(w) = E a^2 + 2F ab + G b^2 .

Classically one writes the line element

ds^2 = E\,du^2 + 2F\,du\,dv + G\,dv^2 ,

meaning the statement about arc length below. In index notation the same object is the symmetric matrix $(g_{ij}) = \begin{pmatrix} E & F \\ F & G\end{pmatrix}$ with $ds^2 = g_{ij}\,dx^i dx^j$ ; Lee calls it the Riemannian metric induced on $S$ by the inclusion into $\mathbb{R}^3$ . The sibling unit The metric tensor develops the index form; Riemannian manifolds drops the ambient space altogether.

A parametrization with $F \equiv 0$ is orthogonal; one with $E = G$ and $F = 0$ is isothermal.

Regions. A domain of $S$ is a connected open subset whose boundary is the image of a circle under a differentiable homeomorphism that is regular except at finitely many points; a region is a domain together with its boundary, and it is bounded if it lies in some ball of $\mathbb{R}^3$ .

Definition(What the first fundamental form measures)AdaptedAdapted from do Carmo (2016), §2-4, Prop. 1, p. 86, §2-5, Eq. (2), pp. 97–98, and §2-5, Def. 2, p. 100

Let $S$ be a regular surface, $\mathbf{x} : U \to S$ a parametrization, $q \in U$ and $p = \mathbf{x}(q)$ .

(Tangent plane.) The set of tangent vectors to $S$ at $p$ is exactly the two-dimensional subspace $d\mathbf{x}_q(\mathbb{R}^2) = \operatorname{span}\{\mathbf{x}_u(q), \mathbf{x}_v(q)\}$ . In particular it does not depend on the parametrization. If $\boldsymbol{\alpha}(t) = \mathbf{x}(u(t), v(t))$ , then $\boldsymbol{\alpha}'(0) = u'(0)\,\mathbf{x}_u + v'(0)\,\mathbf{x}_v$ : the coordinates of a tangent vector in the associated basis are the derivatives of the parameter curve.
(Lengths.) For a smooth curve $\boldsymbol{\alpha}(t) = \mathbf{x}(u(t), v(t))$ on $S$ , the arc length from $0$ to $t$ is $s(t) = \int_0^t \sqrt{I\bigl(\boldsymbol{\alpha}'(\tilde t)\bigr)}\, d\tilde t = \int_0^t \sqrt{E (u')^2 + 2F u'v' + G (v')^2}\; d\tilde t .$
(Angles.) Two curves on $S$ meeting at $t = t_0$ cross at the angle $\theta$ with $\cos\theta = \frac{\langle \boldsymbol{\alpha}'(t_0), \boldsymbol{\beta}'(t_0)\rangle}{\lvert\boldsymbol{\alpha}'(t_0)\rvert\,\lvert\boldsymbol{\beta}'(t_0)\rvert}.$ For the two coordinate curves of $\mathbf{x}$ this reads $\cos\varphi = F/\sqrt{EG}$ ; hence the coordinate curves are orthogonal at every point if and only if $F \equiv 0$ .
(Areas.) $\lvert\mathbf{x}_u\times\mathbf{x}_v\rvert = \sqrt{EG - F^2}$ , and this is strictly positive on $U$ . If $R \subset S$ is a bounded region contained in the coordinate neighbourhood of $\mathbf{x}$ and $Q = \mathbf{x}^{-1}(R)$ is compact, then $A(R) = \iint_Q \sqrt{EG - F^2}\; du\, dv$ is independent of the parametrization and is called the area of $R$ .

Everything in 2–4 is computed from $E$ , $F$ , $G$ alone. This is the sense in which the first fundamental form carries the metric geometry of $S$ : a traveller confined to the surface, able to measure only lengths of paths, angles and areas, has access to exactly these three functions.

Idea of the proofOriginal

Condition 3 in the definition of a regular surface is the whole point. It says the parameter plane is mapped onto the surface without collapsing a direction, and so the two coordinate velocity vectors $\mathbf{x}_u, \mathbf{x}_v$ are independent. Every curve on $S$ through $p$ can be read in the chart as a curve in $U$ , and the chain rule writes its velocity as a combination of $\mathbf{x}_u$ and $\mathbf{x}_v$ ; conversely every such combination is realized by a straight line in $U$ . That gives the tangent plane in one short argument, with both inclusions. Once the tangent plane is known to be a plane sitting inside $\mathbb{R}^3$ , it inherits the ambient dot product, and expanding that dot product in the basis $\{\mathbf{x}_u, \mathbf{x}_v\}$ produces the three numbers $E, F, G$ and nothing else. Lengths, angles and areas are all built from dot products of tangent vectors, so all three reduce to $E$ , $F$ , $G$ .

ProofAdaptedAdapted from do Carmo (2016), §2-4, Prop. 1 and its proof, pp. 86–87, with §2-5, pp. 95, 97–98 and p. 100

Part 1. Two inclusions.

Every tangent vector lies in $d\mathbf{x}_q(\mathbb{R}^2)$ . Let $w = \boldsymbol{\alpha}'(0)$ with $\boldsymbol{\alpha} : (-\varepsilon,\varepsilon) \to \mathbf{x}(U)$ smooth and $\boldsymbol{\alpha}(0) = p$ . Because $\mathbf{x}$ is a homeomorphism onto its image (condition 2), $\boldsymbol{\beta} = \mathbf{x}^{-1}\circ\boldsymbol{\alpha}$ is a well-defined curve in $U$ , and it is smooth: by condition 3, $d\mathbf{x}_q$ is injective, so some $2\times 2$ minor of the Jacobian is invertible, and the inverse function theorem (inverse function theorem) applied to $\mathbf{x}$ followed by the corresponding coordinate projection produces a smooth local inverse that agrees with $\mathbf{x}^{-1}$ on $S$ . Writing $\boldsymbol{\beta}(t) = (u(t), v(t))$ and differentiating $\boldsymbol{\alpha} = \mathbf{x}\circ\boldsymbol{\beta}$ by the chain rule (chain rule),

w = \boldsymbol{\alpha}'(0) = u'(0)\,\mathbf{x}_u(q) + v'(0)\,\mathbf{x}_v(q) = d\mathbf{x}_q\bigl(\boldsymbol{\beta}'(0)\bigr),

which is the coordinate statement at the end of Part 1 as well.

Every element of $d\mathbf{x}_q(\mathbb{R}^2)$ is tangent. Given $\mathbf{z} \in \mathbb{R}^2$ , the segment $\boldsymbol{\gamma}(t) = q + t\mathbf{z}$ stays in the open set $U$ for $\lvert t\rvert$ small, and $\boldsymbol{\alpha} = \mathbf{x}\circ\boldsymbol{\gamma}$ is a smooth curve on $S$ through $p$ with $\boldsymbol{\alpha}'(0) = d\mathbf{x}_q(\mathbf{z})$ .

So the set of tangent vectors equals $d\mathbf{x}_q(\mathbb{R}^2)$ , which by condition 3 is a two-dimensional subspace. The left-hand description makes no mention of $\mathbf{x}$ , so the plane is independent of the chart.

Part 2. By Part 1, $\boldsymbol{\alpha}'(t) = u'\mathbf{x}_u + v'\mathbf{x}_v$ . Expanding the inner product bilinearly,

\lvert\boldsymbol{\alpha}'\rvert^2 = \langle u'\mathbf{x}_u + v'\mathbf{x}_v,\; u'\mathbf{x}_u + v'\mathbf{x}_v\rangle = E(u')^2 + 2Fu'v' + G(v')^2 = I(\boldsymbol{\alpha}'),

the cross terms combining because the inner product is symmetric. Arc length is $\int\lvert\boldsymbol{\alpha}'\rvert$ (line integrals), which is the claimed formula.

Part 3. The angle formula is the definition of the angle between two vectors of $\mathbb{R}^3$ , applied to two tangent vectors. For the coordinate curves $t \mapsto \mathbf{x}(t, v_0)$ and $t\mapsto \mathbf{x}(u_0, t)$ the velocities are $\mathbf{x}_u$ and $\mathbf{x}_v$ , with lengths $\sqrt E$ and $\sqrt G$ , so $\cos\varphi = F/\sqrt{EG}$ . This vanishes exactly when $F = 0$ , and $\sqrt{EG} > 0$ because $\mathbf{x}_u$ and $\mathbf{x}_v$ are nonzero.

Part 4. The identity $\lvert\mathbf{a}\times\mathbf{b}\rvert^2 + \langle\mathbf{a},\mathbf{b}\rangle^2 = \lvert\mathbf{a}\rvert^2\lvert\mathbf{b}\rvert^2$ (vector algebra) with $\mathbf{a} = \mathbf{x}_u$ , $\mathbf{b} = \mathbf{x}_v$ gives $\lvert\mathbf{x}_u\times\mathbf{x}_v\rvert^2 = EG - F^2$ , and the left side is nonzero by condition 3, so $EG - F^2 > 0$ . Independence of the parametrization is the statement already proved in surface integrals and flux: if $\bar{\mathbf{x}}$ is a second parametrization covering $R$ and $h = \mathbf{x}^{-1}\circ\bar{\mathbf{x}}$ is the change of parameters, then $\lvert\bar{\mathbf{x}}_{\bar u}\times\bar{\mathbf{x}}_{\bar v}\rvert = \lvert\mathbf{x}_u\times\mathbf{x}_v\rvert\,\lvert\det Dh\rvert$ , and the change-of-variables theorem for double integrals absorbs the Jacobian factor. $\square$

Remark(Where the hypotheses are used)Original

Condition 3 (regularity). Used three times: to make $T_pS$ two-dimensional at all; to get the smooth local inverse in Part 1 through the inverse function theorem; and to guarantee $EG - F^2 > 0$ , without which the area integrand could vanish and the matrix $(g_{ij})$ fail to be positive definite. The cone $\mathbf{x}(u,v) = (v\cos u, v\sin u, v)$ fails it at $v = 0$ : there $\mathbf{x}_u = \mathbf{0}$ , $EG - F^2 = 0$ , and the vertex has no tangent plane.
Condition 2 (homeomorphism). This is what rules out a parameter plane folded so that it touches itself. The map $\mathbf{x}(u,v) = (\sin u\,,\, \sin 2u\,,\, v)$ on $(0, 2\pi)\times\mathbb{R}$ is smooth and satisfies condition 3, but its image is a self-crossing cylinder whose "tangent plane" at a crossing point is not well defined; the inverse is not continuous there.
Smoothness beyond $C^1$ . The tangent plane, $E, F, G$ and all of Parts 1–4 need only $C^1$ . do Carmo gives the graph of $z = \sqrt[3]{(x^2+y^2)^2}$ as a set that has a tangent plane at the origin yet is not a regular surface in his sense, because $z_{xx}$ does not exist there (do Carmo (2016), §2-4, Example 3, p. 90). The higher derivatives are needed only from the second fundamental form onwards.
$R$ inside one chart. Part 4 assumes the region lies in a single coordinate neighbourhood. That is a restriction on the formula, not on the notion of area: most surfaces have a chart covering everything except a set of curves, which contributes nothing to the integral, as in the example below.
The ambient inner product. $I_p$ is defined by restricting the dot product of $\mathbb{R}^3$ . Nothing in Parts 2–4 uses the ambient space again, which is the observation that Gauss's Theorema Egregium turns into a theorem.

Method(Computing with

E

F

G

)Original

Given a parametrization $\mathbf{x}(u,v)$ of a surface:

Differentiate: $\mathbf{x}_u$ , $\mathbf{x}_v$ .
Form $E = \mathbf{x}_u\cdot\mathbf{x}_u$ , $F = \mathbf{x}_u\cdot\mathbf{x}_v$ , $G = \mathbf{x}_v\cdot\mathbf{x}_v$ . Check $EG - F^2 > 0$ on the whole parameter domain; where it fails, the parametrization is singular and the chart must be changed.
For a curve given in the chart by $t\mapsto (u(t), v(t))$ , its length is $\int \sqrt{E\dot u^2 + 2F\dot u\dot v + G\dot v^2}\,dt$ .
For the angle between two chart directions $(a_1,b_1)$ and $(a_2,b_2)$ , use the polarized form $\langle w_1, w_2\rangle = Ea_1a_2 + F(a_1b_2+a_2b_1) + Gb_1b_2$ in $\cos\theta = \langle w_1,w_2\rangle/(\lvert w_1\rvert\lvert w_2\rvert)$ .
For the area of $R$ , integrate $\sqrt{EG-F^2}$ over $\mathbf{x}^{-1}(R)$ .
Shortcut for a graph. If $S$ is $z = h(x,y)$ with $\mathbf{x}(x,y) = (x, y, h)$ , then $\mathbf{x}_x = (1,0,h_x)$ , $\mathbf{x}_y = (0,1,h_y)$ and $E = 1 + h_x^2, \quad F = h_xh_y, \quad G = 1 + h_y^2, \quad EG - F^2 = 1 + h_x^2 + h_y^2 .$
Sanity checks. $E > 0$ and $G > 0$ always; $F = 0$ means the chart is orthogonal; setting all the surface's parameters to a flat case should return $ds^2 = du^2 + dv^2$ .

Example(The paraboloid

z = \tfrac12(x^2+y^2)

)Original

Parametrize the paraboloid by polar coordinates in the base plane,

\mathbf{x}(r, \theta) = \Bigl(r\cos\theta,\; r\sin\theta,\; \tfrac12 r^2\Bigr), \qquad r > 0, \;\; 0 < \theta < 2\pi .

This chart misses only the vertex and one meridian, a set of zero area.

Step 1: the coefficients.

\mathbf{x}_r = (\cos\theta,\; \sin\theta,\; r), \qquad \mathbf{x}_\theta = (-r\sin\theta,\; r\cos\theta,\; 0),

E = \cos^2\theta + \sin^2\theta + r^2 = 1 + r^2, \qquad F = -r\cos\theta\sin\theta + r\sin\theta\cos\theta + 0 = 0, \qquad G = r^2\sin^2\theta + r^2\cos^2\theta = r^2 .

Hence $EG - F^2 = r^2(1+r^2) > 0$ for $r > 0$ : the chart is regular and orthogonal, and

ds^2 = (1+r^2)\,dr^2 + r^2\,d\theta^2 .

(Check: setting the height to zero would replace $E$ by $1$ and give the Euclidean polar line element $dr^2 + r^2d\theta^2$ , as it must.)

Step 2: a length. Take the spiral that climbs the bowl with $r = \theta$ , from $\theta = 0$ to $\theta = 1$ . Here $\dot r = 1$ and $\dot\theta = 1$ , so

L = \int_0^1 \sqrt{(1+\theta^2)\cdot 1 + \theta^2\cdot 1}\; d\theta = \int_0^1\sqrt{1+2\theta^2}\;d\theta .

With $\int\sqrt{1+2\theta^2}\,d\theta = \tfrac{\theta}{2}\sqrt{1+2\theta^2} + \tfrac{1}{2\sqrt2}\operatorname{arcsinh}(\sqrt2\,\theta)$ ,

L = \frac{\sqrt3}{2} + \frac{\sqrt2}{4}\operatorname{arcsinh}\sqrt2 = 0.8660254 + 0.4052486 = 1.2712739 .

For comparison, the same spiral drawn flat in the base plane has length $\int_0^1\sqrt{1+\theta^2}\,d\theta = 1.1477936$ ; climbing the bowl costs about $10.8\%$ extra.

Step 3: an angle. At the point $r = 1$ , $\theta = 0$ (that is, $(1,0,\tfrac12)$ ), the spiral's chart direction is $(\dot r, \dot\theta) = (1,1)$ and the meridian $\theta = \text{const}$ has direction $(1,0)$ . Since $F = 0$ , the polarized form gives $\langle w_1, w_2\rangle = E\cdot 1\cdot 1 = 2$ , while $\lvert w_1\rvert = \sqrt{E + G} = \sqrt3$ and $\lvert w_2\rvert = \sqrt E = \sqrt2$ . So

\cos\theta_{\text{cross}} = \frac{2}{\sqrt3\sqrt2} = \sqrt{\tfrac23} = 0.8164966, \qquad \theta_{\text{cross}} = 35.2644^\circ .

Step 4: an area. The part of the paraboloid lying over the disc $x^2 + y^2 \le 3$ is $R = \mathbf{x}(Q)$ with $Q = (0,\sqrt3\,]\times(0,2\pi)$ , and

A(R) = \int_0^{2\pi}\!\!\int_0^{\sqrt3} \sqrt{r^2(1+r^2)}\; dr\, d\theta = 2\pi\int_0^{\sqrt3} r\sqrt{1+r^2}\; dr = 2\pi\left[\frac{(1+r^2)^{3/2}}{3}\right]_0^{\sqrt3} = \frac{2\pi}{3}\bigl(4^{3/2} - 1\bigr) .

Since $4^{3/2} = 8$ ,

\boxed{\,A(R) = \frac{14\pi}{3} = 14.66077\,}

against $3\pi = 9.42478$ for the flat disc underneath it. The same answer follows from the graph shortcut: $h = \tfrac12(x^2+y^2)$ gives $\sqrt{1+h_x^2+h_y^2} = \sqrt{1+x^2+y^2}$ , and $\iint_{x^2+y^2\le 3}\sqrt{1+x^2+y^2}\,dx\,dy$ is the same integral in disguise.

Application(Induced metrics in physics)Original

The metric is the data. Almost every surface calculation in physics is run on $E, F, G$ rather than on the embedding. On the sphere of radius $a$ , $\mathbf{x}(\vartheta,\varphi) = a(\sin\vartheta\cos\varphi, \sin\vartheta\sin\varphi, \cos\vartheta)$ gives $E = a^2$ , $F = 0$ , $G = a^2\sin^2\vartheta$ , so $ds^2 = a^2(d\vartheta^2 + \sin^2\vartheta\, d\varphi^2)$ and $\sqrt{EG-F^2} = a^2\sin\vartheta$ . That factor $\sin\vartheta$ is the $d\Omega$ of every spherical-harmonic expansion, every radiation pattern and every solid-angle integral; it comes from the first fundamental form and nothing else.

Membranes and shells. In continuum mechanics the strain of a thin sheet is defined as half the change in its first fundamental form between the reference and the deformed configuration, $\epsilon_{ij} = \tfrac12(g_{ij} - g_{ij}^{(0)})$ . Stretching energy is a quadratic form in that difference. A deformation that leaves $E, F, G$ unchanged costs no stretching energy at all, which is why paper bends freely but tears rather than stretches.

General relativity. The induced metric on a spacelike hypersurface is exactly this construction one dimension up: restrict the ambient metric to the tangent space at each point. It is the "first fundamental form" of the $3{+}1$ initial-value formulation, paired there with a second fundamental form that plays the role of its time derivative.

Surface integrals and transport. The flux and area integrals of surface integrals and flux all carry the factor $\sqrt{EG-F^2}$ . In heat and mass transfer over a curved interface, or in computing the radiating area of a corrugated fin, getting this factor right is the whole calculation: the paraboloid above has $55\%$ more area than its shadow.

Definition

The second fundamental form and curvature

Adapted

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Introduces the Gauss map $N$ , which sends each point of an oriented surface to its unit normal, proves that its differential $dN_p$ is a self-adjoint operator on the tangent plane, defines the second fundamental form $\mathrm{II}_p(v) = -\langle dN_p(v), v\rangle$ and identifies its values on unit vectors as normal curvatures, so that the eigenvalues of $-dN_p$ are the principal curvatures and their product and mean are the Gaussian curvature $K$ and the mean curvature $H$ , with the coordinate formulas in $E,F,G,e,f,g$ .

Prerequisites: Regular surfaces and the first fundamental form, Curves and the Frenet-Serret formulas, Spectral theorem for Hermitian matrices

Definition(Gauss map, second fundamental form, principal, Gaussian and mean curvature)AdaptedAdapted from do Carmo (2016), §3-2, Example 2, p. 140, Defs. 2, 3, 4, 6, 7 and 8, pp. 141–149, and Def. 9, p. 150 (asymptotic direction), with §3-3, pp. 156–157

Let $S$ be a regular surface (regular surfaces and the first fundamental form).

Orientation and Gauss map. $S$ is oriented by a smooth choice of unit normal $N : S \to S^2 \subset \mathbb{R}^3$ ; the map $N$ is the Gauss map. In a parametrization $\mathbf{x}(u,v)$ compatible with the orientation, $N = (\mathbf{x}_u\times\mathbf{x}_v)/\lvert\mathbf{x}_u\times\mathbf{x}_v\rvert$ . Since $T_{N(p)}S^2$ and $T_pS$ are both the plane orthogonal to $N(p)$ , the differential is an operator on one plane, $dN_p : T_pS \to T_pS .$ Its negative $-dN_p$ is the shape operator (Weingarten map).
Second fundamental form. $\mathrm{II}_p(v) = -\langle dN_p(v), v\rangle$ for $v \in T_pS$ . Its coefficients in the basis $\{\mathbf{x}_u, \mathbf{x}_v\}$ are $e = \langle N, \mathbf{x}_{uu}\rangle = -\langle N_u, \mathbf{x}_u\rangle, \qquad f = \langle N, \mathbf{x}_{uv}\rangle = -\langle N_v, \mathbf{x}_u\rangle = -\langle N_u, \mathbf{x}_v\rangle, \qquad g = \langle N, \mathbf{x}_{vv}\rangle = -\langle N_v, \mathbf{x}_v\rangle ,$ so that $\mathrm{II}(a\mathbf{x}_u + b\mathbf{x}_v) = e a^2 + 2f ab + g b^2$ .
Normal curvature. Let $C \subset S$ be a regular curve through $p$ , with curvature $k$ and principal normal $\mathbf{n}$ at $p$ (Frenet-Serret), and let $\cos\theta = \langle \mathbf{n}, N\rangle$ . The normal curvature of $C$ at $p$ is $k_n = k\cos\theta$ : the signed length of the projection of $k\mathbf{n}$ onto $N$ .
Principal curvatures and directions. Write $k_1$ for the largest and $k_2$ for the smallest value that $\mathrm{II}_p$ takes on the unit circle of $T_pS$ ; these are the principal curvatures at $p$ . The unit vectors at which they are attained are the principal directions. A curve whose tangent is a principal direction at each of its points is a line of curvature.
Gaussian and mean curvature. $K = \det(dN_p) = k_1k_2, \qquad H = -\tfrac12\operatorname{tr}(dN_p) = \tfrac{k_1+k_2}{2} .$
Classification of points. $p$ is elliptic if $K = \det(dN_p) > 0$ , hyperbolic if $K < 0$ , parabolic if $K = 0$ with $dN_p \ne 0$ , and planar if $dN_p = 0$ . It is umbilical if $k_1 = k_2$ (planar points included). A direction with $k_n = 0$ is asymptotic.
Sign convention. do Carmo defines $\mathrm{II}$ with the minus sign shown, and computes the unit sphere with the normal $N(x,y,z) = (-x,-y,-z)$ pointing to its centre (his Examples 2 and 7, pp. 140 and 145): then $dN_p = -\mathrm{id}$ and every normal curvature equals $+1$ . With the outward normal instead, $dN_p = +\mathrm{id}$ and every normal curvature is $-1$ . Reversing the orientation sends $N \mapsto -N$ , hence $e,f,g \mapsto -e,-f,-g$ and $dN_p \mapsto -dN_p$ : the unordered pair of principal curvatures becomes $\{-k_1,-k_2\}$ , so with the ordering $k_1 \ge k_2$ kept, $k_1 \mapsto -k_2$ and $k_2 \mapsto -k_1$ . Consequently $H \mapsto -H$ , while $K$ is unchanged because the dimension is even.

Definition(Properties of the shape operator)AdaptedAdapted from do Carmo (2016), §3-2, Prop. 1, p. 142, Prop. 2 (Meusnier), pp. 143–144, Euler's formula, pp. 146–147, Def. 6, p. 148, Prop. 4, p. 149, and §3-3, Eqs. (4)–(6), pp. 157–158

Let $S$ be an oriented regular surface, $p \in S$ .

(Self-adjointness.) $dN_p : T_pS \to T_pS$ is a self-adjoint linear map: $\langle dN_p(w_1), w_2\rangle = \langle w_1, dN_p(w_2)\rangle$ for all $w_1, w_2 \in T_pS$ . Hence $\mathrm{II}_p$ is a quadratic form, $T_pS$ has an orthonormal basis $\{e_1, e_2\}$ of eigenvectors of $dN_p$ , and the eigenvalues are real.
(The second fundamental form is a normal curvature.) If $v \in T_pS$ has $\lvert v\rvert = 1$ and $C$ is any regular curve on $S$ through $p$ with unit tangent $v$ there, then $\mathrm{II}_p(v) = k_n$ , the normal curvature of $C$ at $p$ . In particular (Meusnier) all curves on $S$ through $p$ with the same tangent line have the same normal curvature at $p$ .
(Principal curvatures are the eigenvalues.) With $\{e_1,e_2\}$ as in 1 and $dN_p(e_i) = -k_ie_i$ , the numbers $k_1 \ge k_2$ are the largest and smallest values $\mathrm{II}_p$ takes on the unit circle, that is, the principal curvatures, and $e_1, e_2$ are principal directions. For $v = e_1\cos\vartheta + e_2\sin\vartheta$ , Euler's formula holds: $k_n(\vartheta) = \mathrm{II}_p(v) = k_1\cos^2\vartheta + k_2\sin^2\vartheta .$
(Coordinate formulas.) In a parametrization compatible with the orientation, $K = \frac{eg - f^2}{EG - F^2}, \qquad H = \frac{eG - 2fF + gE}{2\,(EG - F^2)}, \qquad k_{1,2} = H \pm \sqrt{H^2 - K} .$ In particular $H^2 \ge K$ always, with equality exactly at umbilical points, and the functions $k_1 \ge k_2$ are continuous on $S$ and differentiable away from the umbilics.
(All-umbilic surfaces.) If every point of a connected regular surface $S$ is umbilical, then $S$ is contained in a sphere or in a plane.

Idea of the proofOriginal

The Gauss map records how the tangent plane tips as one moves along the surface, so its derivative is the natural measure of curvature. Self-adjointness is the key structural fact and it comes from one line of calculus: the mixed second partial derivatives of $\mathbf{x}$ commute, and differentiating $\langle N, \mathbf{x}_u\rangle = 0$ in $v$ and $\langle N, \mathbf{x}_v\rangle = 0$ in $u$ turns both off-diagonal entries into $-\langle N, \mathbf{x}_{uv}\rangle$ . Self-adjointness buys the spectral theorem, hence an orthonormal eigenbasis and real eigenvalues, and the eigenvalues of a quadratic form on a circle are its extreme values. Identifying $\mathrm{II}_p(v)$ with a normal curvature is another single differentiation: along a curve on the surface the normal is perpendicular to the velocity, so differentiating that relation trades $\langle N', \boldsymbol{\gamma}'\rangle$ for $-\langle N, \boldsymbol{\gamma}''\rangle$ , and $\boldsymbol{\gamma}'' = k\mathbf{n}$ . The coordinate formulas are then nothing but linear algebra: the matrix of $dN$ in the basis $\{\mathbf{x}_u,\mathbf{x}_v\}$ is $-\mathrm{I}^{-1}\,\mathrm{II}$ , and $K$ and $H$ are its determinant and its trace.

ProofAdaptedAdapted from do Carmo (2016), §3-2, proof of Prop. 1, pp. 142–143, with pp. 144, 146–147 and Prop. 4, pp. 149–150, and §3-3, pp. 156–158

Part 1. $dN_p$ is linear, so it suffices to check the symmetry on the basis $\{\mathbf{x}_u, \mathbf{x}_v\}$ . Along a curve $\boldsymbol{\alpha}(t) = \mathbf{x}(u(t), v(t))$ with $\boldsymbol{\alpha}(0) = p$ , the chain rule gives $dN_p(\boldsymbol{\alpha}'(0)) = \frac{d}{dt}N(u(t),v(t))\big|_{0} = N_uu'(0) + N_vv'(0)$ , and taking the coordinate curves,

dN_p(\mathbf{x}_u) = N_u, \qquad dN_p(\mathbf{x}_v) = N_v .

The diagonal entries pose no question, so only $\langle N_u, \mathbf{x}_v\rangle = \langle \mathbf{x}_u, N_v\rangle$ must be shown. Differentiate the two identities $\langle N, \mathbf{x}_u\rangle = 0$ and $\langle N, \mathbf{x}_v\rangle = 0$ with respect to $v$ and $u$ respectively:

\langle N_v, \mathbf{x}_u\rangle + \langle N, \mathbf{x}_{uv}\rangle = 0, \qquad \langle N_u, \mathbf{x}_v\rangle + \langle N, \mathbf{x}_{vu}\rangle = 0 .

By the equality of mixed partial derivatives (here the smoothness of $\mathbf{x}$ is used), $\mathbf{x}_{uv} = \mathbf{x}_{vu}$ , so both left-hand first terms equal $-\langle N, \mathbf{x}_{uv}\rangle$ and are therefore equal to each other. A self-adjoint operator on a two-dimensional real inner-product space has real eigenvalues and an orthonormal eigenbasis (spectral theorem), and $v \mapsto \langle dN_p(v), v\rangle$ is then a quadratic form.

Part 2. Let $\boldsymbol{\gamma}(s)$ be a unit-speed parametrization of $C$ with $\boldsymbol{\gamma}(0) = p$ , $\boldsymbol{\gamma}'(0) = v$ , and write $N(s) = N(\boldsymbol{\gamma}(s))$ . Since $N(s) \perp T_{\boldsymbol{\gamma}(s)}S \ni \boldsymbol{\gamma}'(s)$ we have $\langle N(s), \boldsymbol{\gamma}'(s)\rangle \equiv 0$ ; differentiating,

\langle N'(s), \boldsymbol{\gamma}'(s)\rangle = -\langle N(s), \boldsymbol{\gamma}''(s)\rangle .

At $s = 0$ , $N'(0) = dN_p(v)$ and $\boldsymbol{\gamma}''(0) = k\,\mathbf{n}$ by the first Frenet formula, so

\mathrm{II}_p(v) = -\langle dN_p(v), v\rangle = -\langle N'(0), \boldsymbol{\gamma}'(0)\rangle = \langle N, k\mathbf{n}\rangle = k\cos\theta = k_n .

The left-hand side depends only on $v$ , not on which curve has that tangent; that is Meusnier's statement.

Part 3. By Part 1 write $dN_p(e_i) = -k_ie_i$ in an orthonormal eigenbasis, ordering $k_1 \ge k_2$ . For a unit vector $v = e_1\cos\vartheta + e_2\sin\vartheta$ , expanding bilinearly,

\mathrm{II}_p(v) = -\langle dN_p(v), v\rangle = \langle k_1e_1\cos\vartheta + k_2e_2\sin\vartheta,\; e_1\cos\vartheta + e_2\sin\vartheta\rangle = k_1\cos^2\vartheta + k_2\sin^2\vartheta ,

which is Euler's formula. Writing it as $k_2 + (k_1-k_2)\cos^2\vartheta$ shows the values range over $[k_2, k_1]$ with the endpoints attained at $\vartheta = \pi/2$ and $\vartheta = 0$ ; so $k_1, k_2$ are the extreme normal curvatures and $e_1, e_2$ the directions realizing them.

Part 4. Write the matrix of $dN_p$ in the basis $\{\mathbf{x}_u,\mathbf{x}_v\}$ as $(a_{ij})$ via $N_u = a_{11}\mathbf{x}_u + a_{21}\mathbf{x}_v$ , $N_v = a_{12}\mathbf{x}_u + a_{22}\mathbf{x}_v$ (the Weingarten equations). Taking inner products with $\mathbf{x}_u$ and $\mathbf{x}_v$ and using the definitions of $e, f, g$ and of $E, F, G$ gives four scalar equations which in matrix form are

-\begin{pmatrix} e & f \\ f & g\end{pmatrix} = \begin{pmatrix} a_{11} & a_{21} \\ a_{12} & a_{22}\end{pmatrix} \begin{pmatrix} E & F \\ F & G\end{pmatrix}.

Because $EG - F^2 > 0$ the first-fundamental-form matrix is invertible, and taking determinants of both sides gives

eg - f^2 = \det(a_{ij})\,(EG - F^2) = K\,(EG-F^2),

which is the formula for $K$ . Inverting explicitly,

\begin{pmatrix} a_{11} & a_{21} \\ a_{12} & a_{22}\end{pmatrix} = \frac{-1}{EG-F^2}\begin{pmatrix} e & f \\ f & g\end{pmatrix}\begin{pmatrix} G & -F \\ -F & E\end{pmatrix},

whose diagonal entries are $a_{11} = (fF - eG)/(EG-F^2)$ and $a_{22} = (fF - gE)/(EG-F^2)$ ; hence

H = -\tfrac12(a_{11}+a_{22}) = \frac{eG - 2fF + gE}{2(EG-F^2)} .

Finally $-k_1, -k_2$ are the eigenvalues of $(a_{ij})$ , so $k_1, k_2$ are the roots of $k^2 - (k_1+k_2)k + k_1k_2 = k^2 - 2Hk + K = 0$ , giving $k = H \pm\sqrt{H^2-K}$ . The discriminant is $H^2 - K = \tfrac14(k_1-k_2)^2 \ge 0$ , vanishing exactly when $k_1 = k_2$ .

Part 5. Work in a connected coordinate neighbourhood $V$ . Umbilicity at $q$ means $dN_q = \lambda(q)\,\mathrm{id}$ for a scalar $\lambda(q)$ , which is smooth because $\lambda = -H$ there. Applying this to the basis vectors, $N_u = \lambda\mathbf{x}_u$ and $N_v = \lambda\mathbf{x}_v$ . Differentiate the first in $v$ and the second in $u$ and subtract; the terms $\lambda\mathbf{x}_{uv}$ cancel by equality of mixed partials, leaving

\lambda_v\,\mathbf{x}_u - \lambda_u\,\mathbf{x}_v = \mathbf{0}.

Since $\mathbf{x}_u, \mathbf{x}_v$ are independent, $\lambda_u = \lambda_v = 0$ , so $\lambda$ is constant on the connected set $V$ . If $\lambda = 0$ then $N$ is a constant vector $N_0$ and $\partial_u\langle\mathbf{x}, N_0\rangle = \partial_v\langle\mathbf{x}, N_0\rangle = 0$ , so $\langle \mathbf{x}, N_0\rangle$ is constant and $V$ lies in a plane. If $\lambda \ne 0$ , put $\mathbf{y} = \mathbf{x} - \lambda^{-1}N$ ; then $\mathbf{y}_u = \mathbf{x}_u - \lambda^{-1}N_u = \mathbf{0}$ and likewise $\mathbf{y}_v = \mathbf{0}$ , so $\mathbf{y}$ is a fixed point and $\lvert\mathbf{x} - \mathbf{y}\rvert = 1/\lvert\lambda\rvert$ : $V$ lies on a sphere. A connected $S$ is covered by finitely many such neighbourhoods along any path between two of its points, and overlapping neighbourhoods force the same plane or the same sphere throughout. $\square$

Remark(Where the hypotheses are used)Original

Orientability. Without a globally defined smooth $N$ there is no Gauss map, and $e, f, g$ , $H$ , $k_1$ , $k_2$ have no global sign. $K$ survives: it is unchanged by $N \mapsto -N$ , so a Möbius band has a well-defined Gaussian curvature but no well-defined mean curvature.
Equality of mixed partials. Used twice, in Part 1 and again in Part 5. This is where the smoothness in the definition of a regular surface is spent; a $C^1$ surface has a tangent plane and a first fundamental form but no second fundamental form.
$EG - F^2 > 0$ . The inversion in Part 4 is impossible without it. At a singular point of a parametrization the formulas for $K$ and $H$ return $0/0$ ; the surface may still be perfectly smooth there, as the pole of a sphere is in spherical coordinates, and a different chart must be used.
Connectedness in Part 5. Dropping it makes the conclusion false in the obvious way: the disjoint union of a plane and a sphere is all-umbilic and is neither.
Why $-\langle dN_p(v),v\rangle$ and not $+$ . The sign is chosen so that a sphere carrying the normal that points to its centre has $k_1 = k_2 = 1/a > 0$ and $\mathrm{II}$ positive (with the outward normal the same sphere gives $k_1 = k_2 = -1/a$ , $K = 1/a^2$ and $H = -1/a$ ); it is a convention, and some texts take the opposite one. With the sign fixed, $\mathrm{II}_p(v) > 0$ means the surface bends towards $N$ along $v$ .
$k_1, k_2$ at umbilics. Part 4 gives continuity of $k_1 \ge k_2$ everywhere, but $\sqrt{H^2-K}$ is not differentiable where $H^2 = K$ , so the principal curvatures can fail to be smooth at umbilical points even though $K$ and $H$ are smooth there.

Method(Computing

K

and

H

from a parametrization)Original

Compute $\mathbf{x}_u, \mathbf{x}_v$ and then $E, F, G$ ; check $EG - F^2 > 0$ .
Compute $N = (\mathbf{x}_u\times\mathbf{x}_v)/\sqrt{EG-F^2}$ . Record which way it points; every sign below depends on it.
Compute the second derivatives $\mathbf{x}_{uu}, \mathbf{x}_{uv}, \mathbf{x}_{vv}$ and then $e = \langle N, \mathbf{x}_{uu}\rangle$ , $f = \langle N, \mathbf{x}_{uv}\rangle$ , $g = \langle N, \mathbf{x}_{vv}\rangle$ . The triple-product form $e = \det[\mathbf{x}_u, \mathbf{x}_v, \mathbf{x}_{uu}]/\sqrt{EG-F^2}$ and its analogues avoid writing $N$ out.
$K = (eg-f^2)/(EG-F^2)$ and $H = (eG - 2fF + gE)/\bigl(2(EG-F^2)\bigr)$ .
$k_{1,2} = H \pm\sqrt{H^2-K}$ . Check $k_1k_2 = K$ and $k_1+k_2 = 2H$ .
If $F = f = 0$ (an orthogonal chart whose coordinate curves are lines of curvature) the work collapses: the principal curvatures are $e/E$ and $g/G$ , with $k_1$ the larger of the two and $k_2$ the smaller, and $K = eg/(EG)$ , $H = \tfrac12(e/E + g/G)$ . Which of $e/E$ , $g/G$ is $k_1$ depends on the surface and on the point, so compare them before labelling.
Sanity checks. A plane must give $e=f=g=0$ ; a sphere of radius $a$ with the normal pointing to its centre must give $K = 1/a^2$ and $H = 1/a$ (outward normal: $K = 1/a^2$ still, but $H = -1/a$ ); a cylinder of radius $a$ must give $K = 0$ and $\lvert H\rvert = 1/(2a)$ .

Example(The paraboloid

z = \tfrac12(x^2+y^2)

again)Original

Continue with the chart of the first fundamental form,

\mathbf{x}(r,\theta) = \bigl(r\cos\theta,\; r\sin\theta,\; \tfrac12 r^2\bigr), \qquad E = 1+r^2,\;\; F = 0,\;\; G = r^2 .

Step 1: the normal. With $\mathbf{x}_r = (\cos\theta, \sin\theta, r)$ and $\mathbf{x}_\theta = (-r\sin\theta, r\cos\theta, 0)$ ,

\mathbf{x}_r\times\mathbf{x}_\theta = \bigl(-r^2\cos\theta,\; -r^2\sin\theta,\; r\bigr), \qquad \lvert\mathbf{x}_r\times\mathbf{x}_\theta\rvert = r\sqrt{1+r^2} = \sqrt{EG-F^2},

N = \frac{1}{\sqrt{1+r^2}}\bigl(-r\cos\theta,\; -r\sin\theta,\; 1\bigr).

Its $z$ -component is positive: $N$ points up out of the bowl, into the concave side.

Step 2: the second derivatives and $e,f,g$ .

\mathbf{x}_{rr} = (0,0,1), \qquad \mathbf{x}_{r\theta} = (-\sin\theta, \cos\theta, 0), \qquad \mathbf{x}_{\theta\theta} = (-r\cos\theta, -r\sin\theta, 0),

and therefore

e = \langle N, \mathbf{x}_{rr}\rangle = \frac{1}{\sqrt{1+r^2}}, \qquad f = \frac{r\cos\theta\sin\theta - r\sin\theta\cos\theta}{\sqrt{1+r^2}} = 0, \qquad g = \frac{r^2\cos^2\theta + r^2\sin^2\theta}{\sqrt{1+r^2}} = \frac{r^2}{\sqrt{1+r^2}} .

Both $F$ and $f$ vanish, so the meridians and the horizontal circles are lines of curvature and step 6 of the method applies.

Step 3: the curvatures.

K = \frac{eg - f^2}{EG-F^2} = \frac{r^2/(1+r^2)}{r^2(1+r^2)} = \frac{1}{(1+r^2)^2},

H = \frac{eG + gE}{2(EG-F^2)} = \frac{\dfrac{r^2}{\sqrt{1+r^2}} + \dfrac{r^2(1+r^2)}{\sqrt{1+r^2}}}{2r^2(1+r^2)} = \frac{2+r^2}{2\,(1+r^2)^{3/2}} .

Since $F = f = 0$ , the two coordinate directions are the principal ones, with normal curvatures $e/E = (1+r^2)^{-3/2}$ along the meridian and $g/G = (1+r^2)^{-1/2}$ along the horizontal circle. The second is the larger, so $k_1 = (1+r^2)^{-1/2}$ and $k_2 = (1+r^2)^{-3/2}$ .

Step 4: numbers at $r = 1$ , the point $(1, 0, \tfrac12)$ :

K = \frac14 = 0.25, \qquad H = \frac{3}{2\cdot 2^{3/2}} = \frac{3\sqrt2}{8} = 0.5303301 .

Then $H^2 - K = \tfrac{9}{32} - \tfrac{8}{32} = \tfrac1{32}$ , so

k_{1,2} = \frac{3\sqrt2}{8} \pm \frac{1}{4\sqrt2}, \qquad k_1 = \frac{1}{\sqrt2} = 0.7071068, \qquad k_2 = \frac{1}{2\sqrt2} = 0.3535534 .

Check: $k_1k_2 = \dfrac{1}{\sqrt2}\cdot\dfrac{1}{2\sqrt2} = \dfrac{1}{4} = K$ and $k_1 + k_2 = 0.7071068 + 0.3535534 = 1.0606602 = 2H$ . These agree with the direct values $k_1 = (1+r^2)^{-1/2}$ and $k_2 = (1+r^2)^{-3/2}$ at $r=1$ . Both principal curvatures are positive, so $K > 0$ and every point of the paraboloid is elliptic: near each of its points the surface stays on a single side of the tangent plane there, as a bowl should. Note also that $k_2 = (1+r^2)^{-3/2}$ is exactly the curvature of the generating parabola $z = \tfrac12 r^2$ at $r$ , computed from the plane formula $\kappa = \lvert z''\rvert/(1+(z')^2)^{3/2}$ ; it must be so, because the meridian is a normal section of the surface.

Step 5: limits. As $r \to 0$ , $K \to 1$ and $k_1 = k_2 \to 1$ : the vertex is an umbilical point and the paraboloid osculates the unit sphere there. As $r \to\infty$ , $K \sim r^{-4} \to 0$ and $H \sim \tfrac12 r^{-1} \to 0$ : far out, the bowl flattens.

Application(Surface tension, membranes and shells)Original

Young-Laplace. The pressure jump across an interface of surface tension $\sigma$ is $\Delta p = 2\sigma H = \sigma(k_1 + k_2)$ , with $H$ computed for the normal pointing from the low-pressure side into the high-pressure side, i.e. towards the centre of curvature. The factor comes from the first variation of area: for a graph $z = u(x,y)$ the area functional is $\int(1+\lvert Du\rvert^2)^{1/2}dx$ , whose Euler–Lagrange equation Evans writes as $\operatorname{div}\bigl(Du/(1+\lvert Du\rvert^2)^{1/2}\bigr) = 0$ , the left side being $n = 2$ times the mean curvature of the graph. For a spherical bubble of radius $R$ in air, $k_1 = k_2 = 1/R$ and $\Delta p = 2\sigma/R$ for a single interface, doubled to $4\sigma/R$ for a soap bubble with two. A cylindrical jet has $k_1 = 1/R$ , $k_2 = 0$ and $\Delta p = \sigma/R$ ; comparing the two is the start of the Rayleigh-Plateau analysis of why a jet breaks into drops.

Minimal surfaces. A soap film spanning a wire frame has no pressure difference across it, so $H \equiv 0$ everywhere: the two principal curvatures are equal and opposite and every point is a saddle with $K = -k_1^2 \le 0$ . This is the geometric form of the statement that the film minimizes area: Gelfand and Fomin derive the Euler equation $r(1+q^2) - 2spq + t(1+p^2) = 0$ for the least-area problem and read it off as $H = 0$ using the same coordinate formula for $H$ proved above. It is also why no soap film has an elliptic point.

Lipid membranes. The Canham-Helfrich energy of a biological membrane is $\int \bigl[\tfrac{\kappa_b}{2}(2H - c_0)^2 + \bar\kappa K\bigr]\,dA$ , with $\kappa_b$ a bending modulus and $c_0$ a spontaneous curvature. The Gauss-Bonnet theorem (Gauss-Bonnet) makes the $K$ term a topological constant, so shape equations for red blood cells and vesicles are driven entirely by $H$ .

Structural stiffness. A shell with $K \ne 0$ carries load in membrane stress rather than in bending, which is why a dome, an eggshell and a curved drinks can are strong far out of proportion to their thickness, while the same material rolled with $K = 0$ buckles. The $K = 0$ case is also the one in which the shell can be made from a flat sheet without stretching, a point taken up in Gauss's Theorema Egregium.

Theorem

Gauss's Theorema Egregium

Adapted

Add to PDF

Proves Gauss's remarkable theorem that the Gaussian curvature $K$ , defined through the second fundamental form and so apparently dependent on how the surface sits in $\mathbb{R}^3$ , is in fact computable from $E, F, G$ and their derivatives alone, and is therefore unchanged by any local isometry; the Gauss formula that does it is derived, and the consequences for flat maps of a curved surface are drawn.

Prerequisites: The second fundamental form and curvature, Christoffel symbols and the covariant derivative, Regular surfaces and the first fundamental form

Definition(Local isometry; the Christoffel symbols of a parametrized surface)AdaptedAdapted from do Carmo (2016), §4-2, Defs. 1–2, pp. 221–222, Prop. 1, p. 223 and Def. 3, p. 229, together with §4-3, systems (1) and (2), pp. 235–236

Let $S, \bar S$ be regular surfaces.

A diffeomorphism $\varphi : S \to \bar S$ is an isometry if $\langle d\varphi_p(v_1), d\varphi_p(v_2)\rangle = \langle v_1, v_2\rangle_p$ for every $p\in S$ and all $v_1, v_2 \in T_pS$ : it preserves the first fundamental form. It is a local isometry at $p$ if some neighbourhood of $p$ is mapped isometrically onto an open subset of $\bar S$ . The coordinate criterion is do Carmo's Proposition 1: if $\mathbf{x} : U \to S$ and $\bar{\mathbf{x}} : U \to \bar S$ are parametrizations with the same parameter domain and $E = \bar E$ , $F = \bar F$ , $G = \bar G$ throughout $U$ , then $\varphi = \bar{\mathbf{x}}\circ\mathbf{x}^{-1}$ is a local isometry. By contrast $\varphi$ is conformal if $\langle d\varphi_p(v_1), d\varphi_p(v_2)\rangle = \lambda^2(p)\langle v_1,v_2\rangle_p$ for a nowhere-zero function $\lambda$ ; conformal maps preserve angles but not lengths.
Let $\mathbf{x}$ be a parametrization compatible with the orientation $N$ . The triple $\{\mathbf{x}_u, \mathbf{x}_v, N\}$ is a basis of $\mathbb{R}^3$ at every point of the chart, the analogue for a surface of the Frenet trihedron. Expanding the derivatives of its members in it defines the Christoffel symbols $\Gamma^k_{ij}$ of $S$ in the parametrization $\mathbf{x}$ : $\begin{aligned} \mathbf{x}_{uu} &= \Gamma^1_{11}\mathbf{x}_u + \Gamma^2_{11}\mathbf{x}_v + e\,N, \\ \mathbf{x}_{uv} &= \Gamma^1_{12}\mathbf{x}_u + \Gamma^2_{12}\mathbf{x}_v + f\,N, \\ \mathbf{x}_{vv} &= \Gamma^1_{22}\mathbf{x}_u + \Gamma^2_{22}\mathbf{x}_v + g\,N, \\ N_u &= a_{11}\mathbf{x}_u + a_{21}\mathbf{x}_v, \qquad N_v = a_{12}\mathbf{x}_u + a_{22}\mathbf{x}_v . \end{aligned}$ Taking the inner product of each of the first three lines with $N$ kills the tangential terms and identifies the normal components as $e$ , $f$ and $g$ , the quantities introduced in the second fundamental form and curvature. Since $\mathbf{x}_{uv} = \mathbf{x}_{vu}$ , the symbols are symmetric in their lower indices, $\Gamma^k_{12} = \Gamma^k_{21}$ . The same objects appear index-first in Christoffel symbols and the covariant derivative; here they are indexed by the surface coordinates $u, v$ with $1 \leftrightarrow u$ , $2 \leftrightarrow v$ .

Theorem(Theorema Egregium and the Gauss formula)AdaptedAdapted from do Carmo (2016), §4-3, system (2), p. 235, Eq. (5) and the Theorema Egregium, p. 237, and Exercises 1–2, p. 240

Let $S$ be an oriented regular surface and $\mathbf{x} : U \to S$ a parametrization.

(The Christoffel symbols are intrinsic.) The six functions $\Gamma^k_{ij}$ satisfy the three pairs of linear equations $\begin{aligned} \Gamma^1_{11}E + \Gamma^2_{11}F &= \tfrac12 E_u, &\qquad \Gamma^1_{11}F + \Gamma^2_{11}G &= F_u - \tfrac12 E_v, \\ \Gamma^1_{12}E + \Gamma^2_{12}F &= \tfrac12 E_v, &\qquad \Gamma^1_{12}F + \Gamma^2_{12}G &= \tfrac12 G_u, \\ \Gamma^1_{22}E + \Gamma^2_{22}F &= F_v - \tfrac12 G_u, &\qquad \Gamma^1_{22}F + \Gamma^2_{22}G &= \tfrac12 G_v . \end{aligned}$ Each pair has determinant $EG - F^2 \ne 0$ , so the $\Gamma^k_{ij}$ are determined by $E, F, G$ and their first derivatives alone.
(Gauss formula.) $E\,K \;=\; (\Gamma^2_{11})_v - (\Gamma^2_{12})_u + \Gamma^1_{11}\Gamma^2_{12} + \Gamma^2_{11}\Gamma^2_{22} - \Gamma^1_{12}\Gamma^2_{11} - (\Gamma^2_{12})^2 .$ (do Carmo writes the same identity with both sides negated.)
(Theorema Egregium.) The Gaussian curvature of a surface is invariant under local isometries: if $\varphi : V \subseteq S \to \bar S$ is a local isometry, then $\bar K(\varphi(q)) = K(q)$ for every $q \in V$ .
(Orthogonal and isothermal charts.) If $F \equiv 0$ , $K = -\frac{1}{2\sqrt{EG}}\left[\left(\frac{E_v}{\sqrt{EG}}\right)_v + \left(\frac{G_u}{\sqrt{EG}}\right)_u\right];$ if moreover $E = G = \lambda(u,v)$ , this collapses to $K = -\dfrac{1}{2\lambda}\,\Delta\log\lambda$ with $\Delta = \partial_u^2 + \partial_v^2$ .
(The converse fails.) Two surfaces can have equal Gaussian curvature at corresponding points without that correspondence being an isometry. do Carmo's Exercise 3 (p. 240), which he offers as evidence against any converse to the Gauss theorem, gives the pair $\mathbf{x}(u,v) = (u\cos v, u\sin v, \log u)$ and $\bar{\mathbf{x}}(u,v) = (u\cos v, u\sin v, v)$ for $u > 0$ .

Idea of the proofOriginal

$K$ is built from $e, f, g$ , which are normal components and so involve the ambient space. The trick is to look not at $K$ itself but at an identity that $e, f, g$ are forced to satisfy. The three vectors $\mathbf{x}_u, \mathbf{x}_v, N$ form a frame, and writing the second derivatives of $\mathbf{x}$ in that frame produces tangential coefficients (the Christoffel symbols) and normal coefficients ( $e, f, g$ ). The tangential coefficients are not new data: inner products with $\mathbf{x}_u$ and $\mathbf{x}_v$ turn them into derivatives of $E$ , $F$ and $G$ . Now impose the one constraint that any smooth map must satisfy, $(\mathbf{x}_{uu})_v = (\mathbf{x}_{uv})_u$ , and read off the $\mathbf{x}_v$ component. Everything on one side is made of Christoffel symbols; everything on the other turns out to be $-EK$ after the Weingarten coefficients are substituted. So $EK$ is a formula in $E, F, G$ and their derivatives, and a map that preserves $E, F, G$ preserves it.

ProofAdaptedAdapted from do Carmo (2016), §4-3, pp. 235–238, with the Weingarten coefficients $a_{ij}$ of §3-3, p. 157

Part 1. Take the inner product of the first three expansions with $\mathbf{x}_u$ and with $\mathbf{x}_v$ ; the $N$ terms drop out because $N \perp \mathbf{x}_u, \mathbf{x}_v$ . That gives the six left-hand sides. For the right-hand sides, differentiate the definitions of $E, F, G$ :

E_u = 2\langle \mathbf{x}_{uu}, \mathbf{x}_u\rangle, \qquad E_v = 2\langle\mathbf{x}_{uv},\mathbf{x}_u\rangle, \qquad G_u = 2\langle\mathbf{x}_{uv},\mathbf{x}_v\rangle, \qquad G_v = 2\langle\mathbf{x}_{vv},\mathbf{x}_v\rangle,

and $F_u = \langle\mathbf{x}_{uu},\mathbf{x}_v\rangle + \langle\mathbf{x}_u,\mathbf{x}_{uv}\rangle = \langle\mathbf{x}_{uu},\mathbf{x}_v\rangle + \tfrac12 E_v$ , so $\langle\mathbf{x}_{uu},\mathbf{x}_v\rangle = F_u - \tfrac12 E_v$ ; similarly $F_v = \langle\mathbf{x}_{uv},\mathbf{x}_v\rangle + \langle\mathbf{x}_u,\mathbf{x}_{vv}\rangle$ gives $\langle\mathbf{x}_{vv},\mathbf{x}_u\rangle = F_v - \tfrac12 G_u$ . Substituting produces exactly the six equations stated. Each pair is a $2\times2$ linear system in $(\Gamma^1_{ij}, \Gamma^2_{ij})$ with coefficient matrix $\begin{pmatrix}E&F\\F&G\end{pmatrix}$ , invertible because $EG-F^2 > 0$ (regular surfaces). Hence each $\Gamma^k_{ij}$ is a rational expression in $E, F, G, E_u, E_v, F_u, F_v, G_u, G_v$ .

Part 2. $\mathbf{x}$ is smooth, so the third mixed partial derivatives commute: $(\mathbf{x}_{uu})_v = (\mathbf{x}_{uv})_u$ . Differentiate the two expansions and keep only the $\mathbf{x}_v$ component of each side. On the left,

(\mathbf{x}_{uu})_v = (\Gamma^1_{11})_v\mathbf{x}_u + \Gamma^1_{11}\mathbf{x}_{uv} + (\Gamma^2_{11})_v\mathbf{x}_v + \Gamma^2_{11}\mathbf{x}_{vv} + e_vN + eN_v,

and substituting the expansions of $\mathbf{x}_{uv}$ , $\mathbf{x}_{vv}$ and $N_v$ gives the $\mathbf{x}_v$ coefficient

\Gamma^1_{11}\Gamma^2_{12} + (\Gamma^2_{11})_v + \Gamma^2_{11}\Gamma^2_{22} + e\,a_{22}.

On the right,

(\mathbf{x}_{uv})_u = (\Gamma^1_{12})_u\mathbf{x}_u + \Gamma^1_{12}\mathbf{x}_{uu} + (\Gamma^2_{12})_u\mathbf{x}_v + \Gamma^2_{12}\mathbf{x}_{uv} + f_uN + fN_u,

with $\mathbf{x}_v$ coefficient

\Gamma^1_{12}\Gamma^2_{11} + (\Gamma^2_{12})_u + (\Gamma^2_{12})^2 + f\,a_{21}.

Since $\mathbf{x}_u, \mathbf{x}_v, N$ are linearly independent, the two coefficients are equal. Rearranging,

(\Gamma^2_{11})_v - (\Gamma^2_{12})_u + \Gamma^1_{11}\Gamma^2_{12} + \Gamma^2_{11}\Gamma^2_{22} - \Gamma^1_{12}\Gamma^2_{11} - (\Gamma^2_{12})^2 = f\,a_{21} - e\,a_{22}.

Now insert the Weingarten coefficients computed in the second fundamental form and curvature, namely $a_{21} = (eF - fE)/(EG-F^2)$ and $a_{22} = (fF - gE)/(EG-F^2)$ (do Carmo displays all four on p. 157; in the sibling unit they come from multiplying out the matrix identity for the $a_{ij}$ ):

f a_{21} - e a_{22} = \frac{f(eF - fE) - e(fF - gE)}{EG-F^2} = \frac{efF - f^2E - efF + egE}{EG-F^2} = E\,\frac{eg - f^2}{EG - F^2} = E\,K ,

the two terms in $efF$ cancelling. This is the Gauss formula.

Part 3. Let $\varphi$ be a local isometry at $q \in V$ and let $\mathbf{x} : U \to S$ be a parametrization at $q$ with $\mathbf{x}(U) \subseteq V$ . Then $\bar{\mathbf{x}} = \varphi\circ\mathbf{x}$ is a parametrization of $\bar S$ at $\varphi(q)$ with the same parameter domain $U$ . Because $\varphi$ preserves inner products of tangent vectors and $\bar{\mathbf{x}}_u = d\varphi(\mathbf{x}_u)$ , $\bar{\mathbf{x}}_v = d\varphi(\mathbf{x}_v)$ , we get $\bar E = E$ , $\bar F = F$ , $\bar G = G$ as functions on $U$ . By Part 1 the Christoffel symbols are functions of $E, F, G$ and their derivatives only, so $\bar\Gamma^k_{ij} = \Gamma^k_{ij}$ on $U$ . By Part 2,

\bar K = \frac{1}{\bar E}\Bigl[(\bar\Gamma^2_{11})_v - (\bar\Gamma^2_{12})_u + \cdots\Bigr] = \frac{1}{E}\Bigl[(\Gamma^2_{11})_v - (\Gamma^2_{12})_u + \cdots\Bigr] = K

at corresponding parameter values, which is the assertion. (Division by $E$ is legitimate since $E = \lvert\mathbf{x}_u\rvert^2 > 0$ .)

Part 4. Put $F = 0$ in Part 1. The three pairs decouple and give

\Gamma^1_{11} = \frac{E_u}{2E},\quad \Gamma^2_{11} = -\frac{E_v}{2G},\quad \Gamma^1_{12} = \frac{E_v}{2E},\quad \Gamma^2_{12} = \frac{G_u}{2G},\quad \Gamma^1_{22} = -\frac{G_u}{2E},\quad \Gamma^2_{22} = \frac{G_v}{2G}.

Substituting into the Gauss formula and collecting terms turns the right-hand side into the stated divergence-like expression; this is do Carmo's Exercise 1, and the example below carries the substitution out on a concrete chart. For the isothermal case put $E = G = \lambda$ : then $\sqrt{EG} = \lambda$ and

K = -\frac{1}{2\lambda}\left[\left(\frac{\lambda_v}{\lambda}\right)_v + \left(\frac{\lambda_u}{\lambda}\right)_u\right] = -\frac{1}{2\lambda}\bigl[(\log\lambda)_{vv} + (\log\lambda)_{uu}\bigr] = -\frac{1}{2\lambda}\Delta\log\lambda,

using $(\log\lambda)_u = \lambda_u/\lambda$ .

Part 5. Run the method of the second fundamental form and curvature on do Carmo's pair. Both surfaces give

K(u,v) = -\frac{1}{(1+u^2)^2}

at corresponding parameter values, so the curvatures agree everywhere. Their first fundamental forms, however, are

(E, F, G) = \Bigl(1 + \tfrac{1}{u^2},\; 0,\; u^2\Bigr) \quad\text{and}\quad (\bar E, \bar F, \bar G) = \bigl(1,\; 0,\; 1 + u^2\bigr),

which differ at every $u$ (at $u = 1$ , for instance, $E = 2$ against $\bar E = 1$ ). That the coordinate criterion of the definition is not met proves nothing by itself, since the criterion is only sufficient; but the numbers settle the matter directly. For $\varphi = \bar{\mathbf{x}}\circ\mathbf{x}^{-1}$ we have $d\varphi(\mathbf{x}_u) = \bar{\mathbf{x}}_u$ , so at $u = 1$

\lvert d\varphi(\mathbf{x}_u)\rvert^2 = \bar E = 1 \ne 2 = E = \lvert\mathbf{x}_u\rvert^2 ,

and $\varphi$ does not preserve the first fundamental form: this correspondence is not an isometry, even though it matches the Gaussian curvatures. (It is this particular correspondence that fails; the argument does not by itself rule out some other local isometry between the two surfaces.) $\square$

Remark(Where the hypotheses are used)Original

$EG - F^2 \ne 0$ . Used in Part 1 to solve for the Christoffel symbols and again when dividing by $E$ in Part 3. At a singular point of a chart the symbols blow up, which is why the formula must be applied chart by chart.
Equality of third mixed partials. Part 2 differentiates a second derivative, so $\mathbf{x}$ must be $C^3$ . For a merely $C^2$ surface $e, f, g$ and $K$ exist but the Gauss formula has no meaning, and the theorem is not available.
Local, not global. The theorem compares $K$ at corresponding points of a local isometry. It says nothing about surfaces being globally congruent, and Part 5 shows the converse direction fails even locally.
Only $K$ , not $H$ . The mean curvature is not intrinsic. The plane and a cylinder of radius $a$ are locally isometric (roll the plane up without stretching) and both have $K = 0$ , but $H = 0$ for the plane and $\lvert H\rvert = 1/(2a)$ for the cylinder. The second fundamental form as a whole is extrinsic; only the combination $eg - f^2$ , divided by $EG - F^2$ , survives.
Orientability. The proof uses a chart-compatible $N$ , hence $e, f, g$ . But $K$ and the whole right-hand side of the Gauss formula are unchanged by $N \mapsto -N$ , so the theorem holds on non-orientable surfaces too, applied in each chart.
Isometry versus conformal or area-preserving. The theorem fails for the weaker notions: the sphere is locally conformal to the plane (Mercator's projection) and is locally area-preserving onto the circumscribed cylinder (do Carmo's Exercise 19, pp. 233–234), yet $K \ne 0$ on the sphere and $K = 0$ on both targets. do Carmo's Exercise 18 (p. 233) closes the gap: a map that is both conformal and area-preserving is an isometry.

Method(Deciding whether two surfaces can be locally isometric)Original

Choose convenient charts on each surface and compute $E, F, G$ (first fundamental form).
Compute $K$ for each. Use the extrinsic route $K = (eg-f^2)/(EG-F^2)$ when the embedding is at hand, and the intrinsic route of Part 4 when only the metric is given.
Express $K$ on each surface as a function of a quantity that is itself intrinsic, for example of the geodesic distance from a marked point, so that the comparison does not depend on how the charts are labelled.
If the value sets of $K$ differ, stop: no local isometry exists between the corresponding regions. This is the usual way the theorem is used, as a cheap obstruction.
If the values agree, nothing is proved: by Part 5 the converse is false. One must then construct the isometry, usually by finding charts in which $E, F, G$ match identically and applying the coordinate criterion.
Special case $K \equiv 0$ . A surface with vanishing Gaussian curvature is called developable and is, locally, isometric to a plane; a surface with $K$ constant and nonzero is locally isometric to a sphere ( $K>0$ ) or to the pseudosphere ( $K<0$ ).

Example(The paraboloid, computed without leaving the surface)Original

In the second fundamental form and curvature the paraboloid $z = \tfrac12(x^2+y^2)$ was found to have

K = \frac{1}{(1+r^2)^2}

by way of the Gauss map, the second derivatives of $\mathbf{x}$ and the normal $N$ — all of them references to the ambient $\mathbb{R}^3$ . We now recover the same function from the line element alone,

ds^2 = (1+r^2)\,dr^2 + r^2\,d\theta^2, \qquad E = 1+r^2, \quad F = 0, \quad G = r^2,

writing $u = r$ , $v = \theta$ .

Step 1: Christoffel symbols. Since $F = 0$ , the formulas of Part 4 apply with $E_u = 2r$ , $E_v = 0$ , $G_u = 2r$ , $G_v = 0$ :

\Gamma^1_{11} = \frac{E_u}{2E} = \frac{r}{1+r^2}, \qquad \Gamma^2_{11} = -\frac{E_v}{2G} = 0, \qquad \Gamma^1_{12} = \frac{E_v}{2E} = 0,

\Gamma^2_{12} = \frac{G_u}{2G} = \frac{2r}{2r^2} = \frac1r, \qquad \Gamma^1_{22} = -\frac{G_u}{2E} = -\frac{r}{1+r^2}, \qquad \Gamma^2_{22} = \frac{G_v}{2G} = 0 .

Only $E, F, G$ and their $r$ -derivatives were used.

Step 2: the Gauss formula. Substituting into $E K = (\Gamma^2_{11})_v - (\Gamma^2_{12})_u + \Gamma^1_{11}\Gamma^2_{12} + \Gamma^2_{11}\Gamma^2_{22} - \Gamma^1_{12}\Gamma^2_{11} - (\Gamma^2_{12})^2$ , the first, fourth and fifth terms vanish because $\Gamma^2_{11} = \Gamma^1_{12} = 0$ . What is left is the second, third and sixth,

E K = -\frac{\partial}{\partial r}\!\left(\frac1r\right) + \frac{r}{1+r^2}\cdot\frac1r - \frac{1}{r^2} = \frac{1}{r^2} + \frac{1}{1+r^2} - \frac{1}{r^2} = \frac{1}{1+r^2}.

The two $1/r^2$ terms cancel exactly, which they must: a curvature cannot blow up at $r \to 0$ , where the surface is perfectly smooth. Dividing by $E = 1+r^2$ ,

K = \frac{1}{(1+r^2)^2},

in agreement with the extrinsic computation. At $r = 1$ , $K = \tfrac14$ by both routes.

Step 3: the same by the orthogonal-chart formula. With $\sqrt{EG} = r\sqrt{1+r^2}$ , $E_v = 0$ and $G_u = 2r$ ,

K = -\frac{1}{2r\sqrt{1+r^2}}\left[0 + \frac{\partial}{\partial r}\!\left(\frac{2r}{r\sqrt{1+r^2}}\right)\right] = -\frac{1}{2r\sqrt{1+r^2}}\cdot\frac{\partial}{\partial r}\!\left(\frac{2}{\sqrt{1+r^2}}\right) = -\frac{1}{2r\sqrt{1+r^2}}\cdot\frac{-2r}{(1+r^2)^{3/2}} = \frac{1}{(1+r^2)^2}. \;\checkmark

Step 4: a consequence. Because $K > 0$ everywhere on the paraboloid while $K \equiv 0$ on the plane, no piece of the paraboloid, however small, can be flattened onto a plane without changing lengths. Compare a sphere of radius $a$ , whose metric in geodesic polar coordinates is $ds^2 = d\rho^2 + a^2\sin^2(\rho/a)\,d\theta^2$ : here $E = 1$ , $G = a^2\sin^2(\rho/a)$ , $\sqrt{EG} = a\sin(\rho/a)$ and $G_\rho = 2a\sin(\rho/a)\cos(\rho/a)$ , so

K = -\frac{1}{2a\sin(\rho/a)}\frac{\partial}{\partial\rho}\Bigl(2\cos(\rho/a)\Bigr) = -\frac{1}{2a\sin(\rho/a)}\cdot\left(-\frac{2}{a}\sin(\rho/a)\right) = \frac{1}{a^2},

a number computed without ever mentioning the sphere's embedding. For the Earth, $a = 6371\ \text{km}$ gives $K = 2.4637\times10^{-8}\ \text{km}^{-2}$ — small, but not zero, and that is the whole of the application below.

Application(Why every flat map of the Earth lies)Original

Cartography. A map sheet is a piece of plane, with $K = 0$ . The Earth's surface has $K = 1/a^2 > 0$ . By the Theorema Egregium no map projection, over any region however small, can preserve all distances: the scale bar printed in the corner is valid along some directions and not others. Every projection therefore chooses which property to keep. Mercator's is conformal, so angles and hence compass bearings are right and a rhumb line is straight, at the price of areas inflated towards the poles; Archimedes' radial projection of the sphere onto the circumscribed cylinder — the same map as Lambert's cylindrical equal-area projection — keeps areas, at the price of angles (do Carmo's Exercise 19, pp. 233–234); Gauss's own work on this problem, in the survey of Hanover, is what produced the theorem. Nothing keeps both: by do Carmo's Exercise 18 a projection that was conformal and area-preserving would be an isometry, which is exactly what $K \ne 0$ forbids.

Flat things stay developable. A sheet of paper, a rolled steel plate or a sheet-metal duct has $K = 0$ before deformation, and bending without stretching is a local isometry, so $K = 0$ afterwards. Hence a flat sheet can be formed into a cylinder or a cone (both $K = 0$ ) but never into a sphere or any part of one; dishing a flat blank into a hemisphere requires stretching the material, which is why deep drawing and panel beating thin the metal, and why a world globe is assembled from printed gores rather than from one disc.

Stiffness by curvature. Conversely, a sheet already curved in one direction cannot bend in the other without stretching, because that would change $K$ from $0$ to something nonzero. Folding a slice of pizza across its length so it will not droop, the corrugation of a cardboard sheet or a metal roof, and the strength of a curved can lid are all the same statement: curvature in one direction locks out bending in the other.

Gravitation. The Theorema Egregium is the historical and conceptual starting point for intrinsic geometry, which in turn is the mathematical language of general relativity. There the curvature of spacetime is defined entirely from the metric, with no ambient space to embed into; the Christoffel symbols of Part 1 and the combination of their derivatives in the Gauss formula are the two-dimensional ancestors of the connection coefficients and of the Riemann curvature tensor, for which $K$ is the single independent component in dimension two.

Method

Geodesics

Adapted

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Gives the two equivalent ways of producing the geodesics of a metric — as the curves whose tangent field is parallel along them, and as the critical points of the length (equivalently the energy) functional — derives the geodesic equation $\ddot u^i + \Gamma^i{}_{jk}\dot u^j\dot u^k = 0$ by both routes, and runs the recipe on the sphere, where it returns the great circles and the shortest flight path between two points.

Prerequisites: Christoffel symbols and the covariant derivative, The Euler-Lagrange equation, Regular surfaces and the first fundamental form, The metric tensor, The second fundamental form and curvature, Curves and the Frenet-Serret formulas, Gauss's Theorema Egregium, Existence and uniqueness of solutions

Definition(Geodesic, geodesic curvature)AdaptedAdapted from do Carmo (2016), §4-4, Defs. 8, 8a, 9 and 10 and the discussion on pp. 247–252

Let $S$ be a regular surface (regular surfaces and the first fundamental form) and let $D/dt$ be the covariant derivative along a curve, that is, the tangential part of the ordinary derivative in $\mathbb{R}^3$ (Christoffel symbols and the covariant derivative).

A nonconstant parametrized curve $\gamma : I \to S$ is geodesic at $t$ if its own tangent field is parallel there, $\frac{D\gamma'}{dt}(t) = 0 ,$ and is a parametrized geodesic if this holds for every $t \in I$ . Constant curves are excluded by convention.
Parallel transport preserves inner products, so $\lvert\gamma'(t)\rvert$ is a nonzero constant $c$ along a parametrized geodesic. Hence $s = ct$ is an arc length parameter: a geodesic runs at constant speed, so its parameter differs from arc length only by a constant factor. A geodesic is therefore automatically a regular curve, though it may cross itself.
A regular connected curve $C \subseteq S$ is a geodesic if, near each of its points, its parametrization by arc length is a parametrized geodesic. Seen from outside the surface this says that $\alpha''(s)$ is normal to the tangent plane: a regular curve of $S$ with nowhere-vanishing curvature $k$ is a geodesic exactly when its principal normal $\mathbf{n}$ (the Frenet–Serret formulas) is parallel to the surface normal at every point. Every straight line lying in $S$ is a geodesic as well, and is the case the principal-normal criterion cannot see, $\mathbf{n}$ being undefined where $k = 0$ .
On an oriented surface, a unit field $w$ along $\alpha$ has $Dw/dt$ perpendicular to $w$ and tangent to $S$ , hence $Dw/dt = \lambda\,(N \wedge w)$ for a scalar $\lambda$ . That scalar is the algebraic value of $Dw/dt$ , written $[Dw/dt]$ . For an oriented regular curve $C$ parametrized by arc length, the geodesic curvature is $k_g = \left[\frac{D\alpha'}{ds}\right] ,$ so that the regular geodesics are precisely the curves with $k_g \equiv 0$ . Splitting $\alpha'' = k\mathbf{n}$ into its tangential and normal parts gives, where $k \neq 0$ , $k^2 = k_g^2 + k_n^2 ,$ with $k_n$ the normal curvature of the second fundamental form and curvature. Unlike the geodesic condition itself, the sign of $k_g$ depends on the orientations chosen for $C$ and for $S$ .

Method(The geodesic equations, and what they produce)AdaptedAdapted from do Carmo (2016), §4-4, Eq. (4) and Prop. 5, p. 257, and §4-6, Prop. 4, pp. 296–297; Riley, Hobson & Bence (2006), §26.22, Eq. (26.101), p. 977; Arfken, Weber & Harris (2013), §22.2, Eqs. (22.55)–(22.62), pp. 1103–1104; Gelfand & Fomin (1963), §9, Eq. (3), p. 35 and Example 2, p. 38

(In a surface chart.) Let $\mathbf{x}(u,v)$ parametrize a neighbourhood of a point of $S$ and let a curve be given there by $t \mapsto \mathbf{x}(u(t), v(t))$ . The curve is a parametrized geodesic if and only if $\begin{aligned} u'' + \Gamma^1_{11}(u')^2 + 2\,\Gamma^1_{12}\,u'v' + \Gamma^1_{22}(v')^2 &= 0, \\ v'' + \Gamma^2_{11}(u')^2 + 2\,\Gamma^2_{12}\,u'v' + \Gamma^2_{22}(v')^2 &= 0, \end{aligned}$ with $1 \leftrightarrow u$ , $2 \leftrightarrow v$ and the $\Gamma^k_{ij}$ the Christoffel symbols of the chart. These are do Carmo's differential equations of the geodesics.
(In any coordinates, any dimension.) Let $\mathbf{T} = d\mathbf{r}/ds$ be the unit tangent of a curve $u^i(s)$ parametrized by arc length (Riley, Hobson and Bence write it $\mathbf{t}$ ). The condition $d\mathbf{T}/ds = 0$ is $\frac{d^2u^i}{ds^2} + \Gamma^i{}_{jk}\,\frac{du^j}{ds}\,\frac{du^k}{ds} = 0 ,$ summation implied. Riley, Hobson and Bence derive this for a straight line of Euclidean three-space written in an arbitrary coordinate system, and observe that the derivation carries over with little change to spaces where Euclidean geometry fails; Arfken, Weber and Harris obtain the same equation for a general metric $g_{ij}$ .
(As extremals.) The length of a curve on $S$ is $L[u,v] = \int_{t_0}^{t_1}\sqrt{E(u')^2 + 2Fu'v' + G(v')^2}\;dt ,$ and its Euler equations are the pair written out by Gelfand and Fomin. Arfken, Weber and Harris avoid the square root: a path that extremizes the energy $\mathcal{E}[q] = \int_{A}^{B} g_{ij}\,\dot q^i \dot q^j \,du$ satisfies, through the Euler equations for $n$ unknown functions, exactly the geodesic equation of part 2 with $s$ replaced by $u$ . A parameter admissible here is one related to arc length by $u = as+b$ .
(Existence and uniqueness.) Given $p \in S$ and $w \in T_pS$ with $w \ne 0$ , there are an $\varepsilon > 0$ and a unique parametrized geodesic $\gamma : (-\varepsilon,\varepsilon) \to S$ with $\gamma(0) = p$ and $\gamma'(0) = w$ .
(Local minimisation.) Each $p \in S$ has a neighbourhood $W$ with this property: if $\gamma$ is a parametrized geodesic lying in $W$ with $\gamma(0) = p$ and $\gamma(t_1) = q$ , and $\alpha$ is any regular curve in $S$ joining $p$ to $q$ , then $l_\gamma \le l_\alpha$ , and equality forces the two traces to coincide. The restriction to $W$ is essential. Two points of a sphere are joined by two arcs of a great circle, both of them geodesics, and only the shorter one is the shortest path; Arfken, Weber and Harris nonetheless call the longer arc a geodesic as well, because their definition demands only that no slight deformation of a path shorten it.

Idea of the proofOriginal

Two descriptions of a straight line survive the loss of flatness, and they turn out to give the same curves. "The tangent never turns" becomes "the tangent field is parallel along the curve", and since parallel transport is the covariant derivative set to zero, writing $\gamma' = u'\mathbf{x}_u + v'\mathbf{x}_v$ and demanding $D\gamma'/dt = 0$ produces a second-order system whose only non-trivial coefficients are the Christoffel symbols. "The path is the shortest one" becomes a variational problem; the length integrand carries an awkward square root, which disappears if one extremizes the energy instead, and the Euler equations of the energy are the very same system. The two routes meet because the energy's Euler equations force the speed to be constant, and at constant speed the length and energy problems have identical extremals.

ProofAdaptedAdapted from do Carmo (2016), §4-4, pp. 256–257; Riley, Hobson & Bence (2006), §26.22, pp. 976–977; Arfken, Weber & Harris (2013), §22.2, Eqs. (22.57)–(22.62), pp. 1103–1104

Part 1: from parallelism. Let $w = a(t)\,\mathbf{x}_u + b(t)\,\mathbf{x}_v$ be a tangent field along $t \mapsto \mathbf{x}(u(t),v(t))$ . Differentiating in $\mathbb{R}^3$ and using the chain rule,

\frac{dw}{dt} = a'\mathbf{x}_u + b'\mathbf{x}_v + a\bigl(\mathbf{x}_{uu}u' + \mathbf{x}_{uv}v'\bigr) + b\bigl(\mathbf{x}_{uv}u' + \mathbf{x}_{vv}v'\bigr).

The covariant derivative is the tangential part, so we expand the second derivatives in the frame $\{\mathbf{x}_u,\mathbf{x}_v,N\}$ as in Gauss's Theorema Egregium and discard every $N$ term; what is left of $\mathbf{x}_{uu}$ is $\Gamma^1_{11}\mathbf{x}_u + \Gamma^2_{11}\mathbf{x}_v$ , and similarly for the others. Collecting the $\mathbf{x}_u$ and $\mathbf{x}_v$ coefficients,

\frac{Dw}{dt} = \Bigl[a' + \Gamma^1_{11}au' + \Gamma^1_{12}(av' + bu') + \Gamma^1_{22}bv'\Bigr]\mathbf{x}_u + \Bigl[b' + \Gamma^2_{11}au' + \Gamma^2_{12}(av' + bu') + \Gamma^2_{22}bv'\Bigr]\mathbf{x}_v ,

where the symmetry $\Gamma^k_{12} = \Gamma^k_{21}$ was used to merge the two mixed terms. Now take for $w$ the tangent field of the curve itself, $a = u'$ and $b = v'$ . Since $\mathbf{x}_u, \mathbf{x}_v$ are linearly independent, $D\gamma'/dt = 0$ forces each bracket to vanish separately, and the mixed terms combine into $2\Gamma^k_{12}u'v'$ . That is the system of part 1.

Part 2: the same in any coordinates. Parametrize by arc length and set $\mathbf{T} = d\mathbf{r}/ds = T^i e_i$ , so that $T^j = du^j/ds$ . Requiring the tangent to point always in the same direction means $d\mathbf{T}/ds = 0$ ; this derivative is the absolute derivative of $\mathbf{T}$ along the curve, whose components are $T^i{}_{;k}\,du^k/ds$ (Christoffel symbols and the covariant derivative). Writing the covariant derivative out,

\left(\frac{dT^i}{ds} + \Gamma^i{}_{jk}\,T^j\,\frac{du^k}{ds}\right)e_i = 0 ,

and substituting $T^j = du^j/ds$ gives the geodesic equation of part 2. In the flat space of Riley, Hobson and Bence the curves so produced are the straight lines, which supplies a check: in cylindrical polars the only nonvanishing symbols are $\Gamma^1{}_{22} = -\rho$ and $\Gamma^2{}_{12} = \Gamma^2{}_{21} = 1/\rho$ , and the three equations read $\ddot\rho - \rho\dot\phi^2 = 0$ , $\ddot\phi + (2/\rho)\dot\rho\dot\phi = 0$ , $\ddot z = 0$ — the familiar radial and angular equations of a free particle.

Part 3: from the energy. Put $2T = g_{ij}\dot q^i\dot q^j$ with $\dot q^i = dq^i/du$ , and apply to $\mathcal{E}$ the Euler equations for several unknown functions. For each index $l$ ,

\frac{\partial}{\partial q^l}\bigl(g_{ij}\dot q^i\dot q^j\bigr) - \frac{d}{du}\frac{\partial}{\partial \dot q^l}\bigl(g_{ij}\dot q^i\dot q^j\bigr) = 0 .

The metric depends on the $q^k$ but not on the $\dot q^k$ , so the first term is $(\partial_l g_{ij})\dot q^i\dot q^j$ , while the second differentiates $g_{lj}\dot q^j + g_{il}\dot q^i = 2g_{lj}\dot q^j$ . Expanding $d(g_{lj})/du = (\partial_i g_{lj})\dot q^i$ ,

(\partial_l g_{ij})\,\dot q^i\dot q^j - 2(\partial_i g_{lj})\,\dot q^i\dot q^j - 2 g_{lj}\,\ddot q^j = 0 .

Divide by $2$ , multiply through by $-1$ , and rename the dummy indices in the middle term so that it is symmetric in $i$ and $j$ , which is legitimate because it is contracted with the symmetric $\dot q^i\dot q^j$ ; this turns $-(\partial_i g_{lj})\dot q^i\dot q^j$ into $-\tfrac12(\partial_i g_{lj} + \partial_j g_{li})\dot q^i\dot q^j$ . Hence

g_{lj}\ddot q^j + \tfrac12\bigl(\partial_i g_{lj} + \partial_j g_{li} - \partial_l g_{ij}\bigr)\dot q^i\dot q^j = 0 .

Multiply by the inverse metric $g^{ml}$ and use $g^{ml}g_{lj} = \delta^m_j$ together with the metric formula for the Christoffel symbols (Christoffel symbols and the covariant derivative):

\ddot q^m + \Gamma^m{}_{ij}\,\dot q^i\dot q^j = 0 ,

which is the geodesic equation again.

Part 4: why dropping the square root is legitimate. Write $F = \sqrt{2T}$ , the length integrand, and suppose $F$ never vanishes. Then

\frac{\partial F}{\partial \dot q^l} = \frac{g_{lj}\dot q^j}{F}, \qquad \frac{\partial F}{\partial q^l} = \frac{(\partial_l g_{ij})\dot q^i\dot q^j}{2F},

so the Euler equation for $F$ , multiplied through by $F$ , is

\frac{d}{du}\bigl(g_{lj}\dot q^j\bigr) - \tfrac12(\partial_l g_{ij})\dot q^i\dot q^j \;=\; \frac{1}{F}\frac{dF}{du}\;g_{lj}\dot q^j .

The left-hand side is $-\tfrac12$ times the Euler expression for the energy obtained in part 3, so it vanishes exactly when the energy equation holds. So the two variational problems have the same equations precisely when $dF/du = 0$ , that is, when the parameter runs at constant speed. That condition is not an extra assumption on the energy side: contracting the energy equation $2\,d(g_{lj}\dot q^j)/du = (\partial_l g_{ij})\dot q^i\dot q^j$ with $\dot q^l$ gives $2\,g_{lj}\dot q^l\ddot q^j = -\,(\partial_k g_{lj})\dot q^k\dot q^l\dot q^j$ , and substituting this into

\frac{d}{du}\bigl(g_{ij}\dot q^i\dot q^j\bigr) = (\partial_k g_{ij})\dot q^k\dot q^i\dot q^j + 2 g_{ij}\dot q^i \ddot q^j

makes the right-hand side vanish identically. An extremal of the energy therefore has constant speed of its own accord, and is an extremal of the length; conversely a length extremal reparametrized proportionally to arc length solves the energy equations. This is why Arfken, Weber and Harris may replace the minimisation of $J$ by that of the action without changing the geodesics.

Statement parts 4 and 5. The geodesic equations are a system of two (in general $n$ ) second-order ordinary differential equations whose right-hand sides are smooth functions of $(u,v,u',v')$ on the chart, so the existence and uniqueness theorem for such systems (existence and uniqueness) gives a unique local solution for each initial position and initial velocity, which is do Carmo's Proposition 5; the exclusion of $w = 0$ matches the exclusion of constant curves from the definition. The local minimising property is do Carmo's Proposition 4, proved there with geodesic polar coordinates, and we take it as stated. $\square$

Remark(Where the hypotheses are used)Original

Nonconstant curve, $w \ne 0$ . A constant curve satisfies $D\gamma'/dt = 0$ trivially and would wreck the uniqueness statement, since infinitely many parametrizations of a point would qualify; this is why the definition excludes it and why part 4 of the statement asks for $w \ne 0$ .
Constant speed. Part 1 of the definition gives it for free from the parallelism condition; the energy functional imposes it, as part 4 of the proof shows. It is a genuine restriction: the trace of a geodesic traversed with a varying speed is no longer a solution of the geodesic equation, although it is still a length extremal. Any statement that "geodesics are the critical points of length" is a statement about traces, not about parametrized curves.
$F \ne 0$ in part 4 of the proof. The division by $F$ fails exactly at a stationary point of the parametrization. This is the degeneracy of the length functional: it is invariant under reparametrization, so its Euler equations can never determine the parameter, and one equation of the system is redundant.
Charts. The equations of part 1 hold on one coordinate neighbourhood at a time. A geodesic that leaves the chart is continued by solving the equations of the next chart with matching initial data; nothing in the local theory guarantees that a geodesic can be continued for all time, and on an incomplete surface (the plane with a point deleted, say) it cannot.
Local, not global, minimisation. Part 5 of the statement is confined to a small neighbourhood $W$ , and outside it the comparison with every curve joining $p$ to $q$ breaks down: of the two great-circle arcs between two points of a sphere, both geodesics, only one is the shortest path. Arfken, Weber and Harris make the companion point about their Fig. 22.2 — the Euler equation is necessary and not sufficient, so a solution of it must still be checked against the conditions of the problem. This is exactly why "geodesic" is defined here by an equation rather than by a minimum.
Smoothness. Part 1 of the proof differentiates $\mathbf{x}$ twice, so a $C^2$ surface suffices to write the equations down; the existence theorem then needs the coefficients, hence the $\Gamma^k_{ij}$ , to be Lipschitz, which is a condition on third derivatives of $\mathbf{x}$ .

Method(Finding the geodesics of a given metric)Original

Write the line element. Choose coordinates and read off $g_{ij}$ , or for a surface the coefficients $E, F, G$ of the first fundamental form. Everything below uses only these.
Decide which route is cheaper. If the metric is diagonal and simple, compute the Christoffel symbols and write the equations of part 1 or 2 of the statement. If it is not, go straight to step 3: the energy route needs no Christoffel symbols at all.
Energy route. Form $2T = g_{ij}\dot q^i\dot q^j$ , treat it as a Lagrangian, and write one Euler–Lagrange equation per coordinate. Read the Christoffel symbols off the result if you want them.
Harvest the first integrals before integrating anything. Two come for free:
- Speed. $g_{ij}\dot q^i\dot q^j = \text{const}$ ; set it to $1$ to make the parameter arc length.
- Cyclic coordinates. If $2T$ does not contain $q^m$ explicitly, then $\partial T/\partial\dot q^m = g_{mj}\dot q^j$ is constant along every geodesic.
Surfaces of revolution: use Clairaut. For $\mathbf{x}(u,v) = (f(v)\cos u, f(v)\sin u, g(v))$ the coordinate $u$ is cyclic, and the first equation integrates at once to $f^2u' = c$ . With $r = f$ the radius of the parallel through the point and $\theta$ the angle between the geodesic and that parallel, this is Clairaut's relation $r\cos\theta = \lvert c\rvert = \text{const} ,$ which already settles many questions without any integration: a geodesic cannot reach a parallel of radius smaller than $\lvert c\rvert$ , and it is tangent to the parallel of radius exactly $\lvert c\rvert$ .
Identify the obvious solutions first. A curve fixed by an isometry of the surface is a geodesic: meridians of a surface of revolution, and any curve cut out by a plane of mirror symmetry, qualify. Check them against the equations rather than assuming it.
Count. By part 4 of the statement there is exactly one geodesic through each point in each direction. Once you have produced a family with that many members, you have them all; no further solutions exist.
Check the answer numerically. Substitute the closed-form solution back into both geodesic equations, and verify that the speed is constant.

Example(Great circles on the sphere, and the shortest flight between two points at the same latitude)Original

Take the sphere of radius $a$ with colatitude $\theta$ and longitude $\phi$ , so that the line element is

ds^2 = a^2\,d\theta^2 + a^2\sin^2\theta\,d\phi^2, \qquad g_{\theta\theta} = a^2,\quad g_{\phi\phi} = a^2\sin^2\theta,\quad g_{\theta\phi} = 0 .

Step 1: the equations, by the energy route. With $2T = a^2\dot\theta^2 + a^2\sin^2\theta\,\dot\phi^2$ the two Euler–Lagrange equations are

\frac{d}{ds}\bigl(2a^2\dot\theta\bigr) - 2a^2\sin\theta\cos\theta\,\dot\phi^2 = 0, \qquad \frac{d}{ds}\bigl(2a^2\sin^2\theta\,\dot\phi\bigr) = 0 ,

that is, after dividing by $2a^2$ and expanding the second,

\ddot\theta - \sin\theta\cos\theta\,\dot\phi^2 = 0, \qquad \ddot\phi + 2\cot\theta\,\dot\theta\dot\phi = 0 .

Comparing with part 2 of the statement reads off the Christoffel symbols: $\Gamma^\theta{}_{\phi\phi} = -\sin\theta\cos\theta$ and $\Gamma^\phi{}_{\theta\phi} = \Gamma^\phi{}_{\phi\theta} = \cot\theta$ , all others zero. The second equation is already the conserved quantity of step 4 of the method: $\sin^2\theta\,\dot\phi = \text{const}$ , the sphere's form of Clairaut's relation.

Step 2: the equator solves them. Put $\theta \equiv \pi/2$ and $\phi = s/a$ . Then $\ddot\theta = 0$ and $\sin\theta\cos\theta = 0$ , so the first equation holds; $\ddot\phi = 0$ and $\dot\theta = 0$ , so does the second. The speed is $a^2\sin^2(\pi/2)\,(1/a)^2 = 1$ , so $s$ is arc length. The equator is a geodesic.

Step 3: therefore every great circle is, and nothing else. Every rotation of $\mathbb{R}^3$ maps the sphere to itself preserving distances, so it is an isometry, and the geodesic equations depend only on $E, F, G$ (Gauss's Theorema Egregium), hence are unchanged by it. Any great circle is the image of the equator under a rotation, so it is a geodesic. Through a given point in a given direction there passes exactly one great circle; by part 4 of the statement there is exactly one geodesic. So the geodesics of the sphere are precisely the great circles, with constant-speed parametrizations.

Step 4: a numerical case. Let $A$ be the point at latitude $45^\circ$ N, longitude $0^\circ$ and let $B$ be at latitude $45^\circ$ N, longitude $90^\circ$ E, on a sphere of radius $a = 6371\ \mathrm{km}$ . As unit vectors,

\hat A = \Bigl(\tfrac{1}{\sqrt2},\,0,\,\tfrac{1}{\sqrt2}\Bigr), \qquad \hat B = \Bigl(0,\,\tfrac{1}{\sqrt2},\,\tfrac{1}{\sqrt2}\Bigr), \qquad \hat A\cdot\hat B = \tfrac12 .

The central angle between them is $\arccos\tfrac12 = \pi/3$ , so the geodesic joining them has length

L_{\text{geo}} = a\,\frac{\pi}{3} = 6371 \times 1.047198 = 6671.70\ \mathrm{km}.

Flying along the $45^\circ$ parallel instead — a curve of constant latitude, which is not a geodesic, since it is not a great circle — costs

L_{\text{par}} = a\cos 45^\circ \times \frac{\pi}{2} = 6371 \times 0.707107 \times 1.570796 = 7076.40\ \mathrm{km},

longer by $404.70\ \mathrm{km}$ , about $6\%$ .

Step 5: where the geodesic goes, from Clairaut alone. The unit normal to the plane of the great circle is

\hat n = \frac{\hat A\times\hat B}{\lvert \hat A\times\hat B\rvert} = \frac{(-\tfrac12,\,-\tfrac12,\,\tfrac12)}{\sqrt{3}/2} = \Bigl(-\tfrac{1}{\sqrt3},\,-\tfrac{1}{\sqrt3},\,\tfrac{1}{\sqrt3}\Bigr),

so the largest value of $z$ on the circle is $\sqrt{1 - n_z^2} = \sqrt{2/3} = 0.816497$ and the geodesic climbs to latitude

\lambda_{\max} = \arcsin\sqrt{2/3} = 54.7356^\circ\ \mathrm{N},

reached at longitude $45^\circ$ E by symmetry — a long way north of both endpoints. Clairaut's relation confirms it without the cross product: with $r = a\cos\lambda$ and $\theta$ the angle to the parallel, $\cos\lambda\cos\theta$ is constant, and it equals $\cos\lambda_{\max}$ at the top where $\theta = 0$ . Starting from $\cos\lambda_{\max} = 1/\sqrt3 = 0.577350$ and $\lambda = 45^\circ$ at $A$ ,

\cos\theta_A = \frac{1/\sqrt3}{1/\sqrt2} = \sqrt{2/3} = 0.816497, \qquad \theta_A = 35.2644^\circ ,

so the aircraft leaves $A$ on a heading of $90^\circ - 35.2644^\circ = 54.7356^\circ$ east of north, not due east. Every number in this step is a check on the last: $\cos 45^\circ\cos 35.2644^\circ = 0.707107 \times 0.816497 = 0.577350 = \cos 54.7356^\circ$ , as Clairaut requires.

ApplicationOriginal

Free fall is geodesic motion. General relativity replaces gravitational force by curvature: a particle subject to no non-gravitational force follows a geodesic of the spacetime metric, and the equation it obeys is the one derived above, with the proper time $\tau$ as the affine parameter. The two ingredients of the derivation are exactly the two faces of the equivalence principle — the tangent is parallel-transported (no force), and the worldline extremizes the action (Hamilton's principle, Hamilton's principle). Orbits, the bending of starlight and the precession of Mercury's perihelion are all computed by writing the Schwarzschild metric, taking the cyclic coordinates $t$ and $\phi$ as in step 4 of the method, and integrating the two resulting first integrals. Clairaut's relation is the flat ancestor of the conserved angular momentum that appears there.

Great-circle navigation. Step 4 of the example is why long-haul flight plans and shipping routes curve towards the pole on a Mercator chart: the chart is conformal but not distance-preserving, and the geodesic is not the straight line on the paper. The $404$ km saved on a quarter-turn at latitude $45^\circ$ is roughly half an hour of flight and several tonnes of fuel. The same calculation, with the Earth's oblateness restored, is the geodesy problem solved by Vincenty's algorithm, which integrates the geodesic equations of an ellipsoid of revolution using Clairaut's relation as its first integral.

Optics and wave propagation. Fermat's principle makes a light ray in a medium of refractive index $n(\mathbf{r})$ a geodesic of the optical metric $n^2\delta_{ij}$ , and the geodesic equation becomes the ray equation of gradient-index optics. This is how a graded-index optical fibre confines light — rays oscillate about the axis because the geodesics of the index profile are bound — and it is how mirages and atmospheric refraction are computed. Acoustic ray tracing in the ocean and seismic ray tracing in the Earth use the same equation with $n$ replaced by the reciprocal of the local wave speed.

Shortest paths on manufactured and modelled surfaces. Geodesics on a surface of revolution describe the path of a thread or filament wound under tension on a mandrel, which is what filament-wound pressure vessels and composite pipes are designed around; Clairaut's relation fixes the minimum radius a winding can reach and so the achievable winding angle. In computer graphics and computational anatomy, geodesic distance on a triangulated surface — the discrete version of part 5 of the statement — is the standard measure of separation along a curved object, used for texture mapping, mesh parametrization and cortical surface analysis.

Theorem · advanced

The Gauss-Bonnet theorem

Adapted

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States and proves the local and global Gauss-Bonnet theorems, which tie the total Gaussian curvature of a region, the total geodesic curvature of its boundary and the turning at its corners to a single integer, the Euler-Poincare characteristic $\chi = F - E + V$ of the region; the consequences are that a compact orientable surface has total curvature $2\pi\chi$ whatever its shape, and that a geodesic triangle's angle excess is the curvature it encloses.

Prerequisites: Gauss's Theorema Egregium, Geodesics, The second fundamental form and curvature, Regular surfaces and the first fundamental form, Christoffel symbols and the covariant derivative, Stokes' theorem and Green's theorem in the plane

Definition(Regions, corners and the Euler-Poincare characteristic)AdaptedAdapted from do Carmo (2016), §4-5, pp. 268–271 and 275–277

Let $S$ be an oriented regular surface.

Call a map $\alpha : [0,l] \to S$ simple, closed and piecewise regular when three things hold: it returns to where it started, $\alpha(0) = \alpha(l)$ ; it visits no point twice, so that $\alpha$ is injective on $[0,l)$ ; and for some subdivision $0 = t_0 < t_1 < \cdots < t_{k+1} = l$ it is differentiable with nonvanishing derivative on each open piece $(t_i, t_{i+1})$ . The points $\alpha(t_i)$ are its vertices and the pieces between them its regular arcs. At a vertex both one-sided limits $\alpha'(t_i - 0)$ and $\alpha'(t_i + 0)$ exist and are nonzero.
The external angle $\theta_i$ at $\alpha(t_i)$ is the angle from $\alpha'(t_i-0)$ to $\alpha'(t_i+0)$ of smallest absolute value, taken with the sign of $\det\bigl(\alpha'(t_i-0),\alpha'(t_i+0),N\bigr)$ , so that $-\pi < \theta_i < \pi$ . When $\lvert\theta_i\rvert = \pi$ the vertex is a cusp, and do Carmo fixes the sign by a separate convention using the two branches of the curve in a chart. The interior angle is $\varphi_i = \pi - \theta_i$ .
A simple region $R \subseteq S$ is a connected open set together with its boundary, homeomorphic to a closed disk, whose boundary $\partial R$ is the trace of such a curve. The curve is positively oriented if, at each point of a regular arc, the positive orthonormal basis $\{\alpha'(t), h(t)\}$ has $h(t)$ pointing into $R$ : walking along $\alpha$ with one's head along $N$ , the region stays on the left.
A regular region is a compact connected $R \subseteq S$ whose boundary is a finite union of disjoint simple closed piecewise regular curves. A compact surface counts as a regular region with empty boundary.
A triangle is a simple region with exactly three vertices, all with $\theta_i \ne 0$ . To triangulate a regular region $R$ is to give finitely many triangles that together cover $R$ and overlap only along a shared whole edge or at a shared single vertex. Writing $F$ , $E$ , $V$ for its numbers of triangles, edges and vertices, the Euler-Poincare characteristic of the triangulation is $\chi = F - E + V .$
For a differentiable $f$ on $S$ and a region inside one chart, $\int_R f\,d\sigma := \iint_{\mathbf{x}^{-1}(R)} f\sqrt{EG-F^2}\,du\,dv$ ; this is independent of the chart within one orientation class, and summing over the triangles of a triangulation extends it to any regular region.

Theorem(Gauss-Bonnet, local and global)AdaptedAdapted from do Carmo (2016), §4-5, the local theorem p. 272, Props. 1–4 pp. 275–276, the global theorem p. 277, Eq. (2) p. 274, Cor. 1 p. 279, Cor. 2 p. 280, and application 6 pp. 282–283

Let $S$ be an oriented regular surface, $K$ its Gaussian curvature and $k_g$ the geodesic curvature of a curve in it.

(Topological input, stated without proof.) Every regular region admits a triangulation; it admits one whose triangles each lie in a single chart of a given orientation-compatible family, with adjacent triangles inducing opposite orientations on a shared edge; and $\chi$ is the same for every triangulation of $R$ , so it may be written $\chi(R)$ . A compact connected surface in $\mathbb{R}^3$ has $\chi(S) \in \{2, 0, -2, \dots\}$ , two such surfaces are homeomorphic exactly when their characteristics agree, and $g = (2-\chi(S))/2$ is the genus, the number of handles.
(Local Gauss-Bonnet.) Let $\mathbf{x} : U \to S$ be an isothermal parametrization, $F = 0$ and $E = G = \lambda(u,v) > 0$ , with $U$ homeomorphic to an open disk and $\mathbf{x}$ compatible with the orientation. Let $R \subseteq \mathbf{x}(U)$ be a simple region whose boundary is the trace of a positively oriented $\alpha$ , parametrized by arc length, with vertices $\alpha(s_0),\dots,\alpha(s_k)$ and external angles $\theta_0,\dots,\theta_k$ . Then $\sum_{i=0}^{k}\int_{s_i}^{s_{i+1}} k_g(s)\,ds \;+\; \iint_R K\,d\sigma \;+\; \sum_{i=0}^{k}\theta_i \;=\; 2\pi .$
(Corollary 1.) The same identity is true in every simple region of an arbitrary regular surface; the isothermal chart may be dropped.
(Global Gauss-Bonnet.) Let $R \subseteq S$ be a regular region whose boundary consists of the positively oriented closed curves $C_1,\dots,C_n$ , and let $\theta_1,\dots,\theta_p$ be all of their external angles. Then $\sum_{i=1}^{n}\int_{C_i} k_g(s)\,ds \;+\; \iint_R K\,d\sigma \;+\; \sum_{l=1}^{p}\theta_l \;=\; 2\pi\chi(R) .$
(Corollary 2: total curvature.) If $S$ is a compact orientable surface then $\iint_S K\,d\sigma = 2\pi\chi(S) .$ The left-hand side is unchanged by any deformation of $S$ that does not change its topology.
(Angle excess.) For a geodesic triangle $T$ with interior angles $\varphi_1,\varphi_2,\varphi_3$ , $\sum_{i=1}^{3}\varphi_i - \pi = \iint_T K\,d\sigma ,$ so the angle sum exceeds $\pi$ where $K > 0$ , equals $\pi$ where $K = 0$ and falls short where $K < 0$ . When $K \ne 0$ on $T$ the right-hand side is the area of the spherical image $N(T)$ under the Gauss map, which is how Gauss himself stated it.
(Holonomy.) If $R$ is a simple region without vertices and $w$ is parallel-transported once around $\partial R$ , the angle $\varphi$ it makes with a fixed chart direction changes by $\Delta\varphi = \iint_R K\,d\sigma .$

Idea of the proofOriginal

In the plane, walking once round a closed curve turns the tangent through exactly $2\pi$ , with the corners contributing their external angles: that is the theorem of turning tangents. On a surface the same walk is measured against a moving reference frame — the coordinate direction $\mathbf{x}_u$ — and the mismatch between the turning you feel (the geodesic curvature, which measures turning relative to parallel transport) and the turning the chart records is a term that Green's theorem converts into an area integral. The Theorema Egregium identifies the integrand of that area integral as $-K$ . So the local theorem is nothing but turning tangents plus Green's theorem plus the Gauss formula. The global theorem then comes from arithmetic: triangulate the region, add up the local identity over all $F$ triangles, and watch the boundary integrals over interior edges cancel in pairs while the corner angles reorganize themselves, by a count of faces, edges and vertices, into $2\pi(F-E+V)$ .

ProofAdaptedAdapted from do Carmo (2016), §4-4, Lemma 2 p. 254 and Prop. 3 p. 255; do Carmo (2016), §4-5, pp. 272–273 and 277–279

Lemma (geodesic curvature in an orthogonal chart). Let $\mathbf{x}$ be an orthogonal parametrization ( $F = 0$ ) compatible with the orientation, let $w(t)$ be a differentiable unit tangent field along $t\mapsto\mathbf{x}(u(t),v(t))$ , and let $\varphi(t)$ be a smoothly varying branch of the angle measured from $\mathbf{x}_u$ round to $w(t)$ . Then

\left[\frac{Dw}{dt}\right] = \frac{1}{2\sqrt{EG}}\left(G_u\frac{dv}{dt} - E_v\frac{du}{dt}\right) + \frac{d\varphi}{dt} .

Proof of the lemma. Set $e_1 = \mathbf{x}_u/\sqrt{E}$ and $e_2 = \mathbf{x}_v/\sqrt{G}$ , an orthonormal frame with $e_1\wedge e_2 = N$ . For two unit fields along the curve, the algebraic values of their covariant derivatives differ by the rate of change of the angle between them (do Carmo's Lemma 2), so with $e_1$ as the reference field

\left[\frac{Dw}{dt}\right] = \left[\frac{De_1}{dt}\right] + \frac{d\varphi}{dt} .

For the first term, $[De_1/dt] = \langle de_1/dt,\, N\wedge e_1\rangle = \langle de_1/dt,\, e_2\rangle = \langle (e_1)_u, e_2\rangle\,u' + \langle (e_1)_v, e_2\rangle\,v'$ . Now $F = 0$ gives $\langle \mathbf{x}_u,\mathbf{x}_v\rangle = 0$ , so differentiating it in $u$ yields $\langle\mathbf{x}_{uu},\mathbf{x}_v\rangle = -\langle\mathbf{x}_u,\mathbf{x}_{uv}\rangle = -\tfrac12 E_v$ . Hence

\langle (e_1)_u, e_2\rangle = \left\langle \frac{\mathbf{x}_{uu}}{\sqrt E} - \frac{E_u\,\mathbf{x}_u}{2E^{3/2}},\ \frac{\mathbf{x}_v}{\sqrt G}\right\rangle = \frac{\langle\mathbf{x}_{uu},\mathbf{x}_v\rangle}{\sqrt{EG}} = -\frac{E_v}{2\sqrt{EG}},

the second term dropping out because $\langle\mathbf{x}_u,\mathbf{x}_v\rangle = 0$ ; the same computation in the other variable gives $\langle (e_1)_v, e_2\rangle = \langle\mathbf{x}_{uv},\mathbf{x}_v\rangle/\sqrt{EG} = G_u/(2\sqrt{EG})$ . Substituting proves the lemma. Taking $w = \alpha'$ and $t = s$ turns it into a formula for $k_g$ . $\square$

Part 2: the local theorem. Write the boundary curve in the chart as $u = u(s)$ , $v = v(s)$ , and on each closed arc $[s_i, s_{i+1}]$ pick a smooth branch $\varphi_i(s)$ of the angle that $\alpha'(s)$ makes with $\mathbf{x}_u$ . By the lemma, on each regular arc

k_g(s) = \frac{1}{2\sqrt{EG}}\left(G_u\frac{dv}{ds} - E_v\frac{du}{ds}\right) + \frac{d\varphi_i}{ds} .

Integrate over each $[s_i,s_{i+1}]$ and add. The first group is a line integral $\oint (P\,du + Q\,dv)$ around $\partial R$ with

P = -\frac{E_v}{2\sqrt{EG}}, \qquad Q = \frac{G_u}{2\sqrt{EG}},

so Green's theorem in the $uv$ plane (Stokes' theorem and Green's theorem in the plane) replaces it by $\iint (Q_u - P_v)\,du\,dv$ :

\sum_i \int_{s_i}^{s_{i+1}} k_g\,ds = \iint_{\mathbf{x}^{-1}(R)}\left[\left(\frac{G_u}{2\sqrt{EG}}\right)_{\!u} + \left(\frac{E_v}{2\sqrt{EG}}\right)_{\!v}\right]du\,dv \;+\; \sum_i\int_{s_i}^{s_{i+1}}\frac{d\varphi_i}{ds}\,ds .

Here the hypothesis that the chart is isothermal enters for the first time. With $E = G = \lambda$ we have $\sqrt{EG} = \lambda$ and the bracket is

\frac12\left[\left(\frac{\lambda_u}{\lambda}\right)_{\!u} + \left(\frac{\lambda_v}{\lambda}\right)_{\!v}\right] = \frac{1}{2}\Delta\log\lambda = -K\lambda ,

the last step being the isothermal form of the Gauss formula proved in Gauss's Theorema Egregium, namely $K = -(1/2\lambda)\Delta\log\lambda$ . Since $\sqrt{EG-F^2} = \lambda$ as well, $\iint (-K\lambda)\,du\,dv = -\iint_R K\,d\sigma$ , and therefore

\sum_i \int_{s_i}^{s_{i+1}} k_g\,ds = -\iint_R K\,d\sigma + \sum_i\bigl(\varphi_i(s_{i+1}) - \varphi_i(s_i)\bigr).

The isothermal hypothesis is needed a second time now. Because the chart is conformal, angles measured in the $uv$ plane are the angles on $S$ , so the theorem of turning tangents — a statement about plane curves, which do Carmo quotes from Hopf without proof — may be applied to the image curve:

\sum_i\bigl(\varphi_i(s_{i+1}) - \varphi_i(s_i)\bigr) + \sum_i \theta_i = \pm 2\pi ,

the sign being $+$ for a positively oriented curve, as the case of a circle in the plane shows. Substituting and rearranging gives the local theorem.

Part 4: the global theorem. Triangulate $R$ so that every triangle lies in an isothermal chart compatible with the orientation, with every triangle boundary positively oriented; such a triangulation exists by the topological input. Apply the local theorem to each triangle and add the $F$ identities. Each interior edge belongs to two triangles and is traversed once in each direction, so its $k_g$ integrals cancel; only the boundary curves $C_i$ survive. Writing $\theta_{jk}$ , $k = 1,2,3$ , for the external angles of the triangle $T_j$ ,

\sum_i\int_{C_i} k_g\,ds + \iint_R K\,d\sigma + \sum_{j,k}\theta_{jk} = 2\pi F .

It remains to turn $\sum_{j,k}\theta_{jk}$ into $2\pi F - 2\pi\chi(R)$ plus the boundary corners. Pass to interior angles, $\varphi_{jk} = \pi - \theta_{jk}$ :

\sum_{j,k}\theta_{jk} = 3\pi F - \sum_{j,k}\varphi_{jk}.

Split the edges and vertices into interior and exterior ones, $E = E_i + E_e$ and $V = V_i + V_e$ . Every triangle has three edges; an interior edge is shared by two triangles and an exterior edge by one, so $3F = 2E_i + E_e$ , whence $3\pi F = 2\pi E_i + \pi E_e$ . The boundary curves are closed, so along them edges and vertices alternate and $E_e = V_e$ . Now count the interior angles by the vertex they sit at:

around an interior vertex the angles fill a full turn, contributing $2\pi V_i$ ;
an exterior vertex that the triangulation introduced in the middle of a regular arc has angles filling a half turn, contributing $\pi V_{et}$ , where $V_{et}$ is their number;
an exterior vertex that is a corner $\theta_l$ of some $C_i$ contributes the interior angle $\pi - \theta_l$ of the region there, so these contribute $\sum_l(\pi - \theta_l)$ , and their number is $V_{ec} = p$ , with $V_e = V_{et} + V_{ec}$ .

Therefore

\sum_{j,k}\theta_{jk} = 2\pi E_i + \pi E_e - 2\pi V_i - \pi V_{et} - \sum_l(\pi - \theta_l).

Use $\sum_l \pi = \pi V_{ec}$ and $\pi V_{et} + \pi V_{ec} = \pi V_e$ , then add and subtract $\pi E_e$ and use $E_e = V_e$ once more:

2\pi E_i + \pi E_e = 2\pi E_i + 2\pi E_e - \pi E_e = 2\pi E - \pi V_e ,

so that

\sum_{j,k}\theta_{jk} = 2\pi E - \pi V_e - 2\pi V_i - \pi V_e + \sum_l\theta_l = 2\pi E - 2\pi V + \sum_l \theta_l .

Substituting this into the summed identity and moving the terms,

\sum_i\int_{C_i} k_g\,ds + \iint_R K\,d\sigma + \sum_l\theta_l = 2\pi F - 2\pi E + 2\pi V = 2\pi\chi(R).

Parts 3, 5, 6 and 7. A simple region has $\chi = 1$ — triangulate a disk and count — so part 4 contains part 3, and in particular the local identity holds with no isothermal hypothesis. A compact surface is a regular region with no boundary, so every $k_g$ integral and every $\theta_l$ is absent and part 5 follows. For part 6, a geodesic triangle has $k_g \equiv 0$ on all three arcs and $\chi = 1$ , so $\iint_T K\,d\sigma + \sum_i\theta_i = 2\pi$ ; substituting $\theta_i = \pi - \varphi_i$ gives $\iint_T K\,d\sigma = \sum_i \varphi_i - \pi$ . Part 7 repeats the first half of the proof of part 2 with $w$ a parallel field rather than $\alpha'$ : then $[Dw/ds] = 0$ identically, so integrating the lemma around a vertex-free boundary gives $0 = -\iint_R K\,d\sigma + \Delta\varphi$ . $\square$

Remark(Where the hypotheses are used)Original

Orientability. It is used three times: to give the external angles their signs, to fix what "positively oriented" means, and to define $k_g$ at all, which is a signed quantity. Everything above is therefore stated, as do Carmo states it, for an oriented surface, and part 5 for an orientable compact one; on a non-orientable surface the proof as given has no starting point, since there is no global $N$ with which to sign anything.
Isothermal chart. The proof of part 2 uses it twice, and only for convenience. The curvature step works in any orthogonal chart, because the orthogonal-chart formula of Gauss's Theorema Egregium already gives $\bigl(E_v/2\sqrt{EG}\bigr)_v + \bigl(G_u/2\sqrt{EG}\bigr)_u = -K\sqrt{EG}$ . It is the turning-tangents step that needs conformality, so that angles in the parameter plane are angles on the surface. do Carmo says exactly this, and part 3 removes the hypothesis afterwards.
Compactness of $R$ . A regular region is compact, which is what makes the triangulation finite and the sums above finite. Without it the counting argument has nothing to count: the plane has $\iint K\,d\sigma = 0$ and no finite $\chi$ in this sense.
Simple, non-self-intersecting boundary. Turning tangents fails for a curve that crosses itself — a figure eight turns through $0$ , not $2\pi$ — so a region bounded by a self-intersecting curve is outside the theorem until it is cut into simple pieces.
Finitely many corners, each with $\lvert\theta_i\rvert \le \pi$ . The external angle is only defined where both one-sided tangents exist and are nonzero. A boundary with infinitely many corners, or a curve with a vanishing one-sided tangent, is not piecewise regular and is excluded.
Positive orientation. It fixes the sign $+2\pi$ rather than $-2\pi$ in turning tangents. Reversing the orientation of $\partial R$ flips the signs of every $k_g$ and every $\theta_i$ and would give $-2\pi\chi(R)$ .
What is assumed rather than proved. Three ingredients are quoted: that triangulations exist with the properties in part 1, that $\chi$ does not depend on the triangulation, and the theorem of turning tangents. do Carmo also presents these without proof, referring to Ahlfors and Sario for the first two and to Hopf for the third, and we do the same. The independence of $\chi$ from the triangulation is less mysterious from another direction: Lee defines the Euler characteristic of a smooth manifold as $\sum_p (-1)^p\dim H^p_{dR}(M)$ , manifestly free of any choice of triangulation, and sets it as a problem to show that this number is a homotopy invariant.
Smoothness class. $K$ must be integrable and $k_g$ defined, so $C^3$ as in the Gauss formula; the proof also needs the boundary arcs to be $C^2$ for $k_g$ to exist.

Method(Using Gauss-Bonnet)Original

Decide what you know and what you want. The identity has four slots — boundary turning, enclosed curvature, corners, topology. Three known values give the fourth. Most applications fix the topology and one other slot.
Kill a term by choosing the region well. A boundary made of geodesic arcs kills the $k_g$ integral; a smooth boundary kills the corner sum; a closed surface kills both.
Get the topology right. For a simple region (a disk) $\chi = 1$ ; for an annulus or a region between two disjoint closed boundary curves $\chi = 0$ ; for a sphere $\chi = 2$ ; for a torus $\chi = 0$ ; for a surface of genus $g$ , $\chi = 2 - 2g$ .
As an obstruction. If the identity forces a contradiction, the configuration does not exist. Two examples, both do Carmo's: on a surface with $K \le 0$ two geodesics leaving a point cannot meet again so as to bound a simple region, since $\iint_R K\,d\sigma + \theta_1 + \theta_2 = 2\pi$ with $\theta_i < \pi$ is then impossible; and a compact surface with $K > 0$ has $\chi > 0$ , so it is homeomorphic to a sphere.
As a conservation law. On a compact surface, $\iint K\,d\sigma$ cannot be changed by any deformation. Pushing curvature out of one part of the surface simply piles it up elsewhere.
Check the sign conventions before trusting a number. Orient the boundary so that the region is on the left, and verify on a case you know — a flat disk, where $\oint k_g\,ds = 2\pi$ and $K = 0$ — before applying the formula to a hard case.

Example(Three regions, three checks)Original

Throughout, the sphere of radius $a$ has $K = 1/a^2$ , established intrinsically in Gauss's Theorema Egregium.

(a) A geodesic triangle: the corners carry everything. On the sphere of radius $a = 6371\ \mathrm{km}$ , take the region bounded by the meridian arcs at longitudes $0^\circ$ and $50^\circ$ running from the north pole down to the equator, together with the arc of the equator between them, and call it $T$ . All three sides are great-circle arcs, hence geodesics, so $k_g \equiv 0$ . The interior angles are $50^\circ$ at the pole and $90^\circ$ where each meridian crosses the equator. By part 6,

\iint_T K\,d\sigma = \Bigl(\tfrac{\pi}{2} + \tfrac{\pi}{2} + \tfrac{50\pi}{180}\Bigr) - \pi = \frac{50\pi}{180} = 0.8726646 .

Independently, the region is the slice of the northern hemisphere cut by a $50^\circ$ wedge, of area $(50/360)\times 2\pi a^2$ , so

\iint_T K\,d\sigma = \frac{1}{a^2}\cdot\frac{50}{360}\cdot 2\pi a^2 = \frac{50\pi}{180} = 0.8726646 . \;\checkmark

The two agree for every $a$ , as they must, since the left side is dimensionless. The area itself is $0.8726646\times 6371^2 = 3.5421\times 10^{7}\ \mathrm{km}^2$ , about $7\%$ of the Earth's surface.

(b) A spherical cap: the boundary carries everything. On the unit sphere take the cap $\theta \le \theta_0$ above the parallel of colatitude $\theta_0 = 60^\circ$ . The boundary is smooth, so there are no corners, and the region is a disk, so $\chi = 1$ . The parallel is not a geodesic; its geodesic curvature has absolute value $\cot\theta_0$ , by do Carmo's computation from $k^2 = k_g^2 + k_n^2$ with $k = 1/\sin\theta_0$ (the circle has radius $\sin\theta_0$ ) and $k_n = 1$ ; with the orientation that makes the boundary positively oriented around the cap, the sign is positive for $\theta_0 < \pi/2$ . Its length is $2\pi\sin\theta_0$ . Hence

\oint k_g\,ds = 2\pi\sin\theta_0\cdot\frac{\cos\theta_0}{\sin\theta_0} = 2\pi\cos\theta_0 = 2\pi\times 0.5 = \pi ,

while the cap has area $2\pi(1 - \cos\theta_0) = \pi$ and $K = 1$ , so $\iint_R K\,d\sigma = \pi$ . The total is

\pi + \pi + 0 = 2\pi = 2\pi\chi(R). \;\checkmark

Note how the two terms trade off as $\theta_0$ grows: at $\theta_0 \to 0$ the small circle supplies almost all of the $2\pi$ and the curvature almost none; at $\theta_0 = \pi/2$ the equator is a geodesic and supplies nothing, the hemisphere's curvature supplying the whole $2\pi$ .

(c) A torus: total curvature zero, computed twice. Parametrize the torus of revolution $\Sigma$ with tube radius $r$ and centre-circle radius $A > r$ by

\mathbf{x}(u,v) = \bigl((A + r\cos u)\cos v,\ (A + r\cos u)\sin v,\ r\sin u\bigr), \qquad 0 \le u, v < 2\pi .

Then $E = r^2$ , $F = 0$ , $G = (A + r\cos u)^2$ , and the second fundamental form gives $e = r$ , $f = 0$ , $g = (A+r\cos u)\cos u$ , so

K = \frac{eg - f^2}{EG - F^2} = \frac{\cos u}{r\,(A + r\cos u)}, \qquad d\sigma = \sqrt{EG-F^2}\,du\,dv = r(A + r\cos u)\,du\,dv .

The two factors $r(A + r\cos u)$ cancel exactly, leaving $K\,d\sigma = \cos u\,du\,dv$ and

\iint_\Sigma K\,d\sigma = \int_0^{2\pi}\!\!\int_0^{2\pi}\cos u\,du\,dv = 0 .

This is do Carmo's Exercise 2, done without Gauss-Bonnet. Part 5 predicts the same, since the torus has $\chi = 0$ . With $A = 3$ , $r = 1$ the cancellation is worth watching in detail: on the outer half $\lvert u\rvert < \pi/2$ the curvature is positive, peaking at $K = 1/4$ on the outer equator, and contributes $\int_0^{2\pi}\!\int_{-\pi/2}^{\pi/2}\cos u\,du\,dv = +4\pi$ ; on the inner half it is negative, reaching $K = -1/2$ on the inner equator, and contributes $-4\pi$ . The positive bulge and the negative saddle cancel to the last digit, and they must, for every $A > r$ : the integrand does not remember $A$ or $r$ at all.

ApplicationOriginal

Holonomy and geometric phase. Part 7 says that carrying a vector around a loop on a curved surface and bringing it home leaves it rotated by the enclosed curvature. This is the prototype of every geometric phase in physics. In Berry's phase the "surface" is the space of parameters of a Hamiltonian, the "curvature" is the Berry curvature and the rotation is the phase a quantum state picks up on a cyclic adiabatic excursion; the phase of a spin- $\tfrac12$ in a slowly rotated magnetic field is exactly half the solid angle swept out, which is the sphere's $\iint K\,d\sigma$ . The same structure governs the Aharonov-Bohm phase, the rotation of the polarization of light in a coiled optical fibre (Pancharatnam's phase) and the holonomies of gauge theory, where $K$ becomes a field strength.

The Foucault pendulum. The swing plane of a Foucault pendulum is parallel-transported along the circle of latitude the pendulum rides on as the Earth turns. For latitude $\lambda$ the cap north of that parallel has, by part (b) of the example with $\cos\theta_0 = \sin\lambda$ , total curvature $2\pi(1 - \sin\lambda)$ , so that is the holonomy; the remaining $2\pi\sin\lambda$ is the precession an observer on the ground sees in one sidereal day. In Paris, $\lambda = 48.8566^\circ$ , giving $360^\circ\times\sin\lambda = 271.1^\circ$ per day, or $11.30^\circ$ per hour, the figure the pendulum in the Pantheon keeps.

Topology from a measurement. Corollary 2 turns a geometric integral into an integer. If one can measure curvature over a closed surface — a vesicle, a protein shell, a crystallite — then $\iint K\,d\sigma/2\pi$ counts the handles, and nothing about shape, stretching or local detail can change it. In the elasticity of thin shells this is why the Gaussian-curvature term in the bending energy contributes a constant for a closed shell and can be dropped; it reappears the moment the shell opens a pore or buds off a daughter vesicle, because $\chi$ jumps. Viral capsids and fullerenes are the standard illustration: a closed shell built of hexagons and pentagons has $\chi = 2$ , and the counting of part 1 forces exactly twelve pentagons however large the shell, which is why $\mathrm{C}_{60}$ and icosahedral capsids look the way they do.

Defects and the hairy ball theorem. do Carmo's companion result, Poincare's index theorem, says that the indices of the isolated zeros of a vector field on a compact surface sum to $\chi(S)$ . On a sphere $\chi = 2 \ne 0$ , so no nonvanishing tangent field exists: a wind field on the Earth always has a calm point, and hair on a ball cannot be combed flat. The same count fixes the total topological charge of defects in a two-dimensional nematic liquid crystal or a superfluid film confined to a curved substrate, and in cosmology and condensed-matter field theory it is the two-dimensional ancestor of the index theorems that tie the integral of a curvature to a conserved integer.

Smooth manifolds

Definition · advanced

Smooth manifolds

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Defines a smooth manifold as a topological manifold carrying a maximal atlas of smoothly compatible charts, defines smooth maps and embedded submanifolds, and proves that every regular level set of a smooth map is an embedded submanifold of codimension equal to the dimension of the codomain.

Prerequisites: Implicit and inverse function theorems, Topological and metric spaces, Continuity and homeomorphism, Compactness

Definition(Topological manifold, chart, atlas, smooth structure)AdaptedAdapted from Lee (2013), Ch. 1, pp. 3, 12–13; Hassani (2013), §28.1, pp. 860–861

A topological space $M$ is a topological $n$ -manifold when it has these three properties:

$M$ is Hausdorff: distinct points lie in disjoint open sets.
$M$ is second-countable: its topology has a countable basis.
$M$ is locally Euclidean of dimension $n$ : every point has an open neighbourhood homeomorphic to an open subset of $\mathbb{R}^n$ .

A chart is a pair $(U, \varphi)$ with $U \subseteq M$ open and $\varphi : U \to \widehat U \subseteq \mathbb{R}^n$ a homeomorphism onto an open set. Writing $\varphi(p) = (x^1(p), \dots, x^n(p))$ defines the coordinate functions of the chart; physicists usually name the chart by its coordinates and write $(U, x^i)$ .

If $(U, \varphi)$ and $(V, \psi)$ are charts with $U \cap V \neq \varnothing$ , the transition map

\psi \circ \varphi^{-1} : \varphi(U \cap V) \longrightarrow \psi(U \cap V)

carries one open subset of $\mathbb{R}^n$ to another, and is automatically a homeomorphism. The two charts are smoothly compatible if either $U \cap V = \varnothing$ or that transition map is a diffeomorphism, that is, it and its inverse have continuous partial derivatives of every order in the ordinary sense of calculus on $\mathbb{R}^n$ .

An atlas is a family of charts whose domains cover $M$ ; it is a smooth atlas if its charts are pairwise smoothly compatible, and it is maximal if no chart outside it is smoothly compatible with all of its members. A smooth structure on $M$ is a maximal smooth atlas, and a smooth manifold is a topological manifold together with a smooth structure. A chart belonging to the structure is a smooth chart.

Remark(Why a maximal atlas)AdaptedAdapted from Lee (2013), Ch. 1, Prop. 1.17 and its proof, pp. 13–14

One never exhibits a maximal atlas. One exhibits a small smooth atlas and appeals to the fact that every smooth atlas is contained in exactly one maximal smooth atlas, namely the collection of all charts smoothly compatible with every chart of the given one; and two smooth atlases determine the same maximal atlas exactly when their union is again a smooth atlas. So, for instance, $\{(\mathbb{R}^n, \mathrm{id})\}$ and the atlas of all unit balls with the identity map determine the same smooth structure on $\mathbb{R}^n$ , and any manifold covered by a single chart carries a smooth structure for free, since the compatibility condition is then vacuous.

The smooth structure is extra data, not a property: a topological manifold may admit many inequivalent smooth structures, and some admit none at all.

Definition(Smooth maps and diffeomorphisms)AdaptedAdapted from Lee (2013), Ch. 2, pp. 34–35

Let $M$ and $N$ be smooth manifolds. A map $F : M \to N$ is smooth when every $p \in M$ lies in the domain of a smooth chart $(U, \varphi)$ of $M$ , with $F(p)$ in the domain of a smooth chart $(V, \psi)$ of $N$ , such that

F(U) \subseteq V \qquad\text{and}\qquad \psi \circ F \circ \varphi^{-1} : \varphi(U) \to \psi(V) \ \text{ is smooth}.

The map $\widehat F = \psi \circ F \circ \varphi^{-1}$ is the coordinate representation of $F$ . Requiring $F(U) \subseteq V$ inside the definition, rather than as an afterthought, is what makes every smooth map continuous: on $U$ one has $F = \psi^{-1} \circ \widehat F \circ \varphi$ , a composite of continuous maps. Smoothness of the coordinate representation does not depend on which smooth charts are used, because changing chart composes $\widehat F$ with transition maps, which are diffeomorphisms.

A diffeomorphism is a smooth bijection with smooth inverse. A real-valued function $f : M \to \mathbb{R}$ is smooth in this sense with $N = \mathbb{R}$ and $\psi = \mathrm{id}$ ; the set of smooth real-valued functions on $M$ is written $C^\infty(M)$ .

Definition(Embedded submanifold)AdaptedAdapted from Lee (2013), Ch. 5, p. 98 and pp. 101–102 (Thm 5.8)

Let $M$ be a smooth $n$ -manifold and $0 \le k \le n$ . A subset $S \subseteq M$ satisfies the local $k$ -slice condition if every point of $S$ lies in the domain of a smooth chart $(U, \varphi)$ of $M$ in which $S \cap U$ is cut out by freezing the last $n-k$ coordinates:

\varphi(S \cap U) = \bigl\{\, (x^1,\dots,x^n) \in \varphi(U) : x^{k+1} = c^{k+1},\ \dots,\ x^n = c^n \,\bigr\}

for some constants $c^{k+1}, \dots, c^n$ . Such a chart is a slice chart.

$S$ is an embedded $k$ -dimensional submanifold of $M$ if, in the subspace topology, it is a topological $k$ -manifold carrying a smooth structure for which the inclusion $S \hookrightarrow M$ is a smooth embedding. Lee's local slice criterion says that the two descriptions agree: an embedded $k$ -submanifold satisfies the local $k$ -slice condition, and conversely a subset satisfying that condition is, with the subspace topology and the charts obtained by dropping the frozen coordinates, an embedded $k$ -submanifold. The number $n - k$ is the codimension of $S$ in $M$ . We use the slice condition as the working definition below, since it is the one a calculation can verify.

Definition(Regular level set theorem)AdaptedAdapted from Lee (2013), Ch. 5, Cor. 5.14, p. 106; Hassani (2013), §28.3, Thm 28.3.8, p. 876

Fix smooth manifolds $M$ and $N$ of dimensions $m$ and $k$ , together with a smooth map $\Phi : M \to N$ . At a point $p \in M$ , pick smooth charts around $p$ and around $\Phi(p)$ and form the $k \times m$ Jacobian matrix

D\widehat\Phi(\varphi(p)) = \Bigl( \frac{\partial \widehat\Phi^{\,a}}{\partial x^i}(\varphi(p)) \Bigr)_{a = 1,\dots,k;\ i = 1,\dots,m}

of the coordinate representation. Its rank is unambiguous: by the chain rule, changing either chart multiplies this matrix left and right by the invertible Jacobians of transition maps, which cannot alter rank. We call $p$ a regular point when that rank attains $k$ . That is the largest value a $k \times m$ matrix can have, and reaching it requires $k \le m$ ; when $k > m$ there are no regular points at all, so every level set that meets $M$ is excluded and the theorem below says nothing.

The values we can handle are those at which the rank nowhere drops. So declare $c \in N$ regular when each $p$ with $\Phi(p) = c$ is a regular point. For $c$ outside the image of $\Phi$ there is no such $p$ at all, so the requirement is empty and $c$ is regular by default.

Then, for regular $c$ , the level set $\Phi^{-1}(c)$ is a properly embedded submanifold of $M$ of codimension $\dim N$ . Provided it is not empty, this pins down its dimension:

\dim \Phi^{-1}(c) = m - k \qquad (\Phi^{-1}(c) \neq \varnothing).

An empty level set is an embedded submanifold of every dimension at once, so the formula is to be read only in the non-empty case.

Invariantly, the rank condition says that the differential $d\Phi_p$ of the next unit is surjective; that is how Lee and Hassani state it. For $k = 1$ and $M$ an open subset of $\mathbb{R}^m$ it is just $\nabla \Phi(p) \neq 0$ .

Idea of the proofOriginal

The statement is local, and locally everything happens in Euclidean space, because charts turn $\Phi$ into an ordinary smooth map between open subsets of $\mathbb{R}^m$ and $\mathbb{R}^k$ . There, surjectivity of the derivative says that $k$ of the $m$ coordinates can be solved for in terms of the other $m-k$ : that is exactly the implicit function theorem. Changing to those $m-k$ free coordinates together with the $k$ components of $\Phi$ itself produces a new chart in which the level set is the slice "last $k$ coordinates constant". So the level set satisfies the slice condition, which is what being an embedded submanifold means.

ProofAdaptedAdapted from Lee (2013), Ch. 5, proof of Thm 5.12 and Cor. 5.14, pp. 105–106

Write $S = \Phi^{-1}(c)$ and assume $S \neq \varnothing$ (otherwise there is nothing to prove). Fix $p \in S$ .

Step 1 (pass to coordinates). Choose a smooth chart $(U, \varphi)$ of $M$ with $p \in U$ and a smooth chart $(V, \psi)$ of $N$ with $c \in V$ and $\Phi(U) \subseteq V$ ; shrinking $U$ if necessary, this is possible because $\Phi$ is continuous. Put $\widehat\Phi = \psi \circ \Phi \circ \varphi^{-1}$ , a smooth map from the open set $\widehat U = \varphi(U) \subseteq \mathbb{R}^m$ into $\mathbb{R}^k$ , and let $\widehat c = \psi(c)$ , $a = \varphi(p)$ . Then $\varphi(S \cap U) = \widehat\Phi^{-1}(\widehat c)$ , because $\varphi$ and $\psi$ are bijections onto their images, and by the regular-value hypothesis $D\widehat\Phi(a)$ has rank $k$ .

Step 2 (name the solvable block). Since $D\widehat\Phi(a)$ is a $k \times m$ matrix of rank $k$ , it has $k$ linearly independent columns. Relabel the coordinates of $\mathbb{R}^m$ so that these are the last $k$ , and split a point as $x = (y, w)$ with $y \in \mathbb{R}^{m-k}$ and $w \in \mathbb{R}^{k}$ , $a = (y_0, w_0)$ . Then the $k \times k$ matrix $\partial \widehat\Phi / \partial w$ is invertible at $a$ . Relabelling is a diffeomorphism of $\mathbb{R}^m$ , so it replaces $\varphi$ by another smooth chart.

Step 3 (implicit function theorem). By the implicit function theorem applied to $\widehat\Phi(y, w) - \widehat c = 0$ at $(y_0, w_0)$ , there are open sets $Y \ni y_0$ in $\mathbb{R}^{m-k}$ and $W \ni w_0$ in $\mathbb{R}^{k}$ with $Y \times W \subseteq \widehat U$ , and a smooth map $h : Y \to W$ with $h(y_0) = w_0$ , such that for $(y,w) \in Y \times W$ ,

\widehat\Phi(y, w) = \widehat c \iff w = h(y) .

Thus on $Y \times W$ the level set is the graph of $h$ .

Step 4 (flatten the graph). Define $\Theta : Y \times W \to \mathbb{R}^{m-k} \times \mathbb{R}^{k}$ by

\Theta(y, w) = \bigl(y,\; w - h(y)\bigr).

$\Theta$ is smooth, and it is a diffeomorphism onto its open image because $(y, u) \mapsto (y, u + h(y))$ is a smooth inverse. By Step 3, a point of $Y \times W$ lies in the level set exactly when its last $k$ new coordinates vanish. Hence $(\,\Theta \circ \varphi\,)$ , restricted to $\varphi^{-1}(Y \times W)$ , is a chart of $M$ — it is smoothly compatible with every smooth chart because it differs from $\varphi$ by a diffeomorphism of open subsets of $\mathbb{R}^m$ — and in it $S$ is the slice $\{x^{m-k+1} = \dots = x^m = 0\}$ .

Step 5 (conclude). Step 4 produced a slice chart around an arbitrary $p \in S$ , so $S$ satisfies the local $(m-k)$ -slice condition and is therefore an embedded submanifold of dimension $m-k$ , that is, of codimension $k = \dim N$ . Finally $S = \Phi^{-1}(c)$ is closed in $M$ : $\{c\}$ is closed because $N$ is Hausdorff, and $\Phi$ is continuous. A closed embedded submanifold is properly embedded, because the preimage of a compact set under the inclusion is a closed subset of that compact set, hence compact. $\square$

RemarkOriginal

Where the hypotheses are used.

Surjectivity of $d\Phi_p$ at every point of the level set is used only in Step 2, to produce an invertible $k \times k$ block, and it is indispensable. Take $\Phi(x,y,z) = x^2 + y^2 - z^2$ on $\mathbb{R}^3$ . Every $c \ne 0$ is a regular value and $\Phi^{-1}(c)$ is a smooth surface (a hyperboloid). The value $c = 0$ is not regular, because $\nabla\Phi = (2x, 2y, -2z)$ vanishes at the origin, which lies on $\Phi^{-1}(0)$ ; and indeed the double cone $\Phi^{-1}(0)$ is not a topological $2$ -manifold, since deleting the vertex from a small neighbourhood of it disconnects that neighbourhood, which does not happen in $\mathbb{R}^2$ . Note that the failure is at one point only, yet it is enough to disqualify the value $c = 0$ .
Regularity is a condition on the level set, not on $M$ . The theorem does not ask $\Phi$ to be a submersion everywhere; the set of regular points is open, so the restriction of $\Phi$ to a neighbourhood of a regular level set is a submersion, which is all the proof uses.
Hausdorff and second-countable are used twice. Hausdorffness makes $\{c\}$ closed in Step 5 and makes compact subsets closed; both properties are inherited by subspaces, which is why the level set is again a manifold in the subspace topology rather than merely locally Euclidean. Dropping Hausdorffness admits objects such as the line with two origins, which is locally Euclidean but on which limits are not unique.
Regularity, not smoothness of $\Phi$ , is what does the work. Drop it and both halves of the conclusion go. The lemniscate $\Phi(x,y) = y^2 - x^2 + x^4$ , $\Phi^{-1}(0)$ , crosses itself at the origin, where $\nabla\Phi = (-2x + 4x^3,\, 2y)$ vanishes; it is a figure eight, not a $1$ -manifold. And $\Phi(x,y) = x^2+y^2$ has $\Phi^{-1}(0) = \{(0,0)\}$ , a perfectly good manifold but of dimension $0$ , not of the predicted $m - k = 1$ .
Smooth compatibility of charts is what makes "smooth function on $M$ " well defined at all: if the transition maps were merely continuous, the coordinate representation of $f$ could be smooth in one chart and not in another.

Method(Showing that a subset of

\mathbb{R}^m

is a smooth manifold)Original

Given $S \subseteq \mathbb{R}^m$ , here is how to certify that it is embedded, of dimension $m-k$ , and how to produce charts on it:

Find a defining map. Write $S = \Phi^{-1}(c)$ for a smooth $\Phi : \Omega \to \mathbb{R}^k$ on an open set $\Omega \supseteq S$ . Any smooth extension will do; the choice of $\Omega$ can exclude bad points, which is often how a singular locus is avoided.
Compute the Jacobian $D\Phi = (\partial\Phi^a/\partial x^i)$ , a $k \times m$ matrix of functions.
Check rank $k$ at every point of $S$ — not on all of $\Omega$ . For $k = 1$ this is $\nabla\Phi \ne 0$ ; for $k \ge 2$ , exhibit a $k \times k$ minor that is nonzero, or compute $\det(D\Phi\,D\Phi^{\mathsf T}) \ne 0$ .
Conclude that $S$ is an embedded submanifold of dimension $m-k$ , properly embedded, and compact if it is also bounded.
Charts, if wanted. Around a given point, the nonzero minor names $k$ coordinates to eliminate; the remaining $m-k$ coordinates restricted to $S$ are a chart there, by Steps 3–4 of the proof above. Transition maps between two such charts are smooth because each is built from the same implicit solutions.

Example(The sphere, and an atlas for it)Original

Let $S^2 = \{x \in \mathbb{R}^3 : |x|^2 = 1\}$ .

It is a manifold. Take $\Phi(x,y,z) = x^2+y^2+z^2$ on $\Omega = \mathbb{R}^3$ and $c = 1$ . Then $\nabla\Phi = 2(x,y,z)$ , which has norm $2$ at every point of $S^2$ and in particular never vanishes there, so $1$ is a regular value and $S^2$ is an embedded submanifold of $\mathbb{R}^3$ of codimension $1$ , hence of dimension $2$ . It is closed and bounded, so compact.
A two-chart atlas. Stereographic projection from the north pole $n = (0,0,1)$ and the south pole $s = (0,0,-1)$ gives $\sigma_N(x,y,z) = \frac{(x,y)}{1-z} \quad\text{on } S^2\setminus\{n\}, \qquad \sigma_S(x,y,z) = \frac{(x,y)}{1+z} \quad\text{on } S^2\setminus\{s\} .$ Both are homeomorphisms onto $\mathbb{R}^2$ , with $\sigma_N^{-1}(u,v) = \frac{1}{u^2+v^2+1}\,\bigl(2u,\; 2v,\; u^2+v^2-1\bigr),$ as one checks by verifying that this point has norm $1$ and that applying $\sigma_N$ to it returns $(u,v)$ . The two domains cover $S^2$ , so this is an atlas.
Smooth compatibility. On the overlap, $u^2+v^2 \neq 0$ , and substituting $\sigma_N^{-1}$ into $\sigma_S$ gives the transition map $\sigma_S \circ \sigma_N^{-1}(u,v) = \frac{(u,v)}{u^2+v^2},$ inversion in the unit circle. It is smooth on $\mathbb{R}^2 \setminus \{0\}$ and equals its own inverse, so the two charts are smoothly compatible and the atlas is smooth.
A number. The Jacobian determinant of the transition map is $\det \frac{\partial(\sigma_S \circ \sigma_N^{-1})}{\partial(u,v)} = -\frac{1}{(u^2+v^2)^2},$ obtained by differentiating the two components and cancelling: the numerator is $(v^2-u^2)(u^2-v^2) - 4u^2v^2 = -(u^2+v^2)^2$ . At $(u,v) = (2,1)$ it equals $-1/25 = -0.04$ . It is nonzero everywhere on the overlap, as it must be for a diffeomorphism, and it is negative at every point of the overlap. So no consistent orientation of $S^2$ is obtained by declaring both $\sigma_N$ and $\sigma_S$ positively oriented. Replacing $\sigma_S$ by $\widetilde\sigma_S(x,y,z) = (x,-y)/(1+z)$ repairs this: the new transition map is $(u,v) \mapsto (u,-v)/(u^2+v^2)$ , whose Jacobian determinant is $+1/(u^2+v^2)^2$ , and the two charts then orient $S^2$ .

Example(A torus of revolution as a regular level set)Original

Fix $R = 3$ and $r = 1$ . On the open set $\Omega = \{(x,y,z) : x^2+y^2 > 0\} \subseteq \mathbb{R}^3$ define

\Phi(x,y,z) = \bigl(\rho - R\bigr)^2 + z^2 , \qquad \rho = \sqrt{x^2+y^2} ,

which is smooth on $\Omega$ because $\rho$ is smooth where $\rho > 0$ . Follow the method.

Jacobian. $\partial_x \Phi = 2(\rho-R)x/\rho$ , $\partial_y \Phi = 2(\rho-R)y/\rho$ , $\partial_z \Phi = 2z$ .
Norm of the gradient. Using $x^2+y^2 = \rho^2$ , $|\nabla\Phi|^2 = \frac{4(\rho-R)^2}{\rho^2}\,(x^2+y^2) + 4z^2 = 4\bigl[(\rho-R)^2 + z^2\bigr] = 4\,\Phi .$
Regular value. Hence on the level set $\Phi = r^2 = 1$ we get $|\nabla\Phi| = 2r = 2 \neq 0$ , so $1$ is a regular value and $T = \Phi^{-1}(1)$ is an embedded surface in $\Omega$ , and therefore in $\mathbb{R}^3$ , since $T$ is contained in $\Omega$ and $\Omega$ is open. It is compact: $T$ is bounded ( $2 \le \rho \le 4$ , $|z| \le 1$ ) and closed in $\mathbb{R}^3$ , because any limit point has $\rho \ge R - r = 2 > 0$ and so lies in $\Omega$ , where $\Phi$ is continuous.
Why $\Omega$ and not $\mathbb{R}^3$ . The formula for $\Phi$ is not differentiable on the $z$ -axis. Excluding the axis costs nothing, since the axis misses $T$ by the distance $R - r = 2$ .
Where it fails. For $r = R = 3$ the identity $|\nabla\Phi|^2 = 4\Phi$ still holds, and $|\nabla\Phi| = 2R \ne 0$ on $\Phi^{-1}(R^2)$ within $\Omega$ ; but the point $(0,0,0)$ satisfies $(\rho-R)^2+z^2 = R^2$ with $\rho = 0$ and lies outside $\Omega$ , so it is not covered by the theorem. The set $\{(\rho-3)^2+z^2 = 9\} \subseteq \mathbb{R}^3$ is a pinched torus and is not a manifold at the origin, exactly as the cone in the remark is not one at its vertex.

ApplicationOriginal

Manifolds enter the applied sciences wherever a system has continuous degrees of freedom subject to constraints. The configuration space of a rigid body with one point fixed is the rotation group, a three-dimensional manifold; that of a double pendulum is a torus $S^1 \times S^1$ ; the set of unit quaternions used in spacecraft attitude control is $S^3$ . In each case the manifold is obtained exactly as the torus above was: a constraint $\Phi = c$ is imposed on an ambient space of raw coordinates, and the regular level set theorem both certifies that the constrained set is smooth and tells us its dimension, which is the number of independent degrees of freedom. Nonlinear constrained optimisation and numerical integrators that must stay on a constraint surface (orthogonality-preserving methods, attitude integrators) rely on the same local chart construction from the proof. In general relativity the whole framework is taken as primitive: spacetime is a four-dimensional smooth manifold with no ambient space at all, and the restriction to smoothly compatible charts is precisely the statement that physical laws must not depend on the coordinates used to record them.

Definition · advanced

Tangent vectors and vector fields

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Defines a tangent vector at a point of a smooth manifold as a derivation of smooth functions, identifies the coordinate vectors as a basis of the tangent space, defines smooth vector fields and their action on functions, and proves that the Lie bracket of two vector fields is again a vector field with the coordinate formula for its components.

Prerequisites: Smooth manifolds, Partial derivatives and the chain rule, Vector spaces and bases

Definition(Tangent vector, tangent space)AdaptedAdapted from Lee (2013), Ch. 3, p. 54; Hassani (2013), §28.2, Def. 28.2.3, p. 868

Let $M$ be a smooth manifold and $p \in M$ . A derivation at $p$ is a linear map $v : C^\infty(M) \to \mathbb{R}$ satisfying the product rule at $p$ :

v(fg) = f(p)\, v(g) + g(p)\, v(f) \qquad \text{for all } f, g \in C^\infty(M).

The set of all derivations at $p$ is a real vector space under pointwise operations, written $T_pM$ and called the tangent space to $M$ at $p$ ; its elements are tangent vectors at $p$ .

Two consequences follow at once from the product rule. First, $v(1) = v(1\cdot 1) = 2\,v(1)$ , so $v$ annihilates the constant function $1$ and, by linearity, every constant. Second, if $f(p) = g(p) = 0$ then $v(fg) = 0$ .

Tangent vectors act locally: if $f, g \in C^\infty(M)$ agree on some neighbourhood $W$ of $p$ , then $v(f) = v(g)$ . Indeed, put $h = f - g$ . Then $\{h \neq 0\}$ misses $W$ , so the support of $h$ — the closure of that set — lies in the closed set $M \setminus W$ , and $p$ does not: the support and the point are separated, with $W$ itself an open set between them. Choose a smooth bump function $\chi$ that equals $1$ on $\operatorname{supp} h$ and vanishes on a neighbourhood of $p$ , which the separation permits; then $\chi h = h$ while $\chi(p) = h(p) = 0$ , so $v(h) = v(\chi h) = 0$ by the second consequence above. Consequently $T_pU$ and $T_pM$ may be identified for any open $U \ni p$ , and a tangent vector may be applied to a function defined only near $p$ .

Remark(Two routes to the same space)Original

Hassani reaches the same definition from the physicist's side. Let $\gamma$ be a smooth curve in $M$ with $\gamma(0) = p$ ; then $f \mapsto \frac{d}{du}\big|_{u=0}\, f(\gamma(u))$ is linear and obeys the product rule, by the ordinary product rule for functions of one variable, so it is a tangent vector: the velocity of $\gamma$ at $p$ . Different curves can give the same derivation — only the first-order behaviour of $\gamma$ at $p$ matters — and this is what makes "a tangent vector is an arrow" into "a tangent vector is a directional derivative".

The two books differ in the domain they give the derivation: Lee uses $C^\infty(M)$ , globally defined smooth functions, while Hassani uses $\mathcal{F}^\infty(P)$ , functions defined near $p$ . By the locality property above the resulting vector spaces are canonically isomorphic, so nothing depends on the choice; we follow Lee and write $C^\infty(M)$ , and use locality whenever a function is given only in a chart.

Definition(Coordinate basis)AdaptedAdapted from Lee (2013), Ch. 3, Cor. 3.3, p. 54, and Prop. 3.15, p. 61

Take a smooth $n$ -manifold $M$ , a point $p$ of it, and around $p$ a smooth chart $(U, \varphi)$ whose coordinate functions we write $\varphi = (x^1, \dots, x^n)$ . For $i = 1, \dots, n$ define $\partial/\partial x^i|_p \in T_pM$ by

\frac{\partial}{\partial x^i}\Big|_p f \;=\; \frac{\partial \widehat f}{\partial x^i}\bigl(\varphi(p)\bigr), \qquad \widehat f = f \circ \varphi^{-1} ,

in words: push $f$ down into the chart and differentiate the resulting function of $n$ real variables in its $i$ -th slot, at the image of $p$ . These $n$ coordinate vectors form a basis of $T_pM$ , which therefore has dimension $n = \dim M$ . The components of $v \in T_pM$ in this basis are recovered by letting $v$ act on the coordinate functions:

v = v^i \,\frac{\partial}{\partial x^i}\Big|_p , \qquad v^i = v(x^i) ,

with the summation convention in force. If $(V, \psi)$ , $\psi = (\tilde x^1, \dots, \tilde x^n)$ , is a second chart containing $p$ , the chain rule applied to the transition map gives the change of basis and of components

\frac{\partial}{\partial x^i}\Big|_p = \frac{\partial \tilde x^{\,j}}{\partial x^i}(p)\, \frac{\partial}{\partial \tilde x^{\,j}}\Big|_p , \qquad \tilde v^{\,j} = \frac{\partial \tilde x^{\,j}}{\partial x^i}(p)\, v^i ,

which is the transformation law that makes $v$ a contravariant vector in the sense of Tensors and their transformation laws.

Idea of the proofOriginal

Why should there be no derivations beyond the partial derivatives? Because a smooth function is, to first order, its value plus a linear term, and a derivation kills constants and kills products of functions vanishing at the point. Taylor's theorem with an integral remainder makes this exact: near a point, $f$ equals its value plus $\sum_i (x^i - a^i) g_i(x)$ with $g_i$ smooth and $g_i(a) = \partial_i \widehat f(a)$ . Applying $v$ annihilates the constant, and the product rule leaves only $\sum_i v(x^i)\, \partial_i \widehat f(a)$ . So $v$ is determined by the $n$ numbers $v(x^i)$ , and the coordinate vectors realise every choice of them.

Proof(Coordinate basis)AdaptedAdapted from Lee (2013), Ch. 3, Prop. 3.2 and Cor. 3.3, pp. 53–54, and Prop. 3.15, p. 61

By locality we may work inside the chart, so assume $M$ is an open set $\widehat U \subseteq \mathbb{R}^n$ containing $a = \varphi(p)$ , with the identity chart; the general case follows by transporting along $\varphi$ , which is a diffeomorphism onto $\widehat U$ .

Step 1 (they are tangent vectors, and they are independent). Each $\partial/\partial x^i|_a$ is linear and obeys the product rule, by the ordinary product rule for partial derivatives. If $c^i \,\partial/\partial x^i|_a = 0$ , apply it to the coordinate function $x^j$ : since $\partial x^j/\partial x^i = \delta^j_i$ , we get $c^j = 0$ for each $j$ . So the $n$ vectors are linearly independent.

Step 2 (Taylor with remainder). Let $\widehat f$ be smooth on a ball $B \subseteq \widehat U$ centred at $a$ . For $x \in B$ the segment from $a$ to $x$ stays in $B$ , so we may integrate the derivative of $t \mapsto \widehat f(a + t(x-a))$ from $0$ to $1$ ; the fundamental theorem of calculus and the chain rule then give

\widehat f(x) = \widehat f(a) + \sum_{i=1}^n (x^i - a^i)\, g_i(x), \qquad g_i(x) = \int_0^1 \frac{\partial \widehat f}{\partial x^i}\bigl(a + t(x-a)\bigr)\, dt .

Each $g_i$ is smooth on $B$ , by differentiation under the integral sign, and $g_i(a) = \partial_i \widehat f(a)$ .

Step 3 (apply a derivation). Let $v$ be a derivation at $a$ . The function $x \mapsto \widehat f(a)$ is constant, so $v$ kills it. In each remaining term both factors $(x^i - a^i)$ and $g_i(x) - g_i(a)$ vanish at $a$ , so $v$ kills their product; writing $g_i = g_i(a) + (g_i - g_i(a))$ and using the product rule,

v\bigl((x^i - a^i)\,g_i\bigr) = g_i(a)\, v(x^i - a^i) + (a^i - a^i)\,v(g_i) = \frac{\partial \widehat f}{\partial x^i}(a)\, v(x^i) .

Summing, $v(f) = v(x^i)\,\partial_i \widehat f(a) = \bigl(v(x^i)\,\partial/\partial x^i|_a\bigr) f$ . Since $f$ was arbitrary, $v = v(x^i)\,\partial/\partial x^i|_a$ , so the coordinate vectors span. With Step 1 they are a basis, and $\dim T_pM = n$ .

Step 4 (change of chart). For $f$ smooth near $p$ and the two charts $x$ , $\tilde x$ , the chain rule for the transition map $\psi \circ \varphi^{-1}$ reads $\partial (f \circ \varphi^{-1})/\partial x^i = \bigl(\partial \tilde x^{\,j}/\partial x^i\bigr)\, \partial (f \circ \psi^{-1})/\partial \tilde x^{\,j}$ , evaluated at the corresponding points. That is the stated change of basis; equating the two expansions of a fixed $v$ and using uniqueness of components gives the law for $\tilde v^{\,j}$ . $\square$

Definition(Vector field)AdaptedAdapted from Lee (2013), Ch. 8, pp. 174–175 and Prop. 8.14, p. 180; Hassani (2013), §28.4.1, Def. 28.4.2, p. 877 and p. 878

A vector field $X$ on $M$ assigns to each $p \in M$ a tangent vector $X_p \in T_pM$ . In a smooth chart $(U, x^i)$ it has component functions $X^i : U \to \mathbb{R}$ determined by

X_p = X^i(p)\, \frac{\partial}{\partial x^i}\Big|_p , \qquad X^i = X(x^i) ,

and $X$ is smooth when its component functions are smooth in every smooth chart (equivalently, in the charts of one atlas, since transition maps are smooth and the components transform by the Jacobian of the transition map). The set of smooth vector fields on $M$ is written $\mathfrak{X}(M)$ ; it is a vector space over $\mathbb{R}$ and a module over $C^\infty(M)$ .

A smooth vector field acts on functions: for $f \in C^\infty(U)$ , $U$ open, define $Xf : U \to \mathbb{R}$ by $(Xf)(p) = X_p f$ . In coordinates $Xf = X^i\, \partial \widehat f/\partial x^i$ , so $Xf$ is smooth whenever $f$ is. The resulting map $X : C^\infty(M) \to C^\infty(M)$ is $\mathbb{R}$ -linear and satisfies the derivation identity

X(fg) = f\,(Xg) + g\,(Xf) ,

which is the pointwise product rule read as an identity between functions. Conversely every $\mathbb{R}$ -linear map $D : C^\infty(M) \to C^\infty(M)$ obeying that identity is $f \mapsto Xf$ for a unique smooth $X$ , namely the one with $X_p f = (Df)(p)$ : that recipe defines a derivation at each $p$ , and the field so defined is smooth because $Xf = Df$ is smooth for every smooth $f$ . We therefore use the same letter for the field and for the operator.

A curve $\gamma : I \to M$ is an integral curve of $X$ if its velocity at each time is the value of $X$ there, that is $\dot\gamma(t) = X_{\gamma(t)}$ ; in coordinates this is the system $\dot x^i = X^i(x)$ . Integral curves, the flow they generate and the Lie derivative along it are the subject of Flows and integral curves.

Definition(Lie bracket)AdaptedAdapted from Lee (2013), Ch. 8, p. 186; Hassani (2013), §28.4.1, p. 879

For $X, Y \in \mathfrak{X}(M)$ the Lie bracket $[X, Y]$ is the operator on $C^\infty(M)$

[X, Y] f = X(Yf) - Y(Xf) .

The point of the antisymmetrisation is that neither $XY$ nor $YX$ is itself a vector field, while the difference is.

Definition(The bracket is a vector field, with its coordinate formula)AdaptedAdapted from Lee (2013), Ch. 8, Lemma 8.25 and Prop. 8.26, pp. 186–187; Hassani (2013), §28.4.1, p. 879

Let $X, Y, Z \in \mathfrak{X}(M)$ and let $(U, x^i)$ be a smooth chart.

$[X, Y]$ is a smooth vector field on $M$ , and in the chart, with $X = X^i\,\partial/\partial x^i$ and $Y = Y^j\,\partial/\partial x^j$ ,
$[X, Y] = \Bigl( X^i \frac{\partial Y^j}{\partial x^i} - Y^i \frac{\partial X^j}{\partial x^i} \Bigr) \frac{\partial}{\partial x^j} \;=\; \bigl( X Y^j - Y X^j \bigr)\, \frac{\partial}{\partial x^j} .$
In particular all second derivatives cancel: the bracket is first order, although its definition is not.
$[\,\partial/\partial x^i,\, \partial/\partial x^j\,] = 0$ for all $i, j$ : coordinate vector fields commute.
The bracket is $\mathbb{R}$ -bilinear, antisymmetric, $[X, Y] = -[Y, X]$ , and satisfies the Jacobi identity
$\bigl[[X, Y], Z\bigr] + \bigl[[Z, X], Y\bigr] + \bigl[[Y, Z], X\bigr] = 0 ,$
so $\mathfrak{X}(M)$ with the bracket is a Lie algebra over $\mathbb{R}$ — infinite-dimensional, unlike the finite-dimensional Lie algebras of Matrix Lie groups and Lie algebras.

Idea of the proofOriginal

Both halves of part 1 come from the same cancellation. Applying $XY$ to a product $fg$ produces the terms one wants, $f\,XYg + g\,XYf$ , plus two cross terms $Xf\,Yg + Xg\,Yf$ . Those cross terms are symmetric in $X$ and $Y$ , so they drop out of the difference $XY - YX$ , which is therefore a derivation. In coordinates the same symmetry is the equality of mixed partial derivatives: the second-derivative terms of $XY$ and $YX$ are both $X^iY^j\,\partial_i\partial_j f$ and cancel, leaving only terms in which one derivative has landed on a component function.

ProofAdaptedAdapted from Lee (2013), Ch. 8, Lemma 8.25 and Prop. 8.26, pp. 186–187

Step 1 ( $[X,Y]$ is a derivation). It is $\mathbb{R}$ -linear, being a difference of composites of linear maps, and it sends smooth functions to smooth functions, since $X$ and $Y$ do. For $f, g \in C^\infty(M)$ , expand each term with the derivation identity twice:

X\bigl(Y(fg)\bigr) = X\bigl(f\,Yg + g\,Yf\bigr) = f\,XYg + Xf\,Yg + g\,XYf + Xg\,Yf ,

and the same with $X$ and $Y$ interchanged,

Y\bigl(X(fg)\bigr) = f\,YXg + Yf\,Xg + g\,YXf + Yg\,Xf .

The two cross terms $Xf\,Yg + Xg\,Yf$ are unchanged by the interchange, so subtracting leaves

[X,Y](fg) = f\,\bigl(XYg - YXg\bigr) + g\,\bigl(XYf - YXf\bigr) = f\,[X,Y]g + g\,[X,Y]f .

By the converse half of the definition of a vector field above, $[X,Y]$ is a smooth vector field.

Step 2 (coordinate formula). Since the bracket is now known to be a vector field, its action is local, so we may compute in a single chart, writing $f$ for the coordinate representation and $\partial_i$ for $\partial/\partial x^i$ :

X(Yf) = X^i \partial_i \bigl( Y^j \partial_j f \bigr) = X^i (\partial_i Y^j)(\partial_j f) + X^i Y^j\, \partial_i \partial_j f .

Interchanging $X$ and $Y$ and subtracting, the second-derivative terms are $X^iY^j \partial_i\partial_j f - Y^iX^j \partial_i\partial_j f$ ; relabelling the summed indices $i \leftrightarrow j$ in the second and using $\partial_i\partial_j f = \partial_j\partial_i f$ , which holds because $f$ is smooth, they cancel. What remains is

[X,Y]f = \bigl( X^i \partial_i Y^j - Y^i \partial_i X^j \bigr)\, \partial_j f ,

which is the claimed formula. Part 2 follows by taking $X^i$ , $Y^j$ constant.

Step 3 (algebraic identities). Bilinearity and antisymmetry are immediate from $[X,Y]f = XYf - YXf$ . For the Jacobi identity, expand each double bracket as an operator on $C^\infty(M)$ :

\bigl[[X,Y],Z\bigr] = XYZ - YXZ - ZXY + ZYX .

The two cyclic permutations give the analogous four-term expressions, so the sum has twelve terms. Only six distinct words of length three can occur — the six orderings of $X$ , $Y$ , $Z$ — and each therefore appears exactly twice, once with each sign. For $XYZ$ : it comes with $+$ from $\bigl[[X,Y],Z\bigr]$ and with $-$ from $\bigl[[Y,Z],X\bigr]$ , whose expansion $YZX - ZYX - XYZ + XZY$ contains $-XYZ$ ; the other five cancel in the same way. Hence the sum is zero. $\square$

RemarkOriginal

Where the hypotheses are used.

Smoothness of $f$ is used exactly once, in Step 2, to interchange mixed partial derivatives. It is what makes the bracket first order. For a function with unequal mixed partials the expression $XYf - YXf$ would still make sense but would not be given by the first-order formula.
Antisymmetry is essential, not cosmetic. The composite $XY$ alone is not a vector field. On $\mathbb{R}^2$ take $X = \partial/\partial x$ , $Y = \partial/\partial y$ , $f = x$ , $g = y$ . Then $XY(fg) = \partial_x\partial_y(xy) = \partial_x x = 1$ , whereas $f\,XYg + g\,XYf = x\,\partial_x\partial_y y + y\,\partial_x\partial_y x = 0$ . So $XY$ fails the derivation identity, while $[X,Y] = 0$ here.
The bracket is not $C^\infty(M)$ -bilinear. For $h \in C^\infty(M)$ one computes from the coordinate formula that $[X, hY] = h\,[X,Y] + (Xh)\,Y$ . So the value of $[X,Y]$ at $p$ is not determined by $X_p$ and $Y_p$ alone: it needs the fields near $p$ . This is why the bracket is not a tensor, in contrast with the metric of The metric tensor.
Smoothness of the fields is used so that $Xf$ is again smooth and $Y$ can be applied to it. For merely continuous fields $X(Yf)$ need not exist.
The locality of tangent vectors is what licenses computing in one chart in Step 2 and transporting the answer; it is proved from the product rule and a bump function, so it needs smooth bump functions to exist, which is where second countability of a manifold is eventually used.

Method(Computing a Lie bracket)Original

Write both fields in one chart as $X = X^i \partial_i$ , $Y = Y^j \partial_j$ , listing the components as functions of the coordinates.
For each $j$ , compute the single number-valued function $X Y^j - Y X^j$ — that is, differentiate the $j$ -th component of $Y$ along $X$ , differentiate the $j$ -th component of $X$ along $Y$ , and subtract. Never expand $XY f$ on a test function; the second derivatives only cancel again.
Assemble $[X,Y] = (XY^j - YX^j)\,\partial_j$ .
Check: evaluate at a convenient point, and verify antisymmetry by recomputing $[Y,X]$ or by inspecting the signs. If either field has constant components in the chart, the corresponding term drops.
If the fields are given in another coordinate system, either transform the components first, or use that the bracket is chart-independent and compute in whichever chart makes the components simplest.

Example(A bracket in

\mathbb{R}^3

)Original

On $M = \mathbb{R}^3$ with coordinates $(x,y,z)$ take

X = x\,\frac{\partial}{\partial x} + yz\,\frac{\partial}{\partial z} , \qquad Y = y^2\,\frac{\partial}{\partial x} + x\,\frac{\partial}{\partial y} .

So $X = (X^1,X^2,X^3) = (x, 0, yz)$ and $Y = (y^2, x, 0)$ .

First component. $XY^1 = X(y^2) = 0$ , because $X$ has no $\partial/\partial y$ part. $YX^1 = Y(x) = y^2$ . So $[X,Y]^1 = -y^2$ .
Second component. $XY^2 = X(x) = x$ . $YX^2 = Y(0) = 0$ . So $[X,Y]^2 = x$ .
Third component. $XY^3 = X(0) = 0$ . $YX^3 = Y(yz) = y^2\,\partial_x(yz) + x\,\partial_y(yz) = xz$ . So $[X,Y]^3 = -xz$ .
Result. $[X, Y] = -y^2\,\frac{\partial}{\partial x} + x\,\frac{\partial}{\partial y} - xz\,\frac{\partial}{\partial z} .$
A number. At the point $(2, 3, -1)$ the components are $(-9,\; 2,\; 2)$ .
Check by the definition. Take the test function $f = xz$ . Then $Yf = y^2 z$ and $X(Yf) = y^2\,X(z) = y^2 \cdot yz = y^3 z$ ; while $Xf = x\,z + yz\cdot x = xz + xyz$ and $Y(Xf) = y^2\,\partial_x(xz+xyz) + x\,\partial_y(xz+xyz) = y^2(z + yz) + x\cdot xz = y^2 z + y^3 z + x^2 z$ . Subtracting, $[X,Y]f = -y^2z - x^2 z$ . The coordinate formula gives $[X,Y](xz) = -y^2\cdot z + x\cdot 0 - xz\cdot x = -y^2 z - x^2 z$ . The two agree, and at $(2,3,-1)$ both equal $9 + 4 = 13$ .

Example(The rotation generators)Original

On $\mathbb{R}^3$ the infinitesimal rotations about the three axes are the vector fields

L_1 = y\,\frac{\partial}{\partial z} - z\,\frac{\partial}{\partial y}, \qquad L_2 = z\,\frac{\partial}{\partial x} - x\,\frac{\partial}{\partial z}, \qquad L_3 = x\,\frac{\partial}{\partial y} - y\,\frac{\partial}{\partial x} ,

that is, $L_i = \varepsilon_{ijk}\,x^j\,\partial/\partial x^k$ . Their components are $L_1 = (0, -z, y)$ , $L_2 = (z, 0, -x)$ , $L_3 = (-y, x, 0)$ . By the method,

$[L_1, L_2]^1 = L_1(z) - L_2(0) = y$ ,
$[L_1, L_2]^2 = L_1(0) - L_2(-z) = -\bigl(z\,\partial_x(-z) - x\,\partial_z(-z)\bigr) = -x$ ,
$[L_1, L_2]^3 = L_1(-x) - L_2(y) = 0 - 0 = 0$ ,

so $[L_1, L_2] = y\,\partial_x - x\,\partial_y = -L_3$ , and cyclically $[L_i, L_j] = -\varepsilon_{ijk} L_k$ . The three fields therefore span a three-dimensional Lie subalgebra of $\mathfrak{X}(\mathbb{R}^3)$ , closed under the bracket although $\mathfrak{X}(\mathbb{R}^3)$ itself is infinite-dimensional.

The sign is a convention, not a discrepancy. Quantum mechanics uses the operators $\widehat L_i = -i\hbar\, L_i$ , and then $[\widehat L_1, \widehat L_2] = (-i\hbar)^2 [L_1, L_2] = -\hbar^2(-L_3) = i\hbar\,\widehat L_3$ , the familiar commutation relation of Angular momentum algebra. Each $L_i$ is tangent to every sphere $|x| = r$ , since $L_i(x^2+y^2+z^2) = 0$ , so the same three fields restrict to vector fields on $S^2$ with the same brackets.

ApplicationOriginal

Reading a vector field as a first-order differential operator is what lets the same object be a flow, a symmetry and an observable. In fluid mechanics $X$ is the velocity field and $Xf$ is the advective rate of change of a tracer $f$ ; the bracket $[X,Y]$ then measures the failure of two advections to commute. In mechanics the Hamiltonian vector field of a function is the generator of its flow, and the Lie bracket of two such fields corresponds to the Poisson bracket of the functions (see Symplectic manifolds and Hamiltonian vector fields), which is why conserved quantities form a Lie algebra. In control theory the bracket of two control fields is the direction a system can move by alternating the two controls even when neither points that way — the reason a car can be parked sideways — and the dimension of the Lie algebra they generate decides controllability. In quantum mechanics the generators above become the angular-momentum operators, whose commutators fix the representation theory and hence the spectra of atoms.

Definition · advanced

The Riemann curvature tensor

Adapted

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Defines the Riemann tensor as the obstruction to interchanging two covariant derivatives, derives its coordinate formula and its algebraic symmetries, and builds the Ricci tensor and the scalar curvature from it.

Prerequisites: Christoffel symbols and the covariant derivative, The metric tensor, Tensors and their transformation laws

Definition(The Riemann, Ricci and scalar curvatures)AdaptedAdapted from Riley, Hobson & Bence (2006), Ch. 26, Exercise 26.27, p. 981; Stakgold & Holst (2011), §10.4, p. 728

Let $u^1,\dots,u^n$ be coordinates on a region carrying a metric $g_{ij}$ , with inverse $g^{ij}$ , and let

\Gamma^{l}_{ik} \;=\; \tfrac12\, g^{lm}\Bigl(\frac{\partial g_{mi}}{\partial u^{k}} + \frac{\partial g_{mk}}{\partial u^{i}} - \frac{\partial g_{ik}}{\partial u^{m}}\Bigr)

be the Christoffel symbols of the covariant derivative; they are symmetric in $i$ and $k$ . A semicolon denotes covariant differentiation, and repeated indices are summed. For a covector field $v_i$ we write $v_{i;jk} = (v_{i;j})_{;k}$ , so the outer derivative carries the last index.

The Riemann curvature tensor has components $R^{l}{}_{ijk}$ defined by the requirement that
$v_{i;jk} - v_{i;kj} \;=\; R^{l}{}_{ijk}\, v_{l}$
hold for every covector field $v$ . Part 1 of the statement below shows that this determines $R^{l}{}_{ijk}$ and gives its formula in terms of the $\Gamma$ 's.
The fully covariant form is $R_{lijk} = g_{lm}R^{m}{}_{ijk}$ .
The Ricci tensor is the contraction $R_{ij} = R^{k}{}_{ikj}$ , and the scalar curvature (or Ricci scalar) is $R = g^{ij}R_{ij}$ . Both are obtained from the Riemann tensor by contracting indices.
The region is flat if there are coordinates in which all the $g_{ij}$ are constants.

Stakgold and Holst place the outer derivative on the last index too, but define their curvature tensor by commuting a vector field rather than a covector; theirs is therefore the negative of the one defined here. The remark on conventions below records what that changes.

Definition(Coordinate formula, symmetries and flatness)AdaptedAdapted from Riley, Hobson & Bence (2006), Exercise 26.27, p. 981; Arfken, Weber & Harris (2013), Exercise 4.1.5, p. 214

With the notation of the definition, and with $\partial_j = \partial/\partial u^{j}$ :

Formula. For every covector field, $v_{i;jk} - v_{i;kj} = R^{l}{}_{ijk}v_{l}$ with
$R^{l}{}_{ijk} \;=\; \partial_j \Gamma^{l}_{ik} - \partial_k \Gamma^{l}_{ij} + \Gamma^{m}_{ik}\Gamma^{l}_{mj} - \Gamma^{m}_{ij}\Gamma^{l}_{mk}.$
These $n^4$ functions are the components of a tensor of type $(1,3)$ : one contravariant and three covariant indices.
Symmetries. Writing $R_{lijk} = g_{lm}R^{m}{}_{ijk}$ ,
$R_{lijk} = -R_{likj}, \qquad R_{lijk} = -R_{iljk}, \qquad R_{lijk} = R_{jkli},$
and the first Bianchi identity holds:
$R_{lijk} + R_{ljki} + R_{lkij} = 0 .$
Count. In four dimensions the first two antisymmetries cut the number of independent components from $256$ to $36$ , the pair symmetry $R_{lijk}=R_{jkli}$ cuts it to $21$ , and the Bianchi identity cuts it to $20$ .
Ricci. $R_{ij} = R_{ji}$ , so the Ricci tensor is a symmetric covariant 2-tensor and $R = g^{ij}R_{ij}$ is a scalar.
Flatness. If the region is flat, then every component $R^{l}{}_{ijk}$ vanishes, and it vanishes in every coordinate system on the region, curvilinear ones included.

Idea of the proofOriginal

Two ordinary partial derivatives commute. Two covariant derivatives do not, because each one subtracts a Christoffel correction that itself varies from point to point, and differentiating that correction leaves a residue. The residue is algebraic in $v$ : all the terms in which $v$ is differentiated are symmetric in the two differentiation indices and cancel. What survives multiplies $v_l$ , and the quotient rule then makes the coefficient a tensor. The symmetries are read off once the whole thing is written out in terms of $g$ and its first and second derivatives: the second-derivative part is a four-term combination whose index pattern displays all three symmetries at a glance, and the $\Gamma\Gamma$ part is handled by relabelling the summed indices.

ProofAdaptedAdapted from Riley, Hobson & Bence (2006), Exercise 26.27, p. 981; Riley, Hobson & Bence (2006), Exercise 26.29, p. 982; Arfken, Weber & Harris (2013), Exercise 4.1.5 and its note, p. 214

1. The formula, and the tensor character.

The covariant derivative of a covector is $v_{i;j} = \partial_j v_i - \Gamma^{m}_{ij}v_m$ , and $v_{i;j}$ is a covariant 2-tensor, so

v_{i;jk} \;=\; \partial_k\bigl(v_{i;j}\bigr) - \Gamma^{m}_{ik}\,v_{m;j} - \Gamma^{m}_{jk}\,v_{i;m}.

Subtract the same expression with $j$ and $k$ exchanged, and take the four kinds of term in turn.

The last term, $\Gamma^{m}_{jk}v_{i;m}$ , is unchanged by exchanging $j$ and $k$ , because the Christoffel symbols are symmetric in their two lower indices. It cancels.
Expanding $\partial_k(v_{i;j}) = \partial_k\partial_j v_i - (\partial_k\Gamma^{m}_{ij})v_m - \Gamma^{m}_{ij}\partial_k v_m$ , the term $\partial_k\partial_j v_i$ cancels, since mixed partial derivatives of the smooth function $v_i$ commute.
The two terms in which $v$ is differentiated once, namely $-\Gamma^{m}_{ij}\partial_k v_m$ from the expansion and $-\Gamma^{m}_{ik}\partial_j v_m$ from $-\Gamma^{m}_{ik}v_{m;j}$ , together form an expression symmetric under $j \leftrightarrow k$ . They cancel too.
What is left is undifferentiated $v$ :
$v_{i;jk} - v_{i;kj} = \bigl(-\partial_k\Gamma^{l}_{ij} + \Gamma^{m}_{ik}\Gamma^{l}_{mj}\bigr)v_l - \bigl(-\partial_j\Gamma^{l}_{ik} + \Gamma^{m}_{ij}\Gamma^{l}_{mk}\bigr)v_l,$
which is $R^{l}{}_{ijk}v_l$ with $R^{l}{}_{ijk}$ as stated.

The left-hand side is a covariant 3-tensor for each choice of $v$ , and $v_l$ is an arbitrary covector. By the quotient rule, the coefficients $R^{l}{}_{ijk}$ are the components of a tensor of type $(1,3)$ .

2. Symmetries. Write $\Gamma_{lik} = g_{lm}\Gamma^{m}_{ik}$ for the Christoffel symbols with the first index lowered, so that

\Gamma_{lik} = \tfrac12\bigl(\partial_i g_{lk} + \partial_k g_{li} - \partial_l g_{ik}\bigr), \qquad \Gamma_{lik} = \Gamma_{lki},

and note the identity $\partial_j g_{lm} = \Gamma_{lmj} + \Gamma_{mlj}$ , which is obtained by adding the two half-sums and is the content of the relation $g_{ij;k}=0$ .

Step 2a: lower the index. Using $g_{lm}\partial_j\Gamma^{m}_{ik} = \partial_j\Gamma_{lik} - (\partial_j g_{lm})\Gamma^{m}_{ik}$ and the same for the $k$ -derivative,

R_{lijk} = \partial_j\Gamma_{lik} - \partial_k\Gamma_{lij} - (\partial_j g_{lm})\Gamma^{m}_{ik} + (\partial_k g_{lm})\Gamma^{m}_{ij} + \Gamma^{m}_{ik}\Gamma_{lmj} - \Gamma^{m}_{ij}\Gamma_{lmk}.

Substituting $\partial_j g_{lm} = \Gamma_{lmj}+\Gamma_{mlj}$ kills the terms $\Gamma^{m}_{ik}\Gamma_{lmj}$ and $\Gamma^{m}_{ij}\Gamma_{lmk}$ , leaving

R_{lijk} = \partial_j\Gamma_{lik} - \partial_k\Gamma_{lij} - \Gamma^{m}_{ik}\Gamma_{mlj} + \Gamma^{m}_{ij}\Gamma_{mlk}.

Step 2b: separate second from first derivatives. Inserting the half-sums for $\Gamma_{lik}$ and $\Gamma_{lij}$ , four of the six second-derivative terms survive, and writing $\Gamma^{m}_{ik}\Gamma_{mlj} = g^{mp}\Gamma_{pik}\Gamma_{mlj}$ gives

R_{lijk} \;=\; \tfrac12\bigl(\partial_i\partial_j g_{lk} + \partial_k\partial_l g_{ij} - \partial_j\partial_l g_{ik} - \partial_i\partial_k g_{lj}\bigr) \;+\; g^{mp}\bigl(\Gamma_{pij}\Gamma_{mlk} - \Gamma_{pik}\Gamma_{mlj}\bigr). \tag{$\ast$}

Step 2c: read off the three symmetries. In $(\ast)$ the second-derivative bracket has one term for each of the four pairings, with a plus sign when the pair $\{l,k\}$ or $\{i,j\}$ sits on $g$ and a minus sign otherwise.

Exchanging $j$ and $k$ sends the bracket to its negative (the first and fourth terms swap with sign change, as do the second and third), and in the quadratic part the two products swap; so $R_{lijk} = -R_{likj}$ . The same conclusion is immediate from the original formula, where $j$ and $k$ enter antisymmetrically.
Exchanging $l$ and $i$ sends the bracket to its negative in the same way; in the quadratic part it gives $g^{mp}(\Gamma_{plj}\Gamma_{mik} - \Gamma_{plk}\Gamma_{mij})$ , which becomes the negative of the original after renaming the summed indices $m \leftrightarrow p$ . Hence $R_{lijk} = -R_{iljk}$ .
Replacing $(l,i,j,k)$ by $(j,k,l,i)$ leaves the bracket term for term unchanged; in the quadratic part it gives $g^{mp}(\Gamma_{pkl}\Gamma_{mji} - \Gamma_{pki}\Gamma_{mjl})$ , which equals the original once the symmetry $\Gamma_{lik}=\Gamma_{lki}$ is used and $m\leftrightarrow p$ renamed in the first product. Hence $R_{lijk} = R_{jkli}$ .
For the Bianchi identity, add $(\ast)$ for $(l,i,j,k)$ , $(l,j,k,i)$ and $(l,k,i,j)$ . In the second-derivative part the twelve terms cancel in six pairs, for instance $+\tfrac12\partial_i\partial_j g_{lk}$ from the first against $-\tfrac12\partial_j\partial_i g_{lk}$ from the second. In the quadratic part the six products cancel in three pairs, using only $\Gamma_{pij}=\Gamma_{pji}$ . The sum is zero.

3. Count in four dimensions. By the first two antisymmetries, $R_{lijk}$ is determined by its values on an antisymmetric pair $A=(l,i)$ and an antisymmetric pair $B=(j,k)$ . With $n=4$ each pair takes $\binom{4}{2}=6$ independent values, so $36$ components remain out of $4^4=256$ . The pair symmetry says the $6\times 6$ array indexed by $(A,B)$ is symmetric, leaving $6\cdot 7/2 = 21$ . Finally the Bianchi identity adds information only when all four indices are different, which in four dimensions happens for a single set of values, so one further condition remains and $20$ components are independent.

4. Ricci. Lowering the contracted index, $R_{ij} = R^{k}{}_{ikj} = g^{kl}R_{likj}$ and likewise $R_{ji} = g^{kl}R_{ljki}$ . Apply the pair symmetry to the second: $R_{ljki} = R_{kilj}$ . Now rename the two summed indices, $k \leftrightarrow l$ , and use $g^{kl}=g^{lk}$ :

R_{ji} = g^{kl}R_{kilj} = g^{lk}R_{likj} = g^{kl}R_{likj} = R_{ij}.

Hence the Ricci tensor is symmetric, and $R = g^{ij}R_{ij}$ is a full contraction of tensors, so a scalar.

5. Flatness. In coordinates where the $g_{ij}$ are constants all their derivatives vanish, so every $\Gamma^{l}_{ik}$ is zero and the formula of part 1 gives $R^{l}{}_{ijk}=0$ throughout. By part 1 these are the components of a tensor, and a tensor whose components vanish in one coordinate system has vanishing components in every coordinate system, because the transformation law is linear and homogeneous in them. $\square$

Remark(Where the hypotheses are used)Original

Symmetry of the Christoffel symbols in their lower indices is used twice: to cancel the term $\Gamma^{m}_{jk}v_{i;m}$ in part 1, and in the Bianchi identity in part 2. A connection with torsion fails both: the commutator then acquires an extra term proportional to the first derivative of $v$ , and the cyclic sum of curvatures is no longer zero.
Metric compatibility, in the form $\partial_j g_{lm} = \Gamma_{lmj}+\Gamma_{mlj}$ , is what produces the clean expression $(\ast)$ and hence the two symmetries $R_{lijk}=-R_{iljk}$ and $R_{lijk}=R_{jkli}$ . For a general connection only the antisymmetry in the last two indices and the Bianchi identity survive; there is then no reason for the first two indices to be antisymmetric, and no reason for the Ricci tensor to be symmetric.
Smoothness of $v$ to second order is what lets $\partial_k\partial_j v_i$ cancel; the metric must be of class $C^2$ for $(\ast)$ to make sense pointwise.
Flatness in part 5 is a hypothesis about the existence of special coordinates, not about the components in the coordinates at hand. Euclidean $\mathbb{R}^3$ in spherical polars has six nonzero Christoffel symbols and a non-constant metric, yet $R^{l}{}_{ijk}\equiv 0$ ; see the second example. The converse — that $R\equiv 0$ forces constant-coefficient coordinates to exist locally — is also true, but it is proved by constructing a frame that is parallel in every direction, whose integrability condition is precisely the vanishing of the curvature; that argument is not given here.

Remark(Conventions, and the general count)AdaptedAdapted from Stakgold & Holst (2011), §10.4, p. 728; Arfken, Weber & Harris (2013), Exercise 4.1.5, p. 214

The other sign convention. Stakgold and Holst write the commutator for a vector field,

V^{a}{}_{;bc} - V^{a}{}_{;cb} = R^{a}{}_{dbc}V^{d}, \qquad R^{a}{}_{dbc} = \Gamma^{a}_{bd,c} - \Gamma^{a}_{cd,b} + \Gamma^{a}_{ec}\Gamma^{e}_{bd} - \Gamma^{a}_{eb}\Gamma^{e}_{cd},

with a comma for partial differentiation. Their semicolon convention is the same as ours: $V^{a}{}_{;bc}$ means $(V^{a}{}_{;b})_{;c}$ , so in both conventions the outer derivative carries the last index, and their indices $(a,d,b,c)$ sit in the slots of our $(l,i,j,k)$ . Nothing is transposed; what differs is the field being commuted. The two commutators have opposite signs, because the second covariant derivatives of the scalar $\omega_a V^{a}$ do commute: expanding $(\omega_aV^{a})_{;bc}-(\omega_aV^{a})_{;cb}=0$ and cancelling the cross terms leaves

\bigl(\omega_{a;bc}-\omega_{a;cb}\bigr)V^{a} \;=\; -\,\omega_{a}\bigl(V^{a}{}_{;bc}-V^{a}{}_{;cb}\bigr),

so a covector picks up the curvature with the sign reversed. Hence their $R^{a}{}_{dbc}$ is the negative of the tensor $R^{l}{}_{ijk}$ defined here — equivalently, it is ours with the last two indices transposed. The sign propagates to every contraction: with the same contraction patterns their Ricci tensor $R_{ab}$ , their scalar curvature $R$ and their Einstein tensor $G_{ab}=R_{ab}-\tfrac12Rg_{ab}$ are likewise the negatives of ours, so any formula taken from one convention into the other must be checked for this sign, not only formulas in the Riemann tensor itself. (The field equations quoted in the application below are written with the Ricci tensor and scalar curvature of this unit.) A reliable test is that the unit sphere should come out with positive scalar curvature, as it does below. Arfken's symmetry relations, written for $R_{iklm}$ , match ours after the renaming $(i,k,l,m)\mapsto(l,i,j,k)$ , and are insensitive to the sign.

The count in $n$ dimensions. Repeating the argument of part 3 with $N = \binom{n}{2} = n(n-1)/2$ independent antisymmetric pairs gives $N(N+1)/2$ components after the pair symmetry, and the Bianchi identity removes one for each set of four distinct index values, that is $\binom{n}{4}$ of them. The difference simplifies to

\frac{N(N+1)}{2} - \binom{n}{4} \;=\; \frac{n^{2}(n^{2}-1)}{12},

which is $1$ for $n=2$ , $6$ for $n=3$ , $20$ for $n=4$ and $50$ for $n=5$ . In two dimensions the single component is $R_{1212}$ , so all of the curvature of a surface is carried by one number at each point — this is the tensor counterpart of Gauss's theorema egregium.

Method(Computing the curvature of a given metric)Original

Write down $g_{ij}$ in the chosen coordinates and invert it to get $g^{ij}$ . For a diagonal metric this is just reciprocals.
Compute the Christoffel symbols $\Gamma^{l}_{ik} = \tfrac12 g^{lm}(\partial_i g_{mk} + \partial_k g_{mi} - \partial_m g_{ik})$ , using $\Gamma^{l}_{ik}=\Gamma^{l}_{ki}$ to halve the work.
Compute $R^{l}{}_{ijk} = \partial_j\Gamma^{l}_{ik} - \partial_k\Gamma^{l}_{ij} + \Gamma^{m}_{ik}\Gamma^{l}_{mj} - \Gamma^{m}_{ij}\Gamma^{l}_{mk}$ , but only for $j<k$ and, after lowering, only for pairs $(l,i)$ with $l<i$ and $(j,k)$ with $j<k$ and $(l,i)$ not later than $(j,k)$ . The symmetries supply the rest.
Contract: $R_{ij} = R^{k}{}_{ikj}$ , then $R = g^{ij}R_{ij}$ .
Check. Verify $R_{lijk}=-R_{likj}$ and $R_{lijk}=R_{jkli}$ on one nonzero component; in two dimensions verify $R = 2R_{1212}/\det(g_{ij})$ , and compare $R/2$ with the Gaussian curvature from the second fundamental form when the surface is given in $\mathbb{R}^3$ .

Example(The sphere of radius

a

)Original

Take the surface coordinates $u^1=\theta$ , $u^2=\varphi$ on a sphere of radius $a$ , with

g_{11} = a^2, \qquad g_{22} = a^2\sin^2\theta, \qquad g_{12}=g_{21}=0,

so $g^{11} = a^{-2}$ and $g^{22} = a^{-2}\sin^{-2}\theta$ .

Christoffel symbols. The only non-constant metric component is $g_{22}$ , with $\partial_1 g_{22} = 2a^2\sin\theta\cos\theta$ . Step 2 of the method gives

\Gamma^{1}_{22} = \tfrac12 g^{11}\bigl(-\partial_1 g_{22}\bigr) = -\sin\theta\cos\theta, \qquad \Gamma^{2}_{12} = \Gamma^{2}_{21} = \tfrac12 g^{22}\,\partial_1 g_{22} = \cot\theta,

and all the others vanish.

Curvature. In two dimensions there is one independent component. Take $l=1$ , $i=2$ , $j=1$ , $k=2$ :

R^{1}{}_{212} = \partial_1\Gamma^{1}_{22} - \partial_2\Gamma^{1}_{21} + \Gamma^{m}_{22}\Gamma^{1}_{m1} - \Gamma^{m}_{21}\Gamma^{1}_{m2}.

Here $\partial_1\Gamma^{1}_{22} = -\cos 2\theta$ ; the second term vanishes because $\Gamma^{1}_{21}=0$ ; the third vanishes because $\Gamma^{1}_{11}=\Gamma^{1}_{21}=0$ ; and the fourth is $-\Gamma^{2}_{21}\Gamma^{1}_{22} = \cot\theta\,\sin\theta\cos\theta = \cos^2\theta$ . So

R^{1}{}_{212} = \cos^2\theta - \cos 2\theta = \cos^{2}\theta - (2\cos^{2}\theta - 1) = \sin^{2}\theta, \qquad R_{1212} = g_{11}R^{1}{}_{212} = a^{2}\sin^{2}\theta .

Ricci and scalar. $R_{11} = R^{k}{}_{1k1} = R^{2}{}_{121}$ . Lowering and applying the two antisymmetries, $g_{22}R^{2}{}_{121} = R_{2121} = -R_{1221} = R_{1212} = a^2\sin^2\theta$ , so $R_{11} = a^2\sin^2\theta/g_{22} = 1$ . Similarly $R_{22} = R^{1}{}_{212} = \sin^{2}\theta$ . Hence

R_{ij} = \frac{1}{a^{2}}\,g_{ij}, \qquad R = g^{ij}R_{ij} = \frac{1}{a^{2}}\,g^{ij}g_{ij} = \frac{2}{a^{2}} .

Checks. The metric is positive definite with $\det(g_{ij}) = a^4\sin^2\theta$ , and $2R_{1212}/\det(g_{ij}) = 2a^2\sin^2\theta/(a^4\sin^2\theta) = 2/a^2 = R$ , as step 5 requires. Half the scalar curvature is $1/a^2$ , the Gaussian curvature of a sphere of radius $a$ . For $a=3$ the scalar curvature is $R = 2/9 \approx 0.2222$ , positive and the same at every point, as it must be for a surface no isometry of which distinguishes one point from another.

Example(Flat space in curvilinear coordinates)Original

In spherical polars $(r,\theta,\varphi)$ on $\mathbb{R}^3$ the Euclidean metric is $g = \operatorname{diag}(1,\,r^2,\,r^2\sin^2\theta)$ , which is anything but constant. Step 2 gives six nonvanishing Christoffel symbols,

\Gamma^{r}_{\theta\theta} = -r,\quad \Gamma^{r}_{\varphi\varphi} = -r\sin^{2}\theta,\quad \Gamma^{\theta}_{r\theta} = \frac1r,\quad \Gamma^{\theta}_{\varphi\varphi} = -\sin\theta\cos\theta,\quad \Gamma^{\varphi}_{r\varphi} = \frac1r,\quad \Gamma^{\varphi}_{\theta\varphi} = \cot\theta,

together with those obtained by swapping the two lower indices. Take the component $R^{r}{}_{\theta r\theta}$ :

R^{r}{}_{\theta r\theta} = \partial_r\Gamma^{r}_{\theta\theta} - \partial_\theta\Gamma^{r}_{\theta r} + \Gamma^{m}_{\theta\theta}\Gamma^{r}_{mr} - \Gamma^{m}_{\theta r}\Gamma^{r}_{m\theta} = (-1) - 0 + 0 - \Gamma^{\theta}_{\theta r}\Gamma^{r}_{\theta\theta} = -1 - \frac1r(-r) = 0 .

The third term vanishes because $\Gamma^{r}_{mr}=0$ for every $m$ . Repeating this for the other five independent components $(n=3$ gives six $)$ returns zero every time, so every component of the Riemann tensor is $0$ . This had to happen: Cartesian coordinates make the metric the constant matrix $\delta_{ij}$ , so part 5 applies. It is the reason one may freely exchange the order of two covariant derivatives in vector calculus in curvilinear coordinates, as the differential operators in general coordinates tacitly do.

ApplicationAdaptedAdapted from Stakgold & Holst (2011), §10.4, eq. (10.4.48), pp. 727–728

General relativity replaces the Newtonian gravitational field by the curvature of a four-dimensional spacetime whose metric $g_{ab}$ has Lorentzian signature, so that the curvature computed above is the gravitational field. The field equations are

G_{ab} = \kappa\, T_{ab}, \qquad G_{ab} = R_{ab} - \tfrac12 R\, g_{ab}, \qquad \kappa = \frac{8\pi G}{c^{4}},

where $T_{ab}$ is the stress–energy tensor of the matter and radiation. These are ten equations for the ten independent components of the symmetric tensor $g_{ab}$ . The Einstein tensor $G_{ab}$ is built only from the Ricci tensor and the scalar curvature, so the field equations constrain just the traces of the Riemann tensor; in four dimensions that is $10$ of the $20$ independent components, and the remaining $10$ — the Weyl part — propagate freely. That is why gravitational waves can travel through empty space, where $T_{ab}=0$ , and be detected by interferometers such as LIGO, which measure the change in the relative lengths of two perpendicular arms as a ripple in $g_{ab}$ passes. Stakgold and Holst use the Einstein equations as a source of hard nonlinear elliptic problems: splitting off the time direction turns four of the ten equations into constraints on the initial data, and those constraints are solved numerically for the simulations against which detected waveforms are matched.

Definition · advanced

Riemannian manifolds

Adapted

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Puts an inner product on every tangent space of a smooth manifold, shows that this can always be done, and derives from it the lengths, the volume element, the Levi-Civita connection and the signature conventions that the rest of the geometry runs on.

Prerequisites: Smooth manifolds, The metric tensor, Tangent vectors and vector fields, Partitions of unity

Definition(Riemannian and pseudo-Riemannian manifolds)AdaptedAdapted from Lee (2013), Ch. 13, pp. 327–330, 337–338, 343–344

Let $M$ be a smooth manifold of dimension $n$ .

A Riemannian metric on $M$ is a covariant 2-tensor field $g$ that is smooth, is symmetric, and is positive definite on $T_pM$ for every $p\in M$ ; in other words, $g$ picks out an inner product $g_p$ on each tangent space, and does so smoothly in $p$ . One then calls $(M,g)$ a Riemannian manifold. We write $\langle v,w\rangle_g = g_p(v,w)$ for $v,w \in T_pM$ .
In smooth local coordinates $x^1,\dots,x^n$ the metric is $g = g_{ij}\,dx^i\,dx^j$ , with $g_{ij} = \langle \partial/\partial x^i, \partial/\partial x^j\rangle_g$ . Each $g_{ij}$ is a smooth function, and the matrix they form is symmetric and positive definite at every point of the chart; it is the metric tensor of the coordinate frame, and $g^{ij}$ denotes its inverse. Repeated indices are summed.
The length of $v \in T_pM$ is $|v|_g = \langle v,v\rangle_g^{1/2}$ , and the angle between nonzero $v,w$ is the $\theta \in [0,\pi]$ with $\cos\theta = \langle v,w\rangle_g/(|v|_g|w|_g)$ . A local frame $(E_1,\dots,E_n)$ is orthonormal if $\langle E_i,E_j\rangle_g = \delta_{ij}$ at every point of its domain.
The length of a piecewise smooth curve segment $\gamma:[a,b]\to M$ is $L_g(\gamma) = \int_a^b |\gamma'(t)|_g\,dt$ , and for connected $M$ the Riemannian distance $d_g(p,q)$ is the infimum of $L_g(\gamma)$ over all piecewise smooth $\gamma$ from $p$ to $q$ .
If $F:M\to N$ is smooth and $h$ is a Riemannian metric on $N$ , the pullback $F^{*}h$ is the 2-tensor field $(F^{*}h)_p(v,w) = h_{F(p)}(dF_p v, dF_p w)$ ; it is a Riemannian metric on $M$ exactly when $F$ is an immersion (see pushforwards and pullbacks).
A symmetric 2-tensor on a vector space is nondegenerate if $v \mapsto g(v,\cdot)$ is an isomorphism onto the dual space. A change of basis brings any nondegenerate symmetric 2-tensor to a diagonal matrix of $\pm 1$ ; the pair $(r,s)$ of numbers of $+1$ 's and $-1$ 's is its signature, and is independent of the basis by Sylvester's law of inertia. Relaxing positivity to nondegeneracy, a pseudo-Riemannian metric is a smooth symmetric 2-tensor field whose value at every point is nondegenerate, all these values carrying one and the same signature. When that signature is $(n-1,1)$ — or $(1,n-1)$ , conventions differing — the field is a Lorentz metric.

Definition(Existence, the Levi-Civita connection, and the volume form)AdaptedAdapted from Lee (2013), Prop. 13.3 (p. 329), Props. 15.29 and 15.31 (pp. 389–390); do Carmo (2016), §5-10, p. 447

Let $M$ be a smooth $n$ -manifold.

Existence. $M$ admits a Riemannian metric.
Levi-Civita connection. Let $g$ be a Riemannian metric on $M$ . There is exactly one covariant derivative $D$ , assigning to vector fields $u,w$ a vector field $D_u w$ , with these four properties, for all vector fields $u,v,w$ and all smooth real functions $f,h$ on $M$ :
- $D_{fu + h w}v = f\,D_u v + h\,D_w v$ ;
- $D_u(fv + hw) = f\,D_u v + (uf)\,v + h\,D_u w + (uh)\,w$ , where $uf$ is the derivative of $f$ along $u$ ;
- symmetry: $D_{x_i}x_j = D_{x_j}x_i$ for the coordinate fields $x_i = \partial/\partial u^i$ of any chart;
- compatibility with $g$ : $\dfrac{\partial}{\partial u^{k}}\langle x_i,x_j\rangle_g = \langle D_{x_k}x_i, x_j\rangle_g + \langle x_i, D_{x_k}x_j\rangle_g$ .
Its coefficients $D_{x_i}x_j = \Gamma^{k}_{ij}x_k$ in any chart are the Christoffel symbols of $g$ ,
$\Gamma^{k}_{ij} \;=\; \tfrac12\, g^{kl}\Bigl(\frac{\partial g_{il}}{\partial u^{j}} + \frac{\partial g_{jl}}{\partial u^{i}} - \frac{\partial g_{ij}}{\partial u^{l}}\Bigr).$
Volume form. If in addition $M$ is oriented and $n\ge 1$ , there is exactly one smooth $n$ -form $\omega_g$ on $M$ with $\omega_g(E_1,\dots,E_n)=1$ for every local oriented orthonormal frame $(E_i)$ , and in any oriented chart
$\omega_g = \sqrt{\det(g_{ij})}\;dx^1\wedge\cdots\wedge dx^n .$

Idea of the proofOriginal

All three parts turn a pointwise piece of linear algebra into a global object. For existence, an inner product is available in every chart — copy the Euclidean one — and the set of inner products on a fixed vector space is convex, so a partition of unity can average the local choices without destroying positivity. For the connection, the four required properties are enough to pin down the $n^3$ coefficients $\Gamma^k_{ij}$ : symmetry and compatibility give a linear system whose matrix is $g$ , and $g$ is invertible. For the volume form, the normalisation on orthonormal frames determines the form on each chart, and two oriented orthonormal frames differ by a rotation, whose determinant is $1$ , so the local definitions agree where they overlap.

ProofAdaptedAdapted from Lee (2013), proofs of Prop. 13.3 (p. 329), Props. 15.29 and 15.31 (pp. 389–390); do Carmo (2016), §5-10, pp. 446–447

1. Existence. Cover $M$ by chart domains $U_\alpha$ with coordinate maps $\varphi_\alpha$ , and on each $U_\alpha$ let $g_\alpha = \varphi_\alpha^{*}\bar g$ be the pullback of the Euclidean metric, so that $g_\alpha = \delta_{ij}\,dx^i dx^j$ in those coordinates and is therefore a Riemannian metric on $U_\alpha$ . Take a smooth partition of unity $\{\psi_\alpha\}$ subordinate to $\{U_\alpha\}$ (partitions of unity) and set $g = \sum_\alpha \psi_\alpha g_\alpha$ , each summand read as $0$ off the support of $\psi_\alpha$ .

The sum is smooth and symmetric: local finiteness of the partition makes it a finite sum near each point.
It is positive definite. Fix $p$ and $0 \ne v \in T_pM$ . Then $g_p(v,v) = \sum_\alpha \psi_\alpha(p)\,g_\alpha|_p(v,v)$ is a sum of nonnegative terms, because each $g_\alpha$ is positive definite and each $\psi_\alpha \ge 0$ . The functions $\psi_\alpha$ sum to $1$ , so at least one is strictly positive at $p$ , and its term is strictly positive. Hence $g_p(v,v)>0$ .

2. Levi-Civita connection. Work in one chart with coordinate fields $x_i$ and write $D_{x_i}x_j = \Gamma^{k}_{ij}x_k$ ; the first two properties say that $D$ is determined on all vector fields once these coefficients are known, since $D_{u}(v^j x_j) = u^i\bigl(\partial_i v^k + v^j\Gamma^k_{ij}\bigr)x_k$ . The symmetry property says $\Gamma^k_{ij}=\Gamma^k_{ji}$ .

Lower an index. Put $\Gamma_{lij} = \langle D_{x_i}x_j, x_l\rangle_g = g_{lk}\Gamma^{k}_{ij}$ ; it too is symmetric in $i$ and $j$ . Compatibility reads

\partial_k g_{ij} = \Gamma_{jki} + \Gamma_{ikj}.

Solve. Write that identity three times, cycling the indices, and combine them so that the symmetry in the last two slots of $\Gamma$ cancels all but one term:

\partial_j g_{il} + \partial_i g_{jl} - \partial_l g_{ij} = (\Gamma_{lji}+\Gamma_{ijl}) + (\Gamma_{lij}+\Gamma_{jil}) - (\Gamma_{jli}+\Gamma_{ilj}) = 2\,\Gamma_{lij},

since $\Gamma_{ijl}=\Gamma_{ilj}$ and $\Gamma_{jil}=\Gamma_{jli}$ . So $\Gamma_{lij}$ is forced, and because $\det(g_{ij})\neq 0$ we may raise the index to get $\Gamma^{k}_{ij} = g^{kl}\Gamma_{lij}$ , which is the stated formula. This proves uniqueness. Conversely, defining $\Gamma^k_{ij}$ by that formula in each chart and $D$ by the displayed expression for $D_u(v^jx_j)$ produces an operator with the four properties: the first two hold by inspection, the third because the formula is symmetric in $i$ and $j$ , and the fourth by running the computation above backwards. Two charts give the same $D$ , because $D$ is determined by the four properties, which refer only to $g$ .

3. Volume form. Uniqueness. On the domain of an oriented orthonormal frame $(E_i)$ with dual coframe $(\varepsilon^i)$ , any $n$ -form is $f\,\varepsilon^1\wedge\cdots\wedge\varepsilon^n$ , and the normalisation forces $f=1$ ; so $\omega_g = \varepsilon^1\wedge\cdots\wedge\varepsilon^n$ there, and $M$ is covered by such domains by Gram–Schmidt applied to a coordinate frame.

Existence. Define $\omega_g$ near each point by that formula and check independence of the frame. If $(\tilde E_i)$ is another oriented orthonormal frame, write $\tilde E_i = A^{j}_{\ i}E_j$ . Orthonormality of both frames says $A(p)$ is orthogonal, so $\det A = \pm 1$ , and consistency of orientation forces $\det A = +1$ . Then

\bigl(\varepsilon^1\wedge\cdots\wedge\varepsilon^n\bigr)(\tilde E_1,\dots,\tilde E_n) = \det\bigl(\varepsilon^{j}(\tilde E_i)\bigr) = \det A = 1,

which is the value $\tilde\varepsilon^1\wedge\cdots\wedge\tilde\varepsilon^n$ also takes; two $n$ -forms agreeing on one frame agree everywhere, since $\Lambda^n T_p^{*}M$ is one-dimensional.

Coordinate expression. In an oriented chart write $\omega_g = f\,dx^1\wedge\cdots\wedge dx^n$ and $\partial/\partial x^i = A^{j}_{\ i}E_j$ for an oriented orthonormal frame. Evaluating on the coordinate frame gives $f = \det A$ . On the other hand

g_{ij} = \bigl\langle A^{k}_{\ i}E_k,\; A^{l}_{\ j}E_l\bigr\rangle_g = \sum_k A^{k}_{\ i}A^{k}_{\ j},

the $(i,j)$ entry of $A^{\mathsf T}A$ , so $\det(g_{ij}) = (\det A)^2$ and $f = \pm\sqrt{\det(g_{ij})}$ . Both frames being oriented, $f>0$ , so the sign is $+$ . $\square$

Remark(Where the hypotheses are used)AdaptedAdapted from Lee (2013), p. 344

Positive definiteness is used exactly once in part 1, and it is essential: the proof works because a sum of inner products with nonnegative coefficients, not all zero, is again an inner product. Nondegenerate forms have no such convexity — adding $\operatorname{diag}(1,-1)$ to $\operatorname{diag}(-1,1)$ gives $0$ — and indeed Lee records that not every smooth manifold carries a Lorentz metric, while every one carries a Riemannian metric.
Only nondegeneracy is used in part 2. The derivation of $\Gamma_{lij}$ never uses positivity, and the inversion of $g$ needs only $\det(g_{ij})\neq 0$ . So a pseudo-Riemannian metric also has a unique Levi-Civita connection, with the same formula — which is what general relativity computes with.
Paracompactness is hidden in part 1: it is what makes a partition of unity subordinate to an arbitrary open cover exist. On a manifold whose topology is not paracompact the construction has nothing to average.
Orientability is used in part 3 twice, to force $\det A = +1$ and to fix the sign of $\sqrt{\det(g_{ij})}$ . On a non-orientable manifold there is still a well-defined density $|\omega_g|$ , enough to integrate functions and so to define volume, but no global $n$ -form.
Smoothness of $g$ makes $\Gamma^k_{ij}$ smooth; a merely continuous metric still gives lengths and distances, but no connection and no curvature.

Remark(Distance, geodesics and curvature in this language)AdaptedAdapted from Lee (2013), Thm 13.29, pp. 339–340; do Carmo (2016), §5-10, p. 448

Distance. If $M$ is connected, $d_g$ makes $M$ a metric space whose metric topology is the manifold topology. Symmetry and the triangle inequality come from reversing and concatenating curves, and $L_g$ is unchanged by reparametrisation. The substance is that $d_g(p,q)>0$ for $p\ne q$ and that the two topologies agree; both follow from the comparison $c|v|_{\bar g} \le |v|_g \le C|v|_{\bar g}$ , valid with positive constants $c,C$ on any compact subset of a chart, which holds because $|v|_g$ is continuous and strictly positive on the compact set of $\bar g$ -unit vectors over that subset.

Geodesics. A curve is a geodesic of $(M,g)$ when $D_{\gamma'}\gamma'=0$ , that is $\ddot x^k + \Gamma^{k}_{ij}\dot x^i\dot x^j = 0$ in coordinates. These are exactly the Euler–Lagrange equations of the energy $\tfrac12\int g_{ij}\dot x^i\dot x^j\,dt$ , which is why geodesics are also the stationary curves of length; see Geodesics and the Euler–Lagrange equation.

Curvature. The Riemann tensor of the Levi-Civita connection, built in The Riemann curvature tensor from the same $\Gamma^k_{ij}$ , is the curvature of $(M,g)$ . do Carmo describes the geometric content for a general $n$ : given a 2-plane $\sigma \subset T_pM$ , the geodesics leaving $p$ tangent to $\sigma$ sweep out a surface through $p$ , and the Gaussian curvature of that surface at $p$ is the sectional curvature $K(p,\sigma)$ — the best analogue in $n$ dimensions of the Gaussian curvature of the second fundamental form.

Method(Putting a metric on a manifold and computing with it)Original

Get $g_{ij}$ . If the manifold sits inside $\mathbb{R}^N$ by an immersion $F$ , take $g = F^{*}\bar g$ : substitute the components of $F$ into $dx_1^2+\cdots+dx_N^2$ and expand, which gives $g_{ij} = \partial_i F\cdot \partial_j F$ . If it does not, prescribe $g_{ij}$ in each chart and check that the charts agree, or build one with a partition of unity as in part 1.
Check. Verify that $(g_{ij})$ is symmetric and that its leading principal minors are positive (Riemannian), or compute its signature (pseudo-Riemannian).
Invert to get $g^{ij}$ , and record $\sqrt{\det(g_{ij})}$ .
Lengths and distances. $L_g(\gamma) = \int \sqrt{g_{ij}\dot x^i\dot x^j}\,dt$ . To show a particular curve realises $d_g$ , bound $L_g$ below for every competing curve.
Volumes. $\operatorname{vol}(A) = \int_A \sqrt{\det(g_{ij})}\;du^1\cdots du^n$ over a chart, patched by a partition of unity.
Connection and curvature. Compute $\Gamma^k_{ij}$ from the formula in part 2, then feed them to the recipe in The Riemann curvature tensor.

Example(The hyperbolic upper half-plane)Original

Let $M = \{(x,y)\in\mathbb{R}^2 : y>0\}$ , a smooth 2-manifold with the single chart $(x,y)$ , and put

g = \frac{dx^2+dy^2}{y^2}, \qquad\text{that is}\qquad (g_{ij}) = \frac{1}{y^{2}}\begin{pmatrix}1&0\\0&1\end{pmatrix}.

Step 2. The matrix is symmetric with minors $1/y^2>0$ and $1/y^4>0$ , so $g$ is a Riemannian metric; it is smooth because $y>0$ on $M$ .

Step 3. $(g^{ij}) = y^2 I$ and $\sqrt{\det(g_{ij})} = 1/y^{2}$ .

Step 6: Christoffel symbols. Only $\partial_y g_{11} = \partial_y g_{22} = -2/y^{3}$ is nonzero. The formula of part 2 gives

\Gamma^{x}_{xy} = \Gamma^{x}_{yx} = -\frac1y, \qquad \Gamma^{y}_{xx} = \frac1y, \qquad \Gamma^{y}_{yy} = -\frac1y,

and $\Gamma^{x}_{xx} = \Gamma^{x}_{yy} = \Gamma^{y}_{xy} = 0$ . For instance $\Gamma^{y}_{xx} = \tfrac12 g^{yy}(2\partial_x g_{xy} - \partial_y g_{xx}) = \tfrac12 y^{2}\cdot(2/y^{3}) = 1/y$ .

A geodesic. Take $\gamma(t) = (0,e^{t})$ . Then $|\gamma'|_g = e^{t}/e^{t} = 1$ , so $\gamma$ has unit speed, and the geodesic equations hold: the $x$ -equation is $0=0$ , and the $y$ -equation is $\ddot y + \Gamma^{y}_{yy}\dot y^{2} = e^{t} - (1/e^{t})e^{2t} = 0$ .

Step 4: a distance. The segment of $\gamma$ from $(0,1)$ to $(0,2)$ has length $\int_1^2 dy/y = \ln 2$ . No curve does better: for any piecewise smooth $\sigma(t)=(x(t),y(t))$ joining the two points,

L_g(\sigma) = \int \frac{\sqrt{\dot x^{2}+\dot y^{2}}}{y}\,dt \;\ge\; \int \frac{|\dot y|}{y}\,dt \;\ge\; \left|\int_1^2 \frac{dy}{y}\right| = \ln 2 .

Hence $d_g\bigl((0,1),(0,2)\bigr) = \ln 2 \approx 0.6931$ .

Step 5: an area. The coordinate square $Q = \{1<x<2,\;1<y<2\}$ has

\operatorname{vol}_g(Q) = \int_1^2\!\!\int_1^2 \frac{dx\,dy}{y^{2}} = \int_1^2 \frac{dy}{y^{2}} = \Bigl[-\frac1y\Bigr]_1^2 = \frac12 .

Its Euclidean area is $1$ , so the metric shrinks this square by exactly a factor of two.

Curvature. Feeding the $\Gamma$ 's to the curvature formula, the one independent component is

R^{x}{}_{yxy} = -\partial_y\Gamma^{x}_{yx} + \Gamma^{y}_{yy}\Gamma^{x}_{yx} - \Gamma^{x}_{yx}\Gamma^{x}_{xy} = -\frac{1}{y^{2}} + \frac{1}{y^{2}} - \frac{1}{y^{2}} = -\frac{1}{y^{2}},

where the two terms carrying the vanishing symbols $\Gamma^{x}_{yy}$ and $\Gamma^{y}_{yx}$ have been dropped. Hence $R_{xyxy} = g_{xx}R^{x}{}_{yxy} = -1/y^{4}$ , and in two dimensions the scalar curvature is $2R_{xyxy}/\det(g_{ij}) = -2$ at every point. Half of it, the Gaussian curvature, is $-1$ everywhere — the model of the hyperbolic plane. Note that the curvature is constant although none of $g_{ij}$ , $\Gamma^k_{ij}$ or $\omega_g$ is.

Example(Minkowski space: a signature that is not Riemannian)Original

On $M=\mathbb{R}^4$ with coordinates $(t,x,y,z)$ put $\eta = -c^{2}dt^{2} + dx^{2}+dy^{2}+dz^{2}$ . The matrix $\operatorname{diag}(-c^2,1,1,1)$ has determinant $-c^{2}\neq 0$ , so $\eta$ is nondegenerate, and the basis $(c^{-1}\partial_t, \partial_x,\partial_y,\partial_z)$ makes it $\operatorname{diag}(-1,1,1,1)$ : the signature is $(3,1)$ , a Lorentz metric. It is not Riemannian, since $\eta(\partial_t,\partial_t) = -c^{2}<0$ . The components are constants, so all $\Gamma^{k}_{ij}$ vanish, the Levi-Civita connection is ordinary differentiation, and the curvature is zero: Minkowski spacetime is flat. Lengths of curves are no longer given by a real square root for every tangent direction, which is why relativity separates timelike, null and spacelike curves instead.

ApplicationAdaptedAdapted from Stakgold & Holst (2011), §10.4, p. 727

In general relativity spacetime is a four-dimensional manifold carrying a Lorentz metric, and what is felt as gravity is the geometry of that metric rather than a force field: matter and energy distort $g_{ab}$ , and free particles follow its geodesics. The effect is large enough to matter in engineering — satellite clocks run at a different rate from clocks on the ground because the metric differs between the two, and the global positioning system corrects for it.

Away from relativity, a Riemannian metric is the right setting whenever a problem has a natural notion of length that is not the Euclidean one.

Mechanics. The kinetic energy $\tfrac12 M_{ij}(q)\dot q^i\dot q^j$ of a constrained system is a Riemannian metric on the configuration manifold, and by Hamilton's principle the free motion is a geodesic of that metric. Constraint forces disappear into the Christoffel symbols.
Analysis on curved domains. The Laplace–Beltrami operator $\Delta_g f = \det(g)^{-1/2}\partial_i\bigl(\det(g)^{1/2}g^{ij}\partial_j f\bigr)$ built from $g$ (see differential operators in general coordinates) is what replaces $\nabla^2$ for diffusion, waves and quantum mechanics on a curved surface or in a curved region.
Ray optics and seismology. In a medium of refractive index $n(x)$ the metric $n(x)^2\delta_{ij}$ has light rays as its geodesics, and travel times are its distances; seismic tomography inverts measured travel times for the metric.

Definition · advanced

Tangent and vector bundles

Adapted

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Packages a vector space attached to every point of a manifold into a single smooth manifold that looks locally, but usually not globally, like a product, and identifies the tangent, cotangent and tensor bundles as the examples that carry the rest of differential geometry.

Prerequisites: Tangent vectors and vector fields, Smooth manifolds, Vector spaces and bases

Definition(Vector bundles, sections and frames)AdaptedAdapted from Lee (2013), Ch. 10, pp. 249–250, 255–257

Let $M$ be a topological space.

Take a space $E$ and a continuous surjection $\pi:E\to M$ . Call the preimage $E_p = \pi^{-1}(p)$ the fibre above $p$ . The pair is a real vector bundle of rank $k$ over $M$ when two things hold. First, every fibre is equipped with the structure of a real vector space of dimension $k$ . Second, the bundle is locally a product: around any point of $M$ one can find an open $U$ and a homeomorphism $\Phi$ from $\pi^{-1}(U)$ onto $U\times\mathbb{R}^{k}$ which keeps track of the base point, $\pi_U\circ\Phi = \pi$ with $\pi_U$ the projection to $U$ , and which restricts on each single fibre $E_q$ , $q\in U$ , to a linear isomorphism onto $\{q\}\times\mathbb{R}^k$ . Such a $\Phi$ is a local trivialization of $E$ over $U$ .
$E$ is the total space, $M$ the base, $\pi$ the projection. A bundle of rank $1$ is a line bundle.
If $M$ and $E$ are smooth manifolds, $\pi$ is smooth, and the local trivializations may be taken to be diffeomorphisms, $E$ is a smooth vector bundle and such a trivialization is a smooth local trivialization.
A local trivialization over all of $M$ is a global trivialization; a bundle admitting one is trivial, and a smooth bundle admitting a smooth one is smoothly trivial. The product bundle $M\times\mathbb{R}^k \to M$ is the basic trivial example.
Where two smooth local trivializations $\Phi$ over $U$ and $\Psi$ over $V$ overlap, $\Phi\circ\Psi^{-1}(p,v) = (p,\tau(p)v)$ for a smooth map $\tau: U\cap V\to GL(k,\mathbb{R})$ , the transition function between them.
A section of $E$ is a continuous $\sigma: M\to E$ with $\pi\circ\sigma = \mathrm{Id}_M$ , so that $\sigma(p)\in E_p$ for every $p$ ; a local section is the same thing defined on an open $U\subseteq M$ . The zero section assigns to each $p$ the zero vector of $E_p$ . For a smooth bundle, $\Gamma(E)$ denotes the smooth global sections; they form a vector space under pointwise operations and a module over $C^{\infty}(M)$ under $(f\sigma)(p) = f(p)\sigma(p)$ .
Sections $\sigma_1,\dots,\sigma_k$ over $U$ , taken in that order, make up a local frame for $E$ over $U$ when, at every $p\in U$ , the $k$ values $\sigma_1(p),\dots,\sigma_k(p)$ are a basis of the fibre $E_p$ . The frame is global if $U$ is all of $M$ and smooth if every $\sigma_i$ is.

Definition(The tangent, cotangent and tensor bundles; triviality)AdaptedAdapted from Lee (2013), Prop. 3.18 (p. 66), Prop. 10.4 (p. 252), Lemma 10.5 (pp. 252–253), Prop. 10.19 and Cor. 10.20 (pp. 258–259), Prop. 11.9 (p. 276), Ch. 12 (pp. 316–317)

Let $M$ be a smooth $n$ -manifold.

The tangent bundle. The disjoint union $TM = \coprod_{p\in M} T_pM$ of the tangent spaces, with the projection $\pi(p,v)=p$ , carries a natural topology and smooth structure making it a $2n$ -dimensional smooth manifold; with its fibrewise vector space structure it is a smooth vector bundle of rank $n$ over $M$ . A chart $(U,\varphi)$ of $M$ with coordinate functions $x^i$ yields the smooth local trivialization
$\Phi\Bigl(v^{i}\frac{\partial}{\partial x^{i}}\Big|_p\Bigr) = \bigl(p,\;(v^1,\dots,v^n)\bigr),$
and when two charts overlap, the transition function relating their trivializations sends $p$ to the Jacobian, at $p$ , of the map that converts one set of coordinates into the other.
Frames and triviality. Over an open set $U$ , smooth frames and smooth trivializations of a smooth bundle determine one another: a trivialization $\Phi$ gives the frame $\sigma_i(p) = \Phi^{-1}(p,e_i)$ , and conversely every smooth frame over $U$ arises this way from some smooth trivialization over $U$ . Taking $U=M$ : a smooth bundle can be trivialized smoothly over the whole base precisely when it carries a smooth global frame. For $E=TM$ this says that $TM$ is a product exactly when $M$ is parallelizable.
Cotangent and tensor bundles. The disjoint union $T^{*}M = \coprod_p T_p^{*}M$ of the dual spaces is a smooth vector bundle of rank $n$ over $M$ , for which the coordinate covector fields dual to $(\partial/\partial x^i)$ are smooth local sections. More generally, replacing each fibre by the space $T^{(k,l)}(T_pM)$ of mixed tensors of type $(k,l)$ on $T_pM$ gives a smooth vector bundle $T^{(k,l)}TM$ of rank $n^{k+l}$ . Its sections are the tensor fields of type $(k,l)$ ; the cases $(1,0)$ and $(0,1)$ return $TM$ and $T^{*}M$ , and the case $(0,0)$ returns the trivial line bundle $M\times\mathbb{R}$ , whose sections are the real functions on $M$ .

Idea of the proofOriginal

A bundle is built by prescribing the fibres and the trivializations and then letting the transition functions do the rest. For $TM$ the trivializations come for free: a chart gives a basis of each tangent space in its domain, so listing a tangent vector's $n$ components next to the $n$ coordinates of its base point is a bijection onto an open subset of $\mathbb{R}^{2n}$ . Two charts list the same tangent vector differently, but the chain rule says the two lists differ by the Jacobian of the coordinate change, which is invertible and depends smoothly on the point — exactly what a transition function must be. For part 2 the same observation runs backwards: a frame is a way of writing each fibre element as a list of $k$ numbers, which is what a trivialization is; the only work is checking that the dictionary is smooth in both directions, and that is where inverting a matrix of smooth functions enters.

ProofAdaptedAdapted from Lee (2013), proofs of Prop. 3.18 (pp. 66–67), Prop. 10.4 (p. 252), Lemma 10.5 (pp. 252–253) and Prop. 10.19 (pp. 259–260)

1. The tangent bundle. Given a chart $(U,\varphi)$ with coordinate functions $x^i$ , define $\tilde\varphi:\pi^{-1}(U)\to\mathbb{R}^{2n}$ by

\tilde\varphi\Bigl(v^{i}\frac{\partial}{\partial x^{i}}\Big|_p\Bigr) = \bigl(x^1(p),\dots,x^n(p),\,v^1,\dots,v^n\bigr).

It is a bijection onto an open set. Its image is $\varphi(U)\times\mathbb{R}^n$ , open in $\mathbb{R}^{2n}$ , and the inverse sends $(x,v)$ to $v^i\partial/\partial x^i$ at $\varphi^{-1}(x)$ . This uses that the coordinate vectors are a basis of each tangent space over $U$ .
Two such charts are smoothly compatible. If $(V,\psi)$ is a second chart with coordinates $\tilde x^{j}$ , then on the overlap the chain rule gives $\tilde v^{j} = \bigl(\partial\tilde x^{j}/\partial x^{i}\bigr)v^{i}$ , so $\tilde\psi\circ\tilde\varphi^{-1}(x,v) = \bigl(\psi\circ\varphi^{-1}(x),\,J(x)v\bigr)$ with $J(x)$ the Jacobian of $\psi\circ\varphi^{-1}$ . Both components are smooth, the first because $M$ is a smooth manifold and the second because it is linear in $v$ with smooth coefficients.

These charts therefore give $TM$ a smooth structure of dimension $2n$ , with $\pi$ smooth because in charts it is $(x,v)\mapsto x$ . The map $\Phi = (\varphi^{-1}\times \mathrm{Id})\circ\tilde\varphi$ is then a diffeomorphism $\pi^{-1}(U)\to U\times\mathbb{R}^n$ , is linear on each fibre, and satisfies $\pi_U\circ\Phi = \pi$ : it is a smooth local trivialization. The computation above identifies the transition function as $\tau(p) = J(\varphi(p))$ .

2. Frames and triviality. Let $E\to M$ be a smooth bundle of rank $k$ and $(\sigma_i)$ a smooth local frame over $U$ . Define $\Psi: U\times\mathbb{R}^k\to\pi^{-1}(U)$ by $\Psi(p,v) = v^{i}\sigma_i(p)$ .

$\Psi$ is a bijection, because the $\sigma_i(p)$ form a basis of $E_p$ for each $p$ , and it is linear on each $\{p\}\times\mathbb{R}^k$ .
$\Psi$ is smooth, and so is its inverse. Both are local questions, so fix $q\in U$ and choose a smooth local trivialization $\Phi$ of $E$ over a neighbourhood $V\subseteq U$ of $q$ . Write $\Phi\circ\sigma_i(p) = \bigl(p,\sigma_i^1(p),\dots,\sigma_i^k(p)\bigr)$ with $\sigma_i^j$ smooth. Then $\Phi\circ\Psi(p,v) = \bigl(p, v^{i}\sigma^{1}_{i}(p),\dots,v^{i}\sigma^{k}_{i}(p)\bigr)$ , which is smooth. The matrix $\bigl(\sigma^{j}_{i}(p)\bigr)$ is invertible at each $p$ , because the $\sigma_i(p)$ are a basis and $\Phi$ is a fibrewise isomorphism; matrix inversion is smooth on $GL(k,\mathbb{R})$ , so the entries of the inverse matrix are smooth functions of $p$ , and substituting them gives a smooth formula for $(\Phi\circ\Psi)^{-1}$ .

Hence $\Psi$ is a diffeomorphism and $\Phi' = \Psi^{-1}$ is a smooth local trivialization; its associated frame $\Phi'^{-1}(p,e_i) = \Psi(p,e_i)=\sigma_i(p)$ is the one we started with. Conversely any smooth local trivialization $\Phi$ produces the smooth local frame $\sigma_i = \Phi^{-1}(\cdot,e_i)$ , which is a frame because $\Phi$ restricts to an isomorphism on each fibre. Taking $U = M$ in both directions gives the stated equivalence, and applying it to $E = TM$ , whose smooth global frames are by definition what a parallelizable manifold has, gives the last sentence.

3. Cotangent and tensor bundles. Repeat part 1 with $T_pM$ replaced by $T_p^{*}M$ : a chart gives the dual basis $(\lambda^i|_p)$ of $(\partial/\partial x^i|_p)$ at each $p\in U$ , hence a bijection $\pi^{-1}(U)\to\varphi(U)\times\mathbb{R}^{n}$ recording the $n$ components $\omega_i$ of a covector. The chain rule for the coordinate vector fields transposes into $\omega_i = (\partial\tilde x^{j}/\partial x^{i})\,\tilde\omega_j$ for the components of a covector, so the transition function is again a smooth $GL(n,\mathbb{R})$ -valued map — this time the inverse transpose of the Jacobian — and the same argument as in part 1 applies. For $T^{(k,l)}TM$ , take the basis of $T^{(k,l)}(T_pM)$ built from tensor products of the $\partial/\partial x^i|_p$ and the $\lambda^{j}|_p$ ; there are $n^{k+l}$ of them, the components of a mixed tensor transform by the corresponding product of Jacobians and inverse-transpose Jacobians, which is smooth and invertible, and the construction goes through unchanged. $\square$

Remark(Where the hypotheses are used)AdaptedAdapted from Lee (2013), Ch. 10 (pp. 250–252) and Problem 16-6 (p. 435)

Local triviality is a local hypothesis only, and that is the whole point. The Möbius bundle — the quotient of $\mathbb{R}^2$ by $(x,y)\sim(x+1,-y)$ , fibred over the circle by the first coordinate — satisfies every clause of the definition but is not trivial: a global section is a continuous $f$ on $\mathbb{R}$ with $f(x+1) = -f(x)$ , which must vanish somewhere by the intermediate value theorem, so there is no global frame and part 2 denies a global trivialization. Dropping condition (ii) from the definition would leave a family of vector spaces with no way to compare neighbouring fibres.
Linearity of $\Phi$ on fibres is used in both directions of part 2; without it a trivialization would carry no information about the vector space structure, and the matrix $(\sigma^{j}_i)$ in the proof would not be the change-of-basis matrix it needs to be.
Smoothness of the transition functions is what makes the charts of part 1 compatible. If the coordinate change on $M$ were merely $C^1$ , the Jacobian would be only continuous and $TM$ would be a topological, not a smooth, manifold.
Invertibility of $(\sigma^{j}_i(p))$ at every point — not just at one point — is needed in part 2, because smoothness of $\Psi^{-1}$ is proved by inverting that matrix throughout a neighbourhood. A $k$ -tuple of smooth sections that is a basis at one point only generates a frame on some neighbourhood of that point, not globally.
The equivalence in part 2 is not vacuous. Lee's hairy ball theorem says a nowhere-vanishing vector field exists on $S^{n}$ exactly when $n$ is odd. So $TS^{2}$ has no global frame and is not trivial, even though $S^2$ is as well behaved a manifold as one could want; while $TS^{1}$ and $TS^{3}$ are trivial.

Method(Recognising or building a vector bundle)Original

Given a family $\{E_p\}_{p\in M}$ of $k$ -dimensional vector spaces and $E = \coprod_p E_p$ with the obvious projection $\pi$ :

Cover $M$ by open sets $U_\alpha$ on which a basis $(\sigma^{\alpha}_1,\dots,\sigma^{\alpha}_k)$ of $E_p$ is given for every $p\in U_\alpha$ , varying in whatever way the construction supplies.
Define $\Phi_\alpha:\pi^{-1}(U_\alpha)\to U_\alpha\times\mathbb{R}^k$ by reading off components in that basis. Each $\Phi_\alpha$ is automatically a fibrewise isomorphism.
On each overlap compute the change-of-basis matrix $\tau_{\alpha\beta}(p)$ with $\sigma^{\beta}_i(p) = \tau_{\alpha\beta}(p)^{j}{}_{i}\,\sigma^{\alpha}_j(p)$ , and check that $p\mapsto \tau_{\alpha\beta}(p)$ is smooth into $GL(k,\mathbb{R})$ . This is the only real work.
Declare the maps $(\varphi_\alpha\times\mathrm{Id})\circ\Phi_\alpha$ to be charts on $E$ ; step 3 makes them smoothly compatible, so $E$ is a smooth manifold and a smooth rank- $k$ bundle.
To decide whether the result is trivial, look for a global frame; to show it is not, exhibit an obstruction to one (a section that must vanish, as for the Möbius bundle, or a theorem such as the hairy ball theorem).

Example(The transition function of

TS^{2}

in stereographic charts)Original

Give $S^2\subset\mathbb{R}^3$ the two stereographic charts

\varphi_N(x,y,z) = \frac{(x,y)}{1-z} \quad (z\neq 1), \qquad \varphi_S(x,y,z) = \frac{(x,y)}{1+z} \quad (z\neq -1).

The coordinate transition map. If $u = \varphi_N(p)$ then $|u|^2 = (x^2+y^2)/(1-z)^2 = (1-z^2)/(1-z)^2 = (1+z)/(1-z)$ , so

\varphi_S\circ\varphi_N^{-1}(u) = \frac{(x,y)}{1+z} = \frac{(x,y)}{1-z}\cdot\frac{1-z}{1+z} = \frac{u}{|u|^{2}} \qquad (u\neq 0).

The transition function of the bundle. By part 1 it is the Jacobian of this map. Differentiating $u^{i}/|u|^2$ ,

\tau(u)^{i}{}_{j} = \frac{\delta^{i}_{j}}{|u|^{2}} - \frac{2u^{i}u^{j}}{|u|^{4}} .

A numerical check. Take $u = (1,2)$ , so $|u|^2 = 5$ . Then $(1+z)/(1-z)=5$ gives $z = 2/3$ , $1-z = 1/3$ and $(x,y) = (1-z)u = (1/3,2/3)$ ; the point is $p = (1/3,\,2/3,\,2/3)$ , and indeed $\tfrac19+\tfrac49+\tfrac49 = 1$ . Its second chart image is $\varphi_S(p) = (1/3,2/3)/(5/3) = (1/5,2/5) = u/|u|^2$ , as the formula predicts. The transition matrix there is

\tau(1,2) = \frac15\begin{pmatrix}1&0\\0&1\end{pmatrix} - \frac{2}{25}\begin{pmatrix}1&2\\2&4\end{pmatrix} = \frac{1}{25}\begin{pmatrix}3&-4\\-4&-3\end{pmatrix},

with $\det\tau(1,2) = (-9-16)/625 = -1/25$ , which agrees with the general value $\det\tau(u) = -|u|^{-4} = -1/25$ . So the tangent vector $\partial/\partial u^1$ at $p$ , with components $(1,0)$ in the chart $\varphi_N$ , has components

\tau(1,2)\begin{pmatrix}1\\0\end{pmatrix} = \left(\tfrac{3}{25},\,-\tfrac{4}{25}\right)

in the chart $\varphi_S$ . The determinant is negative at every $u$ , so the two charts induce opposite orientations; and because $\tau$ is nowhere constant and the pair of charts cannot be reduced to one, the two trivializations really do have to be glued. By the hairy ball theorem no third trivialization covers the whole sphere: $TS^2$ is a rank- $2$ bundle over $S^2$ that is not a product.

ApplicationOriginal

Vector bundles are the language in which fields that take values in a space attached to each point are written, and the twisting measured by the transition functions is physical.

Gauge theories. A matter field in electrodynamics or the standard model is a section of a vector bundle over spacetime; the gauge group is the structure group into which the transition functions take values, and the gauge potential is a connection that says how to differentiate sections (the Christoffel symbols of the covariant derivative are the special case $E = TM$ ). A change of gauge is a change of local trivialization, and the field strength is the obstruction to trivializing the connection away.
Band theory and the Berry phase. For a crystal the occupied electronic states form a bundle over the Brillouin torus. The system is an ordinary insulator when that bundle is trivial and a topological insulator when it is not; the integer that distinguishes them, the Chern number, is computed from the transition functions and is what quantizes the Hall conductance.
Mechanics. Velocities live in $TQ$ and momenta in $T^{*}Q$ for a configuration manifold $Q$ , which is why the Lagrangian is a function on the tangent bundle and the Hamiltonian a function on the cotangent bundle; see the canonical equations and symplectic manifolds.
Continuum physics. Stress, strain and conductivity are sections of tensor bundles over the body or the sample, so that a statement such as "the stress is symmetric" is a statement about a section of a subbundle of $T^{(0,2)}TM$ , independent of coordinates; see stress and strain.

Definition · advanced

Pushforwards, pullbacks and immersions

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Gives the one linear map a smooth map induces between tangent spaces, the operation in the opposite direction it induces on covariant tensors and forms, a rule for computing both in coordinates, and the names for maps whose differential is everywhere injective or everywhere surjective.

Prerequisites: Tangent vectors and vector fields, Differential forms, Smooth manifolds, Implicit and inverse function theorems

Definition(The differential and the pushforward)AdaptedAdapted from Lee (2013), Ch. 3, p. 55 and eqs. (3.9)–(3.10), pp. 61–63; do Carmo (2016), §2-4, Props. 1–2, pp. 86–87

Let $M$ and $N$ be smooth manifolds (smooth manifolds) and let $F : M \to N$ be smooth. Tangent vectors are derivations, as in tangent vectors and vector fields.

The differential of $F$ at $p \in M$ is the map $dF_p : T_pM \to T_{F(p)}N, \qquad \bigl(dF_p v\bigr)(f) = v(f \circ F) \quad\text{for } f \in C^\infty(N).$ It is also called the pushforward at $p$ and written $F_{*p}$ or $F_*$ ; do Carmo and much of the physics literature write $d\varphi_p$ or $\varphi_*$ .
In charts $(U, x^i)$ around $p$ and $(V, y^j)$ around $F(p)$ , write $F^j = y^j \circ F$ for the component functions of $F$ . Then $dF_p\left(\frac{\partial}{\partial x^i}\Big|_p\right) = \frac{\partial F^j}{\partial x^i}(p)\,\frac{\partial}{\partial y^j}\Big|_{F(p)} ,$ with the summation convention: the matrix of $dF_p$ in the coordinate bases is the Jacobian matrix of $F$ read in those charts.
A smooth curve $\gamma : J \to M$ has velocity $\gamma'(t) = d\gamma_t(d/dt|_t) \in T_{\gamma(t)}M$ , and $(F \circ \gamma)'(t) = dF_{\gamma(t)}\bigl(\gamma'(t)\bigr)$ . For surfaces in $\mathbb{R}^3$ this is do Carmo's definition: $d\varphi_p(w)$ is the velocity at $0$ of $\varphi \circ \alpha$ for any curve $\alpha$ with $\alpha(0) = p$ , $\alpha'(0) = w$ .

A single tangent vector pushes forward, but a vector field in general does not: $F_*X$ would have to assign a vector to each $q \in N$ , and $F^{-1}(q)$ may be empty or contain several points carrying different images. Pushforward of vector fields is defined when $F$ is a diffeomorphism, by $(F_*X)_{F(p)} = dF_p(X_p)$ . Covariant objects behave better, as the next definition shows.

Definition(Pullback of covariant tensors, forms and metrics)AdaptedAdapted from Lee (2013), eq. (11.13) and Prop. 11.25, pp. 284–285; Lee (2013), Ch. 12, pp. 320–321; Lee (2013), Ch. 13, pp. 330–331; Lee (2013), Lemma 14.16, p. 361; Hassani (2013), Definition 28.5.2, p. 888

With $F : M \to N$ smooth and $p \in M$ :

The pointwise pullback $dF_p^* : T^k\bigl(T^*_{F(p)}N\bigr) \to T^k\bigl(T^*_pM\bigr)$ sends a covariant $k$ -tensor $\alpha$ at $F(p)$ to $\bigl(dF_p^*\alpha\bigr)(v_1, \dots, v_k) = \alpha\bigl(dF_p v_1, \dots, dF_p v_k\bigr), \qquad v_1, \dots, v_k \in T_pM .$ For $k = 1$ this is the dual map $dF_p^* : T^*_{F(p)}N \to T^*_pM$ , also called the cotangent map.
The pullback of a covariant $k$ -tensor field $A$ on $N$ is the $k$ -tensor field $F^*A$ on $M$ with $(F^*A)_p = dF_p^*\bigl(A_{F(p)}\bigr)$ . It is defined at every $p \in M$ , with no condition on $F$ , because the value at $p$ is built from the value of $A$ at the single point $F(p)$ . For a function $u$ (a $0$ -tensor) the convention is $F^*u = u \circ F$ .
A differential $k$ -form is an alternating covariant $k$ -tensor field (differential forms), and pulling back preserves this, so $F^* : \Omega^k(N) \to \Omega^k(M)$ .
If $g$ is a Riemannian metric on $N$ (Riemannian manifolds), $F^*g$ is a symmetric $2$ -tensor field on $M$ ; when it is positive definite it is a metric on $M$ , the pullback metric.

Definition(Rank, immersions, submersions, embeddings)AdaptedAdapted from Lee (2013), Ch. 4, pp. 77–78 and p. 85

Let $F : M \to N$ be smooth, $\dim M = m$ , $\dim N = n$ .

The rank of $F$ at $p$ is the rank of the linear map $dF_p$ , equivalently the rank of the Jacobian matrix in any charts. If it is the same number $r$ at every point, $F$ has constant rank $r$ . Since a linear map's rank is at most the dimension of its domain and of its codomain, the rank is at most $\min\{m, n\}$ ; $F$ has full rank at $p$ when it attains this bound there.
$F$ is a smooth immersion if $dF_p$ is injective for every $p$ (equivalently, $F$ has constant rank $m$ ).
$F$ is a smooth submersion if $dF_p$ is surjective for every $p$ (equivalently, constant rank $n$ ).
$F$ is a smooth embedding if it is a smooth immersion and also a topological embedding, that is, a homeomorphism from $M$ onto $F(M)$ with the subspace topology. Lee emphasises that this is more than a topological embedding that happens to be smooth; do Carmo's Definition 7 for abstract surfaces is the same requirement.
$F$ is a local diffeomorphism if every $p \in M$ has a neighbourhood $U$ with $F(U)$ open and $F|_U : U \to F(U)$ a diffeomorphism.

DefinitionAdaptedAdapted from Lee (2013), Prop. 3.6, p. 55; Lee (2013), Prop. 4.8, p. 80; Lee (2013), Rank Theorem 4.12, p. 81; Lee (2013), Prop. 4.22, p. 87; Lee (2013), Prop. 11.25 and eq. (11.13), pp. 284–285; Lee (2013), Prop. 12.25 and Cor. 12.28, pp. 320–321; Lee (2013), Prop. 13.9, p. 331; Lee (2013), Lemma 14.16, p. 361; Hassani (2013), eq. (28.37), p. 888

Let $F : M \to N$ and $G : N \to P$ be smooth and let $p \in M$ .

Functoriality. $dF_p$ is linear; $d(G \circ F)_p = dG_{F(p)} \circ dF_p$ ; $d(\mathrm{Id}_M)_p = \mathrm{Id}_{T_pM}$ . If $F$ is a diffeomorphism then $dF_p$ is an isomorphism with inverse $d(F^{-1})_{F(p)}$ .
Pullback is contravariant and multiplicative. For covariant tensor fields $A, B$ on $N$ and a function $u$ on $N$ : $F^*(A + B) = F^*A + F^*B, \quad F^*(uB) = (u \circ F)\,F^*B, \quad F^*(A \otimes B) = F^*A \otimes F^*B,$ $(G \circ F)^*B = F^*(G^*B), \qquad (\mathrm{Id}_N)^*B = B .$ $F^*B$ is smooth whenever $B$ is. On forms, $F^*(\omega \wedge \eta) = F^*\omega \wedge F^*\eta$ , and $F^*(du) = d(u \circ F)$ for smooth $u$ .
Coordinate rule. If $y^j$ are coordinates on a neighbourhood of $F(p)$ and $B = B_{i_1 \dots i_k}\,dy^{i_1} \otimes \cdots \otimes dy^{i_k}$ there, then near $p$ $F^*B = (B_{i_1 \dots i_k} \circ F)\, d F^{i_1} \otimes \cdots \otimes d F^{i_k}, \qquad F^j = y^j \circ F,$ and the same substitution with $\wedge$ in place of $\otimes$ computes the pullback of a form. If $\dim M = \dim N = n$ and $x^i$ are coordinates near $p$ , then for a continuous $u$ , $F^*\bigl(u\, dy^1 \wedge \cdots \wedge dy^n\bigr) = (u \circ F)\,(\det DF)\, dx^1 \wedge \cdots \wedge dx^n .$
Rank theorem. If $F$ has constant rank $r$ , then for each $p$ there are charts $(U, \varphi)$ centred at $p$ and $(V, \psi)$ centred at $F(p)$ with $F(U) \subset V$ , in which $\widehat F\bigl(x^1, \dots, x^r, x^{r+1}, \dots, x^m\bigr) = \bigl(x^1, \dots, x^r, 0, \dots, 0\bigr).$ For a submersion this reads $\widehat F(x^1, \dots, x^n, x^{n+1}, \dots, x^m) = (x^1, \dots, x^n)$ , and for an immersion $\widehat F(x^1, \dots, x^m) = (x^1, \dots, x^m, 0, \dots, 0)$ .
Immersion plus submersion. Being a local diffeomorphism is the same thing as being an immersion and a submersion at once. Where the two dimensions agree, $\dim M = \dim N$ , either half of that condition already forces the other, so either half alone is enough.
When an injective immersion is an embedding. An injective smooth immersion $F$ is a smooth embedding if any of the following holds: $F$ is an open map or a closed map; $F$ is proper; $M$ is compact; $M$ has empty boundary and $\dim M = \dim N$ .
Pullback-metric criterion. Let $g$ be a Riemannian metric on $N$ . Then the symmetric $2$ -tensor field $F^*g$ is positive definite at every point, and so a metric on $M$ , exactly in the case that $F$ is an immersion.

Idea of the proofOriginal

Everything here rests on one line: $dF_p$ is defined so that differentiating after composing with $F$ is the same as composing with $F$ after differentiating. Functoriality and the Jacobian formula are both that line applied to a composite. Covariant objects are machines that eat tangent vectors, so feeding them pushed-forward vectors turns a machine on $N$ into a machine on $M$ with no inversion of $F$ anywhere; that is why pullback is unconditional while pushforward of fields is not. Finally, the rank theorem says a constant-rank map is, after a change of coordinates at both ends, exactly its own linearisation, so the local questions about immersions and submersions are answered by linear algebra.

ProofAdaptedAdapted from Lee (2013), Prop. 3.6, p. 55; Lee (2013), eqs. (3.9)–(3.10), pp. 61–63; Lee (2013), Prop. 4.8, p. 80; Lee (2013), Prop. 11.25, p. 285; Lee (2013), Cor. 12.28, p. 320; Lee (2013), Prop. 13.9 (Exercise 13.10), p. 331; Lee (2013), Prop. 14.20, p. 362

We prove 1, 2, 3, 5 and 7, and the immersion and submersion cases of 4 from the general statement; the full rank theorem is proved in Lee by applying the inverse function theorem to a map whose first $r$ components have been made into the first $r$ coordinates.

Functoriality. Linearity of $dF_p v$ in $v$ is immediate from $v \mapsto v(f \circ F)$ being linear. For the chain rule, let $v \in T_pM$ and $f \in C^\infty(P)$ ; then $\bigl(d(G\circ F)_p v\bigr)(f) = v\bigl(f \circ G \circ F\bigr) = \bigl(dF_p v\bigr)(f \circ G) = \Bigl(dG_{F(p)}\bigl(dF_p v\bigr)\Bigr)(f),$ where the middle equality is the definition of $dF_p$ applied to the function $f \circ G \in C^\infty(N)$ . Taking $G = F^{-1}$ and using $d(\mathrm{Id})_p = \mathrm{Id}$ , which holds because $f \circ \mathrm{Id} = f$ , gives the last assertion.
Pullback identities. Additivity and the two product rules are read off pointwise from the defining formula, because $\alpha \mapsto \alpha(dF_pv_1, \dots, dF_pv_k)$ is linear in $\alpha$ and multiplicative on tensor products. Contravariance is the chain rule of part 1 inserted into the definition: $\bigl((G\circ F)^*B\bigr)_p(v_1,\dots,v_k) = B_{G(F(p))}\bigl(dG\,dF\,v_1, \dots\bigr) = \bigl(G^*B\bigr)_{F(p)}\bigl(dFv_1,\dots\bigr) = \bigl(F^*(G^*B)\bigr)_p .$ Since $dF_p^*$ is linear on each fibre, it carries alternating tensors to alternating tensors, and the wedge product being built from $\otimes$ and antisymmetrisation gives $F^*(\omega \wedge \eta) = F^*\omega \wedge F^*\eta$ . For $F^*du = d(u\circ F)$ , evaluate both sides on $v \in T_pM$ : the left side is $du_{F(p)}(dF_pv) = (dF_pv)(u) = v(u \circ F)$ by the definition of the differential of a function and of $dF_p$ , and the right side is $d(u\circ F)_p(v) = v(u \circ F)$ as well.
Coordinate rule. Apply part 2 repeatedly: $F^*$ of a sum of products of coefficient functions and tensor products of the $dy^i$ replaces each coefficient $B_{i_1\dots i_k}$ by $B_{i_1 \dots i_k} \circ F$ and each $dy^i$ by $F^*(dy^i) = d(y^i \circ F) = dF^i$ . In words, substitute the components of $F$ wherever a target coordinate appears and expand. For the top-degree formula, both sides are multiples of $dx^1 \wedge \cdots \wedge dx^n$ , so it is enough to evaluate on $(\partial/\partial x^1, \dots, \partial/\partial x^n)$ . The left side gives $(u\circ F)\;\bigl(dF^1 \wedge \cdots \wedge dF^n\bigr)\!\left(\frac{\partial}{\partial x^1}, \dots, \frac{\partial}{\partial x^n}\right) = (u\circ F)\,\det\left(\frac{\partial F^j}{\partial x^i}\right),$ because a wedge of $n$ covectors evaluated on $n$ vectors is the determinant of the matrix of pairings, and $dF^j(\partial/\partial x^i) = \partial F^j/\partial x^i$ . The right side gives the same, since $dx^1 \wedge \cdots \wedge dx^n$ evaluated on the coordinate frame is $1$ .
Normal forms from the rank. An immersion has constant rank $m$ and a submersion constant rank $n$ by definition, so the two displayed forms are the general one with $r = m$ and $r = n$ .
Local diffeomorphisms. If $F$ is a local diffeomorphism, then near each $p$ it restricts to a diffeomorphism, so $dF_p$ is an isomorphism by part 1; hence $dF_p$ is injective and surjective and $F$ is both an immersion and a submersion. Conversely, if it is both, then $dF_p$ is bijective at every $p$ , and the inverse function theorem for manifolds — the chart form of the inverse function theorem — makes $F$ a diffeomorphism from a neighbourhood of $p$ onto a neighbourhood of $F(p)$ . If $\dim M = \dim N$ , injectivity and surjectivity of the linear map $dF_p$ are equivalent, so one hypothesis gives the other.
Pullback-metric criterion. $F^*g$ is smooth and symmetric for any smooth $F$ , so only positivity is at issue. For $v \in T_pM$ , $(F^*g)_p(v, v) = g_{F(p)}\bigl(dF_pv,\, dF_pv\bigr) \ge 0,$ with equality exactly when $dF_pv = 0$ , because $g_{F(p)}$ is positive definite. So $(F^*g)_p$ is positive definite if and only if $\ker dF_p = 0$ . Requiring this at every $p$ is exactly the definition of an immersion. $\square$

Remark(Where the hypotheses are used)AdaptedAdapted from Lee (2013), Examples 4.18–4.20, pp. 85–86; Lee (2013), Prop. 4.22, p. 87; do Carmo (2016), §5-10, pp. 440–441

Why pushforward of fields needs a diffeomorphism. The formula $(F_*X)_{F(p)} = dF_p(X_p)$ assigns nothing to points off $F(M)$ , and assigns two values at a point with two preimages carrying different vectors. Nothing of the kind can go wrong for $F^*A$ , whose value at $p$ reads $A$ only at $F(p)$ .
Injectivity of $dF_p$ is not injectivity of $F$ , and conversely. The curve $t \mapsto (t^3, 0)$ in $\mathbb{R}^2$ is injective and a topological embedding, but it is not an immersion: its velocity vanishes at $t = 0$ . The lemniscate $t \mapsto (\sin 2t, \sin t)$ on $(-\pi, \pi)$ is an injective immersion whose image is compact while its domain is not, so it is not a homeomorphism onto its image. The line of irrational slope $t \mapsto (e^{2\pi i t}, e^{2\pi i \alpha t})$ in the torus is an injective immersion with dense image, again not an embedding. Part 6 of the statement is what rules these out: the lemniscate's domain is not compact and the map is not closed; the dense line is not proper.
Constant rank, not just rank at a point, is what the normal form needs. Rank can only jump up under a small perturbation, so a point of full rank has a neighbourhood of full rank, and that is why immersions and submersions automatically have constant rank. A map such as $F(x) = x^2$ on $\mathbb{R}$ has rank $1$ off the origin and rank $0$ at it, and no change of coordinates turns it into a linear map near $0$ .
The pullback metric may degenerate, and even when it does not, the image may be badly placed. Where $dF_p$ has a kernel, $F^*g$ has a null direction and is only a semi-definite form. Even for an injective immersion the pullback metric sees only the intrinsic geometry: do Carmo's flat torus carries the pullback of the Euclidean metric of $\mathbb{R}^4$ under an isometric immersion, yet it admits no isometric immersion into $\mathbb{R}^3$ at all, since compactness would force a point of positive Gaussian curvature while the flat torus has curvature $0$ everywhere (the Theorema Egregium).

Method(Computing a pullback in coordinates)Original

To compute $F^*B$ for a covariant tensor field or form $B$ :

Fix coordinates $x^1, \dots, x^m$ on the source and $y^1, \dots, y^n$ on the target, and write $F$ as $n$ functions $y^j = F^j(x^1, \dots, x^m)$ .
Write $B$ in the target coordinates, as a sum of coefficient functions times tensor products (or wedge products) of the $dy^j$ .
In every coefficient, replace each $y^j$ by $F^j(x)$ .
Replace each $dy^j$ by the $1$ -form $dF^j = \dfrac{\partial F^j}{\partial x^i}\,dx^i$ .
Expand, using bilinearity for $\otimes$ and antisymmetry ( $dx^i \wedge dx^i = 0$ , $dx^j \wedge dx^i = -\,dx^i \wedge dx^j$ ) for $\wedge$ . Nothing else is needed: steps 3–5 are exactly part 3 of the statement.

To decide whether $F$ is an immersion, compute the $n \times m$ Jacobian and check that its rank is $m$ at every point; the pullback metric $F^*g$ is then a metric, and its component matrix is $(F^*g)_{ik} = (g_{j\ell} \circ F)\,\dfrac{\partial F^j}{\partial x^i}\dfrac{\partial F^\ell}{\partial x^k}$ .

Example(A saddle as an embedded surface, and the geometry it inherits)Original

Let $F : \mathbb{R}^2 \to \mathbb{R}^3$ be

F(s, t) = \bigl(s + t,\; s - t,\; st\bigr),

with target coordinates $(x, y, z)$ and the Euclidean metric $\bar g = dx^2 + dy^2 + dz^2$ (the notation $dx^2$ means the symmetric product $dx\,dx$ ).

Step 1: $F$ is an embedding. The Jacobian has columns

\frac{\partial F}{\partial s} = (1, 1, t), \qquad \frac{\partial F}{\partial t} = (1, -1, s).

Their first two entries form the matrix $\begin{pmatrix} 1 & 1 \\ 1 & -1\end{pmatrix}$ , of determinant $-2 \ne 0$ , so the two columns are independent at every $(s,t)$ and $F$ is an immersion. It is injective, because $x = s+t$ and $y = s-t$ already determine $s = (x+y)/2$ and $t = (x-y)/2$ ; these formulas also give a continuous inverse on the image, so $F$ is a homeomorphism onto its image and hence an embedding. Substituting them into $z = st$ shows the image is the graph $z = (x^2 - y^2)/4$ , a saddle.

Step 2: the pullback metric. By the method, $dx = ds + dt$ , $dy = ds - dt$ , $dz = t\,ds + s\,dt$ , so

F^*\bar g = (ds + dt)^2 + (ds - dt)^2 + (t\,ds + s\,dt)^2 = (2 + t^2)\,ds^2 + 2st\,ds\,dt + (2 + s^2)\,dt^2 .

In the classical notation of the first fundamental form, $E = 2 + t^2$ , $F_{\mathrm{I}} = st$ , $G = 2 + s^2$ , and

EG - F_{\mathrm{I}}^2 = (2+t^2)(2+s^2) - s^2t^2 = 4 + 2s^2 + 2t^2 .

This is positive everywhere, as part 7 of the statement promises for an immersion. As a cross-check, $\partial_sF \times \partial_tF = (s+t,\; t-s,\; -2)$ , whose squared length is $(s+t)^2 + (t-s)^2 + 4 = 2s^2 + 2t^2 + 4$ : the same number.

Step 3: numbers at one point. At $(s, t) = (1, 2)$ the component matrix is

\bigl(F^*\bar g\bigr)_{(1,2)} = \begin{pmatrix} 6 & 2 \\ 2 & 3\end{pmatrix}, \qquad \det = 18 - 4 = 14 .

So the area element there is $\sqrt{14}\,ds \wedge dt$ with $\sqrt{14} = 3.741657\ldots$ , and the angle $\theta$ between the two coordinate curves satisfies

\cos\theta = \frac{F_{\mathrm{I}}}{\sqrt{EG}} = \frac{2}{\sqrt{18}} = \frac{\sqrt2}{3} = 0.471405\ldots, \qquad \theta = 61.8745^\circ .

Step 4: a $2$ -form. Pull back $\omega = z\,dx \wedge dy$ . Substituting, $F^*\omega = st\,(ds + dt)\wedge(ds - dt) = -2st\,ds\wedge dt$ , using $ds \wedge ds = dt \wedge dt = 0$ and $dt \wedge ds = -\,ds \wedge dt$ . At $(1,2)$ this is $-4\,ds\wedge dt$ .

ApplicationOriginal

Pullback is how a physical field defined on a big space is read on a small one. The induced metric on a surface in $\mathbb{R}^3$ — the first fundamental form — is the pullback of the Euclidean metric by the parametrisation, which is why a parametrisation must be an immersion for the surface to have a sensible geometry. The same construction gives the induced metric on a spacelike hypersurface in general relativity, used to pose the initial-value problem, and the induced metric on a string or brane worldsheet, whose area is the Nambu–Goto action. In mechanics, a holonomic constraint is an embedding of the constraint manifold into configuration space, and the kinetic-energy metric pulled back along it is what produces the reduced Lagrangian (holonomic constraints). A change of coordinates is just the identity map pulled back with different charts on domain and codomain, which is why the same substitution rule reproduces the classical transformation laws of tensors, and why the Jacobian determinant appears in the change of variables for integrals of top-degree forms (integration of forms). Submersions appear as the projection of a bundle onto its base (tangent and vector bundles) and as the momentum or energy map whose regular level sets are the invariant submanifolds of a dynamical system.

Theorem · advanced

Flows and integral curves

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Shows that a smooth vector field on a manifold has a unique maximal integral curve through each point, that these assemble into a single smooth flow whose time- $t$ maps are diffeomorphisms obeying the group law, and that differentiating a second vector field along this flow returns the Lie bracket.

Prerequisites: Tangent vectors and vector fields, Existence and uniqueness of solutions, Smooth manifolds, Dependence on initial data and Gronwall's inequality

Definition(Integral curves, flows and completeness)AdaptedAdapted from Lee (2013), pp. 206–207, 209, 211–212; Hassani (2013), Definition 28.4.5, p. 879 and eqs. (28.20)–(28.21), p. 881

Let $M$ be a smooth manifold and $V$ a smooth vector field on $M$ (tangent vectors and vector fields).

An integral curve of $V$ is a differentiable curve $\gamma : J \to M$ , $J \subset \mathbb{R}$ an interval, with $\gamma'(t) = V_{\gamma(t)} \qquad \text{for all } t \in J .$ If $0 \in J$ , the point $\gamma(0)$ is the starting point. In a chart, writing $\gamma^i$ for the coordinates of $\gamma$ and $V = V^i\,\partial/\partial x^i$ , this is the autonomous first-order system $\dot\gamma^i(t) = V^i\bigl(\gamma^1(t), \dots, \gamma^n(t)\bigr), \qquad i = 1, \dots, n .$ Hassani's streamline picture is the right one: $V$ is a velocity field and the integral curves are the paths of the particles it carries.
A flow domain for $M$ is an open set $\mathcal{D} \subset \mathbb{R} \times M$ such that for each $p \in M$ the slice $\mathcal{D}^{(p)} = \{t : (t, p) \in \mathcal{D}\}$ is an open interval containing $0$ .
A flow on $M$ is a continuous map $\theta : \mathcal{D} \to M$ on a flow domain obeying the group laws $\theta(0, p) = p, \qquad \theta\bigl(t, \theta(s, p)\bigr) = \theta(t + s, p)$ whenever $s \in \mathcal{D}^{(p)}$ , $t \in \mathcal{D}^{(\theta(s,p))}$ and $s + t \in \mathcal{D}^{(p)}$ . We write $\theta_t(p) = \theta^{(p)}(t) = \theta(t, p)$ and set $M_t = \{p : (t, p) \in \mathcal{D}\}$ . When $\mathcal{D} = \mathbb{R} \times M$ , $\theta$ is a global flow, equivalently a (left) action of the additive group $\mathbb{R}$ on $M$ ; the group laws then read $\theta_t \circ \theta_s = \theta_{t+s}$ , $\theta_0 = \mathrm{Id}_M$ , and $\{\theta_t\}_{t\in\mathbb{R}}$ is a one-parameter group of transformations.
A flow is a smooth flow when it is smooth, and not merely continuous, as a map $\mathcal{D} \to M$ . The infinitesimal generator of a smooth flow $\theta$ is the vector field $V_p = \theta^{(p)\prime}(0)$ .
A smooth vector field is called complete when every one of its maximal integral curves has the whole of $\mathbb{R}$ as its domain; equivalently, when the flow it generates is a global one.
A maximal integral curve is one that is not the restriction of an integral curve on a strictly larger open interval; a maximal flow admits no extension to a larger flow domain.

Hassani writes the flow of $X$ as $F_t = \exp(tX)$ , which turns the group laws into $\exp(tX)\exp(sX) = \exp((t+s)X)$ and $\exp(0X) = \mathrm{Id}$ ; the exponential map explains why that notation is not an accident.

Theorem(Fundamental theorem on flows)AdaptedAdapted from Lee (2013), Thm 9.12, pp. 212–213; Lee (2013), Thm 9.16 and Cor. 9.17, p. 216; Lee (2013), Thm 9.38, p. 229; Hassani (2013), Thm 28.4.6, p. 880 and eq. (28.31), p. 886

Throughout, $M$ is a smooth manifold carrying a smooth vector field $V$ .

One flow, and only one. Among all flows on $M$ there is exactly one that is smooth, is maximal, and has $V$ for its infinitesimal generator. Write it $\theta$ , on the flow domain $\mathcal{D}$ .
Its $t$ -slices are the integral curves. Through each point $p$ of $M$ runs one and only one maximal integral curve with starting point $p$ ; it is $t \mapsto \theta(t,p)$ , defined precisely on the interval $\mathcal{D}^{(p)}$ .
Along an orbit the domains are translates. Take $s \in \mathcal{D}^{(p)}$ and set $q = \theta_s(p)$ . Then $\mathcal{D}^{(q)}$ is obtained from $\mathcal{D}^{(p)}$ by subtracting $s$ : $\mathcal{D}^{(q)} = \{t - s : t \in \mathcal{D}^{(p)}\}$ . So all points of one orbit have domains of the same length.
Each time slice is a diffeomorphism. Fix $t \in \mathbb{R}$ . The set $M_t$ of points that survive to time $t$ is an open subset of $M$ ; the map $\theta_t$ carries it bijectively onto $M_{-t}$ ; and $\theta_t$ , $\theta_{-t}$ are mutually inverse diffeomorphisms between these two open sets.
Compact support forces completeness. Should $V$ vanish outside a compact set — in particular should $M$ itself be compact — then $\mathcal{D} = \mathbb{R}\times M$ and $V$ is complete.
Differentiating a second field along the flow. Let $W$ be any smooth vector field on $M$ . The Lie derivative $(\mathcal{L}_VW)_p = \frac{d}{dt}\Big|_{t=0}\; d(\theta_{-t})_{\theta_t(p)}\bigl(W_{\theta_t(p)}\bigr) = \lim_{t\to 0}\frac{d(\theta_{-t})_{\theta_t(p)}\bigl(W_{\theta_t(p)}\bigr) - W_p}{t}$ exists at every $p$ , defines a smooth vector field, and equals the Lie bracket: $\mathcal{L}_VW = [V, W].$ Consequently $\mathcal{L}_VW = -\mathcal{L}_WV$ and $\mathcal{L}_V(gW) = (Vg)W + g\,\mathcal{L}_VW$ for $g \in C^\infty(M)$ .

Idea of the proofOriginal

In a chart the equation $\gamma' = V_\gamma$ is an autonomous ODE system with a smooth right-hand side, so the Picard–Lindelöf theorem gives a solution through each point and forbids two solutions from separating once they meet. Two global facts follow from that local statement by soft arguments. Uniqueness is a connectedness argument: the set where two integral curves on a common interval agree is non-empty, closed by continuity and open by local uniqueness. The group law is the observation that the equation does not mention $t$ explicitly, so a time-shifted integral curve is again one; following the flow for time $s$ and then for time $t$ therefore lands where following it for time $s+t$ does. The one remaining point, that the maximal domain $\mathcal{D}$ is open and $\theta$ is smooth on it, is proved by walking along an integral curve and patching the local solutions supplied by the ODE theorem. For the last part, expanding the flow to first order in $t$ turns the difference quotient into $V^k\partial_kW^i - W^k\partial_kV^i$ , which is the bracket.

ProofAdaptedAdapted from Lee (2013), proof of Thm 9.12, pp. 213–214; Lee (2013), Thm D.1(c), p. 664; Lee (2013), Lemma 9.4 and Prop. 9.6, p. 208; Lee (2013), Lemma 9.15 and Thm 9.16, pp. 215–216; Hirsch, Smale & Devaney (2013), §7.2, pp. 144, 146 and §17.4, pp. 397–398

Local existence. Around any $p$ choose a chart. In it the condition on $\gamma$ is the autonomous system above with smooth right-hand side, hence locally Lipschitz, so the Picard–Lindelöf theorem gives $\varepsilon > 0$ and a solution on $(-\varepsilon, \varepsilon)$ with $\gamma(0) = p$ . Smoothness of $V$ also makes the solution smooth in $t$ , and continuous dependence on initial data makes it depend continuously on the starting point. More than that is true and is what the rest of the proof needs: shrinking $J_0 \ni 0$ and the neighbourhood $U_0$ of $p$ if necessary, the map $(t, x) \mapsto \gamma_x(t)$ sending a starting point $x \in U_0$ to the solution through it is smooth on $J_0 \times U_0$ . That is the smoothness half of the fundamental theorem for autonomous ODEs, which ODE texts usually leave out and which we take from Lee’s Thm D.1(c), p. 664.

Two integral curves that meet once agree. Let $\gamma, \tilde\gamma : J \to M$ be integral curves on a common interval with $\gamma(t_0) = \tilde\gamma(t_0)$ , and let $S = \{t \in J : \gamma(t) = \tilde\gamma(t)\}$ .

$S \ne \varnothing$ , since $t_0 \in S$ .
$S$ is closed in $J$ : it is the preimage of the diagonal, which is closed because $M$ is Hausdorff, under the continuous map $t \mapsto (\gamma(t), \tilde\gamma(t))$ .
$S$ is open in $J$ : if $t_1 \in S$ , work in a chart around $\gamma(t_1)$ ; both curves solve the same ODE with the same value at $t_1$ , so the uniqueness half of Picard–Lindelöf makes them agree on an interval around $t_1$ .

Since $J$ is connected, $S = J$ .

The maximal integral curve. Define $\mathcal{D}^{(p)}$ by taking every open interval around $0$ that carries an integral curve starting at $p$ , and forming their union; set $\theta^{(p)}(t)$ equal to the value at $t$ shared by all those curves; the previous paragraph makes this well defined, and by construction no integral curve starting at $p$ has a larger domain. Put $\mathcal{D} = \{(t,p) : t \in \mathcal{D}^{(p)}\}$ and $\theta(t,p) = \theta^{(p)}(t)$ . This proves part 2 once $\theta$ is known to be a flow.

The group laws and part 3. Because the system is autonomous, if $\gamma$ is an integral curve then so is $t \mapsto \gamma(t + s)$ (the translation lemma): its velocity at $t$ is the velocity of $\gamma$ at $t+s$ , which is $V$ at the same point. Fix $p$ and $s \in \mathcal{D}^{(p)}$ , and set $q = \theta_s(p)$ . The curve $\sigma(t) = \theta^{(p)}(t+s)$ , defined for $t \in \mathcal{D}^{(p)} - s$ , is therefore an integral curve starting at $q$ , so by uniqueness it agrees with $\theta^{(q)}$ where both are defined, and maximality of $\theta^{(q)}$ gives $\mathcal{D}^{(p)} - s \subset \mathcal{D}^{(q)}$ ; this is the second group law, and $\theta(0,p) = p$ is the definition. Since $0 \in \mathcal{D}^{(p)}$ , the inclusion gives $-s \in \mathcal{D}^{(q)}$ and $\theta^{(q)}(-s) = p$ . Running the same argument with $(-s, q)$ in place of $(s,p)$ gives $\mathcal{D}^{(q)} + s \subset \mathcal{D}^{(p)}$ , so the two intervals agree: part 3.

$\mathcal{D}$ is open and $\theta$ is smooth (sketch of Lee's argument). Call a pair $(t,p) \in \mathcal{D}$ tame when some product $J \times U \subset \mathcal{D}$ , with $J$ an interval holding both $0$ and $t$ and $U$ a neighbourhood of $p$ , lies in the region where $\theta$ is already known to be defined and smooth; let $W$ collect the tame pairs. Then $W$ is open and $\theta|_W$ is smooth, so it suffices that $W = \mathcal{D}$ . If not, pick $(\tau, p_0) \in \mathcal{D}\setminus W$ , say with $\tau > 0$ , and let $t_0 = \inf\{t : (t, p_0) \notin W\}$ . The ODE theorem applied in a chart around $p_0$ puts a whole product neighbourhood of $(0, p_0)$ into $W$ , so $t_0 > 0$ ; applied in a chart around $q_0 = \theta^{(p_0)}(t_0)$ it gives $\varepsilon > 0$ and a neighbourhood $U_0$ of $q_0$ with $(-\varepsilon,\varepsilon)\times U_0 \subset W$ . Choosing $t_1 < t_0$ with $t_1 + \varepsilon > t_0$ and $\theta^{(p_0)}(t_1) \in U_0$ , the two smooth formulas $\theta_t$ (for $t$ near and below $t_1$ ) and $\theta_{t - t_1}\circ\theta_{t_1}$ (for $t$ near $t_0$ ) agree where they overlap, by the group law, and together define $\theta$ smoothly past $t_0$ — contradicting the choice of $t_0$ .

Part 4. $M_t$ is a slice of the open set $\mathcal{D}$ , hence open. If $p \in M_t$ , part 3 with $s = t$ gives $\mathcal{D}^{(\theta_t(p))} = \mathcal{D}^{(p)} - t \ni -t$ , so $\theta_t(p) \in M_{-t}$ ; and the group law gives $\theta_{-t}\circ\theta_t = \mathrm{Id}$ on $M_t$ and $\theta_t \circ \theta_{-t} = \mathrm{Id}$ on $M_{-t}$ . Both maps are smooth, so each is a diffeomorphism.

Part 5 (completeness). First, a uniform time lemma: if there is $\varepsilon > 0$ with $(-\varepsilon, \varepsilon) \subset \mathcal{D}^{(p)}$ for every $p$ , then $V$ is complete. Indeed, suppose $b = \sup\mathcal{D}^{(p)} < \infty$ ; pick $t_0 \in (b - \varepsilon, b)$ and $q = \theta^{(p)}(t_0)$ . By the hypothesis $\theta^{(q)}$ is defined on $(-\varepsilon, \varepsilon)$ , and by the group law $\theta^{(q)}(t - t_0) = \theta^{(p)}(t)$ wherever both are defined, so the two formulas splice to an integral curve starting at $p$ on $(-\varepsilon, t_0 + \varepsilon)$ , whose right endpoint exceeds $b$ — contradicting maximality. Now let $K = \operatorname{supp}V$ be compact. Each $p \in K$ has a neighbourhood $U_p$ and $\varepsilon_p > 0$ with $(-\varepsilon_p, \varepsilon_p) \times U_p \subset \mathcal{D}$ ; finitely many $U_{p_1}, \dots, U_{p_k}$ cover $K$ , and $\varepsilon = \min_j \varepsilon_{p_j}$ works for every starting point in $K$ . A point outside $K$ has $V = 0$ there, and the constant curve is its integral curve, defined on all of $\mathbb{R}$ . The uniform time lemma applies.

Part 6 (the Lie derivative is the bracket). The existence and smoothness come from the formula below, which exhibits the quantity as the $t$ -derivative, at $t = 0$ , of a smooth function of $(t, x)$ — so the resulting coefficients are smooth in $x$ as well, and $\mathcal{L}_VW$ is a smooth vector field. Fix $p$ and a chart $x^i$ around $p$ ; for $t$ small, $\theta_t(p)$ stays in the chart. Write $\theta^i(t,x)$ for the components of the flow and set

A(t)^i{}_j = \frac{\partial\theta^i(-t, y)}{\partial y^j}\bigg|_{y = \theta_t(p)}, \qquad w(t)^j = W^j\bigl(\theta_t(p)\bigr),

so that the vector being differentiated is $A(t)^i{}_j\,w(t)^j\,\partial/\partial x^i|_p$ , since the matrix of $d(\theta_{-t})$ in coordinates is the Jacobian of $\theta_{-t}$ . Both $A$ and $w$ are smooth in $t$ , so the derivative at $t = 0$ exists and the product rule gives

(\mathcal{L}_VW)^i_p = \dot A(0)^i{}_j\,W^j(p) + \dot w(0)^i ,

because $A(0) = I$ by $\theta_0 = \mathrm{Id}$ .

$\dot w(0)^i = V^k(p)\,\partial_kW^i(p)$ , by the chain rule and $\partial_t\theta^k(0,p) = V^k(p)$ .
For $\dot A(0)$ , note that $\theta^i(0, y) = y^i$ , so $\partial\theta^i(0,y)/\partial y^j = \delta^i_j$ is constant in $y$ ; hence the motion of the evaluation point contributes nothing at $t=0$ , and $\dot A(0)^i{}_j = \frac{\partial}{\partial t}\bigg|_{t=0}\frac{\partial\theta^i(-t,y)}{\partial y^j}\bigg|_{y=p} = \frac{\partial}{\partial y^j}\bigg|_{y=p}\frac{\partial\theta^i(-t,y)}{\partial t}\bigg|_{t=0} = -\,\partial_jV^i(p),$ the exchange of the two derivatives being legitimate because $\theta$ is smooth in $(t,y)$ jointly.

Therefore $(\mathcal{L}_VW)^i_p = V^k\partial_kW^i - W^j\partial_jV^i$ at $p$ , which is the coordinate expression of $[V,W]$ . The two listed consequences follow from antisymmetry of the bracket and from its derivation property in the second slot. $\square$

Remark(Where the hypotheses are used)AdaptedAdapted from Lee (2013), Examples 9.9–9.10, p. 211; Lee (2013), Thm 9.16, p. 216; Lee (2013), Thm 9.38, p. 229; Lee (2013), commuting flows and Thm 9.44, pp. 232–233

Smoothness of $V$ is used twice. It makes the right-hand side of the ODE locally Lipschitz, which is what uniqueness needs; a merely continuous field still has integral curves but may have several through a point, as $\dot x = x^{2/3}$ shows (existence and uniqueness), and then no flow exists, because $\theta_t$ would not be well defined. It is also what makes $\theta$ jointly smooth in $(t, p)$ , without which $d(\theta_{-t})$ in part 6 would not be differentiable in $t$ .
Nothing forces completeness. On $M = \mathbb{R}^2\setminus\{0\}$ the field $\partial/\partial x$ is as smooth as can be, yet the integral curve starting at $(-1, 0)$ runs into the missing origin and stops at $t = 1$ : the obstruction is a hole in $M$ . Lee’s second example, transplanted from the plane to the line: on $M = \mathbb{R}$ the field $x^2\,\partial/\partial x$ has the integral curve $t \mapsto 1/(1-t)$ from $x = 1$ , which escapes to infinity in finite time, the obstruction now being growth of $V$ rather than a hole. (Lee states it on $M = \mathbb{R}^2$ with the field $x^2\,\partial/\partial x$ and the curve $t \mapsto (1/(1-t), 0)$ from $(1,0)$ ; the second coordinate is inert.) Compact support removes both, which is why part 5 is true; a flow that is only local is exactly why $\mathcal{D}$ , rather than $\mathbb{R}\times M$ , appears throughout.
Why the Lie derivative needs a vector field, not a tangent vector. The naive difference $W_{\gamma(t)} - W_p$ is meaningless because the two vectors live in different tangent spaces. Part 4 is what repairs this: $\theta_{-t}$ is a diffeomorphism near $\theta_t(p)$ , so $d(\theta_{-t})$ transports $W_{\theta_t(p)}$ back to $T_pM$ before the subtraction. That the answer is antisymmetric in $V$ and $W$ — not obvious from the definition, which treats them very differently — is the content of $\mathcal{L}_VW = [V, W]$ .
Flows do not in general commute. $\mathcal{L}_VW = [V,W]$ is the infinitesimal statement of this: the flows of $V$ and $W$ commute precisely when their bracket vanishes (Lee, Thm 9.44, p. 233). For incomplete fields “commute” has to be read in the local sense Lee defines on pp. 232–233: whenever one of $\theta_t\circ\psi_s(p)$ , $\psi_s\circ\theta_t(p)$ is defined for all $(s,t)$ in a rectangle $J \times K$ around the origin, both are and they agree.

Method(Finding the flow and the maximal domains)Original

Write $V$ in a convenient chart and read off the system $\dot x^i = V^i(x)$ .
Solve it with a general initial condition $x(0) = p$ . The solution $\theta_t(p)$ , as a function of $(t, p)$ , is the flow.
For each $p$ , find the largest open interval around $t = 0$ on which the solution stays in $M$ and stays finite. That interval is $\mathcal{D}^{(p)}$ , and the union of $\{p\}\times\mathcal{D}^{(p)}$ is the flow domain.
Check the answer against the group law $\theta_t\circ\theta_s = \theta_{t+s}$ and against $\partial_t\theta_t(p)|_{t=0} = V_p$ .
To differentiate another field along the flow, you do not need the flow: compute $[V, W]^i = V^k\partial_kW^i - W^k\partial_kV^i$ in any chart.

Example(An incomplete flow on the line, with its exact domain)AdaptedAdapted from Hirsch, Smale & Devaney (2013), §7.2, p. 146

The equation and its solution family are the standard blow-up example of Hirsch, Smale and Devaney (p. 146), which existence and uniqueness also reports; what is ours here is the flow written as a function of its starting point, the exact maximal domains and the numbers.

Let $M = \mathbb{R}$ and $V = (1 + x^2)\,d/dx$ , a smooth and nowhere-vanishing field. The system is $\dot x = 1 + x^2$ , which separates:

\int\frac{dx}{1+x^2} = \int dt \quad\Longrightarrow\quad \arctan x(t) = t + \arctan x_0 ,

\theta_t(x_0) = \tan\bigl(t + \arctan x_0\bigr).

Check. $\theta_0(x_0) = x_0$ ; and $\frac{d}{dt}\tan u = \sec^2 u = 1 + \tan^2 u = 1 + \theta_t(x_0)^2$ , as required. The group law holds because $\arctan\theta_s(x_0) = s + \arctan x_0$ , so applying the formula again adds $t$ to the same angle.

Maximal domain. The solution exists exactly while $t + \arctan x_0$ stays in $(-\pi/2, \pi/2)$ , so

\mathcal{D}^{(x_0)} = \Bigl(-\tfrac{\pi}{2} - \arctan x_0,\; \tfrac{\pi}{2} - \arctan x_0\Bigr),

an interval of length $\pi$ for every starting point — consistent with part 3 of the theorem, which says the domains of different points on one orbit are translates of each other. Starting at $x_0 = 1$ gives $\arctan x_0 = \pi/4$ and

\mathcal{D}^{(1)} = \bigl(-3\pi/4,\; \pi/4\bigr) = (-2.356194\ldots,\; 0.785398\ldots),

and for instance $\theta_{0.5}(1) = \tan(0.5 + \pi/4) = 3.408223\ldots$ , while $\theta_t(1) \to +\infty$ as $t \uparrow \pi/4$ . So $V$ is not complete, and $M_t = \varnothing$ for $|t| \ge \pi$ : no single diffeomorphism $\theta_t$ moves the whole line for such $t$ .

Example(A Lie derivative computed twice)Original

On $M = \mathbb{R}^2$ take

V = x\,\frac{\partial}{\partial x} + 2y\,\frac{\partial}{\partial y}, \qquad W = y\,\frac{\partial}{\partial x}.

The flow of $V$ . The system $\dot x = x$ , $\dot y = 2y$ gives $\theta_t(x, y) = (e^tx,\, e^{2t}y)$ , defined for all $t$ : $V$ is complete and $\{\theta_t\}$ is a one-parameter group of linear diffeomorphisms.

By the definition. $\theta_{-t}$ is linear, so $d(\theta_{-t})$ is the matrix $\operatorname{diag}(e^{-t}, e^{-2t})$ at every point. At $p = (x,y)$ , the field $W$ evaluated at $\theta_t(p) = (e^tx, e^{2t}y)$ is $e^{2t}y\,\partial/\partial x$ , and transporting it back,

d(\theta_{-t})_{\theta_t(p)}\bigl(W_{\theta_t(p)}\bigr) = e^{-t}\cdot e^{2t}y\,\frac{\partial}{\partial x}\bigg|_p = e^{t}\,y\,\frac{\partial}{\partial x}\bigg|_p = e^tW_p .

Differentiating at $t = 0$ gives $\mathcal{L}_VW = y\,\partial/\partial x = W$ .

By the bracket. With $V = (x, 2y)$ and $W = (y, 0)$ ,

[V,W]^1 = V^k\partial_kW^1 - W^k\partial_kV^1 = (x\partial_x + 2y\partial_y)(y) - (y\partial_x)(x) = 2y - y = y,

[V,W]^2 = (x\partial_x + 2y\partial_y)(0) - (y\partial_x)(2y) = 0 - 0 = 0 ,

so $[V,W] = y\,\partial/\partial x = W$ . The two routes agree, as part 6 requires.

Numbers. Take $p = (1, 3)$ and $t = \ln 2$ . Then $\theta_t(p) = (2, 12)$ and $W_{\theta_t(p)} = 12\,\partial/\partial x$ . Transporting back multiplies the $x$ -component by $e^{-t} = 1/2$ , giving $6\,\partial/\partial x$ in $T_pM$ — which is $e^{\ln 2}\,W_p = 2\cdot 3\,\partial/\partial x$ , as the formula $e^tW_p$ predicts. Its $t$ -derivative at $t=0$ is $3\,\partial/\partial x = (\mathcal{L}_VW)_p$ . Because the transported field is exactly $e^tW_p$ , the whole family stays parallel to $W_p$ : $W$ is an eigenvector of the flow's action with eigenvalue $e^t$ .

ApplicationOriginal

A flow is the time evolution of an autonomous system, and the theorem is the statement that such evolution is well defined, reversible and smooth in the initial data for as long as it lasts. In Hamiltonian mechanics the Hamiltonian vector field of $H$ has a flow whose time- $t$ maps are canonical transformations preserving phase-space volume (the canonical equations and symplectic manifolds); the group law $\theta_t\circ\theta_s = \theta_{t+s}$ is the statement that the dynamics has no preferred origin of time. In fluid mechanics, $\theta_t$ is the map from a fluid parcel's initial to its current position, so the Lagrangian and Eulerian descriptions are related exactly by the flow, and $\mathcal{L}_V$ is the rate of change seen by an observer carried along with the fluid. Incompleteness is not a technicality: it is the blow-up of a solution in finite time, which is the behaviour the example above displays and which the $\dot x = 1 + x^2$ model shares with the simplest nonlinear reaction equations. One-parameter groups of diffeomorphisms are the geometric form of a continuous symmetry, so a symmetry of a Lagrangian is a flow under which it is invariant, and Noether's theorem (Noether's theorem) converts each such flow into a conserved quantity; when the flow preserves a metric its generator is a Killing field (Killing vector fields).

Definition · advanced

The exponential map

Adapted

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Gives the two constructions that turn a tangent vector at a distinguished point into a point of the manifold — integrating a left-invariant field on a Lie group, and running a geodesic for unit time on a Riemannian manifold — shows that both reduce to the matrix exponential or to normal coordinates in the familiar cases, and says exactly how far each one is a diffeomorphism.

Prerequisites: Flows and integral curves, Matrix Lie groups and Lie algebras, Geodesics, Riemannian manifolds, Functions of matrices and the matrix exponential

Definition(The exponential map of a Lie group)AdaptedAdapted from Lee (2013), Ch. 20, p. 516 and p. 518; Hassani (2013), p. 881 and Thm 29.1.21, p. 925

Let $G$ be a Lie group with identity $e$ and Lie algebra $\mathfrak{g}$ , realised as the left-invariant vector fields on $G$ or, equivalently, as $T_eG$ (matrix Lie groups and Lie algebras).

A one-parameter subgroup of $G$ is a Lie group homomorphism $\gamma : \mathbb{R} \to G$ , where $\mathbb{R}$ carries addition. It is a homomorphism into $G$ , not a subgroup of $G$ .
Left-invariant vector fields are complete, so each $X \in \mathfrak{g}$ has a maximal integral curve $\gamma_X : \mathbb{R}\to G$ with $\gamma_X(0) = e$ (flows and integral curves).
The exponential map of $G$ is $\exp : \mathfrak{g}\to G, \qquad \exp X = \gamma_X(1).$

Hassani writes the flow of any vector field $X$ as $F_t = \exp(tX)$ ; on a Lie group with $X$ left-invariant this notation agrees with the definition above, because the flow of $X$ is right translation by $\exp tX$ .

Definition(The exponential map of a Riemannian manifold)AdaptedAdapted from do Carmo (2016), §4-6, p. 288 and pp. 289–290

Let $(M, g)$ be a Riemannian manifold (Riemannian manifolds) and $p \in M$ . For $v \in T_pM$ let $\gamma(\,\cdot\,, v)$ be the unique maximal geodesic with $\gamma(0, v) = p$ and $\gamma'(0, v) = v$ ; existence and uniqueness come from the geodesic equation being a second-order ODE with smooth coefficients (geodesics).

The exponential map at $p$ is $\exp_p(v) = \gamma(1, v), \qquad \exp_p(0) = p,$ defined for those $v$ for which $\gamma(\,\cdot\,, v)$ reaches $t = 1$ . Geometrically: walk from $p$ a distance $|v|_g$ along the geodesic in the direction of $v$ .
A normal neighbourhood of $p$ is a set $V = \exp_p(U)$ , where $U \subset T_pM$ is a neighbourhood of $0$ on which $\exp_p$ is a diffeomorphism.
Normal coordinates centred at $p$ are the coordinates on $V$ obtained from an orthonormal basis of $T_pM$ through $\exp_p$ : the point $\exp_p(u^ie_i)$ gets coordinates $(u^1, \dots, u^n)$ . Geodesic polar coordinates $(\rho, \theta)$ on a surface come instead from polar coordinates in $T_pS$ ; they omit one radial geodesic, the image of the half-line $\theta = 0$ . The images of circles about $0$ are the geodesic circles, the images of lines through $0$ the radial geodesics.

DefinitionAdaptedAdapted from Lee (2013), Prop. 20.2, pp. 517–518; Lee (2013), Thm 20.1, p. 516; Lee (2013), Prop. 20.5, p. 518; Lee (2013), Prop. 20.8, pp. 519–521; do Carmo (2016), Lemma 1, p. 287; do Carmo (2016), Props. 1–2, p. 289; do Carmo (2016), Prop. 3, p. 291

A. On a Lie group. Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$ .

A curve $\gamma : \mathbb{R}\to G$ is a one-parameter subgroup exactly when, for some left-invariant field $X$ , it is the maximal curve that integrates $X$ and has $\gamma(0) = e$ . For $X \in \mathfrak{g}$ the map $s \mapsto \exp sX$ is the one-parameter subgroup that $X$ generates.
$\exp$ is smooth, and for $X \in \mathfrak{g}$ , $s, t\in\mathbb{R}$ , $n \in \mathbb{Z}$ : $\exp\bigl((s+t)X\bigr) = (\exp sX)(\exp tX), \qquad (\exp X)^{-1} = \exp(-X), \qquad (\exp X)^n = \exp(nX).$
$(d\exp)_0 : T_0\mathfrak{g}\to T_eG$ is the identity, under the canonical identifications of both with $\mathfrak{g}$ ; consequently $\exp$ restricts to a diffeomorphism from a neighbourhood of $0$ in $\mathfrak{g}$ onto a neighbourhood of $e$ in $G$ .
If $\Phi : G \to H$ is a Lie group homomorphism then $\Phi(\exp X) = \exp\bigl(d\Phi_e X\bigr)$ for all $X\in\mathfrak{g}$ .
For $A$ in $\mathfrak{gl}(n,\mathbb{R})$ , the series $e^A = \sum_{k\ge0}A^k/k!$ converges to an invertible matrix, the one-parameter subgroup of $\mathrm{GL}(n,\mathbb{R})$ generated by $A$ is $t\mapsto e^{tA}$ , and the exponential map of $\mathrm{GL}(n,\mathbb{R})$ — and of every Lie subgroup of it, restricted to the corresponding subalgebra — is $\exp A = e^A$ .

B. On a Riemannian manifold. Let $(M,g)$ be Riemannian and $p \in M$ .

Homogeneity. If $\gamma(t, v)$ is defined for $t \in(-\varepsilon,\varepsilon)$ and $\lambda > 0$ , then $\gamma(t, \lambda v)$ is defined for $t \in (-\varepsilon/\lambda, \varepsilon/\lambda)$ and $\gamma(t, \lambda v) = \gamma(\lambda t, v)$ . Hence $\exp_p(tv) = \gamma(t, v)$ whenever either side is defined.
There is $\varepsilon > 0$ such that $\exp_p$ is defined and smooth on the ball $B_\varepsilon \subset T_pM$ of radius $\varepsilon$ about $0$ .
$\bigl(d\exp_p\bigr)_0 = \mathrm{Id}_{T_pM}$ , so $\exp_p$ is a diffeomorphism of some neighbourhood of $0$ onto a normal neighbourhood of $p$ .
In normal coordinates centred at $p$ on a surface, the geodesics through $p$ are the images of the straight lines $u = at$ , $v = bt$ , and the coefficients of the first fundamental form at $p$ itself are $E(p) = G(p) = 1$ , $F(p) = 0$ .
In geodesic polar coordinates $(\rho, \theta)$ on a surface, $E = 1, \qquad F = 0, \qquad \lim_{\rho\to0}\sqrt{G} = 0, \qquad \lim_{\rho\to0}\bigl(\sqrt G\bigr)_\rho = 1 .$

Idea of the proofOriginal

Both constructions answer the same question: a tangent vector at a point is an infinitesimal displacement, so what finite displacement does it name? Each answers by solving an autonomous differential equation for one unit of time — the integral-curve equation on a Lie group, the geodesic equation on a Riemannian manifold. Both equations are homogeneous under rescaling: doubling the initial vector and halving the time gives the same endpoint. That single fact is what makes $t \mapsto \exp(tX)$ a group homomorphism in the first case and $\exp_p(tv)$ a geodesic in the second, and it is what makes $\exp$ differentiable at the origin with derivative the identity, since along a ray the map is just the parametrised solution. The inverse function theorem then does the rest. In the one case where everything can be written down, $G = \mathrm{GL}(n,\mathbb{R})$ , the equation is the linear system $\dot\gamma = \gamma A$ , whose solution is the matrix exponential series.

ProofAdaptedAdapted from Lee (2013), proof of Thm 20.1, p. 516; Lee (2013), proof of Prop. 20.2, pp. 517–518; Lee (2013), proof of Prop. 20.5, p. 519; Lee (2013), proof of Prop. 20.8(b),(e),(f), pp. 520–521; do Carmo (2016), proof of Lemma 1, p. 287 and proof of Prop. 2, p. 289

A1, one-parameter subgroups. Let $X \in \mathfrak{g}$ with maximal integral curve $\gamma$ starting at $e$ ; it is defined on all of $\mathbb{R}$ , because left-invariant fields are complete. Left invariance says $X$ is $L_h$ -related to itself for every $h\in G$ , so $L_h$ carries integral curves of $X$ to integral curves of $X$ . Taking $h = \gamma(s)$ , the curve $t\mapsto \gamma(s)\gamma(t)$ integrates $X$ and begins at the point $\gamma(s)$ ; the translation lemma of flows and integral curves says $t\mapsto\gamma(s+t)$ is too, with the same starting point. By uniqueness of integral curves they coincide:

\gamma(s)\gamma(t) = \gamma(s+t),

which is exactly the homomorphism property. Conversely, a one-parameter subgroup $\gamma$ has $\gamma(0) = e$ , and the left-invariant field $X$ with $X_e = \gamma'(0)$ has $\gamma$ as integral curve, by differentiating $\gamma(t_0 + t) = \gamma(t_0)\gamma(t)$ at $t = 0$ .

A2. Rescaling: if $\gamma$ is an integral curve of $X$ then $t\mapsto\gamma(st)$ is an integral curve of $sX$ , by the chain rule. Hence the one-parameter subgroup generated by $sX$ evaluated at $t=1$ is $\gamma_X(s)$ , i.e. $\exp sX = \gamma_X(s)$ . Then $\exp((s+t)X) = \gamma_X(s+t) = \gamma_X(s)\gamma_X(t)$ by A1, and the inverse and power formulas follow from $t\mapsto\exp tX$ being a homomorphism of $\mathbb{R}$ into $G$ . Smoothness of $\exp$ in $X$ is Lee's argument that the flow of the single vector field $\Theta_{(h,X)} = (X_h, 0)$ on $G\times\mathfrak{g}$ is smooth by the fundamental theorem on flows, and $\exp X$ is a projection of that flow at $t=1$ .

A3. Let $X\in\mathfrak{g}$ and let $\sigma(t) = tX$ , a curve in the vector space $\mathfrak{g}$ with $\sigma'(0) = X$ . Then

(d\exp)_0(X) = (\exp\circ\,\sigma)'(0) = \frac{d}{dt}\Big|_{t=0}\exp(tX) = \gamma_X'(0) = X ,

using $\exp(tX) = \gamma_X(t)$ from A2. So $(d\exp)_0$ is the identity, in particular invertible, and the inverse function theorem gives the local diffeomorphism.

A5, the matrix case. With the Frobenius norm, $|A^k| \le |A|^k$ , so the series $\sum A^k/k!$ is dominated termwise by $\sum c^k/k! = e^c$ on $\{|A|\le c\}$ and converges uniformly there; the same bound applies to the termwise derivative, so term-by-term differentiation of $t\mapsto e^{tA}$ is legitimate and gives $\frac{d}{dt}e^{tA} = e^{tA}A$ . Under the standard identification of $\mathfrak{gl}(n,\mathbb{R})$ with left-invariant fields on $\mathrm{GL}(n,\mathbb{R})$ , the integral curve of $A$ starting at $I$ is the solution of $\dot\gamma = \gamma A$ , $\gamma(0) = I$ ; so $\gamma(t) = e^{tA}$ . Finally $\eta(t) = e^{tA}e^{-tA}$ has $\eta'(t) = e^{tA}Ae^{-tA} - e^{tA}Ae^{-tA} = 0$ and $\eta(0) = I$ , so $e^{tA}$ is invertible with inverse $e^{-tA}$ and the curve does take values in $\mathrm{GL}(n,\mathbb{R})$ . For a Lie subgroup $H$ , a one-parameter subgroup of $H$ is one of $G$ with initial velocity in $T_eH$ , so the exponential of $H$ is the restriction of that of $G$ . (See the matrix exponential for the linear-algebra side of this series.)

B6, homogeneity. Fix $v$ and $\lambda>0$ and set $\alpha(t) = \gamma(\lambda t, v)$ , defined for $|t| < \varepsilon/\lambda$ . Then $\alpha(0) = p$ and $\alpha'(0) = \lambda v$ by the chain rule. The geodesic equation is $D_{\alpha'}\alpha' = 0$ with $D$ the covariant derivative along the curve (Christoffel symbols and the covariant derivative), and reparametrising by a constant factor multiplies it by $\lambda^2$ :

D_{\alpha'(t)}\alpha'(t) = \lambda^2\,\bigl(D_{\gamma'}\gamma'\bigr)(\lambda t) = 0 ,

so $\alpha$ is a geodesic with initial data $(p, \lambda v)$ . By uniqueness of geodesics, $\alpha(t) = \gamma(t, \lambda v)$ . Setting $t = 1$ and renaming gives $\exp_p(tv) = \gamma(t,v)$ : a geodesic is a straight ray in $T_pM$ seen through $\exp_p$ .

B7. Smooth dependence of the geodesic on its initial data gives $\varepsilon_1,\varepsilon_2>0$ and a smooth $\gamma$ on $(-\varepsilon_2,\varepsilon_2)\times B_{\varepsilon_1}$ . The interval of definition shrinks as $|v|$ grows, but by B6 it shrinks in a controlled way: $\gamma(t, (\varepsilon_2/2)v)$ exists for $|t| < 2$ whenever $|v| < \varepsilon_1$ . So for $\varepsilon < \varepsilon_1\varepsilon_2/2$ and $w \in B_\varepsilon$ , $\gamma(1, w) = \exp_p w$ is defined, and it is smooth in $w$ because $\gamma$ is.

B8. Identify $T_0(T_pM)$ with $T_pM$ . For $v \in T_pM$ take the curve $\alpha(t) = tv$ in $T_pM$ ; then $\alpha'(0)=v$ and, by B6,

\bigl(d\exp_p\bigr)_0(v) = \frac{d}{dt}\Big|_{t=0}\exp_p(tv) = \frac{d}{dt}\Big|_{t=0}\gamma(t,v) = v .

So $(d\exp_p)_0$ is the identity and the inverse function theorem applies.

B9. In normal coordinates the curve $t\mapsto\exp_p(t(ae_1 + be_2))$ is, by B6, the geodesic with initial velocity $ae_1+be_2$ , and it has coordinate representation $u = at$ , $v = bt$ . Since $(d\exp_p)_0 = \mathrm{Id}$ and $(e_1, e_2)$ is orthonormal, the coordinate frame at $p$ is $(e_1, e_2)$ , whence $E(p) = \langle e_1,e_1\rangle = 1$ , $F(p) = \langle e_1,e_2\rangle = 0$ , $G(p) = 1$ .

B10. $\rho$ is arc length along each radial geodesic $\theta = \mathrm{const}$ , so $E = \langle x_\rho, x_\rho\rangle = 1$ . For $F$ : differentiating $F = \langle x_\rho, x_\theta\rangle$ in $\rho$ and using that the $\theta$ -curves are geodesics, so the $\rho$ -derivative of $x_\rho$ has no tangential part,

F_\rho = \langle x_{\rho\rho}, x_\theta\rangle + \langle x_\rho, x_{\theta\rho}\rangle = 0 + \tfrac12\,\partial_\theta\langle x_\rho, x_\rho\rangle = \tfrac12\,\partial_\theta(1) = 0 ,

so $F$ does not depend on $\rho$ ; letting $\rho\to0$ , where all the radial geodesics meet at $p$ , forces $F = 0$ . The two limits for $\sqrt G$ say that the geodesic circles shrink to the point $p$ and that their circumference is $2\pi\rho$ to leading order, which is the statement that in the limit the geometry is Euclidean. $\square$

Remark(Where the hypotheses are used, and what fails)AdaptedAdapted from Lee (2013), p. 521; do Carmo (2016), p. 288; Hassani (2013), Thm 29.1.21, p. 925

Completeness is what makes the Lie-group $\exp$ global. The definition $\exp X = \gamma_X(1)$ needs $\gamma_X$ to reach $t=1$ . On a Lie group this is free, because left translation moves any short integral curve anywhere, so left-invariant fields are complete. On a Riemannian manifold nothing of the kind holds, and $\exp_p$ is only defined on a star-shaped neighbourhood of $0$ : on the cone minus its vertex, $\exp_p(v)$ fails to exist for $v$ pointing at the vertex once $|v|$ reaches the distance to it.
$(d\exp)_0 = \mathrm{Id}$ gives only a local diffeomorphism. Globally $\exp$ may fail to be injective and may fail to be surjective. On the unit sphere, $\exp_p$ sends every vector of length $\pi$ to the antipode of $p$ and every vector of length $2\pi$ back to $p$ , so it is injective only on the open ball of radius $\pi$ . On a Lie group, Hassani's Theorem 29.1.21 and Lee's Proposition 20.8(f) both claim only a neighbourhood of the origin.
The group law holds along a line, not across the algebra. $\exp((s+t)X) = \exp sX\exp tX$ is a statement about one $X$ . For $X \ne Y$ one has $\exp(X+Y) \ne \exp X \exp Y$ in general; the two agree for all $X,Y$ in a connected group only when the group is abelian, and the discrepancy at second order is governed by $[X, Y]$ .
The two exponential maps are different constructions. One integrates a first-order equation determined by the group structure, the other a second-order equation determined by the metric. They share a name because both are built by running an autonomous flow for unit time out of a tangent vector, and both have differential the identity at the origin; they are not instances of one definition, and a unit should say which is meant.
The limits in B10 are limits. $F = 0$ and $E = 1$ hold on the whole chart, but $\sqrt G$ is only pinned down at $\rho = 0$ ; how it departs from $\rho$ for $\rho>0$ is where the curvature lives, and is the computation behind the Gauss-Bonnet theorem.

Method(Computing an exponential)Original

On a matrix group. To find $\exp A$ for $A$ in the Lie algebra:

Find a polynomial identity satisfied by $A$ — its characteristic or minimal polynomial.
Use it to reduce every power $A^k$ in the series $\sum A^k/k!$ to a combination of $I, A, \dots, A^{m-1}$ .
Collect the scalar series multiplying each of these; they are elementary functions.
Check the answer: $\exp A$ must lie in the group (orthogonal, unitary, unit determinant, as the case may be), and $\frac{d}{dt}e^{tA}|_{t=0} = A$ .

On a Riemannian manifold. To find $\exp_p(v)$ :

Solve the geodesic equation with $\gamma(0) = p$ , $\gamma'(0) = v/|v|$ , parametrised by arc length $s$ .
Evaluate at $s = |v|$ ; by homogeneity this is $\gamma(1, v) = \exp_p(v)$ .
For normal coordinates, repeat with $v = u^ie_i$ for an orthonormal basis $(e_i)$ and read the result as a parametrisation; for geodesic polar coordinates use $v = \rho(\cos\theta\,e_1 + \sin\theta\,e_2)$ .
Check against B9 and B10: the metric must be the identity at $p$ , and in polar form $E = 1$ , $F = 0$ , $\sqrt G \sim \rho$ .

Example(A rotation from its generator)Original

Let $n = \tfrac13(2, 1, 2)$ , a unit vector, and let $K$ be the matrix with $Kw = n\times w$ :

K = \frac13\begin{pmatrix} 0 & -2 & 1\\ 2 & 0 & -2\\ -1 & 2 & 0\end{pmatrix}, \qquad K^\mathsf{T} = -K ,

so $K \in \mathfrak{so}(3)$ . Because $n$ is a unit vector, $K^2 = nn^\mathsf{T} - I$ and hence $K^3 = -K$ : that is the polynomial identity of step 1. Splitting the series by parity,

e^{\theta K} = I + \Bigl(\theta - \frac{\theta^3}{3!} + \cdots\Bigr)K + \Bigl(\frac{\theta^2}{2!} - \frac{\theta^4}{4!}+\cdots\Bigr)K^2 = I + \sin\theta\,K + (1-\cos\theta)\,K^2 .

Take $\theta = \pi/2$ , so $\sin\theta = 1$ and $1 - \cos\theta = 1$ . With

K^2 = nn^\mathsf{T} - I = \frac19\begin{pmatrix}-5 & 2 & 4\\ 2 & -8 & 2\\ 4 & 2 & -5\end{pmatrix},

we get

R = e^{(\pi/2)K} = \frac19\begin{pmatrix} 4 & -4 & 7\\ 8 & 1 & -4\\ 1 & 8 & 4 \end{pmatrix}.

Checks. $R^\mathsf{T}R = I$ and $\det R = 1$ , so $R \in \mathrm{SO}(3)$ , as statement A5 requires for the exponential of a skew-symmetric matrix. Its trace is $(4+1+4)/9 = 1$ , matching $1 + 2\cos\theta = 1$ for $\theta = \pi/2$ . The axis is fixed: $Rn = n$ , since $Kn = n\times n = 0$ . And $Re_1 = \tfrac19(4, 8, 1)$ , a unit vector making angle $\arccos(4/9) = 63.6122^\circ$ with $e_1$ — the correct angle for a $90^\circ$ rotation about an axis that is not perpendicular to $e_1$ .

Example(Normal and geodesic polar coordinates on the unit sphere)Original

Let $S^2 \subset\mathbb{R}^3$ be the unit sphere with the induced metric and $p = (0,0,1)$ .

The geodesics. For a unit vector $v \in T_pS^2$ put $\gamma(s) = \cos s\,p + \sin s\,v$ . This curve lies on $S^2$ (since $p \perp v$ and both are unit vectors) and has $\gamma'' = -\gamma$ , which is normal to the sphere at $\gamma(s)$ ; so its tangential acceleration vanishes and it is a geodesic, with $\gamma(0)=p$ , $\gamma'(0)=v$ and $|\gamma'| = 1$ . By homogeneity (statement B6), for general $v \ne 0$

\exp_p(v) = \cos|v|\,p + \sin|v|\,\frac{v}{|v|}.

A number. With $v = \tfrac{\pi}{6}(1,0,0)$ , $|v| = \pi/6$ and

\exp_p(v) = \Bigl(\sin\tfrac{\pi}{6},\, 0,\, \cos\tfrac{\pi}{6}\Bigr) = \Bigl(\tfrac12,\, 0,\, \tfrac{\sqrt3}{2}\Bigr) = (0.5,\, 0,\, 0.8660254\ldots),

a point at colatitude $30^\circ$ : the walk of length $\pi/6$ along a meridian, as the definition says.

Geodesic polar coordinates. Taking $v = \rho(\cos\theta, \sin\theta, 0)$ ,

x(\rho,\theta) = \bigl(\sin\rho\cos\theta,\ \sin\rho\sin\theta,\ \cos\rho\bigr),

so $x_\rho = (\cos\rho\cos\theta, \cos\rho\sin\theta, -\sin\rho)$ and $x_\theta = (-\sin\rho\sin\theta, \sin\rho\cos\theta, 0)$ . Hence

E = 1, \qquad F = 0, \qquad G = \sin^2\rho, \qquad ds^2 = d\rho^2 + \sin^2\!\rho\,d\theta^2 .

This matches statement B10 exactly: $\sqrt G = \sin\rho \to 0$ and $(\sqrt G)_\rho = \cos\rho \to 1$ as $\rho\to0$ . In normal coordinates $(u, w) = \rho(\cos\theta,\sin\theta)$ the metric at $p$ itself is $du^2 + dw^2$ , as statement B9 says.

A second number. The area of the geodesic disc of radius $\rho_0$ is

\int_0^{2\pi}\!\!\int_0^{\rho_0}\sqrt{EG - F^2}\;d\rho\,d\theta = 2\pi\bigl(1 - \cos\rho_0\bigr).

For $\rho_0 = \pi/3$ this is $2\pi(1 - \tfrac12) = \pi = 3.14159\ldots$ , whereas the Euclidean value $\pi\rho_0^2 = \pi^3/9 = 3.44514\ldots$ is larger. The deficit is the first visible effect of positive curvature, and $\exp_p$ is what makes the comparison meaningful: both numbers are areas of the same coordinate disc in $T_pS^2$ .

Where it stops being a diffeomorphism. $\exp_p$ is injective exactly on the open disc $|v| < \pi$ ; every vector of length $\pi$ is sent to the south pole, and $\sqrt G = \sin\rho$ vanishes there.

ApplicationOriginal

In quantum mechanics every continuous symmetry acts through an exponential of its generator: the time-evolution operator is $U(t) = e^{-iHt/\hbar}$ , a one-parameter subgroup of the unitary group generated by the Hamiltonian, and a rotation by $\theta$ about $n$ acts as $e^{-i\theta\,n\cdot J/\hbar}$ , the same construction in the rotation group and SU(2). Statement A4 is why a representation of a Lie group is determined near the identity by a representation of its algebra, the working principle behind angular momentum algebra. The $3\times3$ example above is Rodrigues's rotation formula, used in robotics and in attitude control to turn an angular-velocity vector into a rotation matrix. On the Riemannian side, normal coordinates at $p$ are the local inertial frame of general relativity: the metric is $\eta_{\mu\nu}$ at $p$ and the Christoffel symbols vanish there, so a freely falling observer sees special relativity to first order, and the leading correction is the curvature that cannot be transformed away (the Riemann curvature tensor). Geodesic polar coordinates are the natural frame for the local Gauss-Bonnet theorem and for the Bertrand-Puiseux formula, which recovers the Gaussian curvature of a surface from the circumference of a small geodesic circle.

Theorem · advanced

Partitions of unity

Adapted

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Every indexed open cover of a smooth manifold admits a subordinate smooth partition of unity, the standard device for assembling a global metric, integral or field out of pieces defined chart by chart.

Prerequisites: Smooth manifolds, Compactness, Topological and metric spaces

Definition(Support, local finiteness, subordinate partition of unity)AdaptedAdapted from Lee (2013), Ch. 1, p. 9, and Ch. 2, p. 43

Let $M$ be a smooth $n$ -manifold (Hausdorff, second countable and locally Euclidean, with a smooth atlas; see Smooth manifolds).

The support of a function $u : M \to \mathbb{R}$ is $\operatorname{supp} u = \overline{\{p \in M : u(p) \neq 0\}}$ . We say $u$ is supported in a set $U$ if $\operatorname{supp} u \subset U$ , and compactly supported if $\operatorname{supp} u$ is compact.
A family $(A_\alpha)_{\alpha \in A}$ of subsets of $M$ is locally finite if every point of $M$ has a neighbourhood meeting $A_\alpha$ for only finitely many $\alpha$ .
A cover $\mathcal{V}$ refines a cover $\mathcal{U}$ if every member of $\mathcal{V}$ is contained in some member of $\mathcal{U}$ . The space is paracompact if every open cover has a locally finite open refinement.
Let $\mathcal{X} = (X_\alpha)_{\alpha \in A}$ be an open cover of $M$ indexed by a set $A$ . A smooth partition of unity subordinate to $\mathcal{X}$ is a family $(\psi_\alpha)_{\alpha \in A}$ of smooth functions $\psi_\alpha : M \to \mathbb{R}$ , indexed by the same set $A$ , which partitions the constant function $1$ ,
$\sum_{\alpha \in A}\psi_\alpha \equiv 1 ,$
and does so subject to three side conditions: the family of supports $(\operatorname{supp}\psi_\alpha)_{\alpha\in A}$ is locally finite; each support sits inside the matching cover element, $\operatorname{supp}\psi_\alpha \subset X_\alpha$ ; and the shares are fractions, $0 \le \psi_\alpha \le 1$ . The first of these makes the displayed sum a finite sum near each point, so no convergence question arises; in the third, the upper bound is a consequence of the lower one and the displayed sum, since each $\psi_\alpha \le \sum_\beta \psi_\beta = 1$ once all terms are non-negative, but it is convenient to record both.
For $A \subset M$ closed and $U \supset A$ open, a bump function for $A$ supported in $U$ is a smooth $\chi : M \to \mathbb{R}$ with $0 \le \chi \le 1$ , $\chi \equiv 1$ on $A$ and $\operatorname{supp}\chi \subset U$ .

Theorem(Existence of smooth partitions of unity)AdaptedAdapted from Lee (2013), Ch. 2, Thm 2.23, p. 43, Prop. 2.25, p. 44, and Lemma 2.26, pp. 44-45

Fix a smooth manifold $M$ , an index set $A$ , and a family $\mathcal{X} = (X_\alpha)_{\alpha\in A}$ of open sets whose union is all of $M$ . No restriction whatever is placed on the cover: $A$ may be uncountable, the $X_\alpha$ may be unbounded, and they may repeat. Then:

Some family $(\psi_\alpha)_{\alpha\in A}$ , indexed by that same $A$ , is a smooth partition of unity subordinate to $\mathcal{X}$ .
(Bump functions.) For every closed $S \subset M$ and every open $U \supset S$ there is a smooth bump function for $S$ supported in $U$ .
(Extension lemma.) If $S \subset M$ is closed, $U \supset S$ is open, and $f : S \to \mathbb{R}^k$ is smooth on $S$ in the sense that each point of $S$ has a neighbourhood in $M$ on which some smooth map agrees with $f$ , then there is a smooth $\tilde f : M \to \mathbb{R}^k$ with $\tilde f|_S = f$ and $\operatorname{supp}\tilde f \subset U$ .

Lee states all three for smooth manifolds with or without boundary; we use only the boundaryless case.

Idea of the proofOriginal

Three ingredients, each cheap on its own.

First, smoothness allows functions that are positive on a ball and identically zero outside a slightly larger one: $e^{-1/t}$ is flat to infinite order at $t = 0$ , so it can be switched on without leaving a trace in any derivative.

Second, a manifold is second countable and locally compact, and those two facts let any open cover be thinned to a countable, locally finite family of coordinate balls that still refines it. Local finiteness is what lets us add infinitely many bumps.

Third, having put one bump $f_i$ on each ball, the sum $f = \sum_i f_i$ is smooth and strictly positive, so dividing by it normalises the family to sum to $1$ . The only remaining work is bookkeeping: the bumps are indexed by balls, the partition must be indexed by the cover, so the bumps belonging to the same cover element are added together.

ProofAdaptedAdapted from Lee (2013), Ch. 2, Lemmas 2.20-2.22 and Thm 2.23, pp. 41-44

Step 1: a flat-topped bump on $\mathbb{R}^n$ . Put $f(t) = e^{-1/t}$ for $t > 0$ and $f(t) = 0$ for $t \le 0$ .

$f$ is smooth. On $t \neq 0$ this is a composition of smooth functions. For $t > 0$ , induction gives $f^{(k)}(t) = P_k(1/t)\,e^{-1/t}$ with $P_k$ a polynomial: differentiating $P(1/t)e^{-1/t}$ produces $\bigl(P(1/t) - P'(1/t)\bigr)t^{-2}e^{-1/t}$ , which has the same shape with $P_{k+1}(u) = u^2\bigl(P_k(u) - P_k'(u)\bigr)$ . Now argue by induction that $f^{(k)}(0) = 0$ . This holds for $k = 0$ . If it holds for $k$ , the difference quotient of $f^{(k)}$ at $0$ vanishes from the left, and from the right it is $f^{(k)}(t)/t = u\,P_k(u)e^{-u}$ with $u = 1/t \to +\infty$ , which tends to $0$ because $e^{-u}$ beats every power of $u$ . So $f^{(k+1)}(0)$ exists and is $0$ . Each $f^{(k)}$ is therefore differentiable everywhere, hence continuous, and $f \in C^\infty(\mathbb{R})$ .
A cutoff. Fix $0 < r_1 < r_2$ and set $h(t) = f(r_2 - t)\big/\bigl(f(r_2-t) + f(t - r_1)\bigr)$ . At least one of $r_2 - t$ , $t - r_1$ is positive for every $t$ , so the denominator never vanishes and $h$ is smooth; $h \equiv 1$ for $t \le r_1$ , $h \equiv 0$ for $t \ge r_2$ , and $0 < h < 1$ in between.
A bump. Set $H(x) = h(|x|)$ on $\mathbb{R}^n$ . Away from the origin this is a composition of smooth maps, and near the origin $H \equiv 1$ ; so $H$ is smooth, equals $1$ on $\overline{B_{r_1}(0)}$ and vanishes off $B_{r_2}(0)$ .

Step 2: a good refinement of the cover. Each $X_\alpha$ is itself a smooth manifold, so by Lee's Prop. 1.19 it has a countable basis of regular coordinate balls: sets $B$ for which there are a chart $\varphi : B' \to \mathbb{R}^n$ with $\overline{B} \subset B'$ and radii $r < r'$ such that $\varphi(B) = B_r(0)$ , $\varphi(\overline B) = \overline{B_r(0)}$ and $\varphi(B') = B_{r'}(0)$ . The union over $\alpha$ of these bases is a basis $\mathcal{B}$ for $M$ : given $p \in M$ and an open $W \ni p$ , choose $\alpha$ with $p \in X_\alpha$ , note that $W \cap X_\alpha$ is open in the subspace $X_\alpha$ , and take a regular coordinate ball of $X_\alpha$ with $p \in B \subset W \cap X_\alpha$ ; such a $B$ is open in $X_\alpha$ , hence open in $M$ because $X_\alpha$ is. By Lee's Thm 1.15 (manifolds are paracompact, and the refinement may be taken from any prescribed basis) there is a countable, locally finite refinement $(B_i)_{i\in\mathbb{N}}$ of $\mathcal{X}$ with every $B_i \in \mathcal{B}$ ; the closures $\overline{B_i}$ are then locally finite as well. Both of these facts use second countability, through the exhaustion of $M$ by compact sets that proves Thm 1.15 (see Compactness).

Step 3: one bump per ball. Fix $i$ , with chart $\varphi_i : B_i' \to \mathbb{R}^n$ and radii $r_i < r_i'$ as above, and let $H_i$ be the bump of Step 1 built from the pair $(r_i/2,\, r_i)$ , so that $H_i > 0$ exactly on $B_{r_i}(0)$ . Define

f_i = \begin{cases} H_i \circ \varphi_i & \text{on } B_i',\\ 0 & \text{on } M \setminus \overline{B_i}.\end{cases}

The two formulas agree (both give $0$ ) on the overlap $B_i' \setminus \overline{B_i}$ , and those two open sets cover $M$ , so $f_i$ is a well-defined smooth function. Because $\varphi_i(B_i) = B_{r_i}(0)$ , it is strictly positive on $B_i$ and zero off it, so $\operatorname{supp} f_i = \overline{B_i} \subset B_i'$ .

Step 4: normalise. Near any point only finitely many $f_i$ are not identically zero, by local finiteness of $(\overline{B_i})$ , so $f = \sum_i f_i$ is smooth. Every $f_i$ is non-negative, and every $p \in M$ lies in some $B_i$ , where $f_i(p) > 0$ ; hence $f > 0$ everywhere. Therefore $g_i = f_i/f$ is smooth, with $0 \le g_i \le 1$ , $\operatorname{supp} g_i = \overline{B_i}$ and $\sum_i g_i \equiv 1$ .

Step 5: reindex. Since $(B_i)$ refines $\mathcal{X}$ , choose for each $i$ an index $a(i) \in A$ with $B_i' \subset X_{a(i)}$ , which is possible because $B_i$ was drawn from the basis of regular coordinate balls of one of the $X_\alpha$ , and set $\psi_\alpha = \sum_{i : a(i) = \alpha} g_i$ (the empty sum being $0$ ). Each $\psi_\alpha$ is smooth, because near any point the sum is finite. For a locally finite family of sets the closure of the union is the union of the closures, so $\operatorname{supp}\psi_\alpha \subset \bigcup_{a(i)=\alpha}\overline{B_i} \subset X_\alpha$ , and the family of these supports is locally finite because $(\overline{B_i})$ is. Finally $\sum_\alpha \psi_\alpha = \sum_i g_i \equiv 1$ . This proves part 1.

Part 2. Apply part 1 to the two-element cover $U_0 = U$ , $U_1 = M\setminus S$ , getting $\psi_0, \psi_1$ . On $S$ we have $\psi_1 = 0$ , hence $\psi_0 = 1$ ; and $\operatorname{supp}\psi_0 \subset U$ . So $\chi = \psi_0$ is the required bump function.

Part 3. For each $p \in S$ choose an open $W_p \subset U$ containing $p$ and a smooth $f_p : W_p \to \mathbb{R}^k$ agreeing with $f$ on $W_p \cap S$ . The sets $\{W_p\}_{p\in S}$ together with $M \setminus S$ form an open cover; take a subordinate smooth partition of unity $\{\psi_p\}_{p\in S} \cup \{\psi_0\}$ . Each product $\psi_p f_p$ , extended by $0$ off $\operatorname{supp}\psi_p$ , is smooth on $M$ by the same two-open-sets argument as in Step 3. Put $\tilde f = \sum_{p} \psi_p f_p$ , a locally finite sum, hence smooth. For $x \in S$ we have $\psi_0(x) = 0$ and $f_p(x) = f(x)$ whenever $\psi_p(x) \neq 0$ , so $\tilde f(x) = \bigl(\psi_0(x) + \sum_p \psi_p(x)\bigr)f(x) = f(x)$ . And $\operatorname{supp}\tilde f \subset \bigcup_p \operatorname{supp}\psi_p \subset U$ , again because a locally finite union of closed sets is closed. $\square$

Remark(Where the hypotheses are used)Original

Second countability is used exactly once, in Step 2, and it is indispensable. If a space admits a smooth partition of unity subordinate to every open cover then it is paracompact, because the open sets $\{\psi_\alpha \neq 0\}$ form a locally finite open refinement of the given cover. So any Hausdorff locally Euclidean space that is not paracompact — the long line is the standard example — carries no such partitions, and the theorem genuinely needs the countability clause in the definition of a manifold.
Smoothness, not analyticity. Step 1 is where $C^\infty$ is strictly weaker than $C^\omega$ . A real-analytic function that vanishes on a non-empty open subset of a connected real-analytic manifold vanishes identically, so a real-analytic partition of unity subordinate to a cover by two proper open sets of a connected manifold cannot exist. Every gluing argument below is therefore unavailable in the analytic category, and in the holomorphic category, where it is replaced by sheaf cohomology.
Local finiteness is what makes $\sum_\alpha \psi_\alpha$ and every sum built from it a finite sum near each point; without it, "smooth" would require an argument about differentiating a series term by term.
Indexing by the same set $A$ costs Step 5 but is worth it: the sum $\sum_\alpha \psi_\alpha A_\alpha$ over a cover with chosen local objects $A_\alpha$ then has one term per cover element.
Supports need not be compact. They are compact if the cover is by precompact sets; the proof gives compactly supported $g_i$ in any case, and the recombination in Step 5 is what can destroy compactness.

Remark(Two global objects built this way)AdaptedAdapted from Lee (2013), Ch. 13, Prop. 13.3, p. 329, and Ch. 16, eq. (16.2) and Prop. 16.5, pp. 405-406

Every smooth manifold carries a Riemannian metric. Cover $M$ by charts $(U_\alpha, \varphi_\alpha)$ and let $g_\alpha$ be the pullback $\varphi_\alpha^*\bar g$ of the Euclidean metric, a Riemannian metric on $U_\alpha$ . With $(\psi_\alpha)$ subordinate to $(U_\alpha)$ , set $g = \sum_\alpha \psi_\alpha g_\alpha$ , each term read as $0$ outside $\operatorname{supp}\psi_\alpha$ . The sum is a smooth symmetric $2$ -tensor field by local finiteness, and for $v \neq 0$ in $T_pM$ every term $\psi_\alpha(p)\,g_\alpha|_p(v,v)$ is $\ge 0$ while at least one $\psi_\alpha(p)$ is positive, so $g_p(v,v) > 0$ . See Riemannian manifolds. The argument works because positive definiteness is preserved by convex combinations; there is nothing canonical about the resulting $g$ .

The integral of a form is well defined. For a compactly supported $n$ -form $\omega$ on an oriented $n$ -manifold, cover $\operatorname{supp}\omega$ by finitely many oriented chart domains $U_i$ , take a subordinate partition of unity $(\psi_i)$ , and define $\int_M\omega = \sum_i \int_M \psi_i\omega$ , each term being an ordinary integral in one chart. Given a second such cover and partition $(\tilde\psi_j)$ , inserting $\sum_j\tilde\psi_j = 1$ into each term and $\sum_i\psi_i = 1$ into each term of the other sum exhibits both answers as $\sum_{i,j}\int_M \tilde\psi_j\psi_i\omega$ , so they agree. See Integration of forms and the general Stokes theorem.

Method(Gluing local constructions into a global one)Original

To build a global object on $M$ from objects defined chart by chart:

Choose an open cover $(U_\alpha)$ of $M$ on which the local objects $A_\alpha$ exist — usually a cover by chart domains, so that $A_\alpha$ is whatever the construction gives in $\mathbb{R}^n$ .
Take a smooth partition of unity $(\psi_\alpha)$ subordinate to $(U_\alpha)$ .
Form $A = \sum_\alpha \psi_\alpha A_\alpha$ , reading each term as zero outside $\operatorname{supp}\psi_\alpha$ . Local finiteness makes $A$ smooth.
Check that the defining property of $A$ survives the averaging. This is the only step with content, and it succeeds exactly when the admissible objects at each point form a convex set: Riemannian metrics (positive definite forms), connections (an affine space), volume densities, subordinate cutoffs. It fails for properties that are not convex — being nowhere zero, satisfying a nonlinear equation, being a complex structure — and for those one needs agreement on overlaps instead, or a genuinely different argument.
If the local objects do agree on overlaps, skip the partition of unity: the pieces then define $A$ directly, and the result is canonical rather than arbitrary.

Example(An explicit partition of unity on the circle, and a length computed with it)Original

Let $M = S^1$ , with the angle $\theta$ taken modulo $2\pi$ , covered by the two chart domains

U_0 = \{\theta : -\pi < \theta < \pi\}, \qquad U_1 = \{\theta : 0 < \theta < 2\pi\} .

Let $f$ and $h(\,\cdot\,; r_1, r_2)$ be as in Step 1 of the proof, and set

a(\theta) = h\bigl(1 - \cos\theta;\ 1,\ \tfrac32\bigr), \qquad b(\theta) = h\bigl(1 + \cos\theta;\ \tfrac12,\ \tfrac32\bigr).

Both are smooth functions on the circle, because $\cos\theta$ is.

Supports. $a$ vanishes where $1 - \cos\theta \ge \tfrac32$ , i.e. where $\cos\theta \le -\tfrac12$ , so $\operatorname{supp} a \subset \{|\theta| \le 2\pi/3\} \subset U_0$ . Likewise $b$ vanishes where $\cos\theta \ge \tfrac12$ , so $\operatorname{supp} b \subset \{\pi/3 \le |\theta| \le \pi\} \subset U_1$ .

Positivity of the sum. If $\cos\theta > -\tfrac12$ then $1 - \cos\theta < \tfrac32$ and $a(\theta) > 0$ ; if $\cos\theta \le -\tfrac12$ then $1 + \cos\theta \le \tfrac12$ and $b(\theta) = 1$ . So $a + b > 0$ everywhere, and

\psi_0 = \frac{a}{a+b}, \qquad \psi_1 = \frac{b}{a+b}

is a smooth partition of unity subordinate to $(U_0, U_1)$ .

Two exact values. At $\theta = \pi/2$ : $1 - \cos\theta = 1 = r_1$ for $a$ , so $a = 1$ ; and $1 + \cos\theta = 1$ , which is the midpoint of $[\tfrac12, \tfrac32]$ , so $b = f(\tfrac12)/\bigl(f(\tfrac12)+f(\tfrac12)\bigr) = \tfrac12$ . Hence

\psi_0(\pi/2) = \frac{1}{1 + 1/2} = \frac23, \qquad \psi_1(\pi/2) = \frac13 .

At $\theta = 0$ , $(\psi_0,\psi_1) = (1,0)$ ; at $\theta = \pi$ , $(0,1)$ .

A checkable number. The length of $S^1$ , computed the way the integral of a form is defined, is the sum of two chart integrals:

\int_{-\pi}^{\pi}\psi_0(\theta)\,d\theta \;=\; 3.348757\ldots, \qquad \int_{0}^{2\pi}\psi_1(\theta)\,d\theta \;=\; 2.934428\ldots

(evaluated numerically; neither has a closed form, and the two differ because the cutoff radii chosen for $a$ and $b$ differ). Their sum is

\int_{-\pi}^{\pi}\bigl(\psi_0(\theta) + \psi_1(\theta)\bigr)\,d\theta = \int_{-\pi}^{\pi} 1\,d\theta = 2\pi = 6.283185\ldots,

which is the circumference, as it must be: the individual pieces depend on the partition, the total does not. This is exactly the cancellation behind the well-definedness argument in the remark above.

ApplicationOriginal

Partitions of unity are the reason "define it in coordinates, then glue" is a legitimate move in mathematical physics.

In general relativity the metric is given chart by chart; the existence argument above says that the obstruction to a global Lorentzian metric — which does not exist on every manifold, since the local models are not a convex family — is of a different kind from the Riemannian case, where there is none.

In PDE theory a bounded domain is covered by balls and boundary half-balls, the problem is solved or estimated in each, and a subordinate partition of unity reassembles the pieces; this is how local approximation of Sobolev functions by smooth ones is proved (Weak solutions and Sobolev spaces), and the same cutoffs give the interior estimates used throughout elliptic theory.

The finite element method (The finite element method) is a partition of unity made computational: the nodal hat functions sum to one on the mesh, and the global approximation is the sum of the local ones weighted by them. In quantum chemistry the same device appears as the atom-centred weight functions of Becke's numerical integration scheme, which break a molecular integral into single-atom pieces.

Folland proves a continuous version with no manifold in sight: on a locally compact Hausdorff space, a compact set covered by finitely many open sets carries a partition of unity on it by compactly supported continuous functions, built from Urysohn's lemma instead of from $e^{-1/t}$ . That is the form used in measure theory, for instance in the Riesz representation theorem.

Definition · advanced

Symplectic manifolds and Hamiltonian vector fields

Adapted

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Equips an even-dimensional manifold with a closed nondegenerate 2-form, shows that a cotangent bundle carries one canonically and that every such form looks standard in suitable coordinates, and turns each smooth function into a vector field whose flow preserves both the form and the phase-space volume.

Prerequisites: Differential forms, The canonical equations, The exterior derivative, Flows and integral curves, Tangent and vector bundles, Closed and exact forms, Smooth manifolds, Tangent vectors and vector fields

Definition(Symplectic forms, Darboux coordinates and the Liouville volume)AdaptedAdapted from Lee (2013), p. 565, p. 568, p. 571; Hassani (2013), Def. 26.4.1 and Box 26.4.2, p. 801; Hassani (2013), Def. 28.7.1 and Def. 28.7.3, p. 902

Fix $V$ , a real vector space of finite dimension, and let $\omega$ be a $2$ -covector on it (differential forms). Attached to $\omega$ is the linear map

\omega^{\flat} : V \to V^{*}, \qquad \omega^{\flat}(v) = \omega(v, \cdot\,) .

Call $\omega$ nondegenerate when $\omega^{\flat}$ is invertible; equivalently, when the only $v$ with $\omega(v, w) = 0$ for every $w$ is $v = 0$ ; equivalently again, when the matrix $\omega_{ij} = \omega(E_i, E_j)$ in one — hence in every — basis is nonsingular. A nondegenerate $2$ -covector is a symplectic tensor, and $(V,\omega)$ is then a symplectic vector space. The inverse of $\omega^\flat$ is written $\omega^{\sharp} : V^{*} \to V$ .

Now let $M$ be a smooth manifold.

A $2$ -form $\omega$ on $M$ is nondegenerate if $\omega_p$ is a symplectic tensor on $T_pM$ for every $p$ .
A symplectic form on $M$ is a $2$ -form that is both closed ( $d\omega = 0$ , see the exterior derivative) and nondegenerate. The pair $(M,\omega)$ is a symplectic manifold; the form is also called a symplectic structure.
A diffeomorphism $F : (M_1,\omega_1) \to (M_2,\omega_2)$ with $F^{*}\omega_2 = \omega_1$ is a symplectomorphism. In mechanics the same map, from a symplectic manifold to itself, is called a canonical transformation.
Coordinates $(x^1,\dots,x^n,y^1,\dots,y^n)$ on an open set of a $2n$ -dimensional $M$ in which $\omega = \sum_{i=1}^{n} dx^i \wedge dy^i$ are Darboux coordinates, also called symplectic or canonical coordinates. On $\mathbb{R}^{2n}$ this formula in the standard coordinates defines the standard symplectic form.
When $\dim M = 2n$ , the Liouville volume form of $(M,\omega)$ is $\omega^n/n!$ , where $\omega^n = \omega\wedge\cdots\wedge\omega$ has $n$ factors. Part 1 of the statement says it is a volume form, so a symplectic manifold has a canonical notion of phase volume $\int_A \omega^n/n!$ , with no metric anywhere in sight.

Two pieces of notation recur. The interior product of a vector field $X$ with a $k$ -form $\alpha$ is the $(k-1)$ -form

(\iota_X\alpha)(X_1,\dots,X_{k-1}) = \alpha(X, X_1,\dots,X_{k-1}),

also written $X \lrcorner\, \alpha$ ; on functions $\iota_X f = 0$ . And $\mathcal{L}_X$ is the Lie derivative along $X$ (flows and integral curves), extended to tensor fields by $(\mathcal{L}_XA)_p = \frac{d}{dt}\big|_{t=0}\bigl(\theta_t^{*}A\bigr)_p$ for the flow $\theta$ of $X$ .

Definition(Hamiltonian vector fields and the Poisson bracket)AdaptedAdapted from Lee (2013), pp. 574–576, 578; Hassani (2013), Def. 28.7.8 and Prop. 28.7.9, p. 905; Hassani (2013), Def. 28.7.13, p. 908

Let $(M,\omega)$ be a symplectic manifold. Nondegeneracy makes $\omega^{\flat}$ a bundle isomorphism $TM \to T^{*}M$ , so a $1$ -form can be converted into a vector field. Applying this to the differential of a function:

The Hamiltonian vector field of $f \in C^{\infty}(M)$ is the unique smooth vector field $X_f$ with $\iota_{X_f}\omega = df, \qquad\text{equivalently}\qquad \omega(X_f, Y) = Yf \ \text{ for every vector field } Y .$ Hassani writes the same field as $X_f = \omega^{\sharp}(df)$ . It is the symplectic counterpart of the gradient, but, as part 4 below shows, it behaves quite unlike one.
A Hamiltonian system is a symplectic manifold together with a chosen $H \in C^{\infty}(M)$ , the Hamiltonian; the integral curves of $X_H$ are the trajectories of the system, and the flow of $X_H$ is its Hamiltonian flow.
A function $f$ constant on every trajectory is a conserved quantity (a first integral) of the system.
The Poisson bracket of $f, g \in C^{\infty}(M)$ is $\{f, g\} = \omega(X_f, X_g) = df(X_g) = X_g f \in C^{\infty}(M).$ Two functions Poisson commute when $\{f,g\} = 0$ . The last expression is the one to remember: $\{f,g\}$ measures how fast $f$ changes along the Hamiltonian flow of $g$ .
A vector field $X$ on $M$ is symplectic if $\omega$ is invariant under its flow; Hamiltonian if $X = X_f$ for some $f \in C^{\infty}(M)$ ; locally Hamiltonian if every point has a neighbourhood on which $X$ is Hamiltonian.

Note(Signs, orderings and which book does what)AdaptedAdapted from Lee (2013), Prop. 22.11, pp. 569–570 and Prop. 22.8, p. 567; Hassani (2013), Thm 28.7.2, p. 902 and Prop. 28.7.14, p. 908; Gelfand & Fomin (1963), §17, p. 71

Everything below uses $\iota_{X_f}\omega = df$ and $\omega = \sum_i dx^i\wedge dy^i$ in Darboux coordinates. With that pair of choices:

The Poisson bracket comes out as $\{f,g\} = \sum_i(\partial_{x^i}f\,\partial_{y^i}g - \partial_{y^i}f\,\partial_{x^i}g)$ , which is Gelfand and Fomin's (Gelfand & Fomin (1963), §17, p. 71) classical bracket $[\Phi, H]$ with $(y_i, p_i)$ in place of $(x^i, y^i)$ , and Hassani's (Hassani (2013), Prop. 28.7.14, p. 908) bracket in canonical coordinates $(q^i, p_i)$ . The three agree, so the handbook's canonical equations and this unit use one and the same sign.
On a cotangent bundle the canonical form is $\omega = -d\tau$ , not $+d\tau$ , where $\tau$ is the tautological $1$ -form; Lee (Lee (2013), Prop. 22.11, p. 569) is explicit about the minus, and it is exactly what makes $\omega = \sum_i dq^i\wedge dp_i$ rather than its negative.
The ordering inside the Liouville form is a genuine fork. Lee's computation of $\omega^n$ gives $\omega^n/n! = dx^1\wedge dy^1\wedge\cdots\wedge dx^n\wedge dy^n$ , alternating the two families. Hassani (Hassani (2013), Thm 28.7.2, p. 902) instead writes the volume element of a symplectic chart as $\mu_\omega = dx^1\wedge\cdots\wedge dx^n\wedge dy^1\wedge\cdots\wedge dy^n$ , all the $x$ 's first. Sorting one into the other takes $0 + 1 + \cdots + (n-1)$ transpositions, so $\frac{\omega^n}{n!} = (-1)^{n(n-1)/2}\,\mu_\omega .$ For $n = 1$ the two agree; for $n = 2$ they differ by a sign. Nothing in this unit depends on the choice, because every statement here is about a form being preserved, but an orientation or an integral computed from one convention and compared with the other will be out by that sign.
Hassani's Darboux theorem (Thm 28.7.2) is stated for a $2$ -form on a $2n$ -manifold as the equivalence " $d\omega = 0$ if and only if canonical charts exist". Read literally that would make every closed $2$ -form, the zero form included, locally standard, which is false; the nondegeneracy of $\omega$ is carried silently from the surrounding section on symplectic forms. The statement below follows Lee, who asks for a symplectic form outright.

Definition(What a symplectic structure provides)AdaptedAdapted from Lee (2013), Props. 22.7–22.8, pp. 566–567; Lee (2013), p. 568; Lee (2013), Prop. 22.11, pp. 569–570; Lee (2013), Thm 22.13, p. 571; Lee (2013), (22.9) and Props. 22.16–22.17, pp. 574–576; Lee (2013), (22.15)–(22.16), Prop. 22.19 and Cor. 22.20, pp. 578–579; Lee (2013), Prop. 22.21, p. 580; Hassani (2013), Prop. 28.7.9, p. 905; Hassani (2013), Thm 28.7.10, p. 906; Hassani (2013), Prop. 28.7.11 and Cor. 28.7.12, pp. 907–908

Let $(M,\omega)$ be a symplectic manifold and $f, g, h, H \in C^{\infty}(M)$ .

Even dimension, orientability, volume. $M$ has even dimension $2n$ ; the $2n$ -form $\omega^n$ vanishes nowhere; consequently $M$ is orientable and $\omega^n/n!$ is a volume form on it.
Darboux's theorem. Every point of $M$ lies in the domain of Darboux coordinates centred at it. There is thus no local invariant of a symplectic structure: all symplectic manifolds of the same dimension look alike near a point.
Cotangent bundles. Let $Q$ be a smooth manifold and $\pi : T^{*}Q \to Q$ the projection of the cotangent bundle. The tautological $1$ -form $\tau$ on the total space $T^{*}Q$ , defined at the point $(q,\varphi)$ by $\tau_{(q,\varphi)} = d\pi^{*}_{(q,\varphi)}\varphi$ — that is, $\tau_{(q,\varphi)}(v) = \varphi\bigl(d\pi_{(q,\varphi)}(v)\bigr)$ — is a smooth $1$ -form, and the $2$ -form
$\omega = -\,d\tau$
is symplectic; it is the canonical symplectic form of the cotangent bundle. In the natural coordinates $(q^i, p_i)$ induced by coordinates $q^i$ on $Q$ ,
$\tau = p_i\,dq^i, \qquad \omega = \sum_{i=1}^{n} dq^i\wedge dp_i ,$
so natural coordinates are automatically Darboux coordinates.
Hamiltonian vector fields. $X_f$ exists, is unique and is smooth. In any Darboux coordinates
$X_f = \sum_{i=1}^{n}\left(\frac{\partial f}{\partial y^i}\frac{\partial}{\partial x^i} - \frac{\partial f}{\partial x^i}\frac{\partial}{\partial y^i}\right),$
so on $T^{*}Q$ with natural coordinates the trajectories of $H$ are the solutions of Hamilton's equations
$\dot q^i = \frac{\partial H}{\partial p_i}, \qquad \dot p_i = -\frac{\partial H}{\partial q^i} .$
Moreover $f$ is constant along every integral curve of $X_f$ , and at each regular point of $f$ the field $X_f$ is tangent to the level set of $f$ . In particular the energy $H$ is conserved by its own flow.
The Poisson bracket. In Darboux coordinates
$\{f,g\} = \sum_{i=1}^{n}\left(\frac{\partial f}{\partial x^i}\frac{\partial g}{\partial y^i} - \frac{\partial f}{\partial y^i}\frac{\partial g}{\partial x^i}\right),$
and on $T^{*}Q$ in natural coordinates $\{q^i,q^j\} = 0$ , $\{p_i,p_j\} = 0$ , $\{q^i,p_j\} = \delta^i_j$ . The bracket is $\mathbb{R}$ -bilinear, antisymmetric, satisfies the Jacobi identity
$\{\{f,g\},h\} + \{\{g,h\},f\} + \{\{h,f\},g\} = 0 ,$
and relates to the Lie bracket of vector fields by
$X_{\{f,g\}} = -\,[X_f, X_g] .$
Hence $\bigl(C^{\infty}(M), \{\cdot,\cdot\}\bigr)$ is a Lie algebra, and $f \mapsto X_f$ is a Lie algebra antihomomorphism onto the Hamiltonian vector fields.
Conserved quantities. Along the flow of $X_H$ a function evolves by $\dot f = \{f, H\}$ ; so $f$ is a conserved quantity of $(M,\omega,H)$ if and only if $\{f,H\} = 0$ . Conserved quantities are closed under the bracket.
Hamiltonian flows preserve $\omega$ , and Liouville's theorem. If $\theta$ is the flow of $X_H$ then $\theta_t^{*}\omega = \omega$ on the set where $\theta_t$ is defined: every time- $t$ map is a symplectomorphism, a canonical transformation. Consequently $\theta_t^{*}(\omega^n/n!) = \omega^n/n!$ , so the Hamiltonian flow preserves phase volume. This is Liouville's theorem. More generally a smooth vector field is symplectic exactly when it is locally Hamiltonian; and when $H^1_{\mathrm{dR}}(M) = 0$ — this vanishing being exactly what is required — each locally Hamiltonian field turns out to be the Hamiltonian field of a function defined on all of $M$ .

Idea of the proofOriginal

Nondegeneracy is a statement of linear algebra at each point, and all of part 1 and part 2's pointwise shadow come from running Gram–Schmidt with $\omega$ in place of an inner product: pick $A_1 \ne 0$ , find $B_1$ pairing with it, split off the plane they span, and recurse on its $\omega$ -orthogonal complement. Pairs come out two at a time, which is why the dimension is even; and in the resulting basis $\omega^n$ is a nonzero multiple of the top basis covector, which is what makes $\omega^n$ a volume form. The cotangent bundle carries a $1$ -form for a tautological reason — a point of $T^{*}Q$ is a covector, so it can be pulled back to the point itself — and in natural coordinates that form is $p_i\,dq^i$ , whose exterior derivative is the standard symplectic form up to sign. Everything about $X_f$ and $\{f,g\}$ is then bookkeeping with $\omega^{\flat}$ in Darboux coordinates. The one genuinely non-formal input is closedness, and it enters through Cartan's formula $\mathcal{L}_X = \iota_X d + d\,\iota_X$ : for a Hamiltonian field the second term is $d(df) = 0$ and the first contains $d\omega$ , so when $d\omega = 0$ the whole Lie derivative vanishes and the flow moves $\omega$ not at all. Liouville's theorem is then automatic, because a map preserving $\omega$ preserves anything built out of $\omega$ .

Proof(Parts 1, 3, 4, 5, 6 and 7)AdaptedAdapted from Lee (2013), proofs of Props. 22.7–22.8, pp. 567; Lee (2013), proof of Prop. 22.11, pp. 569–570; Lee (2013), the computation of (22.9), pp. 574–575; Lee (2013), proofs of Props. 22.16, 22.17 and 22.19, pp. 575–576, 579; Lee (2013), Thm 14.35, p. 372; Lee (2013), Thm 12.37, p. 324; Hassani (2013), Thm 28.5.10, p. 891; Hassani (2013), proof of Prop. 28.7.11, p. 908

Part 1: a symplectic basis. Work in one tangent space $V = T_pM$ with $\omega = \omega_p$ . We induct on $m = \dim V$ , the case $m = 0$ being vacuous. For $m \ge 1$ , choose $A_1 \ne 0$ ; nondegeneracy supplies $B_1$ with $\omega(A_1,B_1) \ne 0$ , and rescaling $B_1$ makes $\omega(A_1,B_1) = 1$ . Since $\omega$ is alternating, $\omega(A_1,A_1) = 0$ , so $B_1$ is not a multiple of $A_1$ and $S = \operatorname{span}(A_1,B_1)$ is $2$ -dimensional. Let

S^{\perp} = \{v \in V : \omega(v,w) = 0 \ \text{for all } w \in S\} .

Then $\dim S + \dim S^{\perp} = \dim V$ : the map $v \mapsto \omega(v,\cdot)|_S$ from $V$ to $S^{*}$ has kernel $S^{\perp}$ and is onto, because any $\varphi \in S^{*}$ extends to $V^{*}$ and $\omega^\flat$ is surjective, so the rank–nullity theorem applies. Also $S \cap S^{\perp} = \{0\}$ , since $\omega|_S$ has matrix $\left(\begin{smallmatrix}0&1\\-1&0\end{smallmatrix}\right)$ , which is invertible; hence $V = S \oplus S^{\perp}$ and $\omega|_{S^{\perp}}$ is again nondegenerate. By the inductive hypothesis $S^{\perp}$ has even dimension and a basis $(A_2,B_2,\dots,A_n,B_n)$ with $\omega(A_i,A_j) = \omega(B_i,B_j) = 0$ and $\omega(A_i,B_j) = \delta_{ij}$ ; adjoining $A_1, B_1$ gives such a basis of $V$ , and $\dim V = 2n$ .

Let $(\alpha^i,\beta^i)$ be the dual basis. The relations just listed say precisely that $\omega = \sum_i \alpha^i\wedge\beta^i$ . Expanding the $n$ -fold wedge, every term with a repeated index dies, the $n!$ surviving terms are equal because $2$ -forms commute under $\wedge$ , and so

\omega^n = n!\;\alpha^1\wedge\beta^1\wedge\cdots\wedge\alpha^n\wedge\beta^n \ne 0 .

As $p$ varies this is a nowhere-vanishing $2n$ -form on $M$ , which is exactly an orientation together with a volume form; dividing by $n!$ normalises it so that it reads $dx^1\wedge dy^1\wedge\cdots\wedge dx^n\wedge dy^n$ in Darboux coordinates.

Part 3: the cotangent bundle. Let $q^i$ be coordinates on $Q$ and $(q^i,p_i)$ the induced natural coordinates on $T^{*}Q$ , so that the point with these coordinates is the covector $p_i\,dq^i$ at the point with coordinates $q^i$ . In them $\pi(q,p) = q$ , hence $d\pi^{*}(dq^i) = dq^i$ , and therefore

\tau_{(q,p)} = d\pi^{*}_{(q,p)}\bigl(p_i\,dq^i\bigr) = p_i\,dq^i .

Its components are linear, so $\tau$ is smooth. Then

\omega = -d\tau = -\,dp_i\wedge dq^i = \sum_i dq^i\wedge dp_i ,

which is closed because it is exact, and nondegenerate because at each point it is the standard symplectic tensor of part 1 with $(A_i,B_i) = (\partial/\partial q^i, \partial/\partial p_i)$ . The sign in $\omega = -d\tau$ is what turns $dp_i\wedge dq^i$ into $dq^i\wedge dp_i$ .

Part 4: the coordinate formula. Fix Darboux coordinates and write $X_f = \sum_i\bigl(a^i\,\partial/\partial x^i + b^i\,\partial/\partial y^i\bigr)$ . Feeding this into $\omega = \sum_i dx^i\wedge dy^i$ and using $\iota_X(\alpha\wedge\beta) = (\iota_X\alpha)\wedge\beta - \alpha\wedge(\iota_X\beta)$ for $1$ -forms gives

\iota_{X_f}\omega = \sum_i\bigl(a^i\,dy^i - b^i\,dx^i\bigr).

Setting this equal to $df = \sum_i\bigl(\partial_{x^i}f\,dx^i + \partial_{y^i}f\,dy^i\bigr)$ and comparing coefficients of the independent $1$ -forms $dx^i, dy^i$ yields $a^i = \partial f/\partial y^i$ and $b^i = -\partial f/\partial x^i$ , which is the stated formula. Existence and uniqueness of $X_f$ are immediate from nondegeneracy: $\omega^{\flat}$ is a bundle isomorphism, so $X_f = (\omega^{\flat})^{-1}(df)$ is defined and unique, and it is smooth because $\omega^\flat$ and $df$ are. Writing out $\gamma'(t) = X_H|_{\gamma(t)}$ in natural coordinates on $T^{*}Q$ gives Hamilton's equations.

That $f$ is constant along the integral curves of $X_f$ is a one-line consequence of $\omega$ being alternating:

X_f f = df(X_f) = \omega(X_f, X_f) = 0 .

Since the derivative of $f$ along $X_f$ vanishes, $f\circ\gamma$ is constant for every integral curve $\gamma$ ; and where $df_p \ne 0$ the level set of $f$ through $p$ is an embedded hypersurface whose tangent space is $\ker df_p$ , which contains $X_f|_p$ .

Part 5: the bracket. The coordinate formula follows from $\{f,g\} = X_g f$ and part 4. For the three algebraic properties, note first that $f \mapsto X_f$ is $\mathbb{R}$ -linear, so bilinearity and antisymmetry are read off from $\{f,g\} = \omega(X_f,X_g)$ together with the antisymmetry of $\omega$ . For $X_{\{f,g\}} = -[X_f,X_g]$ it is enough, again by nondegeneracy, to check that

\omega\bigl(X_{\{f,g\}}, Y\bigr) + \omega\bigl([X_f,X_g], Y\bigr) = 0

for every vector field $Y$ . The first term is $d\{f,g\}(Y) = Y\{f,g\} = Y X_g f$ . For the second, use that $X_g$ is symplectic (part 7, whose proof below does not use part 5) so that $\mathcal{L}_{X_g}\omega = 0$ , and expand that Lie derivative on the pair $(X_f, Y)$ by the derivation rule for a $2$ -tensor:

0 = X_g\bigl(\omega(X_f,Y)\bigr) - \omega\bigl([X_g,X_f],Y\bigr) - \omega\bigl(X_f,[X_g,Y]\bigr).

The first term is $X_g(df(Y)) = X_gYf$ . The third is $df([X_g,Y]) = X_gYf - YX_gf = X_gYf - \omega(X_{\{f,g\}},Y)$ . Substituting both and cancelling $X_gYf$ leaves $-\omega([X_g,X_f],Y) + \omega(X_{\{f,g\}},Y) = 0$ , which is the displayed identity since $[X_g,X_f] = -[X_f,X_g]$ .

The Jacobi identity now follows without further geometry. Using $\{f,\cdot\}$ twice and the identity just proved,

\{f,\{g,h\}\} = X_{\{g,h\}}f = -[X_g,X_h]f = -X_gX_hf + X_hX_gf = -X_g\{f,h\} + X_h\{f,g\},

that is, $\{f,\{g,h\}\} = \{\{f,g\},h\} - \{\{f,h\},g\}$ . Antisymmetry rewrites the left-hand side as $-\{\{g,h\},f\}$ and the subtracted term as $-\{\{h,f\},g\}$ , so the identity becomes $-\{\{g,h\},f\} = \{\{f,g\},h\} + \{\{h,f\},g\}$ , which is the cyclic sum vanishing.

Part 6. Let $\gamma$ be a trajectory, so $\gamma' = X_H\circ\gamma$ . Then $(f\circ\gamma)'(t) = df(X_H) = X_Hf = \{f,H\}$ at $\gamma(t)$ . Hence $f$ is constant on every trajectory if and only if $\{f,H\}$ vanishes on every trajectory, and since a trajectory passes through each point of $M$ , if and only if $\{f,H\} \equiv 0$ . If $\{f,H\} = \{g,H\} = 0$ , apply the Jacobi identity to the triple $(f,g,H)$ : it gives $\{\{f,g\},H\} = -\{\{g,H\},f\} - \{\{H,f\},g\}$ , and both terms on the right vanish because $\{g,H\} = 0$ and $\{H,f\} = -\{f,H\} = 0$ . So conserved quantities are closed under the bracket. Taking $f = H$ recovers conservation of energy from antisymmetry.

Part 7. By Cartan's magic formula, valid for any vector field and any form,

\mathcal{L}_X\omega = \iota_X(d\omega) + d(\iota_X\omega).

Closedness kills the first term, so $\mathcal{L}_X\omega = d(\iota_X\omega)$ for every $X$ . If $X = X_H$ then $\iota_{X_H}\omega = dH$ and $d(dH) = 0$ , so $\mathcal{L}_{X_H}\omega = 0$ ; and a covariant tensor field has vanishing Lie derivative along $X$ exactly when it is invariant under the flow of $X$ , so $\theta_t^{*}\omega = \omega$ wherever $\theta_t$ is defined. Pullback commutes with $\wedge$ , hence $\theta_t^{*}(\omega^n) = (\theta_t^{*}\omega)^n = \omega^n$ and the Liouville volume is preserved; integrating, $\int_{\theta_t(A)}\omega^n/n! = \int_A\omega^n/n!$ for every domain $A$ on which this makes sense.

The general statement is the same computation read backwards. $X$ is symplectic iff $\mathcal{L}_X\omega = 0$ iff the $1$ -form $\iota_X\omega$ is closed. If $X$ is locally Hamiltonian, then near each point $\iota_X\omega = df$ is exact, hence closed. Conversely if $\iota_X\omega$ is closed then by the Poincaré lemma each point has a neighbourhood on which $\iota_X\omega = df$ for some $f$ , and nondegeneracy upgrades this to $X = X_f$ there. Finally, the closed $1$ -forms are all exact exactly when $H^1_{\mathrm{dR}}(M) = 0$ : in that case $\iota_X\omega = df$ globally, so a symplectic $X$ is globally Hamiltonian; and conversely, given a closed $1$ -form $\sigma$ , the field $X = (\omega^{\flat})^{-1}\sigma$ is symplectic, hence by hypothesis equals some $X_f$ , and then $\sigma = \iota_X\omega = df$ is exact. $\square$

Remark(Where the hypotheses are used)Original

Nondegeneracy is what makes $f \mapsto X_f$ possible at all: it is the invertibility of $\omega^{\flat}$ . Drop it and almost everything collapses. The extreme case $\omega = 0$ on $\mathbb{R}^2$ is closed and alternating, but $\iota_X\omega = df$ has no solution unless $df = 0$ , so no function generates a flow. Nondegeneracy also forces the dimension to be even, by the pairing argument in part 1, and supplies the volume form, since $\omega^n \ne 0$ is equivalent to nondegeneracy. Hassani records the same equivalence as his Prop. 26.4.4.
Closedness is used in exactly two places, and both fail without it. (i) In part 7 the term $\iota_X(d\omega)$ in Cartan's formula drops only because $d\omega = 0$ ; with $d\omega \ne 0$ a Hamiltonian flow need not preserve $\omega$ , and Liouville's theorem is lost. (ii) The Jacobi identity is derived from $X_{\{f,g\}} = -[X_f,X_g]$ , which in turn used $\mathcal{L}_{X_g}\omega = 0$ , that is, part 7. So without closedness the bracket need not be a Lie bracket.

A concrete failure: on $\mathbb{R}^4$ with coordinates $(x^1,y^1,x^2,y^2)$ take
$\omega = dx^1\wedge dy^1 + e^{x^1}\,dx^2\wedge dy^2 .$
It is nondegenerate — its matrix is block diagonal with determinant $e^{2x^1} \ne 0$ — but $d\omega = e^{x^1}\,dx^1\wedge dx^2\wedge dy^2 \ne 0$ . Solving $\iota_X\omega = df$ gives
$X_f = \frac{\partial f}{\partial y^1}\frac{\partial}{\partial x^1} - \frac{\partial f}{\partial x^1}\frac{\partial}{\partial y^1} + e^{-x^1}\left(\frac{\partial f}{\partial y^2}\frac{\partial}{\partial x^2} - \frac{\partial f}{\partial x^2}\frac{\partial}{\partial y^2}\right),$
so $\{f,g\} = \partial_{x^1}f\,\partial_{y^1}g - \partial_{y^1}f\,\partial_{x^1}g + e^{-x^1}\bigl(\partial_{x^2}f\,\partial_{y^2}g - \partial_{y^2}f\,\partial_{x^2}g\bigr)$ . Take $f = y^1$ , $g = x^2$ , $h = y^2$ . Then $\{g,h\} = e^{-x^1}$ while $\{f,g\} = \{h,f\} = 0$ , so the cyclic sum reduces to $\{\{g,h\},f\} = \{e^{-x^1}, y^1\} = -e^{-x^1}$ , which equals $-1$ on the hyperplane $x^1 = 0$ . The Jacobi identity fails, by a margin that is exactly as large as $d\omega$ .
Alternating is used on its own in part 4: $\omega(X_f,X_f) = 0$ is the whole proof that $f$ is conserved by its own flow. A symmetric nondegenerate form — a metric — gives the gradient instead, and a function is emphatically not constant along its own gradient flow.
Darboux's theorem needs both hypotheses and is proved elsewhere. It is the one statement above not proved here. Lee proves it by Moser's method: interpolate between $\omega$ and its constant-coefficient approximation at $p$ by $\omega_t$ , solve $\iota_{V_t}\omega_t = \alpha$ for a time-dependent field, and let its flow carry one to the other. Hassani states it without proof. The contrast with Riemannian geometry is the point: a metric has local invariants (curvature), a symplectic form has none.
$H^1_{\mathrm{dR}}(M) = 0$ is needed only for the last clause of part 7, and it genuinely bites. On $M = \mathbb{R}^2\setminus\{0\}$ with $\omega = dx\wedge dy$ , take
$X = \frac{x\,\partial/\partial x + y\,\partial/\partial y}{x^2+y^2}, \qquad\text{so}\qquad \iota_X\omega = \frac{x\,dy - y\,dx}{x^2+y^2} = d\vartheta ,$
the angle form. It is closed, so $\mathcal{L}_X\omega = d(\iota_X\omega) = 0$ and $X$ is symplectic — its flow pushes points radially outward at the rate $\dot r = 1/r$ , carrying the annulus $a \le r \le b$ to $\sqrt{a^2+2t} \le r \le \sqrt{b^2+2t}$ and preserving its area $\pi(b^2-a^2)$ , as it must. But $d\vartheta$ is not exact on $M$ , where $H^1_{\mathrm{dR}} \cong \mathbb{R}$ , so there is no global $f$ with $X = X_f$ : the only candidate is the polar angle, which is not single-valued. Locally, on any half-plane, a branch of $\vartheta$ does the job.

Method(Working on a symplectic manifold)Original

Get Darboux coordinates. If the manifold is a cotangent bundle $T^{*}Q$ — which it is in mechanics, with $Q$ the configuration space — use the natural coordinates $(q^i,p_i)$ : part 3 says they are already Darboux coordinates. Otherwise find coordinates in which $\omega = \sum_i dx^i\wedge dy^i$ ; Darboux's theorem promises they exist locally.
Check that a given $2$ -form is symplectic. Compute $d\omega$ and confirm it vanishes; then write the matrix $\omega_{ij}$ in a coordinate frame and confirm $\det(\omega_{ij}) \ne 0$ at every point. Equivalently, check $\omega^n \ne 0$ .
Build $X_f$ . Either solve $\iota_{X}\omega = df$ coefficient by coefficient, or, in Darboux coordinates, write down $X_f = \sum_i(\partial_{y^i}f\,\partial_{x^i} - \partial_{x^i}f\,\partial_{y^i})$ directly.
Test a conserved quantity by computing $\{f,H\}$ in coordinates: it vanishes identically if and only if $f$ is constant on every trajectory. This is cheaper than solving the equations of motion, and is how first integrals are found in practice.
Test a canonical transformation. A diffeomorphism $F$ is a symplectomorphism iff $F^{*}\omega = \omega$ . In Darboux coordinates with $F = (X(x,y), Y(x,y))$ this reads $\sum_i dX^i\wedge dY^i = \sum_i dx^i\wedge dy^i$ ; for a linear $F$ with matrix $A$ it is the matrix condition $A^{\mathsf T}JA = J$ , with $J$ the matrix of $\omega$ .
Phase volume. Compute $\omega^n/n!$ once in Darboux coordinates; by part 7 you may then move any region along any Hamiltonian flow without recomputing its volume.

Example(A planar oscillator: energy, angular momentum and phase volume)Original

Take $Q = \mathbb{R}^2$ with coordinates $q = (q^1,q^2)$ , so $M = T^{*}Q = \mathbb{R}^4$ with natural coordinates $(q^1,q^2,p_1,p_2)$ and $\omega = dq^1\wedge dp_1 + dq^2\wedge dp_2$ . Let the Hamiltonian be that of an isotropic oscillator of unit mass and spring constant $k = 4$ :

H = \tfrac12\bigl(p_1^2 + p_2^2\bigr) + 2\bigl((q^1)^2 + (q^2)^2\bigr).

The Hamiltonian vector field. By part 4,

X_H = p_1\frac{\partial}{\partial q^1} + p_2\frac{\partial}{\partial q^2} - 4q^1\frac{\partial}{\partial p_1} - 4q^2\frac{\partial}{\partial p_2},

so Hamilton's equations are $\dot q^i = p_i$ , $\dot p_i = -4q^i$ : two uncoupled oscillators of angular frequency $2$ .

A trajectory. Start at $(q^1,q^2,p_1,p_2) = (1,0,0,3)$ . The solution is

q^1 = \cos 2t,\quad p_1 = -2\sin 2t,\qquad q^2 = \tfrac32\sin 2t,\quad p_2 = 3\cos 2t .

(Check: $\dot q^2 = 3\cos 2t = p_2$ and $\dot p_2 = -6\sin 2t = -4\cdot\tfrac32\sin 2t = -4q^2$ .)

Energy. $H = \tfrac12(4\sin^2 2t + 9\cos^2 2t) + 2(\cos^2 2t + \tfrac94\sin^2 2t) = \tfrac{13}{2}$ , constant, as part 4 requires. Numerically at $t = 0.4$ : $q^1 = 0.6967067$ , $p_1 = -1.4347122$ , $q^2 = 1.0760341$ , $p_2 = 2.0901201$ , and

H = \tfrac12(2.0583 + 4.3686) + 2(0.48540 + 1.15785) = 3.21345 + 3.28655 = 6.5 ,

against $H = \tfrac12\cdot 9 + 2\cdot 1 = 6.5$ at $t = 0$ .

Angular momentum is conserved, without solving anything. Put $L = q^1p_2 - q^2p_1$ . Then $\partial_{q^1}L = p_2$ , $\partial_{q^2}L = -p_1$ , $\partial_{p_1}L = -q^2$ , $\partial_{p_2}L = q^1$ , while $\partial_{q^i}H = 4q^i$ and $\partial_{p_i}H = p_i$ . By part 5,

\{L,H\} = \bigl(p_2p_1 + 4q^1q^2\bigr) + \bigl(-p_1p_2 - 4q^1q^2\bigr) = 0 ,

so by part 6, $L$ is a conserved quantity. Checking against the trajectory: $L = 1\cdot 3 - 0\cdot 0 = 3$ at $t=0$ , and at $t = 0.4$

L = 0.6967067\times 2.0901201 - 1.0760341\times(-1.4347122) = 1.456222 + 1.543778 = 3.000000 .

Phase volume. Here $n = 2$ and $\omega^2/2! = dq^1\wedge dp_1\wedge dq^2\wedge dp_2$ . The time- $t$ map acts on the $(q^1,p_1)$ plane by the matrix

A(t) = \begin{pmatrix}\cos 2t & \tfrac12\sin 2t\\[2pt] -2\sin 2t & \cos 2t\end{pmatrix}, \qquad \det A(t) = \cos^2 2t + \sin^2 2t = 1 ,

and likewise on $(q^2,p_2)$ , so the Jacobian determinant of $\theta_t$ is $1\cdot 1 = 1$ and phase volume is preserved — Liouville's theorem made arithmetic. At $t = 0.4$ the entries are $0.6967067$ , $0.3586780$ , $-1.4347122$ , $0.6967067$ , and $0.6967067^2 + 0.3586780\times 1.4347122 = 0.485400 + 0.514600 = 1.000000$ .

A map that is not canonical. The dilation $(q,p)\mapsto(2q,p)$ pulls $\omega$ back to $2\,dq^1\wedge dp_1 + 2\,dq^2\wedge dp_2 = 2\omega$ , so it is not a symplectomorphism and multiplies phase volume by $2^2 = 4$ . No Hamiltonian flow can reach it.

ApplicationOriginal

Symplectic geometry is the coordinate-free form of Hamiltonian mechanics. The configuration space of a mechanical system is a manifold $Q$ — a circle for a pendulum, $SO(3)$ for a rigid body, a submanifold of $\mathbb{R}^{3N}$ for constrained particles — and part 3 says its phase space $T^{*}Q$ comes with a symplectic form for free, with no choice of coordinates and no metric. The Legendre transformation of the canonical equations is what moves a Lagrangian on $TQ$ to a Hamiltonian on $T^{*}Q$ , and Hamilton's equations are then nothing but the statement that the motion is an integral curve of $X_H$ . Because the formulation is coordinate-free, the "canonical transformations" that classical mechanics uses to simplify a problem — action–angle variables, generating functions, the Hamilton–Jacobi substitution of the Hamilton–Jacobi equation — are exactly the symplectomorphisms, and a change of variables is legitimate precisely when it preserves $\omega$ .

Liouville's theorem is the foundation of statistical mechanics. A probability density $\rho$ on phase space evolves so that $\rho\,\omega^n/n!$ is carried along by the flow; since the flow preserves $\omega^n/n!$ , the density is constant along trajectories, which is Liouville's equation $\partial_t\rho + \{\rho, H\} = 0$ and the reason the microcanonical ensemble — uniform measure on an energy shell — is stationary. In accelerator and plasma physics the same statement is the conservation of emittance: a beam's phase-space area cannot be reduced by any Hamiltonian optics, only reshaped.

The correspondence $\{\cdot,\cdot\} \leftrightarrow \tfrac{1}{i\hbar}[\cdot,\cdot]$ between the Poisson bracket and the commutator is the bridge to quantum mechanics; the relations $\{q^i,p_j\} = \delta^i_j$ of part 5 become the canonical commutation relations, and part 5's statement that $C^\infty(M)$ is a Lie algebra is what makes the comparison meaningful. In numerical work, symplectic integrators (leapfrog, Störmer–Verlet and their higher-order relatives) are designed so that each step is exactly a symplectomorphism; they do not conserve energy exactly, but because they preserve a nearby symplectic structure their energy error stays bounded over times at which an ordinary Runge–Kutta scheme has drifted away, which is why solar-system integrations over millions of years are done with them. Finally, the identification of a conserved quantity with a symmetry — a function $f$ with $\{f,H\} = 0$ generates a flow preserving $H$ , and conversely — is the Hamiltonian face of Noether's theorem, and on a Riemannian manifold the symmetries in question are generated by Killing vector fields.

Definition · advanced

Killing vector fields and symmetries

Adapted

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Identifies the vector fields whose flows move a Riemannian manifold without distorting it, gives the Killing equation in both coordinate and covariant form, shows that they close under the Lie bracket and that each one produces a quantity conserved along every geodesic, and lists them all for Euclidean space and the round sphere.

Prerequisites: Riemannian manifolds, Flows and integral curves, Christoffel symbols and the covariant derivative, Geodesics, Tangent vectors and vector fields, Smooth manifolds

Definition(Isometries, Killing fields and the Killing equation)AdaptedAdapted from Lee (2013), p. 332; Lee (2013), Problem 13-13, pp. 345–346; Lee (2013), (12.11), p. 323

Let $(M,g)$ and $(\widetilde M,\widetilde g)$ be Riemannian manifolds.

A (Riemannian) isometry $F : M \to \widetilde M$ is a diffeomorphism with $F^{*}\widetilde g = g$ ; written out, $\widetilde g_{F(p)}(dF_pv, dF_pw) = g_p(v,w)$ for all $p\in M$ and $v,w \in T_pM$ . A local isometry is a local diffeomorphism with the same property, so that every point has a neighbourhood mapped isometrically onto an open set of $\widetilde M$ . An isometry preserves lengths of curves, angles, distances and volumes, because each of those is built from $g$ alone.
Call $K$ , smooth and defined on all of $M$ , a Killing vector field of the metric $g$ when its flow acts by isometries: for every $t$ , the time- $t$ map $\theta_t$ of the flow (flows and integral curves) satisfies $\theta_t^{*}g = g$ on the open set $M_t$ where it is defined. A tensor field with this property is said to be invariant under the flow.
In coordinates, writing $K = K^k\,\partial/\partial x^k$ and summing repeated indices, the Killing equation is $K^{k}\,\frac{\partial g_{ij}}{\partial x^{k}} \;+\; g_{kj}\,\frac{\partial K^{k}}{\partial x^{i}} \;+\; g_{ik}\,\frac{\partial K^{k}}{\partial x^{j}} \;=\; 0 , \qquad i,j = 1,\dots,n .$ Lowering the index with $K_j = g_{jk}K^k$ and using the Levi-Civita connection $\nabla$ of $g$ (Christoffel symbols and the covariant derivative), the same equation reads $\nabla_{i}K_{j} + \nabla_{j}K_{i} = 0 :$ the covariant derivative of the covector $K_\flat = K_j\,dx^j$ is antisymmetric.

Throughout, $\mathcal{L}_K$ is the Lie derivative along $K$ , defined on a covariant tensor field $A$ by $(\mathcal{L}_KA)_p = \frac{d}{dt}\big|_{t=0}(\theta_t^{*}A)_p$ .

Definition(Killing fields: four descriptions, an algebra, and a constant of the motion)AdaptedAdapted from Lee (2013), Thm 12.37, p. 324; Lee (2013), Example 12.35, p. 323; Lee (2013), Problem 13-13, pp. 345–346; Lee (2013), Problem 13-14, p. 346

Let $(M,g)$ be a Riemannian manifold of dimension $n$ and $K$ a smooth vector field on $M$ .

Four equivalent conditions. The following say the same thing.
- (a) $K$ is a Killing field: the flow of $K$ acts by isometries of $g$ .
- (b) $\mathcal{L}_Kg = 0$ .
- (c) In every smooth chart, $K^{k}\partial_kg_{ij} + g_{kj}\partial_iK^{k} + g_{ik}\partial_jK^{k} = 0$ .
- (d) $\nabla_iK_j + \nabla_jK_i = 0$ , where $K_j = g_{jk}K^k$ .
They form a Lie algebra. The Killing fields of $(M,g)$ are a linear subspace of the smooth vector fields on $M$ and are closed under the Lie bracket; that is, they form a Lie subalgebra of $\mathfrak{X}(M)$ .
A conserved quantity along every geodesic. If $K$ is a Killing field and $\gamma$ is a geodesic of $(M,g)$ (geodesics), then
$g\bigl(K_{\gamma(t)},\ \gamma'(t)\bigr) \quad\text{is constant in } t .$
Each independent Killing field therefore supplies one first integral of the geodesic equations.
Euclidean space, completely. For $\mathbb{R}^n$ with the Euclidean metric $\bar g = \delta_{ij}\,dx^i dx^j$ , the Killing fields are exactly the linear combinations, with constant coefficients, of
$\frac{\partial}{\partial x^{i}}\ \ (1 \le i \le n) \qquad\text{and}\qquad x^{i}\frac{\partial}{\partial x^{j}} - x^{j}\frac{\partial}{\partial x^{i}}\ \ (1 \le i < j \le n) ,$
and these $n + \binom{n}{2} = \tfrac12 n(n+1)$ fields are a basis. They generate the translations and the rotations: the Killing algebra of Euclidean space is the Lie algebra of its group of rigid motions.

Idea of the proofOriginal

A flow preserves a tensor field exactly when the field's derivative along the flow vanishes, so "moves without distorting" becomes the single equation $\mathcal{L}_Kg = 0$ ; writing that Lie derivative out in a chart gives (c), and trading partial derivatives for covariant ones absorbs the term $K^k\partial_kg_{ij}$ into the Christoffel symbols and leaves the clean antisymmetry (d). The conserved quantity in part 3 is then a two-line computation: differentiating $g(K,\gamma')$ along $\gamma$ produces two terms, one killed by the geodesic equation and the other equal to $\nabla_{(i}K_{j)}\gamma'^i\gamma'^j$ , which is the symmetric part of $\nabla K$ contracted with a symmetric pair — zero precisely by (d). For part 2, the bracket of two Killing fields is Killing because $\mathcal{L}_{[V,W]} = \mathcal{L}_V\mathcal{L}_W - \mathcal{L}_W\mathcal{L}_V$ on tensor fields, an identity that falls out of expanding the Lie derivative in Lie brackets and using the Jacobi identity. Part 4 is the observation that in flat coordinates (d) says the matrix $\partial_iK_j$ is antisymmetric, and differentiating that relation once more forces the second derivatives of $K$ to vanish — so a Killing field of Euclidean space is affine, with constant part a translation and linear part an antisymmetric matrix, that is, an infinitesimal rotation.

Proof(Parts 1 to 4)AdaptedAdapted from Lee (2013), Thm 12.37, p. 324; Lee (2013), Cor. 12.33 and Example 12.35, p. 323; Lee (2013), Problems 13-13 and 13-14, pp. 345–346

Part 1, (a) $\Leftrightarrow$ (b). Lee's flow-invariance criterion reads: for smooth $V$ and smooth covariant $A$ , the equation $\mathcal{L}_VA = 0$ holds exactly when the flow of $V$ carries $A$ to itself. Applied with $V = K$ and $A = g$ , invariance is the statement $\theta_t^{*}g = g$ wherever $\theta_t$ is defined; each $\theta_t$ is a diffeomorphism from $M_t$ onto $M_{-t}$ , so this is exactly the statement that $\theta_t$ is an isometry of the open set $M_t$ onto $M_{-t}$ .

Part 1, (b) $\Leftrightarrow$ (c). Take a chart and expand $g = g_{ij}\,dx^i\otimes dx^j$ . The Lie derivative is a derivation that commutes with $d$ on functions, so $\mathcal{L}_K(dx^i) = d(\mathcal{L}_Kx^i) = d(Kx^i) = dK^i$ , and the product rule gives

\mathcal{L}_Kg = (Kg_{ij})\,dx^i\otimes dx^j + g_{ij}\,dK^i\otimes dx^j + g_{ij}\,dx^i\otimes dK^j = \Bigl(K^k\partial_kg_{ij} + g_{kj}\partial_iK^k + g_{ik}\partial_jK^k\Bigr)dx^i\otimes dx^j ,

after relabelling the summation indices in the last two terms. Since the $dx^i\otimes dx^j$ are linearly independent at each point, $\mathcal{L}_Kg = 0$ is equivalent to the vanishing of each coefficient, which is (c).

Part 1, (c) $\Leftrightarrow$ (d). Write $K_j = g_{jk}K^k$ . By the definition of the covariant derivative of a covector field,

\nabla_iK_j + \nabla_jK_i = \partial_i K_j + \partial_j K_i - 2\Gamma^{l}_{ij}K_l ,

the two Christoffel terms coinciding because $\Gamma^l_{ij}$ is symmetric in $i$ and $j$ . Expanding the partial derivatives,

\partial_iK_j + \partial_jK_i = g_{jk}\partial_iK^k + g_{ik}\partial_jK^k + K^k\bigl(\partial_ig_{jk} + \partial_jg_{ik}\bigr).

For the Christoffel term, lowering the index in the formula $\Gamma^{l}_{ij} = \tfrac12 g^{lm}(\partial_ig_{jm} + \partial_jg_{im} - \partial_mg_{ij})$ gives $2\Gamma^{l}_{ij}g_{lk} = \partial_ig_{jk} + \partial_jg_{ik} - \partial_kg_{ij}$ , so that

2\Gamma^{l}_{ij}K_l = 2\Gamma^{l}_{ij}g_{lk}K^k = K^k\bigl(\partial_ig_{jk} + \partial_jg_{ik}\bigr) - K^k\partial_kg_{ij} .

Subtracting, the terms carrying $\partial_i g_{jk}$ and $\partial_j g_{ik}$ cancel and what remains is

\nabla_iK_j + \nabla_jK_i = K^k\partial_kg_{ij} + g_{kj}\partial_iK^k + g_{ik}\partial_jK^k = (\mathcal{L}_Kg)_{ij},

so (c) and (d) are the same equation written two ways. This also records, for later use, the identity $(\mathcal{L}_Kg)_{ij} = \nabla_iK_j + \nabla_jK_i$ .

Part 2. Expanding the Lie derivative of a covariant $2$ -tensor in Lie brackets gives, for any vector fields $V, X, Y$ ,

(\mathcal{L}_VA)(X,Y) = V\bigl(A(X,Y)\bigr) - A\bigl([V,X],Y\bigr) - A\bigl(X,[V,Y]\bigr). \tag{$\ast$}

This is $\mathbb{R}$ -linear in $V$ , so a constant-coefficient combination of Killing fields is Killing and the Killing fields form a linear subspace. For the bracket, apply $(\ast)$ twice. Writing $B = \mathcal{L}_WA$ and expanding $(\mathcal{L}_VB)(X,Y)$ by $(\ast)$ gives nine terms; subtracting the same expression with $V$ and $W$ interchanged, every term in which one of $V, W$ differentiates a function and the other supplies a bracket occurs twice with opposite signs and cancels, as do the two terms $A([V,X],[W,Y])$ and $A([W,X],[V,Y])$ . What survives is

\bigl(\mathcal{L}_V\mathcal{L}_WA - \mathcal{L}_W\mathcal{L}_VA\bigr)(X,Y) = [V,W]\bigl(A(X,Y)\bigr) + A\bigl([W,[V,X]] - [V,[W,X]],\,Y\bigr) + A\bigl(X,\,[W,[V,Y]] - [V,[W,Y]]\bigr).

The Jacobi identity for the Lie bracket turns $[V,[W,X]] - [W,[V,X]]$ into $[[V,W],X]$ , so the second and third groups are $-A([[V,W],X],Y)$ and $-A(X,[[V,W],Y])$ . Comparing with $(\ast)$ for the field $[V,W]$ ,

\mathcal{L}_{[V,W]}A = \mathcal{L}_V\mathcal{L}_WA - \mathcal{L}_W\mathcal{L}_VA .

Now let $V, W$ be Killing: $\mathcal{L}_Vg = \mathcal{L}_Wg = 0$ . Then $\mathcal{L}_{[V,W]}g = \mathcal{L}_V(0) - \mathcal{L}_W(0) = 0$ , so $[V,W]$ is Killing.

Part 3. Let $\gamma$ be a geodesic and put $f(t) = g(K,\gamma')$ along it. The Levi-Civita connection is compatible with $g$ , so differentiating the product along $\gamma$ gives

f'(t) = g\bigl(\nabla_{\gamma'}K,\ \gamma'\bigr) + g\bigl(K,\ \nabla_{\gamma'}\gamma'\bigr).

The second term vanishes because $\gamma$ is a geodesic, $\nabla_{\gamma'}\gamma' = 0$ . The first term is, in coordinates, $(\nabla_iK_j)\,\gamma'^i\gamma'^j$ . Since $\gamma'^i\gamma'^j$ is symmetric under $i \leftrightarrow j$ , only the symmetric part of $\nabla_iK_j$ contributes:

(\nabla_iK_j)\gamma'^i\gamma'^j = \tfrac12\bigl(\nabla_iK_j + \nabla_jK_i\bigr)\gamma'^i\gamma'^j = 0

by the Killing equation in form (d). Hence $f' \equiv 0$ .

Part 4. In the standard coordinates on $\mathbb{R}^n$ the metric has constant components $\delta_{ij}$ , so all Christoffel symbols vanish, indices are raised and lowered by $\delta$ , and (c) becomes simply

\partial_iK_j + \partial_jK_i = 0, \qquad K_j = K^j .

The listed fields satisfy it. For $K = \partial/\partial x^a$ every $\partial_iK_j$ is zero. For $K = x^a\partial/\partial x^b - x^b\partial/\partial x^a$ the components are $K_j = x^a\delta_{jb} - x^b\delta_{ja}$ , so $\partial_iK_j = \delta_{ia}\delta_{jb} - \delta_{ib}\delta_{ja}$ , which is antisymmetric in $i,j$ ; hence $\partial_iK_j + \partial_jK_i = 0$ .

There are no others. Set $S_{ij} = \partial_iK_j$ , so the equation says $S_{ij} = -S_{ji}$ , and let $T_{kij} = \partial_kS_{ij} = \partial_k\partial_iK_j$ . Then $T$ is symmetric in its first two indices, because mixed partial derivatives of the smooth function $K_j$ commute, and antisymmetric in its last two, by the Killing equation differentiated. These two symmetries are incompatible:

T_{kij} = T_{ikj} = -T_{ijk} = -T_{jik} = T_{jki} = T_{kji} = -T_{kij},

using alternately the symmetry in the first pair and the antisymmetry in the last pair; hence $T \equiv 0$ . So all second derivatives of $K$ vanish and $K_j(x) = a_j + S_{ij}x^i$ with $a_j$ and $S_{ij}$ constant and $S$ antisymmetric. Splitting the linear part over the pairs $i<j$ ,

K = a_j\frac{\partial}{\partial x^j} + \sum_{i<j} S_{ij}\left(x^i\frac{\partial}{\partial x^j} - x^j\frac{\partial}{\partial x^i}\right),

which is the asserted combination. The listed fields are linearly independent, because the translations are the fields with $S = 0$ and the rotations are determined by the antisymmetric matrix $S$ , which is zero only if all the coefficients are; so they are a basis, and the dimension is $n + \tfrac12 n(n-1) = \tfrac12n(n+1)$ . $\square$

Remark(Where the hypotheses are used)Original

Smoothness of $K$ is what gives it a flow at all, and the equivalence (a) $\Leftrightarrow$ (b) is a statement about that flow. Note also that the flow need not be complete: $\theta_t$ is only defined on an open set $M_t$ , and "acts by isometries" has to be read there. Removing a point from $\mathbb{R}^2$ leaves $\partial/\partial x$ a Killing field whose flow is defined only for a finite time at each point.
Geodesy is essential in part 3, and only there. Drop $\nabla_{\gamma'}\gamma' = 0$ and the quantity $g(K,\gamma')$ picks up the term $g(K,\nabla_{\gamma'}\gamma')$ , which has no reason to vanish. Concretely, on the round sphere take a unit-speed curve along which $\dot\varphi$ is held constant while $\theta$ varies, so that $\dot\theta^2 = 1 - \sin^2\!\theta\,\dot\varphi^2$ . It is not a geodesic, and $g(\partial_\varphi,\gamma') = \sin^2\!\theta\,\dot\varphi$ changes as $\theta$ does. The symmetric-part argument is what makes the Killing equation enter, so a field satisfying only the weaker conformal Killing equation $\nabla_iK_j + \nabla_jK_i = \lambda\,g_{ij}$ gives a conserved quantity only for null geodesics, where $g(\gamma',\gamma') = 0$ — which is why conformal Killing fields are useful in relativity but not in Riemannian geometry, where no nonzero vector is null.
Symmetry of the connection ( $\Gamma^l_{ij} = \Gamma^l_{ji}$ ) is what let the two Christoffel terms be combined in (c) $\Leftrightarrow$ (d). For a connection with torsion the covariant and coordinate forms of the Killing equation part company.
Flatness is used only in part 4, and twice: it makes the Christoffel symbols vanish so that the Killing equation becomes $\partial_iK_j + \partial_jK_i = 0$ , and it makes $\partial_k\partial_i$ commute on the components. On a curved manifold the same differentiation produces curvature terms instead of zero, the second derivatives of $K$ are not forced to vanish, and the Killing algebra can be anything from $\tfrac12n(n+1)$ -dimensional down to zero. A metric chosen at random has no Killing fields at all: part 4 is a statement about a very special metric. What is general is the mechanism of the counting — in flat space a Killing field is determined by the pair (its value at one point, the antisymmetric matrix $\partial_iK_j$ there), which is $n + \tfrac12n(n-1)$ numbers.
Riemannian versus pseudo-Riemannian. Lee states the definition and the classification for Riemannian metrics. Parts 1, 2 and 3 as proved above use only that $g$ is a smooth symmetric $2$ -tensor field with a symmetric metric-compatible connection; positive definiteness is never invoked. They therefore apply verbatim to a Lorentz metric, which is how Killing fields are used in general relativity — but that extension is ours, not a claim made by the source.

Method(Finding the Killing fields of a metric, and using them)Original

Read off the obvious ones. If, in some chart, none of the components $g_{ij}$ depends on a particular coordinate $x^a$ , then $K = \partial/\partial x^a$ satisfies (c) at once: the first term is $\partial_a g_{ij} = 0$ and the other two vanish because $K$ has constant components. A coordinate that is absent from the metric is a symmetry, and this is the quickest way to spot one.
Solve the Killing equation for the rest. Write (c) as a linear first-order system for the $n$ unknown functions $K^k$ — there are $\tfrac12n(n+1)$ equations, one for each unordered pair $(i,j)$ — and integrate it. Differentiating the system, as in part 4, usually reduces it to a finite-dimensional problem.
Check a candidate. Given a field, form $(\mathcal{L}_Kg)_{ij} = K^k\partial_kg_{ij} + g_{kj}\partial_iK^k + g_{ik}\partial_jK^k$ and confirm every component vanishes. This is a finite computation needing no flow.
Harvest the first integrals. For each Killing field $K$ the function $g(K,\gamma') = g_{ij}K^i\dot x^j$ is constant on every geodesic. With enough independent Killing fields the geodesic equations can be integrated by quadrature instead of solved as a second-order system.
Close the algebra. Compute the brackets of the fields found; by part 2 they stay inside the set, and the structure constants identify the symmetry group.

Example(The round sphere: rotation, and Clairaut's relation as a conserved quantity)Original

Take the unit sphere $S^2$ with the round metric in spherical coordinates, $\theta$ the colatitude and $\varphi$ the longitude:

g = d\theta^2 + \sin^2\!\theta\;d\varphi^2, \qquad (g_{ij}) = \begin{pmatrix}1 & 0\\ 0 & \sin^2\theta\end{pmatrix}.

The Killing field. No component of $g$ depends on $\varphi$ , so by step 1 of the method $K = \partial/\partial\varphi$ is a Killing field. Checking (c) directly: $K^\theta = 0$ , $K^\varphi = 1$ , so every $\partial_iK^k$ vanishes and the only surviving term is $K^k\partial_kg_{ij} = \partial_\varphi g_{ij} = 0$ . Its flow is rotation about the polar axis, which is visibly an isometry.

The conserved quantity. For a geodesic $\gamma(s) = (\theta(s),\varphi(s))$ parametrised by arc length,

L = g(K,\gamma') = g_{\varphi\varphi}\,\dot\varphi = \sin^2\!\theta\;\frac{d\varphi}{ds} ,

and part 3 says $L$ is constant. Let $\psi \in [0,\pi/2]$ be the angle at which the geodesic crosses the parallel of colatitude $\theta$ , whose radius is $r = \sin\theta$ . The unit vector along that parallel is $(1/\sin\theta)\,\partial/\partial\varphi$ , so for a unit-speed curve $\cos\psi = |\sin\theta\,\dot\varphi| = r\,|\dot\varphi|$ and therefore

|L| = r\cos\psi .

That is do Carmo's (do Carmo (2016), §4-4, p. 260) Clairaut relation $r\cos\psi = \text{const}$ , which he obtains for a general surface of revolution from the first integral $f^2u' = c$ of the geodesic equations — with the rotational Killing field here supplying the reason behind it. (do Carmo writes the relation with absolute values, as above; the signed quantity $L$ is the one that part 3 produces.)

Numbers. The geodesics of the round sphere are its great circles. Take the one through $(1,0,0)$ inclined at $\alpha = 30^{\circ}$ to the equator,

\gamma(s) = (\cos s,\ \sin s\cos\alpha,\ \sin s\sin\alpha) \subset \mathbb{R}^3 ,

a unit-speed curve (its velocity is $(-\sin s, \cos s\cos\alpha, \cos s\sin\alpha)$ , of length $1$ ). In coordinates, $\cos\theta = \sin s\sin\alpha$ and $\tan\varphi = \tan s\cos\alpha$ , whence

\sin^2\!\theta = 1 - \sin^2\!s\,\sin^2\!\alpha, \qquad \frac{d\varphi}{ds} = \frac{\cos\alpha}{1 - \sin^2\!s\,\sin^2\!\alpha},

and the product telescopes to $L = \cos\alpha$ , independent of $s$ as promised. With $\alpha = 30^\circ$ , $\cos\alpha = \sqrt3/2 = 0.8660254$ . Spot checks:

$s$	$\sin^2\theta$	$d\varphi/ds$	$L$
$0$	$1.0000000$	$0.8660254$	$0.8660254$
$1$	$0.8229816$	$1.0523022$	$0.8660254$
$1.4$	$0.7572222$	$1.1436873$	$0.8660254$

At $s = 0$ the geodesic sits on the equator ( $r = 1$ ) and crosses it at the angle $\alpha$ , so Clairaut gives $r\cos\psi = 1\cdot\cos 30^\circ$ — the same number. The relation also predicts the turning point: since $\cos\psi \le 1$ , the geodesic cannot reach a parallel of radius $r < L = 0.8660254$ , that is, it stays within colatitude $60^\circ \le \theta \le 120^\circ$ , which is where the great circle inclined at $30^\circ$ indeed lives.

Example(Euclidean three-space: a rotation field and the angular momentum of a straight line)Original

On $\mathbb{R}^3$ with $\bar g = dx^2 + dy^2 + dz^2$ , take the rotation field about the $z$ -axis,

K = x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x},

one of the three rotations in the basis of part 4. Its components are $K^1 = -y$ , $K^2 = x$ , $K^3 = 0$ ; the metric has constant components, so (c) reduces to $\partial_iK^j + \partial_jK^i = 0$ , and the only possibly nonzero entry is $\partial_1K^2 + \partial_2K^1 = 1 + (-1) = 0$ . So $K$ is Killing.

The conserved quantity. The geodesics of Euclidean space are the straight lines $\gamma(s) = p + sv$ with $|v| = 1$ . Along such a line,

\bar g(K,\gamma') = \bigl(-(p_y + sv_y),\ p_x + sv_x,\ 0\bigr)\cdot(v_x,v_y,v_z) = p_xv_y - p_yv_x + s\,(v_xv_y - v_yv_x) = p_xv_y - p_yv_x ,

the $s$ -dependence cancelling identically. The constant is the $z$ -component of $p\times v$ : the Killing field of rotations about an axis yields conservation of angular momentum about that axis.

Numbers. Take $p = (2,1,3)$ and $v = \tfrac13(1,2,2)$ , which has $|v| = \tfrac13\sqrt{1+4+4} = 1$ . The predicted constant is

p_xv_y - p_yv_x = 2\cdot\tfrac23 - 1\cdot\tfrac13 = \tfrac43 - \tfrac13 = 1 .

At $s = 3$ : $\gamma = (3,3,5)$ , $K_\gamma = (-3,3,0)$ , and $K_\gamma\cdot v = -3\cdot\tfrac13 + 3\cdot\tfrac23 = -1 + 2 = 1$ . At $s = -1.5$ : $\gamma = (1.5,\,0,\,2)$ , $K_\gamma = (0,\,1.5,\,0)$ , and $K_\gamma\cdot v = 1.5\cdot\tfrac23 = 1$ . The same value, as part 3 requires.

How many are there. By part 4 the Killing algebra of $\mathbb{R}^3$ has dimension $\tfrac12\cdot3\cdot4 = 6$ : three translations and three rotations. Bracketing two of them stays inside, in agreement with part 2 — for instance, with $K_z = x\partial_y - y\partial_x$ and $K_x = y\partial_z - z\partial_y$ ,

[K_z, K_x] = x\partial_z - z\partial_x = K_y ,

the third rotation of the basis. The six fields close on the Lie algebra of the Euclidean group of rigid motions.

ApplicationOriginal

Killing fields are how symmetry is stated when there are no preferred coordinates, which is why they are the working language of general relativity. A spacetime with a timelike Killing field $\partial/\partial t$ is stationary, and part 3 converts it into the conserved energy of a freely falling particle; an axial Killing field $\partial/\partial\varphi$ gives the conserved angular momentum. In the Schwarzschild metric these two alone reduce the four geodesic equations to a single first-order equation for the radius, which is what makes the perihelion precession and light bending calculable in closed form. The number of independent Killing fields is a coordinate-free measure of how symmetric a spacetime is, and the cosmological models are picked out by demanding a large one.

The same mechanism appears wherever a metric has a symmetry. On a surface of revolution, the rotational Killing field gives Clairaut's relation and with it the qualitative behaviour of geodesics — which parallels they can reach and which they must turn back from, as in do Carmo's (do Carmo (2016), Example 6, pp. 261–262) analysis of the paraboloid, where every non-meridian geodesic is forced to cross itself infinitely often. In optics and acoustics, where ray paths are the geodesics of an effective metric built from the refractive index, a Killing field is a direction in which the medium is uniform and the conserved quantity is Snell's invariant.

Structurally, part 3 is Noether's theorem for the geodesic Lagrangian: the flow of $K$ is a one-parameter group of transformations leaving $\tfrac12g_{ij}\dot x^i\dot x^j$ invariant, and the Noether charge it produces is $g(K,\gamma')$ . Read in the Hamiltonian language of symplectic manifolds, the function $p \mapsto p(K)$ on $T^{*}M$ Poisson-commutes with the kinetic-energy Hamiltonian, and the Lie algebra of part 2 maps to a Lie algebra of such functions under the Poisson bracket. Finally, Killing fields are the practical route to separating variables: a coordinate that a Killing field translates is one in which the metric components do not appear, and it is exactly the coordinate that separates out of the Laplace–Beltrami operator.