Differential forms

Exterior calculus and the general Stokes theorem.

In this chapter 5

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Exterior calculus

Exterior calculus

Definition

Differential forms

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Builds the exterior algebra of alternating tensors on a vector space, defines the wedge product in the determinant convention and the differential $k$ -forms it generates on a manifold, and gives the pullback rule that makes forms the objects a change of coordinates transforms correctly.

Prerequisites: Tensors and their transformation laws, Determinants, Smooth manifolds, Tangent vectors and vector fields, Tangent and vector bundles

Definition(Alternating tensors, the wedge product, the exterior algebra)AdaptedAdapted from Lee (2013), Ch. 14, pp. 350–357; Hassani (2013), §26.3, pp. 795–797

Let $V$ be a real vector space of dimension $n$ . A covariant $k$ -tensor $\alpha$ on $V$ (see tensors and their transformation laws) is alternating if interchanging any two of its arguments reverses its sign. Equivalently — and this is the form used most often below — $\alpha$ vanishes on every $k$ -tuple that has a repeated entry, and equivalently again it vanishes on every linearly dependent $k$ -tuple. The alternating $k$ -tensors form a vector space $\Lambda^k(V^*)$ , whose elements are also called $k$ -covectors. Every $0$ -tensor and every $1$ -tensor is alternating, so $\Lambda^0(V^*) = \mathbb{R}$ and $\Lambda^1(V^*) = V^*$ .

Any covariant $k$ -tensor can be made alternating by averaging over the symmetric group $S_k$ with signs:

(\operatorname{Alt}\alpha)(v_1,\dots,v_k) \;=\; \frac{1}{k!}\sum_{\sigma \in S_k} (\operatorname{sgn}\sigma)\, \alpha\bigl(v_{\sigma(1)},\dots,v_{\sigma(k)}\bigr) .

$\operatorname{Alt}\alpha$ is alternating, and $\operatorname{Alt}\alpha = \alpha$ exactly when $\alpha$ already is.

For $\alpha \in \Lambda^k(V^*)$ and $\beta \in \Lambda^l(V^*)$ the wedge product (or exterior product) is the $(k+l)$ -covector

\alpha \wedge \beta \;=\; \frac{(k+l)!}{k!\,l!}\ \operatorname{Alt}(\alpha \otimes \beta) .

The combinatorial factor is a normalisation, chosen so that the determinant formula of the statement below comes out with no stray factorials; the note records the competing choice. For a $0$ -covector, that is a number $c$ , the product $c \wedge \alpha$ is just $c\alpha$ .

Fix a basis $(E_1,\dots,E_n)$ of $V$ and the dual basis $(\varepsilon^1,\dots,\varepsilon^n)$ of $V^*$ . For a multi-index $I = (i_1,\dots,i_k)$ with entries in $\{1,\dots,n\}$ put

\varepsilon^{I} \;=\; \varepsilon^{i_1}\wedge\cdots\wedge\varepsilon^{i_k}, \qquad \varepsilon^{I}(v_1,\dots,v_k) \;=\; \det\bigl[\varepsilon^{i_a}(v_b)\bigr]_{a,b=1}^{k} .

(That the two descriptions agree is part of the statement below.) Call $I$ increasing if $i_1 < \cdots < i_k$ , and write $\sum'_I$ for a sum over increasing multi-indices only. The direct sum

\Lambda(V^*) \;=\; \bigoplus_{k=0}^{n} \Lambda^{k}(V^*), \qquad \dim \Lambda(V^*) = 2^n ,

with $\wedge$ as its multiplication, is the exterior algebra (or Grassmann algebra) of $V$ .

Note(Three ways the same algebra is set up, and how the conventions differ)AdaptedAdapted from Lee (2013), pp. 357–358, (14.9)–(14.10); Arfken, Weber & Harris (2013), §4.5, (4.80), p. 234; Dummit & Foote (2004), §11.5, pp. 446–448

The library sets up the same object in three ways, and the differences are worth keeping straight before any computation.

The determinant convention, used above and throughout this handbook, carries the factor $(k+l)!/(k!\,l!)$ . Lee (Lee (2013), pp. 357–358) adopts it for the whole of his book, and notes that it is the common choice in introductory differential geometry; Hassani (Hassani (2013), Def. 26.3.1, p. 794) uses the same normalisation, and warns in a footnote there that other authors choose other coefficients.
The Alt convention sets $\alpha \mathbin{\underline{\wedge}} \beta = \operatorname{Alt}(\alpha\otimes\beta)$ with no factor at all. Lee records that with this choice the determinant formula acquires a $1/k!$ , becoming $(\omega^1\mathbin{\underline{\wedge}}\cdots\mathbin{\underline{\wedge}}\omega^k)(v_1,\dots,v_k) = \tfrac{1}{k!}\det[\omega^i(v_j)]$ , and that it is more usual in complex differential geometry. Mixing the two silently is the commonest source of stray factorials in this subject.
The axiomatic route, which is how Arfken, Weber and Harris (Arfken, Weber & Harris (2013), (4.80), p. 234) introduce it for physicists: one simply demands that $\wedge$ be bilinear and associative and that $dx_i \wedge dx_j = - dx_j \wedge dx_i$ , and computes. No normalisation has to be chosen, because the basis products are taken as given rather than built from $\otimes$ .
The algebraic route: Dummit and Foote (Dummit & Foote (2004), §11.5, pp. 446–447) define $\Lambda(M)$ , for a module $M$ over a commutative ring, as the tensor algebra $T(M)$ divided by the two-sided ideal generated by the elements $m \otimes m$ . Anticommutativity is then a consequence rather than an axiom: expanding $0 = (m+m')\wedge(m+m')$ and cancelling the two square terms leaves $m \wedge m' = -m'\wedge m$ . Their Thm 36 gives the universal property — every alternating $k$ -multilinear map out of $M^k$ factors uniquely through $\Lambda^k(M)$ — which is what makes the construction canonical. They also warn that anticommutativity on generators does not mean $ab = -ba$ for arbitrary elements of $\Lambda(M)$ ; the correct general rule is the degree-dependent sign below.

The four descriptions agree on everything in this unit. Where a formula below carries a factorial, it is the determinant convention that put it there.

Definition(Wedge products, and a basis for the alternating tensors)AdaptedAdapted from Lee (2013), Prop. 14.8, p. 353 and Prop. 14.11, pp. 356–357; Hassani (2013), Thm 26.3.3 and (26.22)–(26.23), pp. 796–797

Let $V$ be an $n$ -dimensional real vector space.

Algebra. $\wedge$ is bilinear and associative, and for $\alpha \in \Lambda^k(V^*)$ , $\beta \in \Lambda^l(V^*)$ , $\alpha \wedge \beta \;=\; (-1)^{kl}\, \beta \wedge \alpha .$ In particular $\alpha\wedge\alpha = 0$ whenever $k$ is odd, but not in general when $k$ is even.
Shuffle formula. With the same $\alpha,\beta$ , $(\alpha\wedge\beta)(v_1,\dots,v_{k+l}) \;=\; \sum_{\sigma} (\operatorname{sgn}\sigma)\, \alpha\bigl(v_{\sigma(1)},\dots,v_{\sigma(k)}\bigr)\, \beta\bigl(v_{\sigma(k+1)},\dots,v_{\sigma(k+l)}\bigr),$ the sum running over the $\binom{k+l}{k}$ permutations $\sigma$ of $\{1,\dots,k+l\}$ with $\sigma(1)<\cdots<\sigma(k)$ and $\sigma(k+1)<\cdots<\sigma(k+l)$ (the $(k,l)$ -shuffles).
Determinant formula. For covectors $\omega^1,\dots,\omega^k \in V^*$ and vectors $v_1,\dots,v_k \in V$ , $\bigl(\omega^1\wedge\cdots\wedge\omega^k\bigr)(v_1,\dots,v_k) \;=\; \det\bigl[\omega^{i}(v_j)\bigr]_{i,j=1}^{k} .$
Basis. For $1 \le k \le n$ the $k$ -covectors $\varepsilon^{I}$ with $I$ increasing of length $k$ form a basis of $\Lambda^k(V^*)$ , so $\dim \Lambda^{k}(V^*) = \binom{n}{k} = \frac{n!}{k!\,(n-k)!},$ and $\Lambda^k(V^*) = \{0\}$ for $k > n$ . The expansion of $\alpha \in \Lambda^k(V^*)$ is $\alpha \;=\; {\sum_I}' \alpha_I\, \varepsilon^{I}, \qquad \alpha_I = \alpha(E_{i_1},\dots,E_{i_k}) .$

Idea of the proofOriginal

One identity does all the work. Expanding the definition of $\alpha\wedge\beta$ produces a sum over all $(k+l)!$ permutations, but $\alpha$ and $\beta$ are themselves alternating, so permuting the first $k$ arguments among themselves or the last $l$ among themselves changes nothing but a sign that is immediately cancelled. Each term therefore occurs $k!\,l!$ times, which is exactly the factor the definition divided out; what survives is one term per shuffle. Once the product is written as a sum over shuffles, associativity is visible (a triple product is the sum over shuffles into three blocks, however it is bracketed), anticommutativity is the sign of the permutation that swaps the two blocks past each other, and for $k$ one-forms the shuffles are all of $S_k$ and the sum is Leibniz's expansion of a determinant. The basis claim is then a matter of evaluation: an alternating tensor is determined by its values on increasing tuples of basis vectors, and $\varepsilon^I$ is built to have value $1$ on $I$ and $0$ on every other increasing tuple.

Proof(The shuffle formula, and the four claims that follow from it)AdaptedAdapted from Lee (2013), Prop. 14.8, p. 353; Lee (2013), Prop. 14.11 with Lemma 14.10, pp. 355–357

Step 1: the shuffle formula (claim 2). Write out the definition:

(\alpha\wedge\beta)(v_1,\dots,v_{k+l}) = \frac{1}{k!\,l!}\sum_{\pi \in S_{k+l}} (\operatorname{sgn}\pi)\, \alpha\bigl(v_{\pi(1)},\dots,v_{\pi(k)}\bigr)\beta\bigl(v_{\pi(k+1)},\dots,v_{\pi(k+l)}\bigr),

the $(k+l)!$ of the normalisation having cancelled the $1/(k+l)!$ inside $\operatorname{Alt}$ . Group the permutations by the ordered pair of sets $\bigl(\{\pi(1),\dots,\pi(k)\},\{\pi(k+1),\dots,\pi(k+l)\}\bigr)$ — equivalently, by the first block alone, which determines the second. The order matters: when $k = l$ the two blocks may be interchanged, and reading the pair as unordered would merge two groups into one of $2\,k!\,l!$ permutations containing two shuffles. With the pair read as ordered, each group contains exactly $k!\,l!$ permutations and exactly one shuffle $\sigma$ ; and if $\pi$ lies in the group of $\sigma$ then $\pi = \sigma\circ(\mu\times\nu)$ for a permutation $\mu$ of the first block and $\nu$ of the second. Because $\alpha$ and $\beta$ are alternating, reordering the arguments by $\mu$ and $\nu$ multiplies the product of values by $(\operatorname{sgn}\mu)(\operatorname{sgn}\nu) = \operatorname{sgn}\pi \cdot \operatorname{sgn}\sigma$ , so every term in a group equals the term of its shuffle. The group therefore contributes $k!\,l!$ copies of the shuffle term, and the prefactor cancels them.

Step 2: associativity and anticommutativity (claim 1). Bilinearity is immediate from the definition, since $\otimes$ is bilinear and $\operatorname{Alt}$ is linear. Applying step 1 twice to $(\alpha\wedge\beta)\wedge\gamma$ with $\gamma \in \Lambda^m(V^*)$ gives the sum, over all ways of splitting $\{1,\dots,k+l+m\}$ into an increasing $k$ -block, an increasing $l$ -block and an increasing $m$ -block, of the signed product of the three values; the same computation applied to $\alpha\wedge(\beta\wedge\gamma)$ gives the identical sum, so the two bracketings agree. For anticommutativity, the shuffles for $\beta\wedge\alpha$ are obtained from those for $\alpha\wedge\beta$ by moving the $l$ -block in front of the $k$ -block. That permutation is a composition of $kl$ transpositions — each of the $l$ entries must pass each of the $k$ entries — so each sign changes by $(-1)^{kl}$ .

Step 3: the determinant formula (claim 3). Take $k$ covectors $\omega^1,\dots,\omega^k$ , each of degree $1$ . Iterating step 1, the $(1,1,\dots,1)$ -shuffles are all of $S_k$ (an increasing block of length one is no constraint), so

\bigl(\omega^1\wedge\cdots\wedge\omega^k\bigr)(v_1,\dots,v_k) = \sum_{\sigma\in S_k} (\operatorname{sgn}\sigma)\, \omega^1(v_{\sigma(1)})\cdots\omega^k(v_{\sigma(k)}),

which is the Leibniz expansion of $\det[\omega^i(v_j)]$ of determinants. In particular the two descriptions of $\varepsilon^I$ given in the definition agree, and $\varepsilon^{I}(E_{j_1},\dots,E_{j_k})$ is the determinant of a matrix of Kronecker deltas: it is $0$ if $I$ or $J$ has a repeated entry or if $J$ is not a rearrangement of $I$ , and otherwise it is the sign of the rearrangement. For increasing $I$ and $J$ this says $\varepsilon^I(E_{j_1},\dots,E_{j_k}) = 1$ if $J = I$ and $0$ otherwise.

Step 4: the $\varepsilon^I$ span (claim 4). Let $\alpha \in \Lambda^k(V^*)$ and set $\alpha_I = \alpha(E_{i_1},\dots,E_{i_k})$ for increasing $I$ . Put $\tilde\alpha = \sum'_I \alpha_I \varepsilon^I$ . By the last sentence of step 3, $\alpha$ and $\tilde\alpha$ take the same value on every increasing tuple of basis vectors. They agree on all the others as well: on a tuple with a repeated entry both vanish, since both are alternating; and a tuple of distinct basis vectors is a rearrangement of an increasing one, which multiplies both values by the same sign. Two multilinear maps that agree on all tuples of basis vectors are equal, so $\alpha = \tilde\alpha$ .

Step 5: the $\varepsilon^I$ are independent, and the dimension. Suppose $\sum'_I c_I \varepsilon^I = 0$ . Evaluating at $(E_{j_1},\dots,E_{j_k})$ for an increasing $J$ picks out $c_J$ , so every coefficient vanishes. The increasing multi-indices of length $k$ in $\{1,\dots,n\}$ are the $k$ -element subsets, of which there are $\binom{n}{k}$ ; for $k > n$ there are none, which is why there are no non-zero $k$ -covectors above degree $n$ . $\square$

Definition(Differential forms, and the pullback)AdaptedAdapted from Lee (2013), pp. 359–361, Lemma 14.16; Hassani (2013), Def. 28.5.1–28.5.2 and (28.37), p. 888

Let $M$ be a smooth $n$ -manifold (smooth manifolds). Applying the construction above to each tangent space $T_pM$ (tangent vectors and vector fields) gives the bundle $\Lambda^k T^*M$ whose fibre at $p$ is $\Lambda^k(T_p^*M)$ ; it is a smooth vector bundle of rank $\binom{n}{k}$ (tangent and vector bundles). A differential $k$ -form is a smooth section $\omega$ of this bundle: an assignment, to each point $p$ , of an alternating $k$ -tensor $\omega_p$ on $T_pM$ , depending smoothly on $p$ . The space of smooth $k$ -forms is written $\Omega^k(M)$ , and $\Omega^0(M)$ is just the smooth functions. The wedge product of forms is taken pointwise, $(\omega\wedge\eta)_p = \omega_p \wedge \eta_p$ , so it sends $\Omega^k \times \Omega^l$ into $\Omega^{k+l}$ .

In a chart with coordinates $(x^1,\dots,x^n)$ the differentials $dx^1,\dots,dx^n$ are the coframe dual to $\partial/\partial x^1,\dots,\partial/\partial x^n$ , and writing $dx^{I} = dx^{i_1}\wedge\cdots\wedge dx^{i_k}$ the statement above gives the local expression

\omega \;=\; {\sum_I}'\, \omega_I\, dx^{I}, \qquad \omega_I \;=\; \omega\!\left(\frac{\partial}{\partial x^{i_1}},\dots,\frac{\partial}{\partial x^{i_k}}\right),

with smooth coefficient functions $\omega_I$ . On an open set of $\mathbb{R}^3$ , a $0$ -form is a function, a $1$ -form is $P\,dx + Q\,dy + R\,dz$ , a $2$ -form is $F\,dy\wedge dz + G\,dz\wedge dx + H\,dx\wedge dy$ , and a $3$ -form is $K\,dx\wedge dy\wedge dz$ ; there are no non-zero $4$ -forms.

Let $F : M \to N$ be smooth and $\omega \in \Omega^k(N)$ . The pullback $F^*\omega \in \Omega^k(M)$ is defined by feeding the differential $dF_p : T_pM \to T_{F(p)}N$ (see pushforwards, pullbacks and immersions) each argument in turn:

(F^*\omega)_p(v_1,\dots,v_k) \;=\; \omega_{F(p)}\bigl(dF_p(v_1),\dots,dF_p(v_k)\bigr), \qquad v_1,\dots,v_k \in T_pM ,

and $F^*f = f\circ F$ for a $0$ -form $f$ . No invertibility of $F$ is needed: forms pull back along any smooth map, which is exactly what vector fields do not do.

Definition(Rules for the pullback)AdaptedAdapted from Lee (2013), Lemma 14.16, p. 361 and Prop. 14.20 with Cor. 14.21, pp. 361–362; Hassani (2013), Prop. 26.3.5, p. 796 and (28.37), p. 888

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

$F^* : \Omega^k(N) \to \Omega^k(M)$ is linear, $F^*(\omega\wedge\eta) = (F^*\omega)\wedge(F^*\eta)$ , and $(G\circ F)^* = F^*\circ G^*$ ; the identity map pulls back to the identity.
Computation rule. In charts $(x^i)$ on $M$ and $(y^j)$ on $N$ , writing $F^j = y^j \circ F$ for the components of $F$ , $F^*\Bigl({\sum_J}'\, \omega_J \, dy^{j_1}\wedge\cdots\wedge dy^{j_k}\Bigr) \;=\; {\sum_J}'\, (\omega_J\circ F)\ dF^{j_1}\wedge\cdots\wedge dF^{j_k} .$ In words: substitute, and differentiate each coordinate function.
Top degree. If $\dim M = \dim N = n$ and $u$ is a continuous function on $N$ , then on the overlap of the two charts $F^*\bigl(u \, dy^1\wedge\cdots\wedge dy^n\bigr) \;=\; (u\circ F)\,\bigl(\det DF\bigr)\, dx^1\wedge\cdots\wedge dx^n ,$ where $DF = [\partial F^j/\partial x^i]$ is the Jacobian matrix in these coordinates. Taking $F$ to be the identity read in two charts $(x^i)$ and $(\tilde x^j)$ on $M$ gives the change-of-coordinates rule $d\tilde x^{1}\wedge\cdots\wedge d\tilde x^{n} \;=\; \det\!\left[\frac{\partial \tilde x^{j}}{\partial x^{i}}\right] dx^{1}\wedge\cdots\wedge dx^{n} .$

Proof(The top-degree formula)AdaptedAdapted from Lee (2013), Prop. 14.20, pp. 361–362

Parts 1 and 2 are read off from the definition, which is why the source leaves them as an exercise: each is a statement about the values on $k$ -tuples of tangent vectors, and $dF_p$ is applied to each slot separately. We prove part 3, which is the one that produces a determinant out of nowhere and is used in every change of variables below.

Step 0: the coefficient rides along. Part 3 asks only that $u$ be continuous, while part 2 is a statement about smooth forms. The gap is not real, because the pullback is defined pointwise: for any function $u$ on $N$ and any $k$ -form $\eta$ ,

\bigl(F^*(u\,\eta)\bigr)_p = u\bigl(F(p)\bigr)\,(F^*\eta)_p ,

straight from the definition, as $u(F(p))$ is a scalar multiplying the value of $\eta_{F(p)}$ . So we apply part 2 to the smooth form $\eta = dy^1\wedge\cdots\wedge dy^n$ and put the factor $u\circ F$ back at the end; continuity of $u$ is then needed only to make $F^*(u\,\eta)$ a continuous section rather than a rough one.

Step 1: one evaluation suffices. At each point the fibre $\Lambda^n(T_p^*M)$ is one-dimensional, spanned by $dx^1\wedge\cdots\wedge dx^n$ , by the dimension count $\binom{n}{n} = 1$ . Two $n$ -forms on an $n$ -manifold are therefore equal as soon as they agree on the single frame $(\partial/\partial x^1,\dots,\partial/\partial x^n)$ .

Step 2: the left-hand side. By step 0 and part 2, $F^*(u\,dy^1\wedge\cdots\wedge dy^n) = (u\circ F)\,F^*(dy^1\wedge\cdots\wedge dy^n) = (u\circ F)\,dF^1\wedge\cdots\wedge dF^n$ . Each $dF^j$ is a $1$ -form, so the determinant formula of claim 3 of the statement applies with $\omega^j = dF^j$ and $v_i = \partial/\partial x^i$ :

\bigl(dF^1\wedge\cdots\wedge dF^n\bigr)\!\left(\frac{\partial}{\partial x^1},\dots,\frac{\partial}{\partial x^n}\right) = \det\!\left[dF^{j}\!\left(\frac{\partial}{\partial x^{i}}\right)\right] = \det\!\left[\frac{\partial F^{j}}{\partial x^{i}}\right] = \det DF ,

so the left-hand side returns $(u\circ F)\det DF$ .

Step 3: the right-hand side. By the same determinant formula, $dx^1\wedge\cdots\wedge dx^n$ applied to $(\partial/\partial x^1,\dots,\partial/\partial x^n)$ is the determinant of the identity matrix, namely $1$ . So the right-hand side also returns $(u\circ F)\det DF$ , and by step 1 the two forms are equal. $\square$

Remark(Where the hypotheses are used)Original

Finite dimension enters only at claims 4 of the statement: the count $\binom{n}{k}$ , the vanishing of $\Lambda^k$ above degree $n$ , and hence the one-dimensionality of the top fibre that step 1 of the proof relies on. The algebra of $\wedge$ itself needs no dimension hypothesis.
Alternation of both factors is used exactly once, at step 1 of the proof of the shuffle formula, to say that the $k!\,l!$ permutations within a group all give the same signed term. Drop it and the shuffle formula is false; with a general $\alpha$ the sum does not collapse.
The sign rule is degree-dependent, not uniform. $\alpha\wedge\alpha = 0$ holds for odd $k$ but can fail for even $k$ : in $\mathbb{R}^4$ with $\alpha = dx^1\wedge dx^2 + dx^3\wedge dx^4$ one finds $\alpha\wedge\alpha = 2\,dx^1\wedge dx^2\wedge dx^3\wedge dx^4 \ne 0$ . (This $\alpha$ is also an example of a $2$ -covector that is not a single wedge of $1$ -covectors.) Dummit and Foote (Dummit & Foote (2004), §11.5, p. 447) make the same warning in the algebraic setting.
The normalisation is a hypothesis too. Every factorial-free formula above — the determinant formula and the Jacobian in part 3 — presupposes the determinant convention of the definition. In the Alt convention both acquire factorials: with $\varepsilon^{I}$ still defined by the determinant, Lee records that the concatenation rule becomes $\varepsilon^{I}\mathbin{\underline{\wedge}}\varepsilon^{J} = \frac{k!\,l!}{(k+l)!}\,\varepsilon^{IJ}$ rather than $\varepsilon^{I}\wedge\varepsilon^{J} = \varepsilon^{IJ}$ .
Smoothness is what makes $F^*\omega$ a form rather than a rough section: $dF_p$ must exist at every $p$ , and the coefficients of $F^*\omega$ are built from first derivatives of $F$ and the coefficients of $\omega$ , so $F$ of class $C^r$ and $\omega$ of class $C^r$ give $F^*\omega$ of class $C^{r-1}$ .
No invertibility is needed to pull back, and this asymmetry is the whole reason forms are the natural things to integrate. A vector field can be pushed forward along a diffeomorphism only; a form can be pulled back along a map that collapses dimensions, in which case $F^*\omega$ is simply $0$ in high enough degree. In part 3 the map is between manifolds of equal dimension but need not be invertible: $\det DF$ is allowed to vanish.
Independence test. Hassani (Hassani (2013), Prop. 26.3.8, p. 797) records the converse use of all this: vectors $v_1,\dots,v_p$ are linearly independent if and only if $v_1\wedge\cdots\wedge v_p \ne 0$ . It is the determinant criterion for independence, restated so that no choice of basis or square matrix is needed.

Method(Computing with forms)Original

Reduce to increasing indices. Use anticommutativity of the $1$ -forms, $dx^i\wedge dx^j = -dx^j\wedge dx^i$ , to sort every basis product into increasing order, tracking the sign; any product with a repeated differential is zero. Collect coefficients. In $\mathbb{R}^3$ it is often more convenient to use the cyclic order $dy\wedge dz$ , $dz\wedge dx$ , $dx\wedge dy$ , as Arfken, Weber and Harris (Arfken, Weber & Harris (2013), Example 4.5.1, pp. 234–235) do, because then all the vector-calculus dictionary entries come out with plus signs.
Move functions to the front. Coefficients are $0$ -forms and commute with everything, so $f\,dx \wedge g\,dy = fg\,dx\wedge dy$ .
To evaluate a $k$ -form on $k$ vectors, expand it over increasing multi-indices; every basis term is then a determinant, $dx^{I}(v_1,\dots,v_k) \;=\; \det\bigl[v_b^{\,i_a}\bigr]_{a,b=1}^{k} ,$ where $v_b^{\,i}$ is the $i$ th coordinate component of $v_b$ . In practice: write the $n$ components of each vector in a column, delete every row whose index is absent from $I$ , take the determinant of the square array left behind, and weight it by $\omega_I$ . Summing over the increasing $I$ gives $\omega(v_1,\dots,v_k)$ .
To pull back, substitute the map into the coefficients, replace each $dy^j$ by $dF^j = \sum_i (\partial F^j/\partial x^i)\,dx^i$ , expand, and sort by step 1. Never compute a Jacobian separately: part 3 of the statement says the expansion produces it.
Sanity checks. The degree must be right; a $k$ -form on an $n$ -manifold with $k>n$ is zero; and if the map factors through a manifold of dimension less than $k$ , the pullback is zero.

Example(A pullback computed two ways)Original

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

F(u,v) = \bigl(x,y,z\bigr) = \bigl(u^2 - v^2,\ 2uv,\ u+v\bigr),

and let $\omega = z\,dx\wedge dy + x\,dy\wedge dz$ be a $2$ -form on $\mathbb{R}^3$ . We compute $F^*\omega$ and evaluate it at $(u,v) = (1,2)$ .

Route 1: substitute and expand. The differentials of the components are

dx = 2u\,du - 2v\,dv, \qquad dy = 2v\,du + 2u\,dv, \qquad dz = du + dv .

Wedging, and using $du\wedge du = dv\wedge dv = 0$ and $dv\wedge du = -du\wedge dv$ ,

dx\wedge dy = \bigl(4u^2 + 4v^2\bigr)\,du\wedge dv , \qquad dy\wedge dz = \bigl(2v - 2u\bigr)\,du\wedge dv .

Substituting the coefficients $z = u+v$ and $x = u^2-v^2$ and collecting,

F^*\omega = \Bigl[4(u+v)(u^2+v^2) - 2(u-v)(u^2-v^2)\Bigr] du\wedge dv = 2(u+v)\Bigl[2(u^2+v^2) - (u-v)^2\Bigr] du\wedge dv ,

and the bracket is $u^2 + 2uv + v^2 = (u+v)^2$ . Hence

F^*\omega = 2\,(u+v)^3\, du\wedge dv .

At $(u,v) = (1,2)$ the coefficient is $2\cdot 3^3 = \boxed{54}$ .

Route 2: evaluate through the definition. By the definition of the pullback, the coefficient at $(1,2)$ is $\omega_{F(1,2)}\bigl(dF(\partial_u), dF(\partial_v)\bigr)$ . Here $F(1,2) = (1-4,\,4,\,3) = (-3,4,3)$ , and

dF(\partial_u) = (2u,\,2v,\,1) = (2,4,1), \qquad dF(\partial_v) = (-2v,\,2u,\,1) = (-4,2,1) .

At the point $(-3,4,3)$ the form is $\omega = 3\,dx\wedge dy - 3\,dy\wedge dz$ . Using step 3 of the method, with $a = (2,4,1)$ and $b = (-4,2,1)$ ,

(dx\wedge dy)(a,b) = \det\begin{pmatrix} 2 & -4 \\ 4 & 2\end{pmatrix} = 20, \qquad (dy\wedge dz)(a,b) = \det\begin{pmatrix} 4 & 2 \\ 1 & 1\end{pmatrix} = 2 ,

so $\omega(a,b) = 3(20) - 3(2) = 54$ , as before.

Reading the answer. Why a zero of order three at $v = -u$ ? One factor is forced, the other two are not, and it is worth keeping them apart.

The forced factor is the first. Along $v = -u$ the map sends $(u,v)$ to $\bigl(u^2-v^2,\,2uv,\,u+v\bigr) = (0,-2u^2,0)$ , and the two coefficients of $\omega$ are $z$ and $x$ , both of which vanish there. So $\omega$ is the zero $2$ -covector at every image point of that line, and its pullback must vanish along the line whatever the tangent vectors do. (The two vectors $dF(\partial_u) = (2u,-2u,1)$ and $dF(\partial_v) = (2u,2u,1)$ do stay independent for $u \ne 0$ : the pullback vanishes because of the form, not because the map degenerates.) That argument gives a simple zero and nothing more.

The other two factors are arithmetic. Each of the two terms carries one factor of $u+v$ — the first from $z\circ F = u+v$ , the second from $x\circ F = (u+v)(u-v)$ — so

F^*\omega = (u+v)\Bigl[\,4(u^2+v^2) - 2(u-v)^2\,\Bigr]\,du\wedge dv .

On the line $v = -u$ the two terms inside the bracket are $8u^2$ and $-8u^2$ : they cancel. A quadratic form that vanishes on the line $u + v = 0$ is divisible by $u+v$ , and here the quotient is $2(u+v)$ a second time, since $4(u^2+v^2) - 2(u-v)^2 = 2(u+v)^2$ . Nothing in the general theory predicts that second coincidence; it is special to this $\omega$ and this $F$ . What is general is that $F^*\omega$ at a point depends on $\omega$ only at the image point, so a form can be pulled back through regions where the map is badly behaved and still say something.

ApplicationOriginal

Oriented area and volume elements. The reason $dx\,dy$ in a double integral really behaves like $dx\wedge dy$ is part 3 of the statement: under a change of variables it picks up $\det DF$ , which is the Jacobian of multiple integrals and change of variables — except that the form keeps the sign of the Jacobian, where the classical change-of-variables formula takes its absolute value. The sign is the orientation, and carrying it is what allows forms to be integrated intrinsically, as integration of forms and the general Stokes theorem does. Arfken, Weber and Harris (Arfken, Weber & Harris (2013), §4.5, p. 235) make the same point for physicists: lengths, areas and volumes built from wedge products are oriented quantities.

Cross products and triple products are three-dimensional accidents. For $1$ -forms $A = A_x dx + A_y dy + A_z dz$ and $B$ likewise, expanding $A\wedge B$ gives the three components of the cross product as the coefficients of $dy\wedge dz$ , $dz\wedge dx$ , $dx\wedge dy$ , and $A\wedge B\wedge C$ is the scalar triple product times $dx\wedge dy\wedge dz$ . Both expansions are displayed in the course of Arfken, Weber and Harris's (Arfken, Weber & Harris (2013), Example 4.5.3, pp. 237–238) Example 4.5.3, but the identifications they state as (4.83) and (4.84) have the Hodge star applied: $*(A\wedge B)$ is the $1$ -form whose coefficients are the components of $A\times B$ , and $*(A\wedge B\wedge C) = A\cdot(B\times C)$ as a number. The un-starred reading above is a correspondence of coefficients, not an equality between a $2$ -form and a vector. The wedge product is what survives in other dimensions: $A\wedge B$ exists in every dimension, but only in $\mathbb{R}^3$ does $\binom{3}{2} = 3$ let its result be disguised as a vector. That coincidence, and the reflection behaviour it carries with it, is the subject of pseudotensors and the Levi-Civita tensor, and turning a $2$ -form back into a vector in general requires the extra structure of the Hodge star.

Phase space. On the phase space of a mechanical system with $n$ degrees of freedom, the $2$ -form $\omega = \sum_{i=1}^{n} dp_i \wedge dq^i$ is the object that Hamilton's equations preserve, and $\omega^{\wedge n}/n!$ is the Liouville volume whose conservation is Liouville's theorem. This is the content of symplectic manifolds and Hamiltonian vector fields, which builds on the canonical equations. The antisymmetry is not decoration: it is what makes $\omega$ pair a coordinate with its conjugate momentum rather than with itself.

Electromagnetism. The electromagnetic field is a $2$ -form $F$ on spacetime, with the electric field in the components containing $dt$ and the magnetic field in the purely spatial ones; the whole of Maxwell's theory is two equations about $F$ . Writing it this way is what makes its behaviour under a change of frame automatic, since a Lorentz transformation acts by pullback. See the Hodge star and Maxwell's equations.

Thermodynamics and other state spaces. On a space of thermodynamic states, the work and heat increments are $1$ -forms, not differentials of functions; the distinction is invisible in the notation $\delta W$ but exact in the language of forms, and it is the subject of the exterior derivative. The same remark applies wherever a "differential" appears in an applied subject without a function behind it.

Definition

The exterior derivative

Adapted

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Defines the operator $d$ that raises the degree of a differential form by one, characterises it as the unique antiderivation extending the differential of a function and squaring to zero, proves that it commutes with every pullback, and reads off that gradient, curl and divergence in $\mathbb{R}^3$ are its three instances.

Prerequisites: Differential forms, Divergence, curl and the Laplacian, Gradient and directional derivative, Smooth manifolds, Tangent vectors and vector fields, Pushforwards, pullbacks and immersions, Partitions of unity

Definition(The exterior derivative)AdaptedAdapted from Lee (2013), (14.19)–(14.20), p. 363 and (14.22), p. 365; Hassani (2013), Thm 28.5.3, p. 889

Let $U$ be an open subset of $\mathbb{R}^n$ , or of the closed half-space $\mathbb{H}^n = \{x\in\mathbb{R}^n : x^n \ge 0\}$ — the second case is needed because the charts of a manifold with boundary land there — and let $\omega \in \Omega^k(U)$ be a smooth $k$ -form, written in the increasing-multi-index notation of differential forms as $\omega = {\sum_I}'\,\omega_I\,dx^{I}$ . Its exterior derivative is the $(k+1)$ -form obtained by differentiating each coefficient and wedging the result onto the basis form it belonged to:

d\omega \;=\; {\sum_I}'\, d\omega_I \wedge dx^{I} \;=\; {\sum_I}' \sum_{i=1}^{n} \frac{\partial \omega_I}{\partial x^{i}}\, dx^{i}\wedge dx^{I} .

Here $d\omega_I$ is the ordinary differential of the function $\omega_I$ . For $k = 0$ the formula reads $df = (\partial f/\partial x^{i})\,dx^{i}$ , so $d$ on functions is nothing new; the content of the definition is that the same rule makes sense in every degree.

Now let $M$ be a smooth $n$ -manifold, with or without boundary (smooth manifolds). For $\omega \in \Omega^k(M)$ and a smooth chart $\varphi : U \to \widehat{U} \subseteq \mathbb{R}^n$ , define $d\omega$ on $U$ by carrying the form to $\widehat{U}$ , differentiating there, and carrying the answer back:

d\omega \;=\; \varphi^{*}\Bigl( d\bigl[(\varphi^{-1})^{*}\omega\bigr]\Bigr) .

Part 5 of the statement below — naturality, the fact that $d$ commutes with every pullback — is what makes this independent of the chart: a transition map is a diffeomorphism between open subsets of $\mathbb{R}^n$ or of $\mathbb{H}^n$ , and naturality applied to it converts one chart's recipe into the other's. So the pieces agree on overlaps and define a single $(k+1)$ -form on $M$ . In the coordinates of any chart, $d\omega$ is then given by the displayed formula with $x^i$ the coordinate functions.

A form $\omega$ is closed if $d\omega = 0$ , and exact if $\omega = d\eta$ for some smooth form $\eta$ of one degree lower. Part 3 below says every exact form is closed; whether the converse holds is a question about the shape of $M$ , and is the subject of closed and exact forms.

Definition(What

d

is, and the fact that there is only one)AdaptedAdapted from Lee (2013), Prop. 14.23, p. 364; Lee (2013), Thm 14.24, p. 365; Lee (2013), Prop. 14.26, p. 366; Lee (2013), Prop. 14.29, p. 369; Hassani (2013), Thm 28.5.3, p. 889

Let $M$ be a smooth manifold with or without boundary. There is exactly one family of operators $d : \Omega^k(M) \to \Omega^{k+1}(M)$ , one for each $k \ge 0$ , with the following four properties.

Linearity. $d(a\omega + b\eta) = a\,d\omega + b\,d\eta$ for real constants $a, b$ and forms of equal degree.
Antiderivation. For $\omega \in \Omega^{k}(M)$ and $\eta \in \Omega^{l}(M)$ , $d(\omega\wedge\eta) \;=\; d\omega\wedge\eta \;+\; (-1)^{k}\,\omega\wedge d\eta .$ The sign depends on the degree of the left factor only.
The square vanishes. $d(d\omega) = 0$ for every form $\omega$ .
Functions. For $f \in \Omega^{0}(M) = C^{\infty}(M)$ , $df$ is the usual differential, $df(X) = Xf$ for every smooth vector field $X$ .

That family is the one built chart by chart in the definition. It has two further properties, neither of which has to be assumed:

Naturality. If $F : M \to N$ is smooth then $F^{*}(d\omega) = d(F^{*}\omega)$ for every $\omega \in \Omega^{k}(N)$ : it does not matter whether one differentiates before or after a change of variables.
Invariant formula in degree one. For $\omega \in \Omega^{1}(M)$ and smooth vector fields $X, Y$ , $d\omega(X,Y) \;=\; X\bigl(\omega(Y)\bigr) \;-\; Y\bigl(\omega(X)\bigr) \;-\; \omega\bigl([X,Y]\bigr),$ with $[X,Y]$ the Lie bracket of tangent vectors and vector fields. No coordinates appear on either side.

Definition(Gradient, curl and divergence are the same operator)AdaptedAdapted from Arfken, Weber & Harris (2013), (4.86)–(4.92), pp. 240–241; Lee (2013), Example 14.27 and diagram (14.27), pp. 367–368

Let $V \subseteq \mathbb{R}^{3}$ be open, with Cartesian coordinates $(x,y,z)$ . Attach to a smooth vector field $\mathbf{A} = (A_x, A_y, A_z)$ on $V$ its $1$ -form and its $2$ -form,

\mathbf{A}^{\flat} \;=\; A_x\,dx + A_y\,dy + A_z\,dz, \qquad \beta(\mathbf{A}) \;=\; A_x\,dy\wedge dz + A_y\,dz\wedge dx + A_z\,dx\wedge dy ,

and to a smooth function $g$ its $3$ -form $\sigma(g) = g\,dx\wedge dy\wedge dz$ . Each of $\flat$ , $\beta$ and $\sigma$ is a bijection onto $\Omega^{1}(V)$ , $\Omega^{2}(V)$ , $\Omega^{3}(V)$ respectively, because $\binom{3}{1} = \binom{3}{2} = 3$ and $\binom{3}{3} = 1$ . Then for every smooth $f$ and $\mathbf{A}$ on $V$ :

df \;=\; (\nabla f)^{\flat}, \qquad d\bigl(\mathbf{A}^{\flat}\bigr) \;=\; \beta(\nabla\times\mathbf{A}), \qquad d\bigl(\beta(\mathbf{A})\bigr) \;=\; \sigma(\nabla\cdot\mathbf{A}) ,

with $\nabla f$ , $\nabla\times\mathbf{A}$ and $\nabla\cdot\mathbf{A}$ as defined in the gradient and divergence, curl and the Laplacian. Consequently the two identities

\nabla\times(\nabla f) = \mathbf{0}, \qquad \nabla\cdot(\nabla\times\mathbf{A}) = 0

of vector identities are the two instances of $d\circ d = 0$ available in three dimensions.

Idea of the proofOriginal

In a chart there is only one sensible thing to do to $\omega = {\sum_I}'\,\omega_I\,dx^I$ : differentiate the coefficients, since the basis forms $dx^I$ are constant. Wedging $d\omega_I$ on the left is forced if the Leibniz rule is to hold with a sign that depends on degree, and that same placement makes the second derivative appear as $\partial_i\partial_j\omega_I\,dx^i\wedge dx^j$ — symmetric coefficients contracted against an antisymmetric basis, hence zero. So $d\circ d = 0$ is exactly the equality of mixed partial derivatives, wearing different clothes.

The one thing a coordinate definition owes us is independence of the coordinates. That is not checked by comparing two charts directly; it follows from a stronger statement that is easier to prove, namely that $d$ commutes with the pullback by any smooth map, because a change of chart is the pullback by a transition map. Uniqueness then costs nothing: properties 1–4 force the coordinate formula at every point, once one knows that $d\omega$ near a point depends only on $\omega$ near that point, which a bump function supplies.

The $\mathbb{R}^{3}$ dictionary is then pure bookkeeping. A vector field has three components; so does a $1$ -form and so does a $2$ -form in three variables. Write $\mathbf{A}$ as a $1$ -form and apply $d$ , and the three coefficients that come out are the three components of the curl; write it as a $2$ -form instead and the single coefficient that comes out is the divergence.

Proof(Properties 1–5)AdaptedAdapted from Lee (2013), Prop. 14.23 and Thm 14.24, pp. 364–366; Lee (2013), Prop. 14.26, pp. 366–367; Lee (2013), Prop. 14.29, p. 369

Step 1: a convenience, on non-increasing indices. In $\mathbb{R}^n$ , $d(u\,dx^{I}) = du\wedge dx^{I}$ holds for every multi-index $I$ of length $k$ , not only the increasing ones. If $I$ repeats an index then $dx^I = 0$ and both sides vanish. Otherwise let $\tau$ be the permutation sorting $I$ into an increasing $J$ ; then $dx^{I} = (\operatorname{sgn}\tau)\,dx^{J}$ , and multiplying the definition applied to $u\,dx^{J}$ by $\operatorname{sgn}\tau$ gives the claim. Linearity (property 1) is immediate from the definition, since each $\omega \mapsto \partial\omega_I/\partial x^i$ is linear.

Step 2: the antiderivation rule in $\mathbb{R}^n$ (property 2). By linearity it is enough to take $\omega = u\,dx^{I}$ of degree $k$ and $\eta = v\,dx^{J}$ of degree $l$ . The coefficients are $0$ -forms, so they commute past everything and $\omega\wedge\eta = uv\,dx^{I}\wedge dx^{J}$ . By step 1 and the product rule for the differential of a function,

d(\omega\wedge\eta) = d(uv)\wedge dx^{I}\wedge dx^{J} = (v\,du + u\,dv)\wedge dx^{I}\wedge dx^{J} .

The first piece is $du\wedge dx^{I}\wedge v\,dx^{J} = d\omega\wedge\eta$ . In the second, $dv$ has degree $1$ and $dx^{I}$ degree $k$ , so moving $dv$ to the right of $dx^{I}$ costs $(-1)^{k}$ by the anticommutation rule of differential forms, leaving $(-1)^{k}\,u\,dx^{I}\wedge dv\wedge dx^{J} = (-1)^{k}\,\omega\wedge d\eta$ . It is the degree of the left factor that the sign records, because that is how many $1$ -forms $dv$ must cross.

Step 3: $d\circ d = 0$ in $\mathbb{R}^n$ (property 3). First for a function $u$ . Applying the definition twice,

d(du) = d\Bigl(\frac{\partial u}{\partial x^{j}}dx^{j}\Bigr) = \frac{\partial^{2} u}{\partial x^{i}\partial x^{j}}\,dx^{i}\wedge dx^{j} = \sum_{i<j}\Bigl(\frac{\partial^{2} u}{\partial x^{i}\partial x^{j}} - \frac{\partial^{2} u}{\partial x^{j}\partial x^{i}}\Bigr) dx^{i}\wedge dx^{j} = 0,

where the middle sum was split into $i<j$ and $i>j$ and the names of the summation indices swapped in the second half, and the bracket vanishes because $u$ is $C^{2}$ , so its mixed partials agree. For a general $\omega = {\sum_I}'\,\omega_I\,dx^{I}$ , apply $d$ to $d\omega = {\sum_I}'\,d\omega_I\wedge dx^{I}$ using step 2 with left factor $d\omega_I$ of degree $1$ :

d(d\omega) = {\sum_I}'\Bigl[ d(d\omega_I)\wedge dx^{I} - d\omega_I \wedge d\bigl(dx^{I}\bigr)\Bigr] .

The first term is zero by the case just done. The second is zero because $dx^{I} = dx^{i_1}\wedge\cdots\wedge dx^{i_k}$ is a wedge of differentials of the coordinate functions: repeated use of step 2 writes $d(dx^{I})$ as a sum of terms each containing a factor $d(dx^{i_a}) = 0$ .

Step 4: naturality on open sets (property 5, Euclidean case). First, for a smooth map $F$ and a smooth function $u$ on the target, $F^{*}(du) = d(u\circ F)$ : both are $1$ -forms, and on a tangent vector $v$ at $p$ ,

(F^{*}du)(v) = du_{F(p)}\bigl(dF_p v\bigr) = (dF_p v)u = v(u\circ F) = d(u\circ F)(v),

by the definition of the pushforward in pushforwards, pullbacks and immersions. Now take $\omega = u\,dx^{i_1}\wedge\cdots\wedge dx^{i_k}$ on an open subset of the target. Since pullback preserves wedge products,

F^{*}(d\omega) = F^{*}\bigl(du\wedge dx^{i_1}\wedge\cdots\wedge dx^{i_k}\bigr) = d(u\circ F)\wedge dF^{i_1}\wedge\cdots\wedge dF^{i_k},

writing $F^{j}$ for the $j$ th component of $F$ . On the other side, $F^{*}\omega = (u\circ F)\,dF^{i_1}\wedge\cdots\wedge dF^{i_k}$ , and applying step 2 to this product and then step 3 to each $d(dF^{i_a})$ leaves only

d(F^{*}\omega) = d(u\circ F)\wedge dF^{i_1}\wedge\cdots\wedge dF^{i_k} .

The two agree, and the general case follows by linearity.

Step 5: the definition on a manifold is unambiguous. Let $(U,\varphi)$ and $(V,\psi)$ be charts meeting at $p$ . On $\varphi(U\cap V)$ the transition map $\psi\circ\varphi^{-1}$ is a diffeomorphism between open subsets of $\mathbb{R}^n$ (or of the half-space, at a boundary point), so step 4 applies to it. Pulling the $\psi$ -recipe back by that transition map turns it into the $\varphi$ -recipe, using $(\psi\circ\varphi^{-1})^{*}\circ(\psi^{-1})^{*} = (\varphi^{-1})^{*}$ from the functoriality of pullback. Hence the two charts assign the same $(k+1)$ -form near $p$ , and properties 1–4 transfer from $\mathbb{R}^n$ to $M$ because pullback by a diffeomorphism respects sums, wedge products and degrees. Property 5 on manifolds follows by writing $F$ in charts at $p$ and at $F(p)$ and applying the Euclidean case to its coordinate representative.

Step 6: uniqueness. Suppose $D$ is any family of operators with properties 1–4. First, $D$ is local: if two forms $\omega_1,\omega_2$ agree on an open set $W$ , then $D\omega_1 = D\omega_2$ on $W$ . Indeed, put $\theta = \omega_1 - \omega_2$ and fix $p \in W$ . Choose a smooth bump function $\chi$ that equals $1$ near $p$ and vanishes outside $W$ (partitions of unity). Then $\chi\theta$ is identically zero, so $0 = D(\chi\theta) = d\chi\wedge\theta + \chi\,D\theta$ by properties 2 and 4 with $\chi$ a $0$ -form. Evaluating at $p$ , where $\chi \equiv 1$ on a neighbourhood and hence $d\chi_p = 0$ and $\chi(p) = 1$ , gives $D\theta|_p = 0$ .

Now take any $\omega$ and any chart $(U,x^i)$ around $p$ . Using bump functions again, replace the locally defined functions $\omega_I$ and $x^i$ by globally defined smooth ones agreeing with them near $p$ ; locality says this does not change $D\omega$ at $p$ . Expanding $\omega = {\sum_I}'\,\omega_I\,dx^{i_1}\wedge\cdots\wedge dx^{i_k}$ and applying properties 1, 2 and 3, every term in which $D$ lands on a factor $dx^{i_a} = D x^{i_a}$ dies, and property 4 identifies $D\omega_I$ with $d\omega_I$ . What is left is exactly the coordinate formula of the definition. So $D\omega = d\omega$ at $p$ , and $p$ was arbitrary.

Step 7: the invariant formula (property 6). Both sides are additive in $\omega$ and local in $\omega$ : the left-hand side because the coordinate formula computes $d\omega$ at $p$ from the values of $\omega$ near $p$ alone, the right-hand side because if $\omega$ vanishes on a neighbourhood of $p$ then so do $\omega(X)$ , $\omega(Y)$ and $\omega([X,Y])$ there. Fix $p$ and a chart $(U,x^{i})$ around it. Exactly as in step 6, a bump function equal to $1$ near $p$ and supported in $U$ replaces the chart-defined functions $\omega_i$ and $x^{i}$ by globally defined smooth functions agreeing with them near $p$ , and locality says this changes neither side at $p$ . After that replacement $\omega = \sum_i \omega_i\,dx^{i}$ is a finite sum of globally defined terms of the shape $u\,dv$ with $u,v$ smooth functions, so by additivity it suffices to check $\omega = u\,dv$ . On the left, $d(u\,dv) = du\wedge dv$ by the definition, and the determinant formula for a wedge of two $1$ -forms gives

(du\wedge dv)(X,Y) = du(X)\,dv(Y) - du(Y)\,dv(X) = (Xu)(Yv) - (Yu)(Xv).

On the right, using $\omega(Y) = u\,(Yv)$ and the product rule for the action of a vector field on a function,

X\bigl(u\,Yv\bigr) - Y\bigl(u\,Xv\bigr) - u\,[X,Y]v = (Xu)(Yv) + u\,XYv - (Yu)(Xv) - u\,YXv - u\bigl(XYv - YXv\bigr),

and the four terms carrying a factor $u$ cancel in pairs, leaving $(Xu)(Yv) - (Yu)(Xv)$ . $\square$

Proof(The dictionary in

\mathbb{R}^{3}

)AdaptedAdapted from Arfken, Weber & Harris (2013), (4.86)–(4.92), pp. 240–241

Degree zero. $df = \partial_x f\,dx + \partial_y f\,dy + \partial_z f\,dz$ is the definition of $d$ on a function and the definition of $\nabla f$ side by side, so $df = (\nabla f)^{\flat}$ .

Degree one. Apply $d$ to $\mathbf{A}^{\flat}$ and expand each differential:

d\bigl(\mathbf{A}^{\flat}\bigr) = dA_x\wedge dx + dA_y\wedge dy + dA_z\wedge dz .

In $dA_x\wedge dx$ the term $\partial_x A_x\,dx\wedge dx$ dies, leaving $\partial_y A_x\,dy\wedge dx + \partial_z A_x\,dz\wedge dx$ ; similarly for the other two. Sorting the six surviving products into the cyclic basis $dy\wedge dz$ , $dz\wedge dx$ , $dx\wedge dy$ — which costs a sign exactly when the pair is in the reversed cyclic order — collects them as

d\bigl(\mathbf{A}^{\flat}\bigr) = (\partial_y A_z - \partial_z A_y)\,dy\wedge dz + (\partial_z A_x - \partial_x A_z)\,dz\wedge dx + (\partial_x A_y - \partial_y A_x)\,dx\wedge dy ,

whose three coefficients are the three components of $\nabla\times\mathbf{A}$ . That is $\beta(\nabla\times\mathbf{A})$ .

Degree two. In $d(\beta(\mathbf{A}))$ every term is a $3$ -form, so only one basis element survives in each: of the three summands of $dA_x\wedge dy\wedge dz$ only $\partial_x A_x\,dx\wedge dy\wedge dz$ has no repeated differential, and likewise $dA_y\wedge dz\wedge dx$ contributes $\partial_y A_y\,dy\wedge dz\wedge dx$ and $dA_z\wedge dx\wedge dy$ contributes $\partial_z A_z\,dz\wedge dx\wedge dy$ . Both $dy\wedge dz\wedge dx$ and $dz\wedge dx\wedge dy$ equal $dx\wedge dy\wedge dz$ , since each is a cyclic shift, that is a product of two transpositions. Hence

d\bigl(\beta(\mathbf{A})\bigr) = \bigl(\partial_x A_x + \partial_y A_y + \partial_z A_z\bigr)\,dx\wedge dy\wedge dz = \sigma(\nabla\cdot\mathbf{A}) .

The two identities. Take $\mathbf{A} = \nabla f$ . Then $\mathbf{A}^{\flat} = df$ , so $\beta(\nabla\times\nabla f) = d(df) = 0$ ; since $\beta$ is injective, $\nabla\times\nabla f = \mathbf{0}$ . Take instead any $\mathbf{A}$ and set $\mathbf{B} = \nabla\times\mathbf{A}$ , so that $\beta(\mathbf{B}) = d(\mathbf{A}^{\flat})$ . Then $\sigma(\nabla\cdot\mathbf{B}) = d(\beta(\mathbf{B})) = d(d(\mathbf{A}^{\flat})) = 0$ , and $\sigma$ is injective, so $\nabla\cdot(\nabla\times\mathbf{A}) = 0$ . Both identities need $f$ and $\mathbf{A}$ to be $C^{2}$ , which is where step 3 above used the symmetry of mixed partials. $\square$

Remark(Where the hypotheses are used)Original

Two derivatives, not one. The only analytic input anywhere is the equality of mixed second partial derivatives, used once, in step 3. It is what makes $d\circ d = 0$ , and through that it is what makes the whole theory of closed and exact forms, and both classical identities above, work. For a $C^{1}$ but not $C^{2}$ coefficient the expression $d(d\omega)$ need not even be defined; for functions whose mixed partials exist but are discontinuous and unequal, the two orders of differentiation genuinely disagree, and so would the two ways of computing $d(df)$ .
Smoothness of the map, in naturality. Step 4 differentiates $u\circ F$ and the components $F^{j}$ , so $F$ must be differentiable; no injectivity, surjectivity or equality of dimensions is needed, which is why $d$ transfers along maps that collapse or fold. This generosity is exactly what step 5 cashes in: chart overlaps are a special case.
Bump functions, in the uniqueness proof. Step 6 is the one place that needs a smooth function equal to $1$ near a point and vanishing outside a given open set, and step 7 reuses it. Such functions exist because the manifold is smooth in the $C^{\infty}$ sense; in the real-analytic category there are no cutoffs at all, since a real-analytic function on a connected manifold that vanishes on a non-empty open subset vanishes identically. What that costs is this route to uniqueness, not uniqueness itself. On $\mathbb{R}^{n}$ with its real-analytic structure the coordinate functions $x^{i}$ and the coefficients $\omega_I$ are already defined globally, so nothing has to be cut off and properties 1–4 force the coordinate formula directly. So the dependence on bump functions is a feature of this proof, not of the theorem.
The sign cannot be dropped. Properties 1, 3 and 4 with a signless Leibniz rule $D(\alpha\wedge\eta) = D\alpha\wedge\eta + \alpha\wedge D\eta$ are already contradictory, before any coordinate formula is invoked. Take the $2$ -form $\theta = y\,dx\wedge dz$ in $\mathbb{R}^{3}$ and group it in the two available ways. Note first that $D(dx) = D(Dx) = 0$ and likewise $D(dz) = 0$ , by properties 3 and 4. Grouping as $\theta = (y\,dx)\wedge dz$ gives $D\theta = D(y\,dx)\wedge dz = (dy\wedge dx)\wedge dz = -\,dx\wedge dy\wedge dz$ ; grouping as $\theta = dx\wedge(y\,dz)$ gives $D\theta = dx\wedge D(y\,dz) = dx\wedge dy\wedge dz$ . The same form receives two opposite values. With the factor $(-1)^{k}$ in place both groupings give $-\,dx\wedge dy\wedge dz$ , which is also what the coordinate definition gives for $d(y\,dx\wedge dz) = dy\wedge dx\wedge dz$ . Arfken, Weber and Harris (Arfken, Weber & Harris (2013), Example 4.6.1, p. 239) check the sign differently, by tracking $dx^{\mu}$ through the $p$ differentials of the left factor.
Four properties or five. Lee (Lee (2013), Thm 14.24, p. 365) characterises $d$ on a whole manifold by properties 1–4 and derives naturality from the coordinate construction, which is the route taken above. Hassani (Hassani (2013), Thm 28.5.3, p. 889) instead states existence and uniqueness on a neighbourhood of each point and lists five properties, taking naturality as the fifth axiom. Neither account is contained in the other — a local uniqueness statement with naturality assumed does not by itself say that four properties suffice globally — so the claim above that properties 5 and 6 "need not be assumed" is Lee's, and it is Lee that the statement and proof follow.
Degree matters, not dimension. Nothing above is special to three dimensions except the second statement, which uses $\binom{3}{1} = \binom{3}{2}$ to disguise a $2$ -form as a vector field. In $\mathbb{R}^{n}$ the gradient and the divergence still have form counterparts, but the curl does not: $d$ on $1$ -forms produces a $2$ -form with $\binom{n}{2}$ components, which is a vector field only when $n = 3$ . Turning a $2$ -form back into a vector field in general needs the extra structure of the Hodge star.
Invariant formulas in higher degree. The coordinate-free formula of property 6 generalises. Hassani (Hassani (2013), Thm 28.5.11, p. 891) gives, for a $p$ -form $\omega$ and vector fields $X_1,\dots,X_{p+1}$ , $d\omega(X_1,\dots,X_{p+1}) = \sum_{i}(-1)^{i+1}X_i\,\omega\bigl(\dots \widehat{X_i}\dots\bigr) + \sum_{i<j}(-1)^{i+j}\omega\bigl([X_i,X_j],\dots \widehat{X_i}\dots \widehat{X_j}\dots\bigr),$ the hats marking omitted arguments; Lee (Lee (2013), Prop. 14.32, p. 370) states and proves the same identity. We have proved only the case $p = 1$ , which is the one used in practice.

Method(Computing

d\omega

)Original

Write $\omega$ with its coefficients on the left, as a sum of terms $\omega_I\,dx^{I}$ with $I$ increasing. Anything else is sorted first, as in the method of differential forms.
Differentiate each coefficient and wedge on the left: replace $\omega_I\,dx^{I}$ by $d\omega_I\wedge dx^{I}$ , where $d\omega_I = \sum_i (\partial\omega_I/\partial x^{i})\,dx^{i}$ . Do not differentiate the basis forms; $d(dx^{I}) = 0$ .
Discard every product with a repeated differential, then sort what remains back into increasing order, tracking signs.
Check the degree: $d$ raises it by exactly one, so $d\omega = 0$ automatically if $\omega$ is a top-degree form on an $n$ -manifold.
Check $d(d\omega) = 0$ on the answer. This catches sign errors in step 3 more reliably than re-reading the algebra.
In a non-Cartesian chart, apply the same recipe to the coordinate expression of $\omega$ in that chart. No connection coefficients or scale factors appear: $d$ needs no metric, which is why its formula is the same in every coordinate system, unlike the curl formulas of orthogonal curvilinear coordinates.

Example(A

1

-form in

\mathbb{R}^{3}

, differentiated and tested twice)Original

Take on $\mathbb{R}^{3}$ the smooth $1$ -form

\omega \;=\; x y^{2}\,dx \;+\; y z^{3}\,dy \;+\; z x^{2}\,dz .

Step 1: differentiate. Following the method, $d\omega = d(xy^{2})\wedge dx + d(yz^{3})\wedge dy + d(zx^{2})\wedge dz$ . Expanding each differential and dropping repeats,

d(xy^{2})\wedge dx = 2xy\,dy\wedge dx, \qquad d(yz^{3})\wedge dy = 3yz^{2}\,dz\wedge dy, \qquad d(zx^{2})\wedge dz = 2xz\,dx\wedge dz .

Sorting each into the cyclic basis costs one sign apiece:

d\omega \;=\; -3yz^{2}\,dy\wedge dz \;-\; 2xz\,dz\wedge dx \;-\; 2xy\,dx\wedge dy .

Step 2: check against the curl. The vector field is $\mathbf{A} = (xy^{2},\,yz^{3},\,zx^{2})$ , and

\nabla\times\mathbf{A} = \bigl(\partial_y(zx^{2}) - \partial_z(yz^{3}),\ \partial_z(xy^{2}) - \partial_x(zx^{2}),\ \partial_x(yz^{3}) - \partial_y(xy^{2})\bigr) = \bigl(-3yz^{2},\,-2xz,\,-2xy\bigr),

which are the three coefficients just found, in the cyclic order. The second statement is confirmed on this example.

Step 3: evaluate on a pair of vectors. At $p = (1,2,3)$ the coefficients are $-3\cdot 2\cdot 9 = -54$ , $-2\cdot 1\cdot 3 = -6$ and $-2\cdot 1 \cdot 2 = -4$ , so

d\omega|_{p} = -54\,dy\wedge dz - 6\,dz\wedge dx - 4\,dx\wedge dy .

Feed it the tangent vectors $u = (1,0,-1)$ and $v = (2,1,1)$ . Each basis $2$ -form returns a $2\times 2$ determinant of the corresponding components:

(dy\wedge dz)(u,v) = \begin{vmatrix} 0 & 1\\ -1 & 1\end{vmatrix} = 1, \quad (dz\wedge dx)(u,v) = \begin{vmatrix} -1 & 1\\ 1 & 2\end{vmatrix} = -3, \quad (dx\wedge dy)(u,v) = \begin{vmatrix} 1 & 2\\ 0 & 1\end{vmatrix} = 1 .

Hence

d\omega|_{p}(u,v) = (-54)(1) + (-6)(-3) + (-4)(1) = \boxed{-40} .

The three determinants are the components of $u\times v = (1,-3,1)$ , so this is the scalar $(\nabla\times\mathbf{A})\cdot(u\times v)$ at $p$ — the circulation density of $\mathbf{A}$ through the parallelogram spanned by $u$ and $v$ , which is what a $2$ -form measures.

Step 4: test the invariant formula. Take the vector fields $X = \partial_x$ and $Y = x\,\partial_y$ , whose bracket is $[X,Y] = (\partial_x x)\partial_y = \partial_y$ . The right-hand side of property 6 is

X\bigl(x\,yz^{3}\bigr) - Y\bigl(xy^{2}\bigr) - \omega(\partial_y) = yz^{3} - x\cdot 2xy - yz^{3} = -2x^{2}y .

The left-hand side: of the three basis $2$ -forms, only $dx\wedge dy$ survives on the pair $(1,0,0)$ , $(0,x,0)$ , with value $x$ , so $d\omega(X,Y) = -2xy\cdot x = -2x^{2}y$ . The two agree identically, and at $p = (1,2,3)$ both equal $-4$ . Had the bracket term been dropped, the right-hand side would have been $-2x^{2}y + yz^{3}$ , which is not even a tensor in $X$ and $Y$ .

ApplicationOriginal

Thermodynamics tells closed from exact. On a state space with coordinates $(V,T)$ an increment of heat is a $1$ -form $\delta Q = C_V\,dT + \ell\,dV$ , not the differential of a state function. The question of whether a function $Q$ with $\delta Q = dQ$ exists is the question of whether $\delta Q$ is exact, and $d(\delta Q) = (\partial_V C_V - \partial_T \ell)\,dV\wedge dT$ is the computable obstruction: if it is non-zero, no such $Q$ can exist, because exact forms are closed. Dividing by $T$ can make the obstruction vanish, and the resulting exact form is $dS$ . The whole of the second law's bookkeeping is the distinction between a closed $1$ -form and an exact one, which is closed and exact forms.

Maxwell's homogeneous equations are one line. Assemble the electric and magnetic fields into the spacetime $2$ -form $F = -E_x\,dt\wedge dx - E_y\,dt\wedge dy - E_z\,dt\wedge dz + B_x\,dy\wedge dz + B_y\,dz\wedge dx + B_z\,dx\wedge dy$ (units with $c = 1$ ). Computing $dF$ by the method above produces four coefficients, which are precisely the three components of $\nabla\times\mathbf{E} + \partial_t\mathbf{B}$ and the scalar $\nabla\cdot\mathbf{B}$ . So Faraday's law and the absence of magnetic monopoles together say exactly $dF = 0$ , as Hassani (Hassani (2013), Box 28.5.6, p. 890) records. Because $d$ commutes with pullback, this equation is automatically Lorentz invariant: a change of frame acts on $F$ by pullback, and pullback of $0$ is $0$ . The inhomogeneous pair needs a metric and is treated in the Hodge star and Maxwell's equations.

Gauge potentials. Writing $F = dA$ for a $1$ -form $A$ makes $dF = 0$ automatic, by $d\circ d = 0$ ; and $A$ is determined by $F$ only up to adding $d\chi$ , which is the gauge freedom. The same two sentences describe the vector potential of magnetostatics ( $\mathbf{B} = \nabla\times\mathbf{A}$ forces $\nabla\cdot\mathbf{B} = 0$ , and $\mathbf{A}$ is free up to a gradient) and the connection forms of gauge theories in any dimension.

Conservation laws and integrability. A conserved current in $n$ spacetime dimensions is an $(n-1)$ -form $J$ with $dJ = 0$ ; this single equation is the continuity equation of the divergence theorem with no index juggling. In the same spirit, a system of first-order partial differential equations written as the vanishing of a set of $1$ -forms is integrable only if the exterior derivatives of those forms vanish on the common solution set — the Frobenius condition, and the reason $d$ appears in control theory and in non-holonomic mechanics.

Why $d$ is cheap. Unlike the covariant derivative of covariant differentiation, $d$ needs no metric and no connection, so it is the same operator in Cartesian, spherical or arbitrary curvilinear coordinates. In numerical work this is the basis of mimetic and finite-element exterior-calculus discretisations, which represent $k$ -forms by their integrals over $k$ -dimensional cells of a mesh and define $d$ by signed sums over cell boundaries; $d\circ d = 0$ then holds exactly in floating point, because it becomes a statement about the mesh rather than about derivatives.

Theorem

Integration of forms and the general Stokes theorem

Adapted

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Defines the integral of an $n$ -form over an oriented $n$ -manifold with boundary, proves that the integral of $d\omega$ over the manifold equals the integral of $\omega$ over its boundary, and derives the fundamental theorem of calculus, Green's theorem, the classical Stokes theorem and the divergence theorem as four readings of that one identity.

Prerequisites: The exterior derivative, Stokes' theorem and Green's theorem in the plane, Differential forms, Smooth manifolds, Partitions of unity, Pushforwards, pullbacks and immersions, The divergence theorem, Line integrals and conservative fields, Surface integrals and flux, Multiple integrals and change of variables

Definition(Orientation, boundary, and the integral of a form)AdaptedAdapted from Hassani (2013), Def. 28.6.1 and (28.47), p. 898; Lee (2013), pp. 402–405, (16.1)–(16.2); Lee (2013), Prop. 15.24, p. 386

Throughout, $M$ is a smooth $n$ -manifold with or without boundary, $n \ge 1$ (smooth manifolds), modelled on the closed half-space $\mathbb{H}^{n} = \{x \in \mathbb{R}^{n} : x^{n} \ge 0\}$ . Its boundary $\partial M$ is the set of points carried to $\partial\mathbb{H}^n = \{x^n = 0\}$ by some chart; it is an embedded $(n-1)$ -dimensional submanifold without boundary.

Orientation. $M$ is orientable if it carries a nowhere vanishing $n$ -form. Two such forms $\mu$ and $\mu'$ satisfy $\mu' = h\mu$ for a nowhere vanishing continuous function $h$ , and calling them equivalent when $h > 0$ is an equivalence relation; an orientation is a choice of one of its classes, and a representative $\mu$ is an orientation form. If $M$ is connected then $h$ , being continuous and never zero, keeps one sign, so there are exactly two classes and exactly two orientations. On a manifold with $c$ connected components the sign of $h$ may be chosen on each component separately, so there are $2^{c}$ . That case is not academic here: the boundary of an annulus is two circles and the boundary of a spherical shell is two spheres, and orienting such a $\partial M$ means orienting each piece. Hassani (Hassani (2013), the paragraph after Def. 28.6.1, p. 898) states the two-class count without the connectedness hypothesis. Lee (Lee (2013), Prop. 15.5 with its remark, p. 381, and Exercise 15.4, p. 381) instead defines an orientation as a continuous pointwise choice, and notes that it is connectedness of the domain that forces a frame to be positively or negatively oriented throughout. The count above follows Lee, and so does this unit wherever the two accounts differ. A chart $(U,\varphi)$ is positively oriented if $dx^1\wedge\cdots\wedge dx^n$ is a positive multiple of $\mu$ on $U$ , and negatively oriented if it is a negative multiple.

Interior multiplication and the induced orientation. For a vector $v$ and a $k$ -form $\mu$ , let $\iota_v\mu$ be the $(k-1)$ -form $(\iota_v\mu)(w_1,\dots,w_{k-1}) = \mu(v,w_1,\dots,w_{k-1})$ : the first slot is filled in. Let $N$ be a smooth vector field along $\partial M$ that points out of $M$ at every boundary point. Then the pullback of $\iota_N\mu$ to $\partial M$ is a nowhere vanishing $(n-1)$ -form there, and the orientation it defines does not depend on which outward-pointing $N$ is used. It is the induced or Stokes orientation on $\partial M$ . Concretely: a basis $(e_1,\dots,e_{n-1})$ of $T_p\partial M$ is positively oriented exactly when $(N_p, e_1,\dots,e_{n-1})$ is a positively oriented basis of $T_pM$ — outward vector first.

The integral. Let $\omega$ be a continuous, compactly supported $n$ -form.

On $\mathbb{R}^n$ or $\mathbb{H}^n$ . Any such $\omega$ on an open set $U$ is $f\,dx^1\wedge\cdots\wedge dx^n$ for one continuous compactly supported $f$ , and we set $\int_U \omega = \int_U f\,dV$ , the ordinary multiple integral: erase the wedges.
In one chart. If $\operatorname{supp}\omega$ lies in the domain of a single chart $(U,\varphi)$ , set $\int_M \omega \;=\; \pm \int_{\varphi(U)} \bigl(\varphi^{-1}\bigr)^{*}\omega ,$ with the plus sign for a positively oriented chart and the minus sign for a negatively oriented one.
In general. Cover $\operatorname{supp}\omega$ by finitely many such chart domains $U_i$ , take a subordinate smooth partition of unity $\{\psi_i\}$ (partitions of unity), and set $\int_M\omega = \sum_i \int_M \psi_i\omega$ .

Steps 2 and 3 are independent of the choices made, because a transition map is a diffeomorphism and the change-of-variables formula of multiple integrals produces $\lvert\det DG\rvert$ while the pullback of a top-degree form produces $\det DG$ — the two agree precisely when the transition preserves orientation, which is what positively oriented charts guarantee. For $n = 0$ the integral of a function over an oriented $0$ -manifold is the sum of its values with the sign of each point. Finally, if $S \subseteq M$ is an oriented $k$ -dimensional submanifold with inclusion $\iota_S$ , and $\omega$ is a $k$ -form on $M$ , then $\int_S \omega$ means $\int_S \iota_S^{*}\omega$ .

Theorem(The general Stokes theorem)AdaptedAdapted from Lee (2013), Thm 16.11, p. 411; Hassani (2013), Thms 28.6.2 and 28.6.3, p. 899; Arfken, Weber & Harris (2013), (4.106), p. 245

Take any oriented smooth $n$ -dimensional manifold with boundary $M$ , where $n \ge 1$ , and give $\partial M$ the induced orientation. Suppose $\omega$ is a smooth $(n-1)$ -form whose support is a compact subset of $M$ . Then

\int_{M} d\omega \;=\; \int_{\partial M} \omega ,

the right-hand side being the integral over $\partial M$ of the pullback $\iota_{\partial M}^{*}\omega$ . An empty boundary makes that side zero, so

\int_{M} d\omega = 0 \qquad\text{whenever } \partial M = \varnothing .

In particular, on a compact oriented manifold without boundary no exact top-degree form has non-zero integral, so a nowhere vanishing orientation form on such a manifold is never exact.

Theorem(The classical theorems, as four readings of one identity)AdaptedAdapted from Arfken, Weber & Harris (2013), Examples 4.7.1–4.7.3, pp. 246–248; Lee (2013), Example 16.12, p. 414; Lee (2013), Thm 16.32, p. 424 and Thm 16.34, p. 427

Use the three identifications of the exterior derivative: $\mathbf{A}^{\flat} = A_x dx + A_y dy + A_z dz$ , $\beta(\mathbf{A}) = A_x\,dy\wedge dz + A_y\,dz\wedge dx + A_z\,dx\wedge dy$ and $\sigma(g) = g\,dx\wedge dy\wedge dz$ , for which $d(\mathbf{A}^{\flat}) = \beta(\nabla\times\mathbf{A})$ and $d(\beta(\mathbf{A})) = \sigma(\nabla\cdot\mathbf{A})$ . Then the theorem above specialises as follows.

$n = 1$ , the fundamental theorem of calculus. $M = [a,b]$ with its standard orientation, $\omega = f$ a $0$ -form: $\partial M$ is the two-point set $\{a,b\}$ with $b$ positive and $a$ negative, and $\int_a^b \frac{df}{dx}\,dx = f(b) - f(a) .$ More generally, for a smooth curve $\gamma$ from $p$ to $q$ in a manifold, $\int_\gamma df = f(q) - f(p)$ : the fundamental theorem for line integrals of line integrals and conservative fields.
$n = 2$ in the plane, Green's theorem. $D \subseteq \mathbb{R}^{2}$ a compact region with smooth boundary, oriented by $dx\wedge dy$ , and $\omega = P\,dx + Q\,dy$ : $\oint_{\partial D} \bigl(P\,dx + Q\,dy\bigr) = \iint_{D}\Bigl(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\Bigr)dx\,dy ,$ with $\partial D$ traversed with $D$ on the left.
$n = 2$ in space, the classical Stokes theorem. $S \subseteq \mathbb{R}^{3}$ a compact oriented surface with boundary, with unit normal $\hat{\mathbf{n}}$ , and $\omega = \mathbf{A}^{\flat}$ : $\oint_{\partial S} \mathbf{A}\cdot d\mathbf{r} = \iint_{S} (\nabla\times\mathbf{A})\cdot d\mathbf{S} ,$ $\partial S$ carrying the right-hand traversal relative to $\hat{\mathbf{n}}$ . This is Stokes' theorem.
$n = 3$ , the divergence theorem. $V \subseteq \mathbb{R}^{3}$ a compact region with smooth boundary, oriented by $dx\wedge dy\wedge dz$ , and $\omega = \beta(\mathbf{A})$ : $\oint_{\partial V} \mathbf{A}\cdot d\mathbf{S} = \iiint_{V} (\nabla\cdot\mathbf{A})\,dV ,$ with the outward normal on $\partial V$ . This is the divergence theorem.

In each case the orientation convention stated in the vector-calculus unit is exactly the induced orientation of the definition above; that is the content of the sign bookkeeping in the proof.

Idea of the proofOriginal

Both sides of the identity are additive in $\omega$ , and a partition of unity splits any $\omega$ into pieces each living inside one chart. So it is enough to prove it for a form supported in a single chart, and then — since $d$ commutes with pullback, and the integral was defined by pulling back to a chart — for a form supported in the model half-space $\mathbb{H}^n$ itself.

There the whole theorem is the fundamental theorem of calculus applied once, in one variable. An $(n-1)$ -form on $\mathbb{H}^n$ is a sum of $n$ terms, each missing one differential; taking $d$ differentiates the $i$ th coefficient in the $i$ th direction. Integrating that derivative along the $i$ th axis leaves only boundary values. For the $n-1$ directions parallel to the boundary the support is compact and those boundary values are zero; only the direction transverse to the boundary survives, and it leaves an integral over the face $x^n = 0$ . That is $\int_{\partial M}\omega$ , because the pullback of $dx^n$ to the face vanishes and so kills every other term of $\omega$ in exactly the same pattern.

All the subtlety is in one sign, and it is the same sign twice. The orientation that $\partial\mathbb{H}^n$ inherits from $\mathbb{H}^n$ differs from the standard orientation of $\mathbb{R}^{n-1}$ by $(-1)^{n}$ , so the two agree exactly when $n$ is even. On the other side, the surviving term in $d\omega$ carries a factor $(-1)^{n-1}$ from moving $dx^{n}$ into place and a further factor $-1$ from the fundamental theorem on $[0,\infty)$ , whose product is again $(-1)^{n}$ . The same factor stands in front of both sides and cancels, which is why the theorem has no sign in it.

Proof(The general Stokes theorem)AdaptedAdapted from Lee (2013), Thm 16.11, pp. 411–414; Hassani (2013), Thm 28.6.2, p. 899

Step 1: reduction to one chart. Cover the compact set $\operatorname{supp}\omega$ with finitely many chart domains, each of its charts oriented positively or negatively, and pick a smooth partition of unity subordinate to that cover $\{\psi_i\}_{i=1}^{m}$ , so that $\sum_i\psi_i \equiv 1$ on a neighbourhood of $\operatorname{supp}\omega$ . Each $\psi_i\omega$ is a smooth $(n-1)$ -form compactly supported in one chart domain, and $\omega = \sum_i \psi_i\omega$ . Applying the antiderivation rule of the exterior derivative with the $0$ -form $\psi_i$ on the left,

\sum_{i} d(\psi_i\omega) = \sum_i \bigl(d\psi_i\wedge\omega + \psi_i\,d\omega\bigr) = d\Bigl(\sum_i\psi_i\Bigr)\wedge\omega + d\omega = d\omega ,

because $\sum_i\psi_i$ is constant where $\omega \ne 0$ , so its differential vanishes there. Both sides of the theorem are additive, so it suffices to prove it when $\omega$ is compactly supported in a single chart.

Step 2: reduction to the model. Let $(U,\varphi)$ be a positively oriented chart containing $\operatorname{supp}\omega$ . By the definition of the integral and the naturality of $d$ ,

\int_M d\omega = \int_{\varphi(U)}\bigl(\varphi^{-1}\bigr)^{*}(d\omega) = \int_{\varphi(U)} d\Bigl[\bigl(\varphi^{-1}\bigr)^{*}\omega\Bigr],

so the left-hand side is computed from the form $\widehat\omega = (\varphi^{-1})^{*}\omega$ on an open subset of $\mathbb{H}^n$ (or of $\mathbb{R}^n$ for an interior chart). On the other side, $\varphi$ carries $U\cap\partial M$ into $\partial\mathbb{H}^n$ and, since its differential sends outward-pointing vectors to outward-pointing vectors, it does so preserving the induced orientations. Hence $\int_{\partial M}\omega = \int_{\partial\mathbb{H}^n}\widehat\omega$ . For a negatively oriented chart both sides acquire the same extra minus sign. It therefore suffices to prove the identity for a compactly supported smooth $(n-1)$ -form on $\mathbb{H}^n$ , and for one on $\mathbb{R}^n$ with both sides read as zero on the right.

Step 3: the half-space computation. Write $\widehat\omega = \sum_{i=1}^{n} \omega_i\,dx^{1}\wedge\cdots\wedge\widehat{dx^{i}}\wedge\cdots\wedge dx^{n}$ , the hat marking the omitted factor, with every $\omega_i$ supported in the box $A = [-R,R]^{n-1}\times[0,R]$ for some $R$ . Differentiating, only the term $\partial_i\omega_i\,dx^{i}$ survives in each summand, because any other $dx^{j}$ repeats a factor already present; moving $dx^{i}$ into its place costs $i-1$ transpositions, so

d\widehat\omega = \sum_{i=1}^{n} (-1)^{i-1}\,\frac{\partial\omega_i}{\partial x^{i}}\; dx^{1}\wedge\cdots\wedge dx^{n} .

Erase the wedges and integrate each term in its own variable first, which is legitimate because the integrand is continuous on the compact box.

For $i < n$ the variable $x^{i}$ runs over the whole of $[-R,R]$ , and the fundamental theorem of calculus gives $\int_{-R}^{R}\partial_i\omega_i\,dx^{i} = \omega_i\big|_{x^{i}=-R}^{\,x^{i}=R} = 0$ , because the support was enclosed strictly inside the box. Those $n-1$ terms contribute nothing.

For $i = n$ the variable runs over $[0,R]$ only, and the same theorem gives $\int_0^{R}\partial_n\omega_n\,dx^{n} = \omega_n(x',R) - \omega_n(x',0) = -\,\omega_n(x',0)$ , writing $x' = (x^1,\dots,x^{n-1})$ . Hence

\int_{\mathbb{H}^n} d\widehat\omega = (-1)^{n-1}\cdot(-1)\int_{\mathbb{R}^{n-1}} \omega_n(x',0)\,dx^{1}\cdots dx^{n-1} = (-1)^{n}\int_{\mathbb{R}^{n-1}} \omega_n(x',0)\,dx^{1}\cdots dx^{n-1}.

Step 4: the boundary side, and the sign. On $\partial\mathbb{H}^n$ the function $x^{n}$ is identically zero, so the pullback of $dx^{n}$ to the boundary is zero. Every summand of $\widehat\omega$ except the $i=n$ one contains the factor $dx^{n}$ and therefore pulls back to zero, leaving

\iota^{*}\widehat\omega = \omega_n(x',0)\,dx^{1}\wedge\cdots\wedge dx^{n-1} .

It remains to say with which sign this is integrated. The field $-\partial/\partial x^{n}$ points out of $\mathbb{H}^n$ , so by the outward vector first rule the standard frame $(\partial_1,\dots,\partial_{n-1})$ of $\partial\mathbb{H}^n$ is positively oriented exactly when $(-\partial_n,\partial_1,\dots,\partial_{n-1})$ is positively oriented for $\mathbb{R}^n$ . Moving $\partial_n$ from the front to the back is $n-1$ transpositions, so that frame has the sign $(-1)\cdot(-1)^{n-1} = (-1)^{n}$ relative to $(\partial_1,\dots,\partial_n)$ . Therefore

\int_{\partial\mathbb{H}^n}\widehat\omega = (-1)^{n}\int_{\mathbb{R}^{n-1}}\omega_n(x',0)\,dx^{1}\cdots dx^{n-1},

which is precisely the value found in step 3. For an interior chart the model is $\mathbb{R}^n$ , every variable runs over the full $[-R,R]$ , all $n$ terms vanish as the $i<n$ terms did, and the boundary side is zero because the chart does not meet $\partial M$ . $\square$

Proof(The four classical cases)AdaptedAdapted from Arfken, Weber & Harris (2013), Examples 4.7.1–4.7.3, pp. 246–248; Lee (2013), Example 16.12, p. 414

Each case is the theorem applied to a particular $\omega$ , plus a dictionary entry saying that the integral of that form is the classical integral. The dictionary is three short computations with pullbacks.

(i) Along a curve. Let $\gamma : [a,b]\to\mathbb{R}^{3}$ be smooth. Pulling back a $1$ -form, $\gamma^{*}(\mathbf{A}^{\flat}) = \bigl[\mathbf{A}(\gamma(t))\cdot\gamma'(t)\bigr]dt$ , since $\gamma^{*}(dx) = (dx^1/dt)\,dt$ and likewise for $dy, dz$ . So $\int_{\gamma}\mathbf{A}^{\flat}$ is the line integral $\int\mathbf{A}\cdot d\mathbf{r}$ of line integrals.

(ii) Over a surface patch. Let $\mathbf{r} : D\to\mathbb{R}^{3}$ , $(u,v)\mapsto\mathbf{r}(u,v)$ , be a regular injective patch. The pullback $\mathbf{r}^{*}\bigl(\beta(\mathbf{A})\bigr)$ is a $2$ -form on the two-dimensional parameter domain, so it is determined by its value on the coordinate frame $(\partial_u,\partial_v)$ there; by the definition of the pullback that value is $\beta(\mathbf{A})$ evaluated on the pushed-forward vectors $d\mathbf{r}(\partial_u) = \mathbf{r}_u$ and $d\mathbf{r}(\partial_v) = \mathbf{r}_v$ , which live in $\mathbb{R}^{3}$ . By the determinant formula for wedges of $1$ -forms,

(dy\wedge dz)(\mathbf{r}_u,\mathbf{r}_v) = y_u z_v - y_v z_u = (\mathbf{r}_u\times\mathbf{r}_v)_x ,

with the two cyclic partners giving the $y$ and $z$ components. Hence

\mathbf{r}^{*}\bigl(\beta(\mathbf{A})\bigr) = \bigl[\mathbf{A}(\mathbf{r})\cdot(\mathbf{r}_u\times\mathbf{r}_v)\bigr]\,du\wedge dv ,

and $\int_S\beta(\mathbf{A})$ is the flux $\iint_S\mathbf{A}\cdot d\mathbf{S}$ of surface integrals and flux, for the orientation in which $\mathbf{r}_u\times\mathbf{r}_v$ is the positive normal.

(iii) Over a region of $\mathbb{R}^3$ . By erasing the wedges, $\int_V \sigma(g) = \iiint_V g\,dV$ when $V$ carries the standard orientation.

Case 1. For $n=1$ and $M=[a,b]$ , $d f = f'(x)\,dx$ and the left side is $\int_a^b f'$ . The outward vector at $b$ is $+\partial_x$ and at $a$ is $-\partial_x$ , so the induced orientation of the $0$ -manifold $\{a,b\}$ assigns $+1$ to $b$ and $-1$ to $a$ , and the right side is $f(b)-f(a)$ . For a curve $\gamma$ in a manifold, apply this on $[a,b]$ to $\gamma^{*}f = f\circ\gamma$ and use naturality of $d$ to move the pullback past the derivative.

Case 2. With $\omega = P\,dx + Q\,dy$ on $D$ , the method of the exterior derivative gives $d\omega = (\partial_x Q - \partial_y P)\,dx\wedge dy$ , and both integrals are ordinary double and line integrals after erasing the wedges and using (i). For the orientation: at a boundary point let $\nu$ point out of $D$ and $t$ be the positive tangent; outward first says $(dx\wedge dy)(\nu,t) > 0$ , which means $t$ is $\nu$ rotated by $+90^{\circ}$ — the traversal with $D$ on the left.

Case 3. With $\omega = \mathbf{A}^{\flat}$ on $S$ , $d\omega = \beta(\nabla\times\mathbf{A})$ , so by (ii) the left-hand side $\int_S d\omega$ of the theorem is the flux of the curl and by (i) the right-hand side $\int_{\partial S}\omega$ is the circulation. For the orientation: at a point of $\partial S$ let $\nu$ be tangent to $S$ and pointing out of it and $t$ the positive tangent to $\partial S$ . The orientation form of $S$ is $\beta(\hat{\mathbf{n}})$ restricted to $S$ , whose value on $(\nu,t)$ is $\hat{\mathbf{n}}\cdot(\nu\times t)$ ; outward first demands this be positive, so $\nu\times t$ points along $\hat{\mathbf{n}}$ . That is the right-hand rule exactly as fixed in Stokes' theorem, where $\nu$ is the vector "tangent to $S$ and pointing into it" with the sign reversed and $t$ is the direction of travel.

Case 4. With $\omega = \beta(\mathbf{A})$ on $V$ , $d\omega = \sigma(\nabla\cdot\mathbf{A})$ , so by (iii) the left-hand side is $\iiint_V\nabla\cdot\mathbf{A}\,dV$ . On the boundary, let $\hat{\mathbf{n}}$ be the outward unit normal and $(e_1,e_2)$ a positively oriented basis of the tangent plane; outward first says $\det[\hat{\mathbf{n}}\;e_1\;e_2] > 0$ , that is, $e_1\times e_2$ is a positive multiple of $\hat{\mathbf{n}}$ . By (ii) the right-hand side is therefore the flux taken with the outward normal, which is the convention of the divergence theorem. $\square$

Remark(Where the hypotheses are used)Original

Compact support is not decoration. It is used twice in step 3: to put the support inside a box at all, and to kill the $n-1$ terms whose differentiation runs along a direction parallel to the boundary, where there is no face to collect the boundary values. Drop it and the theorem fails outright. Take $M = (0,1)$ , a $1$ -manifold with empty boundary, and $f(x) = x$ . Then $\int_M df = \int_0^1 dx = 1$ , while the boundary integral is $0$ because there is no boundary. The form $f$ is smooth and bounded but not compactly supported in $M$ : the endpoints are missing from the manifold, and the mass of $df$ escapes through them. On a compact $M$ every form is compactly supported, which is why the hypothesis is invisible in the classical theorems.
Orientability. Without a nowhere vanishing $n$ -form the left-hand side is not even defined, since the chart definition cannot be patched: on the Möbius band two charts disagree on the sign of $dx\wedge dy$ on part of their overlap, and no choice of signs in step 2 of the definition is consistent. Densities, which transform with $\lvert\det DG\rvert$ instead of $\det DG$ , can be integrated on any manifold, but they have no exterior derivative, so there is no Stokes theorem for them in this form.
The induced orientation carries the whole sign. The theorem as stated has no sign only because $\partial M$ is given the Stokes orientation. With the opposite choice every instance reads $\int_M d\omega = -\int_{\partial M}\omega$ . Arfken, Weber and Harris (Arfken, Weber & Harris (2013), p. 247) put this bluntly for the user: the formalism fixes the relative sign of the two sides and nothing more, and matching the result to a physical convention — anticlockwise, outward, right-handed — is the reader's responsibility. This is also why a sign error in a flux is the commonest failure in applications.
The sources differ on simple connectivity, and we follow Lee. Arfken, Weber and Harris (Arfken, Weber & Harris (2013), p. 245) state the theorem for a simply connected region $R$ of a $p$ -dimensional manifold, spelling out that this means one with no holes. Lee (Lee (2013), Thm 16.11, p. 411) and Hassani (Hassani (2013), Thm 28.6.3, p. 899) ask only for an oriented manifold with boundary and a compactly supported form, with no condition on its topology. The statement above follows Lee and Hassani, and nothing in the proof uses simple connectivity: step 1 needs a finite cover and a partition of unity, which any compact support supplies, and steps 3 and 4 are local. The weaker hypothesis matters in practice, since an annulus, a solid torus and a spherical shell are all admissible, their boundaries having two components; the same difference between the sources is logged for the divergence theorem, which is case 4 above.
Smooth boundary, and corners. $M$ must be a manifold with boundary, so a cube or a solid cylinder is formally excluded: at an edge no chart is a diffeomorphism onto an open subset of $\mathbb{H}^n$ . The theorem nevertheless holds for them, by an extension to manifolds with corners, or by exhausting the region with smooth ones. Arfken fills the layer next to the boundary with thin parallelepipeds and lets their thickness shrink, calling the result an indication of the theorem's validity rather than a proof; the cells he uses have corners, so what the argument needs in order to become one is exactly the extension just mentioned.
Only one derivative. Nothing in the proof differentiates twice, so $\omega$ of class $C^1$ suffices; smoothness is assumed only for uniformity with the rest of the chapter. By contrast $d\circ d = 0$ , which the proof never uses, needs two derivatives.
Partitions of unity, and hence smoothness. Step 1 is where the $C^{\infty}$ structure is spent, exactly as in the uniqueness half of the exterior derivative. It also quietly needs the cover to be finite, which compact support supplies.
Not a Riemannian statement. No metric appears anywhere above: $\int_M\omega$ needs an orientation and nothing else. Metrics enter only when one insists on integrating functions rather than forms, which is what the classical $dS$ and $dV$ do. Lee (Lee (2013), Thms 16.32 and 16.34, pp. 424 and 427) re-derives the divergence theorem and the classical Stokes theorem on an arbitrary oriented Riemannian $3$ -manifold in exactly that way, by converting $\langle \mathbf{A},\hat{\mathbf{n}}\rangle\,dA$ back into the pullback of $\beta(\mathbf{A})$ ; those are the general statements that the divergence theorem and Stokes' theorem point to.

Method(Using the theorem)Original

Fix the manifold and its dimension $n$ . The form to be integrated over $M$ must have degree $n$ ; the form on the boundary, degree $n-1$ . If the degrees do not match, the set-up is wrong.
Choose the orientation of $M$ by naming an orientation form (or, in $\mathbb{R}^n$ , a normal or a handedness), then read off the boundary orientation by the outward vector first rule. Do not guess it from a picture.
Decide which side to compute. Going from $\int_{\partial M}\omega$ to $\int_M d\omega$ pays off when $d\omega$ is simpler than $\omega$ — most dramatically when $d\omega$ is a constant multiple of the volume form, so that the boundary integral is a volume. Going the other way pays off when $\omega$ vanishes on most of $\partial M$ .
Pull back and erase the wedges. Parametrise each piece, substitute, sort into increasing order, and read the result as an ordinary multiple integral. A parametrisation that reverses orientation contributes with a minus sign.
Exploit the fact that only $\partial M$ matters. Suppose $\omega$ is closed and $S_1, S_2$ have the same oriented boundary. If the closed surface $S_1 \cup (-S_2)$ obtained by gluing $S_1$ to $S_2$ with reversed orientation is itself the boundary $\partial W$ of a region $W$ on which $\omega$ is defined and closed, then $\int_{S_1}\omega - \int_{S_2}\omega = \int_{\partial W}\omega = \int_W d\omega = 0$ , and the awkward surface may be replaced by a convenient one with the same rim. The bounding hypothesis is doing real work and has to be checked: in $\mathbb{R}^{3}\setminus\{0\}$ the upper and lower hemispheres of the unit sphere share the equator as oriented boundary, but the closed $2$ -form $\beta(\mathbf{r}/\lvert\mathbf{r}\rvert^{3})$ integrates to $2\pi$ over one and $-2\pi$ over the other, because the sphere they form bounds only a region containing the origin, where the form is not defined.
Check. The answer must be independent of the parametrisation; and if $\omega = d\eta$ is itself exact and $\partial M = \varnothing$ , the answer must be $0$ .

Example(Two sides of one identity, in three dimensions and in four)Original

(a) A surface with a rim. Let $S$ be the part of the paraboloid $z = 4 - x^{2} - y^{2}$ with $z\ge 0$ , oriented by the upward normal, so that $\partial S$ is the circle $x^{2}+y^{2} = 4$ in the plane $z = 0$ . Take

\omega = -y^{3}\,dx + x^{3}\,dy + z^{2}\,dz .

The manifold side. Differentiating, $d(-y^{3})\wedge dx = -3y^{2}\,dy\wedge dx = 3y^{2}\,dx\wedge dy$ , $d(x^{3})\wedge dy = 3x^{2}\,dx\wedge dy$ , and $d(z^{2})\wedge dz = 2z\,dz\wedge dz = 0$ , so

d\omega = 3\bigl(x^{2}+y^{2}\bigr)\,dx\wedge dy .

Parametrise $S$ as the graph $G(x,y) = (x,\,y,\,4-x^{2}-y^{2})$ over the disc $D$ of radius $2$ . Then $G^{*}(dx\wedge dy) = dx\wedge dy$ , and $G_x\times G_y = (2x,2y,1)$ points upward, so $G$ is orientation-preserving. In polar coordinates,

\int_S d\omega = \iint_D 3\bigl(x^{2}+y^{2}\bigr)\,dx\,dy = 3\int_0^{2\pi}\!\!\int_0^{2} r^{2}\cdot r\,dr\,d\theta = 3\cdot 2\pi\cdot\frac{2^{4}}{4} = 24\pi .

The boundary side. The outward direction within $S$ at the rim points away from the axis, and outward first then makes the positive traversal anticlockwise seen from above. Parametrise it by $t\mapsto(2\cos t,\,2\sin t,\,0)$ , $t\in[0,2\pi]$ . Since $z \equiv 0$ there, the $dz$ term pulls back to zero, and

\oint_{\partial S}\omega = \int_0^{2\pi}\Bigl[-8\sin^{3}t\cdot(-2\sin t) + 8\cos^{3}t\cdot 2\cos t\Bigr]dt = 16\int_0^{2\pi}\bigl(\sin^{4}t + \cos^{4}t\bigr)dt .

Using $\sin^{4}+\cos^{4} = 1 - \tfrac12\sin^{2}2t$ and $\int_0^{2\pi}\sin^{2}2t\,dt = \pi$ , this is $16\bigl(2\pi - \tfrac{\pi}{2}\bigr) = 24\pi$ . The two sides agree:

\int_S d\omega = \oint_{\partial S}\omega = \boxed{24\pi} \approx 75.3982 .

Note that $d\omega$ is closed, being exact, and is defined on all of $\mathbb{R}^{3}$ . So any other compact embedded surface $S'$ with the same oriented rim gives $24\pi$ too: $S$ together with $S'$ reversed is a closed surface in $\mathbb{R}^{3}$ , every closed surface in $\mathbb{R}^{3}$ bounds a region, and step 5 of the method applies. For the particular family of caps $z = c(4-x^{2}-y^{2})$ one can see it with no topology at all, since $d\omega$ has no $dy\wedge dz$ or $dz\wedge dx$ part and each such cap is a graph over the same disc, so the pullback of $d\omega$ does not involve $c$ .

(b) The same identity where there is no vector calculus. In $\mathbb{R}^{4}$ with coordinates $(x^1,x^2,x^3,x^4)$ , let $B$ be the closed unit ball and $S^{3} = \partial B$ the unit $3$ -sphere, with the induced orientation. Take the $3$ -form

\omega = \tfrac14\Bigl(x^{1}\,dx^{2}\wedge dx^{3}\wedge dx^{4} - x^{2}\,dx^{1}\wedge dx^{3}\wedge dx^{4} + x^{3}\,dx^{1}\wedge dx^{2}\wedge dx^{4} - x^{4}\,dx^{1}\wedge dx^{2}\wedge dx^{3}\Bigr).

Each term differentiates to a single $4$ -form: in the $i$ th one, $dx^{i}$ must be moved past $i-1$ factors, contributing $(-1)^{i-1}$ , which cancels the sign $(-1)^{i-1}$ already written. All four terms therefore give $+dx^{1}\wedge dx^{2}\wedge dx^{3}\wedge dx^{4}$ , and with the factor $\tfrac14$ ,

d\omega = dx^{1}\wedge dx^{2}\wedge dx^{3}\wedge dx^{4} .

By the theorem, $\int_{S^{3}}\omega = \int_{B}d\omega = \operatorname{vol}_4(B)$ . Slicing the ball by $x^{4} = t$ gives a $3$ -ball of radius $\sqrt{1-t^{2}}$ , so

\operatorname{vol}_4(B) = \int_{-1}^{1}\frac{4\pi}{3}\bigl(1-t^{2}\bigr)^{3/2}dt = \frac{4\pi}{3}\int_{-\pi/2}^{\pi/2}\cos^{4}\theta\,d\theta = \frac{4\pi}{3}\cdot\frac{3\pi}{8} = \frac{\pi^{2}}{2}.

Hence

\int_{S^{3}}\omega = \frac{\pi^{2}}{2} \approx 4.93480 .

An independent check. Writing $N = \sum_i x^{i}\,\partial/\partial x^{i}$ for the radial field, the bracket above is exactly $\iota_N(dx^{1}\wedge\cdots\wedge dx^{4})$ . On the unit sphere $N$ is the outward unit normal, so by the definition of the induced orientation the pullback of that bracket is the Riemannian area form of $S^{3}$ . Thus $\int_{S^3}\omega = \tfrac14\lvert S^{3}\rvert = \tfrac14\cdot 2\pi^{2} = \pi^{2}/2$ , as found. The pattern is general: $\tfrac1n\,\iota_N(dx^{1}\wedge\cdots\wedge dx^{n})$ integrates over the unit sphere of $\mathbb{R}^n$ to the volume of the ball, which for $n=3$ is the familiar statement that $\tfrac13\oint \mathbf{r}\cdot d\mathbf{S}$ is the enclosed volume.

ApplicationOriginal

Conservation laws in any dimension. A current in $n$ spacetime dimensions is an $(n-1)$ -form $J$ , and $dJ = 0$ is the continuity equation. Integrating over the slab of spacetime between two time slices, the theorem says that the charge $\int J$ on the later slice equals that on the earlier one, provided nothing leaves through the sides. That single sentence replaces the separate statements of conservation of charge, of energy and of probability, and it does so without a metric: conservation is a statement about a boundary, not about lengths. The same reading gives Gauss's law — the flux of $\star F$ through a closed surface counts the enclosed charge — and makes the consistency condition $d\star J = 0$ of the Hodge star and Maxwell's equations automatic.

Why magnetic flux through a closed surface vanishes. $\mathbf{B}$ corresponds to a closed $2$ -form, and a closed surface has empty boundary, so the flux is $\int_{\partial V}\beta(\mathbf{B}) = \int_V d\beta(\mathbf{B}) = 0$ . Conversely, the magnetic flux through two surfaces with the same rim agrees, which is why the flux in Faraday's law can be attributed to the circuit rather than to a particular spanning surface.

Topology from integrals. If $\omega$ is closed and $M$ compact without boundary, $\int_M\omega$ is unchanged when $\omega$ is altered by an exact form, since $\int_M d\eta = 0$ . So integration pairs cohomology classes with manifolds, and a non-zero value proves that $\omega$ is not exact. That is the machinery behind closed and exact forms and the de Rham theorem, behind the quantisation of magnetic charge (the flux of $F$ through a sphere surrounding a monopole is a topological invariant), and behind the Aharonov–Bohm effect, where the measurable phase is the period $\oint A$ of a locally exact but globally inexact $1$ -form.

Curvature integrals. The Gauss–Bonnet theorem of the Gauss–Bonnet theorem — total Gaussian curvature plus total geodesic curvature of the boundary equals $2\pi\chi$ — is proved by writing the curvature as $d$ of a connection form in a local frame and applying the theorem above. The index theorems of mathematical physics are descendants of the same move.

Mechanics. In the phase space of symplectic manifolds, the Poincaré invariant $\oint_{\gamma}\sum_i p_i\,dq^{i}$ equals $\int_{\Sigma}\sum_i dp_i\wedge dq^{i}$ over any surface with rim $\gamma$ , so its conservation under the Hamiltonian flow of the canonical equations is a statement about the $2$ -form rather than about any coordinates. Iterating the wedge gives Liouville's theorem.

Numerics. Finite-volume and discrete-exterior-calculus schemes store a $k$ -form as its integrals over the $k$ -cells of a mesh and define the discrete $d$ as the signed sum over cell boundaries. The theorem then holds exactly at machine precision, because it has become the statement that each interior face is counted twice with opposite signs. Conservation of mass or charge in such a solver is therefore structural rather than approximate, which is why these schemes are preferred for long-time integration of conservation laws.

Application

The Hodge star and Maxwell's equations

Adapted

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Builds the Hodge star from a metric and an orientation, proves the two sign rules $\star\star = (-1)^{\nu}(-1)^{p(n-p)}$ and the adjointness of the codifferential $\delta$ to $d$ , and uses them to compress the four Maxwell equations into $dF = 0$ and $d\star F = 4\pi \star J$ , with charge conservation falling out of $d \circ d = 0$ .

Prerequisites: The exterior derivative, The metric tensor, Differential forms, Integration of forms and the general Stokes theorem, Divergence, curl and the Laplacian, Surface integrals and flux, Inner products and norms

Note(What the star is for)Original

The exterior derivative of The exterior derivative needs no metric: $d$ is defined on any smooth manifold. That is its strength and its limitation. A $p$ -form can only be differentiated up the ladder $\Lambda^0 \to \Lambda^1 \to \dots \to \Lambda^n$ . In $\mathbb{R}^3$ the three rungs $d: \Omega^0 \to \Omega^1 \to \Omega^2 \to \Omega^3$ do carry the gradient, the curl and the divergence, so all three operators of vector calculus are there; but only the first of them acts on an object $d$ supplies by itself. To feed a vector field to the second rung one must first turn it into a 1-form, and to feed one to the third rung one must turn it into a 2-form, and neither identification exists without extra structure. Nor is there any way back down the ladder, so no Laplacian.

A metric repairs this. The spaces $\Lambda^p$ and $\Lambda^{n-p}$ of a fixed $n$ -dimensional vector space have the same dimension $\binom np = \binom{n}{n-p}$ , so they are isomorphic; a metric and an orientation single out one isomorphism between them, the Hodge star $\star$ . Conjugating $d$ by $\star$ turns it round, giving an operator $\delta$ that lowers degree, and the pair $(d, \delta)$ is enough to write every first-order differential operator of classical field theory.

The payoff this unit is built around is Maxwell's four equations. Two of them are vector equations and two are scalar, eight scalar equations in all; in the language of forms they are two equations, $dF = 0$ and $d\star F = 4\pi\star J$ , and the first of them says only that a 2-form on spacetime is closed.

Definition(Orientation, index, and the inner product on

p

-forms)AdaptedAdapted from Hassani (2013), §26.5.3, Eq. (26.52), p. 819, and §26.6, p. 820; Lee (2013), Problem 16-18(a)–(b), pp. 437–438

Let $M$ be a smooth $n$ -manifold carrying a pseudo-Riemannian metric $g$ : a smooth field of nondegenerate symmetric bilinear forms on the tangent spaces (The metric tensor). At each point $g$ can be diagonalised to $\operatorname{diag}(+1,\dots,+1,-1,\dots,-1)$ , and the number $\nu$ of $-1$ 's is the index of $g$ ; it is the same at every point of a connected $M$ . Hassani writes $\nu$ for this number and we follow him. Two cases matter here: Euclidean $\mathbb{R}^3$ , with $n = 3$ and $\nu = 0$ , and Minkowski spacetime $\mathbb{R}^{1,3}$ with coordinates $(t, x, y, z)$ , signature $(+,-,-,-)$ , $n = 4$ and $\nu = 3$ .

Assume in addition that $M$ is oriented, that is, equipped with a nowhere-vanishing $n$ -form up to positive rescaling. A $g$ -orthonormal coframe on an open set is an $n$ -tuple $(\varepsilon^1,\dots,\varepsilon^n)$ of 1-forms with

\langle \varepsilon^i, \varepsilon^j\rangle \;=\; \epsilon_i\,\delta^{ij}, \qquad \epsilon_i \in \{+1,-1\},

where $\langle\cdot,\cdot\rangle$ is the inner product that $g$ induces on the cotangent space by raising indices, and exactly $\nu$ of the signs $\epsilon_i$ equal $-1$ . The coframe is positively oriented if $\varepsilon^1\wedge\dots\wedge\varepsilon^n$ represents the chosen orientation; that $n$ -form is then the volume form $\mu$ , which Lee writes $dV_g$ and characterises, in the Riemannian case, as the only positively oriented $n$ -form whose norm is $1$ . For indefinite $g$ the corresponding normalisation is $\langle\mu,\mu\rangle = (-1)^\nu$ .

The metric induces an inner product on each $\Lambda^p(T_q^*M)$ by the determinant rule

\bigl\langle \omega^1\wedge\dots\wedge\omega^p,\ \theta^1\wedge\dots\wedge\theta^p \bigr\rangle \;=\; \det\bigl( \langle \omega^a, \theta^b\rangle \bigr)_{a,b=1}^{p} .

In components this is Hassani's $\tilde g(\alpha,\beta) = \tfrac{1}{p!}\,\alpha_{j_1\dots j_p}\beta^{j_1\dots j_p}$ . For $p = 0$ it is ordinary multiplication of functions. Writing $\varepsilon^I = \varepsilon^{i_1}\wedge\dots\wedge\varepsilon^{i_p}$ for an increasing multi-index $I = (i_1 < \dots < i_p)$ , the rule gives at once

\langle \varepsilon^I, \varepsilon^{I'}\rangle = 0 \ \ (I \ne I'), \qquad \langle \varepsilon^I, \varepsilon^{I}\rangle = \epsilon_I := \prod_{i \in I}\epsilon_i ,

because the matrix $(\langle\varepsilon^{i_a},\varepsilon^{i'_b}\rangle)$ is diagonal when $I = I'$ and has a zero row otherwise. So the $\binom np$ forms $\varepsilon^I$ are orthogonal, of norm-squared $\pm 1$ ; in the Riemannian case $\nu = 0$ they are orthonormal, which is how Lee sets the inner product up.

Definition(The Hodge star)AdaptedAdapted from Lee (2013), Problem 16-18(c)–(d), p. 438; Hassani (2013), §26.6, Definition 26.6.1 and Theorem 26.6.4, pp. 820, 822

Let $(M, g)$ be oriented with volume form $\mu$ . The Hodge star is the unique bundle map

\star : \Lambda^p T^*M \longrightarrow \Lambda^{n-p}T^*M, \qquad 0 \le p \le n,

such that

\boxed{\;\alpha \wedge \star\beta \;=\; \langle \alpha, \beta\rangle\,\mu \;} \qquad\text{for all } \alpha, \beta \in \Lambda^p .

Both sides are $n$ -forms, and the equation is required to hold pointwise for every $\alpha$ of the same degree as $\beta$ ; since the pairing $\Lambda^p \times \Lambda^{n-p} \to \Lambda^n$ is nondegenerate, this determines $\star\beta$ completely. Lee takes exactly this as the definition (for a Riemannian metric); Hassani instead writes down the coframe formula and then checks that it is basis-independent, and his Theorem 26.6.4 is the displayed identity. We adopt the identity as the definition because it makes the signs unambiguous, and derive the coframe formula from it below.

Three immediate consequences:

\star 1 = \mu, \qquad \star f = f\,\mu \ \ \text{for a function } f, \qquad \star\mu = \langle \mu, \mu\rangle = (-1)^{\nu} .

The second is Lee's Problem 16-18(d), $\star f = f\,dV_g$ , and the first is the case $f = 1$ of it. The third is not part of that problem, and we read it off the definition instead: $\star\mu$ is a $0$ -form, say the function $c$ , and putting $\alpha = \beta = \mu$ in the defining identity gives $\mu\wedge c = c\,\mu$ on the left and $\langle\mu,\mu\rangle\,\mu$ on the right, so $c = \langle\mu,\mu\rangle = \epsilon_1\cdots\epsilon_n = (-1)^\nu$ .

A caution on generality. Lee states the construction for a Riemannian metric, where $\nu = 0$ . The same characterisation is used verbatim for indefinite metrics by Hassani and by Arfken, who both tabulate the Minkowski case; the only thing that changes is that $\langle\cdot,\cdot\rangle$ is then indefinite, and nondegeneracy, which is all the argument needs, still holds. Where a sign below depends on the index we display $\nu$ explicitly.

Definition(The codifferential)AdaptedAdapted from Hassani (2013), §28.6, Definition 28.6.4, p. 900; Lee (2013), Problem 16-13 with its remark, p. 436, and Problem 16-22, pp. 438–439

Let $(M,g)$ be an oriented pseudo-Riemannian $n$ -manifold of index $\nu$ . The codifferential is

\delta : \Omega^p(M) \to \Omega^{p-1}(M), \qquad \delta\omega \;=\; (-1)^{\nu+1}\,(-1)^{\,n(p+1)}\;\star d \star \omega ,

with $\delta f = 0$ for a $0$ -form $f$ (there are no $(-1)$ -forms). This is Hassani's definition. Lee writes $d^*$ for the same operator and gives $d^*\omega = (-1)^{\,n(k+1)+1}\star d\star\omega$ on $k$ -forms; the two agree, since Lee works with a Riemannian metric, for which $\nu = 0$ and $(-1)^{\nu+1} = -1$ .

The sign is chosen so that $\delta$ is the formal adjoint of $d$ , which is the content of the statement below. Two specialisations worth recording:

Euclidean $\mathbb{R}^3$ ( $n = 3$ , $\nu = 0$ ): $(-1)^{\nu+1} = -1$ and $(-1)^{3(p+1)} = (-1)^{p+1}$ , so $\delta = -\star d\star$ on 1-forms and on 3-forms, and $\delta = +\star d\star$ on 2-forms. The degree-dependence of the sign in odd dimensions is a standing trap.
Minkowski spacetime ( $n = 4$ , $\nu = 3$ ): $(-1)^{\nu+1} = +1$ and $(-1)^{4(p+1)} = +1$ , so $\delta = \star d \star$ in every degree.

The operator $\Delta = d\delta + \delta d$ on $\Omega^p(M)$ is the Hodge Laplacian. On functions in Euclidean $\mathbb{R}^3$ it is $\Delta f = \delta\, df = -\star d \star df = -\nabla^2 f$ , the negative of the Laplacian of vector calculus. Lee notes that there is no agreement in the literature about this sign and that the "geometric" sign used here is the one that makes the eigenvalues non-negative; we flag the convention wherever it is used rather than silently choosing one.

Application(The two sign rules and the adjointness of

\delta

)AdaptedAdapted from Hassani (2013), §26.6, Theorem 26.6.3, p. 822, and §28.6, Theorem 28.6.6, p. 901; Lee (2013), Problem 16-18(e) and Problem 16-22, pp. 438–439

Let $(M,g)$ be an oriented pseudo-Riemannian $n$ -manifold of index $\nu$ .

(a) The coframe formula. Let $(\varepsilon^1,\dots,\varepsilon^n)$ be a positively oriented $g$ -orthonormal coframe, let $I$ be an increasing multi-index of length $p$ and let $J$ be its increasing complement. Then

\star\,\varepsilon^I \;=\; \epsilon_I\,\sigma(I,J)\;\varepsilon^J , \qquad \epsilon_I = \prod_{i\in I}\epsilon_i ,

where $\sigma(I,J) = \pm 1$ is the sign of the permutation carrying $(1,\dots,n)$ to $(I, J)$ , that is, $\varepsilon^I \wedge \varepsilon^J = \sigma(I,J)\,\mu$ .

(b) Applying the star twice. On $p$ -forms,

\star\star \;=\; (-1)^{\nu}\,(-1)^{\,p(n-p)}\,\mathrm{id} .

In particular $\star\star = \mathrm{id}$ on every Euclidean space of odd dimension, and $\star\star = -\,\mathrm{id}$ on $2$ -forms over Minkowski spacetime.

(c) $\delta$ is the formal adjoint of $d$ . Let $\alpha \in \Omega^{p}(M)$ and $\beta \in \Omega^{p+1}(M)$ , and suppose $\alpha\wedge\star\beta$ has compact support. If $\partial M = \varnothing$ , then

\int_M \langle d\alpha, \beta\rangle\,\mu \;=\; \int_M \langle \alpha, \delta\beta\rangle\,\mu .

For a compact oriented Riemannian manifold without boundary this is Lee's $(d^*\omega, \eta) = (\omega, d\eta)$ with respect to the $L^2$ inner product $(\omega,\eta) = \int_M \langle\omega,\eta\rangle\,dV_g$ ; Hassani states it in the form above, for a general index and with the compact-support hypothesis in place of compactness of $M$ .

(d) $\delta \circ \delta = 0$ , as $d\circ d = 0$ and $\star$ is invertible by (b).

Idea of the proofOriginal

All three parts come from one observation: in an orthonormal coframe, the wedge product of basis forms is as rigid as it could be. For increasing multi-indices $I$ and $K$ of complementary lengths, $\varepsilon^I\wedge\varepsilon^K$ is zero unless $K$ is exactly the complement $J$ of $I$ , and then it is $\pm\mu$ . So the defining equation $\alpha\wedge\star\beta = \langle\alpha,\beta\rangle\mu$ , tested on the $\binom np$ basis forms $\alpha = \varepsilon^I$ , is a diagonal system: it determines the coefficient of $\varepsilon^J$ in $\star\varepsilon^I$ and forces every other coefficient to vanish, which is what Step 2 of the proof carries out. That gives (a), and existence and uniqueness of $\star$ with it.

Part (b) is then bookkeeping with two signs. Starring twice multiplies the norm factors $\epsilon_I\epsilon_J$ , which is the product of all the $\epsilon_i$ and hence $(-1)^\nu$ ; and it multiplies the two permutation signs $\sigma(I,J)\sigma(J,I)$ , which differ by the cost of sliding a block of $p$ letters past a block of $n-p$ letters, namely $(-1)^{p(n-p)}$ .

Part (c) is Stokes' theorem wearing a disguise. The Leibniz rule applied to the $(n-1)$ -form $\alpha\wedge\star\beta$ produces the two terms $d\alpha\wedge\star\beta$ and $\pm\,\alpha\wedge d\star\beta$ ; the first is $\langle d\alpha,\beta\rangle\mu$ by definition of $\star$ , and the sign in $\delta$ is rigged precisely so that the second is $-\langle\alpha,\delta\beta\rangle\mu$ . The difference of the two integrands is then an exact $n$ -form, whose integral vanishes.

Proof(Parts (a) and (b))AdaptedAdapted from Hassani (2013), §26.6, Definition 26.6.1 and Theorem 26.6.3, pp. 820, 822; Lee (2013), Problem 16-18(c), (e), p. 438

Step 1 (the candidate satisfies the defining identity). Fix an increasing multi-index $I$ of length $p$ with complement $J$ , and put $\tau = \epsilon_I\,\sigma(I,J)\,\varepsilon^J$ . Let $I'$ be any increasing multi-index of length $p$ . If $I' \ne I$ then $I'$ contains some index belonging to $J$ , so $\varepsilon^{I'}\wedge\varepsilon^J = 0$ because a basis 1-form is repeated; and $\langle\varepsilon^{I'},\varepsilon^I\rangle = 0$ as well. If $I' = I$ then

\varepsilon^{I}\wedge\tau = \epsilon_I\,\sigma(I,J)\,\varepsilon^I\wedge\varepsilon^J = \epsilon_I\,\sigma(I,J)^2\,\mu = \epsilon_I\,\mu = \langle\varepsilon^I,\varepsilon^I\rangle\,\mu ,

using $\sigma(I,J)^2 = 1$ and the value of $\langle\varepsilon^I,\varepsilon^I\rangle$ from the first definition. So $\varepsilon^{I'}\wedge\tau = \langle\varepsilon^{I'},\varepsilon^I\rangle\mu$ for every $I'$ , and by linearity for every $\alpha \in \Lambda^p$ . Hence $\tau$ is a legitimate value for $\star\varepsilon^I$ .

Step 2 (nothing else will do). Suppose $\tau$ and $\tau'$ both satisfy the identity for a given $\beta$ . Then $\alpha\wedge(\tau - \tau') = 0$ for every $\alpha \in \Lambda^p$ . Write $\tau - \tau' = \sum_K c_K\,\varepsilon^K$ over increasing multi-indices $K$ of length $n-p$ . Taking $\alpha = \varepsilon^{I}$ with $I$ the complement of a particular $K_0$ kills every term except $K = K_0$ and leaves $c_{K_0}\,\sigma(I,K_0)\,\mu = 0$ , so $c_{K_0} = 0$ . Thus $\tau = \tau'$ , and $\star$ is well defined on basis forms and extends uniquely by linearity. The formula was derived in one coframe, but the object it defines satisfies a coframe-free characterisation, so it does not depend on the coframe chosen — this is the point Hassani verifies by hand on his p. 821, where he also records the general-basis version

\star\omega \;=\; |G|^{1/2}\,\frac{1}{p!\,(n-p)!}\;\bar\omega_{i_1\dots i_p}\;\mathcal{E}_{i_1 i_2\dots i_n}\;\theta^{i_{p+1}}\wedge\dots\wedge\theta^{i_n} ,

for a dual basis $(\theta^1,\dots,\theta^n)$ that need not be orthonormal; the determinant $|G|$ of the metric in that basis appears because $|G|^{1/2} = 1$ in an orthonormal one. Two notational warnings about this quotation. First, $\mathcal{E}_{i_1\dots i_n}$ is the Levi-Civita symbol: Hassani writes it $\epsilon$ , and we have renamed it because $\epsilon_i$ (the metric signs) and $\varepsilon^i$ (the coframe) are both in use here. Second, the indices on $\bar\omega_{i_1\dots i_p}$ are lowered, exactly as the page prints them, because in his notation they are obtained by evaluating the index-raised form $\tilde\omega = g^{-1}\omega$ on the basis covectors $\theta^{i_1},\dots,\theta^{i_p}$ ; the companion line of his Eq. (26.56) writes the same object with raised $\bar\omega^{i_1\dots i_p}$ , a prefactor $1/p!$ and tensor products in place of the wedge.

Step 3 (starring twice). Apply (a) twice. The complement of $J$ is $I$ , so

\star\star\varepsilon^I \;=\; \epsilon_I\,\sigma(I,J)\;\star\varepsilon^J \;=\; \epsilon_I\,\epsilon_J\;\sigma(I,J)\,\sigma(J,I)\;\varepsilon^I .

Now $\epsilon_I\epsilon_J = \prod_{i=1}^n \epsilon_i = (-1)^{\nu}$ , since $I$ and $J$ partition $\{1,\dots,n\}$ and exactly $\nu$ of the signs are negative — this is where the index enters, and it is the only place it does. For the permutation signs, moving the block $J$ of $n-p$ letters in front of the block $I$ of $p$ letters takes $p(n-p)$ adjacent transpositions, so $\sigma(J,I) = (-1)^{p(n-p)}\sigma(I,J)$ and

\sigma(I,J)\,\sigma(J,I) = (-1)^{p(n-p)}\sigma(I,J)^2 = (-1)^{p(n-p)} .

Hence $\star\star\varepsilon^I = (-1)^\nu(-1)^{p(n-p)}\varepsilon^I$ for every $I$ , and by linearity on all of $\Lambda^p$ . $\square$

Proof(Part (c))AdaptedAdapted from Hassani (2013), §28.6, Theorem 28.6.6 and its proof, p. 901

Let $\alpha$ be a $p$ -form and $\beta$ a $(p+1)$ -form, with $\alpha\wedge\star\beta$ compactly supported.

Step 1 (star the codifferential). Apply the definition of $\delta$ in degree $p+1$ :

\star\,\delta\beta \;=\; (-1)^{\nu+1}(-1)^{\,n(p+2)}\ \star\star\,(d\star\beta) .

Now $\star\beta$ is an $(n-p-1)$ -form, so $d\star\beta$ is an $(n-p)$ -form, and part (b) in degree $n-p$ gives the factor $(-1)^{\nu}(-1)^{(n-p)p}$ . Multiplying the signs,

(-1)^{\nu+1}(-1)^{\nu}\,(-1)^{\,n(p+2)}(-1)^{\,p(n-p)} = -\,(-1)^{\,n(p+2)+p(n-p)} = -\,(-1)^{\,2n(p+1)-p^{2}} = -\,(-1)^{\,p^{2}} = -(-1)^{p},

where $(-1)^{\nu+1}(-1)^{\nu} = -1$ whatever $\nu$ is, the exponent collects as $n(p+2)+p(n-p) = 2n(p+1) - p^2$ , the even part $2n(p+1)$ contributes nothing, and $(-1)^{p^2} = (-1)^p$ . Therefore

\star\,\delta\beta \;=\; -(-1)^{p}\, d\star\beta .

This is the one computation in which the exact constant in the definition of $\delta$ matters; every other sign in the unit is forced.

Step 2 (the Leibniz rule). $\alpha$ is a $p$ -form and $\star\beta$ an $(n-p-1)$ -form, so the antiderivation property of $d$ (The exterior derivative) gives

d(\alpha\wedge\star\beta) \;=\; d\alpha\wedge\star\beta \;+\; (-1)^{p}\,\alpha\wedge d\star\beta .

Step 3 (identify the two terms). By the defining property of $\star$ in degree $p+1$ , $d\alpha\wedge\star\beta = \langle d\alpha,\beta\rangle\mu$ . By Step 1, $(-1)^p\,\alpha\wedge d\star\beta = (-1)^p \alpha \wedge \bigl(-(-1)^p \star\delta\beta\bigr) = -\,\alpha\wedge\star\delta\beta = -\langle\alpha,\delta\beta\rangle\mu$ , the last equality being the defining property of $\star$ in degree $p$ . Hence

\langle d\alpha,\beta\rangle\,\mu \;-\; \langle\alpha,\delta\beta\rangle\,\mu \;=\; d\bigl(\alpha\wedge\star\beta\bigr) .

Step 4 (integrate). On the right-hand side stands $d$ applied to an $(n-1)$ -form of compact support, and $M$ is oriented with empty boundary, so that integral is zero by the general Stokes theorem. Integrating the identity of Step 3 therefore gives

\int_M \langle d\alpha,\beta\rangle\,\mu = \int_M \langle\alpha,\delta\beta\rangle\,\mu . \qquad\square

Remark(Where each hypothesis is used)Original

Nondegeneracy of $g$ . Used to raise indices at all, hence to define $\langle\cdot,\cdot\rangle$ on 1-forms, and used again in Step 2 of the first proof: if the pairing $\Lambda^p\times\Lambda^{n-p}\to\Lambda^n$ were degenerate, $\star\beta$ would not be unique. A degenerate "metric" has no orthonormal coframe in the sense used here and no Hodge star.
Orientability. $\star$ depends on $\mu$ , and reversing the orientation replaces $\mu$ by $-\mu$ and therefore $\star$ by $-\star$ . There is no Hodge star on a non-orientable manifold such as the Möbius band, although $\star\star$ , $\delta$ and the Hodge Laplacian, in each of which $\star$ occurs twice, survive the sign flip and do make sense there.
The index $\nu$ . It enters only through $\epsilon_I\epsilon_J = (-1)^\nu$ in Step 3 of the first proof. Dropping it — using the Riemannian sign rule in Minkowski space, where $\nu = 3$ and the omitted factor is $(-1)^3 = -1$ — would flip the sign of $\star\star$ in every degree, odd and even alike, and with it the sign of $\delta$ and hence of the source term in Maxwell's equations below.
Compact support in part (c). Used only at Step 4, to kill $\int_M d(\alpha\wedge\star\beta)$ . Without it the identity acquires a boundary term $\int_{\partial M}\alpha\wedge\star\beta$ , and the operators are adjoint only up to that term; this is the differential-forms version of the boundary terms in Green's identities. On a manifold with boundary the same proof gives $\int_M\langle d\alpha,\beta\rangle\mu - \int_M\langle\alpha,\delta\beta\rangle\mu = \int_{\partial M}\alpha\wedge\star\beta$ .
$\partial M = \varnothing$ in part (c). Same step, same reason. Hassani assumes only that $\alpha\wedge\star\beta$ has compact support and quotes Stokes' theorem in the form $\int_M d\omega = 0$ ; Lee assumes $M$ compact without boundary instead, which makes the compact-support hypothesis automatic.
Positive orientation of the coframe in part (a). If the coframe is negatively oriented, $\varepsilon^1\wedge\dots\wedge\varepsilon^n = -\mu$ and every sign in (a) reverses. In practice this is the most common source of error, and it is why Arfken insists on fixing a "standard order" of the basis differentials before starting.

Method(Computing

\star

by hand)AdaptedAdapted from Arfken, Weber & Harris (2013), §4.5, pp. 235–237, Eqs. (4.81) and (4.82)

Arfken's recipe, which is part (a) of the statement in practical dress:

Fix the standard order of the basis 1-forms, so that $\mu = \varepsilon^1\wedge\dots\wedge\varepsilon^n$ in that order, and record the metric signs $\epsilon_1,\dots,\epsilon_n$ .
Write the form to be starred as a combination of basis monomials $\varepsilon^I$ with the factors in increasing order; linearity handles the rest.
Take the complement. For each $\varepsilon^I$ , write the wedge $\varepsilon^J$ of all the basis 1-forms not appearing in $\varepsilon^I$ , in increasing order.
Permutation sign. Attach $\sigma(I,J)$ : the sign of the permutation that brings the list (indices of $\varepsilon^I$ , then indices of $\varepsilon^J$ ) back to standard order.
Metric sign. Multiply by $\epsilon_I = (-1)^{m_I}$ , where $m_I$ is the number of indices in $I$ , that is, in the form being starred, whose metric sign is $-1$ (the letter $\mu$ is reserved here for the volume form). In Euclidean space this factor is always $+1$ and step 5 may be skipped.

A note on Arfken's wording in step 5: his prose on p. 236 says to count the negative metric entries among the differentials of the complementary form, but each of his worked cases on pp. 236–237 counts them among the differentials of the form being starred. For $\star 1$ he takes the count to be $0$ on the ground that $\omega = 1$ carries no differentials at all — although its complement $dt\wedge dx\wedge dy\wedge dz$ carries three with negative metric sign — and for $\star(dt\wedge dx)$ he attributes the minus sign to the $dx$ inside the form being starred. The two prescriptions differ by the overall factor $(-1)^\nu$ and cannot both be right. We follow his examples, whose results are Eq. (4.82) and agree term by term with Hassani's Definition 26.6.1 and with the characterisation $\alpha\wedge\star\beta = \langle\alpha,\beta\rangle\mu$ adopted above; the discrepancy in his prose is recorded in verification/discrepancies.md.

In Euclidean $\mathbb{R}^3$ with $\mu = dx\wedge dy\wedge dz$ this gives Arfken's Eq. (4.81):

\star 1 = dx\wedge dy\wedge dz, \quad \star\,dx = dy\wedge dz, \quad \star\,dy = dz\wedge dx, \quad \star\,dz = dx\wedge dy,

\star(dy\wedge dz) = dx, \quad \star(dz\wedge dx) = dy, \quad \star(dx\wedge dy) = dz, \quad \star(dx\wedge dy\wedge dz) = 1 ,

all signs coming out $+$ because the indices have been written in cyclic order. In Minkowski spacetime with $\mu = dt\wedge dx\wedge dy\wedge dz$ , $\epsilon_t = +1$ and $\epsilon_x = \epsilon_y = \epsilon_z = -1$ , one gets Arfken's Eq. (4.82); with $(i,j,k)$ any cyclic permutation of $(x,y,z)$ ,

\star\,dt = dx\wedge dy\wedge dz, \qquad \star\,dx^i = dt\wedge dx^j\wedge dx^k ,

\star(dx^j\wedge dx^k) = dt\wedge dx^i, \qquad \star(dt\wedge dx^i) = -\,dx^j\wedge dx^k ,

\star(dt\wedge dx^i\wedge dx^j) = dx^k, \qquad \star(dx\wedge dy\wedge dz) = dt, \qquad \star\,\mu = -1 .

A worked instance of step 5: for $\star(dt\wedge dx)$ the complement is $dy\wedge dz$ , already in standard order so $\sigma = +1$ ; one index of $I = (t,x)$ , namely $x$ , has metric sign $-1$ , so $\epsilon_I = -1$ and $\star(dt\wedge dx) = -dy\wedge dz$ . These tables are what the Maxwell computation below runs on.

Reading off the vector operators. In Euclidean $\mathbb{R}^3$ , write $X^\flat$ for the 1-form $X_1\,dx + X_2\,dy + X_3\,dz$ obtained from a vector field $X$ by lowering its index, and $\omega^\sharp$ for the inverse operation. Then Arfken's Eqs. (4.86)–(4.90) and Lee's Problem 16-21 read

df = (\nabla f)^\flat, \qquad \star\,d\,X^\flat = (\nabla\times X)^\flat, \qquad \star\,d\star X^\flat = \nabla\cdot X , \qquad \star(X^\flat\wedge Y^\flat) = (X\times Y)^\flat ,

the last being Arfken's Eq. (4.83) and Hassani's Example 26.6.5, and also the reason the cross product exists only in three dimensions: only there do $\Lambda^1$ and $\Lambda^{2}$ have the same dimension. Combining the second and third, $\delta\,df = -\star d\star df = -\nabla^2 f$ , and the vector identities $\nabla\times\nabla f = 0$ and $\nabla\cdot(\nabla\times X) = 0$ of Vector identities are both the single statement $d\circ d = 0$ .

Example(The star of an electromagnetic field, and the two invariants)Original

Work in Minkowski spacetime with coordinates $(t,x,y,z)$ , signature $(+,-,-,-)$ , orientation $\mu = dt\wedge dx\wedge dy\wedge dz$ , and Gaussian units with $c = 1$ , following Hassani. The Faraday 2-form of an electric field $E = (E_x,E_y,E_z)$ and a magnetic field $B = (B_x,B_y,B_z)$ is Hassani's Eq. (28.38),

F \;=\; -E_x\,dt\wedge dx - E_y\,dt\wedge dy - E_z\,dt\wedge dz \;+\; B_z\,dx\wedge dy - B_y\,dx\wedge dz + B_x\,dy\wedge dz .

Take the uniform field

E = (0,\,0,\,3), \qquad B = (2,\,0,\,5)

in those units, so that

F = -3\,dt\wedge dz \;+\; 5\,dx\wedge dy \;+\; 2\,dy\wedge dz .

Step 1: apply the star term by term, using the Minkowski table of the method block.

\star(dt\wedge dz) = -\,dx\wedge dy, \qquad \star(dx\wedge dy) = dt\wedge dz, \qquad \star(dy\wedge dz) = dt\wedge dx .

Hence

\star F = -3\,(-dx\wedge dy) + 5\,(dt\wedge dz) + 2\,(dt\wedge dx) = 3\,dx\wedge dy + 5\,dt\wedge dz + 2\,dt\wedge dx .

Comparing with the general form of $F$ , this is the field with $E' = -B = (-2,0,-5)$ and $B' = E = (0,0,3)$ : indeed $-E'_x\,dt\wedge dx = 2\,dt\wedge dx$ , $-E'_z\,dt\wedge dz = 5\,dt\wedge dz$ and $B'_z\,dx\wedge dy = 3\,dx\wedge dy$ reproduce the three terms. In general

\star F \;=\; E_x\,dy\wedge dz - E_y\,dx\wedge dz + E_z\,dx\wedge dy + B_x\,dt\wedge dx + B_y\,dt\wedge dy + B_z\,dt\wedge dz ,

which is $F$ with $E \mapsto -B$ and $B \mapsto E$ . (Hassani computes the same object in index notation in his Example 28.5.13, pp. 893–894, which the proof of the translation below follows; here it drops out of the Minkowski table of the method block.) Matching coefficients term by term, the $dt\wedge dx^i$ part of $\star F$ carries $B_i$ where $F$ carries $-E_i$ , and the spatial part carries $E$ where $F$ carries $B$ . This is the electromagnetic duality rotation by a quarter turn, and a quarter turn squared is $-1$ , which is the next step.

Step 2: check $\star\star F = -F$ . Starring again, with $\star(dx\wedge dy) = dt\wedge dz$ , $\star(dt\wedge dz) = -dx\wedge dy$ and $\star(dt\wedge dx) = -dy\wedge dz$ ,

\star\star F = 3\,dt\wedge dz - 5\,dx\wedge dy - 2\,dy\wedge dz = -F ,

as part (b) predicts: $n = 4$ , $p = 2$ , $\nu = 3$ give $(-1)^3(-1)^{2\cdot 2} = -1$ .

Step 3: the first invariant, $F\wedge\star F$ . Only terms whose index sets are complementary survive. Writing $\mu = dt\wedge dx\wedge dy\wedge dz$ and reducing each product to $\mu$ by counting transpositions,

(dt\wedge dz)\wedge(dx\wedge dy) = \mu, \qquad (dx\wedge dy)\wedge(dt\wedge dz) = \mu, \qquad (dy\wedge dz)\wedge(dt\wedge dx) = \mu .

Therefore

F\wedge\star F = (-3)(3)\,\mu + (5)(5)\,\mu + (2)(2)\,\mu = (-9 + 25 + 4)\,\mu = 20\,\mu .

Since $F\wedge\star F = \langle F,F\rangle\mu$ by the definition of $\star$ , we read off

\langle F, F\rangle = 20 = \lvert B\rvert^2 - \lvert E\rvert^2 = (4 + 0 + 25) - 9 .

The general identity $\langle F,F\rangle = \lvert B\rvert^2 - \lvert E\rvert^2$ follows the same way with symbols in place of numbers.

Step 4: the second invariant, $F\wedge F$ . Now the surviving products pair an $E$ term with a $B$ term:

F\wedge F = 2\,(-3\,dt\wedge dz)\wedge(5\,dx\wedge dy) = -30\,\mu ,

the factor $2$ because the two cross terms of the square are equal ( $F\wedge F$ is symmetric for a $2$ -form), and the $dy\wedge dz$ term contributing nothing because it shares $dz$ with the $dt\wedge dz$ term. In general $F\wedge F = -2\,(E\cdot B)\,\mu$ , and here $E\cdot B = 3\times 5 = 15$ gives $-30$ , which is what we got.

Step 5: the number. The free-field Lagrangian density of electromagnetism in these units is $\mathcal{L} = (\lvert E\rvert^2 - \lvert B\rvert^2)/8\pi = -\langle F,F\rangle/8\pi$ , so for this field

\mathcal{L} = -\frac{20}{8\pi} = -\frac{5}{2\pi} \approx \boxed{-0.79577} ,

while the energy density is $(\lvert E\rvert^2 + \lvert B\rvert^2)/8\pi = 38/(8\pi) = 19/(4\pi) \approx 1.5120$ . Both invariants are frame-independent: $\langle F,F\rangle = 20 > 0$ says the field is magnetically dominated in every inertial frame, and $E\cdot B = 15 \ne 0$ says that no boost can make either field vanish.

Application(Maxwell's equations in the language of forms)AdaptedAdapted from Hassani (2013), §28.5, Eq. (28.38) and Box 28.5.6, pp. 889–890, with Eqs. (28.43)–(28.46) and Box 28.5.14, pp. 893–894

On Minkowski spacetime with signature $(+,-,-,-)$ , orientation $dt\wedge dx\wedge dy\wedge dz$ and Gaussian units with $c = 1$ , let $F$ be the Faraday 2-form built from $(E, B)$ as above, and let

J = J_\alpha\,dx^\alpha, \qquad J_0 = \rho, \quad (J_x, J_y, J_z) = -(J^x, J^y, J^z),

be the four-current 1-form assembled from the charge density $\rho$ and the current density $\mathbf J = (J^x,J^y,J^z)$ . Then

\boxed{\;dF = 0\;} \quad\Longleftrightarrow\quad \nabla\times E + \frac{\partial B}{\partial t} = 0 \ \text{ and } \ \nabla\cdot B = 0 ,

\boxed{\;d\star F = 4\pi\star J\;} \quad\Longleftrightarrow\quad \nabla\times B = \frac{\partial E}{\partial t} + 4\pi \mathbf J \ \text{ and } \ \nabla\cdot E = 4\pi\rho .

Equivalently, since $\delta = \star d\star$ in this signature and dimension and $\star\star J = J$ on 1-forms, the second equation reads $\delta F = 4\pi J$ . Hassani states the first as his Box 28.5.6 and the second as his Box 28.5.14.

Idea of the proofOriginal

Both equivalences are bookkeeping, and the shape of the bookkeeping is forced by counting. A $2$ -form in four dimensions has six components, which $F$ splits into the three $dt\wedge dx^i$ carrying $E$ and the three spatial ones carrying $B$ ; its exterior derivative is a $3$ -form, which has exactly four components. One of those four carries no $dt$ , so it can only collect spatial derivatives of the spatial part of $F$ — a divergence; the other three carry a $dt$ , so each collects one time derivative of the spatial part and one spatial derivative of the mixed part — a curl. That is why $dF = 0$ can only say " $\nabla\cdot B = 0$ and $\nabla\times E + \partial_t B = 0$ ", with no room for anything else.

The second equivalence is then the first one dualised. By Step 1 of the worked example, $\star$ sends $F$ to the Faraday form of $(-B, E)$ , so $d\star F$ is $dF$ with $E \mapsto -B$ and $B\mapsto E$ , which turns Faraday's law into Ampère's and $\nabla\cdot B = 0$ into Gauss's law. All that remains is to star $J$ and read off the factor $4\pi$ , which is the Gaussian-units convention and nothing more.

Proof(The translation)AdaptedAdapted from Hassani (2013), §28.5, Example 28.5.5, pp. 889–890 and Example 28.5.13, pp. 893–894

Step 1: the homogeneous pair. Differentiate $F$ term by term, using $d(f\,dx^\alpha\wedge dx^\beta) = df\wedge dx^\alpha\wedge dx^\beta$ and discarding every term in which a differential repeats. From $-E_z\,dt\wedge dz$ only $\partial_x E_z$ and $\partial_y E_z$ survive; from $B_z\,dx\wedge dy$ only $\partial_t B_z$ and $\partial_z B_z$ ; and so on. Reordering each surviving product into the standard order and collecting, the result is

dF = \Bigl(\nabla\times E + \frac{\partial B}{\partial t}\Bigr)_{\!z} dt\wedge dx\wedge dy + \Bigl(\nabla\times E + \frac{\partial B}{\partial t}\Bigr)_{\!y} dt\wedge dz\wedge dx + \Bigl(\nabla\times E + \frac{\partial B}{\partial t}\Bigr)_{\!x} dt\wedge dy\wedge dz + (\nabla\cdot B)\,dx\wedge dy\wedge dz .

The four basis $3$ -forms $dt\wedge dx\wedge dy$ , $dt\wedge dz\wedge dx$ , $dt\wedge dy\wedge dz$ and $dx\wedge dy\wedge dz$ are linearly independent, so $dF = 0$ holds if and only if all four coefficients vanish, which is Faraday's law together with the absence of magnetic monopoles. This is Hassani's Example 28.5.5.

Step 2: star $F$ . By Step 1 of the worked example,

\star F = E_x\,dy\wedge dz - E_y\,dx\wedge dz + E_z\,dx\wedge dy + B_x\,dt\wedge dx + B_y\,dt\wedge dy + B_z\,dt\wedge dz .

Step 3: differentiate $\star F$ . The spatial terms contribute $\partial_x E_x + \partial_y E_y + \partial_z E_z$ to $dx\wedge dy\wedge dz$ , together with their time derivatives; the $B$ terms contribute spatial derivatives only. Collecting as in Step 1,

d\star F = \Bigl(\frac{\partial E}{\partial t} - \nabla\times B\Bigr)_{\!z} dt\wedge dx\wedge dy + \Bigl(\frac{\partial E}{\partial t} - \nabla\times B\Bigr)_{\!y} dt\wedge dz\wedge dx + \Bigl(\frac{\partial E}{\partial t} - \nabla\times B\Bigr)_{\!x} dt\wedge dy\wedge dz + (\nabla\cdot E)\,dx\wedge dy\wedge dz ,

which is Hassani's Eq. (28.44). Note that this is the same pattern as Step 1 with $E \mapsto -B$ , $B \mapsto E$ , exactly as the duality in Step 1 of the example predicts: substituting those into Step 1's coefficients $\nabla\times E + \partial_t B$ and $\nabla\cdot B$ gives $\partial_t E - \nabla\times B$ and $\nabla\cdot E$ , the coefficients above.

Step 4: star $J$ . From the Minkowski table, $\star\,dt = dx\wedge dy\wedge dz$ , $\star\,dx = dt\wedge dy\wedge dz$ , $\star\,dy = dt\wedge dz\wedge dx$ and $\star\,dz = dt\wedge dx\wedge dy$ . Hence, using $J_0 = \rho$ and $J_i = -J^i$ ,

\star J = \rho\,dx\wedge dy\wedge dz \;-\; J^x\,dt\wedge dy\wedge dz \;-\; J^y\,dt\wedge dz\wedge dx \;-\; J^z\,dt\wedge dx\wedge dy ,

Hassani's Eq. (28.46).

Step 5: match coefficients. Setting $d\star F = 4\pi\star J$ and equating the coefficients of the four independent basis $3$ -forms gives

\nabla\cdot E = 4\pi\rho , \qquad \Bigl(\frac{\partial E}{\partial t} - \nabla\times B\Bigr)_{\!i} = -4\pi J^i \ \ (i = x,y,z),

the second of which is the Ampère–Maxwell law $\nabla\times B = \partial_t E + 4\pi\mathbf J$ . $\square$

Remark(Conventions, and what changes in SI units)Original

Three conventions have been fixed, and all three are visible in the formulas:

Signature and orientation. $(+,-,-,-)$ with $\mu = dt\wedge dx\wedge dy\wedge dz$ , so $\nu = 3$ . Hassani and Arfken both use this; a reader following a book with signature $(-,+,+,+)$ will find $\nu = 1$ , so $(-1)^\nu = -1$ is the same as ours and every $\star\star$ rule is unchanged, $\star\star = -\mathrm{id}$ on $2$ -forms included. The star itself does change: the count $m_I$ of negative metric signs in $I$ changes parity for $I$ of length $1$ and of length $3$ , so $\star$ reverses sign on $1$ -forms and on $3$ -forms, while on $2$ -forms the two changes cancel and $\star$ — hence $\star F$ — is unaffected. The safest practice is to recompute the short table of the method block in whatever convention is in force before trusting any formula containing a star.
Units. Gaussian with $c = 1$ , which is why $4\pi$ appears on the right of $d\star F = 4\pi\star J$ and the Lagrangian carries $1/8\pi$ . In SI units the same equations read $dF = 0$ and $d\star F = \mu_0\star J$ , where $\mu_0$ — in this sentence only, the letter $\mu$ being the volume form everywhere else in the unit — denotes the vacuum permeability; $c$ is restored by using $x^0 = ct$ and measuring $E/c$ rather than $E$ against $B$ .
$F$ versus $\star F$ . Hassani's $F$ is the one whose $E$ components sit on $dt\wedge dx^i$ . Some treatments start from the dual 2-form instead; the equations then swap, $d\star F = 0$ becoming the homogeneous pair. The physical content is the same, but the statement "the homogeneous equations say $F$ is closed" is not convention-free.

The quantity $\lvert B\rvert^2 - \lvert E\rvert^2 = \langle F,F\rangle$ and the quantity $E\cdot B$ , extracted in the worked example, are the only two independent invariants one can build pointwise from $F$ alone with the metric and the orientation. The count of $4$ -forms proves only half of that: $F\wedge\star F$ and $F\wedge F$ are the only two natural $4$ -forms quadratic in $F$ , so there are exactly two independent quadratic invariants, and nothing in the count excludes an independent invariant of higher degree. That none exists is a separate standard fact which we quote rather than prove; for a $2$ -form in four dimensions it follows from the Cayley–Hamilton theorem, which expresses the trace of every power of the mixed tensor $F^{\alpha}{}_{\beta}$ in terms of $\operatorname{tr}(F^2)$ and $\det F$ — and those two are, up to constants, $\langle F,F\rangle$ and $(E\cdot B)^2$ .

Application(Gauss's law, charge conservation, and the four-potential)Original

Gauss's law is Stokes' theorem. Let a point charge $q$ sit at rest at the origin, so that $E = q\,\mathbf r/r^3$ and $B = 0$ . Then by Step 2 of the translation,

\star F = \frac{q}{r^3}\bigl( x\,dy\wedge dz + y\,dz\wedge dx + z\,dx\wedge dy \bigr)

on the slice $t = \text{const}$ , where $r^2 = x^2+y^2+z^2$ . This is a 2-form, and we integrate it over the sphere $S_R$ of radius $R$ in that slice, oriented outwards. The bracket is the flux 2-form of the position field $\mathbf r$ , whose restriction to $S_R$ is $\mathbf r\cdot\hat{\mathbf n}\,dA = R\,dA$ with $dA$ the area element (Surface integrals). Hence the restriction of $\star F$ to $S_R$ is the constant $q R^{-3}\cdot R = q/R^2$ times $dA$ , and

\int_{S_R} \star F = \frac{q}{R^2}\cdot 4\pi R^2 = 4\pi q ,

independently of $R$ . For $q = 3$ in Gaussian units this is $12\pi \approx 37.699$ . The same number comes out of the field equation: by the general Stokes theorem applied to the ball $B_R$ in the slice,

\int_{S_R}\star F = \int_{B_R} d\star F = 4\pi\int_{B_R}\star J = 4\pi\!\!\int_{B_R}\!\! \rho\,dx\wedge dy\wedge dz = 4\pi q ,

where the three terms of $\star J$ containing $dt$ drop out on restriction to a slice of constant $t$ , and the last step is that $\int_{B_R}\rho\,dx\wedge dy\wedge dz = q$ is the enclosed charge.

This second chain needs one word of care that the first does not. A point charge is not a smooth source: $\star F$ blows up at the origin and $\rho$ is there not a function but the point density $q\,\delta^3$ (The Dirac delta in calculations), so the equation $d\star F = 4\pi\star J$ on $B_R$ , the appeal to Stokes over the whole ball, and the reading $\int_{B_R}\rho\,dx\wedge dy\wedge dz = q$ all hold in the distributional sense only. The surface integral $\int_{S_R}\star F = 4\pi q$ computed first uses nothing but the smooth field on $S_R$ and needs no such reading; and on a shell $\{R_1 \le r \le R_2\}$ , where $\star F$ is smooth with $d\star F = 0$ , Stokes applies literally and already delivers the radius-independence. Gauss's law is thus not an extra postulate but the integrated form of $d\star F = 4\pi\star J$ , and the radius-independence is the statement that $\star F$ is closed where there is no charge.

Charge conservation is $d\circ d = 0$ . Apply $d$ to the inhomogeneous equation. The left-hand side gives $d(d\star F) = 0$ automatically, so $d\star J = 0$ . Expanding $\star J$ as in Step 4 of the translation,

d\star J = \Bigl(\frac{\partial\rho}{\partial t} + \nabla\cdot\mathbf J\Bigr)\,dt\wedge dx\wedge dy\wedge dz ,

so the continuity equation $\partial_t\rho + \nabla\cdot\mathbf J = 0$ is forced by the field equation. Equivalently $\delta J = 0$ , by part (d) of the statement. This is Hassani's Problem 28.24, and it is the cleanest illustration of what the formalism buys: a conservation law that in components needs a short calculation becomes a one-line consequence of a structural identity. The same mechanism — a source coupled to a closed-by-construction object — is how conserved currents arise throughout field theory; the Lagrangian route to the same conclusion is Noether's theorem.

The four-potential, and what it costs. The homogeneous equation says $F$ is closed. On a region of spacetime simple enough for the Poincaré lemma to apply, a closed 2-form is exact, so there is a 1-form $A = A_\alpha\,dx^\alpha$ with $F = dA$ , that is, $F_{\alpha\beta} = \partial_\alpha A_\beta - \partial_\beta A_\alpha$ . Its components are the scalar and vector potentials of electromagnetism, and the gauge freedom $A \mapsto A + d\chi$ is exactly the non-uniqueness of a primitive: $d(A + d\chi) = dA$ because $d\circ d = 0$ . Hassani makes this his Example 28.5.16. In the Lorenz gauge $\delta A = 0$ the remaining equation $\delta F = 4\pi J$ becomes $\delta dA = 4\pi J$ , hence $(d\delta + \delta d)A = 4\pi J$ : the potential satisfies a wave equation with the current as source, which is the starting point of retarded potentials and of Green's functions for the wave and heat equations.

Beyond electromagnetism. The adjointness of part (c) is the reason $\delta$ and not $\star d\star$ is the natural partner of $d$ : it makes $\Delta = d\delta + \delta d$ formally self-adjoint and non-negative on a compact Riemannian manifold, and its kernel — the harmonic forms — is the subject of Hodge theory, which identifies that kernel with the de Rham cohomology of Closed and exact forms. In numerical analysis the same pairing underlies finite element exterior calculus, where discretising $d$ and $\delta$ as genuine adjoints is what makes a scheme for Maxwell's equations preserve $\nabla\cdot B = 0$ exactly rather than approximately.

Theorem · advanced

Closed and exact forms

Adapted

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Proves the Poincare lemma by exhibiting an explicit homotopy operator $K$ on a star-shaped domain with $dK + Kd = \mathrm{id}$ , so that every closed $p$ -form with $p \ge 1$ is exact there; shows by the angle form on the punctured plane and the solid-angle form on punctured space that the conclusion fails without star-shapedness, and reads the failure as the first de Rham cohomology groups.

Prerequisites: The exterior derivative, Differential forms, Integration of forms and the general Stokes theorem, Line integrals and conservative fields, Differentiation under the integral sign, Connectedness and simple connectivity

Note(The question, and a warning about the name)Original

Because $d \circ d = 0$ (The exterior derivative), every form that is a derivative is annihilated by the derivative. The question of this unit is the converse: if $d\omega = 0$ , must $\omega = d\eta$ for some $\eta$ ?

The answer depends on the shape of the domain and not at all on the form. On a ball the answer is yes in every degree; on $\mathbb{R}^2$ with one point removed it is already no in degree $1$ ; on $\mathbb{R}^3$ with one point removed it is yes in degree $1$ and no in degree $2$ . Measuring the gap between "closed" and "exact" is the beginning of de Rham cohomology, and it is the mechanism by which a differential-geometric computation detects a hole.

A warning on the name. Arfken calls the identity $d(d\omega) = 0$ itself "Poincare's lemma", and Hassani does the same, reserving "converse of the Poincare lemma" for the statement that closed forms are exact on a topologically simple region. Lee and most of the mathematical literature use "the Poincare lemma" for the converse. The two usages are exactly opposite. This unit follows Lee: the Poincare lemma below is the statement that closed implies exact on a star-shaped domain, and the easy direction is recorded without a name.

Definition(Closed, exact, and the de Rham cohomology groups)AdaptedAdapted from Lee (2013), Ch. 17, pp. 440–441

Fix a smooth $n$ -dimensional manifold $M$ , whose boundary may be empty or not, and write $\Omega^p(M)$ for its smooth $p$ -forms (Differential forms). Call $\omega \in \Omega^p(M)$

closed if $d\omega = 0$ ;
exact if $\omega = d\eta$ for some $\eta \in \Omega^{p-1}(M)$ , called a primitive or potential for $\omega$ .

Because $d$ is linear, the closed and the exact $p$ -forms are linear subspaces of $\Omega^p(M)$ :

Z^p(M) = \ker\bigl(d: \Omega^p \to \Omega^{p+1}\bigr), \qquad B^p(M) = \operatorname{im}\bigl(d: \Omega^{p-1} \to \Omega^{p}\bigr) ,

with the convention $\Omega^p(M) = 0$ for $p < 0$ and $p > n$ , so that $B^0(M) = 0$ . Since $d\circ d = 0$ we have $B^p(M) \subseteq Z^p(M)$ , and the quotient

H^p_{\mathrm{dR}}(M) \;=\; Z^p(M)\big/B^p(M)

is the $p$ -th de Rham cohomology group of $M$ . It is a real vector space; Lee notes that the word "group" is traditional only because other cohomology theories produce groups rather than vector spaces. Two closed forms differing by an exact form are cohomologous and define the same class $[\omega]$ . By construction,

H^p_{\mathrm{dR}}(M) = 0 \quad\Longleftrightarrow\quad \text{every closed $p$-form on $M$ is exact}.

Finally, call a subset $U$ of a finite-dimensional vector space star-shaped when some centre $c \in U$ sees all of $U$ : the straight segment joining $c$ to any other member of $U$ never leaves $U$ . Convexity is the special case in which every point is a centre, so convex sets are star-shaped; a non-convex example is $\mathbb{R}^2$ with the closed ray $\{(a,0) : a \ge 0\}$ deleted, which is star-shaped about $c = (-1,0)$ : a segment from $c$ to a point with non-zero second coordinate has non-zero second coordinate everywhere except at $c$ itself, and a segment from $c$ to a surviving point of the $x$ -axis runs along the negative axis.

Theorem(The easy direction)AdaptedAdapted from Lee (2013), Proposition 11.44, p. 294; Arfken, Weber & Harris (2013), §4.6, Eq. (4.85) and Example 4.6.1, pp. 239–240

Every exact form is closed: if $\omega = d\eta$ then $d\omega = d(d\eta) = 0$ .

Arfken points out what this rests on. In coordinates, $d(df) = \sum_{i,j}\partial_j\partial_i f\,dx^j\wedge dx^i$ , and the coefficient array $\partial_j\partial_i f$ is symmetric in $(i,j)$ for a $C^2$ function while $dx^j\wedge dx^i$ is antisymmetric, so the sum cancels in pairs. The identity is therefore the equality of mixed partial derivatives in disguise, and it fails for a function whose mixed partials differ. In $\mathbb{R}^3$ the two instances of $d\circ d = 0$ are $\nabla\times\nabla f = 0$ and $\nabla\cdot(\nabla\times A) = 0$ (Vector identities).

In coordinates, closedness of a 1-form $\omega = \omega_i\,dx^i$ is the system

\frac{\partial\omega_j}{\partial x^i} = \frac{\partial\omega_i}{\partial x^j} \qquad\text{for all } i, j ,

which is Lee's criterion and is what one actually checks in practice.

Theorem(The Poincare lemma)AdaptedAdapted from Lee (2013), Theorem 11.49, p. 296 and Theorem 17.14 with Corollaries 17.15–17.16, p. 447; Hassani (2013), §28.5, Theorem 28.5.15, p. 895

Let $U \subseteq \mathbb{R}^n$ be an open star-shaped set and let $p \ge 1$ . Then every closed $p$ -form on $U$ is exact:

\omega \in \Omega^p(U), \quad d\omega = 0 \quad\Longrightarrow\quad \omega = d\eta \ \text{ for some } \eta \in \Omega^{p-1}(U) .

Equivalently $H^p_{\mathrm{dR}}(U) = 0$ for every $p \ge 1$ . The proof below produces $\eta$ by an explicit formula. Lee states the theorem for star-shaped open subsets of $\mathbb{R}^n$ or of the half-space $\mathbb{H}^n$ , which costs nothing extra in the proof.

Two corollaries, both Lee's:

Local exactness. On any smooth manifold, every point has a neighbourhood on which every closed form is exact, since every point has a coordinate ball, and a ball is convex. Whether a given closed form is exact is therefore always a global question.
$H^p_{\mathrm{dR}}(\mathbb{R}^n) = 0$ for all $p \ge 1$ , $\mathbb{R}^n$ being convex.

For $p = 0$ the conclusion is false and is not asserted: a closed $0$ -form is a function with $df = 0$ , that is, a locally constant function, and there are no $(-1)$ -forms for it to be the derivative of. On a connected $U$ , $H^0_{\mathrm{dR}}(U) = \mathbb{R}$ , the constants.

Two stronger statements, and what they cost. Lee's Theorem 17.13 replaces "star-shaped" by "contractible", by way of the homotopy invariance of de Rham cohomology; a star-shaped $U$ is contractible via $H(x,t) = c + t(x - c)$ , so that theorem contains this one. Hassani states the result for a region "contractable to a point", but his definition of that phrase — that every closed curve in the region can be shrunk to a point within it — is the definition of simple connectedness, which is strictly weaker than contractibility and is not enough in degrees above $1$ : the sphere $S^2$ is simply connected but carries a closed $2$ -form that is not exact, as the second example below shows in the equivalent setting of $\mathbb{R}^3 \setminus \{0\}$ . We follow Lee and state the hypothesis as star-shapedness, which is what the proof given here actually uses; the discrepancy with Hassani's phrasing is recorded in verification/discrepancies.md.

Idea of the proofOriginal

The whole proof is one formula. A star-shaped domain comes with a distinguished family of paths: from the centre $c$ , which we may take to be the origin, the segment $t \mapsto tx$ reaches every point $x$ of $U$ without leaving it, and it varies smoothly with $x$ . That lets us average the form along those segments.

Concretely, define an operator $K$ lowering degree by one, which contracts a $p$ -form with the position vector and integrates the result along the segment with a weight $t^{p-1}$ . The claim is that $K$ is a homotopy operator:

d(K\omega) + K(d\omega) = \omega \qquad\text{on } \Omega^p(U),\ p \ge 1 .

Granting this, the theorem is immediate: if $\omega$ is closed, the second term dies and $\omega = d(K\omega)$ , so $K\omega$ is the primitive we wanted — written down, not merely shown to exist.

Why should such an identity hold? Differentiating $K\omega$ produces two kinds of term, one from differentiating the explicit factor $x^{i_a}$ and one from differentiating inside the integrand. The first kind yields $p$ copies of the averaged form; the second kind splits further, and precisely the part of it that does not involve the radial direction is cancelled by $K(d\omega)$ , which is built the same way out of $d\omega$ . What survives is the radial derivative, and the two survivors assemble into $\int_0^1 \frac{d}{dt}\bigl[t^p\,\omega_I(tx)\bigr]\,dt$ , a total derivative in $t$ . The fundamental theorem of calculus evaluates it as $\omega_I(x) - 0$ , the lower limit vanishing because $p \ge 1$ . That last cancellation is the entire role of the hypothesis $p \ge 1$ , and the entire reason the lemma says nothing about functions.

Proof(The Poincare lemma)AdaptedAdapted from Lee (2013), Theorem 11.49 with its proof, pp. 296–297

Lee proves the case $p = 1$ by integrating $\omega$ along the segments $\gamma_x(t) = tx$ and differentiating under the integral sign; we run the same three moves in general degree, with the contraction against the position vector doing in degree $p$ what multiplication by $x^i$ does in degree $1$ .

Step 0 (normalise the centre). Suppose $U$ is star-shaped about $c$ . Translation is a diffeomorphism of $\mathbb{R}^n$ , and pullback along a diffeomorphism commutes with $d$ ; closedness and exactness are therefore both preserved by it (Pushforwards and pullbacks). Translating by $-c$ we may and do assume $c = 0$ , so that $tx \in U$ whenever $x \in U$ and $t \in [0,1]$ .

Step 1 (define the operator). For $p \ge 1$ and $\omega \in \Omega^p(U)$ set, for $x \in U$ and tangent vectors $v_1, \dots, v_{p-1}$ ,

(K\omega)_x(v_1,\dots,v_{p-1}) \;=\; \int_0^1 t^{\,p-1}\,\omega_{tx}\bigl(x,\ v_1,\dots,v_{p-1}\bigr)\,dt .

The integrand is smooth in $t$ and in $x$ jointly, and $[0,1]$ is compact, so $K\omega$ is a smooth $(p-1)$ -form. Writing $\omega = \sum_{I}\omega_I\,dx^I$ over increasing multi-indices $I = (i_1 < \dots < i_p)$ and expanding the determinant $dx^I(x, v_1,\dots,v_{p-1})$ along its first column gives the coordinate form

K\omega \;=\; \sum_{I}\ \sum_{a=1}^{p}\ (-1)^{a-1}\,g_I(x)\,x^{i_a}\;dx^{I \setminus i_a}, \qquad g_I(x) = \int_0^1 t^{\,p-1}\,\omega_I(tx)\,dt ,

where $dx^{I\setminus i_a} = dx^{i_1}\wedge\dots\wedge\widehat{dx^{i_a}}\wedge\dots\wedge dx^{i_p}$ , the hat marking an omitted factor. The same formula, read with $p+1$ in place of $p$ , defines $K$ on $(p+1)$ -forms, and applied to a monomial $f\,dx^{j_0}\wedge\dots\wedge dx^{j_p}$ with arbitrary (not necessarily increasing) indices it gives

K\bigl(f\,dx^{j_0}\wedge\dots\wedge dx^{j_p}\bigr) = \sum_{b=0}^{p}(-1)^{b}\,\Bigl(\int_0^1 t^{\,p}f(tx)\,dt\Bigr)\,x^{j_b}\;dx^{j_0}\wedge\dots\wedge\widehat{dx^{j_b}}\wedge\dots\wedge dx^{j_p},

which is what we shall apply to $d\omega$ . Note $K$ is defined on all forms, closed or not; that is what makes the identity of Step 4 available.

Step 2 (differentiate $K\omega$ ). By the Leibniz rule for $d$ ,

d(K\omega) = \sum_I\sum_{a=1}^{p}(-1)^{a-1}\Bigl[\,g_I\,dx^{i_a} + x^{i_a}\sum_{j}\partial_j g_I\,dx^{j}\,\Bigr]\wedge dx^{I\setminus i_a} .

In the first bracketed term, moving $dx^{i_a}$ back into its place in $dx^I$ costs $a-1$ transpositions, so $(-1)^{a-1}dx^{i_a}\wedge dx^{I\setminus i_a} = dx^I$ ; summing over the $p$ values of $a$ gives

\text{(first part)} \;=\; \sum_I p\,g_I(x)\,dx^I .

For the second, the integrand of $g_I$ is smooth in $(t,x)$ on the compact $t$ -interval, so we may differentiate under the integral sign (Differentiation under the integral sign); the chain rule applied to $x \mapsto \omega_I(tx)$ brings down a factor $t$ , so

\partial_j g_I(x) = \int_0^1 t^{\,p}\,(\partial_j\omega_I)(tx)\,dt \;=:\; h_{I,j}(x) ,

and

\text{(second part)} \;=\; \sum_I\sum_{a=1}^{p}\sum_{j}(-1)^{a-1}\,x^{i_a}\,h_{I,j}\;dx^{j}\wedge dx^{I\setminus i_a} .

This is the only place differentiation under the integral sign is used.

Step 3 (apply $K$ to $d\omega$ ). Since $d\omega = \sum_I\sum_j (\partial_j\omega_I)\,dx^j\wedge dx^I$ , the monomial formula of Step 1 applies to each term with the $(p+1)$ -tuple of indices $(j, i_1, \dots, i_p)$ . The $b = 0$ term of that formula picks out $x^j$ and leaves $dx^I$ ; the term $b = a \ge 1$ picks out $x^{i_a}$ , carries the sign $(-1)^a$ and leaves $dx^j\wedge dx^{I\setminus i_a}$ . The weight is $\int_0^1 t^{\,p}(\partial_j\omega_I)(tx)\,dt = h_{I,j}$ , the same function as in Step 2. Hence

K(d\omega) \;=\; \sum_I\sum_j h_{I,j}\Bigl[\,x^{j}\,dx^I \;+\; \sum_{a=1}^{p}(-1)^{a}\,x^{i_a}\,dx^{j}\wedge dx^{I\setminus i_a}\,\Bigr] .

Step 4 (add, and recognise a total derivative). The double sums in Steps 2 and 3 are termwise opposite, because $(-1)^{a-1} = -(-1)^{a}$ , so they cancel exactly. What is left is

d(K\omega) + K(d\omega) \;=\; \sum_I \Bigl[\,p\,g_I(x) \;+\; \sum_j x^{j}\,h_{I,j}(x)\Bigr]\,dx^I .

Now $\sum_j x^j (\partial_j\omega_I)(tx) = \dfrac{d}{dt}\bigl[\omega_I(tx)\bigr]$ by the chain rule, so the bracket is

\int_0^1 \Bigl[\,p\,t^{\,p-1}\,\omega_I(tx) \;+\; t^{\,p}\,\frac{d}{dt}\omega_I(tx)\Bigr]dt = \int_0^1 \frac{d}{dt}\Bigl[\,t^{\,p}\,\omega_I(tx)\Bigr]dt = \Bigl[\,t^{\,p}\,\omega_I(tx)\Bigr]_{t=0}^{t=1} = \omega_I(x) ,

the lower limit vanishing because $p \ge 1$ makes $t^p \to 0$ while $\omega_I(tx) \to \omega_I(0)$ stays bounded. Therefore

d(K\omega) + K(d\omega) = \sum_I \omega_I(x)\,dx^I = \omega .

Step 5 (conclude). If in addition $d\omega = 0$ , the second term is zero and $\omega = d(K\omega)$ . Setting $\eta = K\omega$ , which is a smooth $(p-1)$ -form on all of $U$ by Step 1, finishes the proof. $\square$

Remark(Where the hypotheses are used)Original

Star-shapedness enters in exactly one place and enters absolutely: the segment $t \mapsto tx$ , $t \in [0,1]$ , must lie in $U$ for every $x \in U$ , or the integrals defining $K$ are not even defined, $\omega_I(tx)$ being meaningless where $tx \notin U$ . Nothing else in Steps 1–4 refers to the shape of $U$ . The first example below shows the conclusion really does fail without it.
$p \ge 1$ is used once, at the lower limit in Step 4. For $p = 0$ the bracket would be $[\,\omega(tx)\,]_0^1 = \omega(x) - \omega(0)$ , and the constant $\omega(0)$ is precisely the obstruction: the closed $0$ -forms on a connected $U$ are the constants, none of which except $0$ is exact.
Smoothness is used for differentiation under the integral sign in Step 2 and for the chain rule in Step 4. $C^1$ coefficients suffice for the argument as written, but the easy direction needs $C^2$ for the equality of mixed partials, so $C^\infty$ is the comfortable hypothesis.
Openness of $U$ is used implicitly throughout, to speak of partial derivatives at every point of $U$ . Lee's version also covers star-shaped open subsets of the half-space $\mathbb{H}^n$ , where one-sided derivatives do the work.
The centre does not matter, but the family of paths does. Step 0 moves the centre to the origin; a different centre gives a different primitive, differing by a closed $(p-1)$ -form. That non-uniqueness is unavoidable — it is gauge freedom, discussed in the application.

What a weaker hypothesis buys. Star-shapedness is far from necessary. Lee obtains the same conclusion for any contractible manifold, as a corollary of a stronger invariance: a homotopy equivalence between two manifolds induces an isomorphism of their de Rham groups. Anything homotopy equivalent to a single point therefore has $H^p_{\mathrm{dR}} = 0$ in every degree $p \ge 1$ . In degree $1$ alone, simple connectedness suffices. The price of the general theorem is the machinery of homotopy operators on $M \times I$ ; the price of the version proved here is a hypothesis one can check by looking at the picture, and a formula one can evaluate.

Method(Finding a primitive)AdaptedAdapted from Lee (2013), Theorem 11.49, pp. 296–297 and Example 11.51, p. 298

Given $\omega$ on a domain $U$ , to decide exactness and produce a potential:

Test closedness. Compute $d\omega$ . If $d\omega \ne 0$ the form is not exact and there is nothing further to do. For a 1-form this is the check $\partial_i\omega_j = \partial_j\omega_i$ .
Check the shape of the domain. Is $U$ star-shaped (in particular, convex)? If so, the lemma applies in every degree $p \ge 1$ . If not, the form may still be exact, but exactness must be established by other means, and the examples below show it can genuinely fail.
Apply the formula. Translate the centre to the origin and evaluate

\eta = K\omega = \sum_{I}\sum_{a=1}^{p}(-1)^{a-1}\Bigl(\int_0^1 t^{\,p-1}\omega_I(tx)\,dt\Bigr)x^{i_a}\,dx^{I\setminus i_a} .

For $\omega_I$ homogeneous of degree $m$ the integral is simply $\omega_I(x)/(m + p)$ , so for polynomial coefficients the whole computation is a division of each monomial by a small integer. 4. Verify. Compute $d\eta$ and confirm it is $\omega$ . This is cheap and catches sign slips, of which there are many opportunities. 5. Or antidifferentiate directly. For a 1-form on a rectangle, Lee's Example 11.51 is usually quicker: integrate $\omega_1$ in $x^1$ holding the other variables fixed, which fixes $f$ up to a function of $(x^2,\dots,x^n)$ ; impose $\partial_2 f = \omega_2$ to pin that function down up to a function of $(x^3,\dots,x^n)$ ; and continue. Closedness is what makes each step consistent. The formula of step 3 is the one to use when a uniform, verifiable answer is wanted in higher degree.

Non-uniqueness. Any two primitives differ by a closed $(p-1)$ -form, since $d(\eta - \eta') = 0$ . For $p = 1$ on a connected domain this is an additive constant; for $p \ge 2$ it is the far larger freedom $\eta \mapsto \eta + d\alpha$ noted by Hassani, with $\alpha$ an arbitrary $(p-2)$ -form.

Example(A closed 2-form on

\mathbb{R}^3

, its potential, and a flux)Original

Take the magnetic-type field $B = (x^2,\ -2xy,\ xy)$ on $\mathbb{R}^3$ , whose divergence is $2x - 2x + 0 = 0$ , and let $\omega$ be its flux 2-form,

\omega = x^2\,dy\wedge dz \;-\; 2xy\,dz\wedge dx \;+\; xy\,dx\wedge dy .

Then $d\omega = (\nabla\cdot B)\,dx\wedge dy\wedge dz = 0$ , so $\omega$ is closed, and $\mathbb{R}^3$ is convex, so the lemma applies with $p = 2$ .

Step 1: increasing-index components. Using $dz\wedge dx = -\,dx\wedge dz$ ,

\omega = 2xy\,dx\wedge dz + xy\,dx\wedge dy + x^2\,dy\wedge dz, \qquad \omega_{13} = 2xy,\quad \omega_{12} = xy, \quad \omega_{23} = x^2 .

Step 2: the radial integrals. Here $p = 2$ , so $g_I(x) = \int_0^1 t\,\omega_I(tx)\,dt$ . Every component is homogeneous of degree $2$ , so $\omega_I(tx) = t^2\omega_I(x)$ and $g_I = \omega_I\int_0^1 t^3\,dt = \omega_I/4$ :

g_{13} = \frac{xy}{2}, \qquad g_{12} = \frac{xy}{4}, \qquad g_{23} = \frac{x^2}{4} .

Step 3: assemble $\eta = K\omega$ . For $I = (i_1 < i_2)$ the formula contributes $g_I\,x^{i_1}\,dx^{i_2} - g_I\,x^{i_2}\,dx^{i_1}$ , with $(x^1,x^2,x^3) = (x,y,z)$ :

\begin{aligned} I = (1,3): &\quad \tfrac{xy}{2}\,x\,dz - \tfrac{xy}{2}\,z\,dx = \tfrac{x^2y}{2}\,dz - \tfrac{xyz}{2}\,dx, \\ I = (1,2): &\quad \tfrac{xy}{4}\,x\,dy - \tfrac{xy}{4}\,y\,dx = \tfrac{x^2y}{4}\,dy - \tfrac{xy^2}{4}\,dx, \\ I = (2,3): &\quad \tfrac{x^2}{4}\,y\,dz - \tfrac{x^2}{4}\,z\,dy = \tfrac{x^2y}{4}\,dz - \tfrac{x^2z}{4}\,dy . \end{aligned}

Adding,

\eta \;=\; -\frac{xy(y + 2z)}{4}\,dx \;+\; \frac{x^2(y - z)}{4}\,dy \;+\; \frac{3x^2y}{4}\,dz .

Step 4: verify. With $P, Q, R$ the three coefficients, the curl components are

\partial_y R - \partial_z Q = \frac{3x^2}{4} - \Bigl(-\frac{x^2}{4}\Bigr) = x^2 ,

\partial_z P - \partial_x R = -\frac{xy}{2} - \frac{3xy}{2} = -2xy ,

\partial_x Q - \partial_y P = \frac{x(y - z)}{2} - \Bigl(-\frac{x(y + z)}{2}\Bigr) = xy ,

since $Q = \tfrac14 x^2(y-z)$ gives $\partial_x Q = \tfrac12 x(y-z)$ and $P = -\tfrac14(xy^2 + 2xyz)$ gives $\partial_y P = -\tfrac12 x(y+z)$ . These three numbers are exactly $(x^2, -2xy, xy) = B$ , so $d\eta = \omega$ .

Step 5: a number both ways. Let $S$ be the unit square $0 \le x \le 1$ , $0 \le y \le 1$ in the plane $z = 1$ , oriented by $+\hat z$ , with boundary traversed counterclockwise. The flux of $\omega$ through it is

\int_S \omega = \int_0^1\!\!\int_0^1 xy \,dx\,dy = \frac14 .

By the general Stokes theorem, this must equal $\oint_{\partial S}\eta$ . On $z = 1$ the relevant coefficients are $P = -\tfrac{xy(y+2)}{4}$ and $Q = \tfrac{x^2(y-1)}{4}$ , and $dz = 0$ . The four sides give

\underbrace{\int_0^1 P(s,0,1)\,ds}_{=\,0} \;+\; \underbrace{\int_0^1 Q(1,s,1)\,ds}_{\int_0^1 \frac{s-1}{4}ds \,=\, -\frac18} \;+\; \underbrace{\int_1^0 P(s,1,1)\,ds}_{\int_1^0 -\frac{3s}{4}\,ds \,=\, +\frac38} \;+\; \underbrace{\int_1^0 Q(0,s,1)\,ds}_{=\,0} \;=\; -\frac18 + \frac38 \;=\; \boxed{\frac14} ,

agreeing with the surface integral. The primitive delivered by the formula is therefore a genuine vector potential for $B$ in the sense of the Helmholtz decomposition.

Example(Two closed forms that are not exact)AdaptedAdapted from Lee (2013), Example 11.43, p. 294, Example 11.48, p. 296 and Example 17.1, p. 441

(a) The angle form on the punctured plane. On $U = \mathbb{R}^2\setminus\{0\}$ put

\omega = \frac{x\,dy - y\,dx}{x^2 + y^2} .

It is closed: writing $\omega = \omega_1 dx + \omega_2 dy$ with $\omega_1 = -y/(x^2+y^2)$ and $\omega_2 = x/(x^2+y^2)$ ,

\partial_x\omega_2 = \frac{(x^2+y^2) - x\cdot 2x}{(x^2+y^2)^2} = \frac{y^2 - x^2}{(x^2+y^2)^2} = \frac{-(x^2+y^2) + y\cdot 2y}{(x^2+y^2)^2} = \partial_y\omega_1 .

It is not exact on $U$ . Suppose $\omega = df$ . Let $\gamma(\theta) = (\cos\theta, \sin\theta)$ , $\theta \in [0, 2\pi]$ , be the unit circle. Then $\gamma^*\omega = (\cos^2\theta + \sin^2\theta)\,d\theta = d\theta$ , so

\oint_\gamma \omega = \int_0^{2\pi} d\theta = 2\pi .

But the integral of an exact 1-form round a closed curve is $f(\gamma(2\pi)) - f(\gamma(0)) = 0$ (Line integrals), and $2\pi \ne 0$ . Hence $H^1_{\mathrm{dR}}(\mathbb{R}^2\setminus\{0\}) \ne 0$ . Restricted to the right half-plane $x > 0$ — which is star-shaped — the same $\omega$ is exact, equal to $d\arctan(y/x)$ ; in polar coordinates $\omega = d\theta$ wherever a single-valued branch of $\theta$ exists. The obstruction is thus not local but the impossibility of choosing one branch of the angle all the way round the hole.

(b) The solid-angle form on punctured space. On $V = \mathbb{R}^3\setminus\{0\}$ , with $r = \lvert\mathbf r\rvert$ , put

\sigma = \frac{x\,dy\wedge dz + y\,dz\wedge dx + z\,dx\wedge dy}{r^3} ,

the flux 2-form of the inverse-square field $\hat{\mathbf r}/r^2$ . It is closed, because $d\sigma = \nabla\cdot(\hat{\mathbf r}/r^2)\,dx\wedge dy\wedge dz$ and that divergence vanishes away from the origin. It is not exact: were $\sigma = d\eta$ , then by Stokes' theorem on the closed surface $S^2$ , which has empty boundary,

\int_{S^2}\sigma = \int_{S^2} d\eta = \int_{\partial S^2}\eta = 0 ,

whereas restricting $\sigma$ to the unit sphere gives the area element and

\int_{S^2}\sigma = \int_0^{2\pi}\!\!\int_0^\pi \sin\theta\,d\theta\,d\varphi = 4\pi \ne 0 .

So $H^2_{\mathrm{dR}}(\mathbb{R}^3\setminus\{0\}) \ne 0$ . Note the degrees: $V$ is simply connected — any loop can be slid off the origin — yet the lemma fails in degree $2$ . This is the counterexample promised above to Hassani's phrasing of the hypothesis, and it shows why "contractible", not "simply connected", is the right word. By contrast $H^1_{\mathrm{dR}}(V) = 0$ : on punctured space every closed 1-form is exact.

The pattern. A closed $p$ -form fails to be exact when there is a $p$ -dimensional closed surface in the domain over which it integrates to something non-zero, because exact forms integrate to zero over such surfaces. The punctured plane has a circle enclosing the hole; punctured space has a sphere enclosing it. The $2\pi$ and the $4\pi$ are not accidents of normalisation: they are the total angle and the total solid angle, and dividing them out gives generators of the cohomology that take integer values on cycles.

Application(Potentials, gauge freedom, and holes that physics can measure)Original

When a potential exists. Three standard moves in physics are instances of the lemma. A force field with $\nabla\times F = 0$ on a star-shaped region has a potential energy $F = -\nabla V$ , which is the case $p = 1$ . A magnetic field with $\nabla\cdot B = 0$ has a vector potential $B = \nabla\times A$ there, which is the case $p = 2$ and is what the worked example constructs. And in thermodynamics a differential $\delta Q = \sum_i X_i\,dY_i$ is an exact differential of a state function exactly when it is closed as a 1-form on the (convex) space of states, which is why $dU$ and $dS$ exist as state functions while heat and work separately do not.

Gauge freedom is non-uniqueness of the primitive. In The Hodge star and Maxwell's equations the homogeneous Maxwell equations are the single statement $dF = 0$ for the Faraday 2-form on spacetime. The lemma turns this into $F = dA$ on any star-shaped region of spacetime, with $A$ the four-potential; Hassani does this computation and reads off $F_{\alpha\beta} = \partial_\alpha A_\beta - \partial_\beta A_\alpha$ . The replacement $A \mapsto A + d\chi$ leaves $F$ unchanged because $d\circ d = 0$ , and that is the whole of electromagnetic gauge invariance: it is the ambiguity of a primitive, no more and no less. Hassani notes that the generalisation of this freedom to non-abelian structure groups is the organising principle of the standard model.

Where the hole is physical. Remove a thin solenoid from space and the region outside it is no longer simply connected. Outside the solenoid $B = 0$ , so $A$ is closed; but $\oint A\cdot d\mathbf{l} = \Phi$ , the enclosed flux, which is not zero, so $A$ is not exact there. By example (a), that non-exactness is a cohomology class, and in the Aharonov–Bohm effect it is measured: the interference pattern of electrons passing on either side of the solenoid shifts by a phase $q\Phi/\hbar$ (SI units) even though the electrons never enter a region with a magnetic field. Likewise, a magnetic monopole of charge $g$ at the origin would make $B$ a closed but non-exact 2-form on $\mathbb{R}^3\setminus\{0\}$ , a constant multiple of the $\sigma$ of example (b), the constant being fixed by the requirement that the flux through a sphere about the origin be the monopole charge; no globally defined vector potential exists, and Dirac's argument that the potential must be built from two patches with a consistent overlap is what quantises the product of the electric and magnetic charges.

Where the hole is a count. Integrating closed forms over cycles is the computational face of de Rham's theorem, which identifies $H^p_{\mathrm{dR}}(M)$ with the real singular cohomology of $M$ and hence makes it a topological invariant. In applied settings this is the engine of topological data analysis and of the discrete theories used for electromagnetic and fluid solvers, where the dimension of a discrete analogue of $H^1$ counts independent current loops in a conductor and determines how many extra constraints a well-posed scheme needs. The counting of the Gauss-Bonnet theorem is the same phenomenon seen through curvature rather than through forms.