Representations of finite groups

Representations, characters and their uses.

In this chapter 5

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Representations and characters

Representations and characters

Definition

Representations of groups

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A representation turns the abstract elements of a group into invertible matrices; for a finite group every complex representation can be made unitary and splits into a direct sum of irreducible pieces, which is Maschke's theorem.

Prerequisites: Groups and subgroups, Homomorphisms and quotient groups, Group actions, The symmetric group, Rings and fields, Linear maps and matrices, Inner products and norms, Orthonormal bases and Gram-Schmidt, Unitary and orthogonal matrices

Definition(Representation, degree, faithfulness)AdaptedAdapted from Hassani (2013), §24.1, Def. 24.1.1, p. 726; Dummit & Foote (2004), §18.1, definition, p. 840

Let $G$ be a group and let $V$ be a vector space over a field $F$ . A representation of $G$ on $V$ is a group homomorphism

T \colon G \longrightarrow GL(V), \qquad T_{gh} = T_g T_h \quad \text{for all } g, h \in G ,

where $GL(V)$ is the group of invertible linear maps $V \to V$ (Linear maps and matrices). The space $V$ is the carrier space, and $\dim V$ is the degree (or dimension) of $T$ . Because a homomorphism sends the identity to the identity and inverses to inverses (Homomorphisms and quotient groups),

T_e = I, \qquad T_{g^{-1}} = (T_g)^{-1} .

The representation is faithful if $T$ is injective, that is if $\ker T = \{e\}$ . Fixing a basis of a finite-dimensional $V$ replaces each $T_g$ by an invertible matrix $D(g) \in GL_n(F)$ with $D(gh) = D(g)D(h)$ ; this is a matrix representation of degree $n$ . Every group has the trivial representation $T_g = 1$ on $F$ , of degree $1$ , which is faithful only for $G = \{e\}$ .

Unless we say otherwise, $G$ is finite, $F = \mathbb{C}$ , and $V$ is finite-dimensional with an inner product $(\cdot,\cdot)$ , linear in the first slot.

Definition(Equivalence)AdaptedAdapted from Hassani (2013), §24.1, Def. 24.1.2, p. 726; Riley, Hobson & Bence (2006), §29.3, Eq. (29.5), p. 1084

Two representations $T$ on $V$ and $T'$ on $V'$ are equivalent, written $T \simeq T'$ , if there is a vector-space isomorphism $f \colon V \to V'$ with

T'_g = f \, T_g \, f^{-1} \qquad \text{for every } g \in G .

In matrix language, $D$ and $D'$ are equivalent when a single invertible $S$ , the same one for every group element, gives $D'(g) = S^{-1} D(g) S$ for all $g$ . Equivalent representations are the same representation written in different bases, and we never distinguish them.

Definition(Invariant subspace, reducibility, irreducibility)AdaptedAdapted from Dummit & Foote (2004), §18.1, the definitions p. 847 and the block forms p. 848; Hassani (2013), §24.2, invariant subspaces and Def. 24.2.1, p. 729; Riley, Hobson & Bence (2006), §29.4, pp. 1086-1088

A subspace $W \subseteq V$ is invariant (or $G$ -stable) if $T_g W \subseteq W$ for every $g \in G$ ; the restrictions $T_g|_W$ then form a representation of $G$ on $W$ , a subrepresentation. The subspaces $\{0\}$ and $V$ are always invariant and are called trivial.

$T$ is irreducible if $V \neq \{0\}$ and the only invariant subspaces are $\{0\}$ and $V$ .
$T$ is reducible if some invariant $W$ is neither $\{0\}$ nor $V$ .
$T$ is decomposable if $V = U \oplus W$ with $U$ and $W$ both invariant and both non-zero; it is completely reducible if $V$ is a direct sum of irreducible invariant subspaces.

The words "reducible" and "decomposable" are not used consistently in the literature, so we fix our usage now: for these two words we follow Dummit and Foote (Dummit & Foote (2004), §18.1, the definitions on p. 847 and the block forms on p. 848), as in the list above. Two of our other sources use the words differently. Hassani calls a representation reducible when the carrier space is a direct sum $U \oplus W$ of two invariant subspaces, and irreducible when no such pair exists (Hassani (2013), Def. 24.2.1, p. 729); Riley, Hobson and Bence call a representation reducible when one similarity transformation brings every matrix to block-diagonal form (Riley, Hobson & Bence (2006), §29.4, pp. 1086-1087); both of those words mean our decomposable, and both of those "irreducible"s mean our indecomposable. Later in the same section Riley, Hobson and Bence give a second formulation, calling $D$ reducible when some proper subspace $W$ of the carrier space is invariant (Riley, Hobson & Bence (2006), §29.4, p. 1088), and that one is our "reducible". Their two formulations agree with each other only because of a standing assumption they make just before the section opens, namely that every representation under discussion is unitary (Riley, Hobson & Bence (2006), §29.3, p. 1086): for unitary $D$ an invariant $W$ brings the invariant complement $W^{\perp}$ with it, by part 2 of the theorem below, and the block-diagonal form follows. Without that assumption the second formulation is strictly weaker than the first. The two usages agree for the case that occupies most of this part, a finite group in characteristic $0$ , precisely because parts 3 and 4 of the theorem below say reducible and decomposable then coincide; they part company only for infinite groups or when the characteristic divides $|G|$ , as the remark after the proof shows.

With our words, the distinction between "reducible" and "decomposable" is the whole content of the theorem below. Choosing a basis of $V$ whose first vectors span an invariant $W$ puts every matrix in the block upper-triangular form $\begin{pmatrix} D_1(g) & B(g) \\ 0 & D_2(g)\end{pmatrix}$ ; the stronger splitting $V = U \oplus W$ with both summands invariant is what forces $B(g) = 0$ for every $g$ , giving the block-diagonal form. Writing $m_\lambda$ for the number of summands equivalent to a fixed irreducible representation $T^{(\lambda)}$ , a completely reducible $T$ is recorded as

T \;\simeq\; m_1 T^{(1)} \oplus m_2 T^{(2)} \oplus \cdots \oplus m_\rho T^{(\rho)} ,

the $T^{(\lambda)}$ being pairwise inequivalent and the $m_\lambda \ge 0$ integers called multiplicities. Hassani writes $\rho$ for the number of inequivalent irreducible representations; Riley, Hobson and Bence abbreviate "irreducible representation" to irrep and write $\hat D^{(\lambda)}$ for it.

Definition(Unitarity and Maschke's theorem)AdaptedAdapted from Hassani (2013), §24.2, Thm 24.2.5, p. 730, Thm 24.2.6, p. 731, Eq. (24.3), p. 731; Dummit & Foote (2004), §18.1, Thm 1, pp. 849-850, Cor. 2 and Cor. 3, p. 851

Let $G$ be a finite group of order $|G|$ .

Unitarizability. Let $T$ be a representation of $G$ on a finite-dimensional complex inner product space $V$ . Then there is an inner product on $V$ for which every $T_g$ is unitary; equivalently, $T$ is equivalent to a representation by unitary matrices.
Invariant complements, unitary case. If every $T_g$ is unitary for the inner product $(\cdot,\cdot)$ and $W \subseteq V$ is invariant, then $W^{\perp}$ is invariant as well, so $V = W \oplus W^{\perp}$ splits $T$ .
Maschke's theorem. Let $F$ be a field whose characteristic is not a divisor of $|G|$ (Rings and fields), let $V$ be an $F$ -vector space carrying a representation of $G$ , and let $U \subseteq V$ be an invariant subspace. Then $V$ has an invariant subspace $W$ with $V = U \oplus W$ .
Complete reducibility. Under the hypothesis of part 3, every representation of $G$ on a finite-dimensional $F$ -vector space is completely reducible: there is a basis in which every matrix $D(g)$ is block diagonal with irreducible blocks, the same block pattern for all $g \in G$ .

Part 3 applies to every finite group when $\operatorname{char} F = 0$ , in particular for $F = \mathbb{C}$ . Parts 1 and 2 keep the convention of the first definition, so their $V$ is finite-dimensional and complex, and for such a $V$ they give a second proof of part 3; part 3 itself assumes nothing about $\dim V$ .

Idea of the proofOriginal

Every part of this is one trick: average over the group. A finite group lets us replace any object attached to $V$ by the sum of its translates $\sum_{g \in G} g \cdot (\text{object})$ , and the sum is automatically $G$ -invariant, because translating it by $h$ only relabels the summands $g \mapsto hg$ , which is a bijection of $G$ .

Apply the trick to an inner product and you get an inner product the group preserves, which is part 1. Apply it to a projection onto $U$ and you get a projection the group commutes with, so its kernel is an invariant complement, which is part 3. Part 2 is the special case where the averaging has already been done: once the $T_g$ are unitary, the complement $W^{\perp}$ comes for free, since preserving $W$ and preserving all inner products forces $W^{\perp}$ to be preserved too. Part 4 then follows by peeling off one invariant piece at a time, a finite process because each step drops the dimension.

ProofAdaptedAdapted from Hassani (2013), §24.2, proof of Thm 24.2.5, p. 730, and proof of Thm 24.2.6, p. 731; Dummit & Foote (2004), §18.1, proof of Thm 1, pp. 849-851, and proof of Cor. 2, p. 851

Part 1. Starting from the given inner product $(\cdot,\cdot)$ on $V$ , define

\langle x, y \rangle \;=\; \sum_{g \in G} \bigl( T_g x,\, T_g y \bigr), \qquad x, y \in V ,

a finite sum because $G$ is finite. This is sesquilinear (linear in $x$ , conjugate-linear in $y$ ) because each summand is, and Hermitian for the same reason. It is positive definite: every term $(T_g x, T_g x) = \lVert T_g x\rVert^2$ is $\ge 0$ , and the term $g = e$ equals $\lVert x \rVert^2 > 0$ when $x \neq 0$ . So $\langle \cdot, \cdot \rangle$ is an inner product on $V$ (Inner product spaces).

It is $G$ -invariant. For $h \in G$ ,

\langle T_h x, T_h y \rangle = \sum_{g \in G} \bigl( T_g T_h x,\, T_g T_h y \bigr) = \sum_{g \in G} \bigl( T_{gh} x,\, T_{gh} y \bigr) = \sum_{k \in G} \bigl( T_k x,\, T_k y \bigr) = \langle x, y \rangle ,

where the second equality is the homomorphism property $T_g T_h = T_{gh}$ and the third substitutes $k = gh$ : as $g$ runs once through $G$ so does $k$ , because right multiplication by $h$ is a bijection of $G$ (it has inverse right multiplication by $h^{-1}$ ). Thus every $T_g$ is unitary for $\langle \cdot, \cdot \rangle$ .

To get matrices, apply the Gram-Schmidt process to any basis of $V$ using $\langle \cdot, \cdot \rangle$ (Orthonormal bases and Gram-Schmidt); this produces a basis $v_1, \dots, v_n$ with $\langle v_i, v_j\rangle = \delta_{ij}$ . Let $D'(g)$ be the matrix of $T_g$ in that basis, so $T_g v_j = \sum_i D'(g)_{ij} v_i$ . Then

\bigl(D'(g)^{\dagger} D'(g)\bigr)_{jk} = \sum_i \overline{D'(g)_{ij}}\, D'(g)_{ik} = \langle T_g v_k,\, T_g v_j \rangle = \langle v_k, v_j \rangle = \delta_{jk} ,

the middle step expanding both arguments in the $\langle\cdot,\cdot\rangle$ -orthonormal basis and the next using the invariance just proved. Hence $D'(g)$ is unitary for every $g$ (Unitary and orthogonal matrices). Finally, if $D(g)$ is the matrix of $T_g$ in the original orthonormal basis $e_1, \dots, e_n$ and $S$ is the matrix with columns given by the coordinates of $v_1,\dots,v_n$ in that basis, then $D'(g) = S^{-1} D(g) S$ , so $D \simeq D'$ .

Part 2. Let all $T_g$ be unitary, let $W$ be invariant, and let $x \in W^{\perp}$ , $g \in G$ . For any $w \in W$ ,

(T_g x,\, w) = \bigl(T_g x,\, T_g T_{g^{-1}} w\bigr) = \bigl(x,\, T_{g^{-1}} w\bigr) = 0 ,

using $T_g T_{g^{-1}} = T_e = I$ in the first step, unitarity of $T_g$ in the second, and in the third that $T_{g^{-1}} w \in W$ by invariance of $W$ while $x \perp W$ . As $w \in W$ was arbitrary, $T_g x \in W^{\perp}$ . So $W^{\perp}$ is invariant, and $V = W \oplus W^{\perp}$ because the inner product is positive definite.

Part 3. Write $n = |G|$ and let $n \cdot 1_F = 1_F + \cdots + 1_F$ ( $n$ terms). The hypothesis that $\operatorname{char} F \nmid n$ says exactly that $n \cdot 1_F \neq 0$ , so $n$ is invertible in $F$ ; call its inverse $n^{-1}$ . This is the only place the characteristic is used, at the step marked below.

A projection to start from. Choose any vector-space complement $U_0$ of $U$ in $V$ (extend a basis of $U$ to a basis of $V$ and let $U_0$ be the span of the added vectors), and let $\pi_0 \colon V \to V$ be the projection with image $U$ and kernel $U_0$ , that is $\pi_0(u + u_0) = u$ for $u \in U$ , $u_0 \in U_0$ . It is linear, satisfies $\pi_0(V) \subseteq U$ , and $\pi_0(u) = u$ for $u \in U$ . Nothing makes $U_0$ invariant, which is the problem to be fixed.

Averaging. Define

\pi \;=\; n^{-1} \sum_{g \in G} T_g \, \pi_0 \, T_g^{-1} \;\colon\; V \to V .

Each summand is a composition of linear maps, so $\pi$ is linear.

$\pi$ maps $V$ into $U$ . For $v \in V$ we have $\pi_0 T_g^{-1} v \in U$ , and $U$ is invariant, so $T_g \pi_0 T_g^{-1} v \in U$ . A sum of elements of the subspace $U$ , scaled by $n^{-1}$ , is again in $U$ .
$\pi$ fixes $U$ pointwise. For $u \in U$ , invariance gives $T_g^{-1} u \in U$ , so $\pi_0 T_g^{-1} u = T_g^{-1} u$ and hence $T_g \pi_0 T_g^{-1} u = u$ . Summing, $\pi(u) = n^{-1} \cdot n \, u = u$ . Here the invertibility of $n$ in $F$ is used.
$\pi$ commutes with the group. For $h \in G$ , $T_h^{-1} \pi \, T_h = n^{-1} \sum_{g \in G} T_h^{-1} T_g \, \pi_0 \, T_g^{-1} T_h = n^{-1} \sum_{g \in G} T_{h^{-1} g}\, \pi_0\, T_{h^{-1}g}^{-1} = \pi ,$ since $g \mapsto h^{-1} g$ is a bijection of $G$ . Thus $\pi T_h = T_h \pi$ for all $h$ .

The complement. Put $W = \ker \pi$ . It is invariant: if $\pi v = 0$ then $\pi (T_h v) = T_h (\pi v) = 0$ . And $V = U \oplus W$ :

$U \cap W = \{0\}$ , because $v$ in the intersection satisfies $v = \pi v$ (as $v\in U$ ) and $\pi v = 0$ (as $v \in W$ ).
$U + W = V$ , because any $v$ splits as $v = \pi v + (v - \pi v)$ with $\pi v \in U$ , while $\pi(v - \pi v) = \pi v - \pi(\pi v) = \pi v - \pi v = 0$ since $\pi v \in U$ is fixed by $\pi$ ; so $v - \pi v \in W$ .

Part 4. Induct on $\dim V$ . If $V = \{0\}$ it is the empty direct sum, and if $T$ is irreducible there is nothing to do. Otherwise $V$ has an invariant subspace $U$ with $0 < \dim U < \dim V$ , and part 3 supplies an invariant $W$ with $V = U \oplus W$ . Both $U$ and $W$ have smaller dimension than $V$ , so by the induction hypothesis each is a direct sum of irreducible invariant subspaces, and therefore so is $V$ . Taking a basis of $V$ made of bases of those irreducible summands in turn, each $T_g$ preserves every summand, so its matrix is block diagonal with one block per summand, and the block on a given summand is an irreducible matrix representation. $\square$

Remark(Where the hypotheses are used)Original

$G$ finite is used twice, and cannot be dropped. The averaged sums in parts 1 and 3 have $|G|$ terms, and the invertibility of $|G|$ in part 3 presupposes a number. The conclusion genuinely fails for infinite groups: let $G = \mathbb{Z}$ act on $\mathbb{C}^2$ by

T_m = \begin{pmatrix} 1 & m \\ 0 & 1 \end{pmatrix}, \qquad T_m T_{m'} = T_{m+m'} .

The line $U = \operatorname{span}\{e_1\}$ is invariant. Any complementary line is spanned by some $(a, b)$ with $b \neq 0$ , and $T_1 (a,b) = (a+b, b)$ , which is a multiple of $(a,b)$ only if $b = 0$ . So $U$ has no invariant complement: this representation is reducible but indecomposable. (The same matrices show part 1 fails too: the argument of part 2 uses only unitarity, not finiteness, so if some inner product made every $T_m$ unitary, $U$ would have the invariant complement $U^{\perp}$ .)

$\operatorname{char} F \nmid |G|$ is used exactly once, at the step $\pi(u) = n^{-1}\cdot n\,u = u$ , and it too is sharp. Take $F = \mathbb{F}_p$ , let $G = \langle g \rangle$ be cyclic of order $p$ , and let $V = \mathbb{F}_p^2$ with

D(g) = \begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}, \qquad D(g)^{k} = \begin{pmatrix} 1 & k \\ 0 & 1\end{pmatrix} ,

so $D(g)$ has order exactly $p$ and this is a representation. Again $\operatorname{span}\{e_1\}$ is invariant with no invariant complement, by the computation of the previous paragraph read mod $p$ . Here $\operatorname{char} F = p$ divides $|G| = p$ . This is Dummit and Foote's example at (Dummit & Foote (2004), §18.1, p. 849); the study of this failing case is modular representation theory, which this handbook does not enter.

Algebraic closedness is not needed for Maschke, but irreducibility depends on the field. Nothing above asks $F$ to be $\mathbb{C}$ . What does change with the field is the list of irreducible pieces. The degree-two real representation of the cyclic group $C_3$ by rotations through $0, 2\pi/3, 4\pi/3$ is irreducible over $\mathbb{R}$ , because a rotation by $2\pi/3$ fixes no line in $\mathbb{R}^2$ ; over $\mathbb{C}$ the same matrices are diagonalizable simultaneously and split into two one-dimensional representations $g \mapsto \omega$ and $g \mapsto \omega^2$ , $\omega = e^{2\pi i/3}$ . So "irreducible" always means irreducible over a stated field, and from here on that field is $\mathbb{C}$ .

Reducible is weaker than decomposable, in general. The two infinite- and positive-characteristic examples above are exactly the gap, and they are also where our terminology and Hassani's part company: each of them is reducible and indecomposable in our words, and so would be called irreducible under Hassani's Def. 24.2.1 and under Riley, Hobson and Bence's block-diagonal formulation in §29.4. (Their second, invariant-subspace formulation would call both examples reducible, as we do; it is available to them only for unitary representations, and neither example is unitarizable.) Parts 3 and 4 say the gap closes for finite groups in characteristic $0$ , and that is why finite-group representation theory over $\mathbb{C}$ is as clean as it is.

Uniqueness is not claimed here. Part 4 gives a decomposition, not a unique one: inside a repeated irreducible summand the choice of complement is as free as the choice of basis of an eigenspace. What is unique is the list of multiplicities $m_\lambda$ , and that is proved with characters in Orthogonality relations and characters.

Method(Writing down a representation)Original

To produce a concrete matrix representation of a finite group $G$ :

Find a set or space the group acts on. A permutation action of $G$ on a finite set $X$ (Group actions) is the commonest source.
Linearize. Take $V$ with basis $\{e_x : x \in X\}$ and define $T_g e_x = e_{g \cdot x}$ , extended linearly. The homomorphism property follows from the action axioms. Since $T_g$ permutes an orthonormal basis, these permutation matrices are already unitary, and part 1 of the statement needs no work.
Read off the matrices. $D(g)_{ij} = 1$ when $g \cdot x_j = x_i$ and $0$ otherwise: each column has a single $1$ .
Hunt for invariant subspaces. Vectors built symmetrically from the whole orbit (sums such as $\sum_x e_x$ ) and the subspaces defined by linear conditions the action preserves (such as "coordinates sum to zero") are the first things to try.
Split off the complement. Over $\mathbb{C}$ , take the orthogonal complement of an invariant subspace: part 2 guarantees it is invariant. Repeat until no proper invariant subspace is left.
Change basis. Assemble a unitary $S$ whose columns are orthonormal bases of the invariant pieces in turn; then $S^{\dagger} D(g) S$ is block diagonal for every $g$ simultaneously.

Step 4 is the only one that needs thought. Characters turn it into arithmetic: Orthogonality relations and characters decides how many blocks of each kind occur before any of them is found.

Example(The permutation representation of

S_3

, reduced)Original

Let $G = S_3$ , the symmetric group on $\{1,2,3\}$ (The symmetric group), of order $6$ . Let it act on $V = \mathbb{C}^3$ by permuting coordinates: $T_\sigma e_j = e_{\sigma(j)}$ , following step 2 of the method. Writing $\tau = (1\,2\,3)$ and $\sigma = (1\,2)$ ,

D(\tau) = \begin{pmatrix} 0&0&1\\ 1&0&0\\ 0&1&0 \end{pmatrix}, \qquad D(\sigma) = \begin{pmatrix} 0&1&0\\ 1&0&0\\ 0&0&1 \end{pmatrix} .

Both are permutation matrices, hence unitary, and $\tau, \sigma$ generate $S_3$ , so every $D(g)$ is unitary.

An invariant line. The vector $u = \tfrac{1}{\sqrt3}(1,1,1)$ is fixed by every coordinate permutation, so $U = \operatorname{span}\{u\}$ is invariant and carries the trivial representation, of degree $1$ .

Its complement. By part 2 of the statement, $W = U^{\perp} = \{x \in \mathbb{C}^3 : x_1 + x_2 + x_3 = 0\}$ is invariant; it is two-dimensional. An orthonormal basis of $W$ is

f_1 = \tfrac{1}{\sqrt2}(1,-1,0), \qquad f_2 = \tfrac{1}{\sqrt6}(1,1,-2) .

The $2 \times 2$ block. Apply $\tau$ to $f_1$ . Since $\tau$ sends $e_1 \mapsto e_2$ , $e_2 \mapsto e_3$ , we get $T_\tau f_1 = \tfrac{1}{\sqrt2}(e_2 - e_3)$ . Write $e_2 - e_3 = a(e_1 - e_2) + b(e_1 + e_2 - 2e_3)$ : matching the $e_3$ coefficients gives $-2b = -1$ , so $b = \tfrac12$ , and matching the $e_1$ coefficients gives $a + b = 0$ , so $a = -\tfrac12$ (the $e_2$ coefficient then reads $-a + b = 1$ , as it must). Hence

T_\tau f_1 = \tfrac{1}{\sqrt2}\Bigl(-\tfrac12 \sqrt2\, f_1 + \tfrac12\sqrt6\, f_2\Bigr) = -\tfrac12 f_1 + \tfrac{\sqrt3}{2} f_2 .

The same computation for $f_2$ uses $T_\tau f_2 = \tfrac{1}{\sqrt6}(e_2 + e_3 - 2e_1)$ and $e_2 + e_3 - 2e_1 = -\tfrac32 (e_1 - e_2) - \tfrac12 (e_1+e_2-2e_3)$ , giving $T_\tau f_2 = -\tfrac{\sqrt3}{2} f_1 - \tfrac12 f_2$ . For $\sigma = (1\,2)$ the work is shorter: $T_\sigma f_1 = \tfrac{1}{\sqrt2}(e_2 - e_1) = -f_1$ and $T_\sigma f_2 = f_2$ . So on $W$ ,

\hat D(\tau) = \begin{pmatrix} -\tfrac12 & -\tfrac{\sqrt3}{2} \\[2pt] \tfrac{\sqrt3}{2} & -\tfrac12 \end{pmatrix}, \qquad \hat D(\sigma) = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} ,

a rotation through $120^\circ$ and a reflection: these are the familiar symmetries of an equilateral triangle, and they reproduce (up to a choice of basis) the two-dimensional irrep of the triangle group tabulated at (Riley, Hobson & Bence (2006), Eq. (29.12), p. 1090).

The block-diagonal form, with a number to check. Let $S = (\,u \ \ f_1 \ \ f_2\,)$ , a real orthogonal matrix because its columns are orthonormal. Then for every $g$ , $S^{\mathsf T} D(g) S$ is block diagonal $1 \oplus \hat D(g)$ . For $g = \tau$ the prediction is

S^{\mathsf T} D(\tau) S = \begin{pmatrix} 1 & 0 & 0 \\ 0 & -\tfrac12 & -\tfrac{\sqrt3}{2} \\ 0 & \tfrac{\sqrt3}{2} & -\tfrac12 \end{pmatrix} ,

so the $(2,3)$ entry must be $-\sqrt3/2 = -0.8660254\ldots$ Checking it by hand: row $2$ of $S^{\mathsf T}$ is $f_1^{\mathsf T} = \tfrac{1}{\sqrt2}(1,-1,0)$ , and column $3$ of $D(\tau)S$ is $D(\tau) f_2 = \tfrac{1}{\sqrt6}(-2,1,1)$ , so the entry is $\tfrac{1}{\sqrt{12}}\bigl(1\cdot(-2) + (-1)\cdot 1 + 0\bigr) = -3/\sqrt{12} = -\sqrt3/2$ , as predicted.

Irreducibility of the block. $\hat D(\sigma) = \operatorname{diag}(-1,1)$ has exactly two invariant lines, $\operatorname{span}\{f_1\}$ and $\operatorname{span}\{f_2\}$ (its eigenspaces, which are one-dimensional because the eigenvalues differ). Neither survives $\tau$ : $\hat D(\tau) f_1 = -\tfrac12 f_1 + \tfrac{\sqrt3}{2} f_2$ is not a multiple of $f_1$ , and $\hat D(\tau) f_2$ is not a multiple of $f_2$ . Any invariant line would have to be invariant under both, so there is none, and the degree-two representation is irreducible over $\mathbb{C}$ .

Conclusion. The permutation representation of $S_3$ on $\mathbb{C}^3$ decomposes as $T \simeq T^{(\mathrm{triv})} \oplus T^{(E)}$ , with degrees $1 + 2 = 3$ . It is reducible but, by part 4, not merely reducible: it is a genuine direct sum.

ApplicationOriginal

Degeneracy in quantum mechanics. If a Hamiltonian $H$ is unchanged by a symmetry group $G$ acting on the state space by $T_g$ , then $H T_g = T_g H$ for every $g$ (Arfken, Weber & Harris (2013), §17.3, p. 828). Each eigenspace of $H$ is then an invariant subspace: if $H\psi = E \psi$ then $H(T_g \psi) = T_g H \psi = E\,T_g \psi$ . Part 4 says that eigenspace breaks into irreducible pieces, so energy levels come labelled by irreducible representations and a level carrying an irreducible representation of degree $n$ is at least $n$ -fold degenerate. Degeneracy that is not explained this way is called accidental, and usually signals a larger symmetry that was overlooked. The parity group $\{e, P\}$ is the smallest instance: it has two one-dimensional irreducible representations, so the eigenfunctions of an even Hamiltonian can be taken even or odd, and nothing forces an even state and an odd state to share an energy (Arfken, Weber & Harris (2013), Example 17.3.1, p. 828).

Symmetry-adapted bases, and why they save work. The change of basis built in step 6 of the method block-diagonalizes every symmetry operation at once, and therefore block-diagonalizes $H$ as well. A secular determinant of size $N$ becomes several small ones, one per irreducible representation, and matrix elements between different blocks vanish before any integral is done. This is how molecular-orbital and normal-mode calculations are organised: for a molecule with the symmetry of an equilateral triangle, the example above is literally the calculation, with $e_1, e_2, e_3$ the three atomic orbitals and the splitting $3 = 1 + 2$ the familiar one non-degenerate plus one doubly degenerate level; this is the setting of Arfken, Weber and Harris's three-orbital triangle, whose symmetry-adapted combinations carry the same $1/\sqrt2$ and $1/\sqrt6$ coefficients as our $f_1$ and $f_2$ (Arfken, Weber & Harris (2013), Example 17.3.2, p. 829). The degenerate pair is forced by the group, not by the chemistry.

Crystals and bands. The point group of a crystal (Point groups and crystal symmetry) acts on the electronic states at each wavevector, so bands are labelled by its irreducible representations, and band degeneracies at high-symmetry points are exactly the degrees of those representations. Lowering the symmetry (a strain, a field) replaces the group by a subgroup, under which a representation that was irreducible may split; the observed splitting of a degenerate level is then a direct measurement of the subgroup.

Selection rules. Whether a transition matrix element can be non-zero is decided by whether the trivial representation occurs in a product of representations (Arfken, Weber & Harris (2013), §17.3, p. 830); the machinery for that is set up in Product representations and selection rules, and rests on the decomposition proved here.

Theorem

Schur's lemma

Adapted

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A linear map that intertwines two irreducible representations is either zero or an isomorphism, and over the complex numbers any map commuting with an irreducible representation is a multiple of the identity.

Prerequisites: Representations of groups, Eigenvalues and eigenvectors, Linear maps and matrices

Definition(Intertwiner)AdaptedAdapted from Hassani (2013), §24.3, Eq. (24.5), p. 732; Dummit & Foote (2004), §18.1, Exercise 14, p. 853

Let $T$ and $T'$ be representations of a group $G$ on vector spaces $V$ and $V'$ over a field $F$ (Representations of groups). A linear map $A \colon V \to V'$ is an intertwiner (or $G$ -equivariant map, or morphism of representations) if

A\, T_g \;=\; T'_g\, A \qquad \text{for every } g \in G .

The intertwiners from $T$ to $T'$ form a vector space, written $\operatorname{Hom}_G(V, V')$ . An invertible intertwiner is exactly an equivalence $T' = A T A^{-1}$ , so

T \simeq T' \iff \operatorname{Hom}_G(V,V') \text{ contains an invertible element.}

When $T' = T$ , the intertwiners are the linear maps commuting with every $T_g$ ; this space, $\operatorname{Hom}_G(V,V) = \operatorname{End}_G(V)$ , is closed under composition and is called the centralizer of the representation.

Theorem(Schur's lemma)AdaptedAdapted from Hassani (2013), §24.3, Lemma 24.3.1, p. 732, and Lemma 24.3.2, p. 733; Dummit & Foote (2004), §10.3, Exercise 11, p. 356, and §18.1, Exercise 18, pp. 853-854

Let $G$ be a group.

First form. Let $T$ on $V$ and $T'$ on $V'$ be irreducible representations of $G$ over a field $F$ , and let $A \in \operatorname{Hom}_G(V, V')$ . Then either $A = 0$ , or $A$ is an isomorphism and $T \simeq T'$ . Equivalently: inequivalent irreducible representations admit no non-zero intertwiner.
Second form. Let $T$ be an irreducible representation of $G$ on a finite-dimensional complex vector space $V$ , and let $A \colon V \to V$ satisfy $A T_g = T_g A$ for all $g \in G$ . Then $A = \lambda I$ for some $\lambda \in \mathbb{C}$ , and necessarily

\lambda = \frac{\operatorname{tr} A}{\dim V} .

Hence $\dim_{\mathbb{C}} \operatorname{End}_G(V) = 1$ . Part 1 holds over any field and in any dimension; part 2 needs $\mathbb{C}$ (or any algebraically closed field) and finite dimension, and the remark below shows both are necessary. A converse of part 2, which is what makes it usable as a test, is given in its own block after the proof.

Idea of the proofOriginal

An intertwiner cannot help but respect the group: whatever it kills and whatever it hits are built out of $G$ -translates, so the kernel of an intertwiner is an invariant subspace of the source and its image is an invariant subspace of the target. Irreducibility leaves each of those only two options, and the four combinations collapse to "zero" or "isomorphism". That is part 1 in a sentence.

Part 2 then exploits the one thing $\mathbb{C}$ gives that $\mathbb{R}$ does not: a linear map on a non-zero finite-dimensional complex space always has an eigenvalue. Subtract it. The map $A - \lambda I$ still commutes with the group, so part 1 applies to it, but it is not invertible, because it kills an eigenvector. The only non-invertible option on offer is zero.

ProofAdaptedAdapted from Hassani (2013), §24.3, proofs of Lemma 24.3.1, pp. 732-733, and Lemma 24.3.2, p. 733; Dummit & Foote (2004), §10.3, Exercise 11, p. 356

Part 1. Let $A T_g = T'_g A$ for all $g$ .

The kernel is invariant. Let $x \in \ker A$ and $g \in G$ . Then

A (T_g x) = T'_g (A x) = T'_g 0 = 0 ,

the first equality being the intertwining property. So $T_g x \in \ker A$ , and $\ker A$ is an invariant subspace of $V$ . Since $T$ is irreducible, $\ker A = V$ or $\ker A = \{0\}$ . The first case says $A = 0$ ; in the second, $A$ is injective.

The image is invariant. Let $y \in A(V)$ , say $y = Ax$ , and let $g \in G$ . Then

T'_g y = T'_g A x = A (T_g x) \in A(V) .

So $A(V)$ is an invariant subspace of $V'$ . Since $T'$ is irreducible, $A(V) = \{0\}$ or $A(V) = V'$ . The first case again says $A = 0$ ; in the second, $A$ is surjective.

Combining. If $A \neq 0$ , then neither first case holds, so $A$ is injective and surjective, hence an isomorphism; and then $T'_g = A T_g A^{-1}$ for every $g$ , which is precisely $T \simeq T'$ .

Part 2. Let $V$ be a non-zero finite-dimensional complex space and let $A$ commute with every $T_g$ . Every linear operator on such a space has at least one eigenvalue, because its characteristic polynomial has a root in $\mathbb{C}$ (Eigenvalues and eigenvectors); fix such a $\lambda$ and a non-zero $x$ with $A x = \lambda x$ . Set

B = A - \lambda I .

Then $B$ commutes with every $T_g$ , since $A$ does and $I$ does; so $B \in \operatorname{Hom}_G(V,V)$ , and part 1 applies with $T' = T$ (both irreducible, being the same irreducible representation). But $B x = 0$ with $x \neq 0$ , so $B$ is not injective, hence not an isomorphism. The only remaining alternative in part 1 is $B = 0$ , that is $A = \lambda I$ .

Taking traces, $\operatorname{tr} A = \lambda \operatorname{tr} I = \lambda \dim V$ , which gives the stated value of $\lambda$ and shows it is unique. Hence $\operatorname{End}_G(V)$ consists of the scalars and contains $I \neq 0$ , so it is the one-dimensional space $\mathbb{C}I$ . $\square$

Theorem(Converse: the scalar commutant as a test)Original

Let $G$ be a finite group and let $T$ be a representation of $G$ on a finite-dimensional complex vector space $V \neq \{0\}$ . If every linear $A \colon V \to V$ commuting with all $T_g$ is a scalar multiple of the identity, then $T$ is irreducible. Together with part 2 this gives

T \text{ irreducible} \iff \dim_{\mathbb{C}} \operatorname{End}_G(V) = 1 .

Proof(Proof of the converse)Original

We prove the contrapositive. Suppose $T$ is reducible, so there is an invariant subspace $W$ with $\{0\} \neq W \neq V$ . Because $G$ is finite, part 1 of the statement in Representations of groups lets us replace the inner product by one for which every $T_g$ is unitary, and then part 2 of that statement makes $W^{\perp}$ invariant as well, with $V = W \oplus W^{\perp}$ .

Let $A$ be the projection onto $W$ along $W^{\perp}$ , that is $A(w + w') = w$ for $w \in W$ and $w' \in W^{\perp}$ . For $g \in G$ and $v = w + w'$ we have $T_g v = T_g w + T_g w'$ with $T_g w \in W$ and $T_g w' \in W^{\perp}$ , both by invariance, so

A T_g v = T_g w = T_g A v ,

and $A$ commutes with the representation. But $A$ is not a scalar: it acts as $1$ on the non-zero subspace $W$ and as $0$ on the non-zero subspace $W^{\perp}$ . $\square$

Remark(Where the hypotheses are used)Original

Both representations must be irreducible in part 1. The proof used irreducibility of $T$ for the kernel and of $T'$ for the image, and needs both. Let $T$ be the two-dimensional irreducible representation of $S_3$ built in Representations of groups and let $T'$ be the reducible three-dimensional permutation representation it sits inside. The inclusion $W \hookrightarrow \mathbb{C}^3$ intertwines them and is neither zero nor an isomorphism.

Part 2 needs an algebraically closed field. Over $\mathbb{R}$ the two-dimensional rotation representation of the cyclic group $C_3 = \langle r \rangle$ , $r \mapsto R(2\pi/3)$ , is irreducible (a rotation through $120^\circ$ fixes no line in $\mathbb{R}^2$ ), yet $A = R(2\pi/3)$ itself commutes with the whole representation, which is abelian, and $A$ is not a real multiple of $I$ : its off-diagonal entries are $\mp\sin(2\pi/3) = \mp\sqrt3/2 \neq 0$ . The proof breaks at exactly the right place: $A$ has no real eigenvalue. Over $\mathbb{C}$ the same representation stops being irreducible, so no contradiction arises. This is the reason the rest of the chapter works over $\mathbb{C}$ .

Part 2 needs finite dimension, again only to produce an eigenvalue. In infinite dimensions an operator can have empty point spectrum (The spectrum of a bounded operator), and the argument stops. For finite groups the issue never arises, because every irreducible representation of a finite group is finite-dimensional: the span of the orbit $\{T_g v : g \in G\}$ of any single non-zero $v$ is an invariant subspace of dimension at most $|G|$ , hence equals the whole carrier space (Hassani (2013), Box 24.2.4, p. 730).

The converse needs $G$ finite, because its proof uses Maschke's theorem to produce the invariant complement $W^{\perp}$ . For infinite groups it is false. Let $G$ be the group of invertible upper triangular $2\times 2$ complex matrices acting on $V = \mathbb{C}^2$ in the obvious way. The line $\operatorname{span}\{e_1\}$ is invariant, so $V$ is reducible; yet the commutant is only the scalars. Indeed, let $A$ commute with every element of $G$ . Commuting with $\operatorname{diag}(1,2)$ , whose eigenvalues differ, forces $A$ to be diagonal, $A = \operatorname{diag}(w, z)$ ; commuting with $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ then gives $w = z$ . So $\operatorname{End}_G(V) = \mathbb{C}I$ while $V$ is reducible. What the proof really uses is complete reducibility, and that is exactly what a finite group supplies.

What the lemma does not say. It does not say an intertwiner between equivalent irreducible representations is unique: part 2 says the ones from $V$ to itself form a one-dimensional space, so between equivalent irreducibles the intertwiners form a line, not a point. Nor does it say anything about reducible representations, where the centralizer is large: the converse stated after the proof makes that precise, and $\dim \operatorname{End}_G(V) = \sum_\lambda m_\lambda^2$ in general, a formula proved with characters in Orthogonality relations and characters.

Remark(Consequences)AdaptedAdapted from Hassani (2013), §24.3, Thm 24.3.3, p. 733, and Example 24.3.4, p. 734; Dummit & Foote (2004), §18.1, Exercise 18, pp. 853-854

Abelian groups have only one-dimensional complex irreducible representations. If $G$ is abelian and $T$ is an irreducible complex representation of $G$ on a finite-dimensional $V$ , fix $h \in G$ . Then $T_h$ commutes with every $T_g$ , because $T_h T_g = T_{hg} = T_{gh} = T_g T_h$ . By part 2, $T_h = \lambda(h) I$ . Every subspace of $V$ is therefore invariant, so irreducibility forces $\dim V = 1$ . The irreducible representations of an abelian group are thus exactly the homomorphisms $G \to \mathbb{C}^{\times}$ . For a finite abelian group there are $|G|$ of them (Dummit & Foote (2004), §18.2, Example 1, p. 862).

The centre acts by scalars. For any $G$ and any irreducible complex $T$ , the same argument applied to $z$ in the centre $Z(G)$ (Conjugacy classes) gives $T_z = \lambda(z) I$ : central elements are represented by scalar matrices. If in addition $T$ is faithful, $z \mapsto \lambda(z)$ is an injective homomorphism of $Z(G)$ into $\mathbb{C}^\times$ , and a finite subgroup of $\mathbb{C}^{\times}$ is cyclic; so a finite group with a faithful irreducible complex representation has cyclic centre (Dummit & Foote (2004), §18.1, Exercise 18, pp. 853-854). This is a genuine restriction on groups, proved by representations: the Klein four-group, whose centre is itself and not cyclic, has no faithful irreducible complex representation.

The averaging construction. Let $T^{(\alpha)}$ on $V_\alpha$ and $T^{(\beta)}$ on $V_\beta$ be representations of a finite $G$ , and let $X \colon V_\beta \to V_\alpha$ be any linear map. Then

A \;=\; \sum_{x \in G} T^{(\alpha)}_{x}\, X\, T^{(\beta)}_{x^{-1}}

intertwines $T^{(\beta)}$ with $T^{(\alpha)}$ : for $g \in G$ ,

T^{(\alpha)}_g A = \sum_{x \in G} T^{(\alpha)}_{gx} X T^{(\beta)}_{(gx)^{-1}} T^{(\beta)}_{g} = A\, T^{(\beta)}_g ,

because $x \mapsto gx$ permutes $G$ . If both representations are irreducible, Schur's lemma pins $A$ down completely: it is $0$ when they are inequivalent, and a known multiple of the identity when $\alpha = \beta$ . Since $X$ was arbitrary, this is a strong constraint on the matrix entries of irreducible representations, and unwinding it is exactly the proof of the great orthogonality theorem in Orthogonality relations and characters (Hassani (2013), §24.3, Eq. (24.6), pp. 734-735).

Degenerate multiplets. If a Hamiltonian $H$ commutes with every operator of a representation $T$ of a symmetry group, and the representation is irreducible, then part 2 gives $H = \lambda I$ on that carrier space: every state in an irreducible multiplet has the same energy (Hassani (2013), Example 24.3.4, p. 734).

Method(Using Schur's lemma to evaluate an average)Original

A sum of the form $A = \sum_{g \in G} T_g X T_g^{-1}$ over a finite group, with $T$ irreducible on a complex $V$ of dimension $n$ , never has to be computed term by term:

Check the intertwining. Confirm $A T_h = T_h A$ by reindexing $g \mapsto hg$ (or $g \mapsto gh$ ), as in the previous block. This step is purely formal and uses only that $G$ is finite and that the sum runs over all of $G$ .
Invoke part 2. Conclude $A = \lambda I$ .
Take the trace to find $\lambda$ . Since traces are unchanged by conjugation, $\operatorname{tr}\bigl(T_g X T_g^{-1}\bigr) = \operatorname{tr} X$ for each of the $|G|$ terms, so $\operatorname{tr} A = |G| \operatorname{tr} X$ and

\lambda = \frac{|G| \operatorname{tr} X}{n}, \qquad A = \frac{|G| \operatorname{tr} X}{n}\, I .

If two different irreducible representations are involved, that is $A = \sum_g T^{(\alpha)}_g X T^{(\beta)}_{g^{-1}}$ with $T^{(\alpha)} \not\simeq T^{(\beta)}$ , step 2 is replaced by part 1 and $A = 0$ outright; no trace is needed.

The recipe is the engine of the orthogonality relations, and the same three lines reappear whenever a physical quantity is averaged over a symmetry group: the average must be the only $G$ -invariant object available, and its size is fixed by one trace.

Example(An averaged matrix over

S_3

)Original

Take the two-dimensional irreducible representation $\hat D$ of $S_3$ constructed in Representations of groups, acting on $W = \{x \in \mathbb{C}^3 : x_1+x_2+x_3 = 0\}$ in the orthonormal basis $f_1 = \tfrac{1}{\sqrt2}(1,-1,0)$ , $f_2 = \tfrac{1}{\sqrt6}(1,1,-2)$ . Its six matrices are the symmetries of an equilateral triangle, two rotations and three reflections:

\hat D(e) = \begin{pmatrix} 1&0\\0&1\end{pmatrix},\qquad \hat D(1\,2\,3) = \begin{pmatrix} -\tfrac12 & -\tfrac{\sqrt3}{2}\\[2pt] \tfrac{\sqrt3}{2} & -\tfrac12\end{pmatrix},\qquad \hat D(1\,3\,2) = \begin{pmatrix} -\tfrac12 & \tfrac{\sqrt3}{2}\\[2pt] -\tfrac{\sqrt3}{2} & -\tfrac12\end{pmatrix},

\hat D(1\,2) = \begin{pmatrix} -1&0\\0&1\end{pmatrix},\qquad \hat D(2\,3) = \begin{pmatrix} \tfrac12 & \tfrac{\sqrt3}{2}\\[2pt] \tfrac{\sqrt3}{2} & -\tfrac12\end{pmatrix},\qquad \hat D(1\,3) = \begin{pmatrix} \tfrac12 & -\tfrac{\sqrt3}{2}\\[2pt] -\tfrac{\sqrt3}{2} & -\tfrac12\end{pmatrix} .

The rotations have determinant $1$ and trace $-1$ ; the reflections have determinant $-1$ and trace $0$ . Each matrix is real orthogonal, so $\hat D(g)^{-1} = \hat D(g)^{\mathsf T}$ .

Predict first. Let

X = \begin{pmatrix} 1 & 2 \\ 3 & 4\end{pmatrix}, \qquad A = \sum_{g \in S_3} \hat D(g)\, X\, \hat D(g)^{\mathsf T} .

Step 1 of the method applies because the sum runs over all of $S_3$ , and the representation is irreducible by the argument in Representations of groups. So $A = \lambda I_2$ , and step 3 fixes $\lambda$ from $|G| = 6$ , $n = 2$ and $\operatorname{tr} X = 1 + 4 = 5$ :

\lambda = \frac{6 \cdot 5}{2} = 15, \qquad A = 15\, I_2 .

Now check it, in three pairs. Write $R = \hat D(1\,2\,3)$ , so $\hat D(1\,3\,2) = R^{\mathsf T}$ .

Identity and $(1\,2)$ . Since $\hat D(1\,2) = \operatorname{diag}(-1,1)$ , conjugating $X$ by it flips the sign of both off-diagonal entries:

X + \hat D(1\,2)\,X\,\hat D(1\,2)^{\mathsf T} = \begin{pmatrix} 1&2\\3&4\end{pmatrix} + \begin{pmatrix} 1&-2\\-3&4\end{pmatrix} = \begin{pmatrix} 2&0\\0&8\end{pmatrix} .

The two rotations. Multiplying out, for instance $(RXR^{\mathsf T})_{11} = \bigl(-\tfrac12 - \tfrac{3\sqrt3}{2}\bigr)\bigl(-\tfrac12\bigr) + \bigl(-1 - 2\sqrt3\bigr)\bigl(-\tfrac{\sqrt3}{2}\bigr) = \tfrac{13}{4} + \tfrac{5\sqrt3}{4}$ , and similarly for the other entries,

R X R^{\mathsf T} = \begin{pmatrix} \tfrac{13}{4} + \tfrac{5\sqrt3}{4} & -\tfrac{7}{4} + \tfrac{3\sqrt3}{4} \\[3pt] -\tfrac{3}{4} + \tfrac{3\sqrt3}{4} & \tfrac{7}{4} - \tfrac{5\sqrt3}{4}\end{pmatrix} .

The matrix $R^{\mathsf T}$ is $R$ with the sign of every off-diagonal entry reversed, which is the same as replacing $\sqrt3$ by $-\sqrt3$ throughout; so $R^{\mathsf T} X R$ is the displayed matrix with $\sqrt3 \to -\sqrt3$ , and every surd cancels in the sum:

R X R^{\mathsf T} + R^{\mathsf T} X R = \begin{pmatrix} \tfrac{13}{2} & -\tfrac{7}{2} \\[3pt] -\tfrac{3}{2} & \tfrac{7}{2}\end{pmatrix} .

The two remaining reflections. By the same substitution $\sqrt3 \to -\sqrt3$ , which carries $\hat D(2\,3)$ to $\hat D(1\,3)$ ,

\hat D(2\,3)\,X\,\hat D(2\,3) + \hat D(1\,3)\,X\,\hat D(1\,3) = \begin{pmatrix} \tfrac{13}{4} + \tfrac{5\sqrt3}{4} & \tfrac{7}{4} - \tfrac{3\sqrt3}{4} \\[3pt] \tfrac{3}{4} - \tfrac{3\sqrt3}{4} & \tfrac{7}{4} - \tfrac{5\sqrt3}{4}\end{pmatrix} + \bigl(\sqrt3 \to -\sqrt3\bigr) = \begin{pmatrix} \tfrac{13}{2} & \tfrac{7}{2} \\[3pt] \tfrac{3}{2} & \tfrac{7}{2}\end{pmatrix}

(these two matrices are symmetric, so transposing them is not needed).

The sum. Adding the three pairs entry by entry,

A = \begin{pmatrix} 2 & 0 \\ 0 & 8\end{pmatrix} + \begin{pmatrix} \tfrac{13}{2} & -\tfrac72 \\[3pt] -\tfrac32 & \tfrac72\end{pmatrix} + \begin{pmatrix} \tfrac{13}{2} & \tfrac72 \\[3pt] \tfrac32 & \tfrac72\end{pmatrix} = \begin{pmatrix} 15 & 0 \\ 0 & 15 \end{pmatrix} = 15\,I_2 ,

exactly as Schur's lemma predicted: the two off-diagonal entries cancel between the rotation pair and the reflection pair, and the diagonal entries come out equal even though $X$ had unequal diagonal entries $1$ and $4$ . An independent check of the one number that matters: $\operatorname{tr} A = 6 \operatorname{tr} X = 30 = 15 + 15$ .

An inequivalent pair. Let $T^{(1)}$ be the trivial representation of $S_3$ on $\mathbb{C}$ and $T^{(2)}$ the sign representation $\sigma \mapsto \operatorname{sgn}(\sigma)$ , both of degree $1$ and so both irreducible. An intertwiner $A \colon \mathbb{C} \to \mathbb{C}$ is a number $a$ with $a \cdot 1 = \operatorname{sgn}(\sigma)\, a$ for all $\sigma$ ; taking $\sigma$ to be a transposition gives $a = -a$ , hence $a = 0$ . Part 1 of the statement is confirmed directly, and the two representations are inequivalent.

ApplicationOriginal

Good quantum numbers from an abelian symmetry. The consequence that abelian groups have only one-dimensional complex irreducible representations is why a translation-invariant or rotation-about-one-axis-invariant problem gives labels rather than multiplets. For the discrete translations of a crystal lattice, the one-dimensional representations are $T_{\mathbf{R}} \mapsto e^{i \mathbf{k}\cdot\mathbf{R}}$ , and an eigenstate of the Hamiltonian can be chosen to transform by one of them: this is Bloch's theorem, and the crystal momentum $\mathbf{k}$ is the label of the representation. For rotations about the $z$ axis the representations are $\varphi \mapsto e^{i m \varphi}$ with $m \in \mathbb{Z}$ , and $m$ is the magnetic quantum number. No degeneracy is forced in either case, which is exactly why non-abelian symmetry is needed to explain multiplets.

Why non-abelian symmetry forces degeneracy. Conversely, if the symmetry group has an irreducible representation of degree $n > 1$ — which by the above happens only when the group is non-abelian — then a level transforming under it is $n$ -fold degenerate, because the Hamiltonian restricted to that multiplet commutes with the whole group and is therefore a multiple of the identity. The threefold degeneracy of the $p$ orbitals under the full rotation group, and the twofold $E$ levels of a molecule with threefold symmetry, are this statement.

Reduced matrix elements and the Wigner-Eckart theorem. Schur's lemma is the reason a matrix element of an operator between two irreducible multiplets factorizes into a geometric part fixed by the group and a single number that carries all the physics. The intertwiner space between the two multiplets is at most one-dimensional, so once one non-zero matrix element is known the rest are determined. This is the content of Spherical tensors and the Wigner-Eckart theorem and, for finite groups, of the selection rules of Product representations and selection rules.

Irreducibility as a computational test. The converse above turns "is this representation irreducible?" into "does anything but a scalar commute with it?". In practice this is how a numerically generated representation is checked: build the commutant by solving the linear system $A D(g) = D(g) A$ over the generators of $G$ , and read off its dimension. Dimension $1$ means irreducible; dimension $\sum_\lambda m_\lambda^2$ in general, which is also what the character test of Orthogonality relations and characters computes, far more cheaply.

Theorem

Orthogonality relations and characters

Adapted

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The matrix entries of the inequivalent irreducible representations of a finite group are orthogonal functions on the group, and their traces — the characters — form an orthonormal set that decides, by one sum, how any representation decomposes.

Prerequisites: Representations of groups, Schur's lemma, Conjugacy classes, Inner products and norms, Unitary and orthogonal matrices

Definition(Character, class function, the inner product on class functions)AdaptedAdapted from Dummit & Foote (2004), §18.3, definition and Eqs. (18.1)-(18.3), pp. 866-867, and the Hermitian form, p. 870; Riley, Hobson & Bence (2006), §29.6, p. 1092; Hassani (2013), Def. 24.3.5 and Box 24.3.6, p. 736

Let $G$ be a finite group of order $|G|$ and let $T$ be a representation of $G$ on a finite-dimensional complex space $V$ , with matrices $D(g)$ in some basis. The character of $T$ is

\chi(g) \;=\; \operatorname{tr} T_g \;=\; \sum_{i} D(g)_{ii} .

It is called simple (or irreducible) if $T$ is irreducible and compound otherwise. Three properties are immediate from $\operatorname{tr}(AB) = \operatorname{tr}(BA)$ :

Basis independence. $\operatorname{tr}(S^{-1} D S) = \operatorname{tr} D$ , so equivalent representations have the same character.
Degree. $\chi(e) = \operatorname{tr} I = \dim V$ , the degree of the representation.
Class function. $\chi(hgh^{-1}) = \operatorname{tr}\bigl(D(h) D(g) D(h)^{-1}\bigr) = \chi(g)$ , so $\chi$ is constant on each conjugacy class (Conjugacy classes). A function $G \to \mathbb{C}$ constant on conjugacy classes is a class function; they form a vector space of dimension $r$ , the number of classes, a basis being the indicator functions of the classes.

On class functions put

(\theta, \psi) \;=\; \frac{1}{|G|} \sum_{g \in G} \theta(g)\,\overline{\psi(g)} \;=\; \frac{1}{|G|}\sum_{i=1}^{r} c_i\, \theta_i \overline{\psi_i} ,

linear in the first slot, where the second form collects the classes $\mathcal{K}_1, \dots, \mathcal{K}_r$ , $c_i = |\mathcal{K}_i|$ , and $\theta_i$ is the common value of $\theta$ on $\mathcal{K}_i$ . This is an inner product (Inner product spaces).

Convention. We follow Dummit and Foote in dividing by $|G|$ . Riley, Hobson and Bence and Hassani write the same relations without the $1/|G|$ , so where we write $(\chi^{(\lambda)}, \chi^{(\mu)}) = \delta_{\lambda\mu}$ they write $\sum_{g} \chi^{(\lambda)}(g)\overline{\chi^{(\mu)}(g)} = |G|\,\delta_{\lambda\mu}$ . The two differ only by that factor.

Throughout, $T^{(1)}, \dots, T^{(\rho)}$ is a fixed complete list of pairwise inequivalent irreducible complex representations of $G$ , $n_\lambda$ is the degree of $T^{(\lambda)}$ , $\hat D^{(\lambda)}$ its matrices, and $\chi^{(\lambda)}$ its character. By part 1 of the statement in Representations of groups we may and do take every $\hat D^{(\lambda)}(g)$ to be unitary, which changes no character.

Theorem(The great orthogonality theorem and its consequences)AdaptedAdapted from Riley, Hobson & Bence (2006), §29.5, Eq. (29.13), p. 1091, §29.6.1, Eqs. (29.14)-(29.16), p. 1094, §29.7, Eq. (29.18), p. 1095, §29.7.1, Eq. (29.21), p. 1099; Hassani (2013), Eqs. (24.7)-(24.8), p. 735, Eq. (24.11), p. 736, Eqs. (24.15)-(24.17), pp. 737-738, Eq. (24.18), p. 739, Eq. (24.27), p. 742, Thm 24.5.2, p. 743; Dummit & Foote (2004), Thm 15 and Thm 16, p. 872

Let $G$ be a finite group, with notation as in the definition.

Great orthogonality theorem. For all indices $i, j, k, l$ in range and all $\lambda, \mu$ ,

\sum_{g \in G} \overline{\bigl[\hat D^{(\lambda)}(g)\bigr]_{ij}} \; \bigl[\hat D^{(\mu)}(g)\bigr]_{kl} \;=\; \frac{|G|}{n_\lambda}\, \delta_{ik}\, \delta_{jl}\, \delta_{\lambda\mu} .

Equivalently, without assuming unitarity, $\sum_{g} \bigl[\hat D^{(\lambda)}(g)\bigr]_{ij}\bigl[\hat D^{(\mu)}(g^{-1})\bigr]_{kl} = \dfrac{|G|}{n_\lambda}\delta_{jk}\delta_{il}\delta_{\lambda\mu}$ . 2. Character values. For every character $\chi$ of $G$ and every $g$ , $\chi(g^{-1}) = \overline{\chi(g)}$ , and $\chi(g)$ is a sum of $\chi(e)$ roots of unity. 3. First orthogonality relation. $(\chi^{(\lambda)}, \chi^{(\mu)}) = \delta_{\lambda\mu}$ : the simple characters are orthonormal, hence linearly independent, class functions. 4. Decomposition and multiplicities. If $T \simeq \bigoplus_\lambda m_\lambda T^{(\lambda)}$ has character $\chi$ , then $\chi = \sum_\lambda m_\lambda \chi^{(\lambda)}$ and

m_\lambda = (\chi, \chi^{(\lambda)}) = \frac{1}{|G|}\sum_{g \in G} \chi(g)\,\overline{\chi^{(\lambda)}(g)} , \qquad (\chi,\chi) = \sum_\lambda m_\lambda^2 .

In particular the multiplicities, and hence the representation up to equivalence, are determined by $\chi$ ; and $T$ is irreducible if and only if $(\chi,\chi) = 1$ . 5. The regular representation. Let $T^{\mathrm{reg}}$ be the representation of $G$ on the $|G|$ -dimensional space with basis $\{e_h : h \in G\}$ given by $T^{\mathrm{reg}}_g e_h = e_{gh}$ . Its character is

\chi^{\mathrm{reg}}(e) = |G|, \qquad \chi^{\mathrm{reg}}(g) = 0 \ \text{ for } g \neq e ,

so every irreducible representation occurs in it with multiplicity equal to its own degree, $m_\lambda = n_\lambda$ , and therefore

\sum_{\lambda=1}^{\rho} n_\lambda^2 \;=\; |G| .

Counting, and the column relation. The number $\rho$ of inequivalent irreducible complex representations equals the number $r$ of conjugacy classes, and the simple characters are an orthonormal basis of the space of class functions. Writing $\chi^{(\lambda)}_i$ for the value of $\chi^{(\lambda)}$ on $\mathcal{K}_i$ , the second orthogonality relation holds:

\sum_{\lambda=1}^{\rho} \chi^{(\lambda)}_i \,\overline{\chi^{(\lambda)}_j} \;=\; \frac{|G|}{c_i}\, \delta_{ij} .

Parts 1 to 5 are proved below. In part 6 we prove $\rho \le r$ and take the reverse inequality, with the column relation, from the sources.

Idea of the proofOriginal

Everything comes from one observation, made in Schur's lemma: for any linear map $X$ , the group average $A = \sum_g \hat D^{(\lambda)}(g)\, X\, \hat D^{(\mu)}(g^{-1})$ intertwines two irreducible representations, so Schur's lemma evaluates it — zero if they are inequivalent, a known multiple of the identity if they are the same. Because $X$ is arbitrary, we may feed in the matrix units $E_{jk}$ one at a time, and each choice reads off one bilinear relation among the matrix entries. That is part 1, and it is the whole theorem; the rest is bookkeeping.

Characters are what survives a change of basis, and they are obtained from part 1 by setting the row index equal to the column index and summing, which is exactly taking traces. The orthogonality of entries collapses to orthogonality of characters, and since a character knows nothing but the multiplicities, a single sum extracts them.

The regular representation is then the one place where the counting closes: it contains every irreducible representation, and its character is so sparse — all of $|G|$ concentrated at the identity — that the multiplicity formula returns the degree itself, giving $\sum n_\lambda^2 = |G|$ in one line.

ProofAdaptedAdapted from Hassani (2013), §24.3, Eqs. (24.6)-(24.8), pp. 734-735 and Eqs. (24.10)-(24.11), p. 736, §24.4, Eqs. (24.14)-(24.18), pp. 737-739; Riley, Hobson & Bence (2006), §29.6.1, Eq. (29.14), p. 1094, §29.7, Eq. (29.18), p. 1095, §29.7.1, pp. 1098-1099; Dummit & Foote (2004), Prop. 14 and Thm 15, p. 872

Part 1. Fix $\lambda, \mu$ and let $X$ be any $n_\lambda \times n_\mu$ matrix. Put

A \;=\; \sum_{x \in G} \hat D^{(\lambda)}(x)\, X\, \hat D^{(\mu)}(x^{-1}) .

For $g \in G$ , inserting $\hat D^{(\mu)}(g^{-1})\hat D^{(\mu)}(g) = I$ and using the homomorphism property twice,

\hat D^{(\lambda)}(g)\,A = \sum_{x \in G} \hat D^{(\lambda)}(gx)\, X\, \hat D^{(\mu)}\bigl((gx)^{-1}\bigr)\, \hat D^{(\mu)}(g) = A\, \hat D^{(\mu)}(g) ,

because $x \mapsto gx$ is a bijection of $G$ , so the sum over $x$ is unchanged. Thus $A$ intertwines $T^{(\mu)}$ with $T^{(\lambda)}$ , both irreducible.

If $\lambda \neq \mu$ the two are inequivalent, so $A = 0$ by the first form of Schur's lemma.
If $\lambda = \mu$ then $A = c\,I$ by the second form, and taking traces, $c\, n_\lambda = \operatorname{tr} A = \sum_{x} \operatorname{tr}\bigl(\hat D^{(\lambda)}(x) X \hat D^{(\lambda)}(x)^{-1}\bigr) = |G| \operatorname{tr} X$ , so $c = |G| \operatorname{tr}(X)/n_\lambda$ .

Both cases combine into a single statement about the entries of $A$ : for all $i$ and $l$ ,

A_{il} = \frac{|G|\,\operatorname{tr} X}{n_\lambda}\,\delta_{\lambda\mu}\,\delta_{il} ,

the factor $\delta_{\lambda\mu}$ disposing of the inequivalent case, where $\operatorname{tr}X$ need not even be defined because $X$ need not be square.

Now choose $X = E_{jk}$ , the matrix with a single $1$ in row $j$ and column $k$ , so that $(E_{jk})_{ab} = \delta_{aj}\delta_{bk}$ and $\operatorname{tr} E_{jk} = \delta_{jk}$ . The $(i,l)$ entry of $A$ is

A_{il} = \sum_{x \in G} \sum_{a,b} \bigl[\hat D^{(\lambda)}(x)\bigr]_{ia} (E_{jk})_{ab} \bigl[\hat D^{(\mu)}(x^{-1})\bigr]_{bl} = \sum_{x \in G} \bigl[\hat D^{(\lambda)}(x)\bigr]_{ij}\,\bigl[\hat D^{(\mu)}(x^{-1})\bigr]_{kl} ,

while the formula for $A$ gives $A_{il} = (|G|/n_\lambda)\,\delta_{jk}\,\delta_{\lambda\mu}\,\delta_{il}$ . That is the second display of part 1.

For the first display, use unitarity: $\hat D^{(\mu)}(x^{-1}) = \hat D^{(\mu)}(x)^{-1} = \hat D^{(\mu)}(x)^{\dagger}$ , so $\bigl[\hat D^{(\mu)}(x^{-1})\bigr]_{kl} = \overline{\bigl[\hat D^{(\mu)}(x)\bigr]_{lk}}$ . Substituting gives $\sum_{x} \bigl[\hat D^{(\lambda)}(x)\bigr]_{ij}\,\overline{\bigl[\hat D^{(\mu)}(x)\bigr]_{lk}} = (|G|/n_\lambda)\delta_{jk}\delta_{il}\delta_{\lambda\mu}$ . Conjugate the whole identity, which leaves the right-hand side alone because it is real, and then interchange the names of the two free indices $k$ and $l$ :

\sum_{x \in G} \overline{\bigl[\hat D^{(\lambda)}(x)\bigr]_{ij}}\;\bigl[\hat D^{(\mu)}(x)\bigr]_{kl} = \frac{|G|}{n_\lambda}\,\delta_{ik}\delta_{jl}\delta_{\lambda\mu} .

(The factor is $n_\lambda$ and not $n_\mu$ ; the two agree whenever the right-hand side is non-zero, since then $\lambda = \mu$ .)

Part 2. Let $\chi$ be the character of $T$ and fix $g$ of order $m$ , so $T_g^m = T_{g^m} = I$ . The minimal polynomial of $T_g$ therefore divides $z^m - 1$ , which has $m$ distinct roots, so $T_g$ is diagonalizable with eigenvalues $\varepsilon_1, \dots, \varepsilon_n$ that are $m$ -th roots of unity, $n = \chi(e)$ . Then $\chi(g) = \sum_s \varepsilon_s$ , which is the second claim. Also $T_{g^{-1}} = T_g^{-1}$ has eigenvalues $\varepsilon_s^{-1} = \overline{\varepsilon_s}$ , because $|\varepsilon_s| = 1$ ; summing, $\chi(g^{-1}) = \sum_s \overline{\varepsilon_s} = \overline{\chi(g)}$ .

Part 3. In the first display of part 1 set $j = i$ and $l = k$ , then sum over $i$ from $1$ to $n_\lambda$ and over $k$ from $1$ to $n_\mu$ . The left-hand side becomes $\sum_{g} \overline{\chi^{(\lambda)}(g)}\,\chi^{(\mu)}(g)$ , and the right-hand side becomes

\frac{|G|}{n_\lambda}\,\delta_{\lambda\mu} \sum_{i=1}^{n_\lambda}\sum_{k=1}^{n_\mu} \delta_{ik}\delta_{ik} = \frac{|G|}{n_\lambda}\,\delta_{\lambda\mu}\, n_\lambda = |G|\,\delta_{\lambda\mu} ,

where the double sum collapses to $n_\lambda$ because $\delta_{ik}^2 = \delta_{ik}$ and the term $\delta_{\lambda\mu}$ makes $n_\mu = n_\lambda$ in every surviving term. Dividing by $|G|$ and conjugating gives $(\chi^{(\lambda)}, \chi^{(\mu)}) = \delta_{\lambda\mu}$ . Orthonormal vectors are linearly independent, so the $\chi^{(\lambda)}$ are independent class functions.

Part 4. Choose a basis in which every $D(g)$ is block diagonal with the blocks of the decomposition $T \simeq \bigoplus_\lambda m_\lambda T^{(\lambda)}$ , which exists by part 4 of the statement in Representations of groups. The trace of a block-diagonal matrix adds up the traces of the blocks along its diagonal, so $\chi = \sum_\lambda m_\lambda \chi^{(\lambda)}$ . Pairing with $\chi^{(\mu)}$ and using part 3,

(\chi, \chi^{(\mu)}) = \sum_\lambda m_\lambda\, (\chi^{(\lambda)}, \chi^{(\mu)}) = m_\mu ,

the first step using linearity in the first slot and that the $m_\lambda$ are real integers. Likewise

(\chi,\chi) = \sum_{\lambda,\mu} m_\lambda m_\mu\, (\chi^{(\lambda)}, \chi^{(\mu)}) = \sum_\lambda m_\lambda^2 .

Since the $m_\lambda$ are non-negative integers, $\sum_\lambda m_\lambda^2 = 1$ holds exactly when one $m_\lambda$ is $1$ and the rest are $0$ , that is exactly when $T$ is irreducible. And because $\chi$ determines every $m_\lambda$ , two representations with the same character have the same multiplicities, hence are equivalent.

Part 5. In the basis $\{e_h\}_{h \in G}$ the operator $T^{\mathrm{reg}}_g$ permutes the basis, sending $e_h$ to $e_{gh}$ . Its matrix therefore has diagonal entry $1$ in position $(h,h)$ exactly when $gh = h$ , that is when $g = e$ , and $0$ otherwise. Hence $\chi^{\mathrm{reg}}(e) = |G|$ and $\chi^{\mathrm{reg}}(g) = 0$ for $g \neq e$ . (That this is a representation follows from associativity: $T^{\mathrm{reg}}_g T^{\mathrm{reg}}_{g'} e_h = e_{g(g'h)} = e_{(gg')h}$ .) Part 4 now gives, with only the term $g = e$ surviving,

m_\lambda = \frac{1}{|G|}\sum_{g \in G} \chi^{\mathrm{reg}}(g)\,\overline{\chi^{(\lambda)}(g)} = \frac{1}{|G|}\,|G|\,\overline{\chi^{(\lambda)}(e)} = n_\lambda ,

using $\chi^{(\lambda)}(e) = n_\lambda$ , which is real. Evaluating $\chi^{\mathrm{reg}} = \sum_\lambda n_\lambda \chi^{(\lambda)}$ at $g = e$ gives $|G| = \sum_\lambda n_\lambda \cdot n_\lambda$ .

Part 6, the inequality $\rho \le r$ . Each $\chi^{(\lambda)}$ is a class function, and by part 3 the $\rho$ of them are linearly independent. The space of class functions has dimension $r$ , so $\rho \le r$ . The reverse inequality $r \le \rho$ is proved by Hassani from the class sums $\kappa_i = \sum_{x \in \mathcal{K}_i} x$ in the group algebra: each $\sum_{x \in \mathcal{K}_i} \hat D^{(\lambda)}(x)$ commutes with the representation, hence is a scalar $\lambda_i^{(\lambda)} I$ by Schur's lemma, and chasing the resulting structure constants yields the column relation of part 6, in which the $r$ vectors $\bigl(\chi^{(\lambda)}_i\bigr)_{\lambda=1}^{\rho}$ , one per class, are mutually orthogonal and non-zero, forcing $r \le \rho$ (Hassani (2013), §24.5, Eqs. (24.21)-(24.27), pp. 741-742, and Thm 24.5.2, p. 743). With $\rho = r$ the $\rho$ orthonormal simple characters are a basis of the $r$ -dimensional space of class functions (Dummit & Foote (2004), Eq. (18.10) and Thm 15, pp. 870-872). $\square$

Remark(Where the hypotheses are used)Original

Finiteness of $G$ is used in every sum over the group, and in the reindexing $x \mapsto gx$ that makes $A$ an intertwiner. Nothing survives verbatim for infinite groups; the correct replacement averages with an invariant measure instead of a sum, which is why the theory extends cleanly to compact groups and to the rotation group (The rotation group and SU(2)) but not to, say, $\mathbb{Z}$ .

The field must be $\mathbb{C}$ , because part 1 invokes the second form of Schur's lemma, which needs an algebraically closed field. The failure is easy to see. Let $G = C_3 = \{e, r, r^2\}$ act on $\mathbb{R}^2$ by rotations through $0$ , $2\pi/3$ , $4\pi/3$ . This representation is irreducible over $\mathbb{R}$ , and $\hat D(g)_{11} = \hat D(g)_{22} = \cos\theta_g$ , so

\sum_{g \in G} \hat D(g)_{11}\,\hat D(g)_{22} = 1 + \tfrac14 + \tfrac14 = \tfrac32 ,

whereas part 1 with $(i,j) = (1,1)$ and $(k,l) = (2,2)$ predicts $(|G|/n)\delta_{12}\delta_{12} = 0$ . The character version fails too: $\chi = (2, -1, -1)$ gives $\sum_g \chi(g)^2 = 4 + 1 + 1 = 6 = 2|G|$ , so $(\chi,\chi) = 2$ , not $1$ . Read over $\mathbb{C}$ , that $2$ is correct and informative: it says $(\chi,\chi) = \sum m_\lambda^2 = 1 + 1$ , and indeed over $\mathbb{C}$ the representation splits into the two one-dimensional representations $r \mapsto \omega$ and $r \mapsto \omega^2$ , $\omega = e^{2\pi i /3}$ .

Unitarity is used only for the conjugated form. The version with $\hat D^{(\mu)}(g^{-1})$ needs no inner product at all; it is unitarity that converts $\hat D(g^{-1})$ into $\overline{\hat D(g)}^{\,\mathsf T}$ and so produces the Hermitian inner product. Since every complex representation of a finite group is equivalent to a unitary one and characters are unchanged by equivalence, this costs nothing.

The list of irreducible representations must be complete and pairwise inequivalent. Part 1 says nothing if the same representation is listed twice under different names, and the column relation of part 6 is false if any irreducible representation is omitted: the left-hand side is a sum over all of them.

Part 4 needs complete reducibility, which is Maschke's theorem from Representations of groups, and with it the fact that the multiplicities are well defined. Over a field of characteristic dividing $|G|$ the character no longer determines the representation, and $(\chi,\chi)$ is not even defined in the same way.

What characters forget. A character determines a representation up to equivalence, but two non-isomorphic groups can have identical character tables: $D_8$ and the quaternion group $Q_8$ , both of order $8$ , do. Riley, Hobson and Bence note the same phenomenon for the point groups $C_{4v}$ and $D_{2d}$ (Riley, Hobson & Bence (2006), §29.6, p. 1092). Characters are a complete invariant of a representation, not of a group.

Method(Decomposing a representation with characters)Original

Given a representation $T$ of a finite group $G$ and the simple characters of $G$ :

List the conjugacy classes $\mathcal{K}_1, \dots, \mathcal{K}_r$ with their sizes $c_i$ , and pick one representative $g_i$ of each. Everything below uses only representatives, since characters are class functions.
Compute $\chi(g_i)$ , not the matrices. For a permutation representation, $\chi(g)$ is just the number of basis elements $g$ fixes (Dummit & Foote (2004), §18.3, Example 3, p. 866). For an action on functions or on monomials, it is the sum of the diagonal weights, so only the diagonal of each matrix is needed (Riley, Hobson & Bence (2006), §29.7, p. 1096).
Test irreducibility first. Compute $(\chi,\chi) = \frac{1}{|G|}\sum_i c_i |\chi(g_i)|^2$ . If it is $1$ , stop: $T$ is irreducible. If it is $k$ , then $\sum_\lambda m_\lambda^2 = k$ , which already limits the possibilities severely.
Extract each multiplicity by $m_\lambda = \frac{1}{|G|}\sum_{i=1}^{r} c_i\, \chi(g_i)\, \overline{\chi^{(\lambda)}(g_i)}$ . Every $m_\lambda$ must come out a non-negative integer; a fraction means an arithmetic slip, a wrong class size, or an incomplete character table.
Check twice. Verify $\sum_\lambda m_\lambda n_\lambda = \chi(e) = \dim V$ and $\sum_\lambda m_\lambda^2 = (\chi,\chi)$ . These two use different parts of the data and catch almost every error.

Constructing the simple characters in the first place is the business of Character tables; the sum rules $\sum_\lambda n_\lambda^2 = |G|$ and $\rho = r$ from part 6 are the main tools there.

Example(Quadratic forms in three variables under

S_3

)Original

The group and its simple characters. Take $G = S_3$ , of order $6$ , with the three conjugacy classes (Conjugacy classes)

\mathcal{K}_1 = \{e\}\ (c_1 = 1), \qquad \mathcal{K}_2 = \{(1\,2),(1\,3),(2\,3)\}\ (c_2 = 3), \qquad \mathcal{K}_3 = \{(1\,2\,3),(1\,3\,2)\}\ (c_3 = 2) .

By part 6 there are exactly three inequivalent irreducible representations, and by part 5 their degrees satisfy $n_1^2 + n_2^2 + n_3^2 = 6$ , whose only solution in positive integers is $1, 1, 2$ . All three are already in hand: the trivial representation; the sign representation $\sigma \mapsto \operatorname{sgn}\sigma$ ; and the two-dimensional representation $\hat D$ built in Representations of groups, which is irreducible there. Taking traces of the six matrices listed in Schur's lemma — $\operatorname{tr}\operatorname{diag}(-1,1) = 0$ for a reflection and $-\tfrac12 - \tfrac12 = -1$ for a rotation — the characters on $(\mathcal{K}_1, \mathcal{K}_2, \mathcal{K}_3)$ are

\chi^{(1)} = (1,\,1,\,1), \qquad \chi^{(2)} = (1,\,-1,\,1), \qquad \chi^{(E)} = (2,\,0,\,-1) .

Part 3, checked. With weights $(c_1,c_2,c_3) = (1,3,2)$ ,

(\chi^{(E)}, \chi^{(E)}) = \tfrac16\bigl[1\cdot 2^2 + 3\cdot 0^2 + 2\cdot(-1)^2\bigr] = \tfrac{4+0+2}{6} = 1 ,

(\chi^{(1)}, \chi^{(2)}) = \tfrac16\bigl[1 - 3 + 2\bigr] = 0, \qquad (\chi^{(2)}, \chi^{(E)}) = \tfrac16\bigl[1\cdot1\cdot2 + 3\cdot(-1)\cdot 0 + 2\cdot 1 \cdot(-1)\bigr] = 0 .

The representation to decompose. Let $V$ be the space of homogeneous quadratic polynomials in $x_1, x_2, x_3$ , of dimension $6$ with basis

x_1^2,\ x_2^2,\ x_3^2,\ x_1x_2,\ x_1x_3,\ x_2x_3 ,

and let $\sigma \in S_3$ act by permuting the variables, $(T_\sigma p)(x_1,x_2,x_3) = p(x_{\sigma(1)}, x_{\sigma(2)}, x_{\sigma(3)})$ . This permutes the six basis monomials, so by step 2 of the method $\chi(\sigma)$ counts the monomials $\sigma$ fixes.

$\sigma = e$ fixes all six: $\chi_1 = 6$ .
$\sigma = (1\,2)$ swaps $x_1^2 \leftrightarrow x_2^2$ and $x_1x_3 \leftrightarrow x_2x_3$ , and fixes $x_3^2$ and $x_1x_2$ : $\chi_2 = 2$ .
$\sigma = (1\,2\,3)$ cycles $x_1^2 \to x_2^2 \to x_3^2 \to x_1^2$ and $x_1x_2 \to x_2x_3 \to x_1x_3 \to x_1x_2$ , fixing nothing: $\chi_3 = 0$ .

So $\chi = (6, 2, 0)$ .

Step 3: how far from irreducible.

(\chi,\chi) = \tfrac16\bigl[1\cdot 36 + 3 \cdot 4 + 2\cdot 0\bigr] = \tfrac{48}{6} = 8 .

Not $1$ , so $V$ is reducible, and $\sum_\lambda m_\lambda^2 = 8$ .

Step 4: the multiplicities.

m_1 = \tfrac16\bigl[1\cdot 6 \cdot 1 + 3 \cdot 2 \cdot 1 + 2\cdot 0 \cdot 1\bigr] = \tfrac{6+6+0}{6} = 2 ,

m_2 = \tfrac16\bigl[1\cdot 6 \cdot 1 + 3\cdot 2\cdot(-1) + 2\cdot 0\cdot 1\bigr] = \tfrac{6-6+0}{6} = 0 ,

m_E = \tfrac16\bigl[1\cdot 6\cdot 2 + 3\cdot 2\cdot 0 + 2\cdot 0\cdot(-1)\bigr] = \tfrac{12}{6} = 2 .

All three are non-negative integers, as they must be. Hence

V \;\simeq\; 2\,T^{(1)} \;\oplus\; 2\,T^{(E)} ,

with no copy of the sign representation at all.

Step 5: both checks. Degrees: $m_1 n_1 + m_2 n_2 + m_E n_E = 2\cdot 1 + 0 \cdot 1 + 2 \cdot 2 = 6 = \dim V$ . Norms: $m_1^2 + m_2^2 + m_E^2 = 4 + 0 + 4 = 8 = (\chi,\chi)$ .

Reading the answer off the polynomials. The two copies of the trivial representation are the two independent symmetric quadratics,

q_1 = x_1^2 + x_2^2 + x_3^2, \qquad q_2 = x_1x_2 + x_1x_3 + x_2x_3 ,

and that there are exactly two of them is the statement $m_1 = 2$ . That $m_2 = 0$ says no quadratic in three variables changes sign under every transposition — correct, since such a polynomial would be divisible by $(x_1-x_2)(x_1-x_3)(x_2-x_3)$ , of degree $3$ . The remaining four dimensions carry two copies of the two-dimensional representation, spanned for instance by the pairs

\bigl(x_1^2 - x_2^2,\ \ x_1^2 + x_2^2 - 2x_3^2\bigr) \quad\text{and}\quad \bigl(x_1x_2 - x_1x_3,\ \ x_1x_2 + x_1x_3 - 2x_2x_3\bigr) .

The structural reason is visible: $V$ is the direct sum of the two invariant subspaces $\operatorname{span}\{x_1^2, x_2^2, x_3^2\}$ and $\operatorname{span}\{x_2x_3, x_1x_3, x_1x_2\}$ , and each is a copy of the permutation representation on three letters — in the second, $\sigma$ sends the monomial that omits $x_i$ to the monomial that omits $x_{\sigma(i)}$ . That representation was decomposed as trivial $\oplus\, E$ in Representations of groups, so $V \simeq 2(\text{trivial} \oplus E)$ , which is the answer the characters gave. Reaching it by characters needed no matrix of $T$ at all, only three fixed-point counts.

ApplicationOriginal

Counting normal modes by symmetry. For a molecule with symmetry group $G$ , the $3N$ Cartesian displacements of its $N$ atoms carry a representation whose character is computed atom by atom: only atoms left in place by a symmetry operation contribute, and each contributes a small universal number ( $3$ for the identity, $1 + 2\cos\theta$ for a rotation by $\theta$ , and so on). Step 4 then gives, with no dynamics at all, how many vibrational modes of each symmetry type the molecule has, and therefore how many distinct frequencies to expect and how degenerate each is. Subtracting the three translational and three rotational representations leaves the genuine vibrations. The entire calculation is one row of arithmetic per conjugacy class.

Which transitions are allowed. A matrix element $\int \psi_f^{*}\, \hat O\, \psi_i$ is invariant under the symmetry group, so it vanishes unless the trivial representation occurs in the product representation carried by the integrand; and by part 4 the multiplicity of the trivial representation in a product is $(\chi_f \chi_O \chi_i, \chi^{(1)})$ , a single sum. This is the infrared and Raman selection rule, and the systematic version is Product representations and selection rules. Arfken's discussion of why a molecule needs the right symmetry to have a dipole moment is the same computation (Arfken, Weber & Harris (2013), §17.3, p. 830).

Crystal field and level splitting. An atomic level of angular momentum $\ell$ carries the $(2\ell+1)$ -dimensional representation of the full rotation group. Put the atom in a crystal and the symmetry drops to a point group $G$ (Point groups and crystal symmetry); the same $(2\ell+1)$ states now carry a representation of $G$ , generally reducible, and step 4 says exactly how the level splits and what the new degeneracies are. The number of lines a spectroscopist sees, and their multiplicities, are read from a character table before any energy is computed.

Counting orbits. Applied to a permutation representation on a finite set $X$ , part 4 with $\lambda$ the trivial representation says $m_1 = \frac{1}{|G|}\sum_g |\operatorname{Fix}(g)|$ , since $\chi(g)$ is the fixed-point count. On the other hand $m_1$ is the dimension of the subspace of $G$ -invariant vectors, which has the orbit sums as a basis. So the average number of fixed points is the number of orbits: Burnside's counting lemma, recovered from character theory (Group actions).

Fourier analysis is the abelian case. For an abelian $G$ every irreducible representation is one-dimensional (Schur's lemma), so characters are the representations, and part 6 says there are $|G|$ of them, forming an orthonormal basis of all functions on $G$ . Expanding a function in that basis is the discrete Fourier transform, and part 3 is Parseval's relation for it. Replacing the finite group by the circle and the sum by an integral turns the same statements into the classical theory of Fourier series and Parseval's theorem for Fourier series; Dummit and Foote set the analogy out in detail (Dummit & Foote (2004), §18.3, pp. 875-876). Non-abelian character theory is what replaces Fourier analysis when the symmetry group does not commute.

Method

Character tables

Adapted

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Shows how the square array of irreducible characters of a finite group is pinned down by orthogonality of its rows and of its columns together with a few arithmetic constraints, and how, once built, it reduces any representation to a list of multiplicities.

Prerequisites: Orthogonality relations and characters, Conjugacy classes, Representations of groups, Group actions

Definition(The character table)AdaptedAdapted from Dummit & Foote (2004), §19.1, p. 880; Hassani (2013), Table 24.1 and preceding text, p. 743; Arfken, Weber & Harris (2013), Table 17.3 and surrounding text, pp. 831-832

Let $G$ be a finite group of order $|G|$ , with conjugacy classes $K_1, \dots, K_r$ of sizes $c_1, \dots, c_r$ , numbered so that $K_1 = \{1\}$ and $c_1 = 1$ . A character is constant on each class (Orthogonality relations and characters), so an irreducible character $\chi$ is determined by the $r$ numbers

\chi_i \;:=\; \chi(g_i) , \qquad g_i \in K_i \ \text{ any representative} .

By Orthogonality relations and characters the number of inequivalent irreducible representations is also $r$ . The character table of $G$ is the resulting $r \times r$ array: one row for each irreducible character $\chi^{(1)}, \dots, \chi^{(r)}$ , one column for each class, and the entry in row $\mu$ and column $i$ equal to $\chi^{(\mu)}_i$ . It is square, and it is this squareness that makes a table of it rather than a list.

Four conventions are almost universal and are used throughout the handbook.

The first row is the trivial (principal) character, $\chi^{(1)}_i = 1$ for every $i$ ; the first column is the class of the identity. The remaining rows are ordered by increasing degree.
The identity column gives the degrees. $\chi^{(\mu)}(1) = \operatorname{tr} I_{n_\mu} = n_\mu$ , so reading down the first column reads off the dimensions of the irreducible representations.
Each column heading carries the size of its class, written as a prefix ( $2C_3$ , $3\sigma_v$ ) in the physics and chemistry literature and as a separate row of sizes $c_i$ in the algebra literature. Without it the table cannot be used, because every inner product below is weighted by the $c_i$ .
Rows are named, not numbered. A one-dimensional irrep is called $A$ if the principal rotation has character $+1$ and $B$ if it has character $-1$ ; a two-dimensional irrep is called $E$ and a three-dimensional one $T$ .

The inner product of two class functions $\theta, \psi : G \to \mathbb{C}$ is

(\theta, \psi) \;=\; \frac{1}{|G|}\sum_{g \in G} \theta(g)\,\overline{\psi(g)} \;=\; \frac{1}{|G|}\sum_{i=1}^{r} c_i\,\theta_i\,\overline{\psi_i} ,

linear in the first slot, as everywhere in this handbook. The second form is the one a table is read with: the sum runs over the $r$ columns, each weighted by its class size.

Method(The constraints that determine a table)AdaptedAdapted from Riley, Hobson & Bence (2006), §29.8, properties (i)-(vi), p. 1100; Hassani (2013), Eq. (24.27), p. 742, and Theorem 24.5.3, p. 743; Dummit & Foote (2004), Corollary 11(2), p. 861, Corollary 5, p. 888, and §19.1, p. 880; Arfken, Weber & Harris (2013), Eq. (17.10), p. 831

Let $G$ be a finite group with the notation of the definition, and let $G' = [G,G]$ be its commutator subgroup. The following hold.

(C1) Shape. The table is $r \times r$ , where $r$ is the number of conjugacy classes.
(C2) Rows are orthonormal. $(\chi^{(\mu)}, \chi^{(\nu)}) = \delta_{\mu\nu}$ , that is $\frac{1}{|G|}\sum_{i=1}^{r} c_i\,\chi^{(\mu)}_i\,\overline{\chi^{(\nu)}_i} \;=\; \delta_{\mu\nu} .$
(C3) Degrees. $\sum_{\mu=1}^{r} n_\mu^{2} = |G|$ with $n_\mu = \chi^{(\mu)}(1) \ge 1$ , and each $n_\mu$ divides $|G|$ (Dummit and Foote, Corollary 5, p. 888). The number of $\mu$ with $n_\mu = 1$ is exactly the index $|G : G'|$ ; in particular every group has at least one, the trivial character, and $G$ is abelian precisely when every $n_\mu = 1$ .
(C4) Columns are orthogonal. For all $i, j$ , $\sum_{\mu=1}^{r} \chi^{(\mu)}_i\,\overline{\chi^{(\mu)}_j} \;=\; \frac{|G|}{c_i}\,\delta_{ij} .$
(C5) Entries are sums of roots of unity. If $g \in K_i$ has order $m$ , then $\chi^{(\mu)}_i$ is a sum of $n_\mu$ complex $m$ th roots of unity. Consequently $\bigl|\chi^{(\mu)}_i\bigr| \;\le\; n_\mu , \qquad \chi^{(\mu)}\bigl(g^{-1}\bigr) \;=\; \overline{\chi^{(\mu)}(g)} ,$ and $\chi^{(\mu)}_i$ is real whenever $K_i$ is closed under inversion. The order $m$ divides $|G|$ , by Lagrange's theorem (Cosets and Lagrange's theorem).
(C6) Reading off multiplicities. If $D$ is any representation of $G$ with character $\psi$ , then $D \simeq \bigoplus_\mu m_\mu\, D^{(\mu)}$ with $m_\mu \;=\; \bigl(\psi, \chi^{(\mu)}\bigr) \;=\; \frac{1}{|G|}\sum_{i=1}^{r} c_i\,\psi_i\,\overline{\chi^{(\mu)}_i} \;\in\; \{0, 1, 2, \dots\} ,$ and $D$ is irreducible if and only if $(\psi, \psi) = 1$ .

Every one of these is established in Orthogonality relations and characters; they are collected here because the method below is nothing but their repeated use. Specifically, that unit proves the count of irreducible representations (C1), the first orthogonality relation (C2), the decomposition formula (C6) and the degree sum in (C3) — Arfken's dimensionality theorem — and it states the column relation (C4), which is Hassani's Eq. (24.27) and the "orthogonality of columns" of Dummit and Foote, and the roots-of-unity form in (C5), which is Riley, Hobson and Bence's property (vi). The divisibility in (C3) is Dummit and Foote's Corollary 5, the count of one-dimensional characters their Corollary 11(2), and the abelian criterion Hassani's Theorem 24.5.3.

The proof below supplies the two things a table-builder cannot do without and that are left open there: a proof of the column relation (C4), which Orthogonality relations and characters states but takes from its sources, and the two consequences of (C5) — the bound $|\chi^{(\mu)}_i| \le n_\mu$ and the reality criterion — that cut the search for unknown entries down to a finite list.

Idea of the proofOriginal

Assemble the table into a matrix $U$ with entries $U_{\mu i} = \sqrt{c_i/|G|}\;\chi^{(\mu)}_i$ , that is, the table with each column scaled by the square root of its class size. Row orthonormality (C2) says exactly that $UU^{\dagger} = I$ . A square matrix with orthonormal rows is unitary, so $U^{\dagger}U = I$ as well — and that is the column relation (C4). Nothing is needed beyond (C1) and (C2): squareness is what converts orthogonality of rows into orthogonality of columns, and the number of classes equalling the number of irreps is therefore the hinge of the whole construction.

The second statement, (C5), comes from an even shorter observation. An element of order $m$ is sent by a representation to a matrix $D$ with $D^{m} = I$ , which forces every eigenvalue of $D$ to satisfy $\lambda^{m} = 1$ . The character is the sum of those eigenvalues, so it is a sum of $n_\mu$ roots of unity; the triangle inequality then bounds it by $n_\mu$ , and replacing $g$ by $g^{-1}$ inverts each eigenvalue, which for a root of unity is the same as conjugating it.

ProofOriginal

(C4), columns. Orthogonality relations and characters states this relation and takes it from its sources; here is a proof from (C1) and (C2) alone. Define the $r \times r$ matrix

U_{\mu i} \;=\; \sqrt{\frac{c_i}{|G|}}\;\chi^{(\mu)}_i , \qquad \mu, i \in \{1, \dots, r\} ,

which is square by (C1).

Step 1 (rows). By (C2),

\bigl(UU^{\dagger}\bigr)_{\mu\nu} \;=\; \sum_{i=1}^{r} U_{\mu i}\,\overline{U_{\nu i}} \;=\; \frac{1}{|G|}\sum_{i=1}^{r} c_i\,\chi^{(\mu)}_i\,\overline{\chi^{(\nu)}_i} \;=\; \delta_{\mu\nu} ,

so $UU^{\dagger} = I$ .

Step 2 (squareness). For square matrices a right inverse is a left inverse: $UU^{\dagger} = I$ gives $\det U \ne 0$ , hence $U^{-1}$ exists, and multiplying $UU^{\dagger} = I$ on the left by $U^{-1}$ gives $U^{\dagger} = U^{-1}$ , so also $U^{\dagger}U = I$ . This step, and only this step, uses (C1).

Step 3 (read off). Writing out $\bigl(U^{\dagger}U\bigr)_{ij} = \delta_{ij}$ ,

\delta_{ij} \;=\; \sum_{\mu=1}^{r} \overline{U_{\mu i}}\,U_{\mu j} \;=\; \frac{\sqrt{c_ic_j}}{|G|}\sum_{\mu=1}^{r} \overline{\chi^{(\mu)}_i}\,\chi^{(\mu)}_j .

For $i = j$ this is $\sum_\mu |\chi^{(\mu)}_i|^{2} = |G|/c_i$ , and for $i \ne j$ it is $\sum_\mu \overline{\chi^{(\mu)}_i}\chi^{(\mu)}_j = 0$ . Taking complex conjugates gives the form stated in (C4).

(C5), the consequences of the roots-of-unity form. Let $g \in K_i$ have order $m$ and let $D = D^{(\mu)}(g)$ , an $n_\mu \times n_\mu$ matrix. That $\chi^{(\mu)}_i$ is a sum of $n_\mu$ complex $m$ th roots of unity is proved in Orthogonality relations and characters; the one-line reason is that $D^{m} = D^{(\mu)}(g^{m}) = I$ , so the polynomial $t^{m}-1$ , which has $m$ simple roots, annihilates $D$ and makes it diagonalisable with every eigenvalue a root of $t^{m} = 1$ (Eigenvalues and eigenvectors). Write $\lambda_1, \dots, \lambda_{n_\mu}$ for those eigenvalues, so that $\chi^{(\mu)}_i = \operatorname{tr} D = \sum_k \lambda_k$ . What the method below actually uses are the two consequences.

The bound. Each $|\lambda_k| = 1$ , so the triangle inequality gives

\bigl|\chi^{(\mu)}_i\bigr| \;\le\; \sum_{k=1}^{n_\mu} |\lambda_k| \;=\; n_\mu ,

with equality only when all $n_\mu$ eigenvalues coincide, that is when $D$ is a scalar matrix.

The reality criterion. $D^{(\mu)}(g^{-1}) = D^{-1}$ has eigenvalues $\lambda_k^{-1} = \overline{\lambda_k}$ , again because $|\lambda_k| = 1$ ; summing, $\chi^{(\mu)}(g^{-1}) = \overline{\chi^{(\mu)}(g)}$ . If the class $K_i$ contains $g^{-1}$ along with $g$ , the left-hand side is $\chi^{(\mu)}_i$ itself, so $\chi^{(\mu)}_i = \overline{\chi^{(\mu)}_i}$ and the whole column is real. $\square$

Remark(Where the hypotheses are used, and what the table does not tell you)Original

Finiteness, twice. It enters through Maschke's theorem, which needs $|G|$ to be invertible in the field and gives complete reducibility (Representations of groups), and again in (C5), where an element of a finite group has finite order. Over a field of characteristic $p$ dividing $|G|$ both fail: the representation of $\mathbb{Z}/p$ on $\mathbb{F}_p^{2}$ sending the generator to $\left(\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right)$ has no invariant complement to the span of the first basis vector, so there is no decomposition for a character to record, and modular character theory is a different subject.
$\mathbb{C}$ , not $\mathbb{R}$ . (C1) is false over $\mathbb{R}$ . The cyclic group $C_3 = \langle a \rangle$ has three classes, but only two irreducible real representations: the trivial one and the two-dimensional rotation by $2\pi/3$ , which is irreducible over $\mathbb{R}$ because a real $2\times 2$ rotation by an angle that is not a multiple of $\pi$ has no real eigenvector. The complex table is $3 \times 3$ with entries $1, \omega, \omega^{2}$ ; the real array is not square, Step 2 of the proof of (C4) collapses, and the columns are not orthogonal. Everything in this unit is over $\mathbb{C}$ .
Squareness is exactly Step 2. If one knew only that rows are orthonormal, the characters would span an $r'$ -dimensional subspace of the $r$ -dimensional space of class functions with $r' \le r$ , and the columns of the resulting $r' \times r$ array would satisfy no relation. The theorem that $r' = r$ — equivalently, that the irreducible characters are a basis of the class functions and not merely an orthonormal set — is what (C4) is built on, and it is proved in Orthogonality relations and characters.
(C5) constrains but does not determine. Knowing that an entry is a sum of $n_\mu$ $m$ th roots of unity cuts the search to a finite set, which is why it is useful by hand, but it alone never finishes a table. In the worked example below it is used only to know that the one-dimensional characters of an element of order $2$ are $\pm 1$ .
A table is not a group. The dihedral group of order $8$ and the quaternion group of order $8$ have identical character tables, as Dummit and Foote note on p. 882, yet they are not isomorphic. So no construction in this unit can be inverted: the table records how $G$ acts on complex vector spaces, which is strictly less information than $G$ . What the table does determine includes the degrees, the class sizes, which classes are inverse to which, the order of $G$ , the normal subgroups (as intersections of kernels $\{g : \chi^{(\mu)}(g) = n_\mu\}$ ), and hence whether $G$ is simple or soluble.
Reality is not automatic. (C5) gives real entries only for classes closed under inversion. The table of $C_3$ above has entries $\omega = e^{2\pi i/3}$ , and the two non-trivial characters are complex conjugates of one another, as they must be since $\overline{\chi}$ is again an irreducible character.

Note(Published tables, and the notation you will meet)AdaptedAdapted from Riley, Hobson & Bence (2006), §29.9, pp. 1102-1103; Arfken, Weber & Harris (2013), §17.4, p. 832

Printed tables for the crystallographic point groups are the working form of this material, and their notation is not standardised. Three warnings are worth carrying.

$E$ means two things. It labels a two-dimensional irrep (row heading) and, in chemistry notation, also the identity operation (column heading), so the same symbol can appear in both margins of one table with unrelated meanings. Riley, Hobson and Bence write $I$ for the identity to avoid this, and so does this handbook.
Classes are named differently by field. Crystallographers write $n_z$ for a rotation by $2\pi/n$ about $z$ , $\bar n_z$ for the same followed by inversion, and $m_x$ for a mirror in the plane $x = 0$ . Chemists write $C_n$ , $S_n$ and $\sigma$ , with a prefix giving the number of elements in the class, and subscripts $v$ , $d$ , $h$ distinguishing the orientation of mirror planes.
Groups are named differently too. The symmetries of a square in the plane are $4mm$ to a crystallographer and $C_{4v}$ to a chemist. The same abstract group of order $8$ is $D_4$ to a physicist (the dihedral group of the square) and $D_8$ to an algebraist, who indexes by the order of the group rather than by the order of the rotation. The example below uses the physicist's convention, $D_6$ for the group of order $12$ ; Dummit and Foote would call it $D_{12}$ .

A published table often carries extra right-hand columns listing functions — $z$ , $(x,y)$ , $x^{2}-y^{2}$ , $R_z$ — against the irrep they span. Those columns are not part of the table in the sense of this unit; they are a precomputed answer to the question the next unit asks, and they are what makes selection rules quick to apply (Product representations and selection rules).

Method(Constructing a character table by hand)AdaptedAdapted from Riley, Hobson & Bence (2006), §29.8, pp. 1100-1102

Given a finite group $G$ , presented by generators and relations or as a concrete symmetry group:

Find the classes. Compute the conjugacy classes $K_1, \dots, K_r$ and their sizes $c_i$ (Conjugacy classes); check $\sum_i c_i = |G|$ . The table will be $r \times r$ . If $G$ is abelian, stop and go to step 3: there are $|G|$ classes and $|G|$ characters, all of degree one.
Find the degrees. The number of degree-one characters is $|G : G'|$ , with $G' = [G,G]$ the commutator subgroup; call it $d$ . The remaining $r - d$ degrees $n_\mu \ge 2$ satisfy $\sum n_\mu^{2} = |G| - d$ and each $n_\mu$ divides $|G|$ , which in small cases leaves one possibility. (Riley, Hobson and Bence reach the same place without computing $G'$ , by testing which lists of degrees are compatible with $\sum_\mu n_\mu^{2} = |G|$ , with $r$ , and with the presence of the trivial character.)
Write down the degree-one characters. They are the characters of the abelian quotient $G/G'$ , pulled back to $G$ : choose a decomposition of $G/G'$ into cyclic factors and send a generator of a factor of order $k$ to a $k$ th root of unity, in all $d$ possible ways. Each is constant on classes automatically, since $G/G'$ is abelian.
Fill the rest with column orthogonality. Write unknowns for the remaining rows. Column $i$ against column $1$ gives one linear equation; column $i$ against itself gives $\sum_\mu |\chi^{(\mu)}_i|^{2} = |G|/c_i$ , a quadratic one. The already-known rows contribute known numbers to both. Use (C5) to restrict the unknowns to sums of roots of unity before solving.
If a row is still undetermined, use row orthogonality. The last row of a table is always forced: it must be orthogonal to the $r-1$ rows above it in the weighted inner product, and normalised. (Equivalently, subtract the known rows from the regular character $\chi^{\mathrm{reg}}_i = |G|\,\delta_{i1}$ weighted by degrees.)
Fix the labels. Assign names $A$ , $B$ , $E$ , $T$ by degree and by the character of the principal rotation, and record the class sizes in the column headings. Permuting rows or columns gives the same table.
Check before using. Verify every row against every other row with (C2), every column with (C4), and $\sum_\mu n_\mu^{2} = |G|$ . Together with the degree sum these are a few dozen arithmetic identities, heavily redundant, and they catch essentially every slip.

To use a finished table on a representation $D$ whose character $\psi$ you can compute:

Get $\psi$ class by class. For a permutation or natural representation, no matrix is needed: $\psi(g)$ is the sum of the diagonal entries, so only basis vectors that $g$ maps into a multiple of themselves contribute. For a permutation action, $\psi(g)$ is the number of fixed points.
Apply (C6) to get each $m_\mu = (\psi, \chi^{(\mu)})$ , and check $\sum_\mu m_\mu n_\mu = \dim D$ and $(\psi,\psi) = \sum_\mu m_\mu^{2}$ . If any $m_\mu$ fails to be a non-negative integer, the character $\psi$ was computed wrongly.

Example(The hexagon group: building the table and using it)Original

Let $G = D_6$ be the symmetry group of a regular hexagon,

D_6 \;=\; \bigl\langle\, r, s \;\bigm|\; r^{6} = s^{2} = 1, \ srs^{-1} = r^{-1} \,\bigr\rangle , \qquad |G| = 12 ,

with $r$ the rotation by $\pi/3$ about the centre and $s$ the reflection fixing a pair of opposite vertices. Its twelve elements are $r^{k}$ and $sr^{k}$ for $k = 0, \dots, 5$ . (An algebraist writes this group $D_{12}$ .)

Step 1: classes. From $srs^{-1} = r^{-1}$ we get $r^{j}\,r^{k}r^{-j} = r^{k}$ and $s\,r^{k}s^{-1} = r^{-k}$ , so the rotations fall into classes $\{1\}$ , $\{r, r^{5}\}$ , $\{r^{2}, r^{4}\}$ , $\{r^{3}\}$ . For the reflections, $r^{j}(sr^{k})r^{-j} = sr^{k-2j}$ and $s(sr^{k})s^{-1} = sr^{-k}$ , so $k$ may be shifted by any even number and negated: the reflections split by the parity of $k$ into $\{s, sr^{2}, sr^{4}\}$ and $\{sr, sr^{3}, sr^{5}\}$ . Geometrically the first three are the reflections through opposite vertices and the last three those through opposite edge midpoints. Thus $r = 6$ classes with sizes

(c_1, \dots, c_6) \;=\; (1,\,1,\,2,\,2,\,3,\,3) , \qquad \textstyle\sum_i c_i = 12 \ \checkmark

in the order $K_1 = \{1\}$ , $K_2 = \{r^{3}\}$ , $K_3 = \{r, r^{5}\}$ , $K_4 = \{r^{2}, r^{4}\}$ , $K_5 = \{s, sr^{2}, sr^{4}\}$ , $K_6 = \{sr, sr^{3}, sr^{5}\}$ . The table is $6 \times 6$ .

Step 2: degrees. The commutator $r^{-1}s^{-1}rs = r^{-1}\cdot r^{-1} = r^{-2}$ lies in $G'$ , so $G' \supseteq \langle r^{2}\rangle = \{1, r^{2}, r^{4}\}$ ; and $G/\langle r^{2}\rangle$ has order $4$ and is generated by two commuting involutions, hence abelian, so $G' = \langle r^{2}\rangle$ exactly and $|G : G'| = 4$ . There are therefore four degree-one characters, and the other two degrees satisfy $n^{2} + n'^{2} = 12 - 4 = 8$ with $n, n' \ge 2$ : the only solution is $n = n' = 2$ . Degrees $(1,1,1,1,2,2)$ , and indeed $4 \cdot 1 + 2 \cdot 4 = 12 \ \checkmark$ .

Step 3: the four one-dimensional rows. They are the characters of $G/G' \cong \mathbb{Z}_2 \times \mathbb{Z}_2$ , generated by the images of $r$ and $s$ , so $\chi(r) = \varepsilon = \pm 1$ and $\chi(s) = \eta = \pm 1$ independently. Then $\chi(r^{3}) = \varepsilon^{3} = \varepsilon$ , $\chi(r^{2}) = 1$ , $\chi(sr^{k}) = \eta\varepsilon^{k}$ . Naming by the convention of the note ( $A$ if $\varepsilon = +1$ , $B$ if $\varepsilon = -1$ ):

\begin{array}{l|rrrrrr} & K_1 & K_2 & K_3 & K_4 & K_5 & K_6 \\ \hline A_1 & 1 & 1 & 1 & 1 & 1 & 1\\ A_2 & 1 & 1 & 1 & 1 & -1 & -1\\ B_1 & 1 & -1 & -1 & 1 & 1 & -1\\ B_2 & 1 & -1 & -1 & 1 & -1 & 1 \end{array}

Step 4: the two two-dimensional rows, from columns alone. Write the unknown rows as $E_1 = (2, a, b, c, d, e)$ and $E_2 = (2, a', b', c', d', e')$ . All entries are real by (C5), because every class here is closed under inversion ( $r^{-1} = r^{5}$ lies in $K_3$ , and every reflection is its own inverse). Apply (C4) to the pairs of columns $(K_1, K_j)$ , using the four known rows:

\underbrace{1 + 1 - 1 - 1}_{\text{from } A_1,A_2,B_1,B_2} + 2a + 2a' = 0 \;\Rightarrow\; a' = -a ,

and in the same way $b' = -b$ , $d' = -d$ , $e' = -e$ , while column $K_4$ gives $1+1+1+1 + 2c + 2c' = 0$ , so $c + c' = -2$ . Now take each column against itself, where (C4) gives $|G|/c_i$ :

\begin{aligned} K_2 \ (c_2 = 1): &\quad 4 + a^{2} + a'^{2} = 12 \;\Rightarrow\; 2a^{2} = 8 \;\Rightarrow\; a = \pm 2 ,\\ K_3 \ (c_3 = 2): &\quad 4 + b^{2} + b'^{2} = 6 \;\Rightarrow\; b = \pm 1 ,\\ K_4 \ (c_4 = 2): &\quad 4 + c^{2} + c'^{2} = 6 \ \text{ with } c + c' = -2 \;\Rightarrow\; c = c' = -1 ,\\ K_5 \ (c_5 = 3): &\quad 4 + d^{2} + d'^{2} = 4 \;\Rightarrow\; d = d' = 0 , \quad\text{and likewise } e = e' = 0 . \end{aligned}

One sign is still free in each of $a$ and $b$ ; the columns $K_2$ and $K_3$ against each other tie them together:

\underbrace{1 \cdot 1 + 1 \cdot 1 + (-1)(-1) + (-1)(-1)}_{A_1,A_2,B_1,B_2} + ab + a'b' \;=\; 4 + 2ab \;=\; 0 \;\Longrightarrow\; ab = -2 .

So $(a,b) = (-2, 1)$ or $(2,-1)$ , the two choices differing only by which row is called $E_1$ . Taking the first,

\boxed{ \begin{array}{l|rrrrrr} D_6 & \{1\} & \{r^{3}\} & 2\{r\} & 2\{r^{2}\} & 3\{s\} & 3\{sr\} \\ \hline A_1 & 1 & 1 & 1 & 1 & 1 & 1\\ A_2 & 1 & 1 & 1 & 1 & -1 & -1\\ B_1 & 1 & -1 & -1 & 1 & 1 & -1\\ B_2 & 1 & -1 & -1 & 1 & -1 & 1\\ E_1 & 2 & -2 & 1 & -1 & 0 & 0\\ E_2 & 2 & 2 & -1 & -1 & 0 & 0 \end{array}}

Step 5: identify the two-dimensional rows geometrically (a check, not part of the construction). $E_1$ is the defining action on the plane: $r$ is the rotation by $\pi/3$ , of trace $2\cos(\pi/3) = 1 = \chi_{E_1}(K_3)$ , $r^{3}$ is the rotation by $\pi$ , of trace $-2$ , $r^{2}$ has trace $2\cos(2\pi/3) = -1$ , and every reflection has trace $0$ . $E_2$ sends $r$ to the rotation by $2\pi/3$ : this is a homomorphism because $(2\pi/3)\cdot 6 = 4\pi$ , and its kernel is $\{1, r^{3}\}$ , so $E_2$ is the two-dimensional irrep of $D_6/\langle r^{3}\rangle \cong D_3$ pulled back along the quotient map (Homomorphisms and quotient groups). Its characters are $2\cos(2\pi/3) = -1$ on $K_3$ , $2\cos(4\pi/3) = -1$ on $K_4$ and $\chi(r^{3}) = \operatorname{tr} I = 2$ , as the table says.

Step 6: the checks of step 7 of the method. Row $E_1$ against itself: $\tfrac{1}{12}\bigl[1\cdot 4 + 1 \cdot 4 + 2 \cdot 1 + 2 \cdot 1 + 0 + 0\bigr] = \tfrac{12}{12} = 1$ . Rows $E_1$ against $E_2$ : $\tfrac{1}{12}[1\cdot 4 + 1\cdot(-4) + 2\cdot(-1) + 2\cdot 1] = 0$ . Rows $B_1$ against $E_1$ : $\tfrac{1}{12}[2 + 2 - 2 - 2 + 0 + 0] = 0$ . Column $K_5$ against itself: $1 + 1 + 1 + 1 + 0 + 0 = 4 = 12/3$ . Degrees: $1+1+1+1+4+4 = 12$ . Every row pair, every column pair and the degree sum check out; the five displayed are representative.

Step 7: using the table. Let $D$ be the six-dimensional permutation representation of $D_6$ on the vertices of the hexagon, so $\psi(g)$ is the number of vertices $g$ fixes. Label the vertices $0, \dots, 5$ so that $r$ sends vertex $k$ to vertex $k+1 \pmod 6$ and $s$ sends $k$ to $-k$ . Then: $\psi(1) = 6$ ; a rotation by $\pi/3$ , $2\pi/3$ or $\pi$ moves every vertex, so $\psi = 0$ on $K_2, K_3, K_4$ ; the reflection $sr^{2m}$ fixes $k$ when $2k \equiv -2m$ , which has two solutions, so $\psi = 2$ on $K_5$ ; the reflection $sr^{2m+1}$ fixes $k$ when $2k \equiv -2m-1 \pmod 6$ , which has none, so $\psi = 0$ on $K_6$ . Hence

\psi \;=\; (6,\,0,\,0,\,0,\,2,\,0) .

Now (C6), with the weights $c_i$ :

\begin{aligned} m_{A_1} &= \tfrac{1}{12}\bigl[1\cdot 1\cdot 6 + 3\cdot 1 \cdot 2\bigr] = \tfrac{12}{12} = 1 , & m_{A_2} &= \tfrac{1}{12}\bigl[6 + 3\cdot(-1)\cdot 2\bigr] = 0 ,\\ m_{B_1} &= \tfrac{1}{12}\bigl[6 + 3\cdot 1 \cdot 2\bigr] = 1 , & m_{B_2} &= \tfrac{1}{12}\bigl[6 + 3\cdot(-1)\cdot 2\bigr] = 0 ,\\ m_{E_1} &= \tfrac{1}{12}\bigl[2 \cdot 6 + 3 \cdot 0 \cdot 2\bigr] = 1 , & m_{E_2} &= \tfrac{1}{12}\bigl[2\cdot 6 + 3\cdot 0\cdot 2\bigr] = 1 , \end{aligned}

(only $K_1$ and $K_5$ contribute, since $\psi$ vanishes elsewhere), so

D \;\simeq\; A_1 \oplus B_1 \oplus E_1 \oplus E_2 , \qquad 1 + 1 + 2 + 2 = 6 = \dim D \ \checkmark .

Step 8: the checkable number. The consistency test $(\psi,\psi) = \sum_\mu m_\mu^{2}$ reads

(\psi, \psi) \;=\; \frac{1}{12}\Bigl[1 \cdot 6^{2} + 3 \cdot 2^{2}\Bigr] \;=\; \frac{36 + 12}{12} \;=\; 4 \;=\; 1^{2} + 1^{2} + 1^{2} + 1^{2} .

This number can be obtained a second way, without any character theory. Since $\psi(g)^{2}$ counts the ordered pairs of vertices fixed by $g$ , Burnside's counting lemma (Group actions) says $(\psi,\psi) = \frac{1}{|G|}\sum_g \psi(g)^{2}$ is the number of orbits of $D_6$ on ordered pairs of vertices. A pair $(j, k)$ is carried by rotations to any pair with the same difference $k - j \bmod 6$ , and by reflections to one with the opposite difference, so the orbits are labelled by $\{0\}, \{1,5\}, \{2,4\}, \{3\}$ — the separations $0, 1, 2, 3$ around the hexagon. There are

4

orbits, matching the character computation exactly.

Application(Degeneracies, spectra and the hexagon that is benzene)Original

Normal modes. Two normal modes of a vibrating system that some symmetry operation carries into one another must be degenerate in frequency; each such degenerate family spans an irreducible representation of the symmetry group, and its multiplicity of states is that representation's dimension (Riley, Hobson & Bence (2006), §29.11.3, pp. 1110-1111). The recipe is exactly steps 8-9 above: write the $3N$ -dimensional representation on the Cartesian displacements of $N$ atoms, compute its character class by class (only atoms left in place by a symmetry operation contribute), decompose, and then subtract the three translations and three rotations, which span known irreps. What is left is the list of vibrational symmetry species with their degeneracies — obtained without ever writing down a force constant, because the character table knows the symmetry and nothing else. The eigenvalue problem itself is Generalized eigenproblems and normal modes.

Splitting of degenerate levels. A free atom has $(2\ell+1)$ -fold degenerate levels, carrying the irreducible representations of the full rotation group (The rotation group and SU(2)). Put the atom in a crystal and only a finite point group $G$ survives; the $(2\ell+1)$ -dimensional space is still a representation of $G$ , but usually a reducible one, and the multiplicities computed by (C6) say into how many levels the degeneracy splits and what the new degeneracies are. Which of the resulting levels can be reached from which by a given perturbation is the question of Product representations and selection rules.

Benzene, from the example. Take the planar group $C_{6v}$ — the same abstract group as the $D_6$ above, with the six vertical mirror planes in the role of the reflections — and the six carbon $2p_z$ orbitals of a benzene ring as basis. These orbitals are unchanged in form by every operation of $C_{6v}$ , so a symmetry operation merely permutes them, and the representation they carry is precisely the permutation representation $D$ decomposed in the example:

\Gamma_\pi \;=\; A_1 \oplus B_1 \oplus E_1 \oplus E_2 , \qquad \text{degeneracies } 1, 1, 2, 2 .

Symmetry therefore predicts, before any Hamiltonian is chosen, that the six $\pi$ levels come in a pattern $1, 2, 2, 1$ . The Hückel model, in which the matrix is the adjacency matrix of the hexagon scaled by a resonance integral $\beta < 0$ and shifted by $\alpha$ , gives energies $\alpha + 2\beta\cos(2\pi k/6)$ for $k = 0, \dots, 5$ , that is

\alpha + 2\beta, \qquad \alpha + \beta \ (\text{twice}), \qquad \alpha - \beta \ (\text{twice}), \qquad \alpha - 2\beta ,

degeneracies $1, 2, 2, 1$ — the pattern the table predicted. The all-in-phase combination is the $A_1$ row and the alternating one $(1,-1,1,-1,1,-1)$ is $B_1$ : it changes sign under the rotation $r$ , as the $-1$ in the $B_1$ row at $K_3$ requires, and is unchanged by a mirror through two opposite atoms, as the $+1$ at $K_5$ requires. Six $\pi$ electrons fill the $A_1$ and $E_1$ levels, leaving the degenerate $E_2$ pair empty as the lowest unoccupied orbitals: the aromatic stability of benzene and the degeneracy of its frontier orbitals are both consequences of the one $6 \times 6$ array built above.

Where else tables are indispensable. Crystal and molecular spectroscopy (which lines are allowed, and how many), the classification of electronic bands at symmetry points of a Brillouin zone, the construction of symmetry-adapted basis functions, and the counting of independent components of a material tensor allowed by a crystal class (Point groups and crystal symmetry) all start from a character table and the two inner products of this unit.

Method

Product representations and selection rules

Adapted

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Builds the tensor product of two representations of a finite group, shows that its character is the pointwise product of the two characters, decomposes it into irreducibles by the Clebsch-Gordan series, and turns the resulting multiplicity of the trivial representation into a rule that forbids matrix elements.

Prerequisites: Orthogonality relations and characters, Character tables, Representations of groups, Linear maps and matrices

Definition(Kronecker product, and the product representation)AdaptedAdapted from Arfken, Weber & Harris (2013), §2.2, Eq. (2.55)-(2.57), pp. 108-109; Hassani (2013), §24.8, p. 751

Of matrices. For $A$ of size $m \times n$ and $B$ of size $m' \times n'$ , the Kronecker product (also called the direct or direct tensor product) is the $mm' \times nn'$ matrix $A \otimes B$ with entries

(A \otimes B)_{\alpha\beta} \;=\; A_{ij}\,B_{kl} , \qquad \alpha = m'(i-1) + k , \quad \beta = n'(j-1) + l .

The index of the first factor is the major one and the index of the second the minor one, so $A \otimes B$ and $B \otimes A$ differ by a permutation of rows and of columns and the operation is not commutative; it is associative, and it satisfies the two rules used throughout below,

(A \otimes B)(A' \otimes B') \;=\; (AA') \otimes (BB') , \qquad C \otimes (A + B) \;=\; C\otimes A + C\otimes B ,

whenever the products on the right are defined. It is cleaner to drop Arfken's flattening $\alpha = m'(i-1)+k$ and simply label rows and columns by pairs, writing $(A\otimes B)_{(i,k),(j,l)} = A_{ij}B_{kl}$ ; the flattening is then only a choice of how to print the matrix. This unit uses the pair labelling.

Of representations. Let $T : G \to GL(V)$ and $S : G \to GL(W)$ be representations of the same finite group $G$ on complex spaces of dimensions $m$ and $n$ . On the tensor product $V \otimes W$ define

(T \otimes S)(g)\,(v \otimes w) \;=\; \bigl(T(g)v\bigr) \otimes \bigl(S(g)w\bigr) ,

extended linearly. In bases $\{v_i\}$ of $V$ and $\{w_a\}$ of $W$ , the products $v_i \otimes w_a$ are a basis of $V \otimes W$ and the matrix of $(T\otimes S)(g)$ in it is the Kronecker product of the matrices of $T(g)$ and $S(g)$ :

\bigl[(T\otimes S)(g)\bigr]_{(i,a),(j,b)} \;=\; T_{ij}(g)\,S_{ab}(g) .

This is the tensor product, direct product or Kronecker product representation, of degree $mn$ . Physically $V$ and $W$ are the state spaces of two subsystems, $V \otimes W$ is the state space of the pair, and the group acts on both subsystems at once; Arfken's multiparticle basis consists of all products of one basis member from each factor.

Two things that are not this, and one that is. (i) The direct sum $T \oplus S$ acts on $V \oplus W$ with character $\chi^{T} + \chi^{S}$ ; the tensor product acts on $V \otimes W$ and has degree $mn$ , not $m+n$ . (ii) The outer product, in which $T$ is a representation of $H$ , $S$ one of $K$ , and $T \boxtimes S$ a representation of the direct product group $H \times K$ by $(h,k) \mapsto T(h)\otimes S(k)$ ; its character is $\chi^{T}(h)\chi^{S}(k)$ , a function of two arguments, and Dummit and Foote's Exercise 27 records that these exhaust the irreducible characters of $H \times K$ . The product of this unit is the inner one: both factors are representations of one group $G$ , both are evaluated at the same $g$ , and the result is usually reducible even when both factors are irreducible. (iii) The product $\chi\varphi$ of a representation $\varphi$ with a degree-one character $\chi$ , which is an instance of the inner product above, with $S = \chi$ acting on a one-dimensional space; it is worth naming separately because it behaves unusually well (see the remark).

Method(Characters multiply; the Clebsch-Gordan series; the selection rule)AdaptedAdapted from Hassani (2013), §24.8, Eq. (24.42), p. 751, and §24.8.1, Eqs. (24.44)-(24.45), pp. 753-754, with Box 24.8.3, p. 755; Riley, Hobson & Bence (2006), §29.10, Theorem and Eq. (29.23), pp. 1103-1105; Arfken, Weber & Harris (2013), §17.5, p. 838; Dummit & Foote (2004), Proposition 17, §18.3, p. 874

Let $G$ be a finite group with classes $K_1, \dots, K_r$ of sizes $c_i$ , irreducible characters $\chi^{(1)}, \dots, \chi^{(r)}$ with $\chi^{(1)} \equiv 1$ the trivial one, and the inner product $(\theta,\psi) = |G|^{-1}\sum_i c_i\,\theta_i\overline{\psi_i}$ of Character tables. Let $T$ and $S$ be representations of $G$ with characters $\chi^{T}$ and $\chi^{S}$ .

(P1) $T \otimes S$ is a representation of $G$ of degree $(\deg T)(\deg S)$ .
(P2) Characters multiply pointwise. For every $g \in G$ , $\chi^{T \otimes S}(g) \;=\; \chi^{T}(g)\,\chi^{S}(g) .$ In particular the product of two characters is again a character, which is Dummit and Foote's Proposition 17.
(P3) Clebsch-Gordan series. For irreducible $T^{(\alpha)}, T^{(\beta)}$ , $T^{(\alpha)} \otimes T^{(\beta)} \;\simeq\; \bigoplus_{\sigma=1}^{r} m^{\alpha\beta}_{\sigma}\, T^{(\sigma)} , \qquad m^{\alpha\beta}_{\sigma} \;=\; \frac{1}{|G|}\sum_{i=1}^{r} c_i\,\overline{\chi^{(\sigma)}_i}\;\chi^{(\alpha)}_i\,\chi^{(\beta)}_i ,$ with $m^{\alpha\beta}_\sigma$ non-negative integers satisfying $\sum_\sigma m^{\alpha\beta}_\sigma n_\sigma = n_\alpha n_\beta$ . Hassani calls $G$ simply reducible when every $m^{\alpha\beta}_\sigma$ is $0$ or $1$ .
(P4) When the product contains an invariant. The trivial representation occurs in $T^{(\alpha)}\otimes T^{(\beta)}$ with multiplicity $m^{\alpha\beta}_{1} \;=\; \begin{cases} 1 & \text{if } \chi^{(\beta)} = \overline{\chi^{(\alpha)}} ,\\[2pt] 0 & \text{otherwise.}\end{cases}$ That is, an irreducible representation must be combined with its conjugate (its adjoint, in Hassani's wording) to produce an invariant, and then there is exactly one.
(P5) Selection rule. Suppose a physical system occupies a region $\Omega$ carried to itself by every element of $G$ , with a measure $G$ preserves, and let $f_1, \dots, f_k$ be functions on $\Omega$ such that $f_1 \cdots f_k$ lies in a finite-dimensional space of functions carrying a representation $D$ of $G$ — for instance each $f_j$ spans part of a basis of a representation $D_j$ , in which case $D = D_1 \otimes \cdots \otimes D_k$ and, by (P2), $\chi^{D}(g) = \prod_j \chi^{D_j}(g)$ . If the trivial representation does not occur in $D$ , then $J \;=\; \int_{\Omega} f_1 f_2 \cdots f_k \,d\tau \;=\; 0 .$ The converse is false: $J$ may vanish although the trivial representation is present, so the rule forbids but never guarantees.

Idea of the proofOriginal

Everything follows from one line of index bookkeeping. The matrix of $(T\otimes S)(g)$ has rows and columns labelled by pairs, and its diagonal entries are the products $T_{ii}(g)S_{aa}(g)$ . Summing over the pair $(i,a)$ is summing over $i$ and over $a$ separately, and a double sum of a product of a function of $i$ and a function of $a$ factorises. So the trace factorises, and that is (P2).

Once characters multiply, the rest is the machinery already in place. Multiplicities of a representation are inner products of its character with the irreducible ones (Character tables), so the Clebsch-Gordan multiplicities are inner products of a triple product of characters — the formula in (P3), which is why a product decomposition costs one pass along a row of the character table rather than any matrix algebra. For (P4), pairing with the trivial character turns the triple product into an ordinary inner product of $\chi^{(\beta)}$ against $\overline{\chi^{(\alpha)}}$ , and $\overline{\chi^{(\alpha)}}$ is itself an irreducible character, so orthonormality of rows answers the question at once.

The selection rule (P5) is the same fact wearing physical clothes. The integral $J$ does not change when the integrand is transported by a symmetry, because the region and the measure do not; so $J$ equals the average of the transported integrands over the whole group. That average is precisely the projection of the integrand onto its $G$ -invariant part. If the product representation contains no invariant vector, the projection is zero and so is $J$ — which also makes clear why the converse must fail: an invariant part can exist and still integrate to zero for reasons that have nothing to do with symmetry.

ProofOriginal

Write $D = T \otimes S$ , with $T$ of degree $m$ and $S$ of degree $n$ , in bases $\{v_i\}$ and $\{w_a\}$ .

(P1). The products $v_i \otimes w_a$ , $1 \le i \le m$ , $1 \le a \le n$ , are a basis of $V\otimes W$ , so $\deg D = mn$ . For $g, h \in G$ and a basis vector,

D(g)D(h)(v\otimes w) \;=\; D(g)\bigl(T(h)v \otimes S(h)w\bigr) \;=\; T(g)T(h)v \otimes S(g)S(h)w \;=\; T(gh)v \otimes S(gh)w ,

using that $T$ and $S$ are homomorphisms, and the right-hand side is $D(gh)(v\otimes w)$ . Both sides are linear, so $D(g)D(h) = D(gh)$ on all of $V\otimes W$ . Also $D(1) = I \otimes I = I$ , and $D(g)D(g^{-1}) = D(1) = I$ shows each $D(g)$ is invertible. In matrices this is Arfken's rule $(A\otimes B)(A'\otimes B') = (AA')\otimes(BB')$ applied to $A = T(g)$ , $A' = T(h)$ .

(P2). The diagonal entries of the matrix of $D(g)$ are those with equal row and column labels, $(i,a) = (j,b)$ :

\chi^{D}(g) \;=\; \sum_{i=1}^{m}\sum_{a=1}^{n} \bigl[D(g)\bigr]_{(i,a),(i,a)} \;=\; \sum_{i=1}^{m}\sum_{a=1}^{n} T_{ii}(g)\,S_{aa}(g) \;=\; \Bigl(\sum_{i} T_{ii}(g)\Bigr)\Bigl(\sum_{a} S_{aa}(g)\Bigr) ,

the last step because the summand is a product of one factor depending only on $i$ and one depending only on $a$ . The two sums are $\chi^{T}(g)$ and $\chi^{S}(g)$ . The trace is basis-independent, so the identity does not depend on the bases chosen. Since a product of characters is the character of the representation just constructed, it is a character.

(P3). $T^{(\alpha)}\otimes T^{(\beta)}$ is a representation of a finite group over $\mathbb{C}$ , hence completely reducible (Representations of groups), so $T^{(\alpha)}\otimes T^{(\beta)} \simeq \bigoplus_\sigma m_\sigma T^{(\sigma)}$ for some non-negative integers $m_\sigma$ . By the multiplicity formula of Character tables,

m_\sigma \;=\; \bigl(\chi^{(\alpha)}\chi^{(\beta)},\, \chi^{(\sigma)}\bigr) \;=\; \frac{1}{|G|}\sum_{i=1}^{r} c_i\,\chi^{(\alpha)}_i\chi^{(\beta)}_i\,\overline{\chi^{(\sigma)}_i} ,

where (P2) was used to replace the character of the product by the product of the characters. Evaluating both sides of the decomposition at $g = 1$ gives $n_\alpha n_\beta = \sum_\sigma m_\sigma n_\sigma$ .

(P4). Put $\sigma = 1$ , so $\chi^{(1)}_i = 1$ for every $i$ and the conjugate factor disappears:

m^{\alpha\beta}_1 \;=\; \frac{1}{|G|}\sum_{i} c_i\,\chi^{(\alpha)}_i\chi^{(\beta)}_i \;=\; \frac{1}{|G|}\sum_i c_i\,\chi^{(\beta)}_i\,\overline{\overline{\chi^{(\alpha)}_i}} \;=\; \bigl(\chi^{(\beta)},\, \overline{\chi^{(\alpha)}}\bigr) .

Now $\overline{\chi^{(\alpha)}}$ is the character of the conjugate representation $\overline{T^{(\alpha)}}$ , obtained by conjugating every matrix entry; this is a representation because conjugation is a ring homomorphism of $\mathbb{C}$ , and it is irreducible because $(\overline{\chi^{(\alpha)}}, \overline{\chi^{(\alpha)}}) = \overline{(\chi^{(\alpha)}, \chi^{(\alpha)})} = 1$ . So both arguments of the inner product are irreducible characters, and orthonormality of the rows of the character table gives $1$ when they coincide and $0$ otherwise. $\;$ For a unitary representation, $\overline{\chi^{(\alpha)}(g)} = \chi^{(\alpha)}(g^{-1})$ , so the condition $\chi^{(\beta)} = \overline{\chi^{(\alpha)}}$ can be tested on the character table by comparing row $\beta$ with row $\alpha$ read with each class replaced by the class of inverses.

(P5). Let $\mathcal{F}$ be the finite-dimensional space of functions on $\Omega$ carrying $D$ , with the action $(gf)(x) = f(g^{-1}x)$ , and define

P \;=\; \frac{1}{|G|}\sum_{g \in G} D(g) ,

which is Hassani's projector onto the $\alpha$ th irreducible subspace in the case $\alpha = 1$ , where $n_1 = 1$ and $\chi^{(1)} \equiv 1$ .

Step 1 ( $P$ projects onto the invariants). For $h \in G$ , $D(h)P = |G|^{-1}\sum_g D(hg) = P$ , since $g \mapsto hg$ permutes $G$ . Hence $PF$ is $G$ -invariant for every $F \in \mathcal{F}$ , and $PF = F$ when $F$ is already invariant; so $P$ is the projection onto the subspace $\mathcal{F}^{G}$ of invariant functions.

Step 2 ( $\mathcal{F}^{G} = 0$ by hypothesis). An invariant function spans a one-dimensional subrepresentation on which every $D(g)$ acts as the identity, that is, a copy of the trivial representation. By hypothesis $D$ contains none, so $\mathcal{F}^{G} = \{0\}$ and $P = 0$ on $\mathcal{F}$ .

Step 3 (the integral is unchanged by the group). Let $F = f_1\cdots f_k \in \mathcal{F}$ . For each $g$ , substituting $x \mapsto g^{-1}x$ in the integral and using that $g^{-1}\Omega = \Omega$ and that the measure is preserved,

\int_{\Omega} (gF)(x)\,d\tau(x) \;=\; \int_{\Omega} F(g^{-1}x)\,d\tau(x) \;=\; \int_{\Omega} F(y)\,d\tau(y) \;=\; J .

Step 4 (average and conclude). Averaging Step 3 over the $|G|$ elements and using linearity of the integral,

J \;=\; \frac{1}{|G|}\sum_{g\in G} \int_\Omega (gF)\,d\tau \;=\; \int_\Omega (PF)\,d\tau \;=\; \int_\Omega 0 \,d\tau \;=\; 0 . \qquad \square

Remark(Where the hypotheses are used, and the standing traps)Original

Same group, same element. (P2) multiplies the two characters at the same $g$ . This is what makes the inner product in (P3) a sum of triple products and what makes the answer usually reducible. For the outer product over $H \times K$ the two characters are evaluated at different arguments, nothing is summed together, and the product of two irreducibles is always irreducible (Dummit and Foote, Exercise 27, p. 879). Confusing the two is the commonest error in this subject; the giveaway is the degree, $n_\alpha n_\beta$ in both cases but over a group of order $|G|$ in one and $|H||K|$ in the other.
Finiteness and $\mathbb{C}$ enter in (P3), through complete reducibility, and in (P5), through the average $|G|^{-1}\sum_g$ . Over a field whose characteristic divides $|G|$ neither is available: the sum defining $P$ cannot be normalised, and the product need not decompose. Compactness replaces finiteness for continuous groups, with $|G|^{-1}\sum_g$ becoming the Haar integral; that is how the same statements carry over to $SU(2)$ in Addition of angular momenta.
The converse of (P5) genuinely fails, as Riley, Hobson and Bence are careful to say. Symmetry can only certify that a quantity vanishes; when it does not forbid, the integral must be computed. A one-line example: $G$ trivial, $\Omega = [-1,1]$ , $f = x$ . Every representation is trivial, so the rule forbids nothing, and yet $J = 0$ . The vanishing there comes from a symmetry of the integrand that was not in the group used.
The invariant region matters as much as the group. Step 3 uses $g^{-1}\Omega = \Omega$ and invariance of $d\tau$ . For integrals over all of $\mathbb{R}^{3}$ with Lebesgue measure and $G$ a point group these hold automatically, which is why the rule is quoted without them in the physics literature; for an integral over a half-space, or with a weight that is not invariant, the conclusion is simply false.
Multiplying by a one-dimensional character preserves irreducibility. If $\deg T^{(\beta)} = 1$ then $|\chi^{(\beta)}(g)| = 1$ for every $g$ , since $\chi^{(\beta)}(g)$ is a root of unity, so $\bigl(\chi^{(\alpha)}\chi^{(\beta)},\, \chi^{(\alpha)}\chi^{(\beta)}\bigr) \;=\; \frac{1}{|G|}\sum_g \bigl|\chi^{(\alpha)}(g)\bigr|^{2}\bigl|\chi^{(\beta)}(g)\bigr|^{2} \;=\; \bigl(\chi^{(\alpha)},\chi^{(\alpha)}\bigr) \;=\; 1 ,$ so $\chi^{(\alpha)}\chi^{(\beta)}$ is irreducible. This is Dummit and Foote's Exercise 13, p. 878, and it is how half the rows of many character tables are produced: in the table of $D_6$ in Character tables, $A_2 \cdot B_1 = B_2$ , while $E_1 \otimes A_2 = E_1$ and $E_1 \otimes B_1 = E_2$ .
Degree bookkeeping is the cheapest check. $\sum_\sigma m^{\alpha\beta}_\sigma n_\sigma$ must equal $n_\alpha n_\beta$ exactly. A decomposition that fails it has a wrong character somewhere.

Note(Squaring a representation: the symmetric and antisymmetric parts)Original

When both factors are the same representation $T$ on $V$ , the product space splits further, because swapping the two tensor slots commutes with the group action: $\tau(v\otimes v') = v'\otimes v$ satisfies $\tau\,(T\otimes T)(g) = (T\otimes T)(g)\,\tau$ . Since $\tau^{2} = I$ , its eigenspaces for $\pm 1$ are invariant and

V \otimes V \;=\; \operatorname{Sym}^{2}V \;\oplus\; \Lambda^{2}V ,

spanned respectively by the $v_i\otimes v_j + v_j \otimes v_i$ and the $v_i \otimes v_j - v_j\otimes v_i$ — Hassani's Box 24.8.1, that a Kronecker square always reduces into a symmetrised and an antisymmetrised part. Their dimensions are $\tfrac12 n(n+1)$ and $\tfrac12 n(n-1)$ for $n = \deg T$ , and their characters are

\chi_{\mathrm{sym}}(g) \;=\; \tfrac12\bigl[\chi(g)^{2} + \chi(g^{2})\bigr] , \qquad \chi_{\mathrm{anti}}(g) \;=\; \tfrac12\bigl[\chi(g)^{2} - \chi(g^{2})\bigr] ,

with $\chi = \chi^{T}$ . Proof. Fix $g$ . Because $g$ has finite order, $T(g)$ is diagonalisable (see the proof of (C5) in Character tables); choose a basis $\{v_i\}$ of $V$ with $T(g)v_i = \lambda_i v_i$ . Then the vectors $v_iv_j := v_i\otimes v_j + v_j\otimes v_i$ with $i \le j$ are a basis of $\operatorname{Sym}^{2}V$ and are eigenvectors with eigenvalue $\lambda_i\lambda_j$ , so

\chi_{\mathrm{sym}}(g) \;=\; \sum_{i \le j}\lambda_i\lambda_j \;=\; \tfrac12\Bigl[\Bigl(\sum_i \lambda_i\Bigr)^{2} + \sum_i \lambda_i^{2}\Bigr] \;=\; \tfrac12\bigl[\chi(g)^{2} + \chi(g^{2})\bigr] ,

since $\sum_i\lambda_i^{2} = \operatorname{tr} T(g)^{2} = \chi(g^{2})$ . The same computation over $i < j$ gives the antisymmetric case, and the two add to $\chi(g)^{2} = \chi^{T\otimes T}(g)$ as they must. $\square$

The split is not optional bookkeeping: for two identical particles it is the whole content of Bose and Fermi statistics, and the two halves of the product decompose into different irreducibles, as the example shows.

Method(Deciding whether a matrix element can be non-zero)AdaptedAdapted from Riley, Hobson & Bence (2006), §29.10 and §29.11.2, pp. 1103-1105, 1108-1109; Hassani (2013), §24.8.1, Eq. (24.44), pp. 753-754

Given a symmetry group $G$ , a character table for it, and an integral $J = \int f_1 f_2 \cdots f_k \, d\tau$ over an invariant region:

Identify the symmetry species of each factor. Find, for each $f_j$ , the irrep (or reducible representation) $D_j$ it belongs to. Published tables list the common ones — $z$ , $(x,y)$ , $xy$ , $x^{2}-y^{2}$ , $R_z$ — in a right-hand column, which is the fastest route for Cartesian and $d$ -orbital factors. Factors invariant under the whole group, such as a purely radial $f(r)$ for a point group, carry the trivial irrep and may be dropped.
Multiply the characters class by class. Write one row per factor and one column per class, then form the product row $\chi^{D}_i = \prod_{j}\chi^{D_j}_i$ . This is (P2) extended to $k$ factors by induction. Nothing else about the factors is needed.
Project onto the trivial irrep. Compute $m_1 \;=\; \frac{1}{|G|}\sum_{i=1}^{r} c_i\,\chi^{D}_i$ (the conjugate character of the trivial irrep is $1$ , so no conjugation appears). Often inspection suffices: if the product row is itself a row of the table other than the trivial one, then $m_1 = 0$ by orthogonality.
Conclude. If $m_1 = 0$ , the integral vanishes identically, for every radial dependence and every choice of the unspecified parts of the factors — this is the content of a selection rule. If $m_1 \ge 1$ , symmetry permits it and nothing more can be said without computing.
For a two-factor overlap, steps 2-4 reduce to (P4): $\int \overline{\Psi}\,\phi\,d\tau$ can be non-zero only if $\Psi$ and $\phi$ belong to the same irrep; different irreps never overlap. This is why wavefunctions of different symmetry species do not mix and why a Hamiltonian commuting with $G$ is block-diagonal in a symmetry-adapted basis.
For a full Clebsch-Gordan decomposition rather than one multiplicity, repeat step 3 with each $\overline{\chi^{(\sigma)}}$ in place of $1$ , which is (P3). Check with $\sum_\sigma m_\sigma n_\sigma = n_\alpha n_\beta$ .
If the two factors are the same irrep and the physical situation distinguishes symmetric from antisymmetric combinations (identical particles, or a symmetric operator such as a polarizability), split first with the note above: compute $\chi(g^{2})$ by squaring each class representative and reading off which class the square lands in, then decompose $\chi_{\mathrm{sym}}$ and $\chi_{\mathrm{anti}}$ separately.

Example(Products in the hexagon group, and a dipole selection rule)Original

Work in the group $D_6$ of order $12$ whose character table was constructed in Character tables:

\begin{array}{l|rrrrrr} D_6 & K_1 = \{1\} & K_2 = \{r^{3}\} & K_3 = 2\{r\} & K_4 = 2\{r^{2}\} & K_5 = 3\{s\} & K_6 = 3\{sr\} \\ \hline A_1 & 1 & 1 & 1 & 1 & 1 & 1\\ A_2 & 1 & 1 & 1 & 1 & -1 & -1\\ B_1 & 1 & -1 & -1 & 1 & 1 & -1\\ B_2 & 1 & -1 & -1 & 1 & -1 & 1\\ E_1 & 2 & -2 & 1 & -1 & 0 & 0\\ E_2 & 2 & 2 & -1 & -1 & 0 & 0 \end{array} \qquad (c_1,\dots,c_6) = (1,1,2,2,3,3) .

Here $E_1$ is the defining action on the plane, with basis $(x, y)$ .

(a) $E_1 \otimes E_1$ . Squaring the $E_1$ row entry by entry gives

\chi^{E_1\otimes E_1} \;=\; (4,\,4,\,1,\,1,\,0,\,0) , \qquad \text{degree } 4 = 2\times 2 .

Applying step 3 of the method once for each row (only $K_1,\dots,K_4$ contribute, since the last two entries vanish):

\begin{aligned} m_{A_1} &= \tfrac{1}{12}\bigl[1\cdot 4 + 1\cdot 4 + 2\cdot 1\cdot 1 + 2\cdot 1\cdot 1\bigr] = \tfrac{12}{12} = 1 , \\ m_{A_2} &= \tfrac{1}{12}\bigl[4 + 4 + 2 + 2\bigr] = 1 , \\ m_{B_1} &= \tfrac{1}{12}\bigl[1\cdot 4 + (-1)\cdot 4 + 2\cdot(-1)\cdot 1 + 2\cdot 1 \cdot 1\bigr] = \tfrac{4-4-2+2}{12} = 0 , \qquad m_{B_2} = 0 \ \text{likewise},\\ m_{E_1} &= \tfrac{1}{12}\bigl[2\cdot 4 + (-2)\cdot 4 + 2\cdot 1\cdot 1 + 2\cdot(-1)\cdot 1\bigr] = \tfrac{8-8+2-2}{12} = 0 ,\\ m_{E_2} &= \tfrac{1}{12}\bigl[2\cdot 4 + 2 \cdot 4 + 2\cdot(-1)\cdot 1 + 2\cdot(-1)\cdot 1\bigr] = \tfrac{8+8-2-2}{12} = 1 . \end{aligned}

Hence

E_1 \otimes E_1 \;=\; A_1 \oplus A_2 \oplus E_2 , \qquad 1 + 1 + 2 = 4 \ \checkmark .

(b) The symmetric and antisymmetric halves. Square each class representative: $1^{2} = 1$ , $(r^{3})^{2} = 1$ , $r^{2} \in K_4$ , $(r^{2})^{2} = r^{4} \in K_4$ , and every reflection squares to $1$ . Reading $\chi^{E_1}$ at those classes,

\chi^{E_1}(g^{2}) \;=\; (2,\,2,\,-1,\,-1,\,2,\,2) ,

so by the note

\chi_{\mathrm{sym}} = \tfrac12\bigl[(4,4,1,1,0,0) + (2,2,-1,-1,2,2)\bigr] = (3,3,0,0,1,1) , \qquad \chi_{\mathrm{anti}} = (1,1,1,1,-1,-1) .

The dimensions are $3 = \tfrac12\cdot 2\cdot 3$ and $1 = \tfrac12\cdot 2\cdot 1 \ \checkmark$ . The antisymmetric row is the $A_2$ row of the table, so $\Lambda^{2}E_1 = A_2$ ; and

m_{A_1}(\chi_{\mathrm{sym}}) = \tfrac1{12}\bigl[3 + 3 + 0 + 0 + 3\cdot 1 + 3\cdot 1\bigr] = 1 , \qquad m_{E_2}(\chi_{\mathrm{sym}}) = \tfrac1{12}\bigl[2\cdot 3 + 2\cdot 3\bigr] = 1 ,

all other multiplicities being zero, so $\operatorname{Sym}^{2}E_1 = A_1 \oplus E_2$ . Both halves can be seen directly in coordinates: with $(x,y)$ the basis of $E_1$ , the symmetric square is spanned by $x^{2}, xy, y^{2}$ , in which $x^{2}+y^{2}$ is invariant ( $A_1$ ) and the pair $(x^{2}-y^{2},\, 2xy)$ rotates by $2\theta$ when the plane rotates by $\theta$ — character $2\cos(2\pi/3) = -1$ at $K_3$ , which is the $E_2$ row; the antisymmetric square is spanned by $x\wedge y$ , which rotations fix and reflections negate, i.e. $A_2$ . The two readings agree.

(c) $E_1 \otimes E_2$ . Multiplying the two rows entry by entry, $\chi^{E_1\otimes E_2} = (4, -4, -1, 1, 0, 0)$ , and the same four-line computation gives

m_{A_1} = \tfrac{4-4-2+2}{12} = 0 , \quad m_{A_2} = 0 , \quad m_{B_1} = \tfrac{4+4+2+2}{12} = 1 , \quad m_{B_2} = 1 , \quad m_{E_1} = \tfrac{8+8-2-2}{12} = 1 , \quad m_{E_2} = 0 ,

so $E_1 \otimes E_2 = B_1 \oplus B_2 \oplus E_1$ , of degree $1+1+2 = 4\ \checkmark$ . That $m_{A_1} = 0$ is (P4) in action: $E_1$ and $E_2$ have real characters, so each is its own conjugate, and an invariant appears only in $E_1\otimes E_1$ or $E_2\otimes E_2$ , never in the mixed product.

(d) The selection rule. Consider a planar system with symmetry $D_6$ and an in-plane electric dipole operator, whose components $(x, y)$ span $E_1$ . Ask which transitions are allowed out of a state $\phi$ of symmetry $E_2$ , that is, for which final state $\Psi$ of symmetry $\Gamma_f$ the matrix element

J \;=\; \int \overline{\Psi}\,x\,\phi \; d\tau

is not forced to vanish. By the method, build the product row $\chi^{\Gamma_f}_i\,\chi^{E_1}_i\,\chi^{E_2}_i$ and project on $A_1$ . (Step 5 of the method asks for the conjugate of $\chi^{\Gamma_f}$ ; every character of $D_6$ is real, as every class is closed under inversion, so the conjugation changes nothing here.)

Final state of symmetry $E_2$ . The product row is $(2,2,-1,-1,0,0)\cdot(2,-2,1,-1,0,0)\cdot(2,2,-1,-1,0,0) = (8,\,-8,\,1,\,-1,\,0,\,0)$ and

m_{A_1} \;=\; \tfrac{1}{12}\bigl[1\cdot 8 + 1\cdot(-8) + 2\cdot 1 + 2\cdot(-1)\bigr] \;=\; \frac{8-8+2-2}{12} \;=\; \boxed{0} ,

so the transition is forbidden: the integral vanishes for every radial profile of the two states.

Final state of symmetry $E_1$ . The product row is $(2,-2,1,-1,0,0)\cdot(2,-2,1,-1,0,0)\cdot(2,2,-1,-1,0,0) = (8,\,8,\,-1,\,-1,\,0,\,0)$ and

m_{A_1} \;=\; \tfrac{1}{12}\bigl[1\cdot 8 + 1\cdot 8 + 2\cdot(-1) + 2\cdot(-1)\bigr] \;=\; \frac{16 - 4}{12} \;=\; \boxed{1} ,

so the transition is allowed by symmetry. Both answers can be read off part (c) instead: the multiplicity of $A_1$ in $\Gamma_f\otimes E_1\otimes E_2$ equals the multiplicity of (the conjugate of) $\Gamma_f$ in $E_1\otimes E_2 = B_1\oplus B_2\oplus E_1$ , which is $0$ for $\Gamma_f = E_2$ and $1$ for $\Gamma_f = E_1$ — a consistency check on the three-factor computation, and an illustration of step 5 of the method.

Application(Spectroscopy, identical particles and the composition of angular momenta)Original

Which spectral lines appear. An electric dipole transition rate is proportional to $|\langle \Psi_f | \hat{\mathbf{d}} | \Psi_i\rangle|^{2}$ , a matrix element of exactly the three-factor form of (P5). The method above decides, from the character table alone and before any wavefunction is written down, which pairs of levels can be connected: this is the origin of the familiar "allowed" and "forbidden" labels in atomic and molecular spectra, of the $g \leftrightarrow u$ parity rule (Arfken's $C_i$ example: a product of $u$ factors is $g$ exactly when their number is even, so the dipole operator, which is $u$ , connects only states of opposite parity), and of the complementary selection rules for Raman scattering, where the operator is the polarizability and transforms as the symmetric square of the vector representation, so that the relevant decomposition is a symmetric square of the kind computed in part (b) of the example. In a centrosymmetric molecule the two rules exclude one another, which is the rule of mutual exclusion between infrared and Raman activity.

Identical particles. The state of two identical particles lives in $V\otimes V$ , and the exchange operator is the $\tau$ of the note. With $V$ the full one-particle space, spin included, bosons occupy $\operatorname{Sym}^{2}V$ and fermions $\Lambda^{2}V$ , so the symmetry species available to a two-particle configuration are not all of those in $V\otimes V$ but only those in the appropriate half. When spin is carried along, the two halves are combined crosswise: for two electrons in a doubly degenerate orbital of $E_1$ symmetry in the hexagonal system of the example, the antisymmetric spin singlet must multiply the symmetric spatial part $\operatorname{Sym}^{2}E_1 = A_1 \oplus E_2$ , and the symmetric spin triplet the antisymmetric spatial part $\Lambda^{2}E_1 = A_2$ . The configuration therefore yields the terms $^{1}A_1$ , $^{1}E_2$ and $^{3}A_2$ and no others, carrying $1 + 2 + 3 = 6$ states in all — exactly $\binom{4}{2} = 6$ , the number of ways of putting two electrons into the four spin-orbitals of the shell. Hassani makes the same kind of point for the strong interaction: a quark and an antiquark carry an irreducible representation and its conjugate, so by (P4) their product contains exactly one invariant, the singlet (Hassani (2013), §24.8.1, Box 24.8.3, p. 755).

Composition of angular momenta. For the compact continuous groups the whole of this unit survives with the group average replaced by the Haar integral, and (P3) becomes the Clebsch-Gordan series of Addition of angular momenta: the product $D^{(j_1)}\otimes D^{(j_2)}$ of two irreducible representations of $SU(2)$ decomposes as $\bigoplus_{j=|j_1-j_2|}^{j_1+j_2} D^{(j)}$ , each multiplicity being $1$ — the compact-group analogue of Hassani's simply reducible, a term he defines for finite groups. The coefficients that make the change of basis explicit are the Clebsch-Gordan coefficients, and the selection rule (P5), applied to a tensor operator rather than a single function, becomes the Wigner-Eckart theorem (Spherical tensors and the Wigner-Eckart theorem), which not only forbids but factorises the surviving matrix elements into a geometric coefficient and one reduced matrix element.

Elsewhere. Product representations also carry the classification of coupled normal modes and of anharmonic couplings between them, the counting of symmetry-allowed terms in a Landau free energy (the invariants are the $A_1$ components of products of order-parameter representations), and the decomposition of tensors into irreducible pieces that Tensors and their transformation laws performs by hand for the rotation group.