5.3 Representations of finite groups

中文

If $G$ is a finite group, it turns out that every finite dimensional representation is equivalent to a representation by unitary transformations on an inner product space – known as a unitary representation. For, let $T$ be a representation on any finite dimensional vector space $V,$ and let $\{e_i\}$ be any basis of $V$ . Define an inner product $(u|v)$ on $V$ by setting $\{e_i\}$ to be an orthonormal set,

$$(u|v)=\sum _{i=1}^n\overline{u^i}{v}^i\text{ where }u=u^i{e}_i,v=v^j{e}_j(u^i,v^j\in \mathbb{C}).\text{\hspace{1em}(5.21)}$$

Of course there is no reason why the linear transformations $T(g)$ should be unitary with respect to this inner product, but they will be unitary with respect to the inner product $⟨u\mid v⟩$

formed by ‘averaging over the group’,

$$⟨u|v⟩=\frac{1}{|G|}\sum _{a\in G}(T(a)u|T(a)v),\text{\hspace{1em}(5.22)}$$

where $|G|$ is the order of the group $G$ (the number of elements in $G$). This follows from

$$\begin{array}{rl}⟨T(g)u|T(g)v⟩& =\frac{1}{|G|}{\sum }_{a\in G}(T(a)T(g)u|T(a)T(g)v)\\ & =\frac{1}{|G|}{\sum }_{a\in G}(T(ag)u|T(ag)v)\\ & =\frac{1}{|G|}{\sum }_{b\in G}(T(b)u|T(b)v)\\ & =⟨u|v⟩\end{array}$$

since, as a ranges over the group $G,$ so does $b=ag$ for any fixed $g\in G$

Theorem 5.7 · Maschke’s theorem

Any finite dimensional representation of a finite group $G$ is completely reducible into a direct sum of irreducible representations.

Proof

Using the above device we may assume that the representation is unitary on a finite dimensional Hilbert space $V$ with inner product $⟨\cdot |\cdot ⟩$ . If $W$ is a vector subspace of $V$ , define its orthogonal complement $W^{\perp }$ to be the set of vectors orthogonal to $W$

$$W^{\perp }=\{u\mid ⟨u\mid w⟩=0,\forall w\in W\}.$$

$W^{\perp }$ is clearly a vector subspace, for if $\alpha$ is an arbitrary complex number then

$$u,v\in W^{\perp }⟹⟨u+\alpha v|w⟩=⟨u|w⟩+\alpha ⟨v|w⟩=0⟹u+\alpha v\in W^{\perp }.$$

By selecting an orthonormal basis such that the first $\dim$ $W$ vectors belong to $W_{;}$ , it follows that the remaining vectors of the basis span $W^{\perp }$ . Hence $W$ and $W^{\perp }$ are orthogonal and complementary subspaces, $H=W\oplus W^{\perp }$ . If $W$ is a $G$-invariant subspace, then $W^{\perp }$ is also $G$-invariant, For, if $u\in W^{\perp }$ then for any $w\in W$ ,

$$\begin{array}{l}0=⟨T(g)u|w⟩=⟨T(g)u|T(g)T(g{)}^{-1}w⟩\\ =⟨u|T(g^{-1})w⟩\text{ since }T(g)\text{ is unitary }\\ =0\end{array}$$

since $T(g^{-1})w\in W$ by the $G$-invariance of $W$ . Hence $T(g)u\in W^{\perp }$

Now pick $W$ to be the $G$-invariant subspace of $V$ ofsmallest dimension, not counting the trivial subspace 0 . The representation induced on $W$ must be irreducible since it can have no proper $G$-invariant subspaces, as they would need to have smaller dimension. If $W=V$ then the representation $T$ is irreducible. If $W\ne V$ its orthogonal complement $W^{\perp }$ is either irreducible, in which case the proofis finished, or it has a non-trivial invariant subspace $W^{\prime }$ Again pick the invariant subspace of smallest dimension and continue in this fashion unti $V$ is a direct sum of irreducible subspaces,

$$V=W\oplus W^{\prime }\oplus W^{\prime \prime }\oplus \dots$$

The representation $T$ decomposes into subrepresentations $T{|}_{W^{(i)}}.$

Orthogonality relations

The components of the matrices of irreducible group representatives satisfy a number of important orthogonality relationships, which are the cornerstone of the classification procedure of group representations. We will give just a few of these relations; others can be found in [4, 5].

Let $T_1$ and $T_2$ be irreducible representations of a finite group $G$ on complex vector spaces $V_1$ and $V_2$ respectively. If $\{e_i\mid i=1,\dots ,n_1=\dim V_1\}$ and $\{f_a|a=1,\dots ,n_2=\dim V_2\}$ are bases of these two vector spaces, we will write the representative matrices as ${\mathsf{T}}_1(g)=$ $[T_{(1)}{}^j{}_i]$ and ${\mathsf{T}}_2(g)=[T_{(2)}{}_a^b]$ where

$$T_1(g)e_i=T_{(1)}{}^j{}_i{e}_j\text{ and }{T}_2(g)f_a=T_{(2)}{}^b{}_a{f}_b.$$

If $A:V_1\to V_2$ is any linear map, define its ‘group average $\tilde{A}:V_1\to V_2$ to be the linear map

$$\tilde{A}=\frac{1}{|G|}\sum _{g\in G}{T}_2(g)A{T}_1(g^{-1}).$$

Then if $h$ is any element of the group $G$ ,

$$\begin{array}{l}{T}_2(h)\tilde{A}{T}_1(h^{-1})=\frac{1}{|G|}{\sum }_{g\in G}{T}_2(hg)A{T}_1((hg{)}^{-1})\\ =\frac{1}{|G|}{\sum }_{g^{\prime }\in G}{T}_2(g^{\prime })A{T}_1((g^{\prime }{)}^{-1})\\ =\tilde{A}.\end{array}$$

Hence $\tilde{A}$ is an intertwining operator,

$$T_2(h)\tilde{A}=\tilde{A}{T}_1(h)\text{ for all }h\in G,$$

and by Schur’s lemma, Theorem 4.6, if $T_1\nsim T_2$ then $\tilde{A}=0$ . On the other hand, from the corollary to Schur’s lemma, 4.7, if $V_1=V_2=V$ and $T_1=T_2$ then $\tilde{A}=c{\mathrm{id}}_V$ . The matrix version of this equation with respect to any basis of $V$ is $\tilde{\mathsf{A}}=c\mathsf{I}$ , and taking the trace gives

$$c=\frac{1}{n}\mathrm{tr}\tilde{\mathsf{A}}.$$

However

$$\begin{array}{l}\mathrm{tr}\tilde{\mathsf{A}}=\frac{1}{|G|}\mathrm{tr}{\sum }_{g\in G}\mathsf{T}(g)\mathsf{A}{\mathsf{T}}^{-1}(g)\\ =\frac{1}{|G|}{\sum }_{g\in G}\mathrm{tr}({\mathsf{T}}^{-1}(g)\mathsf{T}(g)\mathsf{A})\\ =\frac{1}{|G|}{\sum }_{g\in G}\mathrm{tr}\mathsf{A}=\mathrm{tr}\mathsf{A},\end{array}$$

whence

$$c=\frac{1}{n}\mathrm{tr}\mathsf{A}.$$

If $T_1\nsim T_2$ , expressing $A$ and $\tilde{A}$ in terms of the bases $e_1$ and $f_a$ ,

$$A{e}_i=A_i^a{f}_a\mathrm{and}\tilde{A}{e}_i={\tilde{A}}_i^a{f}_a$$

the above consequence of Schur’s lemma can be written

$${\tilde{A}}_i^a=\frac{1}{|G|}\sum _{g\in G}{T}_{(2)b}^a(g)A_j^b{T}_{(1)i}^j(g^{-1})=0.$$

As $A$ is an arbitrary operator the matrix elements ${\mathbf{A}}_j^b$ are arbitrary complex numbers, so that

$$\frac{1}{|G|}\sum _{g\in G}{T}_{(2)b}^a(g)T_{(1)i}^j(g^{-1})=0.\text{\hspace{1em}(5.23)}$$

If $T_1=T_2=T$ and $n=\dim V$ is the degree of the representation we have

$${\tilde{A}}_i^j=\frac{1}{|G|}\sum _{g\in G}{T}_k^j(g)A_l^k{T}_i^l(g^{-1})=\frac{1}{n}{A}_k^k{\delta }_i^j.$$

As ${\mathbf{A}}_l^k$ are arbitrary,

$$\frac{1}{|G|}\sum _{g\in G}{T}_k^j(g)T_i^l(g^{-1})=\frac{1}{n}{\delta }_i^j{\delta }_k^l.\text{\hspace{1em}(5.24)}$$

If $⟨\cdot \mid \cdot ⟩$ is the invariant inner product defined by a representation $T$ on a vector space $V$ by Eq. (5.22), and $\{e_i\}$ is any basis such that

$$⟨e_i|e_j⟩={\delta }_{ji},$$

then the unitary condition $⟨T(g)u\mid T(g)v⟩=⟨u\mid v⟩$ implies

$$\sum _k\overline{T_{ki}(g)}{T}_{kj}(g)={\delta }_{ij},$$

where indices on $T$ are all lowered. In matrices

$${\mathsf{T}}^{†}(g)\mathsf{T}(g)=\mathsf{I},$$

whence

$$\mathsf{T}(g^{-1})=(\mathsf{T}(g){)}^{-1}={\mathsf{T}}^{†}(g),$$

or equivalently

$$T_{ji}(g^{-1})=\overline{T_{ij}(g)}.\text{\hspace{1em}(5.25)}$$

Substituting this relation for $T_1$ in place of $T$ into Eq. (5.23), with all indices now lowered, gives

$$\frac{1}{|G|}\sum _{g\in G}\overline{T_{(1)ij}(g)}{T}_{(2)ab}(g)=0.\text{\hspace{1em}(5.26)}$$

Similarly if $T_1=T_2=T$ , Eqs. Eq. (5.25) and Eq. (5.24) give

$$\frac{1}{|G|}\sum _{g\in G}\overline{T_{ij}(g)}{T}_{kl}(g)=\frac{1}{n}{\delta }_{ik}{\delta }_{jl}.\text{\hspace{1em}(5.27)}$$

The left-hand sides of Eqs. Eq. (5.26) and Eq. (5.27) have the appearance of an inner product, and this is in fact so. Let $\mathcal{F}(G)$ be the space of all complex-valued functions on $G$

$$\mathcal{F}(G)=\{\varphi \mid \varphi :G\to \mathbb{C}\}$$

with inner product

$$(\varphi ,\psi )=\frac{1}{|G|}\sum _{a\in G}\overline{\varphi (a)}\psi (a).\text{\hspace{1em}(5.28)}$$

It is easy to verify that the requirements (IP1)–(IP3) hold for this inner product, namel

$$\begin{array}{c}(\varphi ,\psi )=\overline{(\psi ,\varphi )},\\ (\varphi ,\alpha \psi )=\alpha (\varphi ,\psi ),\\ (\varphi ,\varphi )\ge 0\text{ and }(\varphi ,\varphi )=0\text{ iff }\varphi =0.\end{array}$$

The matrix components $T_{ji}$ of any representation with respect to an o.n. basis form a se of $n^2$ complex-valued functions on $G$ , and Eqs. Eq. (5.26) and Eq. (5.27) read

$$(T_{(1)ij},T_{(2)ab})=0\text{ if }{T}_1\nsim T_2,\text{\hspace{1em}(5.29)}$$

and

$$(T_{ij},T_{kl})=\frac{1}{n}{\delta }_{ik}{\delta }_{jl}.\text{\hspace{1em}(5.30)}$$

Example 5.8

Consider the group $S_3$ with notation as in Example 4.8. The invariant inner product Eq. (5.22) on the space spanned by $e_1,e_2$ and $e_3$ is given by

$$\begin{array}{l}⟨u|v⟩=\frac{1}{6}{\sum }_{\pi }{\sum }_{i=1}^3(\overline{T(\pi )u{)}^i}(T(\pi )v{)}^i\\ =\frac{1}{6}(\overline{u^1}{v}^1+\overline{u^2}{v}^2+\overline{u^3}{v}^3+\overline{u^3}{v}^3+\overline{u^1}{v}^1+\overline{u^2}{v}^2+\dots )\\ =\overline{u^1}{v}^1+\overline{u^2}{v}^2+\overline{u^3}{v}^3.\end{array}$$

Hence $\{e_i\}$ forms an orthonormal basis for this inner product,

$$h_{ij}=⟨e_i|e_j⟩={\delta }_{ij}.$$

It is only because $S_3$ runs through all permutations that the averaging process gives the same result as the inner product defined by Eq. (5.21). A similar conclusion would hold for the action of $S_n$ on an $n$-dimensional space spanned by $\{e_1,\dots ,e_n\}$ , but these vectors would not in general be orthonormal with respect to the inner product Eq. (5.22) defined by an arbitrary subgroup of $S_n$

As seen in Example 4.8, the vector $f_1=e_1+e_2+e_3$ spans an invariant subspace with respect to this representation of $S_3$ . As in Example 4.8, the vectors $f_1,f_2=e_1-e_2$ and $f_3=e_1+e_2-2e_3$ are mutually orthogonal,

$$⟨f_1\mid f_2⟩=⟨f_1\mid f_3⟩=⟨f_2\mid f_3⟩=0.$$

Hence the subspace spanned by $f_2$ and $f_3$ is orthogonal to $f_1$ , and from the proof of Theorem 5.7, it is also invariant. Form an o.n. set by normalizing their lengths to

unity,

$$f_1^{\prime }=\frac{f_1}{\sqrt{3}},f_2^{\prime }=\frac{f_2}{\sqrt{2}},f_3^{\prime }=\frac{f_3}{\sqrt{6}}.$$

The representation $T_1$ on the one-dimensional subspace spanned by $f_1^{\prime }$ is clearly the trivial one, whereby every group element is mapped to the number 1,

$$T_1(\pi )=1\text{ for all }\pi \in S_3.$$

The matrices of the representation $T_2$ on the invariant subspace spanned by $h_1=f_2^{\prime }$ and $h_2=f_3^{\prime }$ are easily found from the $2\times 2$ parts of the matrices given in Example 4.8 by transforming to the renormalized basis,

$${\mathsf{T}}_2(e)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right),{\mathsf{T}}_2({\pi }_1)=\left(\begin{array}{cc}-\frac{1}{2}& -\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}& -\frac{1}{2}\end{array}\right),{\mathsf{T}}_2({\pi }_2)=\left(\begin{array}{cc}-\frac{1}{2}& \frac{\sqrt{3}}{2}\\ -\frac{\sqrt{3}}{2}& -\frac{1}{2}\end{array}\right),$$ $${\mathsf{T}}_2({\pi }_3)=\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right),{\mathsf{T}}_2({\pi }_4)=\left(\begin{array}{cc}\frac{1}{2}& \frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}& -\frac{1}{2}\end{array}\right),{\mathsf{T}}_2({\pi }_5)=\left(\begin{array}{cc}\frac{1}{2}& -\frac{\sqrt{3}}{2}\\ -\frac{\sqrt{3}}{2}& -\frac{1}{2}\end{array}\right).$$

It is straightforward to verify Eq. (5.29):

$$\left(T_{(1)11},T_{(2)11}\right)=\frac{1}{6}(1-\frac{1}{2}-\frac{1}{2}-1+\frac{1}{2}+\frac{1}{2})=0,$$ $$\left(T_{(1)11},T_{(2)12}\right)=\frac{1}{6}\left(0-\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}+0+\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}\right)=0,\text{ etc. }$$

From the exercise following Example 4.8 the representation $T_2$ is an irreducible representation with $n=2$ and the relations Eq. (5.30) are verified as follows:

$$\left(T_{(2)11},T_{(2)11}\right)=\frac{1}{6}\left(1^2+{\left(-\frac{1}{2}\right)}^2+{\left(-\frac{1}{2}\right)}^2+(-1{)}^2+{\frac{1}{2}}^2+{\frac{1}{2}}^2\right)=\frac{3}{6}=\frac{1}{2}{\delta }_{11}{\delta }_{11}=\frac{1}{2}$$ $$\left(T_{(2)12},T_{(2)12}\right)=\frac{1}{6}\left({\left(-\frac{\sqrt{3}}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2+{\left(-\frac{\sqrt{3}}{2}\right)}^2\right)=\frac{1}{2}=\frac{1}{2}{\delta }_{11}{\delta }_{22}$$ $$\left(T_{(2)11},T_{(2)12}\right)=\frac{1}{6}\left(1\cdot 0-\frac{1}{2}\cdot \left(-\frac{\sqrt{3}}{2}\right)-\frac{1}{2}\cdot \frac{\sqrt{3}}{2}-1\cdot 0+\frac{1}{2}\cdot \frac{\sqrt{3}}{2}+\frac{1}{2}\cdot \left(-\frac{\sqrt{3}}{2}\right)\right)$$ $$=0=\frac{1}{2}{\delta }_{11}{\delta }_{12},\text{ etc. }$$

Theorem 5.8 · Finiteness of irreducible representations and the sum-of-squares bound

There are a finite number $N$ ofinequivalent irreducible representations of a finite group, and

$$\sum _{\mu =1}^N(n_{\mu }{)}^2\le |G|,\text{\hspace{1em}(5.31)}$$

where $n_{\mu }=\dim V_{\mu }$ are the degrees of the inequivalent representations.

Proof

Let $T_1:G\to GL(V_1),T_2:G\to GL(V_2),..$ . be inequivalent irreducible representations of $G$. If the basis on each vector space $V_{\mu }$ is chosen to be orthonormal with respect to the inner product for each $\mu =1,2,\dots$ , then Eq. (5.29) and Eq. (5.30) may be

summarized as the single equation

$$\left(T_{(\mu )ij},T_{(\nu )ab}\right)=\frac{1}{n_{\mu }}{\delta }_{\mu \nu }{\delta }^{ia}{\delta }_{jb}.$$

Hence for each $\mu$ the $T_{(\mu )ij}$ consist of $(n_{\mu }{)}^2$ mutually orthogonal functions in $\mathcal{F}(G)$ that are orthogonal to all $T_{(\nu )ab}$ for $\nu \ne \mu$ . There cannot therefore be more than $\dim$ $\mathcal{F}(G)=|G|$ of these, giving the desired inequality Eq. (5.31). Clearly there are at most a finite number $N$ of such representations.

It may in fact be shown that the inequality Eq. (5.31) can be replaced by equality, a very useful identity in the enumeration of irreducible representations of a finite group. Details of the proof as well as further orthogonality relations and applications of group representation theory to physics may be found in [4, 5].

Problems

Problem 5.13

For a function $\varphi :G\to \mathbb{C},$ , if we set $g\phi$ to be the function $(g\varphi )(a)=\varphi (g^{-1}a)$ show that $(g{g}^{\prime })\varphi =g(g^{\prime }\varphi )$ . Show that the inner product Eq. (5.28) is $G$-invariant, $(g\varphi ,g\psi )=(\varphi ,\psi )$ for all $g\in G$

Problem 5.14

Let the character of a representation $T$ of a group $G$ on a vector space $V$ be the function $\chi :G\to \mathbb{C}$ defined by

$$\chi (g)=\mathrm{tr}T(g)=T_i^i(g).$$

(a) Show that the character is independent of the choice of basis and is a member of $\mathcal{F}(G)$ , and that characters of equivalent representations are identical. Show that $\chi (e)=\dim V$

(b) Any complex-valued function on $G$ that is constant on conjugacy classes (see Section 2.4) is called a central function. Show that characters are central functions.

(c) Show that with respect to the inner product Eq. (5.28), characters of any pair of inequivalent irre ducible representations $T_1\nsim T_2$ are orthogonal to each other, $({\chi }_1,{\chi }_2)=0$ , while the character ofany irreducible representation $T$ has unit norm $(\chi ,\chi )=1$

(d) From Theorem 5.8 and Theorem 5.7 every unitary representation $T$ can be decomposed into a direct sum of inequivalent irreducible unitary representations $T_{\mu }:G\to GL(V_{\mu })$ ,

$$T\sim m_1{T}_1\oplus m_2{T}_2\oplus \cdots \oplus m_N{T}_N(m_{\mu }\ge 0).$$

Show that the multiplicities $m_{\mu }$ of the representations $T_{\mu }$ are given by

$$m_{\mu }=(\chi ,{\chi }_{\mu })=\frac{1}{|G|}\sum _{g\in G}\overline{\chi (g)}{\chi }_{\mu }(g)$$

and $T$ is irreducible if and only if its character has unit magnitude, $(\chi ,\chi )=1$ . Show that $T$ and $T^{\prime }$ have no irreducible representations in common in their decompositions if and only if their characters are orthogonal.