4.4 Introduction to group representation theory

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Groups appear most frequently in physics through their actions on vector spaces, known as representations. More specifically, a representation of any group $G$ on a vector space $V$ is a homomorphism $T$ of $G$ into the group of linear automorphisms of $V$ ,

$$T:G\to GL(V).$$

For every group element $g$ we then have a corresponding linear transformation $T(g):V\to$ $V$ such that

$$T(g)T(h)v=T(gh)v\text{ for all }g,h\in G,v\in V.$$

Essentially, a representation of an abstract group is a way ofproviding a concrete model of the elements of the group as linear transformations of a vector space. The representation is said to be faithful if it is one-to-one; that is, if $\ker$ $T=\{e\}$ . While in principle $V$ could be either a real or complex vector space, we will mostly consider representations on complex vector spaces. If $V$ is finite-dimensional its dimension $n$ is called the degree of the representation. We will restrict attention almost entirely to representations of finite degree. Group representation theory is developed in much greater detail in [6–8].

Exercise

Show that any representation $T$ induces a faithful representation of the factor group $G/\ker T$ on $V.$

Two representations $T_1:G\to V_1$ and $T_2:G\to V_2$ are said to be equivalent, written $T_1\sim T_2$ , if there exists a vector space isomorphism $A$ : $V_1\to V_2$ such that

$$T_2(g)A=A{T}_1(g),\text{ for all }g\in G.\text{\hspace{1em}(4.32)}$$

If $V_1=V_2=V$ then $T_2(g)=A{T}_1(g)A^{-1}$ . For finite dimensional representations the matrices representing $T_2$ are then derived from those representing $T_1$ by a similarity transformation. In this case the two representations can be thought ofas essentially identical, since they are related simply by a change of basis.

Any operator $A,$ even if it is singular, which satisfies Eq. (4.32) is called an intertwining operator for the two representations. This condition is frequently depicted by a commutative diagram

$$\begin{array}{ccc}{V}_1& \stackrel{A}{\to }& V_2\\ T_1(g)\downarrow & & \downarrow T_2(g)\\ V_1& \stackrel{A}{\to }& V_2\end{array}$$

Irreducible representations

A subspace $W$ of $V$ is said to be invariant under the action $G$ , or $G$-invariant, if it is invariant under each linear transformation $T(g)$ ,

$$T(g)W\subseteq W\text{ for all }g\in G.$$

For every $g\in G$ the map $T(g)$ is surjective, $T(g)W=W$ , since $w=T(g)(T(g^{-1}w))$ for every vector $w\in W$ . Hence the restriction of $T(g)$ to $W$ is an automorphism of $W$ and provides another representation of $G,$ , called a subrepresentation of $G,$ , denoted $T_W$ $G\to GL(W)$ . The whole space $V$ and the trivial subspace 0 are clearly $G$-invariant for any representation $T$ on $V$ . If these are the only invariant subspaces the representation is said to be irreducible.

If $W$ is an invariant subspace of a representation $T$ on $V$ then a representation is induced on the quotient space $V/W$ , defined by

$$T_{V/W}(g)(v+W)=T(g)v+W\text{ for all }g\in G,v\in V.$$

Exercise

Verify that this definition is independent of the choice of representative from the coset $v+W$ , and that it is indeed a representation.

Let $V$ be finite dimensional, $\dim$ $V=n$ , and $W^{\prime }\cong V/W$ be a complementary subspace to $W$ , such that $V=W\oplus W^{\prime }$ . From Theorem 3.7 and Example 3.22 there exists a basis whose first $r=$ $\dim$ $W$ vectors span $W$ while the remaining $n-r$ span $W^{\prime }$ . The matrices of the representing transformations with respect to such a basis will have the form

$$\mathsf{T}(g)=\left(\begin{array}{cc}{\mathsf{T}}_W(g)& \mathsf{S}(g)\\ \mathsf{O}& {\mathsf{T}}_{W^{\prime }}(g)\end{array}\right).$$

The submatrices ${\mathsf{T}}_{W^{\prime }}(g)$ form a representation on the subspace $W^{\prime }$ that is equivalent to the quotient space representation, but $W^{\prime }$ is not in general $G$-invariant because of the existence of the off-block diagonal matrices $\mathsf{S}(g)$. If $\mathsf{S}(g)\ne 0$ then it is essentially impossible to recover the original representation purely from the subrepresentations on $W$ and $W^{\prime }$ . Matters are much improved, however, if the complementary subspace $W^{\prime }$ is $G$-invariant as well as $W$. In this case the representing matrices have the block diagonal form in a basis adapted to $W$ and $W^{\prime }$,

$$\mathsf{T}(g)=\left(\begin{array}{cc}{\mathsf{T}}_W(g)& \mathsf{O}\\ \mathsf{O}& {\mathsf{T}}_{W^{\prime }}(g)\end{array}\right)$$

and the representation is said to be completely reducible.

Example 4.7

Let the map $T:\mathbb{R}\to GL({\mathbb{R}}^2)$ be defined by

$$T:a\mapsto \mathsf{T}(a)=\left(\begin{array}{cc}1& a\\ 0& 1\end{array}\right).$$

This is a representation since $\mathsf{T}(a)\mathsf{T}(b)=\mathsf{T}(a+b)$ . The subspace of vectors of the form $\left(\genfrac{}{}{0ex}{}{x}{0}\right)$ is invariant, but there is no complementary invariant subspace – for example, vectors of the form $\left(\genfrac{}{}{0ex}{}{0}{y}\right)$ are not invariant under the matrices $\mathsf{T}(a)$ . Equivalently, it follows from the Jordan canonical form that no matrix $A$ exists such that $\mathsf{AT}(1){\mathsf{A}}^{-1}$ is diagonal. The representation $T$ is thus an example of a representation that is reducible but not completely reducible.

Example 4.8

The symmetric group of permutations on three objects, denoted $S_3$ , has a representation $T$ on a three-dimensional vector space $V$ spanned by vectors $e_1,e_2$ and $e_3$ , defined by

$$T(\pi )e_i=e_{\pi (i)}.$$

In this basis the matrix of the transformation $T$($\pi$) is $\mathsf{T}=[T_i^j(\pi )]$ , where

$$T(\pi )e_i=T_i^j(\pi )e_j.$$

Using cyclic notation for permutations the elements of $S_3$ are $e=\mathrm{id},{\pi }_1=(123),{\pi }_2=$ (1 3 2), ${\pi }_3=(12),{\pi }_4=(23),{\pi }_5=(13)$ . Then $T(e)e_i=e_i$ , so that $\mathsf{T}(e)$ is the identity matrix $I$, while $T({\pi }_1)e_1=e_2,T({\pi }_1)e_2=e_3,T({\pi }_1)e_3=e_1$ , etc. The matrix representations of all permutations of $S_3$ are

$$\mathsf{T}(e)=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right),\mathsf{T}({\pi }_1)=\left(\begin{array}{ccc}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right),\mathsf{T}({\pi }_2)=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right),$$ $$\mathsf{T}({\pi }_3)=\left(\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right),\mathsf{T}({\pi }_4)=\left(\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right),\mathsf{T}({\pi }_5)=\left(\begin{array}{ccc}0& 0& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}\right).$$

Let $v=v^i{e}_i$ be any vector, then $T(\pi )v=T_i^j(\pi )v^i{e}_j$ and the action of the matrix $\mathsf{T}(\pi )$ is left multiplication on the column vector

$$\mathbf{v}=\left(\begin{array}{c}{v}^1\\ v^2\\ v^3\end{array}\right).$$

We now find the invariant subspaces of this representation. In the first place any one dimensional invariant subspace must be spanned by a vector $v$ that is an eigenvector ofeach operator $T({\pi }_i)$ . In matrices,

$$\mathsf{T}({\pi }_1)\mathbf{v}=\alpha \mathbf{v}⟹v^1=\alpha v^2,v^2=\alpha v^3,v^3=\alpha v^1,$$

whence

$$v^1=\alpha v^2={\alpha }^2{v}^3={\alpha }^3{v}^1.$$

Similarly $v^2={\alpha }^3{v}^2$ and $v^3={\alpha }^3{v}^3$ , and since $\mathbf{v}\ne 0$ we must have tha ${\alpha }^3=1$ . Since $\alpha \ne 0$ it follows that all three components $v^1,v^2$ and $v^3$ are non-vanishing. A similar argument gives

$$\mathsf{T}({\pi }_3)v=\beta v⟹v^1=\beta v^2,v^2=\beta v^1,v^3=\beta v^3,$$

from which ${\beta }^2=1$ , and $\alpha \beta =1\sec v^1=\alpha v^2=\alpha \beta v^1$ . The only pair of complex numbers $\alpha$ and $\beta$ satisfying these relations is $\alpha =\beta =1$ . Hence $v^1=v^2=v^3$ and the only one dimensional invariant subspace is that spanned by $v=e_1+e_2+e_3$

We shall now show that this representation is completely reducible by choosing the basis

$$f_1=e_1+e_2+e_3,f_2=e_1-e_2,f_3=e_1+e_2-2e_3.$$

The inverse transformation is

$$e_1=\frac{1}{3}{f}_1+\frac{1}{2}{f}_2+\frac{1}{6}{f}_3,e_2=\frac{1}{3}{f}_1-\frac{1}{2}{f}_2+\frac{1}{6}{f}_3,e_3=\frac{1}{3}{f}_1-\frac{1}{3}{f}_3,$$

and the matrices representing the elements of $S_3$ are found by calculating the effect of the various transformations on the basis elements $f_i$ . For example,

$$\begin{array}{c}T(e)f_1=f_1,T(e)f_2=f_2,T(e)f_3=f_3,\\ T({\pi }_1)f_1=T({\pi }_1)(e_1+e_2+e_3)\\ =e_2+e_3+e_1=f_1,\\ T({\pi }_1)f_2=T({\pi }_1)(e_1-e_2)\\ =e_2-e_3=\frac{1}{2}{f}_2+\frac{1}{2}{f}_3,\\ T({\pi }_1)f_3=T({\pi }_1)(e_1+e_2-2e_3)\\ =e_2+e_3-2e_1=-\frac{3}{2}{f}_2-\frac{1}{2}{f}_3,\text{etc.}\end{array}$$

Continuing in this way for all $T({\pi }_i)$ we arrive at the following matrices:

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

The two-dimensional subspace spanned by $f_2$ and $f_3$ is thus invariant under the action of $S_3$ and the representation $T$ is completely reducible.

Exercise

Show that the representation $T$ restricted to the subspace spanned by $f_2$ and $f_3$ is irreducible, by showing that there is no invariant one-dimensional subspace spanned by $f_2$ and $f_3$ .

Schur’s lemma

The following key result and its corollary are useful in the classification of irreducible representations of groups.

Theorem 4.6 · Schur’s lemma

(Schur’s lemma) Let $T_1:G\to GL(V_1)$ and $T_2:G\to GL(V_2)$ be two irreducible representations of a group $G$, and $A:V_1\to V_2$ an intertwining operator such that

$$T_2(g)A=A{T}_1(g)\text{ for all }g\in G.$$

Then either $A=0$ or $A$ is an isomorphism, in which case the two representations are equivalent, $T_1\sim T_2$

Proof

Let $v\in$ $\ker$ $A\subseteq V_1$ . Then

$$A{T}_1(g)v=T_2(g)Av=0$$

so that $T_1(g)v\in$ $\ker$ $A$. Hence $\ker$ $A$ is an invariant subspace of the representation $T_1$ . As $T_1$ is an irreducible representation we have that either $\ker$ $A=V_1$ , in which case $A=0_i$ or $\ker$ $A=\{0\}$ . In the latter case $A$ is one-to-one. To show it is an isomorphism it is only necessary to show that it is onto. This follows from the fact that $\mathrm{im}$ $A\subset V_2$ is an invarian subspace of the representation $T_2$

$$T_2(g)(\mathrm{im}A)=T_2(g)A\left(V_1\right)=A{T}_1(g)\left(V_1\right)\subseteq A\left(V_1\right)=\mathrm{im}A.$$

Since $T_2$ is an irreducible representation we have either $\mathrm{im}$ $A=\{0\}$ or $\mathrm{im}$ $A=V_2$ . In the first case $A=0$ , while in the second $A$ is onto. Schur’s lemma is proved.

Corollary 4.7 · Operators commuting with a representation

Let $T:G\to GL(V)$ be a representation ofafinite group $G$ on a complex vector space $V$ and $A:V\to V$ an operator that commutes with all $T(g)$ ; that is, $AT(g)=$ $T(g)A\forall g\in G$ . Then $A=\alpha {\mathrm{id}}_V$ some complex scalar $\alpha$.

Proof

Set $V_1=V_2=V$ and $T_1=T_2=T$ in Schur’s lemma. Since $AT(g)=T(g)A$ we have

$$(A-\alpha {\mathrm{id}}_V)T(g)=T(g)(A-\alpha {\mathrm{id}}_V)$$

since ${\mathrm{id}}_V$ commutes with all linear operators on $V$. By Theorem 4.6 either $A-\alpha {\mathrm{id}}_V$ is invertible or it is zero. Let $\alpha$ be an eigenvalue of $A$ – for operators on a complex vector space this is always possible. The operator $A-\alpha {\mathrm{id}}_V$ is not invertible, for if it is applied to a corresponding eigenvector the result is the zero vector. Hence $A-\alpha {\mathrm{id}}_V=0$ , which is the desired result.

It should be observed that the proof of this corollary only holds for complex representations since real matrices do not necessarily have any real eigenvalues.

Example 4.9

If $G$ is a finite abelian group then all its irreducible representations are onedimensional. This follows from Corollary 4.7, for if $T:G\to GL(V)$ ) is any representation of $G$ then any $T(h)(h\in G)$ commutes with all $T(g)$ and is therefore a multiple of the identity,

$$T(h)=\alpha (h){\mathrm{id}}_V.$$

Hence any vector $v\in V$ is an eigenvector of $T(h)$ for all $h\in G$ and spans an invariant one dimensional subspace of $V$. Thus, if $\dim$ $V>1$ the representation $T$ cannot be irreducible.