1.5. Lecture 5

1.5.1. Direct sums and reducibility

We start by defining sums of representations:

Definition 1.5.1.

Let $(\pi,V)$ and $(\rho,W)$ be representations of a group $G$ . The direct sum of the two representations is $(\pi\oplus\rho,V\oplus W)$ , where

(\pi\oplus\rho)(g)({\bf v},{\bf w}):=(\pi(g){\bf v},\rho(g){\bf w})

for all $g\in G$ , ${\bf v}\in V$ , ${\bf w}\in W$ .

We also write $(\pi,V)\oplus(\rho,W)$ for the direct sum of the representations.

Example 1.5.2.

Let $(\chi_{1},{\mathbb{C}})$ and $(\chi_{2},{\mathbb{C}})$ be representations of $C_{n}=\langle a\,|\,a^{n}=e\rangle$ given by $\chi_{1}(a^{j})=e^{2\pi ij/n}$ and $\chi_{2}(a^{j})=e^{4\pi ij/n}$ . Then $(\chi_{1}\oplus\chi_{2},{\mathbb{C}}^{2})$ is the representation of $C_{n}$ on ${\mathbb{C}}^{2}$ given by $\chi_{1}\oplus\chi_{2}(a^{j})=(e^{2\pi ij/n},e^{4\pi ij/n})$ .

Conversely, we also have the following definition.

Definition 1.5.3.

If $(\pi,V)$ is a representation with non-trivial subrepresentations $(\pi,U)$ and $(\pi,W)$ such that $V=U\oplus W$ , then we say that $(\pi,V)$ is decomposable or reducible.

Note that if $(\pi,V)$ is decomposable , then

(\pi,V)\cong(\pi,U)\oplus(\pi,W),

with a $G$ -isomorphism from $U\oplus W$ to $V$ given by

({\bf u},{\bf w})\mapsto{\bf u}+{\bf w}

for all ${\bf u}\in U,{\bf w}\in W$ (see Problem 1.5.4 below). We say that the representation $(\pi,V)$ has been decomposed into a direct sum of subrepresentations.

Example 1.5.4.

The permutation representation $(\pi,{\mathbb{C}}^{n})$ decomposes as $(\pi,W_{1})\oplus(\pi,W_{0})$ , where $W_{0}=\{(x_{1},\ldots,x_{n})\in\mathbb{R}^{n}\,|\,x_{1}+\ldots+x_{n}=0\}$ and $W_{1}$ is the linear span of $(1,\ldots,1)$ .

The following example shows that there are representations that are neither decomposable or irreducible:

Example 1.5.5.

Let $(\rho,{\mathbb{C}}^{2})$ be the representation of $({\mathbb{Z}},+)$ given by

\rho(n)=\left(\begin{matrix}1&n\\0&1\end{matrix}\right).

This representation is not irreducible, as ${\mathbb{C}}\left(\begin{matrix}1\\0\end{matrix}\right)$ is a non-trivial subrepresentation.

The representation is not decomposable , as then there would be two non-zero vectors ${\bf v}_{1}$ and ${\bf v}_{2}$ such that ${\mathbb{C}}^{2}={\mathbb{C}}{\bf v}_{1}\oplus{\mathbb{C}}{\bf v}_{2}$ , and $(\rho,{\mathbb{C}}{\bf v}_{1})$ , $(\rho,{\mathbb{C}}{\bf v}_{2})$ are subrepresentations. The vectors ${\bf v}_{1}$ and ${\bf v}_{2}$ would then be two linearly independent eigenvectors of all the matrices $\left(\begin{smallmatrix}1&n\\0&1\end{smallmatrix}\right)$ , which is not possible - these matrices are not diagonalisable.

However, for finite groups, we have the following:

Theorem 1.5.6 (Maschke’s Theorem, 1899).

Let $(\pi,W)$ be a non-trivial subrepresentation of a finite-dimensional complex representation of a finite group $G$ . Then there exists a subrepresentation $(\pi,W^{\prime})$ of $(\pi,V)$ such that

(\pi,V)=(\pi,W)\oplus(\pi,W^{\prime}).

Before proving this, we introduce another important notion:

1.5.2. Unitary representations

Definition 1.5.7.

Let $(\pi,V)$ be a representation of a group $G$ on an inner product space $V$ , with inner product $\langle\cdot,\cdot\rangle$ . The representation is said to be unitary if

\langle\pi(g){\bf v},\pi(g){\bf u}\rangle=\langle{\bf v},{\bf u}\rangle

(1.5.8)

for all ${\bf v},{\bf u}\in V$ and $g\in G$ .

We say that the inner product $\langle\cdot,\cdot\rangle$ is $G$ -invariant if (1.5.8) holds for all ${\bf v},{\bf u}\in V$ and $g\in G$ . Other ways of writing (1.5.8) include

\langle\pi(g){\bf v},{\bf u}\rangle=\langle{\bf v},\pi(g)^{*}{\bf u}\rangle

and $\pi(g)^{*}=\pi(g^{-1})=\pi(g)^{-1}$ .

Proposition 1.5.9.

Let $(\pi,W)$ be a subrepresentation of a unitary representation $(\pi,V)$ . Then $(\pi,W^{\perp})$ is also a subrepresentation, and hence

(\pi,V)=(\pi,W)\oplus(\pi,W^{\perp}).

Consequently, all unitary representations with a nontrivial subrepresentation are decomposable.

Proof.

We simply need to verify that $W^{\perp}$ is $G$ -invariant. Recall that

W^{\perp}=\{{\bf v}\in V\,|\,\langle{\bf v},{\bf w}\rangle=0\text{ for all }{%\bf w}\in W\}.

Given $g\in G$ , ${\bf v}\in W^{\perp}$ and any ${\bf w}\in W$ , we have $\langle\pi(g){\bf v},{\bf w}\rangle=\langle{\bf v},\pi(g^{-1}){\bf w}\rangle$ . However, $(\pi,W)$ is a subrepresentation, so $\pi(g^{-1}){\bf w}\in W$ . Since ${\bf v}\in W^{\perp}$ , we thus have $\langle{\bf v},\pi(g^{-1}){\bf w}\rangle=0$ , hence $\pi(g){\bf v}\in W^{\perp}$ . ∎

The following shows that every finite-dimensional representation of a finite group may be “made” unitary:

Proposition 1.5.10.

Let $(\pi,V)$ be a finite-dimensional complex representation of a finite group $G$ . Then there exists a $G$ -invariant inner product on $V$ .

Proof.

Since $V$ is finite-dimensional, it has an inner product $\langle\cdot,\cdot\rangle$ (see Problem 1.5.4 below). We now define $\langle\cdot,\cdot\rangle_{G}:V\times V\rightarrow{\mathbb{C}}$ by the formula

\langle{\bf v},{\bf u}\rangle_{G}=\frac{1}{|G|}\sum_{h\in G}\langle\pi(h){\bf v%},\pi(h){\bf u}\rangle,

and claim that this is a $G$ -invariant inner product. We leave it as an exercise for the reader to verify that this is an inner product (see Problem 1.5.4). The $G$ -invariance is shown as follows:

	$\displaystyle\langle\pi(g){\bf v},\pi(g){\bf u}\rangle_{G}$	$\displaystyle=\frac{1}{\|G\|}\sum_{h\in G}\langle\pi(h)\pi(g){\bf v},\pi(h)\pi(g%){\bf u}\rangle$
		$\displaystyle=\frac{1}{\|G\|}\sum_{h\in G}\langle\pi(hg){\bf v},\pi(hg){\bf u}\rangle$
		$\displaystyle=\frac{1}{\|G\|}\sum_{\widetilde{h}\in G}\langle\pi(\widetilde{h}){%\bf v},\pi(\widetilde{h}){\bf u}\rangle=\langle{\bf v},{\bf u}\rangle_{G}.$

∎

We note that as a consequence of the spectral theorem for unitary matrices, we have

Proposition 1.5.11.

Let $(\pi,V)$ be a finite-dimensional unitary representation of a group $G$ . Then for any $g\in G$ , $\pi(g)$ is diagonalisable, i.e. has an eigenbasis.

Proof.

See Exercise 1.5.4 below ∎

1.5.3. Decompositions into irreducibles

Note that Theorem 1.5.6 now follows directly from Propositions 1.5.9 and 1.5.10. In fact, we can give a stronger statement:

Theorem 1.5.12.

Let $(\pi,V)$ be a representation of $G$ . Then there exist irreducible subrepresentations $(\pi,W_{i})$ such that

(\pi,V)=(\pi,W_{1})\oplus(\pi,W_{2})\oplus\ldots\oplus(\pi,W_{n}).

Moreover, the number of times each isomorphism class of an irreducible representation shows up in the above decomposition is independent of the exact choice of decomposition.

The existence of such a decomposition follows as a corollary from repeated application of Maschke’s Theorem (Theorem 1.5.6). To complete the proof of the second part of the theorem we need the following two lemmas:

Lemma 1.5.13.

Let $(\pi_{1},U)$ , $(\pi_{2},V)$ , and $(\pi_{3},W)$ be representations of a group $G$ . Then

(i)

$\mathrm{Hom}_{G}(U\oplus V,W)\cong\mathrm{Hom}_{G}(U,W)\oplus\mathrm{Hom}_{G}(%V,W)$
(ii)

$\mathrm{Hom}_{G}(W,U\oplus V)\cong\mathrm{Hom}_{G}(W,U)\oplus\mathrm{Hom}_{G}(%W,V)$

Proof.

Exercise: Problem 1.5.4 below. ∎

Lemma 1.5.14.

Let $(\pi,V)=\bigoplus_{i=1}^{n}(\pi,W_{i})$ be a decomposition of $(\pi,V)$ into irreducible subrepresentations. For any irreducible representation $(\rho,U)$ , the number of $(\pi,W_{i})$ that are isomorphic to $(\rho,U)$ is $\operatorname{dim}\operatorname{Hom}_{G}(V,U)$ .

Proof.

By Lemma 1.5.13, $\operatorname{dim}\operatorname{Hom}_{G}(V,U)=\sum_{i}\operatorname{dim}%\operatorname{Hom}_{G}(W_{i},U)$ . By Corollary 1.4.2 to Schur’s lemma, $\operatorname{dim}\operatorname{Hom}_{G}(W_{i},U)=1$ if $(\pi,W_{i})\cong(\rho,U)$ , and zero otherwise. ∎

Proof of Theorem 1.5.12.

The existence of such a decomposition follows from repeated application of Theorem 1.5.6. The second part of the statement follows from Lemma 1.5.14. ∎

1.5.4. Exercises

Problem 31. Let $(\pi,U)$ and $(\pi,W)$ be subrepresentations of a representation $(\pi,V)$ such that $V=U\oplus W$ . Verify that the map

({\bf u},{\bf w})\mapsto{\bf u}+{\bf w}

is a $G$ -isomorphism from $(\pi,U)\oplus(\pi,W)$ to $(\pi,V)$ .

Problem 32. Verify that

\langle\pi(g){\bf v},\pi(g){\bf u}\rangle=\langle{\bf v},{\bf u}\rangle\qquad%\forall{\bf v},{\bf u}\in V,\;g\in G

is equivalent to

\langle\pi(g){\bf v},{\bf u}\rangle=\langle{\bf v},\pi(g^{-1}){\bf u}\rangle%\qquad\forall{\bf v},{\bf u}\in V,\;g\in G.

Problem 33. Let $V$ be a finite-dimensional complex vector space with a basis $\{{\bf b}_{i}\}_{i=1,\ldots,n}$ , where $n=\operatorname{dim}V$ . Show that

\left\langle\sum_{i=1}^{n}x_{i}{\bf b}_{i},\sum_{j=1}^{n}y_{j}{\bf b}_{j}%\right\rangle:=\sum_{i=1}^{n}x_{i}\overline{y_{i}},

for $x_{i},\,y_{j}\in{\mathbb{C}}$ , defines an inner product on $V$ .

Problem 34. Fill in the remaining details of the proof of Proposition 1.5.10, i.e. show that $\langle\cdot,\cdot\rangle_{G}$ is linear in the first argument, conjugate-symmetric, and positive-definite.

Problem 35. Let $(\pi,V)$ be a finite-dimensional unitary representation of a group $G$ , with $G$ -invariant inner product $\langle\cdot,\cdot\rangle$ . Show that the matrix of any $\pi(g)$ with respect to an orthonormal basis for $\langle\cdot,\cdot\rangle$ is unitary (recall that a matrix is unitary if $M^{*}(=\overline{M}{\,{}^{\mathrm{t}}\!}\,)=M^{-1}$ ).

Problem 36. Prove Lemma 1.5.13.

	$\displaystyle\langle\pi(g){\bf v},\pi(g){\bf u}\rangle_{G}$	$\displaystyle=\frac{1}{\|G\|}\sum_{h\in G}\langle\pi(h)\pi(g){\bf v},\pi(h)\pi(g%){\bf u}\rangle$
		$\displaystyle=\frac{1}{\|G\|}\sum_{h\in G}\langle\pi(hg){\bf v},\pi(hg){\bf u}\rangle$
		$\displaystyle=\frac{1}{\|G\|}\sum_{\widetilde{h}\in G}\langle\pi(\widetilde{h}){%\bf v},\pi(\widetilde{h}){\bf u}\rangle=\langle{\bf v},{\bf u}\rangle_{G}.$