Lecture 5 ‣ Representation theory exercise solutions for Michaelmas

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 $(\pi,V)$ .

Solution: Let $\phi$ be the map in question. It should be clear from prior knowledge that this is an isomorphism of vector spaces. Clearly

\phi(({\bf u}_{1}+\lambda{\bf u}_{2},{\bf w}_{1}+\lambda{\bf w}_{2}))={\bf u}_% {1}+\lambda{\bf u}_{2}+{\bf w}_{1}+\lambda{\bf w}_{2}=\phi(({\bf u}_{1},{\bf w% }_{1}))+\lambda\phi(({\bf u}_{2},{\bf w}_{2})),

for all ${\bf u}_{1},{\bf u}_{2}\in U$ , ${\bf w}_{1},{\bf w}_{2}\in W$ and $\lambda\in k$ . Also for $(u,w)$ to be in $\ker\phi$ both $u$ and $w$ must be zero, as $U\cap W=\{0\}$ . As the dimensions of both $V$ and $U\oplus W$ are the same, by definition, this must be a bijection and hence an isomorphism of vector spaces.

Now let $g\in G$ and $({\bf u},{\bf w})\in U\oplus w$ , then

\phi((\pi(g){\bf u},\pi(g){\bf w}))=\pi(g){\bf u}+\pi(g){\bf w}=\pi(g)({\bf u}% +{\bf w})=\pi(g)\phi(({\bf u},{\bf w})).

Hence it is a $G$ -homomorphism and thus a $G$ -isomorphism.

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.

Solution: We apply $\pi(g)^{-1}$ to the second term on both sides:

	$\displaystyle\langle\pi(g){\bf v},\pi(g){\bf u}\rangle$	$\displaystyle=\langle{\bf v},{\bf u}\rangle,$
	$\displaystyle\langle\pi(g){\bf v},\pi(g)^{-1}\pi(g){\bf u}\rangle$	$\displaystyle=\langle{\bf v},\pi(g)^{-1}{\bf u}\rangle,$
	$\displaystyle\langle\pi(g){\bf v},{\,\mathrm{Id}}{\bf u}\rangle$	$\displaystyle=\langle{\bf v},\pi(g)^{-1}{\bf u}\rangle,$
	$\displaystyle\langle\pi(g){\bf v},{\bf u}\rangle$	$\displaystyle=\langle{\bf v},\pi(g^{-1}){\bf u}\rangle,$

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

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 $V$ .

Solution: Consider $\sum_{i=1}^{n}x_{i}{\bf b}_{i},\sum_{i=1}^{n}y_{i}{\bf b}_{i},\sum_{i=1}^{n}z_% {i}{\bf b}_{i}\in V$ and $\lambda\in{\mathbb{C}}$ , then

	$\displaystyle\left\langle\sum_{i=1}^{n}x_{i}{\bf b}_{i}+\lambda\sum_{i=1}^{n}y% _{i}{\bf b}_{i},\sum_{j=1}^{n}z_{j}{\bf b}_{j}\right\rangle$	$\displaystyle=\sum_{i=1}^{n}(x_{i}+\lambda y_{i})\overline{z_{i}}$
		$\displaystyle=\sum_{i=1}^{n}x_{i}\overline{z_{i}}+\lambda\sum_{i=1}^{n}y_{i}% \overline{z_{i}}$
		$\displaystyle=\left\langle\sum_{i=1}^{n}x_{i}{\bf b}_{i},\sum_{j=1}^{n}z_{j}{% \bf b}_{j}\right\rangle+\lambda\left\langle\sum_{i=1}^{n}y_{i}{\bf b}_{i},\sum% _{j=1}^{n}z_{j}{\bf b}_{j}\right\rangle,$

so it is linear in the first term. We have

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

so it is conjugate symmetric. Lastly

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

and recall that $x_{i}\overline{x_{i}}>0$ for $x_{i}\neq 0$ so it is also positive definite. Thus we have an inner product.

Problem 34. Fill in the remaining details of the proof of Proposition 5.10, i.e. show that $\langle\cdot,\cdot\rangle_{G}$ is linear in the first argument, conjugate-symmetric, and positive-definite.

Solution: We repeatedly use the fact that $\langle\cdot,\cdot\rangle$ is an inner product. Consider ${\bf u},{\bf v},{\bf w}\in V$ and $\lambda\in{\mathbb{C}}$ , then

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

so it is linear in the first term. We have

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

so it is conjugate symmetric. Lastly

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

and each $\langle\pi(h){\bf v},\pi(h){\bf v}\rangle$ is greater than zero for $\pi(h){\bf v}\neq 0$ and zero otherwise. Thus $\langle\cdot,\cdot\rangle_{G}$ is positive definite and therefore an inner product.

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 a orthonormal basis for $\langle\cdot,\cdot\rangle$ is unitary (recall that a matrix is unitary if $M^{*}(=\overline{M}{\,{}^{\mathrm{t}}\!}\,)=M^{-1}$ ).

Solution: Let $\{{\bf e}_{1},\ldots,{\bf e}_{n}\}$ be an orthonormal basis of $V$ and let $M$ be the matrix of $\pi(g)$ with respect to this basis, for some $g\in G$ . As $\langle\cdot,\cdot\rangle$ is $G$ -invariant we have

\langle M{\bf e}_{i},M{\bf e}_{j}\rangle=\langle{\bf e}_{i},{\bf e}_{j}\rangle% =\begin{cases}1\quad&\text{if }i=j,\\ 0\quad&\text{if }i\neq j.\end{cases}

So $\{M{\bf e}_{i}\}_{1\leq i\leq n}$ is also an orthonormal basis for $V$ . Note that $M{\bf e}_{i}$ is a column of the matrix $M$ , hence the columns of $M$ are an orthonormal basis of $V$ . Thus $M\overline{M^{t}}={\,\mathrm{Id}}$ and we have that our matrix is unitary.

Problem 36. Show the following:
Lemma 5.13. Let $(\pi_{1},U)$ , $(\pi_{2},V)$ , and $(\pi_{3},W)$ be representations of a group $G$ . Then

(i)

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

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

Solution: Given $R\in\operatorname{Hom}(U\oplus V,W)$ , we we define $\varphi(R)=(\varphi_{1}(R),\varphi_{2}(R))\in\mathrm{Hom}(U,W)\oplus\mathrm{% Hom}(V,W)$ by

\varphi_{1}(R)u=R(u,0),\quad\varphi_{2}(R)v=R(0,v)\qquad\forall u\in U,\,v\in V.

Similarly, define a map $\widetilde{\varphi}:\operatorname{Hom}(U,W)\oplus\operatorname{Hom}(V,W)% \rightarrow\operatorname{Hom}(U\oplus V,W)$ by

\widetilde{\varphi}(S,T)(u,v)=Su+Tv\qquad\forall(u,v)\in U\oplus V.

Routine linear algebra verifications show that all the maps above are linear in the relevant variables, and that $\varphi\widetilde{\varphi}(S,T)=(S,T)$ and $\widetilde{\varphi}\varphi(R)=R$ , giving the identity

\mathrm{Hom}(U\oplus V,W)=\mathrm{Hom}(U,W)\oplus\mathrm{Hom}(V,W).

It remains to show that $\varphi(\operatorname{Hom}_{G}(U\oplus V,W))\subset\mathrm{Hom}_{G}(U,W)\oplus% \mathrm{Hom}_{G}(V,W)$ , and $\widetilde{\varphi}\big{(}\mathrm{Hom}_{G}(U,W)\oplus\mathrm{Hom}_{G}(V,W)\big% {)}\subset\mathrm{Hom}_{G}(U\oplus V,W)$ . Given $R\in\operatorname{Hom}_{G}(U\oplus V,W)$ , we verify that $\varphi(R)$ intertwines with the $G$ -actions: for all $u\in U$ ,

\displaystyle\varphi_{1}(R)(\pi_{1}(g)u)=R(\pi_{1}(g)u,0)=R(\pi_{1}(g)u,\pi_{2% }(g)0)=R\big{(}\pi_{1}\oplus\pi_{2}(g)(u,0)\big{)}=\pi_{3}(g)R(u,0)=\pi_{3}% \varphi_{1}(R)u,

showing that $\varphi_{1}(R)\in\operatorname{Hom}_{G}(U,W)$ . A similar computation gives $\varphi_{2}(R)\in\operatorname{Hom}_{G}(V,W)$ , hence $\varphi(R)=(\varphi_{1}(R),\varphi_{2}(R))\in\mathrm{Hom}_{G}(U,W)\oplus% \mathrm{Hom}_{G}(V,W)$ . In the other direction, let $S\in\operatorname{Hom}_{G}(U,W)$ and $T\in\operatorname{Hom}_{G}(V,W)$ . Then for all $(u,v)\in U\oplus V$ ,

	$\displaystyle\widetilde{\varphi}(S,T)\pi_{1}\oplus\pi_{2}(g)(u,v)=\widetilde{% \varphi}(S,T)(\pi_{1}(g)u,\pi_{2}(g)v)=S\pi_{1}(g)u+T\pi_{2}(g)v$
	$\displaystyle=\pi_{3}(g)Su+\pi_{3}(g)Tv=\pi_{3}(g)(Su+Tv)=\pi_{3}(g)\widetilde% {\varphi}(S,T)(u,v),$

showing that $\widetilde{\varphi}(S,T)\in\operatorname{Hom}_{G}(U\oplus V,W)$ . (ii) is proved analogously.