Lecture 14

Solution:

1. (a)
   ​

   Let $\{{\bf e}_{1},\ldots,{\bf e}_{m}\}$ be a basis of $V$ and let $\{r_{1},\ldots,r_{n}\}$ be a set of representatives of left cosets of $H\leq G$. Then define the function $f_{i,j}\in C_{\pi}(G,V)$ to be $f_{i,j}(r_{i})={\bf e}_{j}$ and zero on all other coset representatives. This is well defined as, for any $g\in G$, we can write $g=r_{k}h$, for some unique $r_{k}$ a representative and some unique $h\in H$, and

   $$f_{i,j}(g)=f_{i,j}(r_{k}h)=\pi(h^{-1})f_{i,j}(r_{k})=\begin{cases}\pi(h^{-1}){\bf e}_{j}&\text{ if }i=k\\ 0&\text{ otherwise.}\end{cases}.$$

   It is easy to see that if $i\neq i^{\prime}$ or $j\neq j^{\prime}$ then $f_{i,j}$ and $f_{i^{\prime},j^{\prime}}$ are linearly independent. Additionally if $\gamma\in C_{\pi}(G,V)$, then for any $g=r_{i}h\in G$ we have

   $$\gamma(g)=\pi(h^{-1})\gamma(r_{i})=\pi(h^{-1})\sum_{1\leq j\leq m}\mu_{j}{\bf e}_{j}=\pi(h^{-1})\sum_{1\leq j\leq m}\mu_{j}f_{i,j}(r_{i})=\sum_{1\leq j\leq m}\mu_{j}f_{i,j}(g),$$

   for some $\mu_{j}$s in ${\mathbb{C}}$. Thus if we range through all $g\in G$ we get

   $$\gamma(g)=\sum_{\begin{subarray}{c}1\leq i\leq n\\ 1\leq j\leq m\end{subarray}}\mu_{i,j}f_{i,j}(g),$$

   and we have a basis of $C_{\pi}(G,V)$. The dimension of $C_{\pi}(G,V)$ is therefore ${nm=[H:G]\operatorname{dim}V}$.

2. (b)
   ​

   Let $g\in G$, then

   	$\displaystyle[\lambda(g)(f+\mu f^{\prime})](g_{0})$	$\displaystyle=(f+\mu f^{\prime})(g^{-1}g_{0})$
   		$\displaystyle=f(g^{-1}g_{0})+\mu f^{\prime}(g^{-1}g_{0})=[\lambda(g)(f)](g_{0})+\mu[\lambda(g)(f^{\prime})](g_{0})$

   for all $g_{0}\in G$, all $f,f^{\prime}\in C_{\pi}(G,V)$ and all $\mu\in{\mathbb{C}}$. Thus the image of $\lambda$ is in $\operatorname{GL}(C_{\pi}(G,V))$. Let $g,\tilde{g}\in G$, then

   $$[\lambda(g\tilde{g})(f)](g_{0})=f(\tilde{g}^{-1}g^{-1}g_{0})=[\lambda(\tilde{g})f](g^{-1}g_{0})=[\lambda(g)[\lambda(\tilde{g})f]](g_{0}),$$

   for all $g_{0}\in G$ and all $f\in C_{\pi}(G,V)$. Thus $\lambda$ is a group homomorphism and hence $\big{(}\lambda,C_{\pi}(G,V)\big{)}$ is a $G$-representation.

3. (c)
   ​

   Let $f,f^{\prime}\in C_{\pi}(G,V)$ and $\mu\in{\mathbb{C}}$ then

   	$\displaystyle T_{{\bf r}}(f+\mu f^{\prime})$	$\displaystyle=r_{1}(f+\mu f^{\prime})(r_{1})+\ldots+r_{n}(f+\mu f^{\prime})(r_{n})$
   		$\displaystyle=r_{1}(f)(r_{1})+\ldots+r_{n}(f)(r_{n})+\mu\left(r_{1}(f^{\prime})(r_{1})+\ldots+r_{n}(f^{\prime})(r_{n})\right)$
   		$\displaystyle=T_{{\bf r}}(f)+\mu T_{{\bf r}}(f^{\prime}),$

   and so $T_{{\bf r}}$ is a linear map. Let $f\in\ker T_{\bf r}$ then

   $$T_{{\bf r}}(f)=r_{1}f(r_{1})+\ldots+r_{n}f(r_{n})=0,$$

   so each $f(r_{i})=0$. As we can write any $g\in G$ as $g=r_{i}h$ for some $h\in H$ and $f(r_{i}h)=\pi(h^{-1})f(r_{i})$ we have that $f(g)=0$ for all $g\in G$. Thus $\ker T_{\bf r}=0$. As $\operatorname{dim}V_{\bf r}=[G:H]\operatorname{dim}V=\operatorname{dim}C_{\pi}(G,V)$ it must be an isomorphism of vector spaces. Lastly let $g\in G$, then for all $f\in C_{\pi}(G,V)$

   	$\displaystyle T_{\bf r}(\lambda(g)f)$	$\displaystyle=r_{1}[\lambda(g)f](r_{1})+\ldots+r_{n}[\lambda(g)f](r_{n})$
   		$\displaystyle=r_{1}f(g^{-1}r_{1})+\ldots+r_{n}f(g^{-1}r_{n})$
   		$\displaystyle=r_{j(g,1)}f(g^{-1}r_{j(g,1)})+\ldots+r_{j(g,n)}f(g^{-1}r_{j(g,n)})$
   		$\displaystyle=r_{j(g,1)}f(r_{j(g^{-1},j(g,1))}h_{(g^{-1},j(g,1))})+\ldots+r_{j(g,n)}f(r_{j(g^{-1},j(g,n))}h_{(g^{-1},j(g,n))})$
   		$\displaystyle=r_{j(g,1)}f(r_{j(e,1)}h_{(e,1)}h_{(g,1)}^{-1})+\ldots+r_{j(g,n)}f(r_{j(e,n)}h_{(e,n)}h_{(g,n)}^{-1})$
   		$\displaystyle=r_{j(g,1)}f(r_{1}h_{(g,1)}^{-1})+\ldots+r_{j(g,n)}f(r_{n}h_{(g,n)}^{-1})$
   		$\displaystyle=r_{j(g,1)}\pi(h_{(g,1)})f(r_{1})+\ldots+r_{j(g,n)}\pi(h_{(g,n)})f(r_{n})$
   		$\displaystyle=\mathrm{Ind}_{H}^{G}\pi(g)\left(r_{1}f(r_{1})+\ldots+r_{n}f(r_{n})\right)$
   		$\displaystyle=\mathrm{Ind}_{H}^{G}\pi(g)T_{\bf r}(f),$

   so $T_{\bf r}$ is a $G$-homomorphism. Thus it is a $G$-isomorphism and the two representations are isomorphic.