Lecture 2

Solution: Let $w\in W_{1}\cap W_{2}\subset W_{1}$. Since $W_{1}$ is a subrepresentation of $(\pi,V)$, $\pi(g)w\in W_{1}$ for all $g\in G$. Similarly, $\pi(g)w\in W_{2}$ for all $g\in G$, and hence $\pi(g)w\in W_{1}\cap W_{2}$. $(\pi,W_{1}\cap W_{2})$ is therefore a subrepresentation of $(\pi,W_{1})$ and also $(\pi,W_{2})$. Since $(\pi,W_{1})$ is irreducible, we then have $W_{1}\cap W_{2}=0$ or $W_{1}$. If $W_{1}\cap W_{2}=0$, we are done. If not, then $W_{1}\cap W_{2}=W_{1}$, hence $W_{1}\subset W_{2}$. However, $(\pi,W_{2})$ is also irreducible, so the same applies: $W_{1}\cap W_{2}=0$ or $W_{2}$, with the latter case giving $W_{2}\subset W_{1}$. Since we assumed $W_{1}\cap W_{2}\neq 0$, we then have $W_{1}\subset W_{2}$ and $W_{2}\subset W_{1}$, i.e. they are equal.

Solution: Since $T$ is an invertible linear map, $T^{-1}\in\operatorname{Hom}(W,V)$. It remains to check that in fact $T^{-1}\in\operatorname{Hom}_{G}(W,V)$. For any $w\in W$ there exists $v\in V$ such that $w=Tv$. This gives

as required.

Solution: Let ${\mathcal{B}}=\{v_{1},\ldots,v_{n}\}$ be a basis of $V$, and given $g\in G$, define $\widetilde{\pi}(g)\in{\mathbb{C}}^{n\times n}$ by

i.e. the $i$-th column is the vector $[\pi(g)v_{i}]_{{\mathcal{B}}}\in{\mathbb{C}}^{n}$; this is the coordinate vector of $\pi(g)v_{i}$ w.r.t. the basis ${\mathcal{B}}$. Since $\widetilde{\pi}(g)$ is a matrix, the map $z\mapsto\widetilde{\pi}(g)z$ is linear on ${\mathbb{C}}^{n}$. We wish to show that $(\widetilde{\pi},{\mathbb{C}}^{n})$ is a representation that is isomorphic to $(\pi,V)$. To do this, we consider the linear map $T:V\rightarrow{\mathbb{C}}^{n}$ defined on the basis elements by $Tv_{i}=e_{i},$ where $\{e_{i}\}$ is the standard basis of ${\mathbb{C}}^{n}$. By construction, $T$ is an invertible linear map. We verify that it is in fact an intertwining operator: let $v=\sum_{i}a_{i}v_{i}$. Then for any $g\in G$,

On the other hand,

This shows that $\widetilde{\pi}(g)T=T\pi(g)$ for all $g\in G$. We now use this to prove that $\widetilde{\pi}$ is a homomorphism; this will conclude the proof, as the above shows that $T$ is an invertible intertwining operator. Given $z\in{\mathbb{C}}^{n}$, we let $z=Tv$. Then for any $g,h\in G$,

showing that $(\widetilde{\pi},{\mathbb{C}}^{n})$ is a representation, with $T\in\operatorname{Hom}_{G}(V,{\mathbb{C}}^{n})$ being invertible.

Remark: This result actually follows from a stronger result in linear algebra: for any two operators $S,T$ on $V$, we have $[ST]_{{\mathcal{B}}}=[S]_{{\mathcal{B}}}[T]_{{\mathcal{B}}}$. This shows that $\operatorname{GL}(V)$ is isomorphic to $\operatorname{GL}_{n}({\mathbb{C}})$ as groups, and that the map $v\mapsto[v]_{{\mathcal{B}}}$ is an intertwining operator between the natural representations of $\operatorname{GL}(V)$ on $V$ and $\operatorname{GL}_{n}({\mathbb{C}})$ on ${\mathbb{C}}^{n}$.

Solution: Let ${\bf v}_{1}={\bf e}_{1}+\ldots+{\bf e}_{n}$. Then $\pi(\sigma){\bf v}_{1}={\bf v}_{1}$ for all $\sigma\in S_{n}$, so $(\pi,{\mathbb{C}}{\bf v}_{1})$ is a subrepresentation of $(\pi,{\mathbb{C}}^{n})$. Being 1-dimensional, it is irreducible.

Denote by $W_{0}$ the subspace of ${\mathbb{C}}^{3}$ consisting of all vectors $(z_{1},\ldots,z_{n})\,^{\mathrm{t}}\!$ such that $z_{1}+\ldots+z_{n}=0$. Then since $S_{n}$ permutes the coordinates of a vector, it doesn’t change their sum; so $\pi(\sigma)W_{0}\subseteq W_{0}$, and $W_{0}$ is a subrepresentation (of dimension $n-1$).

We claim that $W_{0}$ is irreducible. Indeed, suppose that $U\subseteq W_{0}$ is a nonzero subrepresentation; we have to show that $U=W_{0}$. Let $(x_{1},\ldots,x_{n})\,^{\mathrm{t}}\!\in U$ be nonzero. As $x_{1}=\ldots=x_{n}$ can’t happen, we can apply an element of $S_{n}$ to permute the coordinates so that $x_{1}\neq x_{2}$. Then applying $\pi((12))$, we have $(x_{2},x_{1},x_{3},\ldots,x_{n})\,^{\mathrm{t}}\!\in U$. Taking the difference, $(x_{1}-x_{2},x_{2}-x_{1},0,\ldots,0)\,^{\mathrm{t}}\!\in U$; scaling, $(1,-1,0,\ldots 0)\,^{\mathrm{t}}\!\in U$. Applying $\pi((2i))$, for $3\leq i\leq n$, we get a family of vectors with $1$ as their first entry, $-1$ as the $i$th entry and $0$s elsewhere, each in $U$. But these vectors span $W_{0}$ (e.g. because they are linearly independent and $\operatorname{dim}W_{0}=n-1$), so $U=W_{0}$ as required.

Solution: Let $\phi$ be the map

This map is well defined as all elements of ${\mathbb{C}}(X)$ have only finitely many non-zero terms. This map is linear as

for all $\sum_{x\in X}z_{x}x,\sum_{x\in X}y_{x}x\in{\mathbb{C}}(X)$, $w\in X$ and $\mu\in k$. It is a $G$-homomorphism as

and

for all $w\in X$, $\sum_{x\in X}z_{x}x\in{\mathbb{C}}(X)$ and $g\in G$.

Suppose $\sum_{x\in X}z_{x}x\in\ker\phi$, then $\phi\left(\sum_{x\in X}z_{x}x\right)(w)=0$ for all $w\in X$ and thus all coefficients $z_{x}=0$. Therefore $\sum_{x\in X}z_{x}=0$ and $\ker\phi$ is trivial. Now let $g\in C_{0}(X)$ and consider $\sum_{x\in X}g(x)x\in{\mathbb{C}}(X)$. Then, for all $w\in X$,

Thus $\phi\left(\sum_{x\in X}g(x)x\right)=g$ and as $g$ was arbitrary $\phi$ is a bijection. Thus $\phi$ is a $G$-isomorphism and the two representations are isomorphic.