424 Theorems

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1 Finite groups
2 Compact groups
- 2.1 Haar measure

Finite groups

Dimension of simple representation is less than or equal to order of the group

Take any $v\neq 0$ in $V$ and consider the subspace spanned by the transforms of $v$ , $U = \lbrace gv: g \in G \rbrace$ . Each element of $G$ permutes these transforms, as $g(hv) = (gh)v$ , hence $g$ sends $U$ into itself, so $U$ is stable and hence $V = U$ as our representation is simple. But $U$ is spanned by the $||G||$ transforms of $v$ , so $\mbox{dim}\,V = \mbox{dim}\,U \leq ||G||$

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Simple representation of a finite abelian group over C is necessarily one-dimensional

Let $G$ be abelian, and $g \in G$ with eigenvalue $\lambda$ and corresponding eigenspace $E_{\lambda} = \lbrace v \in V | gv = \lambda v\rbrace$ . Then $E_{\lambda}$ is stable under $G$ , as $(gh)v = (hg) v = h (gv) = h( \lambda v) = \lambda (hv) \Rightarrow hv \in E_{\lambda}$ . As our representation is simple, it follows that $E_{\lambda} = V$ and that $g$ acts as a scalar multiple $\lambda I$ of the identity, hence every subspace of $V$ is stable under $g$ and so under $G$ , and as our representation is simple this means that $V$ has no proper subspaces, true only if $\mbox{dim}\,V = 1$ .

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If V is a sum of simple subspaces then V is semisimple

We suppose that $V$ is a sum (not necessarily direct) of simple subspace, $V = S_1 + \dots + S_r$ .

Now, since $S_2$ is simple, $S_1 \cap S_2 = 0$ or $S_2$ , i.e. either they do not intersect or else $S_2$ is a subspace of $S_1$ . Hence $S_1 + S_2 = S_1 \oplus S_2$ or $S_1$ . Similarly, $(S_1+S_2)\cap S_3 = 0$ or $S_3$ , so $S_1 + S_2 + S_3 = (S_1+S_2) \oplus S_3$ or $S_1+S_2$ , i.e. $S_1 + S_2 + S_3 = S_1 \oplus S_2 \oplus S_3$ or $S_1 \oplus S_3$ or $S_1 \oplus S_2$ or $S_1$ .

Continuing in this way we find that $V = S_1 + \dots + S_r = S_{i_1} \oplus \dots \oplus S_{i_s}$ where $\lbrace S_{i_1}, \dots, S_{i_s} \rbrace$ are a subset of $\lbrace S_1, \dots, S_r \rbrace$ .

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A representation is semisimple iff each stable subspace has at least one complementary stable subspace

First let $V$ be semisimple, i.e. $V = S_1 \oplus \dots \oplus S_r$ , and let $U \subset V$ be stable, then $U \cap S_1 = 0$ or $S_1$ , so that $U+S_1 = U \oplus S_1$ or $U$ . Similarly, $(U+S_1)\cap S_2 = 0$ or $S_2$ , and continuing in this way we find $V = U \oplus S_{i_1} \oplus \dots \oplus S_{i_s}$ , then define $W = S_{i_1} \oplus S_{i_s}$ to be the complementary stable subspace.

Conversely let any stable subspace $U \subset V$ have at least one complementary subspace. As $V$ is finite dimensional we can find a stable subspace $S_1$ of minimal dimension, and then by hypothesis $V= S_1 \oplus W_1$ . We then can find a stable subspace $S_2 \subset W_1$ of minimal dimension, with $S_2$ simple. We have $S_1 \cap S_2 \subset S_1 \cap W_1 = 0$ , implying $S_1 + S_2 = S_1 \oplus S_2$ . Now we can find a stable complement $W_2$ such that $V = S_1 \oplus S_2 \oplus W_2$ and continuing in this way we conclude that $V = S_1 \oplus \dots \oplus S_r$ , so $V$ is semisimple.

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Maschke's theorem: representation of a finite group over R or C is semisimple

We will show that for any stable subspace $U$ there exists a complementary stable subspace. First take any positive definite quadratic (or hermitian) form $Q(u,v)$ and average over the group $G$ to obtain an invariant form:

$P(u,v) = \frac{1}{||G||}\sum_{g \in G} Q(gu, gv)$

(Invariant as $P(hu,hv) = \frac{1}{||G||}\sum_{g \in G} Q(ghu, ghv)$ and $gh$ runs over the group as $g$ does.)

Given a stable subspace $U$ , define a complementary subspace $W$ by

$W = \lbrace v \in V : P(u,v) = 0 \, \forall u \in U \rbrace$

This subspace is stable as

$P(u,v) = 0 \Rightarrow P(gu, gv) = P(u,v) = 0 \Rightarrow P(u, gv) = P (g^{-1}u, v) = 0$ (this following from invariance of $P(u,v)$ and stability of $U$ ).

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Intertwining number of simple representations

We want to show that for two simple representations $\alpha, \beta$

$I(\alpha, \beta) = \begin{cases} 0 & \alpha \neq \beta \\ \geq 1 & \alpha = \beta\end{cases}$

Recall that $I(\alpha, \beta) = \mbox{dim}\,\mbox{hom}^G(U,V)$ , ie the dimension of the space of $G$ -maps from $U$ to $V$ . First suppose that $\mbox{dim}\,\mbox{hom}^G(U,V) \geq 1$ , i.e. we have a map $t: U \rightarrow V$ such that $t (gu) = g (tu)$ .

Consider $\mbox{ker}\, t = \lbrace u \in U : tu = 0 \rbrace$ and $\mbox{im}\,t = \lbrace v \in V: \exists u \in U, tu = v \rbrace$ . Both of these subspaces are stable under $G$ , as for $g \in G$ we have

$u \in \mbox{ker}\, t \Rightarrow t (gu) = g (tu) = 0 \Rightarrow gu \in \mbox{ker}\, t$

and

$v \in \mbox{im}\, t \Rightarrow g v = g (tu) = t (gu) \Rightarrow gv \in \mbox{im}\, t$

It follows that $\mbox{ker}\,t = 0$ or $U$ and $\mbox{im}\,t=0$ or $V$ . But $\mbox{ker}\,t=U \Rightarrow t=0$ , and $\mbox{im}\,t =0 \Rightarrow t = 0$ , so if $t \neq 0$ then we have $\mbox{ker}\,t=0$ and $\mbox{im}\,t = V$ , which implies $t$ is an isomorphism between $U$ and $V$ preserving the action of $G$ , and thus that $\alpha = \beta$ .

Conversely, if $\alpha = \beta$ , then there exists a $G$ -map between $U$ and $V$ , and so $\mbox{dim}\,\mbox{hom}^G(U,V) \geq 1$ .

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Intertwining number of simple representation over C

We want to show that for a simple representation $\alpha$ over $\mathbb{C}$ , $I(\alpha, \alpha) = 1$ .

Let $t$ be a $G$ -map with eigenvalue $\lambda$ and corresponding eigenspace $E_{\lambda} = \lbrace v \in V: tv = \lambda v\rbrace$ . Then $t (gv) = g (tv) = g ( \lambda v ) = \lambda (gv) \Rightarrow gv \in E_{\lambda}$ so $E_{\lambda}$ is stable under $G$ , so $E_{\lambda} = V$ . Thus $t$ acts on $V$ as a scalar multiple of the identity, and so the space of $G$ -maps has dimension one, proving the result.

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Formula for intertwining number

We want to prove that $I(\alpha, \beta) = \frac{1}{||G||}\sum_{g \in G} \chi_{\alpha} (g^{-1}) \chi_{\beta} (g) = \frac{1}{||G||}\sum_{g \in G} \chi_{\alpha^*} (g) \chi_{\beta} (g)$ .

Now the left-hand side can be written $I(\alpha, \beta) = I (1, \alpha^* \beta)$ while the right-hand side can be written

$\frac{1}{||G||}\sum_{g \in G} \chi_{\alpha^*} (g) \chi_{\beta} (g) = \frac{1}{||G||}\sum_{g \in G} \chi_{\alpha^* \beta } (g) = \frac{1}{||G||}\sum_{g \in G} \chi_1(g) \chi_{\alpha^* \beta } (g)$

which means that it suffices to prove the result when $\alpha = 1$ .

So we want to show that $I(1, \beta) = \frac{1}{||G||}\sum_{g \in G} \chi_{\beta } (g)$ . Now, by definition $I(1, \beta) = \mbox{dim}\,\mbox{hom}^G (k, V)$ . But $\mbox{hom}\,(k,V) = V$ so $\mbox{hom}^G (k, V) = V^G = \lbrace v \in V : gv =v \rbrace$ . We want to show that $\mbox{dim}\,V^G = \frac{1}{||G||}\sum_{g \in G} \chi_{\beta } (g)$

To do this we define the map $\pi = \frac{1}{||G||} \sum_{g\in G} \beta(g)$ , i.e. $\pi v = \frac{1}{||G||} \sum_{g \in G} g v$ . It is clear that

$h \pi v = \frac{1}{||G||} \sum_{g \in G} hg v = \frac{1}{||G||} \sum_{g \in G} g v = \pi v$

hence $\pi v \in V^G$ , and also that if $v \in V^G$ then $\pi v = v$ , so $\pi$ is a projection onto $V^G$ .

We now use the fact that the trace of a projection equals the dimension of the subspace it projects onto (can easily be seen by taking a basis and considering matrix form of the projection). Hence $\mbox{dim}\,V^G = \mbox{tr}\,\pi = \frac{1}{||G||} \sum_{g \in G} \mbox{tr}\,\beta(g) = \frac{1}{||G||} \sum_{g \in G} \chi_{\beta}(g)$ , proving the result.