1 Schur Lemma

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Definition

A representation of group $G$ is a mapping $\rho :G\to {\mathrm{End}}_{k}V$ which satisfies the following properties

• for any $g_1$, $g_2\in G$, $\rho (g_1{g}_2)=\rho (g_1)\rho (g_2)$,
• $\rho (e)=\mathrm{Id}$ where $\mathrm{Id}:V\to V,v\mapsto v$,

where $k$ is a field and ${\mathrm{End}}_{k}V$ is the set of linear transformation on $V$.

In other words, representation of $G$ on $V$ is a homomorphism $\rho :G\to \mathrm{GL}(V)$. Also we call $V$ a $G$-module.

Definition

Let $V$ and $W$ be two $G$-modules, and let $\rho$ and $\sigma$ be representations in $V$ and $W$, respectively. A map between two $G$-modules is a linear transformation $T:V\to W$ such that for any $g\in G$ the diagram

$$\text{require}AMScd\begin{array}{ccc}V& \stackrel{T}{\to }& W\\ \rho (g)\downarrow & & \downarrow \sigma (g)\\ V& \underset{T}{\to }& W\end{array}$$

is commutative. That is, $\sigma (g)\circ T=T\circ \rho (g)$ for all $g\in G$. If $T$ is invertible, we say that $\rho$ and $\sigma$ are isomorphic or equivalent.

dual representation

Suppose $V$ is a $G$-module. Let $V^{\ast }={\mathrm{Hom}}_k(V,k)$, and let ${\rho }^{\ast }$ is a representation of $G$ on $V^{\ast }$. For any $f\in {\mathrm{Hom}}_k(V,k)$, define ${\rho }^{\ast }(g)f=f\circ \rho (g^{-1})$. Then ${\rho }^{\ast }$ is a representation of $G$ in $V^{\ast }$.

\begin{proof} See here. \end{proof}

Definition

We call a $G$-module $V$ irreducible, if $V$ and $\{0\}$ are only $G$-invariant subspaces.

For example, let $G=S_n$ and $\sigma :G\to {\mathrm{GL}}_n(k),\sum {\alpha }_i{e}_i\mapsto \sum {\alpha }_i{e}_{\sigma (i)}$. Then $D:=\{(\alpha ,\cdots ,\alpha ):\alpha \in k\}$ and $W:=\{({\alpha }_1,\cdots ,{\alpha }_n):\sum {\alpha }_i=0\}$ are two $G$-invariant subspaces, and $V=D\oplus W$.

Schur lemma

A morphism between irreducible representations of groups is either zero or isomorphism.

If $k$ is algebraically closed field, $\rho$ is a irreducible representation of $G$ in $V$, $\dim V<\infty$ and $S:V\to V$ morphism of $G$-modules, then there exists $\lambda \in k$ such that $S=\lambda \mathrm{Id}$.

\begin{proof} i) Let $T:V\to W$ be a morphism with $V,W$ irreducible. Since $\ker T$ and $\mathrm{Im}T$ are submodules of $V$ and $W$, respectively, then one of the following holds:

• $\ker T=V$, $\mathrm{Im}T=0$
• $\ker T=0$, $\mathrm{Im}T=W$

Thus either $T$ is zero or isomorphism.

ii) Let $\lambda$ be a root of the characteristic polynomial $p_S(\lambda )$. Then $\ker (S-\lambda I)\ne \{0\}$ and $S-\lambda I$ is also a morphism. By i) we have that $\ker (S-\lambda I)=V$. Therefore, $S=\lambda I$. \end{proof}

Lemma

Any irreducible representation of finite group is of finite dimension.

\begin{proof} Let $V$ be a irreducible representation of $G$. For any nonzero $v\in V$, define $W:=⟨v⟩=\{gv:g\in G\}$. Then $\dim W<\infty$ by $|G|<\infty$. Since $W$ is $G$-invariant, we have that $W=V$ and so $V$ is of finite dimension. \end{proof}