6 Inner product & Schur's orthogonality relations

Definition

Define the following inner product on $F_C(G,\mathbb{C})$ as

$$⟨f_1,f_2⟩=\frac{1}{|G|}\sum _{g\in G}{f}_1(g)\overline{f_2(g)}.$$

Remark. Since it is on $F_C(G,\mathbb{C})$, we can also write

$$⟨f_1,f_2⟩=\frac{1}{|G|}\sum _C|C|f_1(C)\overline{f_2(C)},$$

where $C$ runs over all conjugacy classes of $G$.

Schur's orthogonality relations

The irreducible characters of $G$ form an orthonormal basis with respect to inner product introduced above.

To prove theorem first we talk about dual representation and tensor product of representation. The proof is written here.

Dual Representation

Let $V$ be a vector space of dimension $n$, then the dual space $V^{\ast }=\mathrm{Hom}(V,\mathbb{C})$ is also vector space of dimension $n$. Let $v_1,\cdots ,v_n$ be a basis of $V$, it is easy to prove that $v_1^{\ast },\cdots ,v_n^{\ast }\in V^{\ast }$ defined as $v_i^{\ast }(v_j)={\delta }_{ij}$ is a basis of $V^{\ast }$. If $V$ is a $G$-module, then $V^{\ast }$ also has a structure of $G$-module defined as

$$(gf)(v)=f(g^{-1}(v)),\text{mbox}forv\in V,f\in \mathrm{Hom}(V,\mathbb{C}),g\in G.$$

The relations between matrices of $V$ and $V^{\ast }$ is the following: Suppose $g\in G$ has matrices as $M(g)=[a_{ij}(g)]$ and $M^{\ast }(g)=[a_{kl}^{\ast }(g)]$ relative to these two bases. Then we have

$$a_{kl}^{\ast }(g)=(g{v}_l^{\ast })(v_k)=v_l^{\ast }(g^{-1}{v}_k)=a_{lk}(g^{-1}),$$

and so $M^{\ast }(g)=M(g^{-1}{)}^T=(M(g{)}^{-1}{)}^T$. In particular,

$${\chi }_{V^{\ast }}(g)=\mathrm{tr}(M^{\ast }(g))=\mathrm{tr}(M(g{)}^{-1})={\chi }_V(g^{-1}).$$

Lemma

${\chi }_V(g^{-1})=\overline{{\chi }_V(g)}$.

\begin{proof} Let $M(g)$ be matrix of $g$, and let ${\lambda }_1,\cdots ,{\lambda }_n$ be eigenvalues . Since $G$ is finite, then all $\lambda$ has to be root of unity. Therefore, matrix $M(g^{-1})$ has eigenvalues ${\lambda }_1^{-1},\cdots ,{\lambda }_n^{-1}$ where ${\lambda }_i^{-1}=\overline{{\lambda }_i}$. Hence, ${\chi }_V(g^{-1})=\overline{{\chi }_V(g)}$. \end{proof}

Corollary

On the characters, inner product is symmetric.

\begin{proof} Note that

$$⟨{\chi }_V,{\chi }_W⟩=\frac{1}{|G|}\sum _{g\in G}{\chi }_V(g){\chi }_W(g^{-1})=\frac{1}{|G|}\sum _{h\in G}{\chi }_V(h^{-1}){\chi }_W(h)=⟨{\chi }_W,{\chi }_V⟩.$$

Therefore, the inner product is symmetric. \end{proof}

Tensor Product

Definition

Given two vector spaces $V$ and $W$, we define their tensor product a vector space $V\otimes W$ which consists of linear combinations $\sum c_{ij}{v}_i\otimes w_j$ subject to

• $(c_1{v}_1+c_2{v}_2)\otimes w=c_1(v_1\otimes w)+c_2(v_2\otimes w)$, and
• $v\otimes (b_1{w}_1+b_2{w}_2)=b_1(v\otimes w_1)+b_2(v\otimes w_2)$.

Remarks.

• If $A$ and $B$ are two algebras, then $A\otimes B$ is also an algebra, where $(a\otimes b)(a^{\prime }\otimes b^{\prime })=a{a}^{\prime }\otimes b{b}^{\prime }$.
• $\mathrm{End}(V)\otimes \mathrm{End}(W)=\mathrm{End}(V\otimes W)$,
• $\mathrm{Hom}(V,W)=W\otimes V^{\ast }$.
• ${\chi }_{\mathrm{Hom}(V,W)}=\overline{{\chi }_V}{\chi }_W$.

If $v_1,\cdots ,v_n$ is a basis of $V$ and $w_1,\cdots ,w_m$ is a basis of $W$, we claim that $v_i\otimes w_j$ is a basis of $V\otimes W$. Since $\mathrm{Hom}(V\otimes W,\mathbb{C})=(V\otimes W{)}^{\ast }$. Define ${\phi }_{ij}$ by ${\phi }_{ij}(v_k\otimes w_{\ell })={\delta }_{ik}\cdot {\delta }_{j\ell }$, then ${\phi }_{ij}$ is a basis of $(V\otimes W{)}^{\ast }$ and so $\dim (V\otimes W)=nm$. Therefore, $v_i\otimes w_j$ is a basis.

tensor product of algebras

Let $A$, $B$ be algebras. Define the algebra $A\otimes B$ as $(a\otimes b)(a^{\prime }\otimes b^{\prime })=a{a}^{\prime }\otimes b{b}^{\prime }$. Then $\mathrm{End}V\otimes \mathrm{End}W\simeq \mathrm{End}(V\otimes W)$, where $(\phi \otimes \psi )(v\otimes w)=\phi (v)\otimes \psi (w)$. Furthermore, $\mathrm{End}(V\otimes W)$ is also an algebra, defined by $(\phi \otimes \psi )({\phi }^{\prime }\otimes {\psi }^{\prime })=\phi {\phi }^{\prime }\otimes \psi {\psi }^{\prime }$.

Definition

If $A,B$ are square matrices of $n\times n$ and $m\times m$, respectively. Then tensor product $A\otimes B$ is a block matrix

$$\left[\begin{array}{ccc}{a}_{11}B& \cdots & a_{1b}B\\ & & \\ ⋮& & ⋮\\ & & \\ a_{n1}B& \cdots & a_{nn}B\end{array}\right].\text{\hspace{1em}(*)}$$

Proposition

Let $\phi \in \mathrm{End}V$ with matrix $A$ with respect to basis $v_1,\cdots ,v_n$ and $\psi \in \mathrm{End}W$ with matrix $B$ with respect to basis $w_1,\cdots w_m$, then $\phi \otimes \psi \in \mathrm{End}(V\otimes W)$ will have matrix $A\otimes B$ with respect to basis $v_i\otimes w$ written in lexicographical order $v_1\otimes w_1,v_1\otimes w_2,\cdots ,v_n\otimes w_m$.

Definition

Let $V$ be a $G$-module and $W$ be a $H$-module, and let $(\rho ,V)$ and $(\pi ,W)$ be representations of $G,H$, respectively. Then the (outer) tensor product of $V$ and $W$, will be $G\times H$-module $V\otimes W$, where $(g,h)(v\otimes w)=(gv)\otimes (hw)$.

If $G=H$, then the (inner) tensor product of $V$ and $W$ is $g(v\otimes w)=(g,g)(v\otimes w)=gv\otimes gw$.

Proposition

Let $(\rho ,V)$ and $(\pi ,W)$ be representations of $G$ and $H$, respectively. Then ${\chi }_{\rho \otimes \pi }(g,h)={\chi }_{\rho }(g){\chi }_{\pi }(h)$.

\begin{proof} Easy by $(\ast )$. \end{proof}

Now consider the representation of $\mathrm{Hom}(V,W)$.

Proposition

Let $V$ and $W$ be $G$-modules, then $\mathrm{Hom}(V,W)$ has a structure of $G$-module defined as $(g\phi )(v):=g\phi (g^{-1}v)$. Furthermore, $\mathrm{Hom}(V,W)\simeq V^{\ast }\otimes W$.

\begin{proof} To check this, first we show that $\mathrm{Hom}(V,W)\simeq V^{\ast }\otimes W$ as vector space. Let $v_1,\cdots ,v_n$ be a basis of $V$, and let $w_1,\cdots ,w_m$ be a basis of $W$. Define $v_i^{\ast }(v_j)={\delta }_{ij}$, and define

$$\Phi :V^{\ast }\otimes W\to \mathrm{Hom}(V,W),v_i^{\ast }\otimes w_j\to {\phi }_{ij}$$

where ${\phi }_{ij}(v_k)={\delta }_{ik}{w}_i$. It is easy to check that $\Phi$ is an isomorphism. Therefore, $\mathrm{Hom}(V,W)\simeq V^{\ast }\otimes W$.

Now we claim that $V^{\ast }\otimes W$ has structure of $G$-module. Define $g(f\otimes w)=(gf)\otimes (gw)$ where $gf:v\mapsto f(g^{-1}v)$. Using our isomorphism $\Phi$, define $\phi =\Phi (f\otimes w)$. If $\mathrm{Hom}(V,W)\simeq V^{\ast }\otimes W$ as modules, then we have $g\Phi (f\otimes w)=\Phi (g(f\otimes w))$ and so

$$g\phi (v)=g\Phi (f\otimes w)(v)=\Phi (gf\otimes gw)(v)=(gf)(v)(gw)=f(g^{-1}v)(gw)=g(f(g^{-1}v)w)=g\phi (g^{-1}v)$$

by $\phi (v)=f(v)w$. \end{proof}

For characters, ^65b04d gives

$${\chi }_{\mathrm{Hom}(V,W)}(g)={\chi }_{V^{\ast }\otimes W}(g)={\chi }_{V^{\ast }}(g){\chi }_W(g)=\overline{{\chi }_V(g)}{\chi }_W(g).$$

Definition

For any $G$-module $V$, define $V^G=\{v\in V:gv=v\text{mbox}forallg\}$.

By Maschke’s theorem, for $V=V_1\oplus \cdots \oplus V_r$ with irreducible $V_i$‘s, $V^G=V_1^G\oplus \cdots \oplus V_r^G$ is a submodule. Since each $V_i$ is irreducible, $V_i^G$ is either $0$ or $\mathbb{C}$. Note that $V_i^G\simeq \mathbb{C}$ iff $V_i\cong \mathbb{C}$ iff $V_i$ is trivial representation. It follows that $V^G$ says how many isotypical component isomorphic to trivial representation are in $V$. Define ${\psi }_1(g)(v)=\frac{1}{|G|}{\sum }_{g\in G}gv$ for all $g\in G$. Note that ${\psi }_1(g)$ is a projection of $V$ onto $V^G$, which is zero on any non-trivial irreducible submodule and identity on $V^G$. Hence, $\dim V^G=\mathrm{tr}{\psi }_1(g)=\frac{1}{|G|}{\sum }_{g\in G}{\chi }_V(g)=⟨{\psi }_1,\mathrm{id}⟩$.

Remark. Note that ${\chi }_V(g)=\mathrm{tr}(\rho (g))$ where $\rho :G\to \mathrm{GL}(V)$ satisfies $gv:=\rho (g)v$.

Now we are ready to prove ^rld430.

\begin{proof} For any $g\in G$ and $\phi \in \mathrm{Hom}(V,W)$ with $g\phi =\phi$, there is $\phi (v)=(g\phi )(v)=g\phi (g^{-1}v)$ and so $\phi (g^{-1}v)=g^{-1}\phi (v)$. It follows that $\phi \in {\mathrm{Hom}}_{\mathbb{C}G}(V,W)$. Note that

$$\dim {\mathrm{Hom}}_{\mathbb{C}G}(V,W)=\dim \mathrm{Hom}(V,W{)}^G=\frac{1}{|G|}\sum _{g\in G}{\chi }_{\mathrm{Hom}(V,W)}(g)=\frac{1}{|G|}\sum _{g\in G}\overline{{\chi }_V(g)}{\chi }_W(g)=⟨{\chi }_W,{\chi }_V⟩.$$

By Schur’s lemma, $\dim {\mathrm{Hom}}_{\mathbb{C}G}(s_i,s_j)={\delta }_{ij}$ for irreducible representations $s_i$ and $s_j$. Now we finish the proof. \end{proof}

Corollary

A $G$-module $V$ is simple iff $⟨{\chi }_V,{\chi }_V⟩=1$.

\begin{proof} By Maschke’s theorem, if $V={\alpha }_1{S}_1\oplus \cdots \oplus {\alpha }_r{S}_r$, then $⟨{\chi }_V,{\chi }_V⟩={\alpha }_1^2+\cdots +{\alpha }_r^2$. Hence $V$ is simple iff ${\alpha }_1^2+\cdots +{\alpha }_r^2=1$ iff $⟨{\chi }_V,{\chi }_V⟩=1$ iff there is the unique ${\alpha }_i=1$ and all others equal $0$. \end{proof}