10.9 Characteristical Map

For a finite group $G$, its irreducible characters form basis for the space of class functions $F_c(G,\mathbb{C})$. Define $R_n=F_c(S_n,\mathbb{C})$ and $R={\oplus }_{n⩾0}{R}_n$ with $R_0=\mathbb{C}$.

We introduce on the $R$ product. For any $\varphi \in R_n$ and $\psi \in R_m$, where $\varphi$ (rep. $\psi$) is a character of $S_n$ (rep. $S_m$), define $\varphi \cdot \psi \in R_{n+m}$ as

$$\varphi \cdot \psi =(\varphi \otimes \psi ){\uparrow }_{Sn\times S_m}^{S_{n+m}}.$$

Theorem

The product $\cdot$ on $R$ is commutative and associative.

\begin{proof} Define $\rho =\left(\begin{array}{cccccc}1& \cdots & n& n+1& \cdots & m\\ m+1& \cdots & n+m& 1& \cdots & m\end{array}\right)$. Then there is a bijection

$$S_n\times S_m\to S_m\times S_n,\omega \mapsto {\rho }^{-1}\omega \rho .$$

For associativity, use $\left(W{\uparrow }_K^H\right){\uparrow }_H^G=W{\uparrow }_K^G$. \end{proof}

Furthermore, we can equip the algebra $R$ with an inner product

$$⟨f,g⟩=\sum _{n⩾0}{⟨f_n,g_n⟩}_{S_n}.$$

Definition

Define the character map

$$\mathrm{ch}:R\to \Lambda ,f\mapsto \frac{1}{n!}\sum _{\omega \in S_n}f(\omega )p_{\omega }.$$

Remark. For any $f\in R_n$ and any conjugacy class $C_p$ with representative $p$, we can write the character map as

$$\mathrm{ch}:f\to \frac{1}{n!}\sum _{p⊢n}f(C_p)p_p|C_p|=\sum _{p⊢n}{z}_p^{-1}f(C_p)p_p.$$

Theorem

The characteristic map $\mathrm{ch}:R\to \Lambda$ is an isometry and an algebra isomorphism. Moreover, $\mathrm{ch}({\chi }^{\lambda })=s_{\lambda }$ and $\mathrm{ch}({\varphi }^{\lambda })=h_{\lambda }$, where ${\varphi }^{\lambda }$ is the character of permutation module $M^{\lambda }$.

\begin{proof}

A Generalization

Let $G$ be a finite group, and let $A$ be an associative algebra. If $\Phi ,\Psi :G\to A$ are functions, we may define a bilinear form with values in $A$

$$⟨\Phi ,\Psi ⟩=\frac{1}{|G|}\sum _{g\in G}\Phi (g)\Psi (g^{-1}).$$

If $\Phi$ is a class function on $H$ and $\Psi$ is a class function on $G$, the generalized Frobenius reciprocity still holds, that is

$$⟨\Phi {\uparrow }^G,\Psi ⟩=⟨\Phi ,\Psi {\downarrow }_H⟩$$

where $\Phi {\downarrow }_H(h)=\Phi (h)$ and $\Psi {\uparrow }^G(g)=\frac{1}{|H|}{\sum }_{x\in G}\Psi (x^{-1}gx)$ with the convention that $\Psi (g^{\prime })=0$ if $g^{\prime }\notin H$.

For any $f,g\in R_n$, there is

$$\begin{array}{rl}⟨\mathrm{ch}(f),\mathrm{ch}(g)⟩& =⟨\sum _{p⊢n}{z}_p^{-1}f(C_p)p_p,\sum _{\lambda ⊢n}{z}_{\lambda }^{-1}g(C_{\lambda })p_{\lambda }⟩\\ & =\sum _{\lambda ⊢n}{z}_{\lambda }^{-1}{z}_{\lambda }^{-1}f(C_{\lambda })g(C_{\lambda })z_{\lambda }\\ & =\sum _{\lambda ⊢n}|C_{\lambda }|\frac{1}{n!}f(C_{\lambda })g(C_{\lambda })\\ & =⟨f,g⟩\end{array}$$

Also note that $g({\omega }^{-1})=g(\omega )$.

Since $p_{\omega }=p_{{\omega }^{-1}}$ for $\omega \in S_n$ and $\mathrm{ch}(f)=⟨f,p⟩$, we have

$$\begin{array}{rl}\mathrm{ch}(fg)& =⟨fg,p⟩=⟨(f\otimes g){\uparrow }_{S_n\times S_m}^{S_{m\times n}},p⟩\\ & =\frac{1}{n!m!}\sum _{\sigma \in S_n,\tau \in S_m}(f\otimes g)(\sigma \tau )p_{\sigma \tau }\\ & =\frac{1}{n!m!}\sum _{(\sigma ,\tau )\in S_n\times S_m}f(\sigma )g(\tau )p_{\sigma }{p}_{\tau }\\ & =\left(\frac{1}{n!}\sum _{\sigma \in S_n}f(\sigma )p_{\sigma }\right)\left(\frac{1}{m!}\sum _{\tau \in S_m}g(\tau )p_{\tau }\right)\\ & =\mathrm{ch}(f)\mathrm{ch}(g)\end{array}$$

and so $\mathrm{ch}$ is a homomorphism.

Let ${\eta }_n$ be the trivial representation of $S_n$, then

$$\mathrm{ch}({\eta }_n)=⟨{\eta }_n,p⟩=\sum _{\lambda ⊢n}{z}_{\lambda }^{-1}{p}_{\lambda }=h_{\lambda }.$$

Recall that the permutation module $M^{\lambda }=1{\uparrow }_{S_{\lambda }}^{S_n}$ with $S_{\lambda }=S_{{\lambda }_1}\times \cdots \times S_{{\lambda }_k}$. Then ${\varphi }^{\lambda }={\eta }_{{\lambda }_1}\cdots {\eta }_{{\lambda }_k}$ yields that

$$\mathrm{ch}({\varphi }^{\lambda })=\mathrm{ch}({\eta }_{{\lambda }_1})\cdots \mathrm{ch}({\eta }_{{\lambda }_k})=h_{{\lambda }_1}\cdots h_{{\lambda }_k}=h_{\lambda }.$$

To prove that $\mathrm{ch}({\chi }^{\lambda })=s_{\lambda }$, note that

\eta_{\lambda_1} & \cdots & \eta_{\lambda_1+m-1} \\ \vdots & & \vdots \\ \eta_{\lambda_m-m+1} & \cdots & \eta_{\lambda_m} \end{matrix}\right|$$ and $\mathrm{ch}(\widetilde {\chi^\lambda})=\mathscr s_\lambda$. Now we finish the proof, as $\widetilde {\chi^\lambda}$ is induced $\chi^\lambda$. `\end{proof}` > [!corollary] > > The character table $\chi_{\mathscr p}^\lambda$ is the transition matrix between bases $p_\lambda$ and $\mathscr s_\lambda$ of $\Lambda^n$. ^7ed3d1 `\begin{proof}` It suffices to show $p_{\mathscr p}=\sum_{\lambda}\chi_{\mathscr p}^\lambda \mathscr s_\lambda$. By [[#^492b55|^492b55]] we have

\mathscr s_\lambda=\mathrm{ch}(\chi^\lambda)=\sum_{\mathscr p}z_{\mathscr p}^{-1}\chi_{\mathscr p}^\lambda p_{\mathscr p}

and so $\chi_{\mathscr p}^\lambda=\left\langle \mathscr s_\lambda, p_\mathscr p\right\rangle$. It deduces that

\left\langle \mathscr s_\lambda,p_\mu\right\rangle =\sum_{\mathscr p} z_\mathscr p^{-1}\chi_{\mathscr p}^{\lambda}\left\langle p_\mathscr p,p_\mu\right\rangle =z_{\mu}^{-1}\chi_{\mu}^\lambda z_{\mu}=\chi_{\mu}^\lambda.

Now we finish the proof. `\end{proof}` > [!corollary] Frobenius character formula > > If $\rho=(\rho_1,\cdots,\rho_m)$ with $\rho_1\geqslant \rho_2\geqslant\cdots\geqslant \rho_m\geqslant 1$, then for any $k\geqslant\ell(\lambda)$ the value $\chi_\rho^\lambda$ is the coefficient of $x_1^{\ell_1}\cdots x_k^{\ell_k}$ in the monomial > > $$ > \prod_{i=1}^m(x_1^{\rho_i}+\cdots+x_k^{\rho_i})\prod_{1\leqslant i< j\leqslant k}(x_i-x_j) > $$ > > where $\ell_i=\lambda_i+k-i$ for $i=1,\cdots,k$. **Remark.** It has been proved in [[10.6 Frobenius Character Formula#^4b468a|^4b468a]]. `\begin{proof}` Recall that $p_{\rho}=\prod_{i=1}^m(x_1^{\rho_i}+\cdots+x_k^{\rho_i})$ and ![[10.8 Symmetric Polynomials#^i6y60w|^i6y60w]] By [[#^7ed3d1|^7ed3d1]], we have $p_{\mathscr p}=\sum_{\lambda}\chi_{\mathscr p}^\lambda \mathscr s_\lambda$ and it deduces that $$\mathrm{LHS}=\sum_{\mu}\chi_{\rho}^\mu\left|\begin{matrix} x_1^{\mu_1+k-1} & ... & x_k^{\mu_1+k-1} \\ \vdots & & \vdots \\ x_1^{\mu_l} & \cdots & x_k^{\mu_k} \end{matrix}\right|.$$ Therefore, the coefficient of $x_1^{\ell_1}\cdots x_k^{\ell_k}$ with $\ell_i=\lambda_i+k-i$ of right hand side is $\chi_{\rho}^\lambda$. `\end{proof}`