Value of Irreducible Character in Quotient Algebra

Let $$G$$ be a finite group an $$F(G)$$ the algebra of functions on $$G$$. Let $$N\lhd G$$ be a normal subgroup and consider the ideal: $$J_N=\{f\in F(G)\,|\,\forall\,n\in N,\,f(n)=0\}.$$ Consider the quotient algebra $$F(G)/J_N$$ and the quotient map: $$\theta_N:F(G)\rightarrow F(G)/J_N\cong F(N),$$ $$f\mapsto f+J_N=:[f]$$.

Let $$\rho$$ be an irreducible representation of dimension $$d$$ of $$G$$. Where $$\chi=\sum_{i=1}^d\rho_{ii}\in F(G),$$ is the character of $$\rho$$, I believe that

$$\frac{1}{|N|}\sum_{t\in N}[\chi](t)$$

is equal to $$d$$ or zero.

Question 1: Is this true? It is a Frobenius-type result but I am interested, if possible, in a more direct proof.

Question 2: Does this fail for $$N$$ non-normal?

• I am sure I am failing to see something obvious, but how do $[\chi](t)$ and $\chi(t)$ compare for $t \in N$? – Andreas Caranti Jun 19 at 9:01
• Can't you identify $F(G)/J_N$ with $F(N)$ and hence just use Frobenius and $\operatorname{Ind}^G_N 1=1^{\oplus [G:N]}$ (since $N$ normal)? – user10354138 Jun 19 at 9:06
• @AndreasCaranti my instinct is that the thing to compare is $\sum_{t\in N}[\chi](t)$ and $\sum_{t\in N}\chi(t)$. If you are asking what is $[\chi](t)$ I guess I am invoking an isomorphism $F(G)/J\cong F(N)$. I suppose that $\theta_N(\delta_g)=\delta_g$ for $g\in N$, and zero otherwise. – JP McCarthy Jun 19 at 9:07
• @user10354138 I had in a previous edit said that this is a Frobenius type result but I was interested in a more direct proof if possible. – JP McCarthy Jun 19 at 9:08
• I have taken some of these comments back into the question. – JP McCarthy Jun 19 at 9:10

Here is a proof in the language of representations of quantum groups. This is work of Piotr Podlés and later Shouzou Wang.

Let $$\rho=(\rho_{ij})_{i,j=1}^{d_\rho}$$ be an irreducible representation and $$N$$ a subgroup of $$N$$. Let $$\theta_N:F(G)\rightarrow F(N)$$ be the map: $$\delta_g\mapsto \begin{cases}\delta_g&\text{ if }g\in N\\ 0 & \text{ otherwise}\end{cases}.$$

Note that where $$\Delta:F(G)\rightarrow F(G)\otimes F(G)$$ is the comultiplication: $$\delta_g\mapsto \sum_{t\in G}\delta_{gt^{-1}}\otimes\delta_t,$$ and $$\Delta_N:F(N)\rightarrow F(N)\otimes F(N)$$ the comultiplication in $$F(N)$$:

$$\delta_g\mapsto \sum_{n\in N}\delta_{gn^{-1}}\otimes \delta_n,$$ the map $$\pi$$ has the property that: $$(\pi\otimes\pi)\circ \Delta=\Delta_N\circ \pi.$$

That $$\rho$$ is a representation of $$G$$ on $$V$$ implies that $$(\rho_{ij})_{i,j=1}^{d_\rho}\in M_{d_\rho}(F(G))$$ is a corepresentation matrix for $$F(G)$$.

That $$N$$ is a subgroup of $$G$$ implies that $$(\pi(\rho_{ij}))_{i,j=1}^{d_\rho}$$ is a corepresentation matrix for $$F(N)$$.

The trivial corepresentation of $$F(N)$$ is the map $$\kappa_{\tau_N}$$ $$\lambda\mapsto \lambda\otimes \mathbf{1}_N$$. Let $$n_\rho$$ be the multiplicity of $$\kappa_{\tau_N}$$ in $$\pi(\kappa_\rho)$$. Podlés/Wang claim that $$n_\rho$$ is either $$d_\rho$$ or zero. Assume $$1.

Let $$F(N\backslash G)$$ be the algebra of functions constant on the right cosets of $$N$$ in $$G$$ and $$F(G/N)$$ the algebra of functions on the right cosets.

If $$N$$ is normal, then these algebras coincide.

Let $$E$$ be the projection $$F(G)\mapsto F(G/N)$$ that maps a function $$f$$ to $$E(f)$$ where $$E(f)(Ng)=\frac{1}{|N|}\sum_{n\in N}f(ng),$$ the average of $$f$$ on the coset $$Ng$$.

Let $$n_N=\frac{1}{|N|}\sum_{n\in N}\delta^n$$ be the averaging state on $$F(N)$$. We claim that $$E(\rho_{ij})=E_{N\backslash G}(\rho_{ij})=\sum_{k=1}^{d_\rho}h_N(\pi(\rho_{ik}))\,\rho_{kj},$$ and $$E(\rho_{ij})=E_{G/N}(\rho_{ij})=\sum_{k=1}h_N(\pi(\rho_{kj}))\rho_{ik}.$$ This follows from the fact that $$E=\underbrace{(h_N\pi\otimes I_{F(G)})\circ \Delta}_{E_{N\backslash G}}=\underbrace{(I_{F(G)}\otimes h_N\pi)\circ \Delta}_{E_{G/N}},$$ as can be seen by showing that this maps $$\delta_g\mapsto \frac{1}{|N|}\mathbf{1}_{Ng}.$$

Now choose a basis such that the $$n_\rho$$ trivial representations of $$N$$ in $$\pi(\rho)$$ occur in the top left hand corner so that:

$$\pi(\rho)=\left(\begin{array}{cccc}\mathbf{1}_N & \dots & 0 & 0 \\ \vdots & \ddots & 0 & 0\\ 0 & 0 & \mathbf{1}_N & 0 \\ 0 & 0 & 0 & \varrho\end{array}\right),$$ where $$\varrho$$ is a sum of non-trivial representations of $$N$$.

Now, beyond this point we have matrix elements $$\pi(\rho_{ij})$$ of non-trivial representations of $$N$$. From corepresentation theory, we have that the average of a non-trivial representation is zero, and so, outside the $$n_\rho\times n_\rho$$ upper corner, $$n_N(\pi(\rho_{ij}))=0$$.

Beyond this square there must be a non-zero matrix element $$\rho_{ij}$$

Recall that the map $$E$$ is a projection from functions on $$G$$ onto functions on the quotient group.

Note that, from the above discussion $$E_{N\backslash G}(\rho_{ij})=\begin{cases}\rho_{ij}&1\leq i\leq n_\rho,\,1\leq j\leq d_\rho\\0 & \text{otherwise}\end{cases},$$ and $$E_{G/N}(\rho_{ij})=\begin{cases}\rho_{ij}&1\leq i\leq d_\rho,\,1\leq j\leq n_\rho\\0 & \text{otherwise}\end{cases}$$

Take an element $$\rho_{ij}$$ such that $$j>n_\rho$$.

Now $$0\neq \rho_{ij}=E_{N\backslash G}(\rho_{ij})=E_{G/N}(\rho_{ij})=0.$$

This implies that there is no such $$j$$ and so $$n_\rho=d_\rho$$. Alternatively $$n_\rho=0$$.

The result above should follow.