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?

  • $\begingroup$ I am sure I am failing to see something obvious, but how do $[\chi](t)$ and $\chi(t)$ compare for $t \in N$? $\endgroup$ – Andreas Caranti Jun 19 '19 at 9:01
  • $\begingroup$ 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)? $\endgroup$ – user10354138 Jun 19 '19 at 9:06
  • $\begingroup$ @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. $\endgroup$ – JP McCarthy Jun 19 '19 at 9:07
  • $\begingroup$ @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. $\endgroup$ – JP McCarthy Jun 19 '19 at 9:08
  • $\begingroup$ I have taken some of these comments back into the question. $\endgroup$ – JP McCarthy Jun 19 '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 $1<n_\rho<d_\rho$.

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.