3
$\begingroup$

Consider the regular representation of a finite group $G$ and let $X_{reg}$ be its character. Let $(\pi, V)$ be any finite dimensional representation of $G$ with character $X$.

Show that $<X_{reg}, X>=dimV$


The regular representation of $G$ is the permutation representation of the action "acting on itself by left multiplication."

$<X_{reg}, X>=\frac{1}{|G|}\sum_{g \in G}X_{reg}(g)X(g^{-1})$

I have the fomrula, but am not sure how to use this in practice. Would appreciate your help, thanks

$\endgroup$
  • $\begingroup$ Can you first show it when $X$ is irreducible ? $\endgroup$ – Clément Guérin Apr 4 '16 at 13:01
  • 1
    $\begingroup$ The following might help. $$X_{reg}(g)=\begin{cases}\mid G\mid&g=1_G\\0&\text{otherwise}\end{cases}$$ Also $X(1_G)=\text{dim} V.$ $\endgroup$ – awllower Apr 4 '16 at 13:05
  • $\begingroup$ @ClémentGuérin I think you meant to use that $k[G]\cong\oplus_i V_i^{\text{dim}V_i}.$ But I also think that this decomposition is usually proved by the formula we want to prove here. Do you know a way of proving this decomposition of the left regular representation without using the characters? If so, I would be glad to know what that method is, thanks in advance. $\endgroup$ – awllower Apr 4 '16 at 13:29
  • $\begingroup$ So if $g=1_G$ then $X_{reg}(g)=|G|$ and $<X_{reg}, X>=<|G|, X>$. So how can this relate to $dimV$? $\endgroup$ – thinker Apr 4 '16 at 20:33
3
$\begingroup$

Let $V_{\rm reg}=\bigoplus_{g\in G}\Bbb Cg$ be the vector space with basis the elements of $G$. Then the regular representation associates to each $g\in G$ the matrix $M_g$ that permutes the basis according to the multiplication table of $G$. Since $gx\neq x$ for all $x\in G$ and for all $1\neq g\in G$, the matrices $M_g$ have trace zero for all $g\neq 1$.

Thus $X_{\rm reg}(1)=|G|$ and $X_{\rm reg}(g)=0$ for $g\neq1$.

At this point the formula you want should be fairly obvius.

$\endgroup$
  • $\begingroup$ Thanks for this. How would this relate to $dimV$ ? $\endgroup$ – thinker Apr 4 '16 at 20:29
  • 1
    $\begingroup$ @thinker: If $V$ is just any representation, what is $X(1_G)$ where $X$ is the character of $V$? $\endgroup$ – Andrea Mori Apr 5 '16 at 10:06

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.