# Character regular representation

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

• Can you first show it when $X$ is irreducible ? – Clément Guérin Apr 4 '16 at 13:01
• The following might help. $$X_{reg}(g)=\begin{cases}\mid G\mid&g=1_G\\0&\text{otherwise}\end{cases}$$ Also $X(1_G)=\text{dim} V.$ – awllower Apr 4 '16 at 13:05
• @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. – awllower Apr 4 '16 at 13:29
• So if $g=1_G$ then $X_{reg}(g)=|G|$ and $<X_{reg}, X>=<|G|, X>$. So how can this relate to $dimV$? – thinker Apr 4 '16 at 20:33

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 $g\neq1$.
• Thanks for this. How would this relate to $dimV$ ? – thinker Apr 4 '16 at 20:29
• @thinker: If $V$ is just any representation, what is $X(1_G)$ where $X$ is the character of $V$? – Andrea Mori Apr 5 '16 at 10:06