# questions about the paper: Affine quivers and canonical bases

I am reading the paper Affine quivers and canonical bases. I have a question on page 114 of the paper. In the proof of property (b), line 6 of page 114, why "for each $\gamma \neq 1$, $tr(\gamma, M)=0$" implies "$M=n\mathbf{r}$, where $\mathbf{r}=\mathbf{C}[\Gamma]$"? Thank you.

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A complex representation of a finite group $G$ whose character vanishes on all non-unit elements of $G$ is a multiple of the regular representation. That follows immediately by standard character theory, as set up in, for example, Serre's book on the Representation Theory of Finite Groups.

Later: You asked for an proof of this...

The hypothesis you have is that the character $\chi$ of your module $M$ is supported on $\{1_G\}\subseteq G$. Now, the character $\chi_{\mathrm{reg}}$ of the regular representation is also supported on $\{1_G\}$, and its value at $1_G$ is $|G|$. It follows that $\chi=p\chi_{\mathrm{reg}}$ for some $p\in\mathbb Q$, because the value of a character at $1_G$ is always an integer. Now $\chi$ is a linear combination of irreducible characters with integer coefficients, and since the trivial representation of $G$ occurs with multiplicity one occurs exactly once in the regular representation, it occurs with multiplicity $p$ in $M$. It follows immediately that $p$ must be an integer and, since a representation is determined by its character, that $M=p\mathbb C[G]$.

This argument, again, assumes the basic character theory of finite groups which you can find in lots of places (the book Harris+Fulton of on representation theory, Feit's or Huppert's on characters, Serre's, and many others) Google should find other expositions—I happen to have just taught a minicourse on the subject, and you'll find the notes, with most proofs, here, but in Spanish.

Aside: I think it would be extremely useful for you to study the basics of the representation theory of finite groups a bit. Lots of the things that one wants to do in representation theory, and which in particular Lusztig does, can be seen, in their simplest incarnation, in that context, so learning it has immense practical value!

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thanks. Do you know how to prove this property? I do not have the book of Serre. Or where could I find a proof? –  LJR May 25 '11 at 19:37
thanks. –  LJR May 26 '11 at 15:45