all. I am reading a paper and I could not understand the dimension of the vector space $Hom_{\Gamma}(\rho_i\otimes \rho, \rho_j)$. The paper is http://pages.uoregon.edu/njp/Lusztig92.pdf, section 1.1, page 113. Here $\rho$ is a 2-dimensional C-vector space. $\Gamma$ is a finite subgroup of the special linear group $SL(\rho)$. Then $\rho$ is a $\Gamma$-module. Let $I$ be the set of isomorphism classes of simple $\Gamma$-modules. For each $i\in I$, $\rho_i$ is a simple $\Gamma$-module in class $i$. It is said that $dim Hom_{\Gamma}(\rho_i\otimes \rho, \rho_j)$ is 0, 1, or 2. I don't know why. Could you explain this? Thank you very much.
I have another question about this paper. On page 114, equation (c), it is said that (M, M')=|\Gamma|^{-1}\sum_{\gamma\in \Gamma}tr(\gamma, M)tr(\gamma^{-1}, M')(2-tr(\gamma, \rho)). Here (M, M') is defined on page 113, section 1.2. (M, M')=(M\otimes C^2: M')-(M\otimes \rho: M'), (M, M')=dim Hom_{\Gamma}(M, M'). Why the equation (c) is true? What does $tr(\gamma, \rho)$ mean? We could define trace of a linear map of a matrix. But here $(\gamma, \rho)$ is not a matrix or a map. Thank you very much.