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I am having trouble understanding part of a lecture in class.

Let $G$ be a group, $\phi$ a representation of $G$ into $GL_n(C)$, and let $E^{ij}$ denote the matrix with 1 in the $i$th row and $j$th column and zero everywhere else.

Define the matrix $M = \sum_{g \in G} \phi(g) E^{ij} \phi(g)^{-1}$. We showed that $M$ commutes with $\phi(g)$ for every $g$. So by Schur's lemma, we know that $M = \alpha I_n$ for some nonzero complex number $\alpha$.

However, if you compute the trace of $M$, it comes out to be $|G| \delta_{ij}$, where $\delta_{ij}$ is the Kronecker delta function. Since $M = \alpha I_n$, the trace should simply be a fixed number $\alpha n$.

Where is the mistake?

1 Answers 1

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Schur's lemma doesn't tell you that $\alpha$ is nonzero, and you have also neglected to make it clear with your notation that $M = M^{ij}$ and $\alpha = \alpha^{ij}$ both depend on $i$ and $j$. You then get $n \alpha^{ij} = |G| \delta_{ij}$ and there is no contradiction here.