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Let $G$ be a finite group and $H\le G$. Let $\lbrace{\rho_1,\rho_2,\ldots,\rho_t}\rbrace$ be the set of all inequivalent, irreducible representations of $G$. Consider the sum $$T:=\sum_{g\in G\setminus H}\rho_i(g)\otimes\rho_j(g).$$ I was trying to compute the eigenvalues of $T$ when $H$ is a normal subgroup of $G$.

All I can think of is that if one of the representations is the trivial one, the sum reduces to $T=\sum_{g\in G\setminus H}\rho_i(g)$. Now if $H$ is normal in $G$, then it can be seen that $T\rho_i(g)=\rho_i(g)T$, for all $g\in G$, which implies $T\in\operatorname{Hom}_G(\rho_i,\rho_i)$ and by Schur's lemma, this yields the scalar matrix $T=\frac{Tr(T)}{deg(\rho_i)}I=\frac{\sum_{g\in G\setminus H}\chi_i(g)}{deg(\chi_i)}I$, where $\chi_i$ is the irreducible character of $G$ corresponding to $\rho_i$. Then the eigenvalues of $T$ are just $\frac{\sum_{g\in G\setminus H}\chi_i(g)}{deg(\chi_i)}$. Moreover, the eigenvalues of $T$ are all integers in this case. But I cannot think of the case when none of $\rho_i$ and $\rho_j$ is the trivial representation. Can the eigenvalues of $T$ be computed in terms of the irreducible characters of $G$ in general?

Any help would be appreciated.

darij grinberg
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SPDR
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  • Now re-answered on https://mathoverflow.net/questions/500961/eigenvalues-of-a-sum-of-tensor-product-of-representation-matrices – darij grinberg Sep 30 '25 at 22:06

1 Answers1

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Decompose $\rho_i\otimes\rho_j=\bigoplus_{\ell=1}^tm^{ij}_\ell\rho_\ell$ where $m^{ij}_1,\ldots,m^{ij}_t$ are multiplicities. Then your sum can be written as $$T=\bigoplus_{\ell=1}^t\left(\sum_{g\in G\setminus H}\rho_\ell(g)\right)^{\oplus m^{ij}_\ell}.$$ Using your computation, the eigenvalues of this consist exactly of $m^{ij}_\ell\dim\rho_\ell$ copies of $\frac{\sum_{g\in G\setminus H}\chi_\ell(g)}{\dim\rho_\ell}$ for each $\ell$.

Carl Schildkraut
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    Okay. Another thing is how can we decompose $\rho_i\otimes\rho_j$ into the irreducible representations of $G$? Isn't $\rho_i\otimes\rho_j$ an irreducible representation of $G\times G$? Please pardon me if I get anything wrong. I am quite new in this topic. – SPDR Aug 18 '25 at 14:07
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    @SPDR You can think of $\rho_i\otimes\rho_j$ as a representation of $G\times G$, but the diagonal homomorphism $G\to G\times G$ given by $g\mapsto (g,g)$ makes $\rho_i\otimes\rho_j$ into a representation of $G$ itself: we have $$\rho_i\otimes\rho_j\colon g\mapsto \rho_i(g)\otimes\rho_j(g).$$ You can read more about this on the Wikipedia page for tensor products of representations. – Carl Schildkraut Aug 25 '25 at 23:14
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    Okay. So that means for irreducible representations $\rho_i$ and $\rho_j$ of $G$, the tensor product $\rho_i\otimes\rho_j$ is irreducible in $G\times G$ but it need not be irreducible in $G$. – SPDR Aug 27 '25 at 14:11
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    @SPDR Yes, exactly. – Carl Schildkraut Aug 27 '25 at 18:41