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Let $ \tau $ be an irrep of finite group $ G $. Let $ \rho: G \to GL(n,\mathbb{C}) $ be a matrix representation of $ G $. The formula for the projection of $ \rho $ onto the $ \tau $ isotypic component is $$ \Pi_\tau = \frac{deg(\chi_\tau)}{|G|}\sum_{g \in G} \overline{\chi_\tau(g)}\rho(g) $$ Recently I have been constructing some projectors and I noticed that the matrix $$ |G|\Pi_\tau=deg(\chi_\tau) \sum_{g \in G} \overline{\chi_\tau(g)}\rho(g) $$ has integer entries. What are some sufficient conditions on $ \rho $ or $ \tau $ for $ |G|\Pi_\tau $ to have integer entries?

In particular I was looking at tensor powers of a matrix irrep $ \tau: G \to GL(n,\mathbb{C}) $ and calculating the projection of $ \tau^{\otimes n} $ onto the $ \tau $ isotypic component. I noticed that $$ |G|\Pi_\tau =deg(\chi_\tau) \sum_{g \in G} \chi_\tau(g)\tau(g)^{\otimes d} $$ is a matrix with integer entries for all small values of $ d $.

In other words, let $ \tau: G \to GL(n,\mathbb{C}) $ be an irrep of a finite group $ G $. Is it true for all $ d $ that the matrix $$ |G|\Pi_\tau =deg(\chi_\tau) \sum_{g \in G} \chi_\tau(g)\tau(g)^{\otimes d} $$ has integer entries? Certainly for $ d=1 $ this is true since the above expression simplifies to just $ |G| $ times the identity matrix.

Certainly the conjecture is true for all $ d $ if $ \tau $ has degree $ 1 $, it just follows from the identity for sum over all powers of a root of unity https://math.stackexchange.com/a/3524168/758507

I've also check some degree $ 2 $ irreps $ \tau $ for pretty high tensor powers $ d $ and its true so far.

Ian Gershon Teixeira
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    The question doesn't really make sense--it doesn't make sense to ask whether $|G|\Pi_\tau$ has integer entries unless you pick a basis to represent it as a matrix, and the answer will certainly be different for different bases (any two projection matrices of the same rank are related by a change of basis). – Eric Wofsey Mar 09 '23 at 02:54
  • Good point that this is ambiguous for a generic representation $ \rho: G \to GL(V) $. But it's pretty common to think about matrix representations $ \rho: G \to GL(n,\mathbb{C}) $ in which case the question is well posed. I'll edit to specify that it is a matrix representation. – Ian Gershon Teixeira Mar 09 '23 at 15:39
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    But the point is that normally you care about representations only up to isomorphism, and the answer to your question is not preserved by isomorphisms. – Eric Wofsey Mar 09 '23 at 15:45

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This already fails for larger tensor powers $ n=11 $. For example let $ G\cong 2.S_4 $ be the $ 48 $ element binary octahedral subgroup of $ SU(2) $ generated by $$ H:= \frac{i}{\sqrt{2}}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} $$ and $$ P:= \begin{bmatrix} \zeta_8 & 0 \\ 0 & \overline{\zeta_8} \end{bmatrix} $$ Let $ \tau: G \to SU(2) $ be the natural representation. Then $$ |G|\Pi_\tau= deg(\chi_{\tau})\sum_{g \in G} \overline{tr(g)} g^{\otimes 11 } $$ is not an integer matrix. Indeed, its entries are exactly the elements of the set $ \{ -\frac{1}{2},0,\frac{1}{2}, \frac{15}{2}.\frac{17}{2} \} $

Moreover, as pointed out by Eric Wofsey in the comments, since $ \Pi_\tau $ has all eigenvalues $ 1,0 $ , which are integers, then $ \Pi_\tau $ will always be conjugate to some integer matrix. Same is true for $ |G|\Pi_\tau $ which has integer eigenvalues $ |G|,0 $.

Ian Gershon Teixeira
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