I want to understand the (complex) representation theory of $\mathrm{GL}_n(\mathbb{Z})$, the general linear group of the integers. I have gone through several representation theory texts but all of them present theory relevant to compact groups. What are the irreducible complex representations of $\mathrm{GL}_n(\mathbb{Z})$?
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4$\begingroup$ There is rather a different story for $n=2$ and $n \ge 3$. For the related case of $SL_n(\mathbb Z)$ see e.g. here mathoverflow.net/questions/419795/… and here www3.nd.edu/%7Eandyp/notes/RepTheorySLnZ.pdf $\endgroup$– Kevin CastoCommented Nov 10, 2023 at 19:22
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For $n=2$, these groups are very close to free groups, so their representation theory is not very tractable. For $n \geq 3$, however, there is a beautiful story arising from Margulis superrigidity. I wrote this up for the special linear group here.
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$\begingroup$ Thank you for your reply, could you please tell me how the representation theory of the subgroup $SL_n(\mathbb{Z})$ relates to that of $GL_n(\mathbb{Z})$? I am studying for a Physics PhD and I am sorry that it does not seems straight forward for me. $\endgroup$– KenjiCommented Nov 10, 2023 at 19:52
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1$\begingroup$ It’s a degree 2 extension, so there shouldn’t be a huge difference. I would have to think a bit to figure out exactly what changes, but my guess is that there is a similar description of the proalgebraic completion. You have to be a little careful since the Zariski closure of GL(n,Z) is not GL(n,R), but rather the group of matrices over the reals whose determinant is 1 or -1. $\endgroup$ Commented Nov 10, 2023 at 20:01
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2$\begingroup$ (sorry that this is probably not very understandable to physicists — these groups are significantly deeper and more complicated than the compact or semisimple Lie groups that you all tend to study). $\endgroup$ Commented Nov 10, 2023 at 20:02
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8$\begingroup$ If $n$ is odd, then $\text{GL}_n(\mathbb{Z}) \cong \text{SL}_n(\mathbb{Z}) \times \{ \pm \text{Id} \}$, so every irrep of $\text{GL}_n$ is an irrep of $\text{SL}_n$ tensored with one of the two one-dimensional reps of $\{ \pm 1 \}$. $\endgroup$ Commented Nov 10, 2023 at 21:06
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1$\begingroup$ I don't see any similarly easy trick for $n$ even. $\endgroup$ Commented Nov 10, 2023 at 21:07