I want to consider the crossed module: $H \xrightarrow{t} Aut(H)$ for the case where $H = GL_n(\mathbb{Z}) = Aut(T^n)$ is the automorphism group of the $n$torus. Any suggestions on how to understand this automorphism group, $Aut(GL_n(\mathbb{Z}))$, explicitly?

4$\begingroup$ I'm not sure of the meaning of your first sentence, but for the second one, the automorphism group for $n\ge 3$ is reduced to the "obvious" one, namely $\mathrm{PGL}_n(\mathbf{Z})\rtimes\{\tau\}$ where $\tau$ is the inversetranspose involutive automorphism. (Essentially, Mostow rigidity reduces to compute the normalizer in the real automorphism group, i.e., to show that $\mathrm{SL}_n(\mathbf{Z})$ equals its own normalizer in $\mathrm{SL}_n(\mathbf{R})$.) However it's very plausible that this was known earlier in this case by a direct algebraic approach. $\endgroup$– YCorJan 6, 2019 at 17:31

$\begingroup$ @YCor Thank you! My first sentence was just a motivation in case it was helpful. Would you like to rewrite this as an answer so I can select it? It would be helpful if you added a reference or added a few more details about the semidirect product with tau. $\endgroup$– cheyneJan 6, 2019 at 18:24

$\begingroup$ I'm lazy at the moment to fill in the details, this is why a comment is maybe better to start with. $\endgroup$– YCorJan 6, 2019 at 18:26

$\begingroup$ I completely understand. Thanks again! $\endgroup$– cheyneJan 6, 2019 at 18:45
1 Answer
Hua and Reiner described this in their paper "Automorphisms of the unimodular group" (for them "unimodular" means determinant of absolute value 1, hence it's ${\rm GL}_n(\mathbf Z)$ rather than just ${\rm SL}_n(\mathbf Z)$) that appeared in 1951 here. The end result (their Theorem 4) is that the automorphisms of ${\rm GL}_n(\mathbf Z)$ are generated by inner automorphisms, the conjugatetranspose map $A \mapsto (A^{1})^\top$, the map $A \mapsto (\det A)A$ (they include it "for even $n$ only," which must mean they have a way of getting it from the other generators for odd $n$  I haven't read the paper closely to see where that is indicated [Edit: see comment below]), and one additional automorphism when $n=2$.

2$\begingroup$ For odd $n$ the endomorphism $A\mapsto \det(A)A$ is not injective (it maps $I$ to $I$). $\endgroup$– YCorJan 7, 2019 at 0:54

