# Automorphisms of $GL_n(\mathbb{Z})$

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?

• 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 inverse-transpose 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. – YCor Jan 6 at 17:31
• @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 semi-direct product with tau. – cheyne Jan 6 at 18:24
• I'm lazy at the moment to fill in the details, this is why a comment is maybe better to start with. – YCor Jan 6 at 18:26
• I completely understand. Thanks again! – cheyne Jan 6 at 18:45

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 conjugate-transpose 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 $$n=2$$.
• For odd $n$ the endomorphism $A\mapsto \det(A)A$ is not injective (it maps $-I$ to $I$). – YCor Jan 7 at 0:54