# Lie Algebra of Aut(GL(n,R))

What is the Lie Algebra of $Aut(Gl(n,F))$ when $F$ is either $\mathbb{R}$ or $\mathbb{C}$?

Is it enough to consider the injection via Hochschild: $Aut(GL(n)) \to Aut(\mathfrak{gl}(n))$?

Edit: The issue is, if I consider $der(\mathfrak{gl}(n))$ then I know I have elements in $Aut(GL(n))$ of the form $(M \to BMB^{-1})$ and $(M \to |det(M)|^cM)$ from some $c \ne -\frac{1}{n}$. I could then find elements of the form $(M \to [B,M])$ and $(M \to c\cdot tr(M)M)$ in $der(\mathfrak{gl}(n))$.

What I don't know is why (if I can) think of those elements as sitting in $Aut(GL(n))$.

• As far as I know, the connected component of the identity is just $PGL_n$ in both cases. (I assume that in the complex case you mean holomorphic automorphisms.) – Qiaochu Yuan Feb 14 '16 at 0:23
• I also don't understand how the connected component of $I$ in $GL(n)$ is $PGL_n$, since in the projective group, you have quotiented by the center, but there are still commuting elements in $GL(n)$. Am I missing something? – cheyne Feb 14 '16 at 0:34
• Beware that if $n$ is odd $GL(n,\mathbf{R})\simeq SL(n,\mathbf{R})\times\mathbf{R}$, so the automorphism group is isomorphic to $PGL_n(\mathbf{R})\times\mathbf{R}^*$, in contradiction with some claims. – YCor Feb 14 '16 at 4:25
• You start by asking for the derivations of the Lie algebra. You remember (Knapp, Lie Groups Beyond p.102) that the derivations of a semisimple Lie algebra are inner. So you see what the derivations of sl(n) are. Then I think it is not too tricky. – Ben McKay Feb 14 '16 at 9:00
• @YCor I think you have helped me answer my question, albeit cryptically and condescendingly, so thanks, I guess. – cheyne Feb 15 '16 at 18:06