Let $G$ be a connected Lie group with a maximal torus $T$. Suppose $\sigma$ is an automorphism of $G$ so that $\sigma(T)=T$. Then can we conclude that $\sigma$ is an inner automorphism of $G$? (i.e. is there an element $g\in G$ so that $\sigma= Ad_g : G\rightarrow G$).
$\begingroup$
$\endgroup$
4
-
1$\begingroup$ @Li: Did you try to find a counterexample by looking at $SO(2)$ (which is its own maximal torus) or $SO(3)$ in the nonabelian case? $\endgroup$– MishaApr 27, 2013 at 4:08
-
$\begingroup$ Try $G=SL(2,{\mathbb R})$ and $\sigma(x)=x^{-t}$, so $sigma$ is the composition of taking inverses and transposition. $\endgroup$– user1688Apr 27, 2013 at 9:40
-
2$\begingroup$ Assume $G$ compact. If your result was true then every automorphism would be inner. Indeed if $\mu$ was an automorphism you can find $g$ such that $Ad_g \mu (T) = T$. But $Aut(G)/Ad(G)$ is the non-trivial group of automorphisms of the Dynkin diagram. $\endgroup$– Michael MurrayApr 27, 2013 at 13:12
-
$\begingroup$ I interpret Michael Murray's answer as saying that the answer is "this is as false as possible". $\endgroup$– Allen KnutsonMay 4, 2013 at 13:37
Add a comment
|