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Let $G$ be a compact lie group and $\alpha:\mathbb{R}\to \text{Aut}(G)$ be a group homomorphism with continuous orbit maps $t\mapsto \alpha_t(g)$, for $g\in G$. Why is there always an $X\in \textbf{L}(G)$ such that $\alpha_t(g)=\text{exp}(tX)\cdot g\cdot\text{exp}(-tX)$?

Tobias Simon
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  • If $G=S^1$, then any $\mathbb{R}$-action on it is trivial according to what you say. This is not true, as there is this rotation action $(\theta, z)\mapsto e^{i\theta}z$. – Alex Fok Aug 20 '21 at 19:51
  • @user953376 He did not say $\mathbb{R}$ must act by group homomorphism. – Alex Fok Aug 20 '21 at 19:58
  • @user953376 If what he meant was that $\mathbb{R}$ acts by Lie group automorphisms, then it is true that the automorphisms must be inner automorphisms by a certain one-parameter subgroup. That is because by the standard structure theory of compact Lie groups, $\text{Aut}(G)=\text{Inn}(G)\rtimes \text{Out}(G)$ where $\text{Out}(G)$ is a finite group. So the $\mathbb{R}$-action must be a 1-parameter subgroup of the connected component $\text{Inn}(G)$. – Alex Fok Aug 20 '21 at 20:09
  • @user953376 By the way there is nothing wrong with his grammar. It is the precision of his mathematical language that is at issue here. – Alex Fok Aug 20 '21 at 20:12
  • A correction: $\text{Out}(G)$ is discrete, but not finite in general. If $G$ is an $n$-dimensional torus, then $\text{Out}(G)=\text{GL}_n(\mathbb{Z})$. $\text{Out}(G)$ is finite if and only if $G$ is semisimple. – Alex Fok Aug 20 '21 at 20:48

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