Assume that $G$ is a connected Lie group and that $\alpha:G\rightarrow G$ is an automorphism of $G$. Furthermore let $\alpha_*:\mathfrak{g}\rightarrow\mathfrak{g}$ be the corresponding tangent map at the identity (arrow between the corresponding Lie algebra).
One can easily show that $\alpha_*$ is a Lie algebra automorphism since $\alpha$ is a (local diffeomorphism) which implies that $\alpha_*$ is bijective. Ready.
Now I want to show two things:
- $\alpha^2=Id_G$ $\Longleftrightarrow$ $\alpha_*^2=Id_{\mathfrak{g}}$
- If we assume that $\sigma^2=Id$ then $H=\{g\in G:\alpha(g)=g\}$ is a closed subgroup of $G$ and its Lie algebra $\mathfrak{h}$ is given by $\mathfrak{h}=\{X\in\mathfrak{g}:\alpha_*(X)=X\}$
For the first part I can only give a prove of the implication from the left to the right:
Assume $\alpha^2=Id$, then it follows from the observation we made before that $\alpha_*^2=(\alpha^2)_*=(Id_G)_*=Id_{\mathfrak{g}}$ which proves the implication.
For the other implication i think you have to use that $G$ is connected and that therefore the connected component of the identity is equal to the whole Lie group $G$, but i don't know how to use it precisely. Someone an idea?
For the second point one can easily prove that $H$ is a subgroup of $G$. To conclude that $H$ is closed it would be enough to show that $H$ is a submanifold of $G$ at the identity, but i don't now how to prove this, maybe by a local diffeomorphism? Or is there another way to conclude the result? Is the conclusion of how $\mathfrak{h}$ looks like a direct consequence?
Can someone help me with this problems? 🙂
Best Answer
For the first part, you want to prove the general result that if $G$ is connected and $\phi, \psi : G \to \tilde G$ are two Lie group homomorphism with $\tilde G$ an arbitrary Lie group then $\phi_* = \psi_*$ implies $\phi = \psi$. One way to prove this is by using the following two results:
Because of these two results it is sufficient to show that $\phi$ and $\psi$ agree on elements of the form $\exp X, X \in \mathfrak g$. But this follows from naturality of the exponential map: $$ \phi(\exp X) = \exp(\phi_* X) = \exp(\psi_* X) = \psi(\exp X). $$
For the second part, $H$ is closed because it is defined by equations. More explicitly, consider the map $$ F = \alpha \times Id : G\to G\times G, g\mapsto (\alpha(g), g). $$ The diagonal $\Delta = \{(g,g) \} \subset G\times G$ is closed (this follows from $G$ being Hausdorff) and $H = F^{-1} (\Delta)$ is closed since it is the inverse image of a closed set under a continuous map.
Let me know if you would like me to expand on any of the results I used.