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Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$, and let $\sigma \colon G \rightarrow G$ be a Lie group automorphism such that $\sigma^2 = \text{id}_G$. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$ such that $d_e \sigma (\mathfrak{h}) \subseteq \mathfrak{h}$.

Then, we know from general Lie theory that there exists a unique connected immersed Lie subgroup of $G$, $\iota \colon H \rightarrow G$, with $Lie(H) = \mathfrak{h}$.

Question: Does it follow that we get a restriction of $\sigma$ to $H$? More specifically, is it true that there exists a Lie group automorphism $\sigma' \colon H \rightarrow H$ such that $\sigma \circ \iota = \iota \circ \sigma'$?

My attempt: We have the restriction of $d_e \sigma$ to $\mathfrak{h}$, which is a Lie algebra homomorphism $\mathfrak{h} \rightarrow \mathfrak{h}$. I would like to lift this to a Lie group homomorphism $H \rightarrow H$, but a priori this result requires $H$ to be simply connected, which I don't have.

Context: I am studying Riemannian symmetric spaces, and this comes up when you try to prove that a totally geodesic submanifold of a Riemannian symmetric space naturally defines a Riemannian symmetric pair. (See Helgason, Remark after Theorem 7.2, page 224).

Diego Artacho
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    Note that $H$ is connected, and then you can use the fact that any element of $H$ is a product of exponentials to lift from the lie algebra – Holmes Mar 15 '22 at 23:26

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Hint: It is easy to see that it suffices to show that $\sigma(H)\subset H$ (and then defined $\sigma'=\sigma|_H$). Now show first that for $X\in\mathfrak h$ you get $\sigma(\exp(X))\in H$ and then use that $H$ is connected.

Andreas Cap
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