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Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$, and $H \subset G$ be a Lie subgroup of $G$ with Lie algebra $\mathfrak{h}$. Suppose that there is some inner-product $\left<\cdot,\cdot\right>$ on $\mathfrak{g}$ and define $\mathfrak{m} = \mathfrak{h}^\perp$ with respect to this inner product, so that $\mathfrak{g} = \mathfrak{h} \oplus \mathfrak{m}$. Then, is it the case that $\mathfrak{h}$ is Ad$_H$ invariant (that is, $\text{Ad}_h(\mathfrak{h}) \subset \mathfrak{h}$ for all $h \in H$)?


Since $\mathfrak{h}$ is Lie subalgebra of $\mathfrak{g}$, we have by definition that $[\mathfrak{h}, \mathfrak{h}] \subset \mathfrak{h}$. Let $\xi, \eta \in \mathfrak{h}$ and $\sigma \in \mathfrak{m}$. Then,

\begin{align*} 0 = \left<[\xi, \eta], \sigma\right> &= \left<\frac{d}{dt}\Big{\vert}_{t=0} \text{Ad}_{\exp(t\xi)}(\eta), \sigma \right>, \\ &= \frac{d}{dt}\Big{\vert}_{t=0}\left< \text{Ad}_{\exp(t\xi)}(\eta), \sigma \right>. \end{align*} It follows that the map $t \mapsto \left< \text{Ad}_{\exp(t\xi)}(\eta), \sigma \right>$ is constant. In particular, since $\left< \text{Ad}_{\exp(t\xi)}(\eta), \sigma \right>\big\vert_{t=0} = \left<\eta, \sigma\right> = 0$, we have $\left< \text{Ad}_{\exp(t\xi)}(\eta), \sigma \right> = 0$ for all $\xi, \eta \in \mathfrak{h}, \sigma \in \mathfrak{m}, t \in \mathbb{R}$. Since the exponential map is a local diffeomorphism, and $\mathfrak{h} = \{X \in \mathfrak{g} \ \vert \ \exp(tX) \in H \text{ for all } t \in \mathbb{R}\}$, it follows that there exists an open subset $V \subset G$ containing $e$ such that $\text{Ad}_g(\mathfrak{h}) \subset \mathfrak{h}$ for all $g \in V \cap H$. I was thinking that, from here, I could try to show that for every $h \in H$, there exists some finite collection of points $g_1, \dots, g_n$ such that $g_1\cdots g_n = h$, so that, for all $\xi \in \mathfrak{h}$, $\text{Ad}_h(\xi) = \text{Ad}_{g_1} \circ \cdots \circ \text{Ad}_{g_n}(\xi) \in \mathfrak{h}$. However, I'm not sure if such a collection of points exists or if I've made some mistake in my reasoning.

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2 Answers 2

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If you take the Lie subgroup H to be connected then this is true, for any $h\in H$ there exists $X_1,X_2,\dots,X_m\in\mathfrak{h}$ such that $h=e^{X_1}e^{X_2}\dots e^{X_m}$. See Corollary 3.47 of Brian C Hall's book Lie groups, Lie algebras and Representations for proof.

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    $\begingroup$ In fact, connectedness of the subgroup is not needed for the result and neither is the exponential map. $\endgroup$ Apr 5, 2023 at 0:54
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As noted in the comment by @Moishe_Kohan, connectedness is not needed. The result follows easily from the definition of $Ad$ as the derivative of conjugation: For $g\in G$ and $X\in\mathfrak g$, you can simply define $Ad(g)(X)$ as the derivative at $t=0$ of $g\exp(tX)g^{-1}$. If $g\in H$ and $X\in\mathfrak h$, then $\exp(tX)\in H$ for all $t$, so the curve has values in $H$ and hence it's derivative at $t=0$ lies in $\mathfrak h$.

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