# Is the Lie algebra of a Lie subgroup Ad-invariant?

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.

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.
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 $$\mathfrak h$$.