Let $G$ be a connected lie group with finite center. Let $H<G$ be a connected lie subgroup with lie algebra $\mathfrak{h}<\mathfrak{g}$ isomorphic to $\mathfrak{sl}(2,\mathbb{R})$. Prove that $H$ has finite center.
If required, you may assume $G$ is semisimple
My attempts:
I'm not quite sure how to go with it, but there are two facts I have found probably relevant:
- I know that in general for connected groups, $\ker Ad_G=Z(G)$ and therefore $G/Z(G)=Ad_G(G)$.
- Since $\mathfrak{h}=\mathfrak{sl}(2,\mathbb{R})$, $Ad_G(H)=Ad_H(H)$ is isomorphic to $GL(\mathfrak{sl}(2,\mathbb{R}))$ and so is a matrix lie algebra, which I think should have finite center? [still, doesn't say anything about $H$ itself)
I don't mind references to books proving relevant propositions.
The kernel of the map $\operatorname{Ad}\colon G \to GL(\mathfrak{g})$ is $Z(G)$ . The image of $H$ under this map ( let's call it $H_1$) is isomorphic to $SL(2, \mathbb{R})$ or $PSL(2,\mathbb{R})$ ( this because any linear representation of $sl(2, \mathbb{R})$ comes from a representation of $SL(2, \mathbb{R})$). Therefore we have a covering map $H \to H_1$ with kernel $Z(G) \cap H$ and this gives an exact sequence $$0 \to Z(G) \cap H \to Z(H)\to Z(H_1) \to 0$$ and now the conclusion follows.
This holds more generally for $H$ semi simple, since $H_1$, a linear semisimple connected Lie group will have a finite center.