# Prove this lie subgroup has finite center

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:

1. I know that in general for connected groups, $\ker Ad_G=Z(G)$ and therefore $G/Z(G)=Ad_G(G)$.
2. 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.

• You're on the right track here, all that needs doing is proving that the derivative of the adjacency equals zero for the subgroup $H \subseteq G$ – Cppg Aug 19 '17 at 12:10
• @Cppg: Do you mean $[Ad(H),Ad(H)]=0$? Why is it enough? Doesn't it just show $Ad(H)$ is abelian? – The way of life Aug 19 '17 at 12:15
• $\mathcal{D}G \sim g(n+1) - g(n)$, and similarly $\mathcal{D}H \sim h(p+1) - h(p)$, so if $p < n$ for $\mathcal{D}G = 0$, then $\bar{H} \subseteq G$ thus $H$ has a finite centre about $n_{k}$. – Cppg Aug 19 '17 at 12:35
• @Cppg: I'm sorry, but most of the things here are unclear to me. I am probably lacking some context/basic knowledge in these derivatives. Any chance for some clarifications/relevant links? – The way of life Aug 19 '17 at 12:45
• Hint: Use the fact that the universal cover of $SL(2,R)$ is not isomorphic to any matrix group. – Moishe Kohan Aug 19 '17 at 13:10

## 1 Answer

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.

• Thank you. Still need two clarifications: 1. Isn't the map $Z(H)\to Z(H_1)$ simply the zero map? as the adjoint of a central element is the derivative of the identity which is zero? 2. I didn't quite get why the kernel is $Z(G)\cap H$ and not something perhaps larger - it is easier to be in $Z(H)$ than in $Z(G)$ – The way of life Aug 20 '17 at 7:57
• @The way of life: We don't know apriori if $H_1$ itself is an adjoint group ( like $PSL(2,\mathbb{R})$) or larger ( like $SL(2, \mathbb{R})$). It may be that an adjoint group contains semisimple subgroups with non-trivial centers. Now, about the kernel, it's what acts trivially on $\mathfrak{g}$, so trivially on $G$ ( $G$ connected), so it has to be in the center of $G$. – Orest Bucicovschi Aug 20 '17 at 8:04
• I get the part about the kernel now (an element of $H_1$ isn't an automorphism of $\mathfrak{h}$ but of $\mathfrak{g}$). This also makes the question abuot the map $Z(H)\to Z(H_1)$ okay. But one last question is why $Z(H)\to Z(H_1)$ surjective? in general, $\phi(Z(H))\subset Z(\phi (H))=Z(H_1)$ but couldn't it possiblt be a proper subset? Thanks a lot! – The way of life Aug 20 '17 at 9:19
• @The way of life: this is true for covering maps of connected Lie groups. If $h$ maps to an element in the center of $H_1$ then the image acts trivially on $\mathfrak{h}_1$. But then $h$ acts trivially on $\mathfrak{h}$ so it is in the center of $H$. Of course it's not true for arbitrary morphisms of (Lie ) groups. – Orest Bucicovschi Aug 20 '17 at 9:27