1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $ $\endgroup$ – Cppg Aug 19 '17 at 12:10
  • $\begingroup$ @Cppg: Do you mean $[Ad(H),Ad(H)]=0$? Why is it enough? Doesn't it just show $Ad(H)$ is abelian? $\endgroup$ – The way of life Aug 19 '17 at 12:15
  • $\begingroup$ $ \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} $. $\endgroup$ – Cppg Aug 19 '17 at 12:35
  • $\begingroup$ @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? $\endgroup$ – The way of life Aug 19 '17 at 12:45
  • $\begingroup$ Hint: Use the fact that the universal cover of $SL(2,R)$ is not isomorphic to any matrix group. $\endgroup$ – Moishe Kohan Aug 19 '17 at 13:10
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $Z(G)$ $\endgroup$ – The way of life Aug 20 '17 at 7:57
  • $\begingroup$ @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 $G$. $\endgroup$ – Orest Bucicovschi Aug 20 '17 at 8:04
  • $\begingroup$ 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! $\endgroup$ – The way of life Aug 20 '17 at 9:19
  • $\begingroup$ @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. $\endgroup$ – Orest Bucicovschi Aug 20 '17 at 9:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.