I was reading "Ergodic Theory and Topological Dynamics of Group Actions on Homogeneous Spaces." They prove a certain theorem for Lie groups locally isomorphic to $\operatorname{SL}_2(\mathbb{R})$ which are connected and have finite center.

But they only consider three cases: $\operatorname{SL}_2(\mathbb{R}),$ $\operatorname{PSL}_2(\mathbb{R}),$ and finite covers of $\operatorname{PSL}_2(\mathbb{R}).$ Are these cases exhaustive of all Lie groups locally isomorphic to $\operatorname{SL}_2(\mathbb{R})$?

  • 1
    $\begingroup$ There are also disconnected groups, as well as the universal cover of $SL(2,R)$. $\endgroup$ Jul 30 at 23:17
  • $\begingroup$ @MoisheKohan Ah sorry, I forgot to say (just added) that I'm assuming my group is connected w/ finite center. $\endgroup$
    – user147556
    Jul 30 at 23:20
  • 1
    $\begingroup$ Then there is nothing left, of course, since $\pi_1(PSL(2, {\mathbb R}))\cong {\mathbb Z}$ and the universal cover has infinite center. $\endgroup$ Jul 30 at 23:49
  • $\begingroup$ See for instance here. $\endgroup$ Jul 31 at 0:07

Here are some basic facts about Lie groups.

  1. Every connected Lie group $G$ has a unique (up to an isomorphism) universal covering $$ \pi: \tilde{G}\to G, $$ which is both a covering map and a Lie group homomorphism, necessarily with discrete kernel $\Lambda$.

  2. The subgroup $\Lambda$ is central in $\tilde{G}$. Indeed, for each $h\in \Lambda, g\in \tilde G$, the commutator $c_h(g)=[g,h]$ lies in the kernel of $\pi$, i.e. in the discrete subgroup $\Lambda< \tilde G$. By continuity and connectivity of $\tilde{G}$, the map $c_h$ is constant. Since $c_h(e)=e$, it follows that $c_h(g)=e$ for all $g\in \tilde G$, hence, $h$ is central in $\tilde G$.

  3. If $G_1, G_2$ are two locally isomorphic connected Lie groups (equivalently, groups with isomorphic Lie algebras), then their universal covering groups are isomorphic. This is the only nontrivial fact I will be using, one of Lie's theorems.

  4. From 1 and 2 it follows that for every connected Lie group $H$ locally isomorphic to $G$, there exists a discrete subgroup $\Lambda_H$ of the center $Z_{\tilde G}$ of $\tilde G$ such that $H\cong \tilde G/\Lambda_H$.

  5. Suppose now that the center of $\tilde G$ is a discrete subgroup $\Lambda$, equivalently, the Lie algebra of $G$ has trivial center. For instance, the universal covering group of $SL(2, {\mathbb R})$ has this property. (Since $sl(2, {\mathbb R})$ is centerless.)

Then every connected Lie group $H$ locally isomorphic to $G$ is isomorphic to a quotient $\tilde G/\Lambda_H$ for some subgroup $\Lambda_H< \Lambda$. The quotient $\tilde G/\Lambda$ is the unique connected Lie group locally isomorphic to $\tilde G$ which has trivial center. For instance, in the case of the Lie algebra $sl(2, {\mathbb R})$ this smallest quotient will be the adjoint group of $SL(2, {\mathbb R})$, namely, $PSL(2, {\mathbb R})$.

  1. Now, to the specific case of connected groups $G$ with the Lie algebra $sl(2, {\mathbb R})$. All these groups appear as covering groups of $PSL(2, {\mathbb R})$. Since $\pi_1(PSL(2, {\mathbb R})\cong {\mathbb Z}$, every such $G$ will be either a finite cyclic covering group of $PSL(2, {\mathbb R})$ or the unique infinite cyclic covering, the universal covering $\widetilde{SL}(2, {\mathbb R})$.

The conclusion is that in the book you are reading, they exhausted the list of all the connected groups with finite center locally isomorphic to $SL(2, {\mathbb R})$.


Your Answer

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