Christian Lange - Lattices (not) acting on S1

Abstract

Talk

Definition

A Lie Gruppe $G$ is called (semi)simple, if it does not have nontrivial connected (abelian) normal subgroups This is true iff. $\mathfrak{g}=Lie(G)$ has no nontrivial (abelian) ideals

Example

An easy simple Lie group is $SL(n,\mathbb{Z})$. This can be shown by looking at the Lie algebra, which are the traceless matrices. We look at Root spaces and look at common Eigenraum.

Definition

We define the Rank (Lie group), which is a set containing dimension of specific subspaces.

Definition

A Lattice (Lie group) in $G$ is a discrete subgroup $\Gamma \subseteq G$ s.t. $G/\Gamma$ has finite Haar measure.

Example

0. ${\mathbb{Z}}^n\subseteq {\mathbb{R}}^n$
1. $M^n$ some manifold. Then its Fundamental group is a discrete subgroup of the Isometric group $S{O}_0(n,1)$ of $\mathbb{H}$.
2. $SL(n,\mathbb{Z})\in SL(n,\mathbb{R})$. We need to make sure, the quotient has finite volume. This can be shown by letting $SL(2,\mathbb{Z})$ act on $\mathbb{H}$. The fundamental domain has finite volume and some identifications give us $G/\Gamma$ finite volume as well.
3. Example for a Lie group of non-trivial rank: $SL(2,\mathbb{Z}[\sqrt{2}])$. Funnily enough, if this group is embedded into $SL(2,\mathbb{Z})$ then the embedding is dense but if you embed it into $SL(2,\mathbb{Z}{)}^2$ then the embedding is discrete.

Theorem (Witte)

Let $\Gamma \subseteq SL(n,\mathbb{Z})$, $n\ge n$. Be a finite index subgroup. Then any homomorphism $\varphi :\Gamma \to Home{o}_{+}(S^1)$ has finite image.

Theorem (Witte)

The same group is not left-orderable.

Lemma

If two subgroups have finite index then their intersection subgroup also has finite index.

Proof of Theorem 2: By the Lemma we can reduce the size f the matrices to $n=3$. Assume there is a left order on $\Gamma$. There is a $k$ such that some elementary matrices sit inside $\Gamma$. I was too tired for the rest.

Theorem

A normal subgroup $N$ in a lattice $\Gamma$ of a higher rank simple Lie group with finite center is either finite or has finite index in $\Gamma$.

Proof of Thm 1