Panos P. - Actions on R and Orderings II

Abstract

Prove Theorem 6.15. Discuss the examples SLp2, Zq and SLpn, Zq, n ě 3 in terms of commutator lengths. For the latter look at the paper of Carter-Keller [CK] and [Cal, §5.2.2], giving some indication of the boundedness of commutator length on SLpn, Zq, n ě 3. Explain why the commutator subgroup of the Thompson’s group F has bounded commutator length using the fact that there are arbitrarily large conjugates that commute (following [Tsu, Theorem 2.1]). Prove vanishing of bounded cohomology of amenable groups (Theorem 6.16) and discuss the example of finitely generated (free) abelian groups or more generally solvable or nilpotent groups. Discuss Propositions 6.17 and 6.18 as well as the examples that conclude §6.6.

Talk

Theorem

Let $\Gamma$ be a countable group and $c\in H_b^2(\Gamma ,\mathbb{Z})$. Then there exists a homomorphism $\varphi :\Gamma \to Home{o}_{+}(S^1)$ with ${\varphi }^{\ast }(eu)=c$ if and only if $c$ is a $\{0,1\}$-valued cocycle.

What is the Euler class?

We did a long proof with a hard-to-understand end.

We make the remark that there is an element of the Spezielle Lineare Gruppe $SL(2,\mathbb{Z})$ whose Stable commutator length is greater than $0$.

Theorem (CK)

For each $n\ge 3$ every $g\in SL(2,\mathbb{Z})$ can be written as a uniformly long product of “elementary elements”.

This apparently has some connection to the fact, that for $n\ge 3$ the special linear group is not hyperbolic.

We look at the Thompson’s group F and its commutator length.

Theorem

Let $f\in [F,F]$. Then $f$ has slope $1$ at the beginning and at the end, because the slopes at that point multiply and for $f$ which can be written as a product of commutators, these multiply out to $1$. We will then show that the commutator length is $2$.

Definition

A topological group is amenable if there is a linear map $m$ from ${\mathcal{C}}_b^0(\Gamma )$ (bounded, continuous function from $\Gamma$ to $\mathbb{R}$) to $\mathbb{R}$ s.t.

1. It is nonnegative for nonnegative functions
2. It is one for the constand function one.
3. It is invariant under left-multiplication of all entries.

$m$ can be understood as an averaging operation on the bounded function.

This is not very appealing but it classifies groups which are nice.

Theorem (Johnson)

If $\Gamma$ is amenable, then the $k$th bounded cohomology is zero for all $k\ge 1$