Lilly Sandberger - Stable commutator length and quasi homomorphisms

Abstract

Talk

1. scl % genus of bounding surfaces

We motivate the subject with an element of the Fundamental group. When a fundamental group of $X$ has nontrivial commutator, then this implies there is a handle.

Definition

Let $G$ be a group and $a\in [G,G]$ be an element of the commutator subgroup. The commutator length is minimal count of commutators needed to write an element. The Stable commutator length is obtained by stabilising. Among other things it is interesting because it is subadditive.

We present some covers. These covers describe some sort of possible preimages for my boundary curve of a handle.

Proposition

Let $X$ be a topological space. Let $G$ be a fundamental group of some space $X$ and $a$ an element inside. Then $scl(a)={\inf }_{S\to X\text{, admissable}}\frac{-{\chi }^{-}(S)}{2n(S)}$ where $-{\chi }^{-}(S)$ is some sum over Euler characteristics. An admissable map is a map where the boundary of $S$ maps to $a$. $n(S)$ is the boundary

2. Quasi morphisms and scl

Definition

We define a Quasimorphism.

Trivial examples

Homomorphisms are qms with defect $0$. Bounded functions are quasimorphisms.

Proposition Bavard's Duality

Let $G$ be a group, $a\in [G,G]$. Then $scl(a)={\sup }_{\varphi \in QM(G)/Hom(G,\mathbb{R})}\frac{|\varphi (a)|}{2D(\varphi )}$ Where $QM(G)$ is the Homogeneous quasi-homomorphism space.

This directly tells us that if $scl$ is positive, then there exists a non-trivial qm with $|\varphi (a)|>0$.

Lemma Bavard's commutator estimate

Let $\varphi :G\to \mathbb{R}$ be a homogenous qm. Then ${\sup }_{a,b\in G}|\varphi ([a,b])|=D(\varphi )$

Example by Barge-Ghys

We examine a hyperbolic surface $S$. We create a quasi-morphism by integrating a 1-form along a geodesic. Concatenating geodesics is almost additive, giving us a quasimorphism. We can prove the defect by using Stokes Satz.

Example

We defined the Counting quasimorphism and present some numerical examples. We note an important property. Because copies of $w$ and $w^{-1}$ are always disjoint, we can construct an unbounded Quasimorphism $h_w(w^n)$ by letting $n$ g to infinity.