## Quasimorphisms and laws

A basic reference for the background to this post is my monograph.

Let $G$ be a group, and let $[G,G]$ denote the commutator subgroup. Every element of $[G,G]$ can be expressed as a product of commutators; the commutator length of an element $g$ is the minimum number of commutators necessary, and is denoted $\text{cl}(g)$. The stable commutator length is the growth rate of the commutator lengths of powers of an element; i.e. $\text{scl}(g) = \lim_{n \to \infty} \text{cl}(g^n)/n$. Recall that a group $G$ is said to satisfy a law if there is a nontrivial word $w$ in a free group $F$ for which every homomorphism from $F$ to $G$ sends $w$ to $\text{id}$.

The purpose of this post is to give a very short proof of the following proposition (modulo some background that I wanted to talk about anyway):

Proposition: Suppose $G$ obeys a law. Then the stable commutator length vanishes identically on $[G,G]$.

The proof depends on a duality between stable commutator length and a certain class of functions, called homogeneous quasimorphisms

Definition: A function $\phi:G \to \mathbb{R}$ is a quasimorphism if there is some least number $D(\phi)\ge 0$ (called the defect) so that for any pair of elements $g,h \in G$ there is an inequality $|\phi(x) + \phi(y) - \phi(xy)| \le D(\phi)$. A quasimorphism is homogeneous if it satisfies $\phi(g^n) = n\phi(g)$ for all integers $n$.

Note that a homogeneous quasimorphism with defect zero is a homomorphism (to $\mathbb{R}$). The defect satisfies the following formula:

Lemma: Let $f$ be a homogeneous quasimorphism. Then $D(\phi) = \sup_{g,h} \phi([g,h])$.

A fundamental theorem, due to Bavard, is the following:

Theorem: (Bavard duality) There is an equality $\text{scl}(g) = \sup_\phi \frac {\phi(g)} {2D(\phi)}$ where the supremum is taken over all homogeneous quasimorphisms with nonzero defect.

In particular, $\text{scl}$ vanishes identically on $[G,G]$ if and only if every homogeneous quasimorphism on $G$ is a homomorphism.

One final ingredient is another geometric definition of $\text{scl}$ in terms of Euler characteristic. Let $X$ be a space with $\pi_1(X) = G$, and let $\gamma:S^1 \to X$ be a free homotopy class representing a given conjugacy class $g$. If $S$ is a compact, oriented surface without sphere or disk components, a map $f:S \to X$ is admissible if the map on $\partial S$ factors through $\partial f:\partial S \to S^1 \to X$, where the second map is $\gamma$. For an admissible map, define $n(S)$ by the equality $[\partial S] \to n(S) [S^1]$ in $H_1(S^1;\mathbb{Z})$ (i.e. $n(S)$ is the degree with which $\partial S$ wraps around $\gamma$). With this notation, one has the following:

Lemma: There is an equality $\text{scl}(g) = \inf_S \frac {-\chi^-(S)} {2n(S)}$.

Note: the function $-\chi^-$ is the sum of $-\chi$ over non-disk and non-sphere components of $S$. By hypothesis, there are none, so we could just write $-\chi$. However, it is worth writing $-\chi^-$ and observing that for more general (orientable) surfaces, this function is equal to the function $\rho$ defined in a previous post.

We now give the proof of the Proposition.

Proof. Suppose to the contrary that stable commutator length does not vanish on $[G,G]$. By Bavard duality, there is a homogeneous quasimorphism $\phi$ with nonzero defect. Rescale $\phi$ to have defect $1$. Then for any $\epsilon$ there are elements $g,h$ with $\phi([g,h]) \ge 1-\epsilon$, and consequently $\text{scl}([g,h]) \ge 1/2 - \epsilon/2$ by Bavard duality. On the other hand, if $X$ is a space with $\pi_1(X)=G$, and $\gamma:S^1 \to X$ is a loop representing the conjugacy class of $[g,h]$, there is a map $f:S \to X$ from a once-punctured torus $S$ to $X$ whose boundary represents $\gamma$. The fundamental group of $S$ is free on two generators $x,y$ which map to the class of $g,h$ respectively. If $w$ is a word in $x,y$ mapping to the identity in $G$, there is an essential loop $\alpha$ in $S$ that maps inessentially to $X$. There is a finite cover $\widetilde{S}$ of $S$, of degree $d$ depending on the word length of $w$, for which $\alpha$ lifts to an embedded loop. This can be compressed to give a surface $S'$ with $-\chi^-(S') \le -\chi^-(\widetilde{S})-2$. However, Euler characteristic is multiplicative under coverings, so $-\chi^-(\widetilde{S}) = -\chi^-(S)\cdot d$. On the other hand, $n(S') = n(\widetilde{S})=d$ so $\text{scl}([g,h]) \le 1/2 - 1/d$. If $G$ obeys a law, then $d$ is fixed, but $\epsilon$ can be made arbitrarily small. So $G$ does not obey a law. qed.

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### 3 Responses to Quasimorphisms and laws

1. Tereez says:

Hmm, veeery interesting-reminds me of a danish I once ate. qed.