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

Let be a group, and let denote the commutator subgroup. Every element of can be expressed as a product of commutators; the *commutator length* of an element is the minimum number of commutators necessary, and is denoted . The *stable commutator length* is the growth rate of the commutator lengths of powers of an element; i.e. . Recall that a group is said to satisfy a *law* if there is a nontrivial word in a free group for which every homomorphism from to sends to .

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 obeys a law. Then the stable commutator length vanishes identically on .

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

**Definition:** A function is a *quasimorphism* if there is some least number (called the* defect*) so that for any pair of elements there is an inequality . A quasimorphism is *homogeneous* if it satisfies for all integers .

Note that a homogeneous quasimorphism with defect zero is a homomorphism (to ). The defect satisfies the following formula:

**Lemma: **Let be a homogeneous quasimorphism. Then .

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

**Theorem:** (Bavard duality) There is an equality where the supremum is taken over all homogeneous quasimorphisms with nonzero defect.

In particular, vanishes identically on if and only if every homogeneous quasimorphism on is a homomorphism.

One final ingredient is another geometric definition of in terms of Euler characteristic. Let be a space with , and let be a free homotopy class representing a given conjugacy class . If is a compact, oriented surface without sphere or disk components, a map is *admissible* if the map on factors through , where the second map is . For an admissible map, define by the equality in (i.e. is the degree with which wraps around ). With this notation, one has the following:

**Lemma:** There is an equality .

Note: the function is the sum of over non-disk and non-sphere components of . By hypothesis, there are none, so we could just write . However, it is worth writing and observing that for more general (orientable) surfaces, this function is equal to the function 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 . By Bavard duality, there is a homogeneous quasimorphism with nonzero defect. Rescale to have defect . Then for any there are elements with , and consequently by Bavard duality. On the other hand, if is a space with , and is a loop representing the conjugacy class of , there is a map from a once-punctured torus to whose boundary represents . The fundamental group of is free on two generators which map to the class of respectively. If is a word in mapping to the identity in , there is an essential loop in that maps inessentially to . There is a finite cover of , of degree depending on the word length of , for which lifts to an embedded loop. This can be compressed to give a surface with . However, Euler characteristic is multiplicative under coverings, so . On the other hand, so . If obeys a law, then is fixed, but can be made arbitrarily small. So does not obey a law. qed.

## 3 comments

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June 7, 2009 at 7:19 pm

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

June 8, 2009 at 3:03 pm

Combable functions « Geometry and the imagination[...] quasimorphism for (i.e. a quasimorphism achieving equality under generalized Bavard duality; see this post) for which there is with homogenization satisfying . Consequently, if every point in the boundary [...]

July 16, 2009 at 1:32 am

van Kampen soup and thermodynamics of DNA « Geometry and the imagination[...] boundary, taken over all configurations, is precisely the stable commutator length of ; see e.g. here for a [...]