Stable Commutator Length

A reminder on scl

If $w \in G'$ is an element of the commutator subgroup of a group, we can define $\operatorname{cl}(w)$ as its minimal length with respect to the generating set of $G'$ consisting of all commutators. It is not hard to see that the quantity

$$\operatorname{scl}(w) = \lim_{n \to \infty} \frac{\operatorname{cl}(w^n)}{n} = \inf_n \frac{\operatorname{cl}(w^n)}{n}$$

is also well-defined. It is called the stable commutator length of $w$.

It turns out that scl is closely related to another class of objects: quasimorphisms on the group. We say that a function $f \colon G \to \mathbb{R}$ is a quasimorphism if

$$D(f) = \sup_{g,h \in G} |f(gh)-f(g)-f(h)| < \infty.$$

Every quasimorphism is, in a certain sense, equivalent to a homogeneous one—that is, one satisfying $f(x^n)=nf(x)$. Denote the space of homogeneous quasimorphisms by $Q^h$.

Here are a few non-obvious facts about $\operatorname{scl}(w)$ that we use:

• If $G$ is a free group, then the scl of every nontrivial element is at least $1/2$.

• If $w=[x,y]$ is a single commutator, then $\operatorname{scl}(w) \leq 1/2$.

• The problem of computing $\operatorname{scl}(g)$ in a free group is decidable: there is an algorithm called scallop, based on linear programming.

• Bavard duality says that

  $$\operatorname{scl}(g)=\frac12\sup_{f\in Q^h}\frac{|f(g)|}{D(f)}.$$

Brooks quasimorphisms (reference) are a family of functions parametrized by an element $v \in F$. They are defined as follows:

$$f_v(g) = \bigl(\text{number of occurrences of }v\text{ in }g\bigr) - \bigl(\text{number of occurrences of }v\text{ in }g^{-1}\bigr).$$

It is claimed that if $v$ is cyclically minimal, then $f_v$ is a homogeneous quasimorphism.

All of this can be put to use in group theory. For example, one can ask: how are the properties of a one-relator group $\langle S \mid r\rangle$, with its relator in the commutator subgroup, related to $\operatorname{scl}(r)$? This is studied to some extent in this beautiful paper.

Let me also recall that the general isomorphism problem for one-relator groups is still open (there is a survey of the theory here). Using a computer search (and Claude), I recently found a counterexample to this conjecture (paper).

A counterexample

Let me recall the statement: suppose that $r \in F(S)'$ and $r' \in F(S')'$, and

$$\langle S \mid r\rangle \cong \langle S' \mid r'\rangle.$$

Must we then have $\operatorname{scl}(r)=\operatorname{scl}(r')$?

Here is a counterexample. Take $S=S'=\lbrace a,b\rbrace$, and let

$$r=\mathtt{aabABabABBAbaabABBAb}, \qquad r'=\mathtt{aabABabABabABBAbaBAb}.$$

It came out of a search for the first potential counterexample. The search was fairly simple.

1. First, observe that if $\langle S \mid r\rangle \cong \langle S' \mid r'\rangle$, then

   $$\langle S \mid r\rangle_{\mathrm{ab}} \cong \mathbb{Z}^{|S|} \cong \langle S' \mid r'\rangle_{\mathrm{ab}} \cong \mathbb{Z}^{|S'|}.$$

   Thus, for our counterexample we need to take $S=S'$. We choose the first nontrivial case, $S=\lbrace a,b\rbrace$, so we work with the group $F=F(a,b)$.

2. Next, observe that if $\varphi \colon F \to F$ is an automorphism, then $\operatorname{scl}(\varphi(r))=\operatorname{scl}(r)$. Therefore, we need to look for $r$ and $r'$ in different orbits of the tautological action of $\operatorname{Aut}(F)$.

Then, with Claude's help, we construct unique representatives of the $\operatorname{Aut}(F)$-orbits of elements of $F'$. For each representative we compute various invariants—also with Claude in my case—such as the number of homomorphisms to certain finite groups. If a pair $r,r'$ cannot be distinguished by our tests, we try to construct an isomorphism. This turns out not to be too difficult to do on a computer. Finally, for pairs $r,r'$ that we managed to prove define isomorphic groups, we compute $\operatorname{scl}(r)$ and $\operatorname{scl}(r')$ using scallop—and victory: we have found a counterexample!

The reproducible code is available on GitHub.

All that remains is to see, at least a little more humanly, that the scl values of these words really are different. For this example, that turns out not to be difficult:

1. By Bavard duality,

   $$\operatorname{scl}(r) \geq \frac{f(r)}{2D(f)}$$

   for every homogeneous quasimorphism $f$. Take $f=f_v$, the Brooks quasimorphism associated with $v=\mathtt{bA}$. It turns out that $D(f_v)=2$ and $f_v(r)=3$, so the formula gives $\operatorname{scl}(r) \geq 3/4$.

2. For $r'$, everything is simpler: it turns out that $r'$ is conjugate to a commutator:

   $$r'=a[ab,ABabABabA]A.$$

   Therefore, $\operatorname{scl}(r') \leq 1/2$ by the observation from the previous part. To be precise, we can also use the first observation from that part, which gives $\operatorname{scl}(r') \geq 1/2$. Thus $\operatorname{scl}(r')=1/2$.