MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $F$ be a free group, and $w$ an element of $F$. In any group $G$, a $w$-word is the image of $w$ or $w^{-1}$ under a homomorphism from $F$ to $G$. The subgroup of $G$ generated by $w$-words is denoted $G(w)$.

For any $g \in G(w)$, the $w$-length of $g$, denoted $l(g|w)$, is the minimum number of $w$-words in $G$ whose product is $g$, and the stable $w$-length of $g$, denoted $sl(g|w)$, is the limit $sl(g|w) = lim_{n \to \infty} l(g^n|w)/n$.

If $w$ is not in the commutator subgroup of $F$, the stable $w$-length of every element in any group is trivial. Otherwise, one has a universal inequality $$1/2 \le sl_F(w|w) \le 1$$ (where the subscript $F$ indicates that stable $w$-length is being calculated in the free group $F$ containing $w$ itself.)

The lower bound of $1/2$ is realized e.g. by the word $w=xyx^{-1}y^{-1}$ (i.e. a standard commutator) in $F_2$ but I don't know how to compute (or even approximate!) $sl(w|w)$ in (essentially) any other case.

What values are achieved by $sl(w|w)$? Are they all rational? Are they dense? Is $1$ ever achieved? Is $1/2$ ever achieved for a word other than $xyx^{-1}y^{-1}$?

(Added:) After reading FC's answer, it is probably worth pointing out that the lower bound $1/2 \le sl_F(w|w)$ comes from the inequality $scl_G(g) \le sl_G(g|w)(scl_F(w)+1/2)$ for any $g$ in any $G$, so one gets a better lower bound on $sl_F(w|w)$ if one knows $scl_F(w)>1/2$ (the estimate $scl_F(w)\ge 1/2$ is always true). Upper bounds can be established by exhibiting identities (like FC's identity below). Does a blind computer search yield any interesting examples?

share|cite|improve this question
An interesting question! Can you give some indication of the argument showing that the stable $w$-length of every element in any group is trivial if $w$ is not in the commutator subgroup of $F$? – Andy Putman Nov 11 '09 at 3:30
Hi Andy - the argument is actually trivial: if $w$ is not in the commutator subgroup of $F$, then there is a homomorphism from $F$ to $Z$ sending $w$ to some nonzero $n$. This means that the $n$th power of any element (and therefore the $nm$th power for any $m$) in any group is a $w$-word, so the stable $w$ length of anything is zero. – Danny Calegari Nov 11 '09 at 3:39
up vote 4 down vote accepted

Here are some weak observations that don't quite answer any of your questions. Let $g$ be a positive integer, and consider the free group $F_{2g}$ generated by $a_k$ and $b_k$ for $k = 1$ to $g$. Consider the word:

$$w_g = [a_1,b_1][a_2,b_2][a_3,b_3] \ldots [a_g,b_g].$$

Suppose that $\lambda_g = sl(w_g,w_g)$. I claim that for any $x$ in the commutator of $F_2$ with $cl(x) = g$, the stable commutator length $scl(x)$ of $x$ is $\le g \cdot \lambda_g$. Suppose otherwise. First of all, note that for large $n$ we can write $w^n_g$ as the product of (roughly) $n \cdot \lambda_g \cdot g$ commutators. Since the commutator length of $x$ is $g$, there exists a map from $F_{2g}$ to $F_{2}$ such that the image of $w_g$ is $x$. On the other hand, we see that the image of $w^n_g$ is $x^n$, and thus the commutator length of $x^n$ is (asymptotically) at most by $n \cdot \lambda_g \cdot g$, and thus $scl(x) \le \lambda_g \cdot g$.

Example: $cl([x,y]^3) = 2$ and $scl([x,y]^3) = 3/2$, and thus $\lambda_3 \ge 3/4$. In general, the fact that $scl([x,y]) = 1/2$ implies that that $\lambda_g$ tends to one as $g$ increases.

I think one can promote this example to a word in $F_2$. Consider the characteristic homomorphism $\phi_n:F_2 \rightarrow \mathbf{Z} \oplus \mathbf{Z} \rightarrow \mathbf{Z}/n\mathbf{Z} \oplus \mathbf{Z}/n\mathbf{Z}$. Suppose that $n$ is odd, and write $2g = n^2 + 1$. The kernel of $F_2$ is free of rank $2g$. Pick generators for $\ker(\phi_n)$ once and for all, and call them $a_k$ and $b_k$ for $k = 1$ to $g$. We may think of $a_k$ and $b_k$ as elements in $F_2$, but also as formal words. Since $\ker(\phi_n) = F_{2g}$ is characteristic, the formal words $a_k$ and $b_k$ always yield elements of $F_{2g}$ (alternatively, the images of $a_k$ and $b_k$ in $\mathbf{Z} \oplus \mathbf{Z}$ are divisible by $n$, and this will be so for any substitution of elements of $F_2$ for the generators). Let

$$w_g = [a_1,b_1][a_2,b_2] \ldots [a_g,b_g].$$

The argument proceeds as above. If $sl(w_g,w_g) = \mu_g$, then we can write $w^n_g$ (for large $n$) as the product of $n \cdot \mu_g \cdot g$ commutators, each of which is the commutator of a pair of elements of $F_{2g}$ (by the characteristic property of the words $a_k$ and $b_k$ described above). Hence, choosing an appropriate map from $F_{2g}$ to $F_2$, we may deduce that for any $x \in F_2$ with $cl(x) = g$ that $scl(x) \le g \cdot \mu_g$. Thus we have found words $w_g$ in $F_2$ such that $sl(w_g,w_g)$ tends to $1$ as $g$ goes to infinity. Of course, this says nothing about whether $sl(w_g,w_g)$ actually equals $1$ for any $g$.

Finally, a random other example. If $w = [a,b^2]$, then

$$w^3 = [ab^2a^{-1},b^{-2} ab^2a^{-2}][b^{-2} a b^2,b^4] = [b^{-2} a b^2 a^{-2},(aba^{-1})^2]^{-1}[b^{-2} a b^2,(b^2)^2],$$

so $sl(w,w) \le 2/3$.

I wrote this on a very old computer that was too slow for previewing LaTeX, but hopefully this can still be read.

share|cite|improve this answer
Hi FC - very nice answer! Let $a_1,b_1,\cdots,a_g,b_g$ generate a free group $F_g$. Let $w_g = [a_1,b_1]\cdots[a_g,b_g] \in F_g$. Then $scl(w_g) = g-1/2$ exactly in $F_g$, so $sl(w_g|w_g)=1-1/2g$ exactly, so that proves rationality for these elements. I know where you found your formula for $w^3$, but I don't see a pattern . . . – Danny Calegari Nov 13 '09 at 4:45
You two crack me up. – Greg Kuperberg Nov 13 '09 at 5:00
I don't consider this question "answered", but FC's answer is evidently the best answer that is likely to come along. – Danny Calegari Nov 19 '09 at 18:54

In case any one is still thinking about this question, it turns out one can say a lot. For example, $sl(w|w)=1/2$ whenever $w$ is a word of the form $[a,b^n]$ (and several other examples), one has $2/3 \le sl(w|w) \le 4/5$ when $w=[a,b]^2$, and if $\gamma_n$ is the iterated commutator $[x_1,[x_2,\cdots[x_{n-1},x_n]\cdots]$ one has $sl(\gamma_n|\gamma_n)\le 1-2^{1-n}$. In fact, I would explicitly conjecture that $sl(w|w)<1$ (i.e. strict inequality) for every $w$.

Getting systematic lower bounds on $sl(w|w)$, other than $scl(w)/(scl(w)+1/2)$ seems difficult; one imagines that the (currently nonexistent) theory of nonabelian Bavard duality might do the trick.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.