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?



Stable w-length
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.


