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2 Corrected LaTeX errors

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

$w_g$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^{-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],$$so sl(w,w) \le 2/3. I wrote this on a very old computer that was too slow for previewing LaTeX, but hopefully this can still be read. 1 Here are some weak observations that don't quite answer any of your questions. Let g be a positive integer, and consider the free group F_{2g} generated by a_k and b_k for k = 1 to g. Consider the word:$$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.