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http://arxiv.org/pdf/math/0611889v4.pdf (page 13)

In the above paper by Danny Calegari he says that the result $\text{scl}(g) \geq 1/2$ (i.e. a stable commutator length $\text{scl}(g) := \displaystyle{\lim_n}\ cl(g^n) / n$, where $\text{cl}(g)$ is the smallest $k$ such that $g$ is a product of $k$ commutators) follows (apart from the Howie-Duncan result) from the paper by Comerford-Edmunds.

"Products of commutators and products of squares in a free group" by Leo P. Comerford , Jr. , Charles C. Edmunds, 1994.

Although I am familiar with that paper, I do not immediately see how that result follows from the description of solutions of quadratic equations described in Comerford&Edmunds.

Here is a reference to the paper:

http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.233.9038

Could anyone clarify the link between the two results?

thanks!

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    $\begingroup$ I think it would be helpful if you would define notation which is not standard, like ${\rm scl}(g)$. Also I think adding a little more details would be in order. $\endgroup$
    – Stefan Kohl ♦
    Commented Apr 1, 2013 at 19:04
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    $\begingroup$ @Stefan: you're right. $scl(g)=\lim cl(g^n)/n$ where $cl(g)$ is the smallest $k$ such that $g$ is a product of $k$ commutators. The quantification "for all nontrivial $g$ in the derived subgroup of a free group" is probably the other missing part. Anyway, even if I'm correct, the post should be edited accordingly. $\endgroup$
    – YCor
    Commented Apr 1, 2013 at 21:48
  • $\begingroup$ I think that $\text{scl}(g)$ (for "stable commutator length of $g$) is pretty standard at this point. $\endgroup$
    – Andy Putman
    Commented Apr 1, 2013 at 22:08
  • $\begingroup$ Andy: I think it depends on the subfield of the group theory. It is standard for people in geometric group theory, but not otherwise. $\endgroup$
    – Misha
    Commented Apr 1, 2013 at 22:34
  • $\begingroup$ ok. I've made some changes to the question. thanks. $\endgroup$
    – Alexey Kvashchuk
    Commented Apr 2, 2013 at 0:49

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