Let $G=F_2$ be the free group of rank $2$. Is there a constant $c>0$ such that the word length $|[u,v]|$ of every commutator $[u,v]=uvu^{-1}v^{-1}$ where $u,v\in G$, $|u|,|v|>0$ is at least $c(|u|+|v|)$ unless $[u,v]=[u_1,v_1]$ for some $u_1,v_1$ with $|u_1|+|v_1|<|u|+|v|$?
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$\begingroup$ Let $c_0$ be the smallest such constant (so $0\le c_0<2$ and the question is whether $c_0>0$): what is the best upper bound you know on $c_0$? And do you conjecture some precise value for $c_0$? $\endgroup$– YCorCommented Jul 25, 2021 at 18:16
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$\begingroup$ I might be missing something in my argument but to me it seems that every $[u,v]$ has a minimal representative of the form $[u_1,v_1]$, where $|[u_1,v_1]|=2(|u_1|+|v_1|)$, is this incorrect? $\endgroup$– user127776Commented Jul 25, 2021 at 19:52
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$\begingroup$ @user127776: It might be true. How did you prove that? $\endgroup$– markvsCommented Jul 25, 2021 at 19:55
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$\begingroup$ I had an argument using covering spaces. I think you can prove that the covering space corresponding to the commutator subgroup of wedge of two circles is a lattice (\mathbb{Z}^2). Then you lift a commutator to this lattice, it becomes a loop. The important thing is that loops in a lattice are very rigid, there is not much wiggle room, for every loop in the lattice you can assign a minimal loop (by cancelling the parts of loop that for example goes from vertex $x$ to $y$ then immediately goes from $y$ to $x$). Now I think it is not hard to see the minimal loop itself is a commutator and has $\endgroup$– user127776Commented Jul 25, 2021 at 20:02
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$\begingroup$ the minimum perimeter among the representatives of the chosen loop in the lattice. Its image in the wedge of circles can be generated by a word of length $2(|u_1|+|v_1|)$ i.e. the perimeter. Now I think it is not hard to check that this is the minimal word length. (if there is a shorter representative its lift to the lattice will give us a loop with smaller perimeter which is a contradiction.)(I might be missing some details ...) $\endgroup$– user127776Commented Jul 25, 2021 at 20:05
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Victor Guba sent me a proof. The proof is based on the old result by Wicks, Wicks, N. J. Commutators in free products. J. London Math. Soc. 37 (1962), 433–444., which describes all words which are commutators in a free group (Lemma 5 in the paper). By that result, a word is a commutator iff it is a conjugate of a reduced word of the form $ABCA^{-1}B^{-1}C^{-1}$. This immediately implies that one can take $c=1/4$ (possibly bigger).