Generators for commutator subgroup of surface group Let $\pi_1(\Sigma_g) = \langle\text{$x_1,\ldots,x_{2g}$ $|$ $[x_1,x_2]\cdots[x_{2g-1},x_{2g}]$}\rangle$ be a surface group.  Can anyone tell me an explicit free basis for the commutator subgroup of $\pi_1(\Sigma_g)$?  I would prefer one consisting of conjugates of the elementary commutators $[x_i,x_j]$.  
 A: Before I answer this old question, a confession: the user "Linda" is actually me!  I asked this basically to make sure that something I proved was not already known.  This is such a classical subject that you can never be sure!
Anyway, I finally got around to writing up a paper that among other things answers this question.  It can be downloaded here.  Theorem B of it shows that the commutator subgroup of a surface group is freely generated by the set
$$\{\text{$[x_i,x_j]^{x_i^{k_i} \cdots x_{2g}^{k_{2g}}}$ $|$ $1 \leq i<j \leq 2g$, $(i,j) \neq (1,2)$, and $k_i,\ldots,k_{2g} \in \mathbb{Z}$}\}$$
Here I'm using superscripts to indicate conjugation: $a^b = b^{-1} a b$.
This should be compared to a theorem of Tomaszewski that says that the commutator subgroup of a free group $F_n$ on $n$ generators $\{x_1,\ldots,x_n\}$ is freely generated by the set
$$\{\text{$[x_i,x_j]^{x_i^{k_i} \cdots x_{n}^{k_{n}}}$ $|$ $1 \leq i<j \leq n$, and $k_i,\ldots,k_{n} \in \mathbb{Z}$}\}$$
In other words, when you impose the surface relation to go from $F_{2g}$ to the surface group, you have to just omit the conjugates of $[x_1,x_2]$.  I give a new proof of Tomaszewski's theorem in my paper as well (see Theorem A).
