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 $[x_i,x_j]$.