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 $[x_i,x_j]$.
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- 2$\begingroup$ See the comments here. $\endgroup$Dietrich Burde– Dietrich Burde2018-03-23 19:48:36 +00:00Commented Mar 23, 2018 at 19:48
- 3$\begingroup$ @DietrichBurde: I hadn't seen that question, but the comments don't look like they contain any nontrivial information. $\endgroup$Linda– Linda2018-03-23 20:26:18 +00:00Commented Mar 23, 2018 at 20:26
- 2$\begingroup$ @HJRW: I think you must have a typo or something in your 1st sentence: the conjugates of commutators of basis elements generate $F'$, but they are not a free basis -- if they were, then they would project to a $\mathbb{Z}$-basis for the abelianization $[F,F]^{\text{ab}}$ and it would follow that this abelianization is a free $\mathbb{Z}[F_n]$-module (while in fact the $F_n$-action descends to an action of $\mathbb{Z}^n$; what is more, it is not even a free $\mathbb{Z}[\mathbb{Z}^n]$-module). Can you give me the precise reference in Serre's book so I can figure out what you are trying to say? $\endgroup$Linda– Linda2018-03-24 15:41:52 +00:00Commented Mar 24, 2018 at 15:41
- 4$\begingroup$ @HJRW: I see! It is true that for $n=2$ these form a basis, but I don't think this is true for higher $n$. The issue is that $[F,F]^{\text{ab}}$ is not a free $\mathbb{Z}[\mathbb{Z}^n]$-module for $n \geq 3$. There are a lot of ways to see this; one easy one is to use the fact that $H_k(\mathbb{Z}^n;[F_n,F_n]^{\text{ab}}) \cong \wedge^{k+2} \mathbb{Z}^n$ together with the fact that if $M$ if a free $\mathbb{Z}[G]$-module, then $H_k(G;M)=0$ for $k \geq 1$. $\endgroup$Linda– Linda2018-03-24 16:44:45 +00:00Commented Mar 24, 2018 at 16:44
- 4$\begingroup$ nb: to see that $H_k(\mathbb{Z}^n;[F_n,F_n]^{\text{ab}}) \cong \wedge^{k+2} \mathbb{Z}^n$, simply use the fact that the spectral sequence for the extension $1 \rightarrow [F_n,F_n] \rightarrow F_n \rightarrow \mathbb{Z}^n \rightarrow 1$ has to degenerate to give the homology of $F_n$, so all of the differentials have to be isomorphisms. $\endgroup$Linda– Linda2018-03-24 16:46:08 +00:00Commented Mar 24, 2018 at 16:46
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1 Answer
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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).
answered Jan 11, 2021 at 20:23