1
$\begingroup$

Let $G$ be a (non-abelian) group, and let $G_2$ denote its commutator subgroup. Then the abelianization $G_2^{ab} = H_1(G_2,\mathbb{Z})$ is a module over the group ring $\mathbb{Z}[G^{ab}]$. The action of $G^{ab}$ on $G_2^{ab}$ is induced by the conjugation action of $G$ on $G_2$. I am interested in understanding the structure of this module when $G$ is a finitely generated free group, $F$, or the fundamental group of a compact Riemann surface $S_g$ of genus $g\geq 2$. In either case, I find it convenient to think of $\mathbb{Z}[G^{ab}]$ as a ring of Laurent polynomials with integer coefficients.

$\textbf{What I know}$:

1) The group structure. In case $G = F_g$ or $G = \pi_1(S_g,*)$ with $g\geq 2$, the commutator subgroup $G_2$ is a free group of infinite rank. Consequently, $H_1(G_2,\mathbb{Z})$ is a free abelian group of infinite rank.

2) Commutators of pairs of generators of $G$ form a generating set for $H_1(G_2,\mathbb{Z})$ as a module over $\mathbb{Z}[G^{ab}]$. For example, if $G = \langle x_1,..., x_n \rangle$ then $H_1(G_2,\mathbb{Z})$ is generated over $\mathbb{Z}[x_1^{\pm 1},..., x_n^{\pm 1}]$ by the classes of the elements $[x_i,x_j]$ for $i < j$. I am fairly certain that $H_1((F_2)_2,\mathbb{Z})$ is a free module of rank $1$ over $\mathbb{Z}[x_1^{\pm1}, x_2^{\pm 1}]$ with the single generator $[x_1,x_2]$.

3) If $G = \pi_1(S_g,*) = \langle x_1,...,x_{2g} | [x_1,x_2]\cdots [x_{2g-1},x_{2g}] = 1\rangle$, the $[x_i,x_j]$ generate $H_1(G_2,\mathbb{Z})$ over $\mathbb{Z}[x_1^{\pm 1},..., x_{2g}^{\pm 1}]$, but there is definitely a relation of the form $\sum_{i=1}^g [x_{2i-1},x_{2i}] = 0$.

$\textbf{My Question}$:

1) For what values of $n\geq 2$ is $H_1((F_n)_2,\mathbb{Z})$ freely generated over $\mathbb{Z}[x_1^{\pm 1},...,x_n^{\pm 1}]$ by the commutators $[x_i,x_j]$ with ${i<j}$?

2) In case $H_1((F_{2g})_2,\mathbb{Z})$ is a free $\mathbb{Z}[x_1^{\pm 1},...,x_{2g}^{\pm 1}]$ module, let $\langle \Theta \rangle$ be the submodule spanned by $\Theta = \sum_{i=1}^g [x_{2i-1},x_{2i}]$. Is there a module isomorphism $$ H_1((F_{2g})_2,\mathbb{Z})/\langle \Theta \rangle\simeq H_1((\pi_1(S_g,*))_2, \mathbb{Z})?$$

My apologies if the level of this question is too low. I feel that the answer must be well-known, but I have been unable to find it despite an extensive search.

$\endgroup$
  • 2
    $\begingroup$ Look up Alexander module/polynomial and Fox calculus. $\endgroup$ – Alex Degtyarev Jan 18 '14 at 19:44
1
$\begingroup$

To answer your Q1, the group in question is $H_1$ of the complex $$0\to\Lambda a_1\oplus\ldots\oplus\Lambda a_n\overset\partial\to\Lambda\to0,$$ where $\Lambda$ is your ring of Laurent polynomials and $\partial\colon a_i\mapsto(x_i-1)$. (Clearly, the $a_i$ are in a correspondence with the generators of $F_n$.) So, it's very unlikely that $\ker\partial$ is a free module; in any case, you suggest too many generators!

For the other group, you replace the leftmost $0$ with another copy of $\Lambda$, mapping its generator to $$(1-x_2)a_1+(x_1-1)a_2+(1-x_4)a_3+(x_3-1)a_4+\ldots,$$ and still compute $H_1$ of the resulting complex.

The elements $(1-x_j)a_i+(x_i-1)a_j$ are the commutators (up to sign, depending on the definitions), so, to an extent, the answer to your Q2 is "yes". As to Q1, I would say that the module is generated by the commutators (probably, a bit of algebra is needed), but definitely not freely; e.g., there are obvious relations for each triple of indices $(i,j,k)$.

All statements are immediate from the explicit CW-complex realizing the group: a graph in the former case and a graph with a single $2$-cell in the latter.

$\endgroup$
  • $\begingroup$ Thanks, Alex. I guess I just didn't know where to look for this stuff. $\endgroup$ – Kevin Jan 19 '14 at 0:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.