Let $G$ be a group. It acts canonically on its derived subgroup by conjugation. Can on describe the orbits of this action when $G$ is the fundamental group of a compact orientable surface of genus $g \geq 2$ ? Especially, is the number of orbits finite ?

$\textbf{Edit : Another question has arisen : what about the dynamic of $\pi_1(S)$ on $D\pi_1(S) / D^2\pi_1(S)$ ?} $

  • $\begingroup$ And by the way, does anyone know a free set of generators of $D\pi_1(S)$ ? $\endgroup$
    – Selim G
    Nov 21, 2013 at 16:35

3 Answers 3


The number of the orbits is infinite.

Consider the upper central series, that is a sequence of derived subgroups: $G^1=[G,G]$ and $G^{i+1}=[G^{i},G^{i}]$. All subgroups $G^i$ are normal in the group $G$. Since $G^1$ is free of infinite rank, the sequence $\{G^i\}_{i=1,\ldots,\infty}$ is a sequence of free groups of countable rank that does not stabilize, i.e. $G^{i+1}<G_i$. For any element $x\in [G,G]$ define its depth as $d(x)=\max\{i\mid x\in G_i\}$. Obviously, $d(x)$ is constant on the conjugation orbit, but taking $y_i\in G^{i+1}\setminus G^{i}$ we get an infinite set of elements in $[G,G]$ with different depth, hence they form different orbits.

  • $\begingroup$ Thanks for the answer. Actually I am realizing I'm interested in the action of $G$ on $G^1/G^2$ ! $\endgroup$
    – Selim G
    Nov 21, 2013 at 10:56

I don't know a simple description of the orbits, but I claim that $[G,G]/[G,[G,G]] \cong (\wedge^2 \mathbb{Z}^{2g})/\mathbb{Z}$, which implies that there are infinitely many orbits. Here $\mathbb{Z}^{2g}$ is the abelianization of $G$ and $\mathbb{Z}$ is embedded in $\wedge^2 \mathbb{Z}^{2g}$ as follows. For $x \in G$, let $[x] \in \mathbb{Z}^{2g}$ be its image in the abelianization. Let $a_1,b_1,\ldots,a_g,b_g \in G$ be the usual basis satisfying $[a_1,b_1]\cdots[a_g,b_g]=1$. Then the image of $\mathbb{Z}$ in $\wedge^2 \mathbb{Z}^{2g}$ is generated by $[a_1] \wedge [b_1] + \cdots + [a_g] \wedge [b_g]$ (it's a fun exercise to show that this does not depend on the choice of basis).

Here's how this works. First, let $F$ be the free group on the generators $a_1,b_1,\ldots,a_g,b_g$. Below I will prove that $[F,F]/[F,[F,F]] \cong \wedge^2 \mathbb{Z}^{2g}$. The surjection $F \rightarrow G$ induces a surjection $[F,F]/[F,[F,F]] \rightarrow [G,G]/[G,[G,G]]$, and it will be clear from our description below of $[F,F]/[F,[F,F]]$ that the kernel is generated by $[a_1] \wedge [b_1] + \cdots + [a_g] \wedge [b_g]$ (this comes from the surface relation).

It remains to show that $[F,F]/[F,[F,F]] \cong \wedge^2 \mathbb{Z}^{2g}$. We have a short exact sequence $$1 \longrightarrow [F,F] \longrightarrow F \longrightarrow \mathbb{Z}^{2g} \longrightarrow 1.$$ The usual five-term exact sequence in group homology (see, e.g., Brown's book on group homology) associated to this short exact sequence is of the form $$H_2(F) \rightarrow H_2(\mathbb{Z}^{2g}) \rightarrow (H_1([F,F]))_{\mathbb{Z}^{2g}} \rightarrow H_1(F) \rightarrow H_1(\mathbb{Z}^{2g}) \rightarrow 0.$$ Since $F$ is free, we have $H_2(F)=0$. Also, the map $H_1(F) \rightarrow H_1(\mathbb{Z}^{2g})$ is an isomorphism. Next, we have $H_2(\mathbb{Z}^{2g}) \cong \wedge^2 \mathbb{Z}^{2g}$. Finally, the coinvariants $(H_1([F,F]))_{\mathbb{Z}^{2g}}$ are easily seen to be isomorphic to $[F,F]/[F,[F,F]]$. We conclude that $[F,F]/[F,[F,F]] \cong \wedge^2 \mathbb{Z}^{2g}$, as desired.

  • $\begingroup$ I thought the question referred to the action of $G$ on the second derived quotient of $G$, that is, $H_1([G,G])$, rather than on the second nilpotent quotient of $G$, no? $\endgroup$
    – Alex Suciu
    Nov 21, 2013 at 12:26
  • $\begingroup$ Alex is correct (according to the questioner's comment on Al Tai's answer). However, since $G_1/G_2$ surjects to $G_1/[G,G_1]\cong \textstyle{\bigwedge}^2 \mathbb{Z}^{2g}/\mathbb{Z}$, Andy's argument does still prove that there are infinitely many orbits. $\endgroup$
    – Tom Church
    Nov 21, 2013 at 13:38
  • $\begingroup$ Andy answered the right question, I've just edited my post this morning :) $\endgroup$
    – Selim G
    Nov 21, 2013 at 14:54
  • $\begingroup$ @SelimGhazouani, you should phrase your edit so it's clear what the first version of the question was. $\endgroup$
    – HJRW
    Nov 22, 2013 at 10:22
  • 1
    $\begingroup$ Yes you are right, I'll correct that. $\endgroup$
    – Selim G
    Nov 22, 2013 at 10:47

I'm answering the edit: what is the action of $G=\pi_1(S)$ on $G'/G''$? A first thing to notice is that $G'$ acts trivially on $G'/G''$, so the action factors through to an action of $G/G'$ on $G'/G''$. The $\mathbb{Z}[G/G']$-module $G'/G''$ is known as the first Alexander module of $G$. (If $G$ is the fundamental group of $S^3-K$ for a knot, then this module is where the Alexander polynomial comes from.)

Topologically, $G/G'=H_1(S)$ and $G'/G''=H_1(\overline{S})$, where $\overline{S}$ is the universal abelian cover of $S$. That is, it is the cover associated to the abelianization homomorphism $G\to H_1(G)$, namely the cover with $\pi_1(\overline{S})=G'$ by the correspondence between subgroups and covers. A way to imagine the action of $G/G'$ on $G'/G''$ is that $H_1(S)$ acts on $H_1(\overline{S})$ as the group of deck transformations. We will characterize $H_1(\overline{S})$ as a $\mathbb{Z}[H_1(S)]$-module.

If $S$ is a compact surface with nonempty boundary, its fundamental group is the free group $F_k$ on some number of generators $k$. If $F_k=\langle g_1,g_2,\dots,g_k\rangle$ is the free generating set, the abelianization is a map $F_k\to \mathbb{Z}^k$ sending $g_i$ to a generator $t_i$. The ring $\mathbb{Z}[\mathbb{Z}^k]$ is isomorphic to $\mathbb{Z}[t_1^{\pm 1},\dots,t_k^{\pm 1}]$, the Laurent polynomial ring in $k$ variables with integer coefficients. I can't really explain the following calculations in enough detail here, other than what I am doing is calculating with $C_i(\overline{X_G})\cong \mathbb{Z}[H_1(G)]\otimes C_i(X_G)$ for a presentation complex $X_G$ of $G=\pi_1(S)$. For example, $t_ig_1$ represents the copy of $g_i$ (a distinguished lift of the $1$-cell for $g_i$ to $\overline{X_G}$) shifted by the deck transformation for $t_i$. (I might post a link to some notes about this at some point.) The complex only has $0$- and $1$-cells, so we need only identify the kernel of $\partial_1$, where $\partial_1(g_i)=t_i-1$. Since these are all pairwise coprime polynomials, $\ker\partial_1$ is minimally generated by $(t_j-1)g_i-(t_i-1)g_j$ for all $1\leq i<j\leq k$. Then $$H_1(\overline{S}) \cong \bigoplus_{i=1}^{\binom{k}{2}} \mathbb{Z}[t_1^{\pm 1},\dots,t_k^{\pm 1}]$$ as a free $\mathbb{Z}[t_1^{\pm 1},\dots,t_k^{\pm 1}]$-module. As a special case, $k=1$ is when $S$ is an annulus, and we can see the above matches the observation that $H_1(\overline{S})=0$ since $\pi_1(S)$ is abelian.

In a more topological approach, we can replace $S$ with a wedge of $k$ circles $X=\bigvee_{i=1}^k S^1$. A model for $\overline{X}$ is to take $\mathbb{Z}^k\subset \mathbb{R}^k$ as the $0$-cells and take the length-$1$ line segments between these points as the $1$-cells. The kernel elements identified above are squares spanned by pairs of standard basis vectors.

If $S$ is a closed compact surface of genus $g$, then instead we have $$G=\langle e_1,f_1,\dots,e_g,f_g \mid [e_1,f_1]\cdots[e_g,f_g]=1\rangle.$$ The abelianization is $\pi_1(G)\to \mathbb{Z}^{2g}$ with $e_i\mapsto s_i$ and $f_i\mapsto t_i$ for generators $s_1,t_1,\dots,s_g,t_g$. \begin{align} \partial_1 e_i &= s_i - 1\\ \partial_1 f_i &= t_i - 1\\ \partial_2([e_1,f_1]\cdots[e_g,f_g]) &= (1-t_1)e_1-(1-s_1)f_1 + \dots + (1-t_g)e_g-(1-s_g)f_g \end{align} The situation is very similar to the case of a surface with boundary in that the kernel of $\partial_1$ is the same, but now there is a single relation. This relation lets us write the element $(t_1-1)e_1-(s_1-1)f_1$ in terms of other elements of the generating set. Hence $$H_1(\overline{S}) \cong \bigoplus_{i=1}^{\binom{2g}{2}-1} \mathbb{Z}[s_1^{\pm 1},t_1^{\pm 1},\dots,s_g^{\pm 1},t_g^{\pm 1}]$$ as a free $\mathbb{Z}[s_1^{\pm 1},t_1^{\pm 1},\dots,s_g^{\pm 1},t_g^{\pm 1}]$-module.

Therefore, the orbits of $G=\pi_1(S)$ on $G'/G''$ are in one-to-one correspondence with $\left(\binom{2g}{2}-1\right)$-tuples of Laurent polynomials in $2g$ variables with integer coefficients, modulo multiplication by $s_i$ and $t_i$ for all $1\leq i\leq g$.


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