Let $G=\pi(X,x)$ be the fundamental group of a compact orientable surface of genus $g\ge 2$. It is well known that a presentation of $G$ is $$G=\langle x_1,y_1,\dots,x_g,y_g \ | \ [x_1,y_1]\cdots [x_g,y_g]=1\rangle$$ (where $[x,y]=xyx^{-1}y^{-1}$ is the commutator).

Denote by $F$ be the free group with $2g$ generators $x_1,y_1,\dots,x_g,y_g$ and by $R$ be the normal closure of the relation $r=[x_1,y_1]\cdots [x_g,y_g]$, so $G=F/R$.

It is clear that $r\in [F,F]$.

Question: Is there an elementary proof that $r=[x_1,y_1]\cdots [x_g,y_g]\not\in [F,[F,F]]$?

This result appears when one considers the Stallings exact sequence associated to $$1\to [G,G]\to G\to G^{ab}\to 1$$ to get $$ H_2(G,\mathbb{Z})\to H_2(G^{ab},\mathbb{Z})\to [G,G]/[G,[G,G]]\to H_1(G,\mathbb{Z})\to H_1(G^{ab},\mathbb{Z}) \to 0$$

Since $H_1(G,\mathbb{Z})\cong H_1(G^{ab},\mathbb{Z})\cong G^{ab}$ we obtain a short exact sequence $$ H_2(G,\mathbb{Z})\to H_2(G^{ab},\mathbb{Z})\to [G,G]/[G,[G,G]]\to 0$$ which should be injective at the left (see at the end some argument why).

Now, Hopf's formula gives $$H_2(F/R,\mathbb{Z})=(R\cap [F,F])/[R,F]=R/[R,F]$$ since $R\subset [F,F]$, hence $H_2(F/R,\mathbb{Z})$ is cyclic and the generator is given by (the class of) $r$. So the map $\psi:H_2(G,\mathbb{Z})\to H_2(G^{ab},\mathbb{Z})$ is either injective or zero. But $$H_2(G^{ab},\mathbb{Z})\cong [F,F]/[F,[F,F]]$$ since $G^{ab}\cong F/[F,F]$, and so the map $\psi$ is given by the natural map $$\psi:R/[R,F]\to [F,F]/[F,[F,F]]$$ coming from the inclusion $R\hookrightarrow [F,F]$, hence $\psi$ is injective if and only if $r\not\in [F,[F,F]]$.

One possibility is to use another description of the map $\psi$ as $$H_2(G,\mathbb{Z})\to \bigwedge H_1(G^{ab},\mathbb{Z})$$ that should correspond to the dual of the cup product in cohomology via Poincaré duality (i.e. dual universal coefficient theorem) (but I am not sure if to consider this approach as really elementary).

  • 7
    $\begingroup$ Yes it's elementary: consider the group $H$ of upper triangular integral matrices with 1 on the diagonal (integral Heisenberg group): it's 2-nilpotent. Write $e_{ij}(t)=1+tE_{ij}$. Consider the homomorphism $F\to H$ mapping $x_i$ to $e_{12}(1)$ and $y_i$ to $e_{23}(1)$. Hence it maps $[x_i,y_i]$ to $e_{13}(1)$, and hence $r=\prod [x_i,y_i]$ to $e_{13}(g)$. So $r\notin [F,[F,F]]$, since otherwise it would be in the kernel of every homomorphism to every 2-nilpotent group. $\endgroup$
    – YCor
    Nov 22, 2018 at 17:50
  • $\begingroup$ (or do the same with killing $x_i,y_i$ for $i\ge 2$, again with $x_1\mapsto e_{12}(1)$, $y_1\mapsto e_{23}(1)$) $\endgroup$
    – YCor
    Nov 22, 2018 at 18:24
  • $\begingroup$ @YCor I like your proof: it is really elementary! $\endgroup$
    – Xarles
    Nov 22, 2018 at 19:03

2 Answers 2


Probably the easiest way to see that the map $\psi\colon H_2(G) \rightarrow H_2(G^{\text{ab}})$ is injective is as follows. Since we're dealing with a surface group, the surface $\Sigma_g$ itself is an Eilenberg-MacLane space. Let $\{a_1,b_1,\ldots,a_g,b_g\}$ be the usual collection of oriented simple closed curves that one draws whose homology classes form a basis for $H_1(\Sigma_g)$. Thus $a_i$ intersects $b_i$ once, and otherwise the curves are disjoint. Let $f\colon G^{\text{ab}} \rightarrow \mathbb{Z}^2$ be the map whose kernel is spanned by $\{[a_2],[b_2],\ldots,[a_g],[b_g]\}$ and which takes $[a_1]$ and $[b_1]$ to the usual basis for $\mathbb{Z}^2$. To prove that $\psi$ is injective, it is enough to prove that the composition $$\phi\colon H_2(G) \stackrel{\psi}{\longrightarrow} H_2(G^{\text{ab}}) \stackrel{f_{\ast}}{\longrightarrow} H_2(\mathbb{Z}^2)$$ is injective. But $\phi$ is easy to understand geometrically: the surface $\Sigma_g$ is an Eilenberg-MacLane space for $G$, a torus $T$ is an Eilenberg-MacLane space for $\mathbb{Z}^2$, and $\phi$ is induced by the map $\Phi\colon \Sigma_g \rightarrow T$ that collapses a genus $(g-1)$-subsurface with one boundary component to a point. This subsurface contains $a_2,b_2,\ldots,a_g,b_g$. The point here is that it is obvious that $\Phi_{\ast}$ takes the fundamental class of $\Sigma_g$ to the fundamental class of $T$, and thus induces an isomorphism on $H_2$.

You can soup up this argument to show that $\psi$ takes the fundamental class of $\Sigma_g$ to the element $a_1 \wedge b_1 + \cdots + a_g \wedge b_g$ of $H_2(\mathbb{Z}^{2g}) \cong \wedge^2 \mathbb{Z}^{2g}$. For more details, see Theorem 2.7 of the lecture notes from my Park City course on the Torelli group, which are available here.


The dual to the map $\psi\colon H_2(G,\mathbb{Z}) \to H_2(G^{\operatorname{ab}},\mathbb{Z})$ is the cup-product map $\cup\colon H^1(G,\mathbb{Z})\wedge H^1(G,\mathbb{Z}) \to H^2(G,\mathbb{Z})$; see e.g. Lemma 1.10 in arXiv:math/9812087. Clearly, the latter map is surjective; hence, the former map must be injective.

  • $\begingroup$ But, why the dual map is the cup-product map? $\endgroup$
    – Xarles
    Nov 23, 2018 at 8:38
  • 1
    $\begingroup$ Added a reference. $\endgroup$
    – Alex Suciu
    Nov 23, 2018 at 12:49

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