# A question on the fundamental group of a compact orientable surface of genus >1

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).

• 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. – YCor Nov 22 '18 at 17:50
• (or do the same with killing $x_i,y_i$ for $i\ge 2$, again with $x_1\mapsto e_{12}(1)$, $y_1\mapsto e_{23}(1)$) – YCor Nov 22 '18 at 18:24
• @YCor I like your proof: it is really elementary! – Xarles Nov 22 '18 at 19:03

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.