Qfyilyi

Let $G=pi(X,x)$ be the fundamental group of a compact orientable
surface of genus $gge 2$. It is well known that a presentation of
$G$ is
$$G=langle x_1,y_1,dots,x_g,y_g | [x_1,y_1]cdots
[x_g,y_g]=1rangle$$ (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 $rin [F,F]$.

Question: Is there an elementary proof that $r=[x_1,y_1]cdots [x_g,y_g]notin [F,[F,F]]$?

This result appears when one considers the Stallings exact
sequence associated to
$$1to [G,G]to Gto G^{ab}to 1$$
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 $$H_2(F/R,mathbb{Z})=(Rcap
[F,F])/[R,F]=R/[R,F]$$ since $Rsubset [F,F]$, 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 $Rhookrightarrow [F,F]$, hence $psi$ is
injective if and only if $rnotin [F,[F,F]]$.

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).

Probably the easiest way to see that the map $psicolon H_2(G) rightarrow H_2(G^{text{ab}})$ 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 $fcolon G^{text{ab}} rightarrow mathbb{Z}^2$ 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
$$phicolon H_2(G) stackrel{psi}{longrightarrow} H_2(G^{text{ab}}) stackrel{f_{ast}}{longrightarrow} H_2(mathbb{Z}^2)$$
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 $Phicolon Sigma_g rightarrow T$ 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
$psicolon H_2(G,mathbb{Z}) to H_2(G^{operatorname{ab}},mathbb{Z})$ is the cup-product map $cupcolon H^1(G,mathbb{Z})wedge H^1(G,mathbb{Z}) to H^2(G,mathbb{Z})$; see e.g. Lemma 1.10 in arXiv:math/9812087. Clearly, the latter map is surjective; hence, the former map must be injective.