Prove these are not surface groups

Question

For $g,n \geq 1$, let $\Gamma_{g,n}$ be the group with the following presentation: $$\langle \text{$a_1,b_1,\ldots,a_g,b_g$ $|$ $[a_1,b_1]^n [a_2,b_2] \cdots [a_g,b_g]=1$} \rangle.$$ For $n = 1$, these are the fundamental groups of close genus $g$ surfaces. I expect that they are not isomorphic to fundamental groups of closed surfaces for $n \geq 2$, but I can't figure out a proof (except for the trivial case $g=1$, where they have torsion). Note that from the abelianization you can see that if they are surface groups they are also genus $g$ surface groups. Can anyone prove this?

Answer 1

Here is an argument inspired by but not actually using one-relator group theory. Let's write $x_0$ instead of $a_1$ and $t$ instead of $b_1$ and then Tietze transform the presentation to $$ \Gamma_{g, n} \cong \langle x_0, x_1, a_2, b_2, \dots, a_n, b_n, t \mid (x_0^{-1} x_1)^n [a_2, b_2] \dots [a_n, b_n] = 1, x_0^t = x_1 \rangle. $$ This expresses the group as an HNN extension over $\mathbb{Z}$. The base group however cannot be a subgroup of a surface group because its abelianization has torsion.

Answer 2

One can do this using group cohomology, specifically the symplectic form $H^1(G,\mathbb Z)\times H^1(G, \mathbb Z) \to H^2(G,\mathbb Z)$ arising from the cup product. I guess this is closely related to the two-step nilpotent completion suggested by YCor, the advantage is that the classification of symplectic forms is more transparent.

Given a one-relator group whose relator has a trivial abelianization, there is a natural map $H^2(G, \mathbb Z) \to \mathbb Z$ which sends a central extension to the class of the relator (after lifting the generators, and this doesn't depend on the choice of a lift). It's not so hard to check this is an isomorphism.

I'll change your notation slightly so the generators are $a_1,\dots, a_{2g}$ and the relator is $[a_1,a_{g+1}]^n [a_2, a_{g+2}]\dots [a_g, a_{2g}]$.

$H^1(G,\mathbb Z)$ consists of linear forms on the abelianization, i.e. it is freely generated by $c_i$ that sends $a_i$ to $1$ and all other $a_j,a_j$ to $0$.

To calculate $c_i \vee c_j$, we just need to find the corresponding central extension and look at the image of the relation in it. The central extension is pulled back from $\mathbb Z \times \mathbb Z$, where it is clearly the unique nontrivial central extension, i.e. the one with relators $[[a_i,a_j],a_i]$ and $[[a_i,a_j],a_j]$. The center is generated by $[a_i,a_j]$.

The image of $[a_1,a_{g+1}]^n [a_2, a_{g+2}]\dots [a_g, a_{2g}]$ is given by the sum for each appearence of $a_j$ of the number of copies of $a_i$ that appear before it minus the number of copies of $a_i^{-1}$ that appear before it. This gives a $2g\times 2g$ symplectic matrix which is a block sum of $g-1$ copies of $J_2$, the $2 \times 2$ standard symplectic form, plus one copy of $n \cdot J_2$.

The integer $n$ is an invariant of the isomorphism class of this symplectic form, as we can tell by putting it in Smith normal form (invariant under a coarser notion of isomorphism), which is a diagonal matrix with two $n$s followed by $2g-2$ ones.

Answer 3

Here's an argument that uses only basic combinatorial group theory (Reidemeister-Schreier).

Let $n \geq 2$, and let $G = \langle a_1, b_1, \dots, a_g, b_g | [a_1, b_1]^n [a_2, b_2] \cdots [a_g, b_g] \rangle$. I claim that $G$ has an index $2$ subgroup $H$ with presentation $$ K = \langle x, b_1, \dots, a_g, b_g, \: b'_1, \dots, a'_g, b'_g \: | \: \\(b_1'^{-1}b_1)^n [a_2', b_2'] \cdots [a_g', b_g'] =1 \\ (b_1^{-1}xb_1'x^{-1})^n [a_2, b_2] \cdots [a_g, b_g] = 1 \rangle. $$ Indeed, let $H \leq G$ be the subgroup generated by $$ H = \langle a_1^2, b_1, a_2, b_2, \dots, a_g, b_g, a_2^{a_1}, a_3^{a_1}, \dots, a_g^{a_1}, b_g^{a_1} \rangle $$ where $g^h = h^{-1}gh$ denotes conjugation. Then $H$ is clearly a (normal) subgroup of index $2$ in $G$.

Using the Reidemeister-Schreier method to get a $2$-relator presentation for $H$, we find exactly $K$; explicitly, we have that $K \cong H$ by the isomorphism mapping $x \mapsto a_1^2, a_i \mapsto a_i, b_i \mapsto b_i$ for all $1 \leq i \leq g$ (except $a_1$), and finally $a_i' \mapsto a_1^{-1}a_i a_1$ and $b_i' \mapsto a_1^{-1}b_i a_1$. This is obtained by rewriting the two conjugates of the relator of $G$ in the usual (easy, in this case) manner.

The presentation for $K$ now clearly shows that $K^{\operatorname{ab}}$ has torsion (of order $n$), and hence $K$ cannot be a surface group when $n>2$. But every subgroup of a surface group is a surface group, so $G$ is also not a surface group when $n>2$. When $n=2$, we can use the fact that orientable surface groups have no finite index subgroup isomorphic to a non-orientable surface group to get the same contradiction.

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Note: when $g=1$, i.e. we consider $\langle a, b | [a, b]^n = 1 \rangle$, then this is virtually a surface group of genus $g' > 1$ when $n \geq 2$. So even though there is torsion, it is not very far at all from being a surface group.

Answer 4

Here is an answer that is inspired by but not using Gromov's simplicial norm considerations. If we use the Gromov norm, we can distinguish all $\Gamma_{g,n}$; See the second part of the answer.

The group of interest has a surface subgroup $G$ of infinite index, whereas subgroups of infinite index in a surface group must be free groups. To find such a surface subgroup $G$ of infinite index, think of $\Gamma_{g,n}=F_2\star_{\mathbb{Z}}F_{2g-2}$ as an amalgam, where the image of a generator of $\mathbb{Z}$ in $F_2$ (resp. $F_{2g-2}$) is $[a_1,b_1]^{-n}$ (resp. $[a_2,b_2]\cdots[a_g,b_g]$). If $n$ is odd, there is a subgroup $H$ of $F_2$ of index $n$ containing the $\mathbb{Z}$ subgroup, such that $G=H\star_\mathbb{Z} F_{2g-2}$ is a closed surface group, which is easy to see to be of infinite index in $\Gamma_{g,n}$. To find $H$, think of $F_2$ as $\pi_1(S)$, where $S$ is a torus with one boundary. For $n$ odd, it has a degree $n$ cover $S_n$ homeomorphic to a surface of genus $(n+1)/2$ and one boundary, so that the boundary of $S_n$ covers the boundary of $S$ by a degree $n$ map. Gluing the boundary with the genus $g-1$ surface with one boundary gives the closed surface representing the surface group $G$.

When $n$ is even, a similar construction works, except that we need to take a degree $2n$ cover of $S$ that has genus $n$ and two boundary components, each covering the boundary of $S$ by a degree $n$ map. Close up these two boundary components by gluing each component with a copy of the genus $g-1$ surface with one boundary. This closed surface represents the desired surface subgroup $G$.

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A less elementary (but maybe better?) proof shows that $\Gamma_{g,n}$ are non-isomorphic for different pairs of $(g,n)$. This uses the Gromov norm and stable commutator length (scl). The computation below uses some simple scl values of elements in free groups; See the book of Calegari for references. Note that $H_2(\Gamma_{g,n})\cong\mathbb{Z}$, and one can compute the Gromov norm $\|\sigma\|_1$ of a generator $\sigma$ (sitting inside $H_2(\Gamma_{g,n};\mathbb{R})$). As a graph of groups with free vertex groups and cyclic edge groups, Calegari showed in https://arxiv.org/pdf/0803.4137.pdf (page 8) that $$\|\sigma\|_1=4[\mathrm{scl}_{F_2}([a_1,b_1]^n)+\mathrm{scl}_{F_{2g-2}}([a_2,b_2]\cdots[a_g,b_g])]=4(\frac{n}{2}+\frac{2g-3}{2}),$$ which is distinct for distinct $n\ge1$.

Any isomorphism would take $\sigma$ to a generator of $H_2$, and the Gromov norm is invariant under isomorphisms. Hence the groups $\Gamma_{g,n}$ are not isomorphic fixing $g$ varying $n$. But of course they are not isomorphic when we have different $g$ by considering the abelianization as noted in the post.

The surface subgroups exhibited in the first part are actually fundamental groups of extremal surfaces for the Gromov norm of $\sigma$, i.e. they achieve the infimum in the definition of Gromov norm using surface maps; See e.g. page 11 of the book above for such a definition. Such extremal surfaces are always $\pi_1$-injective.