Why isn't $\langle x,y,z|xyzx^{-1}y^{-1}z^{-1}\rangle$ a hyperbolic surface group?

Question

The group mentioned in the title, $\langle x,y,z|xyzx^{-1}y^{-1}z^{-1}=1\rangle$, is in between the torus fundamental group $\langle x,y|xyx^{-1}y^{-1}=1\rangle$ and the two-holed torus fundamental group $\langle x,y,z,w|xyzx^{-1}y^{-1}z^{-1}w^{-1}=1\rangle$.

It is not the hexagonal presentation of the torus because the homology is not the same (gluing up a hexagon with that pattern gives you two vertices instead of one, giving you only two loops in the one-skeleton, so this is not the torus group).

Its Cayley graph can be realized as an infinite tiling of hexagons, each of which meet six to a vertex.

Such a tiling can be embedded in the hyperbolic plane, making the Cayley graph quasi-isometric to hyperbolic space, which means that the group is delta hyperbolic with a circle at infinity, implying that the group is Fuchsian, by the work of Gabai and others.

So it has a finite index surface subgroup. But this group is a subgroup of the RAAG with defining graph the diamond graph, I.e. $F_2\times F_2$. This can be seen by letting $a,b$ generate the first free group, $c,d$ generate the second subgroup, and letting $x=ab^{-1},y=bc^{-1}$, and $z=cd^{-1}$.

*Edit:*I meant to say that $a,c$ generate the first group and $b,d$ generate the second.\

So this implies that the diamond graph contains a hyperbolic subgroup. But in all the RAAG references, they say that RAAG's contain surface subgroups if they contain 5-cycles. So why does it seem as if a four-cycle (the diamond graph) contains a surface subgroup?

Answer 1

I think Lee's and Steve's comments pretty much answer this question. Let me try to summarize, and clear up a couple of misconceptions that seem to be lurking. For convenience, I'll denote your group by $G$.

The map $G\to F_2\times F_2$.

Actually, I don't think the map $G\to F_2\times F_2$ that you describe is an injection. The images of $x$ and $z$ commute, so the image also satisfies the relation $[x,z]=1$. Your relator then becomes $[z^{-1}x,y]=1$. It follows from these two relations that the image is isomorphic to $F_2\times\mathbb{Z}$.

Indeed, the theorem of Baumslag and Roseblade that I alluded to in my comments says more or less (there are some difficulties in getting the statement exactly right) that every fp subgroup of a free-cross-free group is virtually free-cross-free, so we could have guessed it would be of this form.

The group $G$.

If you have a presentation in which every generator (or its inverse) appears exactly twice then it is, indeed, natural to guess that it might be a surface group. And it nearly is. But there's one more thing you need to check: the link of the 1-vertex of the corresponding presentation complex.

This is called the Whitehead graph of the relations. It's necessarily a union of cycles (in the case when each generator appears exactly twice), but the presentation complex is a surface exactly when the Whitehead graph consists of just one cycle. Otherwise, the presentation complex is a surface with some points identified, and so the group is a free product of a surface group and a free group.

The Whitehead graph is easy to calculate; in this case, it turns out to consist of two cycles, and so we have two points identified, and $H_1$ tells us that the surface is in fact a torus.

Therefore, $G\cong\mathbb{Z}*\mathbb{Z}^2$, as Lee and Steve correctly said. But I hope the above gives you some idea of how one might calculate this, rather than just pluck it from thin air.

The Convergence Group Theorem.

By the way, here's another way to see that something must have been wrong. It follows from the Convergence Group Theorem of Tukia, Casson--Jungreis and Gabai, which you mention, that any torsion-free group which is virtually a surface group is in fact a surface group. Therefore, if $G$ really were a subgroup of $F_2\times F_2$ and virtually a surface group, it would have had to have been a surface group. Of course, there's only one candidate, which can be ruled out by looking at $H_1$.