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I believe that there are four ways to glue (all) the edges of a regular Euclidean hexagon to get a locally CAT(0) space: Gluings of a hexagon The first two give the torus and the Klein bottle, respectively. What are the last two? In particular, do their fundamental groups have another name? Do they have the same fundamental group?

EDIT: HJRW points out that I missed the several ways to glue a hexagon to get a non-orientable surface of Euler characteristic -1. I count 8 possibilities up to symmetries.

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    $\begingroup$ Maybe I'm missing something obvious here, but how do you glue three edges together? $\endgroup$
    – Steven Stadnicki
    Commented Oct 31, 2018 at 23:12
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    $\begingroup$ You just glue them. It's not a surface. $\endgroup$
    – Lee Mosher
    Commented Nov 1, 2018 at 1:52
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    $\begingroup$ My motivation was very simple: I wanted an example for my class on Metric Geometry. But then I was surprised I hadn't seen these examples. $\endgroup$
    – Dylan Thurston
    Commented Nov 2, 2018 at 19:25
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    $\begingroup$ Wait, what about the surface of Euler characteristic -1? Shouldn’t that be on your list? $\endgroup$
    – HJRW
    Commented Nov 3, 2018 at 7:50
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    $\begingroup$ @HJRW, indeed I missed those! There are several gluings that give that surface. $\endgroup$
    – Dylan Thurston
    Commented Nov 4, 2018 at 3:16

2 Answers 2

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The third example is a spine of the Gieseking manifold. The existence of a spine of this sort follows from a result of Iain Aitchison and the fact that the Gieseking is make of a single regular ideal tetrahedron. I think one could get this directly from the combinatorial description, but I also checked that it's a 1-relator group that fibers with fiber with the same monodromy as the Gieseking manifold.

The group has a presentation $\langle a, b | a^2b^2a^{-1}b^{-1} \rangle$. Letting $u=ab$, we can change to a presentation $ \langle a , u | a^2ua^{-1}ua^{-1} u^{-1} \rangle$. This has abelianization $\mathbb{Z}$ with $a\to 1, u\to 0$. Hence the kernel is generated by $u_i=a^i u a^{-i}$, and the relator shows that $u_0=a(u_1 u_0)a^{-1}, u_1=au_0a^{-1}$, hence this is a free-by-cyclic group with the same monodromy as the Gieseking.

The fourth complex has fundamental group a 1-relator group $\langle a, b| a^2b^2a^{-1}b\rangle$. One may see also that this is a rank-3-free-by-cyclic group. It also has an automorphism given by $(a,b) \mapsto (a, a^{-1}ba )$. Making the substitution $u=b^3a$, one gets a presentation $\langle b,u| bub^{-3}ub^2u^{-1}\rangle$. If $u_i=b^iub^{-i}$, then one has the relations $u_0=b^{-1}(u_0 u_{-2}^{-1})b, u_{-1}=b^{-1}u_0b, u_{-2}=b^{-1}u_{-1}b$, and hence this is the mapping torus of the free group automorphism $(u_0,u_{-1},u_{-2}) \mapsto (u_0u_{-2}^{-1},u_0,u_{-1})$.

Since the ranks of the kernels of the maps to $\mathbb{Z}$ are different, the groups are not isomorphic.

I'm guessing that you can find these examples in the literature on 1-relator groups, but I'm not sure if they have names in that context.

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    $\begingroup$ @HJRW I don’t think that it thickens to be a 3-manifold, since it would then have to admit an ideal triangulation, and hence is the Gieseking. $\endgroup$
    – Ian Agol
    Commented Nov 1, 2018 at 7:28
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    $\begingroup$ Oh sorry, right, the relation is a one-sided curve on the boundary of a non-orientable handlebody. $\endgroup$
    – HJRW
    Commented Nov 1, 2018 at 8:59
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    $\begingroup$ Thanks, that's great. We might get another description of the third example from the theory of arithmetic groups: the group will be commensurable with $\mathit{PSL}(2; \mathbb{Z}[\omega])$ where $\omega^3 = 1$. $\endgroup$
    – Dylan Thurston
    Commented Nov 1, 2018 at 13:33
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    $\begingroup$ Okay, now I actually tried carrying out the development of flats, and they are obstructed (there are no complete flats). By Bridson's flat plane theorem, this implies that $\pi_1(C)$ is $\delta$-hyperbolic. $\endgroup$
    – Ian Agol
    Commented Nov 1, 2018 at 23:02
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    $\begingroup$ OK, I thought about it a little more: the obstruction to extending a flat is actually the same obstruction to thickening this complex to a 3-manifold. If you take a hexagon in $\widetilde{C}$ and look at a neighborhood of its boundary (outside of the hexagon), you get a Möbius strip rather than an annulus. On the one hand this is the obstruction to thickening a simple complex to a 3-manifold, and on the other hand it means that when you try to develop a flat you get some inconsistency. $\endgroup$
    – Dylan Thurston
    Commented Nov 2, 2018 at 20:22
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The fourth example was studied by Brady and Crisp in their CMH paper CAT(0) and CAT(-1) dimensions of torsion-free hyperbolic groups, so it would be reasonable to call its fundamental group the "Brady--Crisp group". (Brady and Crisp also note that it belongs to a family studied by Haglund and Ballmann--Brin.)

They study a one-parameter family of CAT(0) metrics on the complex, and prove the very nice fact that any CAT(-1) model for this group has to have dimension at least 3. (And they exhibit a 3-dimensional CAT(-1) model.)

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