# Contractible Rips complex from non-hyperbolic group

I heard that the Rips complexes associated to the Cayley graphs of hyperbolic groups are contractible for a sufficiently large radius. Is the converse true? Namely, if a group is non-hyperbolic, then is its Rips complex never asymptotically" contractible?

For example, we can ask if the non-hyperbolic group $$\mathbb{Z}^2$$ has contractible Rips complex associated to its Cayley graph for large radius.

I don't know much geometric group theory, and quick Google searches for this question weren't enough to fetch me an easy answer for this. This seemed like a natural question that would have been addressed at some point, so if anyone has a standard answer for this, it would be nice to know :)

• Yes, $\mathbf{Z}^d$, and direct products of hyperbolic groups, have this property. Also CAT(0) groups. Also groups with polynomial growth. (And more generally groups that act properly cocompactly on a contractible geodesic proper space.)
– YCor
Sep 30 '20 at 20:01
• Those are delightful. May I know the references or Google-able keywords for those facts? Sep 30 '20 at 22:08
• Also, could you tell me if the converse statement stated in the first paragraph is true? Sep 30 '20 at 22:11
• This paper gives a criterion for a group to have a contractible Rips complex: arxiv.org/abs/1812.10976 See Theorem 6.5 Oct 1 '20 at 0:25
• So, I was too optimistic: probably most cases I mentioned are just open: it's unknown for $\mathbf{Z}^2$ with standard metric according to Zaremsky's paper linked by Ian (see the discussion after Theorem 6.5). Note it's even sensitive (a priori) to the choice of word metric.
– YCor
Oct 1 '20 at 7:18

Another source of Cayley graphs with contractible Rips complexes comes from Helly graphs.

Proposition: Rips complexes of uniformly locally finite Helly graphs are contractible.

See Lemma 5.28 and Theorem 4.2(v) from the preprint arXiv:2002.06895.

One construction of Helly graphs is the following: Given a CAT(0) cube complex $$X$$, the graph obtained from $$X^{(1)}$$ by adding an edge between any two vertices which belong to a common cube is a Helly graph. And there exist many groups admitting one-skeleta of CAT(0) cube complexes as Cayley graphs. For instance:

Corollary: Let $$\Gamma$$ be a finite simplicial graph. Let $$G$$ denote the Cayley graph of the right-angled Artin group $$A(\Gamma)$$ constructed from the generating set $$\{ u_1\cdots u_n \mid u_1, \ldots, u_n \in V(\Gamma) \text{ pairwise adjacent} \}$$. Then all the Rips complexes of $$G$$ are contractible.

Notice that $$A(\Gamma)$$ is hyperbolic if and only if $$\Gamma$$ has no edges, so most of these examples are not hyperbolic. For instance, the corollary includes $$\mathbb{Z}^2$$ with the generating set $$\{(1,0), (1,1), (0,1)\}$$. Other groups having one-skeleta of CAT(0) cube complexes as Cayley graphs include right-angled Coxeter groups and right-angled mock reflection groups.

Edit: I should also mention arxiv:1904.09060, which proves that Garside groups (such as braid groups) and Artin groups of type FC also admit Cayley graphs that are Helly, and so whose Rips complexes are contractible.

Here is a very ad hoc proof that $$Rips_2(\mathbb{Z}^2)$$ is contractible, which occurred to me at some point in discussions with Brendan Mallery, about a year after I wrote the paper that Ian Agol linked to above.

Consider the maximal simplex $$\{(n,m),(n+1,m),(n-1,m),(n,m+1),(n,m-1)\}$$. This has $$\{(n+1,m),(n-1,m),(n,m+1),(n,m-1)\}$$ as a free face (meaning it is the only simplex properly containing that face), so we can delete the simplex and this free face without changing the homotopy type of the complex. Do this for every $$n$$ and $$m$$. Within the resulting subcomplex, the (now maximal) simplex $$\{(n,m),(n+1,m),(n-1,m),(n,m+1)\}$$ has $$\{(n+1,m),(n-1,m),(n,m+1)\}$$ as a free face, so these can be deleted without changing the homotopy type. Similarly we can "pair up" $$\{(n,m),(n+1,m),(n-1,m),(n,m-1)\}$$ with $$\{(n+1,m),(n-1,m),(n,m-1)\}$$ and delete them, and then also the version where you use $$n+1,m+1,m-1$$ and the version where you use $$n-1,m+1,m-1$$. Finally, pair up $$\{(n,m),(n+1,m),(n-1,m)\}$$ with $$\{(n+1,m),(n-1,m)\}$$ and $$\{(n,m),(n,m+1),(n,m-1)\}$$ with $$\{(n,m+1),(n,m-1)\}$$.

After deleting all these pairs of simplices for all $$m$$ and $$n$$, the homotopy type has never changed and now we have removed all simplices containing edges of the form $$\{(n+1,m),(n-1,m)\}$$ or $$\{(n,m+1),(n,m-1)\}$$. The subcomplex of what's left over (assuming I didn't forget any cases) is a "quilt of tetrahedra", that is, the flag complex of the graph obtained from the standard Cayley graph of $$\mathbb{Z}^2$$ by adding in every edge of the form $$\{(n,m),(n+1,m+1)\}$$ and $$\{(n,m),(n+1),(m-1)\}$$. This is visibly contractible, so $$Rips_2(\mathbb{Z}^2)$$ is contractible. (By the way, this remove-lots-of-pairs-of-simplices thing can all be phrased using Forman's discrete Morse theory.)

Like I said, this is very ad hoc, and it seems like it would be a big mess to try and generalize to $$Rips_k(\mathbb{Z}^2)$$ for $$k>2$$ (I guess your actual question was more about "large radius"), but at least here is a concrete example of a non-hyperbolic group with a contractible Rips complex using a word metric. In general, surprisingly little is known about Rips complexes of groups using word metrics!

• Another direction of generalization would be $\mathbf{Z}^2$ with an arbitrary word metric.
– YCor
Oct 1 '20 at 15:24