Amalgamated product acting on CAT(0) cube complex

Question

I was reading the following result from the book Metric spaces of non-positive curvature by Bridson and Haefliger.

Result: Let $F_0,F_1$ and $H$ be groups acting properly by isometries on complete $CAT(0)$ spaces $X_0, X_1$ and $Y$ respectively. Suppose that for $j=0,1 $ there exists a monomorphism $\phi_j:H \to F_j$ and a $\phi_j$-equivariant isometric embedding $f_j:Y\to X_j$. Then

(1) the amalgamated product $G=F_0\ast_H F_1$ associated to the maps $\phi_j$ acts properly by isometries on a complete $CAT(0)$ space $X$;

(2) if the given action of $F_0$, $F_1$ and $H$ are cocompact, then the action of $G$ on $X$ is cocompact.

I'd be happy to see proof of the above result for $CAT(0)$ (finite dimensional if needed) cube complexes with the same hypothesis as above. If it is false what may go wrong?

From a quick google search I found the paper "Cubulating malnormal amalgams" by Hsu and Wise (https://link.springer.com/article/10.1007/s00222-014-0513-4) and saw a similar result (see Theorem A) but with extra assumsions like resulting group is hyperbolic relative to free abelian and the edge group hyperbolic, quasi-convex, malnormal etc.

What may go wrong if we simply try to extend the above result for $CAT(0)$ cube complexes without assuming these conditions of Theorem A (by Hsu & Wise)?

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I think the edge subgroup must act on a convex subset (that sits inside both the bigger cube complexes) such a way that hyperplanes are "merged" in both the other spaces while patching or some sort of wall compatibility condition must be satisfied.

Answer 1

To extend the gluing result from Bridson--Haefliger to non-positively curved cube complexes, it is important to work in the correct category.

If we want the result to also be a non-positively curved cube complex, then it is not enough just to assume that the attaching maps are local isometries. This is shown by Leary--Minasyan groups. These are HNN extensions

$G=\mathbb{Z}^2 *_{H_1\sim H_2}$

where the $H_i$ correspond to isometric finite-sheeted covers of the torus, so the result is a CAT(0) group by the gluing theorem mentioned in the question. However, for carefully chosen monodromy (the idea is it should come from a non-integer matrix in $SL_2(\mathbb{Q})\cap SO(2)$, i.e. from a Pythagorean triple), Leary and Minasyan can prove that $G$ is not biautomatic. In particular, $G$ cannot be cocompactly cubulated.

The point is that, while the attaching maps in the Leary--Minasyan examples are local isometries, they do no respect the cubical structure on $\mathbb{Z}^2$.

However, there is an 'obvious' generalisation of the Bridson--Haefliger result if we work in the category of non-positively curved cube complexes, with maps being combinatorial local isometries. ('Combinatorial' means they map $n$-cubes to $n$-cubes homeomorphically.) The statement is something like:

Theorem: Suppose that $X$ is a (not necessarily connected) non-positively curved cube complex, and $\partial_\pm:Y\to X$ a pair of combinatorial local isometries. Then the 'mapping torus'

$M=(X\sqcup (Y\times [-1,1]))/\sim$

where $(y,\pm1)\sim\partial_\pm(y)$ is a non-positively curved cube complex.

The proof isn't very difficulty -- you just check locally that $M$ satisies the link condition, using the fact that a combinatorial local isometry has to map links of vertices in $Y$ to full subcomplexes of links of vertices of $X$. Probably this has been written down many times in the literature, although I don't know a reference off the top of my head.

However, it's important to emphasise that this theorem isn't very useful, because it relies on knowing that the two cube-complex structures that $Y$ inherits from $X$ via the two maps $\partial_+$ and $\partial_-$ are the same. As the Leary-Minasyan examples show, this isn't always true. This is why the Hsu--Wise result is much deeper.

In fact, if you combine the Hsu--Wise result with more work of Haglund--Wise and Wise, and finally Agol's theorem, you get a very satisfactory and deep theorem:

Theorem(Agol, Wise et al.): Let $\Gamma$ be an acylindrical graph of groups. If the vertex groups are all cocompactly cubulated and hyperbolic, and the edge groups are all quasiconvex in the vertex groups, then the fundamental group is cocompactly cubulated and hyperbolic.

Answer 2

I assume that isomorphic embedding really meant isometric embedding in the assumptions and thus $Y$ is really a convex subset of $X_i$ for all $i$.

Then the CAT(0)-space on which the amalgamated product acts may be thought of as a thickened version of the Bass-Serre tree.

Usually the Bass-Serre tree has as vertex set $G/F_0 \amalg G/F_1 = G\times_{F_0}pt \amalg G\times_{F_1} pt$ and as edges $G/H = G\times_{H} pt$.

Then the whole tree is constructed by identifying $(gH,0)\in G/H\times [0,1]$ with its coset in $gF_0\in G/F_0$ and $(gH,1)\in G/H\times [0,1]$ with its coset in $gF_1\in G/F_1$ (using the inclusions $H\to F_i$).

Now the big space on which the amalgamated product acts can be constructed similarly. The idea is to thicken the points, i.e. take $G\times_{F_0}X_0\amalg G\times_{F_1}X_1\amalg G\times_{H}Y\times [0,1]$ subject to the identifications $(g,y)\sim (g,f_i(y))$ for $i=1,2$.

Now there are a couple of questions:

1. Is that space still contractible after forgetting the $G$-action. Yes (one can also use this contruction to build a model for $E(F_0\ast_HF_1)$ out of models for $EF_i$ and $EH$ without the metric assumptions and embeddings.

2. Is it still locally CAT(0) ? Yes, this should be the usual glueing lemma for CAT(0)-spaces along convex subspaces.

Thus the result is (globally) CAT(0). If $Y$ and each $X_i$ is a CAT(0) cube complex and the inclusions of $Y\to X_i$ are inclusions of $CAT(0)$-cube complexes, the whole construction should also give a CAT(0)-cube complex.

The same idea also works for HNN-Extensions, although in this case it might be easy to forget to check all assumptions (like for example for $BS(1,2)$).

Let me just stress how crucial the assumption that each $Y\to X_i$ is an isometric embedding really is. For example, if one drops this assumption it is in general unknown whether the result is a CAT(0) group. Usually this construction does not give a CAT0 space but sometimes one can find a better space. For example for free by cyclic groups I asked this question a long time ago When is a extension of $\mathbb{Z}$ by a free group a CAT(0) group?.