What are some potential approaches to classifying/characterising (finitely presented?) groups that act (without fixed points and potentially more conditions) on CAT(0)-cube complexes? As so many groups act on CAT(0)-cube complexes, is this sort of classification/characterisation even possible?
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4$\begingroup$ Classifying, certainly not in any reasonable sense. Maybe you mean characterizing (in some nontrivial way). Then it's still somewhat open-ended: yes there are many "potential additional conditions", on the complex (proper, finite-dimensional, etc) and on the action (proper, conditions on stabilizers). $\endgroup$– YCorCommented Apr 14, 2018 at 7:20
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$\begingroup$ Regarding classification, I think it’s known, via work of Bridson and Miller, that the isomorphism problem is unsolvable for finitely presented groups acting freely on proper CAT(0) cube complexes. The isomorphism problem when the groups act cocompactly is open (but surely unsolvable). $\endgroup$– HJRWCommented Apr 15, 2018 at 21:26
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$\begingroup$ @HJRW: It sounds interesting. Could you be more specific about the reference? Bridson and Miller wrote several articles together. Alternatively, even if I am not sure that such an approach works, Guba and Sapir's formalism of diagram groups associates to any semigroup presentation a group acting freely and properly on a CAT(0) cube complex. Starting from a tricky semigroup presentation might (or not) also prove this statement. $\endgroup$– AGenevoisCommented Apr 23, 2018 at 8:50
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$\begingroup$ @AGenevois: The reference is Bridson & Miller, Recognition of subgroups of direct products of hyperbolic groups, Proc. Amer. Math. Soc., together with the observation that, using a suitable version of the Rips construction (eg Haglund--Wise), one can take the group $\Gamma$ to be cubulated. (By the way I deleted an earlier, incorrect, comment.) $\endgroup$– HJRWCommented Apr 23, 2018 at 10:38
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$\begingroup$ Recently there was a talk by Ben Stucky from the University of Oklahoma which I was (unfortunately) unable to attend because I was busy with conference organizing. I spoke with him between sessions and he told me he an interesting result where he claims that you can cubulate one-relator quotients of free products $A*B$ which aren't necessarily small-cancellation, as long as the relator is of length at least 4(?). $\endgroup$– Christopher PerezCommented May 2, 2018 at 12:55
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The question is probably too broad to get an answer, but at least Niblo and Roller's article Groups acting on cubes and Kazhdan's property (T) should be mentioned. Therein, they proved:
Theorem: A group acts fixed-point-freely on a CAT(0) cube complex if and only if it contains a codimension-one subgroup.
(To be precise, they proved this statement with connected cubes instead of general CAT(0) cube complexes, but a CAT(0) cube complex always embeds isometrically and equivariantly into a connected cube, so the two statements turn out to be equivalent.)
The motivation comes from Sageev's thesis, where he characterised codimension-one subgroups as subgroups arising as stabilisers of essential hyperplanes for actions on CAT(0) cube complexes.
answered Apr 14, 2018 at 11:59
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$\begingroup$ Can you please explain how a group G acting fixed-point freely on a CAT(0) cube complex is equivalent to G admitting an action on a connected cube that is transitive on the set of hyperplanes and has no fixed points? $\endgroup$– user114152Commented Apr 23, 2018 at 7:18
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1$\begingroup$ Let $G$ be a group acting fixed-point freely on a CAT(0) cube complex $X$. Given a hyperplane $J$, the orbit $G \cdot J$ induces a wallspace structure on $X$, which can be cubulated to get a new CAT(0) cube complex $Y$. Notice that the induced action of $G$ on $Y$ has a single orbit of hyperplanes. The claim is that choosing carefully $J$ the action is moreover fixed-point free. Next, labelling the halfspaces delimited by any hyperplane by $0$ and $1$, you get a natural isometric and equivariant embedding $Y \to \{0,1\}^{\mathcal{H}}$ where $\mathcal{H}$ is the set of hyperplanes of $Y$. $\endgroup$ Commented Apr 23, 2018 at 8:38