Cubulable Kahler groups
Joint with Thomas Delzant.
Kahler group means fundamental group of a compact Kahler manifold, i.e. a complex
manifold admitting a Kahler metric, i.e. a Hermitian metric whose imaginary part is
a closed 2-form. This includes complex projective manifolds. Finding which nitely
presented groups are Kahler, or merely restrictions on Kahler group, is a classical
problem (Serre, Gromov, Simpson, Toledo,…). More recently, Delzant and Gromov
introduced ideas from geometric group theory in the area.
Finding interesting examples is also part of the problem : Toledo, Dimca-Papadima-
Suciu, Llosa Isenrich, Bridson,…
Today, I will explain restrictions on actions of Kahler groups on finite dimensional
cubical complexes. This is to be expected: as Sageev pointed out, such
actions are related to codimension 1 subgroups , especially to the number of
ends of the relative Schreier graph. Gromov observed that innite Kahler
groups are 1-ended. Napier and Ramachandran showed that if a Kahler group G
has a subgroup H with , then G virtually surjects onto a surface group.
Delzant and Gromov expanded this: if a Kahler group has “many” subgroups with
relative ends, then G embeds in a direct product of surface groups. This lead
Dimca-Papadima-Suciu to investigate which subgroups of products of surface groups
Say a group is cubulable if it properly discontinuously on a nite dimensional
Theorem 1 A cubulable Kahler group is virtually a direct product of copies of and surface groups.
Furthermore, the product structure relates to the irreducible decomposition of the
cubical complex on which acts essentially (i.e. orbits not contained in a bounded
neighborhood of a hyperplane). The action is a product action.
What can one say about Kahler manifolds whose fundamental group are cubulable?
Theorem 2 Let be a closed projective manifold. Assume that X has the same homotopy type as where is a cubical complex and G acts freely co- compactly properly discontinuously on Y . Then X has a finite cover which is biholo- morphic to a product of a complex torus and Riemann surfaces.
The conclusion in the more general Kahler case is slightly weaker.
Caprace-Sageev?s work on cubical complexes. Bridson-Howie-Miller-Short?s work on subgroups of direct products of surface groups or free groups: can be only if itself is a direct product of subgroups of factors.
4. One more result
4.1. The case of irreducible cube complexes
Theorem 3 Assume that a Kahler group acts on an irreducible cubical complex which is locally finite. Assume that
- the G-action is essential ;
- no invariant flats ;
- no fixed points in the visual boundary.
Then virtually surjects onto a surface group, contains a convex -invariant set on which the action virtually factors trough a surface group.
We would like to remove the local finiteness assumption on , but we are unable to do so now. If we could do so, we could remove the no fixed point assumption as well, thanks to results by Caprace-Chatterji-Fernos.
Theorems 1 and 2 follow from Theorem 3.
Harmonic maps are of no help, since the needed vanishing theorem is not available for cube complex targets.
We shall merely use harmonic functions.
Let act on . If is a half-space, we denote by the corresponding hyperplane. Let denote the subgroup which preserves and each of the compo- nents of .
Fix a half-space . Let denote the signed distance to the hyperplane . Let be an arbitrary continuous -equivariant map. Define
If is locally finite, we can show that is proper, so has at least 2 ends. The set of ends splits in two subsets according to the sign of . We produce a harmonic function tending to 1 (resp. ?1) in such ends.
The following can be found in Kapovich?s paper on Gromov?s proof of Stallings? theorem.
Theorem 4 (Li-Tam, Woess-Kaimanovich, Ramachandran-Kapovich) Let be a bounded geometry Riemannian manifold. Assume satisfies a linear isoperimetric inequality. Then any continuous function
has a continuous harmonic extension to of finite energy.
Under the assumptions of Theorem 3, one can pick such that contains a non-abelian free group such that
We deduce that satisfies a linear isoperimetric inequality (otherwise, would
have almost invariant vectors on , which is not true).
Now we have a harmonic proper finite energy function . Stokes and
a cut-off argument imply that is pluriharmonic.
We must show that the associated (singular) foliation is in fact a fibration. The point
is to find a psh function which is not a function of .
Once this is done, we get a proper holomorphic map a Riemann surface. Let denote the kernel of the corresponding morphism on fundamental groups. We prove that has fixed points on . The trick (due to Behrstock and Charney) is to choose so that its -orbit contains 2 strongly separated hyperplanes.