** Cubulable Kahler groups **

Joint with Thomas Delzant.

**1. Introduction **

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

are Kahler.

**2. Results **

Say a group is cubulable if it properly discontinuously on a nite dimensional

cubical complexes.

Theorem 1A 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 2Let 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.

**3. Ingredients **

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 3Assume 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.

**5. Comments **

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.

**6. Proof **

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.