-cube complexes and the median class
Joint with Talia Fernos and Alessandra Iozzi.
The following corollary.
Theorem 1 A cocompact, irreducible lattice in is not cubical.
Completed by Fernos, Caprace, Lecureux in order to prove that anay such lattives, when acting isometrically on a cube complex, must have a fixed point.
Consider lattices in semi-simple Lie groups .
If has property (T), they can’t be cubical. If , they are cubical (Bergeron-Wise, using Kahn-Markovic, unobvious). For , some lattices are cubical, as Anne Giralt explained, but for the other ones we don’t know.
2. Main result
Theorem 2 Let act isometrically on a cube complex in a non elementary manner (no fixed point on nor on the visual boundary ). Then a certain bounded cohomology class vanishes.
Corollary 3 (Superrigidity) Let be a cocompact irreducible lattice in a product of locally compact groups . Let act essentially and non-elementarily on a cube complex. Then the action extends continuously to , factoring via one of the factors.
The fact that this follows from the theorem is due to Shalom and Burger-Monod.
3. The median class
3.1. Case of trees
I explain the case when is a tree.
Let be the set of oriented paths of length 2 in the tree. Let be the obvious action of on . Let be defined by
where denotes the set of which are between and . This is unbounded, but the coboundary is bounded. Indeed, cancellations leave us only with paths that touch the median.
In the case of trees, this is not surprising (classical fact that generalizes to hyperbolic metric spaces).
3.2. Median metric spaces
In a metric, the side is the set of points for wich the triangle inequality is an equality. A metric space is median if given 3 points, there is a unique common point to the 3 sides.
cube complexes equipped with the metric which is on cubes are median.
3.3. Case of CCC
cube complexes have half-spaces: start cutting s cube in equal parts and continue for ever in contiguous cubes. Say half-spaces are tightly nested if any half-space that sits in between must be one of them. Define as the set of pairs of tightly nested half-spaces. The same formula defines a 1-cochain . The same cancellations show that only involves pairs touching the median point. Therefore it is bounded. However, the bound depends on the dimension of . Pull-back on via an orbit.
3.4. Non vanishing
Burger-Monod show that
where is a Poisson boundary. We use the Roller compactification, defined as follows. embeds in the set of subsets of (a point is mapped to the set of half-space pairs that contain it). Take the closure of the image of that embedding. Then (Zimmer), there is an equivariant map of to the set of probability measures on . One shows that the image is in , this gives the image of as a nonzero cocycle on .