## Wednesday, June 16, 2010

### Bounded groups

Today's post arises from a paper by my Penn State colleague Dima Burago and his collaborators, Conjugation-invariant norms on groups of geometric origin. (October 7, 2007). http://arxiv.org/abs/0710.1412.

The basic definition is a very simple one: say that a group $G$ is bounded if it has finite diameter with respect to any bi-invariant metric.
(Of course it is enough to consider the distance from a group element $g$ to the identity element, and this function $|g|$ then becomes a \emph{conjugation invariant norm} on the group, hence the title of the paper. A group is then bounded if every conjugation-invariant norm is bounded from above. One can also ask whether every conjugation-invariant norm is bounded from below (away from the identity element). When a norm is bounded from above and below and can be said to be trivial: it is equivalent to the norm which equals 1 on all non-identity elements. If every conjugation-invariant norm is trivial the group is called meager.

The basic phenomenon studied in the paper is the surprising fact that conjugation-invariant norms on large groups of diffeomorphisms tend to be trivial. In fact the main result is that the identity component of the diffeomorphism group of $M$ is meager whenever $M$ is a sphere or a closed connected 3-manifold. However the paper begins by proving some of the basic facts about bounded groups, and this part was already very interesting to me.

The groups $SL(n,R)$, $n\ge 2$, and $SL(n,Z)$, $n\ge 3$, are bounded. In the real case this follows from "a suitable version of the Gauss elimination process" to quote the paper... you have to show that any element of $SL(n,R)$ can be written as the product of a bounded number of conjugates of a bounded number of generators (namely, the elementary matrices with a 1 off the diagonal). Takes a moment's thought. Over the integers one needs also the bounded generation of $SL(n,Z)$ for $n\ge 3$ (Carter, David, and Gordon Keller. “Bounded Elementary Generation of SLn(O).” American Journal of Mathematics 105, no. 3 (June 1983): 673-687.)

An abelian group is bounded if and only if it is finite. This seems obvious, but there is no finite generation restriction here, so some care is needed in applying structure theory.

If $G$ surjects onto an unbounded group $H$, then $G$ is unbounded The basic idea is simple enough - take a conjugation-invariant norm on $H$ and pull it back via the surjection - but unfortunately the result need not be a norm (it vanishes on the whole kernel not just on the identity), so one needs to develop a more flexible theory of unbounded "quasi-norms" and prove that the existence of unbounded norms and quasi-norms are equivalent.

#### 1 comment:

JohnR said...

There is a new paper on the arXiv today that follows up on this: Michalik, Ilona, and Tomasz Rybicki. On the structure of the commutator subgroup of certain homeomorphism groups. 1006.3269 (June 16, 2010). http://arxiv.org/abs/1006.3269.