*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:

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.Post a Comment