This week I had the pleasure of attending a seminar talk Nick Vlamis gave at CUNY, where he taught us a very pretty trick due to Anderson which one can use, as Anderson did, to prove that the groups of orientation-preserving homeomorphisms of the - and -spheres are simple. The purpose of this post is to reproduce Nick’s exposition of the trick.
Suppose is a homeomorphism of a surface (for me a surface is a Hausdorff and second countable -manifold) whose support
is contained in the interior of some closed disk in . Since is closed, there is in fact some closed disk properly contained in and containing . In fact, we can find a countable sequence of disjoint disks with , each contained in the interior of . Since a sequence of points, one from each disk must have a limit point in , why don’t we arrange the to Hausdorff converge to a point as and a point as . Here’s the picture.
Here’s how we’ll use these disks. There is a homeomorphism which shifts each to and holds fixed. Since the are pairwise disjoint, if we choose distinct and , the homeomorphisms and have disjoint support and thus commute. Thus the infinite product makes sense as a homeomorphism whose support is contained in .
Proposition.
- .
- Suppose is a homeomorphism of satisfying and such that there exists a homeomorphism exchanging and and satisfying and . Then any normal subgroup of containing contains .
Proof. From the definition of \varphi\sigma^{-1}\varphi^{-1} = \prod_{n=1}^\infty \varphi^n g^{-1} \varphi^{-n}$$ from which it follows that $\sigma(\varphi\sigma^{-1}\varphi^{-1}) = g$ . We claim that $g = [\sigma,f]\psi [\sigma,f]\psi^{-1}.$ Expanding, we see that this is a product of conjugates of $f$ and $f^{-1}$ and is thus contained in any normal subgroup containing $f.$
To see that this product is really , observe that since is supported on , for , we have . On the other hand, for , we have . Thus if ,
which is in , so further applying we obtain .
Next, if , we have , applying yields and further applying yields , since .
Finally, if , note that is not in , so we have and in fact the whole product fixes . Thus we have shown that .
Simplicity of .
Now, all of cannot be simple, since it has an action on the -element set of orientations of . On the other hand, let’s show that is simple. To do so, we want to apply the proposition, for which we need to express any given orientation-preserving homeomorphism as being built out of homeomorphisms supported on disks.
Suppose at first that fixes some (topological) circle on . Then is the product of and , which are supported in the two hemispheres (thanks Schoenflies!) determined by the circle. If does not fix a given circle , say one disjoint from some fixed point of , there is again a homeomorphism of fixing taking back to (thanks Schoenflies!) so that the composition fixes , and we can use the argument from above. It looks like this expresses as the product of three homeomorphisms supported in disks, but Nick observed that we can actually combine with one of or , so only two are needed.
Anyway, now we have where each of and are supported in disks and . If displaces (i.e. ) and is an orientation-preserving homeomorphism of taking to , (one exists! I think this is still thanks Schoenflies!) we have that . Indeed, if our original had a , then our new has one.
It follows from the proposition, assuming we can construct , that and are both products of conjugates of , so so is . But now itself can be arbitrary so long as it’s nontrivial, for if , we can find an open neighborhood small enough that it is moved entirely off itself, and this open neighborhood contains some disk that therefore displaces. Therefore the smallest normal subgroup of containing a nontrivial element is the whole group—we’ve shown it is simple.
Other results
From here it’s actually not too hard to show that is coarsely bounded. Originally I was going to give some argument here, but I have to turn in an application tomorrow, so I’ll sign off and work on that instead.