# Anderson's Trick

30 Oct 2022

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 $$2$$- and $$3$$-spheres are simple. The purpose of this post is to reproduce Nick’s exposition of the trick.

Suppose $$g$$ is a homeomorphism of a surface $$S$$ (for me a surface is a Hausdorff and second countable $$2$$-manifold) whose support

$\operatorname{Supp}(g) = \overline{\{x \in S : g(x) \ne x\}}$

is contained in the interior of some closed disk $$D$$ in $$S$$. Since $$\operatorname{Supp}(g)$$ is closed, there is in fact some closed disk $$\Delta$$ properly contained in $$D$$ and containing $$\operatorname{Supp}(g)$$. In fact, we can find a countable sequence of disjoint disks $$\{\Delta_n\}_{n \in \mathbb{Z}}$$ with $$\Delta = \Delta_0$$, each contained in the interior of $$D$$. Since a sequence of points, one from each disk must have a limit point in $$D$$, why don’t we arrange the $$\Delta_n$$ to Hausdorff converge to a point $$x_{+\infty}$$ as $$n \to +\infty$$ and a point $$x_{-\infty}$$ as $$n \to -\infty$$. Here’s the picture.

Here’s how we’ll use these disks. There is a homeomorphism $$\varphi\colon D \to D$$ which shifts each $$\Delta_n$$ to $$\Delta_{n+1}$$ and holds $$x_{\pm\infty}$$ fixed. Since the $$\Delta_{n}$$ are pairwise disjoint, if we choose distinct $$n$$ and $$m$$, the homeomorphisms $$\varphi^n g \varphi^{-n}$$ and $$\varphi^m g \varphi^{-m}$$ have disjoint support and thus commute. Thus the infinite product $$\sigma = \prod_{n = 0}^\infty \varphi^n g\varphi^{-n}$$ makes sense as a homeomorphism whose support is contained in $$D$$.

Proposition.

1. $$[\sigma,\varphi] = \sigma\varphi\sigma^{-1}\varphi^{-1} = g$$.
2. Suppose $$f$$ is a homeomorphism of $$S$$ satisfying $$f(D) \cap D = \varnothing$$ and such that there exists a homeomorphism $$\psi\colon S \to S$$ exchanging $$D$$ and $$f(D)$$ and satisfying $$\psi|_D = f$$ and $$\psi|_{f(D)} = \varphi f^{-1}$$. Then any normal subgroup of $$\operatorname{Homeo}(S)$$ containing $$f$$ contains $$g$$.

Proof. From the definition of $$\sigma$$ we have $$\varphi\sigma^{-1}\varphi^{-1} = \prod_{n=1}^\infty \varphi^n g^{-1} \varphi^{-n}$$ 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 $$g$$, observe that since $$\sigma$$ is supported on $$D$$, for $$x \in D$$, we have $$\sigma f \sigma^{-1}f^{-1}(x) = \sigma(x)$$. On the other hand, for $$x \in f(D)$$, we have $$\sigma f \sigma^{-1}f^{-1}(x) = \sigma^{-1}(x)$$. Thus if $$x \in D$$,

$\psi\sigma f \sigma^{-1}f^{-1}\psi^{-1}(x) = \psi\sigma f \sigma^{-1} f^{-1} f\varphi^{-1}(x) = \psi f \sigma^{-1} \varphi^{-1}(x) = \varphi \sigma^{-1} \varphi^{-1}(x)$

which is in $$D$$, so further applying $$\sigma f\sigma^{-1} f^{-1}$$ we obtain $$\sigma \varphi \sigma^{-1} \varphi^{-1}(x) = g(x)$$.

Next, if $$x \in f(D)$$, we have $$\psi^{-1}(x) = f^{-1}(x) \in D$$, applying $$\psi [\sigma,f]$$ yields $$f \sigma f^{-1}(x),$$ and further applying $$\sigma f \sigma^{-1} f^{-1}$$ yields $$\sigma(x) = x$$, since $$x \notin D$$.

Finally, if $$x \notin D \cup f(D)$$, note that $$f^{-1}\psi^{-1}(x)$$ is not in $$D$$, so we have $$\psi[\sigma, f]\psi^{-1}(x) = x$$ and in fact the whole product fixes $$x$$. Thus we have shown that $$g = [\sigma, f]\psi[\sigma,f]\psi^{-1}$$. $$\blacksquare$$

### Simplicity of $$\operatorname{Homeo}_+(S^2)$$.

Now, all of $$\operatorname{Homeo}(S^2)$$ cannot be simple, since it has an action on the $$2$$-element set of orientations of $$S_2$$. On the other hand, let’s show that $$\operatorname{Homeo}_+(S^2)$$ is simple. To do so, we want to apply the proposition, for which we need to express any given orientation-preserving homeomorphism $$h\colon S^2 \to S^2$$ as being built out of homeomorphisms supported on disks.

Suppose at first that $$h$$ fixes some (topological) circle on $$S^2$$. Then $$h$$ is the product of $$h^+$$ and $$h^-$$, which are supported in the two hemispheres (thanks Schoenflies!) determined by the circle. If $$h$$ does not fix a given circle $$C$$, say one disjoint from some fixed point $$p$$ of $$h$$, there is again a homeomorphism $$h'$$ of $$S^2$$ fixing $$p$$ taking $$h(C)$$ back to $$C$$ (thanks Schoenflies!) so that the composition $$h'h$$ fixes $$C$$, and we can use the argument from above. It looks like this expresses $$h$$ as the product of three homeomorphisms supported in disks, but Nick observed that we can actually combine $$h'$$ with one of $$h^+$$ or $$h^-$$, so only two are needed.

Anyway, now we have $$h = h^+h^-$$ where each of $$h^+$$ and $$h^-$$ are supported in disks $$D^+$$ and $$D^-$$. If $$f$$ displaces $$D^+$$ (i.e. $$f(D^+) \cap D^+ = \varnothing$$) and $$\iota_+^-$$ is an orientation-preserving homeomorphism of $$S^2$$ taking $$D^+$$ to $$D^-$$, (one exists! I think this is still thanks Schoenflies!) we have that $$\iota_+^- f (\iota_+^-)^{-1} (D^-) \cap D^- = \varnothing$$. Indeed, if our original $$f$$ had a $$\psi$$, then our new $$\iota_+^- f (\iota_+^-)^{-1}$$ has one.

It follows from the proposition, assuming we can construct $$\psi$$, that $$h^+$$ and $$h^-$$ are both products of conjugates of $$f$$, so so is $$h$$. But now $$f$$ itself can be arbitrary so long as it’s nontrivial, for if $$f(x) \ne x$$, we can find an open neighborhood small enough that it is moved entirely off itself, and this open neighborhood contains some disk that $$f$$ therefore displaces. Therefore the smallest normal subgroup of $$\operatorname{Homeo}_+(S^2)$$ 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 $$\operatorname{Homeo}_+(S^2)$$ 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.