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.