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 22 - and 33 -spheres are simple. The purpose of this post is to reproduce Nick’s exposition of the trick.

Suppose gg is a homeomorphism of a surface SS (for me a surface is a Hausdorff and second countable 22 -manifold) whose support

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

is contained in the interior of some closed disk DD in SS . Since Supp(g)\operatorname{Supp}(g) is closed, there is in fact some closed disk Δ\Delta properly contained in DD and containing Supp(g)\operatorname{Supp}(g) . In fact, we can find a countable sequence of disjoint disks {Δn}nZ\{\Delta_n\}_{n \in \mathbb{Z}} with Δ=Δ0\Delta = \Delta_0 , each contained in the interior of DD . Since a sequence of points, one from each disk must have a limit point in DD , why don’t we arrange the Δn\Delta_n to Hausdorff converge to a point x+x_{+\infty} as n+n \to +\infty and a point xx_{-\infty} as nn \to -\infty . Here’s the picture.

The countable disjoint union of disks and their two limit points

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

Proposition.

  1. [σ,φ]=σφσ1φ1=g[\sigma,\varphi] = \sigma\varphi\sigma^{-1}\varphi^{-1} = g .
  2. Suppose ff is a homeomorphism of SS satisfying f(D)D=f(D) \cap D = \varnothing and such that there exists a homeomorphism ψ ⁣:SS\psi\colon S \to S exchanging DD and f(D)f(D) and satisfying ψD=f\psi|_D = f and ψf(D)=φf1\psi|_{f(D)} = \varphi f^{-1} . Then any normal subgroup of Homeo(S)\operatorname{Homeo}(S) containing ff contains gg .

Proof. From the definition of σ\sigmawehave 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 gg , observe that since σ\sigma is supported on DD , for xDx \in D , we have σfσ1f1(x)=σ(x)\sigma f \sigma^{-1}f^{-1}(x) = \sigma(x) . On the other hand, for xf(D)x \in f(D) , we have σfσ1f1(x)=σ1(x)\sigma f \sigma^{-1}f^{-1}(x) = \sigma^{-1}(x) . Thus if xDx \in D ,

ψσfσ1f1ψ1(x)=ψσfσ1f1fφ1(x)=ψfσ1φ1(x)=φσ1φ1(x)\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 DD , so further applying σfσ1f1\sigma f\sigma^{-1} f^{-1} we obtain σφσ1φ1(x)=g(x)\sigma \varphi \sigma^{-1} \varphi^{-1}(x) = g(x) .

Next, if xf(D)x \in f(D) , we have ψ1(x)=f1(x)D\psi^{-1}(x) = f^{-1}(x) \in D , applying ψ[σ,f]\psi [\sigma,f] yields fσf1(x),f \sigma f^{-1}(x), and further applying σfσ1f1\sigma f \sigma^{-1} f^{-1} yields σ(x)=x\sigma(x) = x , since xDx \notin D .

Finally, if xDf(D)x \notin D \cup f(D) , note that f1ψ1(x)f^{-1}\psi^{-1}(x) is not in DD , so we have ψ[σ,f]ψ1(x)=x\psi[\sigma, f]\psi^{-1}(x) = x and in fact the whole product fixes xx . Thus we have shown that g=[σ,f]ψ[σ,f]ψ1g = [\sigma, f]\psi[\sigma,f]\psi^{-1} . \blacksquare

Simplicity of Homeo+(S2)\operatorname{Homeo}_+(S^2) .

Now, all of Homeo(S2)\operatorname{Homeo}(S^2) cannot be simple, since it has an action on the 22 -element set of orientations of S2S_2 . On the other hand, let’s show that Homeo+(S2)\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 ⁣:S2S2h\colon S^2 \to S^2 as being built out of homeomorphisms supported on disks.

Suppose at first that hh fixes some (topological) circle on S2S^2 . Then hh is the product of h+h^+ and hh^- , which are supported in the two hemispheres (thanks Schoenflies!) determined by the circle. If hh does not fix a given circle CC , say one disjoint from some fixed point pp of hh , there is again a homeomorphism hh' of S2S^2 fixing pp taking h(C)h(C) back to CC (thanks Schoenflies!) so that the composition hhh'h fixes CC , and we can use the argument from above. It looks like this expresses hh as the product of three homeomorphisms supported in disks, but Nick observed that we can actually combine hh' with one of h+h^+ or hh^- , so only two are needed.

Anyway, now we have h=h+hh = h^+h^- where each of h+h^+ and hh^- are supported in disks D+D^+ and DD^- . If ff displaces D+D^+ (i.e. f(D+)D+=f(D^+) \cap D^+ = \varnothing ) and ι+\iota_+^- is an orientation-preserving homeomorphism of S2S^2 taking D+D^+ to DD^- , (one exists! I think this is still thanks Schoenflies!) we have that ι+f(ι+)1(D)D=\iota_+^- f (\iota_+^-)^{-1} (D^-) \cap D^- = \varnothing . Indeed, if our original ff had a ψ\psi , then our new ι+f(ι+)1\iota_+^- f (\iota_+^-)^{-1} has one.

It follows from the proposition, assuming we can construct ψ\psi , that h+h^+ and hh^- are both products of conjugates of ff , so so is hh . But now ff itself can be arbitrary so long as it’s nontrivial, for if f(x)xf(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 ff therefore displaces. Therefore the smallest normal subgroup of Homeo+(S2)\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 Homeo+(S2)\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.