# Equivalence of Definitions of Quasiconformal Surfaces?

I have been reading John H. Hubbard's book Teichmüller Theory vol. 1 and I am a little bit concerned with his definition of Quasiconformal Surface.

Definition: A Quasiconformal surface $$S$$ is a topological surface with a Riemann-surface structure; two Riemann surface structures on $$S$$ define the same quasiconformal structure if the identity map between them is quasiconformal.

If $$S_1,$$ $$S_2$$, are two quasiconformal surfaces, a map $$f:S_1 \to S_2$$ is quasiconformal if it is a quasiconformal homeomorphism for one, hence all, analytic structures on each $$S_1$$ and $$S_2$$. This implies that all quasiconformal maps are isomorphisms.

I do understand the definition and one it's advantage is that it is easy to prove that if two compact quasiconformal surfaces $$S_1$$ and $$S_2$$ are homeomorphic, then they are isomorphic as quasiconformal surfaces.

However, I am use to define structure like that in a similar way as we define manifolds. Hence, if I had to give a definition of Quasiconformal Surface, I would say that it is a topological surface $$S$$ together with a maximal atlas that contains all the charts for which the transition maps are quasiconformal maps.

My question is are those 2 definitions equivalent (can I define a specific Riemann surface structure from a maximal atlas which is unique up to quasiconformal maps?), my first idea to solve this problem was to use the measurable Riemann mapping theorem which allow us to find local quasiconformal maps that will satisfy a Beltrami equation. The problem is that I am not sure if this argument will work for any surface, we might have to do the process on infinitely many charts (I want to consider surfaces with puncture and boundaries as well.)

• At the very least I think your definition would have to require the existence of a constant $K$ such that all overlap maps in the atlas are $K$-quasiconformal. Jun 9 '15 at 18:59

I believe the problem is exactly this. A composition of $K$-quasiconformal maps is not necessarily $K$-quasiconformal, which makes them difficult to work with. And a locally quasiconformal map is not necessarily globally quasiconformal. Normally when you define a type of manifold in terms of a class of permitted overlap maps the class of maps should be defined in terms of a local property and closed under composition. There's no way to do that so as to get structures that are then related by global quasiconformal maps.

I think - without having checked all the details - that you are right, and the definitions are equivalent. However, as has already been mentioned, there should be a requirement that all the transition maps in the atlas are $K$-qc for some $K$.

Suppose that we are given a surface $S$ with a "qc atlas" in this sense. Our goal is to find a Riemann surface structure on $S$ such that all the coordinates in the atlas are $K'$-qc, for some $K'$. Note that this structure will, by definition, be non-canonical.

First, we can assume that the covering corresponding to the atlas is uniformly locally finite (i.e., any point belongs to at most finitely many, let's say three, elements of the cover), and that all the sets in the covering are discs. (The latter doesn't seem necessary, but I think it makes it easier to think about these things.) [If we modify our atlas to one having these properties and solve the problem there, then any chart in the original atlas locally is a composition of a $K'$-qc map and a $K$-qc map, so the desired conclusion holds with a possibly worse constant also for the original atlas.]

So now just enumerate the elements of the cover. (I guess I am assuming countable topology - of course, this is true for Riemann surfaces by Rado's theorem; I won't worry whether this is true for your "qc" surfaces also, and just make it part of the assumptions.)

One by one, we modify the charts, using the Measurable Riemann Mapping theorem to ensure that the transition maps to any earlier charts become holomorphic. You might worry about the compatibility where a point is in more than one earlier chart, but in that case those two charts are already holomorphically compatible by construction, which means that the pullback of the corresponding complex structures under the different transition maps agree.

So this procedure results in a Riemann surface structure. The qc constant of the maps constructed may increase, but by the assumption on the cover only ever finitely many times, so this increase in dilatation is controlled.

This argument works for any surface; I see no issue with extending it to surfaces with boundaries. Apologies if I misunderstood anything.