# 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 its 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, 2015 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.

• Hi Jeremy, sorry to resuscitate an ancient answer. Since we never end up composing the change-of-charts maps, except to get another change-of-charts map that is also $K$-qc by assumption, does the problem of dilatation increasing under compositions actually arise? I thought that this is OK and one does indeed get a qc surface with this alternative definition also. Dec 18, 2022 at 22:53

EDIT. I am returning to this many years after the fact. The essence of the answer remains unchanged, but I am writing it now as a more detailed proof.

Suppose that we are given a surface $$S$$ with a countable "$$K$$-qc atlas", where the change-of-charts maps are $$K$$-qc. (As already mentioned in the comments, it is important that the $$K$$ is uniform across $$S$$. I am assuming countable topology - this is true for Riemann surfaces by Rado's theorem; it should be true for your "qc" surfaces also, but I will just make it part of the definition.)

More precisely, the surface $$S$$ is covered by open sets $$(U_i)_{i=1}^{\infty}$$, and there are homeomorphisms $$\phi_j\colon U_i\to D_i\subset \newcommand{\C}{\mathbb{C}}\C$$ such that the transition maps $$\phi_j\circ \phi_k^{-1}$$ (when and where defined) are all $$K$$-qc.

We may assume for simplicity that the closures of the U_i are closed topological discs, every D_i is an actual Euclidean disc, and that any $$U_i$$ intersects at most finitely many other $$U_k$$. We won't actually use these, but they may simplify the mental picture.

CLAIM. There exist open sets $$W_j\subset\C$$ and $$K$$-quasiconformal homeomorphisms $$\theta_i\colon D_i\to W_i$$ such that the charts $$\psi_i\colon D_i\to W_j; \quad \psi_i := \theta_i\circ \phi_i$$ define a Riemann surface structure. (I.e., the transition maps $$\psi_i\circ \psi_k^{-1}$$ are all holomorphic.)

PROOF OF THE CLAIM. We construct the maps $$\theta_i$$ inductively, with the additional property that the dilatation of $$\theta_i$$ is trivial on $$\phi_i\left(U_i\setminus \bigcup_{k=1}^{i-1} U_k\right).$$ For $$i=1$$, this means that $$\theta_1$$ is conformal; we set $$\theta_1 = \operatorname{id}$$ and $$W_1=D_1$$. Suppose $$i>1$$ and the maps are defined up to $$i-1$$ with the desired properties. We define a Beltrami differential $$\mu_i$$ on $$D_i$$ as follows: for points in $$\phi_i(U_k)$$ with $$k, it is the pullback of the standard complex structure under $$\phi_i\circ \psi_k^{-1}$$. At other points, it is the standard complex structure. By the induction hypothesis, $$\psi_{k'}^{-1}\circ \psi_k$$ is holomorphic for $$k'\leq k, which means that the differential is well-defined.

Moreover, the dilatation of the differential is bounded by $$K$$. Indeed, for any point $$z\in D_i$$, let $$k$$ be minimal such that $$z\in \phi(U_k)$$. If $$k=i$$, then the dilatation is $$1$$ at $$z$$ by definition. Otherwise, it is given by the pullback of the standard dilatation under $$(\phi_i\circ \phi_k^{-1})\circ \theta_k^{-1}$$. Now $$\theta_k$$ is holomorphic at $$\phi_k(z)$$ by our additional inductive hypothesis, and the change of charts map is $$K$$-qc by assumption. So the dilatation is indeed bounded by $$K$$.

Now we can solve the Beltrami equation and find $$W_i$$ and a quasiconformal map $$\theta_i\colon D_i \to W_i$$ such that $$\mu_i$$ is the pullback of the standard complex structure on $$\theta_i$$. (If $$U_i$$ is connected and simply-connected, we can let $$W_i$$ be a disc.) This means that $$\psi_i\circ \psi_k^{-1}$$ is holomorphic for $$k, as desired. This completes the inductive construction and the proof of the claim.

Observe that the map from our original "qc surface" to this new Riemann surface is, in charts, given by the $$\theta_i$$. So it is $$K$$-quasiconformal, and we have found the desired Riemann surface structure.

The argument works for any surface; I see no issue with extending it to surfaces with boundaries.

Note that, if we choose a different enumeration of our charts, then we will get a different Riemann surface structure. So this structure is not canonical, as mentioned in the original version of my answer.