I'm re-reading a paper of Bers and for the second time, and I am yet again confused about the claim in the title, which Bers declares to be easy to prove.

For context, I'll lay out some terminology. We call a pair $(S, \alpha)$ a marked Riemann surface when $S$ is a Riemann surface and $\alpha$ is an equivalence class of generators of $\pi_1S$, each of which $\{a_i\}_{i=1}^{2g}$ satisfies the standard one relation $1= \prod_{j=1}^{2g} a_{2j-1} a_{2j} a_{2j-1}^{-1}a_{2j}^{-1}$. A map between marked Riemann surfaces $f:(S, \alpha) \rightarrow (S',\alpha')$ is a homeomorphism that respects the markings. Finally, call a pair of marked Riemann surfaces $(S, \alpha), (S', \alpha')$ to be similarly oriented if there is an orientation-preserving homeomorphism $f: S \rightarrow S'$.

Now, Bers goes on to make the following claim:

If $(S, \alpha)$ and $(S', \alpha')$ are similarly oriented marked Riemann surfaces, then there is a quasiconformal mapping $f: (S, \alpha) \rightarrow (S', \alpha')$.

Ok, this sounds plausible enough to me. On the other hand, it is by no means obvious. Here, I should mention Bers has earlier stated the equivalence of a few (standard) definitions of quasiconformality for a mapping $f: S \rightarrow S'$, including the following:

(i) $\hat{f}_\bar{z} = \mu \hat{f}_{z}$, in terms of weak derivatives, for any coordinate representation $\hat{f}$, where $\mu \in L^{\infty}$

(ii) any coordinate representation $\hat{f}$ has bounded dilatation across all quadrilaterals.

Specifically, Bers goes on to say

Had we demanded that the homeomorphism $f$ be continuously differentiable everywhere the proof would be somewhat laborious. Since we use a very general definition of quasiconformality, the proof presents no difficulties and may be omitted.

My question is thus, what is the idea that Bers' has in mind for the proof? By hypothesis, we've got our hands on a homeomorphism $f$ that respects the desired marking, but a priori $f$ might have unbounded dilatation. So, we need some clever way to homotope $f$ so that its dilatation becomes bounded across the whole surface, at which point we are done. However, I see no intuitive way to make this work. Technical details aside, I am very interested to see what ideas anyone has, or better yet, what idea they imagine Bers thought would easily yield a proof.

Bers, Lipman, Quasiconformal mappings and Teichmüller’s theorem, Princeton Math. Ser. 24, 89-119 (1960). ZBL0100.28904.

  • $\begingroup$ There are various ways to do this, e.g., one could triangulate $S$, smooth the image of the triangulation under $f$, and then map corresponding triangles by smooth diffeomorphisms which agree on the edges. By piecewise smoothness and compactness you then get quasiconformality. $\endgroup$ Jan 8, 2020 at 21:10
  • $\begingroup$ If by smooth you mean $C^{\infty}$ as a map of real manifolds, then smooth-ness is insufficient for quasiconformality. If $f$ is smooth but has $f_{z} = 0$ and $f_{\bar{z}} \neq 0$ at any point $p$, then the dilatation will be infinite at $p$. $\endgroup$
    – P7E
    Jan 9, 2020 at 15:49
  • 2
    $\begingroup$ In that case $f$ would be locally orientation-reversing. If you make sure that the map is piecewise smooth ($C^1$ should be enough) and orientation-preserving, then $f$ is quasiconformal. $\endgroup$ Jan 9, 2020 at 20:05
  • $\begingroup$ Ah, that explains things. Do you have a reference for this result? $\endgroup$
    – P7E
    Jan 10, 2020 at 16:42

1 Answer 1


These things are easier said than done. The relevant theorem is:

Theorem 1. Let $S$ be a closed connected oriented smooth surface. Then every orientation-preserving homeomorphism $f: S\to S$ is homotopic (equivalently, isotopic) to an (orientation-preserving) diffeomorphism $g: S\to S$. Alternatively, $g$ can be taken to be a piecewise-linear homeomorphism provided that $S$ is equipped with a PL structure.

If, in addition, $S$ is equipped with the structure of a Riemann surface, then $g$ is quasiconformal. (This is the easy part.)

The part about piecewise-linear homeomorphism is proven in Appendix to

D. B. A. Epstein, Curves on 2-manifolds and isotopies. Acta Math. 115 (1966), 83–107.

From that result, one obtains (by one more isotopy) a diffeomorphism, according to

J. Munkres, Obstructions to the smoothing of piecewise-differentiable homeomorphisms. Ann. of Math. (2) 72 (1960), 521–554.

One can prove more:

Theorem 2. Let $S$ be as above. Then every automorphism $\phi: \pi_1(S)\to \pi_1(S)$ is induced by a diffeomorphism $S\to S$.

In fact, one can also detect (group-theoretically) if $\phi$ is "orientation-preserving," but I will not get into this.

There are several other arguments (neither one is easy) for proving this theorem, you can find these in

B. Farb, D. Margalit, A primer on mapping class groups. Princeton Mathematical Series, 49. Princeton University Press, Princeton, NJ, 2012. Chapter 8

and (an analytical proof)

J. Hubbard, Teichmüller theory and applications to geometry, topology, and dynamics. Vol. 1. Matrix Editions, Ithaca, NY, 2006. Section 5.1.

With more work, these proofs apply to compact Riemann surfaces with finitely many punctures as long as certain conditions on $\phi$ are satisfied (these conditions are automatically satisfied in Theorem 1). The proof given in Hubbard's book requires very little extra work.

  • 2
    $\begingroup$ Indeed, with Riemann surface structure, the quasiconformality is easy. I think the relevant fact I forgot to use is that the Jacobian $J(z)$ of a mapping $w$ in local coordinate $z$ is $J(z) = |w_{z}|^2- |w_{\bar{z}} |^2$. Hence, if $w$ is orientation-preserving, then $|w_{z}| >0$ everywhere so that one obtains the local bound on dilatation in the $z$ chart via $k_{w}(z) \leq \frac{ \sup_{z} |w_{\bar{z}|} }{\inf_z |w_{z}| } < \infty$. By compactness of our surface, this is enough for quasiconformality everywhere. $\endgroup$
    – P7E
    Jan 10, 2020 at 18:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .