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Let $(A,g)$ be a compact surface with boundary, diffeomorphic to the standard annulus $\{z\in\mathbb{C}:1\le|z|\le 2\}$, equipped with a smooth metric $g$.

Does there always exist a conformal (bijective) map $\phi:A\to\{z\in\mathbb{C}:1\le|z|\le R\}$, for some $1<R<\infty$?

A possible idea is to use the well-known fact that $A\setminus\partial A$, as a Riemann surface (whose conformal structure is induced by $g$), is biholomorphic to either $\{1<|z|<R\}$ (for some $1<R<\infty$), $\{0<|z|<1\}$ or $\mathbb{C}\setminus\{0\}$. Is there a way to rule out the latter two cases and to prove that the biholomorphism extends nicely to $\partial A$?

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An alternate proof that interior of A is not biholomorphic with punctured disc or punctured plane is to notice that that there is a harmonic function on the interior of A which is 0 and 1 on the different boundary components respectively . Such functions do not exist on the punctured disc or the punctured plane. In fact this harmonic function has no critical points and the corresponding holomorphic 1 form has its periods an imaginary multiple of integers .Hence composing the integral of a multiple of the holomorphic one form with the exponential function gives the uniformizing map.

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  • $\begingroup$ Dear Mohan, could you please add a reference to a proof that this harmonic function has no critical points? $\endgroup$
    – Amitai Yuval
    Dec 13, 2018 at 9:01
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Yes. I assume that by "compact surface diffeomorphic to a closed annulus" you mean a bordered surface. Then by definition of a bordered surface, each boundary component has more than one point. Now an open Riemann surface homeomorphic to an annulus is conformally equivalent to $\{ z: r<|z|<R\}$, where $0\leq r<R\leq+\infty$. If each boundary component is not a point, we have the honest annulus with $r=1,R<\infty$.

That a neighborhood of a puncture is not conformally equivalent to a neighborhood of the non-degenerate boundary component is a consequence of the elementary theorem on the removable singularity.

EDIT. Let me add some detail. The punctured annuli are characterized by the property that the extremal length of the family of all homotopically non-trivial curves is zero. On the other hand, if we have a bordered Riemann surface, homemorphic to a non-degenerate annulus, we can estimate this extremal length from below. To do this, we put on our annulus some conformal metric, in which the length of the boundary components is not zero. In your case, the conformal structure is defined by a diffeomorphism. Just pull back the Euclidean metric on $1\leq |z|\leq R$ by your diffeomorphism.

For the definition and properties of extremal length, see any of the two books of Ahlfors, Conformal invariants or Lectures on quasiconformal mappings.

An alternative proof can be obtained by referring to Kwak's theorem, which is the generalization of the removable singularity theorem to maps between Riemann surfaces. (Ann. Math., 90 (1969) 9-22, or S. Lang, Introduction to complex hyperbolic spaces, Springer, 1987.)

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  • $\begingroup$ Yes, I mean a bordered surface, i.e. $A$ is diffeomorphic to a closed annulus as a manifold with boundary. $A\setminus\partial A$ is naturally a Riemann surface, with the conformal structure induced by $g$ (plus a fixed orientation, of course). I was wondering if we can exclude that $A\setminus\partial A$ is conformally equivalent to a punctured disk or plane, and whether the map provided by the uniformization theorem extends to the boundary. $\endgroup$
    – Mizar
    Oct 29, 2016 at 19:00
  • $\begingroup$ Then my answer answers your question. $\endgroup$
    – Alexandre Eremenko
    Oct 30, 2016 at 1:34
  • $\begingroup$ You can remove "diffeomorphic" and say "homeomorphic". This is enough, and the proof is in my answer. $\endgroup$
    – Alexandre Eremenko
    Oct 30, 2016 at 1:36
  • $\begingroup$ Sorry, I do not understand. Could you explain better why $A\setminus\partial A$ is not conformally equivalent to a punctured disk or plane? The removable singularity theorem does not hold for maps taking values in Riemann surfaces... $\endgroup$
    – Mizar
    Oct 30, 2016 at 1:52
  • $\begingroup$ @Mizar: it does. $\endgroup$
    – Alexandre Eremenko
    Oct 30, 2016 at 14:57

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