Question

Every open set in the complex plane homeomorphic to an annulus is biholomorphic to exactly one annulus whose inner radius is 1 and whose out radius is $r>1$. Each value of $r$ gives a different complex manifold. This number r is called the "modulus" of the annulus. You could say that the set of real number $(0,\infty)$ give a moduli space of complex annuli.

If I give you a parameterization of the inner boundary and the outer boundary of a topological annulus, is there some explicit formula for the modulus of the annulus it is conformally equivalent to? I am guessing there is some kind of integral formula. I looked through a few papers, and have not found such a formula, but it must certainly be classical.

My question is very similar to

http://mathoverflow.net/questions/11239/conformal-maps-of-doubly-connected-regions-to-annuli

but that question has an accepted answer which does not answer my question. The advice on Meta was just to ask a new question, but explain the link to the old one.

Answer 1

I think that there is no formula. The best one can do is to estimate. Here is a simpler problem of the same sort: suppose you have a parametrization of the boundary of a simply connected region, and suppose that 0 is inside. Consider the Riemann mapping f of this region sending 0 to 0. The problem is to find |f'(0)|. There is no formula in any reasonable case.

Of course this is not a theorem, because one cannot define what a "formula" is. Both quantities, the modulus of a ring, and |f'(0)| in the simplified problem are solutions of certain extremal problems. So one can write a "formula" involving sup over some class of functions.

Added on 9.19: I don't know why the question about a "formula" is important. There are reasonably good converging algorithms for finding moduli of rings, of course. The closest thing to a "formula" for a conformal map of a simply connected region that I know is described in the papers of Wiegmann and Zabrodin, for example, MR1785428. Perhaps this can be modified to make a formula for the modulus of a ring.

Added on the same day: Here is a "formula". Let $\mu$ and $\nu$ be two probability measures, one sitting on each boundary component. Let $\rho=\mu-\nu$. Then $$\log r=-2\pi\sup\int\int\log|z-w|d\rho(z)d\rho(w),$$ where the $\sup$ is taken over all such measures. If your boundaries are smooth, the measures are also smooth, and can be described by smooth densities.

Explanation. Think of the boundaries as bases of metal cyinders, and put unit charges on them, one positive another negative. Then allow the charges to flow according to Coulomb Law. they will occupy the equilibrium position (minimizing the energy). This minimal energy is $\log r/(2\pi)$ and it is conformally invariant. It is the so-called capacity of a condenser.

This was given as an example of what I meant by a formula containing a sup over a set of functions.

Answer 2

You can read all about numerical approximation schemes to compute this in this Diplomarbeit. link text

Answer 3

I also think there is no "explicit formula" for the conformal modulus, even in simple cases. However, as mentioned in Igor Rivin's answer, there are methods for approximating the conformal map and the conformal modulus. These methods are generelizations to doubly-connected domains of the well-known Bergman Kernel Method for simply connected domains.

For an introduction to these methods, I suggest you take a look at this document, sections 1.3 and 2.5.