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Hello all,

I am interested in the following question. Suppose a,b,c,z are points in the complex sphere. Consider the family of curves g through a,b,c, and for each g let U be the complement of g in the sphere. Can we find g so that p_U(z) is minimal for this family of curves, where p_U represents the hyperbolic(or Poincare) metric of U? Actually more important to me is the value of p_U, or at least a good, nontrivial lower bound. We can assume g is as smooth as you like if it helps. This problem is not too hard with only two points, a and b, but I am stumped with 3 points. I thought about mapping z to infinity by a Mobius transformation and then using capacity or some such thing, but I can't make it work. I don't know whether this can be approached using some techniques from free boundary problems, etc. Anyway, and help or ideas would be most appreciated.

Greg

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I guess you wanted to say that p_U(z) is the conformal factor for the hyperbolic metric (?); I guess the one with curvature $=-1$ (?). –  Anton Petrunin Mar 8 '11 at 1:22
    
Is the infimum always achieved by a curve, or sometimes by a 1-complex which looks like a Y? –  Douglas Zare Mar 8 '11 at 3:13
    
@Petrunin - I'm not sure what you mean by "factor", but I probably should've said density rather than metric. Yes, I mean the one with constant curvature. Thanks. @Zare - Well, I guess it is a Y, considering the answer given below. That's fine by me, it's still a curve, or close enough. –  Greg Markowsky Mar 8 '11 at 7:38
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1 Answer

up vote 2 down vote accepted

If I understand your question correctly, the "complex sphere" is the same as the Riemann sphere, or the one point compactification of $\mathbb C$.

There is a well-developed theory of questions like this. Some of the keywords are Jenkins-Strebel differential, measured foliations, extremal length, fatgraphs, Schwarz-Christoffel mappings.

A key statement: any surface with at least one puncture is conformally equivalent to a surface obtained from half-infinite cylinder $[0,\infty) \times S^1$ by partitioning the boundary component into intervals and pairing them in a way so as to preserve arc-length. There can be vertices of order 1, but these should correspond only to punctures; the infinite end of the cylinder corresponds to the desginated puncture. This representation is unique up to isomorphism. (This is a special case of a much more general structure theory for quadratic differentials and their relationship to measured foliations, which are a topological concept. Uniqueness is an elementary argument in the theory of extremal length). The infinite cylinder is really $\log(w), where w ranges over a punctured disk, so the Poinare metric is immediate.

In your case: the boundary of the cylinder will fold into a Y, with its shape determined by the positions of three "folding" points on a circle whose pairwise distances must satisfy the triangle inequality. Given the glued-up cylinder, you can then solve for the uniformizing map to the sphere using the Schwarz-Christoffel technique generalizing the method of finding Riemann mappings to polygons; in general, you'd need to solve for the Schwarzian derivative. Then the problem is to match the crossratio of the four points a,b,c,z to the four points that result. This may sound complicated, but it should work numerically quite quickly once it's programmed: it's at least a finite-dimensional problem, and I think the solutions are numerically well-behaved if you do it right.

I suspect that for this particular case you can express the solution in terms of elliptic functions, but I'm not up to working out the formula.

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Hmm, I see, thank you. I'm going to have to do some reading before I can understand your reply. And yes, complex sphere=Riemann sphere. I'll hit the books. –  Greg Markowsky Mar 8 '11 at 7:41
    
@Greg Markowsky: I'm sorry my answer isn't more directly helpful: I left you with a lot of terms that might take a lot of energy to chase down. If your main aim is just to get the answer rather than to learn something about these various topics, it might not be worth it, and your best bet might be to find a local person knowledgable enough and interested enough to help work it out. They might be physicists. –  Bill Thurston Mar 8 '11 at 22:15
    
No, I think it is definitely worth my time to learn these things. Thanks, and I may have to make more use of this board as I stumble my way through. –  Greg Markowsky Mar 9 '11 at 2:53
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