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Assume we have a Riemann surface, the underlying topological surface of which is a sphere with (possibly uncountably many) points removed. Can we always conformally embed this Riemann surface in the Riemann sphere? If not, can someone suggest a counter example?

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Could you please edit your question to describe the kind of embedding you're interested in? –  Ryan Budney May 12 '12 at 18:30
    
Isometrically? Conformally? Topologically? Topologically obviously yes, isometrically obviously no - just take a very bumpy sphere with nonconstant curvature. –  Will Sawin May 12 '12 at 18:31

2 Answers 2

up vote 10 down vote accepted

See e.g. here:

Theorem 3.2.7. Any planar connected Riemann surface is biholomorphic to an open subset of $S^2$.

The proof is very straightforward: Exhaust a genus $0$ surface $S$ by relatively compact domains $D_n$ each of which necessarily has genus $0$. For each $D_n$ find a conformal embedding $f_n$ to $S^2$. Now, normalize the family of mappings $f_n$ to to send a point $x\in D_1$ to a fixed point $z\in {\mathbb C}$ and to have unit derivative (in a chart) at $x$. Then use normality of the family of maps $f_n$ to get the limit (for a subsequence). Lastly, check that the limit is injective. This is the same argument Caratheodory used in his proof of uniformization theorem.

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Unless I am confused, why not add the points back, map the resulting surface to the Riemann sphere conformally by unifomization, then remove the images of the offending points?

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@Igor: It is not as simple. The assumption is the surface embeds topologically to $S^2$. The conclusion is that there exists a conformal embedding. –  Misha May 12 '12 at 19:26
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I think the problem with this is that you are assuming the sphere had a complex structure before the points were removed, which might not be the case. For example assume we take a topological sphere $X$ and remove one point $x$. Then topologically $D = X \setminus \{ x \}$is a disc and we can give it the conformal structure of a disc. However there is no homeomorphism from $X$ to the Riemann sphere which embeds $D$ conformally. This is because the image of $D$ must be the whole of the Riemann sphere minus one point, and thus be conformally equivalent to the plane by the uniformization theorem. –  uncooltoby May 12 '12 at 19:40
    
@Igor: I think, I understood the source of confusion. Indeed, if the surface has finite typological type, it is easy to "fill in the holes": Take annuli around the holes, embed them conformally in $S^2$ (since every annulus is conformal to a round annulus or the punctured disk) and then attach disks conformally along the annuli ("fill in the holes"). This is all fine, but it does not work if your surface is homeomorphic to, say, $S^2$ minus Cantor set. Then one has to use the "proof by exhaustion" argument I outlined above. –  Misha May 12 '12 at 20:01
    
Yes, all true. I guess the point is that I was not sure what the assumptions were... –  Igor Rivin May 12 '12 at 20:55

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