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The uniformization theorem in complex analysis says that

T1. Any Riemann surface of genus $0$ is conformally equivalent to the unit sphere.

The uniformization theorem in differential geometry says that

T2. Any smooth Riemannian metric on $S^2$ is conformal to the round metric.

T2 implies that any metric $g_{ij}$ on a sphere has the form $e^\sigma (g_0)_{ij}$, where $g_0$ is the standard metric of the unit sphere. In particular, any two metrics are conformal to each other. Here are a few paradoxical statements that seem to follow from this:

Cor1. Any diffeomorphism $f:S^2\to S^2$ is a holomorphic map.

This is because we can use this map to define a new metric $\,f_\ast g$, but the new metric must be conformal to the old metric, therefore $f$ is a conformal map. As far as I understand, being conformal in the sense of Riemannian geometry is the same as being conformal in the sense of complex analysis (?).

Obviously, this is nonsense because the only holomorphic automorphisms of $S^2$ are the Möbius transformations.

Cor2. Any coordinate chart on $S^2$ is conformal for any metric.

This is because the metric is proportional to some other metric, which is diagonal in these coordinates, therefore is itself diagonal.

This is also obviously nonsense because locally the matrix of the metric is an arbitrary symmetric positive $2\times 2$ matrix.

What am I missing, and what is the relationship between T1 and T2? If I want to deform Riemannian metrics on the sphere (no complex structure), is it indeed enough to look at just the conformal variations, or are there nontrivial quasiconformal variations?

Alex Bogatskiy
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    Possibly relevant / helpful is Mike Miller's (very clear) answer here: https://math.stackexchange.com/questions/1729690/constant-curvature-metric-and-biholomorphic-equivalence – Jesse Madnick May 01 '17 at 19:38

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The statement of Corollary 1 presumes a given holomorphic structure on $S^2$, but in your proof you changed the holomorphic structure on the range. So you did not prove what you stated in Corollary 1. Instead you proved:

  • For any diffeomorphism $f : S^2 \to S^2$, one can change the holomorphic structure on the range so that $f$ is holomorphic.

Which is, of course, trivial to prove using just the method you gave. Note that your proof did not even use T2, so one might be a tad suspicious about this proof solely on that basis.

Similarly, the statement of Corollary 2 presumes a given conformal structure on the chart, and your proof changes the holomorphic structure.

Lee Mosher
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  • So what is the exact statement of T2? When I see it in differential geometric sources, there are no complex structures around. And I think my proof did use T2, because I used the fact that any two metrics are conformal. Is this not what T2 is saying? – Alex Bogatskiy May 01 '17 at 19:02
  • Also, isn't the complex structure on the sphere unique? Forgive my confusion, I just can't find a source that discusses and compares all the different sides of uniformization. – Alex Bogatskiy May 01 '17 at 19:06
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    The complex structure is unique up to diffeomorphism. The means that if you have two complex structures on the sphere, then there is diffeomorphism of the sphere which takes one complex structure to the other one. It does not mean those two complex structures are "the same". – Lee Mosher May 01 '17 at 20:10
  • Agreed. So is the full statement of T2 then: "given a fixed holomorphic structure, any conformal class of metrics compatible with this structure contains a complete metric of constant curvature"? And to go from one complex structure to another I can use non-holomorphic (say, quasiconformal) maps? – Alex Bogatskiy May 01 '17 at 20:18
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    Delete the words "given a fixed holomorphic structure" and it's good. The reason you see "no complex structures around" is because a conformal class of Riemannian metrics determines a unique complex structure --- the words "complex structure" and "holomorphic structure" and "conformal structure" and "conformal class of Riemannian metrics" are all basically equivalent concepts on a surface. – Lee Mosher May 01 '17 at 23:24
  • Ah, makes sense. Are there still many conformal classes on the sphere? It would seem that they are classified by the radius of the round metric they contain. But all round metrics are conformal to each other, so there's only one conformal class... And I come back to my initial question, which is that how can it be that I can't vary the metric non-conformally? – Alex Bogatskiy May 02 '17 at 01:42