Uniformization of metrics vs. uniformization of Riemann surfaces

Question

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?

Answer 1

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.