For globally conformally flat surfaces, does radial symmetry of conformal factor imply the surface is a sphere?

Question

We know that for the unit sphere in $\mathbb{R}^3$, the standard metric on such a sphere can be written as $$g=\frac{4}{(1+x_1^2+x_2^2)^2}(dx_1^2+dx_2^2)$$by a stereographic projection map from the sphere to the complex plane.

I'm curious about the following question:

Given a closed Riemannian surface in $\mathbb{R}^3$, if the metric is globally conformal to $\mathbb{R}^2$, that is, $g=e^{2u}(dx_1^2+dx_2^2)$, and if we further assume that $u$ is radially symmetric, then does the surface have to be a sphere? Are there any other possible shapes except spheres? Can anyone give some specific examples?

Any comments or ideas would be really appreciated.

Answer 1

This looks like a purely topological question to me because (orientable) closed surfaces are classified by their genus; the conformal hypothesis looks like a red herring. Please correct me if I misunderstood the premise of your question.

Let $(\Sigma,g)$ be the closed surface in question. Under the hypotheses imposed, there is a point $p \in \Sigma$ and a coordinate chart $\psi: \Sigma \setminus \{ p \} \to \mathbf{R}^2$ so that $\psi^* g_e = e^{-2u} g$ for some function $u: \mathbf{R}^2 \to \mathbf{R}$. Here I write $g_e = \mathrm{d} x_1^2 + \mathrm{d} x_2^2$ for the euclidean metric. In particular $\Sigma \setminus \{ p \}$ and $\mathbf{R}^2$ are homeomorphic. Therefore the Euler characteristic of $\Sigma \setminus \{ p \}$ is $1$, and $\chi(\Sigma) = 2$. But then $\Sigma$ has genus zero, and thus is a sphere.

The slight hiccup in the argument is that it assumes orientability. It was not clear whether you are willing to impose this or not; certainly this would be the case if the surface is an embedded submanifold of $\mathbf{R}^3$. Without this assumption, $\Sigma$ could also be a Klein bottle. However returning to the homeomorphism above, one sees that $\Sigma \setminus \{ p \}$ is contractible, which is not the case for a Klein bottle.

Edit: It looks like I misunderstood your question, and you are instead asking whether a surface $(\Sigma,g)$ with properties as above is necessarily a round sphere. I suspect there exist simple examples that demonstrate this not to be the case; regardless here is my attempt.

By the above we know that $\Sigma$ is diffeomorphic to $\mathbf{S}^2$. Next, by Nirenberg's solution of the Weyl embedding problem, we know that the metric $g$ can be induced from an isometric embedding into $\mathbf{R}^3$ provided it has positive Gauss curvature.

Therefore it is enough to find $u$ for which the Gauss curvature $K = K_g$ is positive but not constant. In terms of $u$ this is $K = - e^{-2u} \Delta u$. Let $u_0 = \frac{1}{2} \ln \frac{4}{(1 + r^2)^2}$ be the `Euclidean' conformal factor, which we perturb by a compactly supported rotationally invariant function $\varphi$, meaning $u = u_0 + \delta \varphi$ for some small $\delta > 0$. (Doing this only changes the metric near the south pole.)

Then $K = -e^{-2u_0 -2 \delta \varphi} (\Delta u_0 + \delta \Delta \varphi) = e^{-2\delta \varphi}(1 - \delta e^{-2u_0} \Delta \varphi)$. This is positive provided $\delta$ is small enough in terms of $\varphi$. `Basically any' perturbation should do the trick, for example picking $\varphi(r) = r^2 - 1$ one finds $K = -e^{-2\delta(r^2-1)}(1 - 4\delta e^{-2u_0})$, which is not constant. (Technically one would need to change $\varphi$ so as to make the metric smooth, but this can be done without changing the curvature near the south pole.)

Answer 2

Let $\gamma(s)$ be a curve in the plane parametrized by arclength. Assume $\gamma(0) = (0,0)$ and $\gamma'(0) = (1,0)$. Rotate $\gamma$ around the $y$-axis and you get a surface of revolution parametrized by $(s,\theta) \in (0,S)\times \mathbb{S}^1$.

The induced metric is

$$ g = ds^2 + \gamma_1(s)^2 d\theta^2 $$

where $\gamma_1$ is the $x$-component of $\gamma$. By construction we have $\lim_{s\to 0} \gamma_1(s) / s = 1$.

Let $r$ be a function that solves $$\tag{#} \frac{d}{ds} \ln r(s) = \frac{1}{\gamma_1(s)} $$ One can check that necessarily $r(0) = 0$.

Using the $(r,\theta)$ coordinates instead we find

$$ g = \Big(\frac{\gamma_1(s)}{r}\Big)^2 (dr^2 + r^2 d\theta^2) $$

is conformally flat, with conformal factor that is radially symmetric (independent of $\theta$).

Note that this works for any surface of revolution.

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In terms of uniformization: it is of course known by the uniformization theorem that that any topological sphere embedded in $\mathbb{R}^3$ has its induced metric conformal to $\mathbb{S}^2$ and hence is conformally flat. You can regard this as trying to solve an elliptic PDE (for the conformal factor) on the surface. If you assumed that your surface is rotationally symmetric, then this symmetry descends to the PDE you are trying to solve and reduce it to an ordinary differential equation. This is essentially equation (#) above.

Answer 3

Any rotationally symmetric metric will do, so any compact surface of revolution of genus zero embedded in 3-dimensional Euclidean space.

On the other hand, suppose we have any such surface as in the problem. To have an open subset which is conformal to the plane, an oriented connected surface with Riemannian metric must fail to be Brody hyperbolic, and clearly its universal covering space has to be the sphere or the plane. So the surface, if compact, is diffeomorphic to the torus or the sphere. But in the case of the torus, the open set in the universal covering space which is conformal to the plane is already the entire covering space (no Fatou Bieberbach domains in complex dimension one), so in the quotient to the torus, there is no isometric symmetry vector field commuting with the covering map, as the fixed points would have to be invariant under the translations. So no rotational symmetry of the metric. So we are down to the sphere with its standard conformal structure. In the universal covering space, a connected open set conformal to the plane must omit precisely one point. So the symmetry vector field is defined except perhaps at that point. Its flow being by isometries, it will extend smoothly to fix that remaining point. So the question reduces to whether a surface of revolution of genus zero has an isometric embedding which is not rotationally symmetric. For positive curvature surfaces, it follows from Nirenberg's theorem that there is only one isometric embedding up to rotation, so it is rotationally symmetric. But I don't know the general case.