Hopf's theorem on CMC surfaces

Question

I got stuck reading the proof of the following theorem:

Theorem (Heinz Hopf) Let $X: S^2\to \mathbb R^3$ be a constant mean curvature immersion. Then $X(S^2)$ is a round sphere.

Proof: Let $g_{S^3}$ denote the round metric on $S^2$ (such that the area is $4\pi$). By the uniformization theorem there exists a map $\phi: S^2 \to S^2$ such that $(X\circ \phi)^\ast g_{\mathbb R^3}$ is conformal to $g_{S^2}$. Hence we may assume that $X$ is a conformal map $(S^2, g_{S^2}) \to (\mathbb R^3, g_{\mathbb R^3})$. Let $\pi: (\mathbb R^2, g_{\mathbb R^2})\to (S^2\setminus \{pt\}, g_{S^2})$ be stereographic projection. Let $Y = X\circ \pi$. Then $Y$ is a conformal immersion from $\mathbb R^2$ to $\mathbb R^3$. If $\nu = \nu^\alpha e_\alpha$ denotes the unit normal vector field along this immersion, then $\Delta \nu^\alpha + |h|^2 \nu^\alpha = 0$, where $\Delta = \frac{\partial^2}{\partial u^2} + \frac{\partial^2}{\partial v^2}$ is the usual Laplacian and where $|h|^2$ is the length of the second fundamental form tensor along the immersion.

Edit: Here is the remainder of the proof (from memory). In particular, this shows that $\Delta \nu$ is parallel to $\nu$. Identifying $\mathbb R^2\cong \mathbb C$ and using complex notation ${\partial_z} = \frac 12 \left(\frac{\partial }{\partial u} - i\frac{\partial}{\partial v}\right)$, $\overline{\partial_z} = \frac 12 \left(\frac{\partial }{\partial u} + i\frac{\partial}{\partial v}\right)$, we have $\Delta \nu = 4 \partial_z\overline{\partial_z}\nu$. Note that $\partial_z\nu \perp \nu$, whence $\partial_z\nu \perp \Delta \nu$. It follows that $$\overline{\partial_z} (\partial_z\nu)^2 = 2 \partial_z \nu \cdot \partial_z\overline{\partial_z}\nu = \frac 12 \partial_z \nu \cdot \Delta \nu = 0.$$ This implies that the complex-valued function $z\mapsto (\partial_z\nu)^2$ is holomorphic. Here $$(\partial_z \nu )^2 = (\partial_u\nu - i \partial_v \nu)^2 = (|\partial_u\nu|^2 - |\partial_v\nu|^2) - 2i \partial_u \nu \cdot \partial_v\nu.$$

It now follows from the conformal invariance of the Dirichlet energy $E(u) = \int_M |\nabla u|^2 \, \mathrm{dvol_g}$ that $\int_{\mathbb R^2} |(\partial_z \nu)^2| \, < \infty$. Indeed, the Dirichlet energy of $\nu$ over $\mathbb R^2$ (with respect to the Euclidean metric), bounds $\int_{\mathbb R^2} |(\partial_z \nu)^2|$. Now $\nu$ can be pulled-back to a vector field on the sphere via stereographic projection (which is conformal), and the Dirichlet energy can be calculated there (with respect to the round metric). But the pull-back of $\nu$ extends to a smooth vector field on all of $S^2$, and $S^2$ is compact. Therefore its Dirichlet energy must be finite.

It follows from $\int_{\mathbb R^2}|(\partial_z \nu)^2|<\infty$, that the (holomorphic) function $(\partial_z\nu)^2$ is identically $0$. This is equivalent to $|\partial_u \nu| = |\partial_v\nu|$ and $\partial_u \nu \cdot \partial_v \nu = 0$.

Write $\partial_u \nu = h^u_u \partial_u Y + h^v_u \partial_v Y$, $\partial_v \nu = h^u_v \partial_u Y + h_v^v\partial_v Y$. It follows from the conformality of $Y$ together with the above, that $h^u_v H = h^u_v (h^u_u + h^v_v) = 0$, $|h^u_u| = |h^v_v|$. This is only possible if $h^u_u = h^v_v = H/2$ and $h^u_v = 0$.

But then $\partial_u \nu = H/2 \partial_u Y$ and $\partial_v \nu = H/2\partial_v Y$, imply that $Y = c + 2/H \nu$ for some constant vector $c$. This shows that $Y$, and by continuity also $X$, map into a sphere. $X$ is onto, because it is open (being an immersion) and closed (being continuous, mapping from a compact set).

I don't see how to show that $\Delta \nu^\alpha + |h|^2\nu^\alpha = 0$?

Thanks for your help! Any ideas are welcome.

Answer 1

First of all we note that $ Y $ is a c.m.c conformal immersion (note that $ \pi^\ast g_{S^2}=\mu g_{R^2} $). The crucial point is that it is a constant mean curvature immersion. In fact for constant mean curvature immersions it holds the following result:

Let $ M $ be an oriented hypersurface immersed in $ \mathbb{R}^{n} $ with constant mean curvature and let $ \nu $ be the unit normal vector field along $ M $. For any $ a \in \mathbb{R}^{n} $ it holds that

$ \Delta \langle a,\nu\rangle +|B|^{2}\langle a,\nu\rangle=0 $

where $ B $ is the second fundamental form of $ M $.

For a proof see Xin 'Minimal Submanifolds and relatated topics', Proposition 1.3.5.

Answer 2

Consider a smooth deformation $\{\phi_t\}_{|t|<\epsilon}$ of the immersion $X$, i.e. $\phi: S^2\times (-\epsilon, \epsilon)\to \mathbb R^3$ is a smooth map, with $\phi_t := \phi(\cdot, t):S^2\to \mathbb R^3$, such that $\phi_t$ is a smooth immerison for fixed $t$ and $\phi_0 = X$. For each $t$, we can calculate the mean curvature $H_t$ for $\phi_t$. We are interested in how $H_t$ varies with $t$.

Let us assume that $\phi_t$ is in fact of the simple form $\phi_t(x) = \phi_0(x) + t\psi(x) \nu(x)$, for some smooth function $\psi: S^2 \to \mathbb R^3$. Here $\nu:S^2\to \mathbb R^3$ denotes the unit normal to $\phi_0$.

If one does the calculation, then one eventually obtains that (I haven't yet checked this in complete detail though!) $$ \frac{d}{dt}\Big|_{t=0} H_t = \Delta_g\psi + |h|_g^2 \psi.$$

In our particular case, we are interested in a variation of the form $\phi_t(x) = x+te_\alpha$. In this case, the function $\psi(x)$ is given by $$\psi =t^{-1} \langle \phi_t- \phi_0, \nu\rangle = \langle e_\alpha, \nu\rangle = \nu^\alpha.$$ But $\phi_t$ is just a translation in direction $e_\alpha$, so the mean curvature clearly doesn't change. It follows that $ \Delta_g\nu^\alpha + |h|_g^2 \nu^\alpha = 0$. To get the claim in the notes, one has to use the fact that in dimension $2$, the Laplacian obeys a conformal invariance.