Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
Can you tell us where you found this proof? –  treble Apr 25 '13 at 2:38
@treble: It is from a set of lecture notes (not available online). But I believe I know how to do it by now. I will write it up as an answer as soon as I find the time... (the answer is really geometric) –  Sam Apr 25 '13 at 23:19
Since this is evidently not one of Hopf's original proofs, it would be nice to see it in full. I would guess that earlier in the notes, there is a discussion of the properties of conformal immersions, e.g. there are some formulas that reduce to the one you want in the case of a conformal immersion from $\mathbb R^2$ to $\mathbb R^3$. Otherwise you just have to compute. I would be interested to read the notes. –  treble Apr 25 '13 at 23:24
If you find the time to write your solution, then you know that at least one person (me) will be happy to read it. :) –  treble Apr 25 '13 at 23:28
@treble: I have included the rest of the argument now. :) I hope what I wrote is intelligible. And unfortunately, the author of these notes specifically asked that they not be circulated for the time being. –  Sam Apr 27 '13 at 3:22

2 Answers 2

up vote 2 down vote accepted

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.

share|improve this answer

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.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.