# There are no compact minimal surfaces

This is one of the exercises of 'Do Carmo' (Section 3.5, 12)

How do you prove that there are no compact (i.e., bounded and closed in $\mathbb{R}^3$) minimal surfaces?

Thanks!

• Any thoughts of your own? Commented Oct 29, 2013 at 13:15
• Take a look at the harmonic definition of a minimal surface here en.wikipedia.org/wiki/Minimal_surface Commented Oct 29, 2013 at 13:24

Or can we say the following?

Since a compact surface must have an elliptic point, if there was a compact minimal surface, then the existence of an elliptic point $p$ would mean $\kappa_1(p)\kappa_2(p)>0$, where $\kappa_1(p)=-\kappa_2(p)$, which would imply $\kappa_1(p)\kappa_2(p)=-\kappa_1(p)^2<0$, contradiction.

Well actually this "proof" uses the crucial fact that compact surfaces have elliptic points, whose proof is similar to the ones given in above. Since Lazywei is asking this question as an exercise in do Carmo's book, and the crucial fact is also an exercise in do Carmo's book,

• The existence of an elliptic point means only that the sign of both principal curves is the same. Commented Sep 1, 2018 at 9:12
• @TheWanderer You are right. But since it is a minimal surface, we have $\kappa_1(p)=-\kappa_2(p)$, which contradicts the fact that $\kappa_1$ and $\kappa_2$ have the same sign at $p$. Or did I miss something here? Commented Sep 1, 2018 at 13:29

If a surface $S$ is compact, then every linear functional, such as $f(x)=x_1$, will attain its maximum $M$ somewhere on it. In some neighborhood of the maximum point, $S$ is the image of a (conformal) harmonic embedding $h:U\to \mathbb R^3$, where $U$ is a domain in $\mathbb R^2$. It follows that $f(h)$ attains interior maximum, and is therefore constant. It follows that the intersection of $S$ with the plane $x_1=M$ is both open and closed in $S$. Hence, $S$ is contained in a plane, which quickly leads to a contradiction.

• h is conformal doesn't imply that f(h) is. Commented Oct 29, 2013 at 14:15
• @XipanXiao But $h$ being harmonic implies that $f\circ h$ is, and this is what is used here. Commented Nov 3, 2013 at 5:53
• where did you use the fact that S is minimal? Commented Nov 3, 2013 at 21:07
• @XipanXiao A surface is minimal if and only if it has a conformal harmonic parametrization. Really, only the harmonicity of parametrization is used here. So the result applies to any surface which can be parametrized by a harmonic map. Commented Nov 3, 2013 at 21:28
• Got it. Thank you very much. added my +1. Commented Nov 4, 2013 at 0:29

I just found another proof.

proof:

Suppose $S$ is compact. Consider $f:S \rightarrow \mathbb{R}$, with $f(p) = |p|$.

Since $f$ is continuous, $\exists p_0 \in S$ s.t. $f(p_0)$ is maximum.

Now, consider all normal section of $p_0$ for all direction.

Fact: $|k(p_0)|=|k_n(p_0)|\geq \frac{1}{|p_0|}$ (Consider plane curve to prove this fact.)

Since $k_n$ is continuous, $k_n(p_0) \geq \frac{1}{|p_0|} > 0$ or $k_n(p_0) \leq -\frac{1}{|p_0|} < 0$. Therefore, $k_1 k_2 > 0$ and $k_1 + k_2 \neq 0$ leads to a contradiction.

• Could you please explain why $|(k(p_0)|=|k_n(p_0)|\ge \dfrac{1}{|p_0|}$? Commented May 28, 2014 at 9:33

If you know that $$\Delta_\Sigma x=\vec{H}=0$$, it says that the coordinates of position vector of surface are harmonic functions: $$\Delta_\Sigma x_i=0$$ for $$i=1,2,3$$. By maximum principle, harmonic function attains maximum on the boundary, but for closed surface $$\Sigma$$, there is no boundary which is a contradiction.

We can solve this problem with principal curvature. Note that

• A minimal surface has zero mean curvature $$H=0$$.
• A compact surface always has a point with positive Gaussian curvature $$K>0$$.

Consider a compact surface $$S$$, we can always find a point $$p$$ with principal curvature $$\kappa_1, \kappa_2$$ such that $$K=\kappa_1 \kappa_2>0$$. This implies $$\kappa_1, \kappa_2$$ are the same sign and non-zero at $$p$$.

Assume for the sake of contradiction, that $$S$$ is a compact minimal surface. It has a mean curvature $$K= \frac{1}{2}(\kappa_1+ \kappa_2)=0$$, which is impossible when $$\kappa_1, \kappa_2$$ are the same sign. It raises a contradiction. Therefore, there is no compact minimal surface.

If the surface is compact, every point is an elliptic point, so the principal curvatures have the same sign. Hence, the average of the principal curvature is nonzero and the surface can't be minimal.

• This is incorrect. Commented Jan 2, 2023 at 23:25