# 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!

## 5 Answers

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. – TheWanderer Sep 1 '18 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? – GRR Sep 1 '18 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. – Xipan Xiao Oct 29 '13 at 14:15
• @XipanXiao But $h$ being harmonic implies that $f\circ h$ is, and this is what is used here. – user103402 Nov 3 '13 at 5:53
• where did you use the fact that S is minimal? – Xipan Xiao Nov 3 '13 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. – user103254 Nov 3 '13 at 21:28
• Got it. Thank you very much. added my +1. – Xipan Xiao Nov 4 '13 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 $|(k(p_0)|=|k_n(p_0)|\ge \dfrac{1}{|p_0|}$? – Arsenaler May 28 '14 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.

Since 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.