There are no compact minimal surfaces

Question

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!

Answer 1

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.

Answer 2

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.

Answer 3

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.