|
|
gr.inc/question/show-that-there-is-no-closed- … nimal-surface-in-r3/
Suppose for contradiction that there exists a closed (i.e. compact without boundary) minimal surface $\Sigma$ immersed in $\mathbb{R}^3$. Let $x = (x_1, x_2, x_3)$ be the position vector of the immersion. A key fact about minimal surfaces is that the coordinate functions of the immersion are harmonic. In other words, for each $i = 1, 2, 3$, the function
\[
x_i : \Sigma \to \mathbb{R}
\]
satisfies the Laplace equation
\[
\Delta x_i = 0
\]
where $\Delta$ is the Laplace–Beltrami operator on $\Sigma$. This property follows from a standard computation showing that for an immersion $x: \Sigma \to \mathbb{R}^3$ one has
\[
\Delta x = 2H n
\]
where $H$ is the mean curvature and $n$ is the unit normal field. Since $\Sigma$ is minimal, $H = 0$, and therefore $\Delta x = 0$ componentwise.
Now, because $\Sigma$ is closed, each coordinate function $x_i$ is a continuous function on a compact set. By the extreme value theorem, $x_i$ attains a maximum and minimum. However, by the strong maximum principle for harmonic functions, if a harmonic function on a connected open set attains a non-constant maximum or minimum, then it must be constant. Since $\Sigma$ is compact and without boundary, every harmonic function must be constant.
Thus, each coordinate function $x_1, x_2, x_3$ is constant on $\Sigma$. This implies that the immersion $x: \Sigma \to \mathbb{R}^3$ is constant, meaning that the image of $\Sigma$ consists of a single point. This contradicts the assumption that $\Sigma$ is a nontrivial surface.
Therefore, there is no closed (nondegenerate) minimal surface in $\mathbb{R}^3$. |
|
