📂Geometry

Differentiable Manifolds in Compact Surfaces

Definition¹

• Let us say $\overline{B}_{r} = \left\{ p \in \mathbb{R}^{3} : \left| p \right| \le r \right\}$. If for a surface $M \subset \mathbb{R}^{3}$, there exists $r$ that satisfies the following, then $M$ is said to be bounded. $$M \subset \overline{B}_{r}$$

• If every sequence of points $\left\{ p_{n} \right\}$ on $M$ satisfies the following equation, in other words, converges to point $p$ on $M$, then $M$ is said to be closed.

$$\exist \lim\limits_{n \to \infty}p_{n} = p,\quad p \in M$$

• If the surface $M$ is bounded and closed, it is said to be compact.

Explanation

The above definitions simply redefine boundedness, closedness, and compactness, which were defined in $\mathbb{R}$, for surfaces.

Lemma

Let $M$ be a compact surface, say $r = \min \left\{r | M \subset \overline{B_{r}}\right\}$. Then, there exists $p \in M$ such that $\left| p \right| = r$. In other words,

$$\exists p \in M \text{ such that } \left| p \right| = r \quad \text{and} \quad M\cap S_{r} \ne \varnothing$$

Here, $S_{r}$ is a sphere with radius $r$.

Proof

Let’s say $n \gt 0$ for $r_{n} = r - \dfrac{1}{n}$. Then,

$$M - \overline{B_{r_{n}}} \ne \varnothing$$

Now, let’s say $p_{j} \in M - \overline{B_{r_{j}}}$. Then, since $M$ is bounded and $p_{j} \in M$, there exists a convergent subsequence to some point $p \in \mathbb{R}^{3}$.

$$\exists \text{subsequnce } \left\{ p_{n_{j}} \right\} \text{ such that } \lim\limits_{j\to \infty} p_{n_{j}} = p \in \mathbb{R}^{3}$$

Since $M$ is closed,

$$p \in M,\quad \left| p \right| = r$$

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Theorem

Let $M$ be a compact surface. Then, there exists a point $p$ with positive Gaussian curvature.

$$\exists p \in M \text{ such that } K(p) \gt 0$$

Proof

By the lemma,

$$r = \min\left\{ r : M \subset \overline{B_{r}}\right\} \implies S_{r} \cap M \ne \varnothing$$

Claim: $K(p) \gt 0$ for any $p \in S_{r}\cap M$

Let’s say $p \in S_{r}\cap M$. Then, $S_{r}$ and $M$ have the same unit normal vector at $p$. Therefore, the tangent plane at point $p$ is the same.

$$T_{p}S_{r} = T_{p}M$$

Now, let’s say $\mathbf{X} \in T_{p}S_{r} \cap T_{p}M$. And let’s consider the plane generated by $\Pi$ as $\left\{\mathbf{n}, X\right\}$. Then, the normal curvature of $M$ in the direction $\mathbf{X}$, $\kappa_{n}$, is the same as the curvature of the curve $\Pi \cap M$. However, this has the same sign as the normal curvature of $S_{r}$ at $p$. Therefore, $\left| \kappa_{n} \right| \ge \frac{1}{r}$, and if $\kappa_{1}, \kappa_{2}$ is called the principal curvature,

$$K = \kappa_{1}\kappa_{2} \ge \dfrac{1}{r^{2}} \gt 0$$

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