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I was trying to prove that $S^n$ is geodesically complete, which it is direct if I can prove that $S^n$ is compact. The problem is that I don't know how to do this. I know that every closed submanifold of an Euclidean space is complete and that $S^n$ is a closed submanifold of $\mathbb{R}^{n+1}$ because it is the inverse image of the regular value $0$ of the function $h: \mathbb{R}^{n+1} \longrightarrow \mathbb{R}$ defined by $h(x_1,\cdots,x_{n+1}) := \sum\limits_{i=1}^{n+1} x_i^2 - 1$. Therefore $S^n$ is complete as metric space and it is compact from Hopf-Rinow's theorem once that $S^n$ is a closed and bounded subset of itself, but how to prove that $S^n$ is compact if I don't consider an immersion of $S^n$ into $\mathbb{R}^{n+1}$?

Thanks in advance!

$\textbf{EDIT:}$

After some discussions on the comments, I am trying to be more specific about my doubt: how to prove that $S^n$ compact with respect to the natural topology of a smooth manifold, i.e., $A$ is an open subset of $M$ if $x_{\alpha}^{-1}(A \cap x_{\alpha}(U_{\alpha}))$ is an open set of $\mathbb{R}^n$ for each $\alpha$, where $x_{\alpha}: U_{\alpha} \subset \mathbb{R}^n \longrightarrow M$ are local charts of $M$? My doubt is because I don't know if this topology on $M = S^n$ agree with the subspace topology of $S^n$ inherited by the topology of $\mathbb{R}^{n+1}$. If these topologies agree, then this answer my question, but if it is not, then I can not see easily why $S^n$ is compact.

George
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    Note that $S^n$ is a subspace of $\mathbb{R}^{n + 1}$ that is both closed and bounded. By the Heine-Borel theorem, $S^{n}$ is compact. – Mark Saving Aug 31 '21 at 20:05
  • Compactness of $S^n$ comes from knowing that $S^n$ is a closed and bounded subset of $\mathbb R^{n+1}$ (not from knowing that $S^n$ is a closed and bounded subset of itself). – Lee Mosher Aug 31 '21 at 20:05
  • @MarkSaving, I do not know if my question is clear, but my question is if there is a way to prove the compactness of the smooth manifold without see $S^n$ as subset of $\mathbb{R}^{n+1}$? Because I know that the natural topology of a smooth manifold coincides with the topology generated by the metric defined by $d(p,q) := \inf { l(c_{pq}) \ ; \ c_{pq} \ \text{is a piecewise differentiable curve joining p to q}$, but I don't know if the topology of a smooth manifold coincides with the topology inherited by $\mathbb{R}^{n+1}$. – George Aug 31 '21 at 20:13
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    A submanifold is automatically a subspace. So $S^n$ is automatically a subspace of $\mathbb{R}^{n + 1}$ because it's a submanifold. – Mark Saving Aug 31 '21 at 20:17
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    At this point, I am getting confused about what you are asking. Ordinarily, when confronted with a subset of a topological space, one uses the subspace topology. And in the special case of a subset of $\mathbb R^n$, one applies the Heine-Borel theorem, which says that a subset of $\mathbb R^n$ is compact with respect to the subspace topology if and only if the subset is closed and bounded. If you want to disallow all of that, then you'll have to tell us what alternate topology on $S^n$ you are using. Otherwise, your question is rather unclear. – Lee Mosher Aug 31 '21 at 20:31
  • @LeeMosher, I am considering the natural topology of a smooth manifold $M^n$, i.e., $A$ is an open subset of $M$ if $x_{\alpha}^{-1}(A \cap x_{\alpha}(U_{\alpha}))$ is an open set of $\mathbb{R}^n$ for each $\alpha$, where $x_{\alpha}: U_{\alpha} \subset \mathbb{R}^n \longrightarrow M$ are local charts of $M$. My doubt is because I don't know if this topology on $M = S^n$ coincides with the subspace topology. If these topologies coincide, then this answer my question, but if it is not, then I can not see easily why $S^n$ is compact. – George Aug 31 '21 at 20:44
  • @George: A smooth manifold is a topological space with additional structure. How are you defining the topological space $S^n$ if not as the unit sphere in $\mathbb{R}^{n+1}$? – Michael Albanese Aug 31 '21 at 21:01
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    Okay, so now that's a valid question. But I strongly suggest that you edit your post to point out that this latest comment is the real question. – Lee Mosher Aug 31 '21 at 21:04

2 Answers2

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Use stereographic projection, which defines two coordinate charts, one centered at the North Pole and omitting the South Pole, and the other with the poles switched. Let $\overline{B_N}$ be the closure of the unit ball in the coordinate chart centered at the North Pole. It is compact. Let $B_S$ be the complement of $\overline{B_N}$ in the coordinate chart centered at the South Pole. $\overline{B_S}$ is compact. The union of the two compact sets is the sphere. Therefore, the sphere is compact.

More generally, if a manifold is covered by a finite number of coordinate charts $\phi_k: U_k \rightarrow M$, where each $U_k \subset \mathbb{R}^n$ is a bounded open set and each $\phi_k$ extends to a continuous map $\phi_k: \overline{U}_k \rightarrow M$, then $M$ is compact.

Deane
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Answer is combining these 2 results:

1)How to prove a quotient space is again compact and Hausdorff

and

2)The n -disk $D^n$ quotiented by its boundary $S^{n−1}$ gives $S^n$

without using immersion of $S^n\subset \mathbb{R}^{n+1}$

Oğuzhan Kılıç
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