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I am trying to understand how to do the following two questions concerning compact manifolds:

Show that$\ M\ $is a compact manifold in $\mathbb{R}^{n},\ $then $\partial\ M\ $is also compact; if also $\ M\ $is $\ n\ $-dimensional, then $\partial\ M=\ $bdry$\ M\ $

Show that a compact manifold cannot be represented by a (single) parametric equation.

My confusion is what it means for a manifold to be compact. In some books having to do with advanced calculus or theory of manifolds, it states that the notion of compactness when describing manifolds is distinct from the topological notion of compactness. I also come across questions where it asks to prove properties about compact manifolds without boundary. This just makes it more confusing. It is like saying a closed interval, circle or sphere has no boundary points or an empty boundary, but yet it is closed and bounded. On Wikipedia, compact manifold is discussed in the topic of closed manifolds. Again, I am encountering topological notions associated with compactness. But textbooks says otherwise.

Can someone please help me with some clarifications over my confusions please. Thank you in advance

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  • $\begingroup$ I observe that the only one-dimensional compact manifold, $S^1$, satisfies the equation $x^2 + y^2 = 1$ in $\mathbb{R}^2$. Whether that is a single parametric equation depends on how you expect to represent points in $\mathbb{R}^2$. For instance $$\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}\cos\theta\\ \sin\theta\end{pmatrix}, \theta\in[0,2\pi] \text{.}$$ $\endgroup$ – Eric Towers Apr 6 '18 at 0:15
  • $\begingroup$ @Sou, I am not sure i understand your question? $\endgroup$ – Seth Mai Apr 6 '18 at 0:17
  • $\begingroup$ #Eric Towers, I got the questions from a text titled: Analysis in Vector Spaces. I am not sure if it could meant equivalently that a k-manifold can not be represented by a single chart $\endgroup$ – Seth Mai Apr 6 '18 at 0:24
  • $\begingroup$ @Sou, can you explain briefly what it means then for a compact manifold to not have a boundary. It just seem really weird the way the word "compact" is used in this context. Thanks $\endgroup$ – Seth Mai Apr 6 '18 at 0:54
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    $\begingroup$ It just means that for every $p\in M$ there are no charts $(U,\varphi)$ contain $p$ such that $\varphi(p) \in \partial \mathbb{H}^n$. $\endgroup$ – Sou Apr 6 '18 at 1:01
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We know that $M$ is compact manifold in $\mathbb{R}^n$ and the boundary (of any manifold with boundary) $\partial M$ is closed in $M$. Since every closed subset of a compact space is compact, then $\partial M$ is compact. For the problem that any compact manifold cannot be represent as single parametric equation, just note that if we can, then $M$ must be homeomorphic to an open subset of $U \subset \mathbb{R}^{\text{dim }M}$. That is, there exists homeomorphism $\varphi : M\to U = \varphi(M)\subset \mathbb{R}^{\text{dim }M}$. This implies that $U$ is compact (closed and bounded) and also open. Since $U \subset \mathbb{R}^{\text{dim }M}$ is open and closed and $\mathbb{R}^{\text{dim }M}$ connected, this means $U=\mathbb{R}^{\text{dim }M}$. But this is impossible since $U=\varphi(M)$ is compact by continuity.

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  • $\begingroup$ Thank you for your response. My apologizes for replying back late. I am typing out the definition of boundary of M from textbook using LaTex and I still have not gotten use to typing it fast since I am still so new at it. Anyways, from Andrew Browder's text: We sau that$\ p\ $lies on the boundary of M, and write$\ p\in \partial M, \ $if$\ p \in\ M\ $and there exists a local coordinate $\alpha \ $at$\ p\ $such that$\ V_{\alpha} \subset \mathbb{R}_{+}^{k}, \ $and$\ p\ $=$\ \alpha(u)\ $ with $u$=$\ (u_1,...u_{k-1}, 0) \ $. $\endgroup$ – Seth Mai Apr 6 '18 at 0:50
  • $\begingroup$ thank you for the reference. I am guessing all the standard quadric surfaces that we encounter in a typical Analytic Geometry/vector calculus text can be consider closed manifold. And when they mean the boundary of those surfaces, they meant the topological boundary. Is that correct? $\endgroup$ – Seth Mai Apr 6 '18 at 0:59
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    $\begingroup$ 1)If they are level set the it must be closed 2) Yes $\endgroup$ – Sou Apr 6 '18 at 1:08
  • $\begingroup$ thank you for the examples. :) $\endgroup$ – Seth Mai Apr 6 '18 at 1:11

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