Compact manifold cannot be represented by (single) parametric equation

[elias001]

TL;DR

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

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.

I asked online about two portion questions exercise, and I would like to know if the solutions displayed below is correct?

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.

Thank you in advance

 

[dextercioby]

Sorry, why would $U$ need to be open?

 

[elias001]

@dextercioby I am not sure if the entire solution is correct. I am not sure if $U$ need to be open myself.

 

[mathwonk]

actually the parametric mapping t-->e^it represents the compact manifold S^1 rather well.

 

[weirdoguy]

Sorry, why would U need to be open?

All definitions of parametrization I were taught were having an open subset as a domain. I checked lecture notes of my superivsor (link to pdf, but it's in polish, so beware...) and it says on page 9 that "parametrization has to be an injection..." and something more about differentiability, but I guess this thread is about topological manifolds only, so this doesn't matter.

 

[elias001]

@weirdoguy, @mathwonk, @dextercioby I did more googling, and this is what I found, someone asked online "I am learning about differentiable manifold, and I got a question asking to give a reason why I need at least two charts to cover a compact manifold." His idea is:

If $M$ is a compact manifold covered by a single chart $(U,\phi)$, then by the definition (the locally Euclidean), $U$ is homeomorphic to the open subset $\phi(U) \subset R^{n}$ for some $n$.

But the only connected open compact subset of $R^{n}$ is empty set for $n \geq 1$, if $n=0$ we also have $R^{0}$ itself. Then, since $\phi(U)$ is not empty, it cannot be compact.

This idea seems right in the first place, but afterwards a confusion arose. Yes, I prove that $\phi(U)$ is not compact, but it does not contradict the homeomorphism, since $M$ is compact manifold, but $U$ is not necessarily compact. Or a compact manifold is naturally covered by a compact covering? or the covering itself is naturally compact?

After much discussion with others online, he got the key insight:

Since $U$ is a chart, $U\subset M$. Since $U$ covers $M$, $M\subset U$. So $M=U$

he gave the answer as follows:

Since $U$ is a chart, then $U \subset M$, but since $U$ covers $M$, $M \subset U$. Thus, $M=U$.

Thus the chart now is actually $(M,\phi)$.

By the definition of local Euclidean, $M$ is homeomorphic to an open subset $\phi(M) \subset R^{n}$.

But the only connected open compact subsets of $R^{n}$ are empty set and $R^{0}$ if $n=0$, but $\phi(M)$ is not empty, thus $\phi(M)$ is not compact, which contradicts the homeomorphism.