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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. I know there is a question here also related to this, I got some idea from that question but also generated one more confusions.

My 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?

Please feel free to point out where I got wrong

Thank you!

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    $\begingroup$ Since $U$ is a chart, $U\subseteq M$. Since $U$ covers $M$, $M\subseteq U$. So $M = U$. Does that fix your objection? $\endgroup$ – Jason DeVito Feb 10 '18 at 21:55
  • $\begingroup$ Oh! I see, so if there is only one covering, the cover must be itself. Thank you so much!@JasonDeVito $\endgroup$ – JacobsonRadical Feb 10 '18 at 22:15
  • $\begingroup$ You could post this answer and I could give you an approval! ^ ^ @JasonDeVito $\endgroup$ – JacobsonRadical Feb 10 '18 at 22:16
  • $\begingroup$ You could post it yourself and accept it - I don't need any more reputation points ;-) $\endgroup$ – Jason DeVito Feb 11 '18 at 0:19
  • $\begingroup$ @JasonDeVito Thank you! $\endgroup$ – JacobsonRadical Feb 11 '18 at 1:41
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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.

Also, I appreciate the help from the comment, thank you so much!

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