Let $X=M \times [0,T]$, where $M$ is a smooth and closed compact Riemannian manifold.

I want to know if: $X$ is smooth compact manifold, and if $\partial X$ is smooth compact manifold?

  1. I am not sure if $X$ is compact. I think it is compact iff it has finite diameter and is geodesically complete. But take two points $a =(m,0)$ and $b=(m,T)$ where $m \in M$. Then the geodesic cannot be extended forever, right? Or have I misunderstood??

Also I guess $X$ is smooth since $M$ is.

  1. Well $\partial X = M \times \{0, T\}$ which again I think is smooth for the same reason. I don't know about compactness.
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    $\begingroup$ The product of two compact topological spaces is compact, right...? And a closed subset of a compact space is compact. Sometimes it's easier to go back to the basics. $\endgroup$ – Najib Idrissi Jul 16 '15 at 10:25

Compactness is a topological property and should not depend on the metric at all. $X$ is compact, because is topologically a product of two compact spaces. Similarly for $\partial X=M\times \{0,T\}$.

  • $\begingroup$ Ok, but doesn't this suggest that $M \times (0,T)$ is not compact? Which is strange. $\endgroup$ – Upin Jul 16 '15 at 10:29
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    $\begingroup$ Yes, but I don't see why is it strange: open intervals are usually not compact.. intuitively, the "upper" and "lower" boundary are missing. $\endgroup$ – Peter Franek Jul 16 '15 at 10:30
  • $\begingroup$ Thanks. Am i right that $M \times \{0\}$ is an embedded submanifold of $M \times [0,T]$? Since it is just the inclusion map. $\endgroup$ – Upin Jul 16 '15 at 13:07
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    $\begingroup$ It is an embedded submanifold, more precisely, it is embedded into the boundary of $X$. Note also that if $M$ had a boundary (which it has not in your setting) than $X$ would be a so-called manifold with corners (the corners being in $\partial M\times \{0,T\}$.) $\endgroup$ – Peter Franek Jul 16 '15 at 13:17

The product $X = M \times [0, T]$ is compact, but it's not a manifold. (It's a manifold-with-boundary, which is a special type of non-manifold.) That's the resolution of the apparent problem with extending geodesics. Even if $M$ is a point, $X \simeq [0, T]$ isn't geodesically complete; this doesn't contradict your claim in 1. because $[0, T]$ isn't a manifold.

  • $\begingroup$ So $X$ is a manifold with boundary, I agree. I assume your definition of "manifold" means closed manifold. So what you're saying is "compact manifold = geodesically + finite diamater" only applies for compact manifold WITHOUT boundary, I think. $\endgroup$ – Upin Jul 16 '15 at 11:17
  • $\begingroup$ A "manifold" is "locally modeled by an open subset of $\mathbf{R}^{n}$ at each point". :) A manifold with boundary equipped with a Riemannian metric (a notion requiring a non-trivial definition, incidentally) is never geodesically complete, essentially because "a geodesic starting at a boundary point and heading away from the interior leaves the universe immediately". $\endgroup$ – Andrew D. Hwang Jul 16 '15 at 11:43

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