Timeline for Construct compact submanifold containing non-compact Nash embedded submanifold
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Mar 19 at 9:15 | comment | added | DavideL | I see @DeaneYang, agreed, thanks. I reformulated the question from a topological point of view including boundedness here: math.stackexchange.com/questions/4883160/… | |
| Mar 18 at 18:10 | comment | added | Deane Yang | First, the embedding can be extended to an embedding of a compact manifold only if the embedding is bounded. If the embedding is bounded, the question is whether there exists an extension of the open submanifold to a compact submanifold. That's a purely topological question. If such an extension exists, then the Riemannian metric induced on the compact submanifold is an extension of the Riemannian metric on the open subset. That step is automatic. | |
| Mar 18 at 13:20 | comment | added | DavideL | @MoisheKohan ok, thanks for the tip. I'll think more about it adding the assumption that $\mathring{\mathcal{X}}$ have finite $g$-volume, which should give a bounded $N(\mathring{\mathcal{X}})$; and post on Math Stack Exchange if needed. | |
| Mar 18 at 12:21 | review | Close votes | |||
| Apr 18 at 3:02 | |||||
| Mar 18 at 11:57 | comment | added | Moishe Kohan | Can you think of a Riemannian metric on $(0,1)$ which has infinite total length? Can you induce it by an embedding $f: (0,1)\to {\mathbb R}$? What will be the image? All in all, it is not a research-level question and is more appropriate for Math Stack Exchange. I am voting to close. | |
| Mar 18 at 11:38 | comment | added | DavideL | @MoisheKohan I can't quite think of a 1-dimensional counterexample. 1-dimensional convex sets are precisely open intervals. So let $\mathring{\mathcal{X}}$ be an open interval; no matter how weird the metric $g$ on $\mathring{\mathcal{X}}$ might be, a Nash embedding will give a non-intersecting smooth curve without endpoints $N(\mathring{\mathcal{X}})$ in some Euclidean space, inheriting the Euclidean metric from the ambient space. Connecting the endpoints with another curve is always possible and gives the compact extension, diffeomorphic to $S^1$. Is there any obstruction to this procedure? | |
| Mar 18 at 11:31 | comment | added | DavideL | If the metric didn't matter, one could always extend the initial $n$-dimensional open convex manifold $\mathring{\mathcal{X}}$ to the $(n+1)$-sphere $S^{n+1}$ (identify $\mathring{\mathcal{X}}$ with an open $n$-ball, and identify the ball with an open hemisphere of $S^{n+1}$. ) But nothing would guarantee that the metric $g$ on $\mathring{\mathcal{X}}$ extends to $S^{n+1}$, since $g$ is not induced by an ambient Euclidean metric. | |
| Mar 18 at 11:31 | comment | added | DavideL | @DeaneYang I'd argue that the Riemannian metric plays a role in this way: the open submanifold $N(\mathring{\mathcal{X}})$ that needs be extended to a compact submanifold is the image of the isometric embedding $N$, that does depend on the original metric $g$ on $\mathring{\mathcal{X}}$. | |
| Mar 15 at 17:23 | comment | added | Deane Yang | It seems to me that your question is really a purely topological one. I don't see where the Riemannian metric plays a role. If the open submanifold can be extended to a compact submanifold, then the Riemannian metric induced by the Euclidean metric extends the Riemannian metric of the original open submanifold to the compact submanifold. | |
| Mar 15 at 15:59 | comment | added | Moishe Kohan | Hint: Look for a 1-dimensional counter example. | |
| Mar 15 at 15:37 | history | asked | DavideL | CC BY-SA 4.0 |