Timeline for Find a circle that intersects the image of $[0,1]$ in a manifold $\mathcal{M}$ at only 1 point
Current License: CC BY-SA 4.0
9 events
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| Feb 11, 2020 at 23:46 | history | edited | Piotr Hajlasz | CC BY-SA 4.0 |
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| Feb 11, 2020 at 23:26 | vote | accept | Rahul Sarkar | ||
| Feb 11, 2020 at 23:18 | comment | added | Rahul Sarkar | @AnthonyQuas My apologies for not being precise with the terminology. The precise version of the question is given in the answer above by Piotr Hajlasz (though it should also add "there exists an r"). I meant radius of the circle in the coordinate system containing $\gamma(0)$. So yes, upstairs in the manifold it is indeed a topological circle. | |
| Feb 11, 2020 at 22:48 | history | edited | Piotr Hajlasz | CC BY-SA 4.0 |
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| Feb 11, 2020 at 22:39 | comment | added | Anthony Quas | @Wowoju: I don't think so: notice that the OP talks about the radius of the circle... | |
| Feb 11, 2020 at 22:38 | comment | added | Wojowu | @AnthonyQuas $\mathcal M$ is just a manifold, so doesn't have an intrinsic notion of a circle (like Riemannian manifolds, say). I suspect OP means topological circles. | |
| Feb 11, 2020 at 22:36 | comment | added | Christian Remling | I think you are assuming that $\gamma$ is a homeomorphism, but the OP doesn't say this. Clearly the claim is false under the assumptions the OP stated. (I'm also wondering why $\dim \mathcal M =2$, though perhaps the use of the word circle suggests this.) | |
| Feb 11, 2020 at 22:35 | comment | added | Anthony Quas | I'm not sure why the last sentence is right "Then the result is obvious". The problem is that your coordinate chart doesn't preserve circles. | |
| Feb 11, 2020 at 22:30 | history | answered | Piotr Hajlasz | CC BY-SA 4.0 |