Timeline for Non-calibrated area-minimising surface
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 16, 2023 at 14:24 | answer | added | Zhenhua Liu | timeline score: 2 | |
| Oct 11, 2020 at 17:50 | vote | accept | Leo Moos | ||
| Oct 11, 2020 at 17:25 | comment | added | Leo Moos | @Robert You're right to point out the inaccurate statement about $\Sigma$, and the typo in the 'more precise' part. I guess what I was trying to say is that a surface $\Sigma'$ in the same homology class and with $\Sigma' \subset U$ would have larger area. | |
| Oct 11, 2020 at 17:19 | history | edited | Leo Moos | CC BY-SA 4.0 |
added 29 characters in body
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| Oct 10, 2020 at 8:47 | answer | added | Robert Bryant | timeline score: 8 | |
| Oct 10, 2020 at 1:19 | comment | added | Robert Bryant | By the way, because you have changed the definition of 'calibrated' (you now only require that it be calibrated in an open neighborhood $U$), it is no longer true that a calibrated submanifold need be area-minimizing in its homology class in $M$. Also, I think that, in your 'more precise' question, you meant to write '$\mathrm{Area}(T+\partial S) \ge \mathrm{Area}(T)$' instead of '$\mathrm{Area}(T+\partial S) \le \mathrm{Area}(T)$', didn't you? | |
| Oct 9, 2020 at 21:18 | comment | added | Robert Bryant | $\mathbb{RP}^2$ obviously can't be calibrated because it's not orientable, so maybe one should require orientability in the question. But it can still be homologically minimizing: Let $M^3$ be the quotient of the standard product $S^2\times \mathbb{R}$ by the involution $(u,t)\mapsto (-u,-t)$. The image of $S^2\times\{0\}$ in $M$ is an $\mathbb{RP}^2$ that is minimizing in its homology class: Anything it its homology class must meet all the images of the lines ${u_0}\times\mathbb{R}$, and the projection $M\to\mathbb{RP}^2$ given by $[u,t]\mapsto [u]$ is clearly area decreasing on any surface. | |
| Oct 9, 2020 at 18:16 | comment | added | Otis Chodosh | I might be mistaken but I think you should get some milage out of $RP^2\subset RP^3$. More precisely, I think it should be true that if it were calibrated, then so would the double cover $S^2 \to RP^3$. But this is not even stable. | |
| Oct 9, 2020 at 16:01 | comment | added | Leo Moos | @Ben I'm aware of this result, but I believe it's slightly different from what I am asking here. If I am not mistaken it says something like "if $\Sigma$ is a (regular) minimal surface, then for every point $p \in \Sigma$ there is $r > 0$ so that $\Sigma \cap B_r(p)$ is calibrated." However the local calibration forms cannot be patched to a global $n$-form. | |
| Oct 9, 2020 at 15:28 | comment | added | Ben McKay | I think all minimal surfaces are locally calibrated, probably a result of Robert Bryant. | |
| Oct 9, 2020 at 14:24 | history | asked | Leo Moos | CC BY-SA 4.0 |