Timeline for Why Calabi-Yau manifolds should be complex?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Oct 28, 2017 at 12:44 | comment | added | Xige Yang | Thank you all! I know that the vacuum solutions to Einstein's field equation yield Einstein manifolds, which are basically in \mathbb{R}. Is the idea of Calabi-Yau has something to do with Einstein manifolds? | |
| Oct 27, 2017 at 10:05 | comment | added | José Figueroa-O'Farrill | My answer to another MO question might be of interest: mathoverflow.net/a/43603/394 | |
| Oct 27, 2017 at 10:03 | answer | added | Ben McKay | timeline score: 4 | |
| Oct 27, 2017 at 9:52 | answer | added | diverietti | timeline score: 5 | |
| Oct 27, 2017 at 1:32 | comment | added | Aaron Bergman | To be a little more precise, the physics of supersymmetry requires a covariantly constant spinor. This implies the SU(3) holonomy, which implies the complex structure. | |
| Oct 27, 2017 at 1:15 | history | edited | Suvrit | CC BY-SA 3.0 |
corrected spelling in title; minor etc
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| Oct 27, 2017 at 0:04 | comment | added | JJJ | Since you seem to be looking for physics motivation, the easiest answer is holonomy. If you compactify superstring theory on a manifold with SU(3) holonomy, then a good bit of the supersymmetry is preserved (1/4 of it). Turns out then only manifolds with SU(3) holonomy are Calabi-Yau 3-folds. Also note every complex n-manifold does have an underlying smooth 2n-manifold. | |
| Oct 27, 2017 at 0:02 | comment | added | dhy | Here is my understanding of the story: if you are doing 4-dimensional GR, then indeed the relevant manifold is not a Calabi-Yau but an Einstein Lorentzian manifold (I might not have the terminology quite right.) However if you are doing some form of 10d supergravity/string theory the 6d manifold you are compactifying on gets forced to be a Calabi-Yau manifold for some reason. I really have no understanding of the physics so take this with a large grain of salt. | |
| Oct 26, 2017 at 23:45 | comment | added | paul garrett | Superficially, "algebraic geometry" works best over algebraically-closed fields, and $\mathbb C$ is the obvious choice when the underlying "arithmetical analysis" involves $\mathbb R$. Is this the sort of thing you're asking about? | |
| Oct 26, 2017 at 23:16 | review | First posts | |||
| Oct 26, 2017 at 23:40 | |||||
| Oct 26, 2017 at 23:11 | history | asked | Xige Yang | CC BY-SA 3.0 |