Timeline for Why Donaldson's Four-Six Conjecture?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Sep 14, 2023 at 1:01 | history | edited | Chris Gerig | CC BY-SA 4.0 |
added 90 characters in body
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| Aug 12, 2016 at 13:35 | history | edited | Chris Gerig | CC BY-SA 3.0 |
added 273 characters in body
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| Nov 15, 2012 at 20:56 | vote | accept | Chris Gerig | ||
| Nov 15, 2012 at 20:55 | history | edited | Chris Gerig | CC BY-SA 3.0 |
edited body
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| Nov 15, 2012 at 12:52 | comment | added | HJRW | @Chris: no it doesn't. begthequestion.info | |
| Nov 15, 2012 at 9:25 | comment | added | Jonny Evans | I think you should definitely edit it to say 4-manifolds: Donaldson never made the conjecture for 2n-manifolds, which is wrong (e.g. there exist diffeomorphic, non-symplectomorphic six manifolds which give eight manifolds which are distinguished by Gromov-Witten invariants). This is because it's much easier for high-dimensional manifolds to be diffeomorphic than for four-manifolds. | |
| Nov 15, 2012 at 7:04 | answer | added | Jonny Evans | timeline score: 14 | |
| Nov 15, 2012 at 6:19 | answer | added | YangMills | timeline score: 7 | |
| Nov 15, 2012 at 1:42 | comment | added | Chris Gerig | But my typo then begs the question, why just 4-manifolds and not $2n$-manifolds? | |
| Nov 15, 2012 at 0:42 | comment | added | Chris Gerig | Oops. It's because I forgot to write "$4-$manifolds" :-) | |
| Nov 15, 2012 at 0:15 | history | edited | ARupinski | CC BY-SA 3.0 |
Added missing $'s for TeX
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| Nov 14, 2012 at 22:37 | comment | added | Qfwfq | And... why is it called "four-six"? (ok, if $X_1$, $X_2$ are symplectic $4$-folds, then $X_i\times S^2$ is a $6$-fold, but the theorem as in the question doesn't seem to impose any dimension restriction on $X_i$) | |
| Nov 14, 2012 at 21:22 | history | asked | Chris Gerig | CC BY-SA 3.0 |