Timeline for Is the Hodge Conjecture an $\mathbb{A}^1$-homotopy invariant?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 2, 2017 at 10:30 | vote | accept | CommunityBot | ||
| Jan 30, 2017 at 13:36 | history | edited | Mikhail Bondarko |
edited tags
|
|
| Jan 30, 2017 at 13:36 | answer | added | Mikhail Bondarko | timeline score: 15 | |
| Jan 28, 2017 at 14:56 | answer | added | Donu Arapura | timeline score: 4 | |
| Jan 28, 2017 at 14:36 | comment | added | Donu Arapura | Colin, I believe (1) should be true, but it might take some effort to flesh out since the formalism is pretty sophisticated, cf Berner's answer. | |
| Jan 28, 2017 at 14:13 | answer | added | Joe Berner | timeline score: 5 | |
| Jan 28, 2017 at 1:53 | comment | added | user94803 | Thank you for your reference, Donu. Does being in the same class in the Grothendieck group a kind of weak homotopy equivalence, in the sense that (1) two varieties which are A^1-homotopy equivalence lie in the same class, (2) if two varieties lie in the same class, then they are not necessarily A^1-homotopy equivalent? Is there some condition which guarantees that two varieties lying in the same class in the Grothendieck group is in fact A^1-homotopy equivalent? | |
| Jan 27, 2017 at 17:54 | comment | added | Donu Arapura | Kang and I (arxiv.org/abs/math/0506210) proved that if two smooth projective varieties $X_i$ have the same class in the Grothendieck group of varieties with $\mathbb{A}^1$ inverted, then HC for $X_1$ is equivalent to HC for $X_2$. This is not exactly what you asked but certainly related. | |
| Jan 27, 2017 at 13:52 | history | edited | Francesco Polizzi | CC BY-SA 3.0 |
added 13 characters in body
|
| Jan 27, 2017 at 10:50 | history | asked | user94803 | CC BY-SA 3.0 |