Timeline for $l$-dependence of the group of homologically zero cycles
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Dec 19, 2015 at 16:07 | history | edited | Will Sawin | CC BY-SA 3.0 |
added 196 characters in body
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| Dec 19, 2015 at 16:04 | comment | added | Will Sawin | @MikhailBondarko However over function fields it could be fine as then the map will factor through a usual etale cohomology group and then independence of $\ell$ would follow from the Numerical=Homological standard conjecture - although one should check that the kernel of the map from etale to Galois cohomology is not problematic. | |
| Dec 19, 2015 at 16:04 | comment | added | Will Sawin | @MikhailBondarko Hmm, maybe this does not work. I was thinking of something like the Beilinson-Lichtenbaum conjecture given an isomorphism between etale and motivic cohomology, but it looks like that does not apply to $H^p(X, \mathbb Q_\ell(q))$ for $p>q$ which is what we need as here $p=2q$. | |
| Dec 19, 2015 at 14:59 | comment | added | Mikhail Bondarko | Do you have a reference for a conjecture of this sort (in this case)? | |
| Dec 19, 2015 at 14:50 | comment | added | Will Sawin | @MikhailBondarko I was only thinking of finitely generated base fields. | |
| Dec 19, 2015 at 14:23 | comment | added | Mikhail Bondarko | This conjecture is clearly wrong in general: possibly true if the base field is finitely generated (yet I am not sure; I may be false for trivial reasons in this case also). | |
| Nov 20, 2015 at 10:50 | vote | accept | SashaP | ||
| Nov 19, 2015 at 15:55 | comment | added | Daniel Litt | Oops, didn't see that you'd written number field. You're right of course! | |
| Nov 19, 2015 at 15:23 | comment | added | Will Sawin | @DanielLitt Isn't that only true for $H^2(E_{\overline{K}}, \mathbb Z_\ell(1))$, not $H^2(E_K, \mathbb Z_\ell(1))$? The exact sequence $H^1(E_K, \mathbb G_m) \to H^1(E_K, \mathbb G_m) \to H^1(E_K, \mu_{\ell^n})$ shows that the class of an $\ell$-torsion point is nonvanishing mod $\ell^n$ for each $n$, hence nonvanishing $\ell$-adically. | |
| Nov 19, 2015 at 5:30 | comment | added | Daniel Litt | Just a note: I don't think your elliptic curve example works, since $H^2$ will be torsion-free in that case. The Enriques example does, though. | |
| Nov 19, 2015 at 0:53 | history | answered | Will Sawin | CC BY-SA 3.0 |