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Consider the class map $$cl:CH^i(X)\to H^{2i}_{cont}(X,\mathbb{Z}_l(i))$$ where the RHS is the continuous etale cohomology(defined by Jannsen in his paper "Continuous etale cohomology"). In this paper he mentiones that the kernel $CH^i(X)_l^0$ might depend on $l$.

What is known about this issue? Is there an example of $X$ and two primes $l_1,l_2$ such that $CH^i(X)_{l_1}^0\neq CH^i(X)_{l_2}^0$?

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the way it's stated, with integral cohomology, it certainly can depend on $\ell$. This is because of the occurrence of torsion classes Clearly a $p$-torsion class is sent to $0$ by this map for $\ell\neq p$, so it is sufficient to find an $\ell$-torsion class that is not sent to zero under this map.

But the Kummer exact sequence shows that for $i=1$, the kernel of the map is $\ell$-divisible. Hence any non-divisible torsion class, like a torsion point on an elliptic curve over a number field, or the canonical bundle of an Enriques surface, will do.

However with $\mathbb Q_\ell$ it may still be independent of $\ell$, depending on the base field. For algebraically closed fields it should be independent, because of the standard conjecture D which says the kernel would just be the group of cycles numerically equivalent to $0$. For arbitrary fields there are counteredamples as Mikhail Bondarko pointed out. For finitely generated fields I'm not sure.

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  • $\begingroup$ 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. $\endgroup$
    – Daniel Litt
    Commented Nov 19, 2015 at 5:30
  • $\begingroup$ @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. $\endgroup$
    – Will Sawin
    Commented Nov 19, 2015 at 15:23
  • $\begingroup$ Oops, didn't see that you'd written number field. You're right of course! $\endgroup$
    – Daniel Litt
    Commented Nov 19, 2015 at 15:55
  • $\begingroup$ 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). $\endgroup$
    – Mikhail Bondarko
    Commented Dec 19, 2015 at 14:23
  • $\begingroup$ @MikhailBondarko I was only thinking of finitely generated base fields. $\endgroup$
    – Will Sawin
    Commented Dec 19, 2015 at 14:50

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