X is a smooth projective variety over a number field k. Write $\bar{X}$ for $X_{\bar{k}}$. Consider the cycle class map on Neron-Severi group of divisors: $cl_l: NS(\bar{X})\hookrightarrow H^2_{et}(\bar{X}, Q_l(1))$. Then Tate's conjecture 1 predicts $cl_l(NS(\bar{X}))\otimes_Q Q_l=H^2_{et}(\bar{X}, Q_l(1)$ . Edit: should be $cl_l(NS(\bar{X}))\otimes_Q Q_l=H^2_{et}(\bar{X}, Q_l(1)$. (See e.g. Tate, "conjectures on algebraic cycles in l-adic cohomology", Motives, Jannsen et, proceeding of symposia vol55.)

Comment for edit: Consider a product of two modular elliptic curves $E_1$ and $E_2$ over Q. Then it was proved in the following cited paper by Rosen and Silverman that their product satisfies both Tate's conjectures. However, we know the $NS(\bar{E_1}\times \bar{E_2})=2, 3 or 4$. On the other hand, by Kunneth formula $H^2(E_1\times E_2(C), C)=C^6$, so by the comparison theorem the algebraic cycles have no hope of generating the whole space in $H^2_{et}(\bar{E_1}\times \bar{E_2}, Q_l(1))$. One will have to look at $H^2_{et}(\bar{E_1}\times \bar{E_2}, Q_l(1))^{G_Q}$ instead.

Moreover, there is a stronger conjecture let's call it Tate's 2:

consider the Hasse-Weil L function defined by the Galois representation of $G_k:=Gal(\bar{k}, k)$ on $H^2_{et}(\bar{X}, Q_l(1))$. For notational convenience we denote the latter as $V_l(1)$. Conjecture 2 says that: $-ord_{s=2}L_2(\bar{X}, s)=dim_{Q_l}Hom_{Gal(k/Q)}(1, V_l(1))$.

Question 1. (Edit: to keep track my original notation I'll leave question 1 and 2 still separated, but actually after editting I only have one question now.)

If X is a smooth projective variety over a finite field then conjecture 2 follows from conjecture 1, however over a number field I don't know if it still follows. In the article by Rosen and Silverman "on the rank of an elliptic surface" MR1626465, they proved conjecture 2 on rational surfaces over a number field. (Here conjecture 1 is known and I have no problem with that.) However I do not understand their proof.

Here is roughly how their argument goes:

Edit: for the speacial case when $Q_l\otimes cl_l(NS(\bar{X}))=H^2_{et}(\bar{X}, Q_l(1))$ before taking the Galois invariants of the latter,

Then $L_2(\bar(X),V_l(1), s)=L(G_k, NS(\bar{X})\otimes_Z Q, s)$. The latter is a classical (rather than motivic) Artin L function, and the order of poles at s=1 equals the Q-dimension of $G_k$-invariant of Neron-Severi space. My confusion is, if this line of argument would work, then isn't it ture in general that conjecture 1 would imply Edit: in this special case we would get conjecture 2 over a number field?

Question 2:

Note here we are only talking about divisors, i.e. 1-cocyles, so we don't need to worry about the difference between numerical equivelance and homological equivalence. Tate has shown $cl_l(NS(\bar{X}))\otimes Q_l\cong Q_lNS(\bar{X})$. (see prop 2.9, 2.10 from Tate's article in the above cited "motive" book. The following argument is mine, which might contain mistakes:) Therefore $dim_Q NS(\bar{X})=dim_{Q_l}V_l(1)$. Moreover, $dim_{Q_l} V_l(1)=dim_{Q_l}V_l$, by the compasison theorem, $dim_{Q_l}V_l=dim_C H^2_{singular}(\bar{X}(C), C)$. Therefore if I know $dim_Q NS(\bar{X})= H^2_{singular}(\bar{X}(C), C)$, then I would know conjecture 1 is true. (e.g. this happends when $H^1(X(C))=0$. Edit: and say $H^2(X, O_X)=0$.) This seems a very strong result, but I've never seen it anywhere, so my feeling is that there has to be something wrong with this.

  • 1
    $\begingroup$ Concerning your second question, if $X$ is a variety over a number field, and $H^{2r}(X(\mathbb{C}),\mathbb{Q})(r)$ is spanned by algebraic classes, then it is indeed obvious that Tate's Conjecture 1 is true for $X$. $\endgroup$ – mephisto Apr 29 '11 at 1:55
  • $\begingroup$ @mephisto: Really? That will be great, and $dim_Q H^{2r}(X(C), Q)(r)=dim_Q H^{2r}(X(C), Q)$, right? The reason I ask question 2 is because whenever I check some reference on "cases where Tate's conjecture is known", I never see such comments on these trivial cases, so I assumed it might be not correct. $\endgroup$ – Ying Zhang Apr 29 '11 at 2:14

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