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Let $k$ be a finite field, say with $q=p^a$ elements. Honda-Tate theory states that there is a bijection between isogeny classes of simple abelian varieties over $k$ and $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$-conjugacy classes of algebraic integers in $\mathbb{C}$ all of whose conjugate have absolute value $p^{a/2}$ (these are the so-called "Weil numbers"). For a beautiful survey, see Tate's Bourbaki seminar .

Given a simple abelian variety $A/k$, the associated Weil number $\pi_A$ (or rather its conjugacy class) corresponds to the Frobenius endomorphism, view as an element of $\mathrm{End}_k(A)$. Let now $C$ be a smooth, projective, irreducible curve over $k$ and let $\mathcal{A}/C$ be an abelian scheme, therefore a family of abelian varieties parametrized by points of $C$. There is a Frobenius operator acting on $\mathcal{A}$ giving the Frobenius operator on each fiber $A_v$ for all $v\in C$, and these fibers are abelian varieties over finite fields for all closed $v$. My question is how Honda-Tate behaves in families, so if we can find a bijection between isogeny types of $\mathcal{A}/C$ and "polynomials" in $\mathcal{O}_C[T]$ (or may be in $\Gamma(C,\mathcal{O}_C)[T]$?) whose specialization at every closed point has roots that are Weil numbers.

EDIT: As Piotr Achinger observes, it seems reasonable that the coefficients be in characteristic $0$. I do not know if hoping the coefficients to live in the Witt vectors of $\mathcal{O}_C$ is enough for giving some sense to "a family parametrized by $C$.

I am tempted to think the answer should be "yes", being so fiber-wise. On the other hand, if we pick a "random" collection of Weil polynomials building an element of $\mathcal{O}_C[T]$, I would be surprised if we can build an abelian scheme over $C$ having the "right" fibers (because, for instance, abelian schemes must have good reduction everywhere, and it seems too strong a condition to be simply controlled by a "nice" collection, if what I write ever makes any sense).

If the answer to my question is "yes" (or if it can be made to be "yes" after some modification of my question...), is this "polynomial" in two variables related to the Hasse-Weil function of the abelian scheme? After all, they are both constructed by looking at Frobenius acting on Tate modules, so I would expect a connection.

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I think your "polynomial" should at least have characteristic zero coefficients, while $\mathcal{O}_C$ has characteristic $p$. But maybe there is such a polynomial in $W(\mathcal{O}_C)[T]$, over the Witt vectors of $\mathcal{O}_C$? –  Piotr Achinger Sep 27 '12 at 17:37
    
@ Piotr: Oh yes, you're perfectly right! I'll correct that. –  Filippo Alberto Edoardo Sep 27 '12 at 17:55
    
I think the right way to think about this is to take the trivial $l$-adic sheaf $\mathbb{Q}_l$ on $A$ and compute $G:=R^1\pi_* \mathbb{Q}_l$ -- a constructible $l$-adic sheaf on $C$. Then at every point $x$ of $C$ the action of the Frobenius on $G_x$ has characteristic polynomial whose roots are the Weil numbers corresponding to the fiber at $x$. –  Piotr Achinger Sep 27 '12 at 18:26
    
I certainly agree, you are globalising the Tate modules. But Honda-Tate tells you more: if two varieties have the same Weil numbers then they are isogenous; and if you pick a Weil number you'll find a variety with that number as Frobenius. How do you connect this bijection with your global interpretation of the Tate module? –  Filippo Alberto Edoardo Sep 27 '12 at 18:30
    
One certainly needs that the local monodromy of the sheaf is integral, and that it is symplectically self-dual to $Q_l(-1)$, to lift it to an abelian scheme. To get a lift one wants some kind of description of the moduli stack of abelian varieties of genus $g$ as a classifying space, so as to establish a bijection between lisse sheaves / fundamental group representations and abelian schemes / maps to the moduli stack. –  Will Sawin Sep 27 '12 at 21:27

1 Answer 1

up vote 5 down vote accepted

I think your intuition that Weil numbers generalize to functions is wrong, or rather only partially correct. A Weil "number" is in fact a sheaf on $\operatorname{Spec}\mathbb F_p$: The sheaf corresponding to the Galois representation that sends $Frob_p$ to a matrix whose characteristic polynomial is the minimal polynomial of $\pi_A$.

Lisse and constructible sheaves, or rather their derived categories / Grothendieck groups, are often the appropriate scheme-theoretic notion of function. One has addition and multiplication of functions, and even integration in the form of higher pushforward maps. Thus scheme-theoretic analogues of convolution, Fourier analysis, etc. often exist.

I will thus discuss the sheaf version of the problem. I would guess that the Witt vector version of the problem can be solved by thinking of the sheaf of Tate modules using crystalline cohomology, but as I don't know anything about crystalline cohomology you should take this with a grain of salt.

One can split the sheaf version of the problem into two parts. First, what are necessary and sufficient conditions on a constructible $l$-adic sheaf on $C$ for it to be the sheaf of Tate modules of an abelian scheme over $C$? Second, if two abelian varieties $A_1$ and $A_2$ have the same sheaf of Tate modules, are they isogenous?

I don't have much to say about the first part, but the second part is a special case of the Tate conjecture. An isomorphism between $R^1 \pi_{1*} \mathbb Q_l$ and $R^1 \pi_{2*} \mathbb Q_l$ produces a Galois-invariant cohomology class in $H^0(C,R^1 \pi_{1*} \mathbb Q_l ^\vee \otimes R^1 \pi_{2*} \mathbb Q_l)$, which is a quotient of $H^2( A_1^\vee \times _C A_2, \mathbb Q_l)$. If the Tate conjecture is true, then that cohomology class is a $\mathbb Q_l$-linear combination of classes induced by algebraic cycles. At least one of those cycles must correspond to a line bundle on $A_1^{\vee} \times_C A_2$ that induces a nontrivial morphism $A_2 \to A_1$. For $A_1$ and $A_2$ simple abelian schemes this must be an isogeny. It is not too hard to see that this means we can also get an isogeny for products of simple abelian schemes.

I have no idea whether this special case of the Tate conjecture has a solution. A similar problem that naively seems about as hard, the Tate conjecture for $H^1 (R^1 \pi_* \mathbb Q_l)$ of an elliptic surface, is the function field BSD conjecture.

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Will, thanks for your very insightful answer. A part from the "existence" issue, which as you mention must depend upon some $\pi_1$ analysis, can you say a bit more about how Tate implies the existence of algebraic cycles? When you say that at least one of those must correspond to a line bundle, are you using that $H^2$ subjects onto $H^0$ (which is kind of Leray, right?)? Finally, how do you connect this to BSD - Expecially because it is known (up to finiteness of Sha), by Kato-Trihan? Thanks again! –  Filippo Alberto Edoardo Sep 27 '12 at 23:57
    
I mean, Tate directly states that algebraic cycles exist whose classes span the Galois-invariant part of $H^{2n}(X,\mathbb Q_l)(-n)$. I might need some kind of semisimplicity assumption to show that the Galois invariant part of $H^0(C,R^1 \pi_{1*} \mathbb Q_l^\vee \otimes R^1 pi_2_* \mathbb Q_l)$ lifts to $H^2(A_1^\vee \otimes_C A_2, \mathbb Q_l)$. The surjection from the one to the other is indeed Leray, plus Kunneth, plus the fact that the Tate module of $A_1^\vee$ is dual to the Tate module of $A_1$. There's a Tate twist somewhere in there that's not important. –  Will Sawin Sep 28 '12 at 1:33
    
BSD comes from the fact that the $L$-function can be written in terms of the characteristic polynomial of Frobenius acting on $H^1(R^1 \pi_* \mathbb Q_l)$. The order of vanishing at $s=1$ comes from the dimension of the eigenspace of Frobenius with eigenvalue $p$, i.e., fixed up to a Tate twist. This part of the cohomology comes from algebraic cycles by Tate. Algebraic cycles are the zero section + fibral divisors + the rank, roughly. The class of a fiber lives in $H^2(R^0 \pi_* \mathbb Q_l)$ and the zero section and the other fibral divisors live in $H^0(R^2 \pi^* \mathbb Q_l)$. –  Will Sawin Sep 28 '12 at 1:41
    
so the algebraic cycles that live in $H^1 (R^1 \pi_* \mathbb Q_l)$ are exactly the rank. Tate conjecture says that the dimension of this space of algebraic cycles is the dimension of the eigenspace, so the rank is the dimension of the eigenspace. –  Will Sawin Sep 28 '12 at 1:42
    
Great, I see your point. This is exactly what I was looking for. –  Filippo Alberto Edoardo Sep 28 '12 at 2:14

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