Invent. Math. 5 1968 63--84.

Comments

In my thesis, I proved results about pairs of abelian varieties, sometimes assuming that one was a Jacobian (because that was all I need for the applications I had in mind). As Tate pointed out, I never really used that the abelian variety was a Jacobian. So I extracted the results from my theses on abelian varieties and publishing them separately.

Let $A$ and $B$ be abelian varieties over a finite field $k$. Let $(\alpha_i)$ (resp. $(\beta_j)$) be a $\mathbb{Z}$-base for $\mathrm{Hom}(A,B)$ (resp. $\mathrm{Hom}(B,A)$. Then, to quote Tate (Bourbaki Seminar 1968, page 352-15), I prove that the group $\mathrm{Ext}^1(A,B)$ is finite and the product of its order with $\mathrm{det}(\alpha_i\beta_j)$ is calculated by "une belle formule" in terms of the characteristic polynomials of the Frobenius elements $\pi_A$ and $\pi_B$.

The proof of the formula uses the Tate conjecture for divisors on abelian varieties over finite fields (proved by Tate in 1966). Sometime in the 2000s, Niranjan Ramachandran suggested to me that we generalize the formula to motives. This led to a series of articles in which we generalize it to a conjectural formula for motives or complexes of motives (in the sense of Voevodsky completed at $p$), and we proved that the formula is implied by the Tate conjecture. In fact, the conjecture could be true even if the Tate conjecture fails: all it needs is a sufficiently good theory of rational structures on the $l$-adic spaces of Tate classes.

Erratum

p.81 line 2 from bottom. The formula should read $z(g)=\left\vert q^{m_{1}n_{2}} \prod\left( 1-\frac{a_{i}}{b_{j}}\right) \right\vert _{p}$.

pp.78,79 Lemma 4. Because of the change on p81, the lemma should state that% \[ z(g)=\left\vert q^{d(G^{t})\cdot d(H)}\prod_{a_{i}\neq b_{j}}\left( 1-\frac{a_{i}}{b_{j}}\right) \right\vert _{p}=z(g_{1}), \] not $\cdots q^{d(G)\cdot d(H^{t})}\cdots$.

Examples

Recall that the main theorem (Theorem 3) says that, if $A$ and $B$ are abelian varieties over $\mathbb{F}{}_{q}$, then \[ q^{d(A)d(B)}\prod_{a_{i}\neq b_{j}}\left( 1-\frac{a_{i}}{b_{j}}\right) =[\mathrm{Ext}_{k}^{1}(A,B)]|\det(\langle\alpha_{i},\beta_{j}\rangle)|. \]

Example 1.

Assume that the characteristic polynomials $c_{A}(T)$ and $c_{B}(T)$ have no common root. Then $\mathrm{Hom}(A,B)=0$ and $|\det(\langle\alpha _{i},\beta_{j}\rangle)|=1$. For a fixed $i$, \[ \prod\nolimits_{j}\left( 1-\frac{a_{i}}{b_{j}}\right) =\frac{\prod \nolimits_{j}(b_{j}-a_{i})}{\prod\nolimits_{j}b_{j}}=q^{-d(B)}\prod \nolimits_{j}(b_{j}-a_{i}), \] and so \[ \prod\nolimits_{i,j}\left( 1-\frac{a_{i}}{b_{j}}\right) =q^{-d(A)d(B)} \prod\nolimits_{i,j}(b_{j}-a_{i}). \] Hence \[ \lbrack\mathrm{Ext}_{k}^{1}(A,B)]=q^{-d(A)d(B)}\prod\nolimits_{i,j}(b_{j} -a_{i})=q^{-d(A)d(B)}\cdot\mathrm{Resultant}(c_{A},c_{B}). \] For example, if $A$ and $B$ are elliptic curves, then \begin{align*} \lbrack\mathrm{Ext}^{1}(A,B)] & =q^{-1} \begin{vmatrix} 1 & n_{1}-q-1 & q & 0\\ 0 & 1 & n_{2}-q-1 & q\\ 1 & n_{2}-q-1 & q & 0\\ 0 & 1 & n_{2}-q-1 & q \end{vmatrix} \\ & =q^{-1} \begin{vmatrix} 0 & n_{1}-n_{2} & 0 & 0\\ 0 & 0 & n_{1}-n_{2} & 0\\ 1 & n_{2}-q-1 & q & 0\\ 0 & 1 & n_{2}-q-1 & q \end{vmatrix} \\ & =q^{-1} \begin{vmatrix} n_{1}-n_{2} & 0\\ 0 & n_{1}-n_{2} \end{vmatrix} \begin{vmatrix} 1 & 0\\ 0 & q \end{vmatrix} \\ & =(n_{1}-n_{2})^{2}. \end{align*} For higher dimensional abelian varieties, it is a priori obvious that $[\mathrm{Ext}^{1}(A,B)]$ can be expressed in terms of $n_{1}^{(1)},\,n_{1} ^{(2)},\,\ldots$ and $n_{2}^{(1)},\,n_{2}^{(2)},\,\ldots$ where $n_{1} ^{(i)}=[A(\mathbb{F}{}_{q^{i}})]$ and $n_{2}^{(i)}=[B(\mathbb{F}{}_{q^{i}})]$, but not in terms of $n_{1}$ and $n_{2}$ alone.

Example 2.

Suppose that $A=B$ and $c_{A}(T)$ is irreducible over $\mathbb{Q}{}$ (hence without multiple roots). Let $d=d(A)$. For fixed $i$, \[ \prod_{j\neq i}\left( 1-\frac{a_{i}}{a_{j}}\right) =\frac{\prod \nolimits_{j\neq i}(a_{j}-a_{i})}{\prod\nolimits_{j\neq i}a_{j}}\text{.} \] Now \begin{align*} \prod_{i,\,j\,i\neq j}\left( 1-\frac{a_{i}}{a_{j}}\right) & =\frac {\prod\nolimits_{i\neq j}(a_{j}-a_{i})}{(\prod\nolimits_{i\neq j}a_{j} )^{2d-1}}\\ & =q^{-2d^{2}+d}\prod\nolimits_{i\neq j}(a_{j}-a_{i})\\ & =q^{d-d^{2}}\mathrm{disc}(c_{A}(T)). \end{align*} Let $E=\mathrm{End}(A)$ and $F=\mathbb{Z}{}[\pi]\subset E$, where $\pi$ is the Frobenius endomorphism of $A$ over $\mathbb{F}{}_{q}$. An endomorphism $\alpha$ of $A$ has a characteristic polynomial as an element of $E\otimes\mathbb{Q}{}$ over $\mathbb{Q}{}$ and as an endomorphism of $A$. These are both monic of degree $2d$, and hence equal. Let $\alpha_{1},\ldots,\alpha_{r}$ be a basis for $E/\mathbb{Z}{}$. Then \begin{align*} |\det\langle\alpha_{i},\alpha_{j}\rangle| & =\mathrm{disc}(E/\mathbb{Z}{})\\ \mathrm{dis\mathrm{c}}(c_{A}(T)) & =\mathrm{disc}(\mathbb{Z}{}[\pi ]/\mathbb{Z}{}) \end{align*} and so \[ \lbrack\mathrm{Ext}^{1}(A,A)]=q^{d-d^{2}}\frac{\mathrm{disc}(\mathbb{Z}{} [\pi]/\mathbb{Z}{})}{\mathrm{disc}(E/\mathbb{Z}{})}=q^{d-d^{2}}(E\colon \mathbb{Z}{}[\pi])^{2}. \] For example, if $A$ is an elliptic curve, then \[ \lbrack\mathrm{Ext}^{1}(A,A)]=(E\colon\mathbb{Z}{}[\pi])^{2}. \]

Example 3.

$A=B$ and $c_{A}$ is a power of a linear polynomial. In this case, \[ \lbrack\mathrm{Ext}^{1}(A,A)]=\left( \frac{q}{p^{2}}\right) ^{d^{2}}. \] For example, if $A$ is a supersingular elliptic curve, then \[ \lbrack\mathrm{Ext}^{1}(A,A)]=\frac{q}{p^{2}}. \]