Weil pairing and Galois descent

There is an interesting way in which Weil pairing on Abelian Varieties is nothing more than Galois descent for a particular Galois extension. I have not seen this connection used before in the literature but I have also not seen a lot of literature…

Let $A/k$ be an abelian variety over a field $k$ of characteristic $p$ . Suppose $m$ is an integer coprime to $p$ and let $[m]: A \to A$ denote the multiplication by $m$ map with kernel $A[m]$ .

Recall that there is a dual abelian variety $A^\vee$ representing the Picard functor for $A$ . In particular, $A^\vee[m]$ is the group of line bundles $\mathcal L$ on $A$ such that $[m]^*\mathcal L$ is trivial. $A$ and $A^\vee$ are finite abelian groups and the Weil pairing is a perfect pairing of the form:

$\langle-,-\rangle: A[m] \times A^\vee[m] \to \mu_m.$

It’s definition goes as follows: Let $D$ be a divisor corresponding to $\mathcal L$ and let $g(x)$ be a rational function on $A$ such that $\mathrm{div}\ g(x) = [m]^*D$ . Then, for $a \in A[m]$ , we define $\langle a,\mathcal L\rangle = g(x+a)/g(x)$ .

If $t_a$ denotes translation by $a$ , then $g(x+a)$ is the pull back of $g(x)$ along $t_a$ . Since $[m]\circ t_a = [m]$ , $g(x+a)$ and $g(x)$ have the same divisor and so $g(x+a)/g(x)$ is regular everywhere on $A$ and hence constant by properness of $A$ . A little more work shows that this rational functions is in $\mu_m$ . We will see below that this is an immediate conclusion of our alternate viewpoint.

This is the standard picture. However, there is also this alternate way of looking at things:

Consider the etale Galois extension $[m]: A \to A$ . It’s galois group is canonically identified with $A[m]$ . Moreover, $A^\vee[m]$ is precisely the set of line bundles on $A$ trivialized by this etale cover. In other words, it is the set of Galois twists of $\mathcal O_A$ (or even for any line bundle). Since descent is effective for this extension, this will immediately imply that:

$A^\vee[m] \cong H^1(\mathrm{Gal}([m]), \mathcal O_A^\times) = H^1(A([m]), k^\times) = \mathrm{Hom}\ (A[m],k^\times).$

The final equality is because $A[m]$ acts trivially on $\mathcal O_A^\times = k^\times$ . We also see immediately that $\mathrm{Hom}\ (A[m],k^\times) = \mathrm{Hom}\ (A[m],\mu_m)$ since $A[m]$ is m-torison as an Abelian group. Therefore, this gives us a pairing:

$A[m]\times A^\vee[m] \to \mu_m.$

Explicitly, the map goes as follows: Given a line bundle $\mathcal L$ , we pick an isomorphism $g: O_A \to [m]^*\mathcal L$ . Then, the corresponding 1-cocycle for $a \in A[m]$ is defined by $a \to (t_a^*g)^{-1}g \in \mathrm{Aut}(\mathcal O_A)$ . It is easily seen that this is the same explicit construction as in the standard viewpoint.

Conclusion:

I find the Galois descent viewpoint conceptually satisfying. The standard treatments of the Weil pairing can seem arbitrary and it not clear why such a pairing should exist or be useful. On the other hand, $[m]:A \to A$ is a perfectly natural Galois extension connected to $A$ and $\mathcal O_A$ torsors are clearly an interesting thing to consider.

Asvin G

Wir müssen wissen, wir werden wissen.

Weil pairing and Galois descent

Conclusion:

Leave a Reply Cancel reply