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Maths > Abelian varieties > Polarisations, dual abelian varieties and the Weil pairing

Weil pairings: definition

Posted by Martin Orr on Monday, 29 August 2011 at 17:27

Recall that for an abelian variety A over the complex numbers, H_1(A^\vee, \mathbb{Z}) is dual to H_1(A, \mathbb{Z}) (this is built in to the analytic definition of A^\vee). Since T_\ell A \cong H_1(A, \mathbb{Z}) \otimes_\mathbb{Z} \mathbb{Z}_\ell, this tells us that T_\ell(A^\vee) is dual to T_\ell A (as \mathbb{Z}_\ell-modules). We would like to show that this is true over other fields as well, which we will do by constructing the Weil pairings.

The pairing over \mathbb{C}

Let us first unpack the duality between T_\ell(A^\vee) and T_\ell A over the complex numbers.

Recall that the isomorphism T_\ell A \cong H_1(A, \mathbb{Z}) \otimes_\mathbb{Z} \mathbb{Z}_\ell is obtained as the inverse limit of isomorphisms A[\ell^n] \cong \Lambda_\mathbb{Z} \otimes_\mathbb{Z} \mathbb{Z}/\ell^n\mathbb{Z} where \Lambda_\mathbb{Z} = H_1(A, \mathbb{Z}). So the pairing T_\ell A \times T_\ell (A^\vee) \to \mathbb{Z}_\ell is obtained as the inverse limit of pairings \psi_{\ell^n} : A[\ell^n] \times A^\vee[\ell^n] \to \mathbb{Z}/\ell^n\mathbb{Z}.

The finite-level pairings \psi_m (where m = \ell^n above, but actually could be any integer) are defined as follows:

Let \pi : V \to A(\mathbb{C}) and \pi^\vee : V^\vee \to A^\vee(\mathbb{C}) be the analytic universal covering maps.

Suppose that x \in A[m] and y \in A^\vee[m]. Choose v \in V and f \in V^\vee such that \pi(v) = x and \pi^\vee(f) = y.

Under the isomorphism A[m] \cong \Lambda_\mathbb{Z} \otimes_\mathbb{Z} \mathbb{Z}/m\mathbb{Z} , x corresponds to mv \bmod {m\Lambda_\mathbb{Z}} and y to mf \bmod {m\Lambda_\mathbb{Z}} so we have \psi_m(x, y) = mf(mv) = m^2 f(v). Because x and y are m-torsion, this is an integer and independent (mod m) of the choice of v and f.

The pairing over \mathbb{C} in terms of line bundles

We will now describe this pairing in terms of line bundles, still over the complex numbers.

When we work over other fields, it will be more natural to take the pairing as mapping into the group \mu_m of m-th roots of unity, rather than \mathbb{Z}/m\mathbb{Z}. Over \mathbb{C}, we identify these groups via a \mapsto \exp(2\pi ia/m). So we replace the pairing \psi_m by e_m(x, y) = \exp(2\pi imf(v)).

Recall that under the analytic construction of the isomorphism A^\vee(\mathbb{C}) \cong \operatorname{Pic}^0(A), y corresponds to the line bundle \mathcal{L} = \mathcal{L}(0, \exp(2\pi if(-)) The sections of \mathcal{L} over an open set U are holomorphic functions \theta : \pi^{-1}(U) \to \mathbb{C} satisfying \theta(z + \lambda) = \exp(2\pi if(\lambda)) \, \theta(z) \text{ for all } z \in \pi^{-1}(U), \lambda \in \Lambda_\mathbb{Z}.

So e_m(x, y) is the factor by which sections of \pi^* \mathcal{L} are multiplied when we translate them by mv.

This "translation by mv" happens in the universal cover, so cannot be generalised directly to other fields. But really everything happens within the finite cover V/m\Lambda_\mathbb{Z} \to V/\Lambda_\mathbb{Z}, or equivalently the finite cover [m] : A \to A.

Translating \pi^* \mathcal{L} by mv corresponds to translating [m]^* \mathcal{L} by x, so informally e_m(x, y) is the factor by which sections of [m]^* \mathcal{L} are multiplied when we translate them by x.

General definition of the Weil pairing

We now define a pairing e_m : A[m](k) \times A^\vee[m](k) \to \mu_m(k) for an abelian variety over an arbitrary field k, for m an integer not divisible by \operatorname{char} k.

Let A be an abelian variety over an arbitrary field k and let x \in A[m], y \in A^\vee[m]. Let \mathcal{L} be a line bundle on A corresponding to y under the isomorphism A^\vee[m] \cong \operatorname{Pic}^0(A).

The idea, as in the last paragraph on the complex case, is to look at what happens when we translate [m]^* \mathcal{L} by x. Since mx = 0, we have [m] \circ t_x = [m] and so [m]^* \mathcal{L} = t_x^* [m]^* \mathcal{L}. This is a genuine equality, not just an isomorphism of line bundles.

However, we no longer have a concrete description of sections of \mathcal{L} as in the complex case. So instead of looking at what happens to sections when we translate, we will look at what happens to isomorphisms \mathcal{O} \to [m]^* \mathcal{L}. Of course, before doing this, we need to know that such an isomorphism exists. It does, using my = 0 and applying the following theorem which can be proved by the Theorem of the Cube.

Theorem. Let \mathcal{L} be a line bundle on A homologically equivalent to 0. Then [m]^* \mathcal{L} \cong \mathcal{L}^{\otimes m}.

Choose an isomorphism u : \mathcal{O} \to [m]^*\mathcal{L}. We can translate this to get an isomorphism t_x^* u : t_x^* \mathcal{O} \to t_x^* [m]^* \mathcal{L}. By the equalities \mathcal{O} = t_x^* \mathcal{O} and [m]^* \mathcal{L} = t_x^* [m]^* \mathcal{L}, we can view t_x^* u as an isomorphism \mathcal{O} \to [m]^* \mathcal{L}.

Now the composition \usepackage[matrix, arrow]{xy} \xymatrix{ [m]^* \mathcal{L} \ar[r]^-{(t_x^* u)^{-1}} & \mathcal{O} \ar[r]^-u & [m]^*\mathcal{L} } is an automorphism of [m]^* \mathcal{L}, but the only automorphisms of a line bundle on a complete variety are scalars.

So we define e_m(x, y) to be the scalar u \circ (t_x^* u)^{-1}. A little calculation shows that this defines a bilinear pairing A[m](k) \times A^\vee[m](k) \to k^\times, and we deduce that the image is in \mu_m(k) because A[m](k) is m-torsion.

Another calculation shows that e_{mn}(x, y)^m = e_n(mx, my) and so we can take the inverse limit of the pairings e_{\ell^n} to get a pairing e_\ell : T_\ell A \times T_\ell A^\vee \to \lim_\leftarrow \mu_{\ell^n}.

Any of e_\ell or e_m are called Weil pairings.

Remarks

If you check carefully for k = \mathbb{C}, you will find that the pairing e_m defined in the previous section is the reciprocal of the one defined analytically. This disagreement of signs seems to be the standard (anti-)convention.

The pairing e_\ell gives a map T_\ell(A^\vee) \to (T_\ell A)^\vee. In order to show that these modules are isomorphic, we need to show that the pairing is nondegenerate. This can be done observing that the correspondence \{ \text{line bundles on } A \} \xrightarrow{[m]^*} \{ \text{line bundles on } A \text{ with } A[m]\text{-action} \} specialises to \{ \text{elements of } \operatorname{Pic}^0(A)[m] \} \xrightarrow{[m]^*} \{ A[m]\text{-actions on } \mathcal{O}_A \} and that A[m]-actions on \mathcal{O}_A biject with characters A[m] \to \mu_m.

In the above construction, we could replace [m] by any separable isogeny f : A \to B. The construction then shows that \ker (f^\vee) is dual to \ker(f).

Tags abelian-varieties, alg-geom, hodge, maths, number-theory

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