Martin's Blog

Weil pairings: the skew-symmetric pairing

Posted by Martin Orr on Tuesday, 06 September 2011 at 13:52

Last time, we defined a pairing  e_\ell : T_\ell A \times T_\ell (A^\vee) \to \lim_\leftarrow \mu_{\ell^n}. By composing this with a polarisation, we get a pairing of T_\ell A with itself. This pairing is symplectic; the proof of this will occupy most of the post.

We will also see that the action of the Galois group on this pairing is given by the (inverse of the) cyclotomic character, as I promised a long time ago (in the comments). This tells us that the image of the \ell-adic Galois representation of A is contained in \operatorname{GSp}_{2g}(\mathbb{Q}_\ell). This is the end of my series on Mumford-Tate groups and \ell-adic representations attached to abelian varieties.

Definition of the pairing

Let \phi : A \to A^\vee be a polarisation. We define e_m^\phi : A[m] \times A[m] \to \mu_m by  e_m^\phi(x, y) = e_m(x, \phi(y)), where e_m : A[m] \times A^\vee[m] \to \mu_m is the Weil pairing defined last time.

Taking the inverse limit over powers of \ell, we get the \ell-adic pairing  e_\ell^\phi : T_\ell A \times T_\ell A \to \lim_\leftarrow \mu_{\ell^n}.

The theta group

We would like to show that the pairings e_m^\phi are symplectic. The elegant way to do this is by introducing the theta group of a line bundle.

Let \mathcal{L} be a line bundle on A. We define the theta group of \mathcal{L} to be  \mathcal{G}(\mathcal{L}) = \{ (x, f) | x \in A(k) \text{ and } f \text{ is an isomorphism } \mathcal{L} \to t_x^* \mathcal{L} \}. This is a group under the operation  (x, f).(y, g) = (x+y, t_y^* f \circ g). The inverse is given by  (x, f)^{-1} = (-x, t_{-x}^* f^{-1}). (The formula for \mathcal{G}(\mathcal{L}) is functorial in k, so really this is not just a group but the functor of points of a group scheme.)

Projection onto the first factor gives a surjection \pi : \mathcal{G}(\mathcal{L}) \to K(\mathcal{L}) where K(\mathcal{L}) = \{ x \in A(k) | \mathcal{L} \cong t_x^* \mathcal{L} \}. We are mainly interested in non-degenerate line bundles i.e. those for which K(\mathcal{L}) is finite; we have already proved that ample line bundles are non-degenerate.

The kernel of \pi is the group of automorphisms of \mathcal{L}, which is the scalars. So we have a short exact sequence of group schemes  1 \to \mathbb{G}_m \to \mathcal{G}(\mathcal{L}) \xrightarrow{\pi} K(\mathcal{L}) \to 1.

This short exact sequence has the following two properties:

  1. \mathbb{G}_m is contained in the centre of \mathcal{G}(\mathcal{L}).
  2. K(\mathcal{L}) is commutative.

These two properties imply that the commutator [x, y] = xyx^{-1}y^{-1} of any x, y \in \mathcal{G}(\mathcal{L}) lies in \mathbb{G}_m, and that [x, y] depends only on \pi(x) and \pi(y). So the commutator induces a function e^{\mathcal{L}} : K(\mathcal{L}) \times K(\mathcal{L}) \to \mathbb{G}_m, and a calculation shows that this is a bihomomorphism.

An explicit calculation of the commutator gives  e^{\mathcal{L}}(x, y) = f^{-1} \circ t_x^* g^{-1} \circ t_y^* f \circ g, where (x, f) and (y, g) are in \mathcal{G}(\mathcal{L}).

The commutator pairing is obviously symplectic. So in order to show that e_m^\phi is symplectic, it will suffice to relate it to e^\mathcal{L}.

Theorem. Let \phi = \phi_\mathcal{L} be a polarisation. If x, y \in A[m] and z \in A[m^2] such that y = mz, then  e_m^{\phi}(x, y) = e^{[m]^* \mathcal{L}}(x, z).

Proof. We must first check that x, z \in K([m]^* \mathcal{L}) so that e^{[m]^* \mathcal{L}}(x, z) is defined. Since [m] \circ t_z = t_y \circ [m], we have  [m]^*\mathcal{L}^{-1} \otimes t_z^* [m]^* \mathcal{L}
   \,\cong\, [m]^* \mathcal{L}^{-1} \otimes [m]^* t_y^* \mathcal{L}
   \,\cong\, [m]^* \phi(y)
   \,\cong\, \phi(y)^m
   \,\cong\, \mathcal{O}. So z \in K([m]^* \mathcal{L}). Likewise [m] = [m] \circ t_x so that x \in K([m]^* \mathcal{L}).

Choose an isomorphism g : [m]^* \mathcal{L} \to t_z^* [m]^* \mathcal{L}. Thanks to the canonical isomorphisms  \operatorname{Hom}([m]^* \mathcal{L}, t_z^* [m]^* \mathcal{L})
   \,\cong\, [m]^* \mathcal{L}^{-1} \otimes t_z^* [m]^* \mathcal{L}
   \,\cong\, [m]^* \phi(y)
   \,\cong\, \operatorname{Hom}(\mathcal{O}, [m]^* \phi(y)), we can also interpret g as an isomorphism \mathcal{O} \to [m]^* \phi(y). Hence, by the definition of the Weil pairing, we have  e_m^\phi(x, y) = t_x^* g^{-1} \circ g.

Since the line bundles [m]^* \mathcal{L} and t_x^* [m]^* \mathcal{L} are not just isomorphic but equal, we have (x, \mathrm{id}) \in \mathcal{G}(\mathcal{L}).

Substituting (x, \mathrm{id}) and (z, g) in the commutator formula gives  e^{[m]^* \mathcal{L}}(x, z) = t_x^* g^{-1} \circ g.

The Galois action

All the pairings we have defined are compatible with the Galois action in that if \sigma \in \operatorname{Gal}(\bar{k}/k), then  e_m({^\sigma x}, {^\sigma y}) = {^\sigma e_m(x, y)}.

Now \lim\limits_\leftarrow \mu_{\ell^n} is a free \mathbb{Z}_\ell-module of rank 1, traditionally written as \mathbb{Z}_\ell(1). The Galois action on this module is given by a character \chi_\ell : \operatorname{Gal}(\bar{k}/k) \to \mathbb{Z}_\ell^\times called the \ell-adic cyclotomic character. Concretely  {^\sigma \zeta} = \zeta^{\chi_\ell(\sigma) \bmod \ell^n} \text{ for } \zeta \in \mu_{\ell^n}(\bar{k}).

Meanwhile the natural action of \operatorname{Gal}(\bar{k}/k) on the space \bigwedge^2 (T_\ell A)^\vee of symplectic forms on T_\ell A is given by  ({^\sigma e})(x, y) = e({^{\sigma^{-1}} x}, {^{\sigma^{-1}} y}). So we see that  ({^\sigma e})(x, y) = \chi_\ell(\sigma)^{-1} e_m(x, y).

The representation of \operatorname{Gal}(\bar{k}/k) on T_\ell A \cong \mathbb{Z}_\ell^{2g} preserves the non-degenerate symplectic pairing e_\ell up to a scalar, so the image of this representation is contained in \operatorname{GSp}_{2g}(\mathbb{Q}_\ell). We saw long ago that the Mumford-Tate group is contained in \operatorname{GSp}_{2g,\mathbb{Q}}, so this was expected.

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

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