The Weil Pairing

A nice thing about elliptic curves is the wealth of information which is tied up in the isomorphism class of the group. Over the complex numbers, every elliptic curve is \mathbf{C}/\Lambda where \Lambda is a lattice of rank 2 contained in the complex numbers. The Mordell-Weil Theorem tells us that when we restrict to a number field we get a finitely generated abelian group.  The Birch and Swinnerton-Dyer conjecture tells us how important the rank is and today we get a glimpse of what the torsion can tell us. The Weil Pairing is a canonical identification of \bigwedge^2 E[m] with the m-th roots of unity (here we work over a field of characteristic not dividing m).We begin with a lemma concerning rational functions on an elliptic curve.

Lemma: A divisor D = \sum_P n_P P is div(f) for some rational function f if and only if \sum_P n_P=0(i.e.  \deg(D) = 0) and in elliptic curve addition, \sum_P [n_P]P = O (i.e. \sigma([D]) = O where \sigma : Pic^0(E) \stackrel{\sim}{\to}E and [D] denotes the class of D in Pic^0(E).)

The reason is, if f is a rational function, \deg(div(f)) =0 and [div(f)] = [0] so \sigma[0] = O. On the other hand if \deg(D) = 0 then it makes sense to speak of [D]\in Pic^0(E). If \sigma([D]) = O then [D] = [0] so there is some rational function f for which D = div(f).

We need this to say that if Q\in E[m]  there is some rational function f_Q such that div(f_Q) = m(Q - O) = mQ - mO. Consider f_Q\circ [m]. Since f_Q is only zero at Q and will have multiplicity m at each point, the divisor of zeros is m([m]^*Q) and likewise the divisor of poles is m([m]^*O). This is to say that div(f_Q\circ [m]) = m[m]^*(Q-O).

Note that f_Q\circ [m] is constant on the addition of any element of E[m]. Thus if R is any element of E[m] then \frac{f_Q\circ[m]( X +R)}{f_Q\circ [m] (X)} = 1 for any X \in E. We now show the m-th root of f_Q\circ [m] is a rational function.

Consider [m]^*(Q-O) = [m]^*Q - [m]^*O = [m]^*Q - \sum_{P\in E[m]} P. Meanwhile we know that unless [m] is a inseparable map(i.e. when the characteristic of k does not divide m, for which I still need to give a proof) then deg([m]^*D) = m^2 deg(D) and \# E[m] = m^2. Thus if [m]Q' = Q then [m]^{-1}(Q) = \{ Q' \oplus P: P \in E[m]\}(we use \oplus to make clear that this is the chord and tangent addition in E) and so [m]^*Q = \sum_{P\in E[m]} Q' \oplus P. Thus [m]^*(Q-O) = \sum_{P\in E[m]} Q' \oplus P - P .

Now consider for any P \in E[m], the divisor Q' \oplus P - P. Clearly it has degree zero, but in the addition law, it is not O but Q', so we subtract Q' and add O to get something of degree zero which adds up to O. So by our lemma, there is a rational function f such that div(f) = Q'\oplus P - P -Q' + O, so [m]^*(Q-O) = \sum_{P\in E[m]} Q' - O = m^2 Q' - m^2 O. Then since [m]^2Q' = [m][m]Q' = [m]Q = O, there is a rational function g_Q such that div(g_Q^m) = div(f_Q\circ [m]).

Definition: The Weil Pairing e_m(R,Q) := \frac{g_Q(X + R)}{g_Q(X)} for some X\in E.

We have already shown that e_m(R,Q)^m = 1 since f_Q(X + R) = f_Q(X). Thus it is a map E[m]\times E[m] \to \mu_m. Moreover, e_m is

  • alternating (e_m(Q,Q) = 1 so by bilinearity, e_m(R,Q) = e_m(Q,R)^{-1})
  • bilinear(e_m(Q+R,S+T) = e_m(Q,S)e_m(Q,T)e_m(R,S)e_m(R,T))
  • nondegenerate(if e_m(R,Q) = 1 for all R\in E[m] then R=O)
  • Galois invariant(if s \in Gal(\overline{k}/k) then s(e_m(R,Q)) = e_m(s(R),s(Q))
  • Surjective, (there exist Q,R\in E[m] such that e_m(Q,R) is a primitive m-th root of unity)

These properties alone are enough to show that e_m induces an isomorphism of Galois modules E[m] \wedge E[m] \to \mu_m and thus if E[m] is k-rational, then so is \mu_m.

There is also a corresponding pairing for an abelian variety A, where we match A[m] with A^\vee[m] : = \{ L \in A^\vee | L^{\otimes m} \cong \mathcal{O}, the m-torsion line bundles (which, since every abelian variety is smooth could just be thought of in terms of Weil Divisors). The proof that such a pairing exists, instead of relying on the extensive knowledge we have about E[m], must instead come from a homomorphism End(A) \to End(A^\vee), wherein we show that if L \in A^\vee = Pic^0(A) then [m]^*L \cong L^{\otimes m} (corresponding in Divisors to mD linearly equivalent to [m]^*D).

Tune in next time when we talk about Modular Curves and the geometric role that modular forms play.

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3 Responses to The Weil Pairing

  1. X says:

    Hello Jim,

    Is it clear that the Weil Pairing is well-defined? (regardless of the choice of “some X \in E”)

  2. Jim Stankewicz says:

    Oh yes. The point is that g_Q(X + R) = g_Q(X) for any X \in E, and we can see that because they have the same divisor as functions of X.

  3. Pingback: Symposium “Abel Prize 2010” « Disquisitiones Mathematicae

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