## 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.

This entry was posted in Abelian Varieties, AG From the Beginning, Curves, Uncategorized. Bookmark the permalink.

### 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 $X$.