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Let $E$ be an elliptic curve over a field $K$, then $G_K:=G_{K^s/K}$ acts on the Tate module of $E$. This is a 2-dim representation of $G_K$. I have heared that we can prove the determinant representation of this representation is the cyclotomic character $\chi$ and this is because the Weil pairing is bilinear, alternating and Galois-invariant.

$\textbf{Questions 1}$: I don't know how to prove this fact, and I even don't know how Weil pairing relates to this fact. Moreover, can we choose a basis of this 2-dim representation such that this representation is $$ \left[ \begin{matrix} 1 & f(g) \\ 0 & \chi(g) \end{matrix} \right] $$

under this basis, where $f(g)$ is some function of $G_K$?

$\textbf{Questions 2}$: Can we prove these facts for an abelian variety ?

$\textbf{Remark}$: If $K$ is a finite extension of $Q_p$, this is very easy in the elliptic curve case because we can use Tate's curve to construct a basis directly, so I want to know how to prove these when $K$ is a general field other than a p-adic field.

Thanks for any answers!

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I'm going to approach your questions in reverse order. For an abelian variety of dimension $g$, it is well known that the determinant representation of the Tate module is $\chi^g$, i.e. the $g$-th power of the cyclotomic character (the argument I'll give below involves matrices so won't be nice to give in the general case but hopefully it will convince you of why it should work).

For question 1, I'm going to instead look at the Galois action on the $l$-torsion points $E[l]$ for some prime $l$ (different to the characteristic of $K$); this is a quotient of the Tate module representation and the arguments will commute with the profinite limits.

Let's fix a basis $P,Q$ for this representation $\rho$. If $\rho$ looks like $\left( \begin{smallmatrix} 1 & \psi \\ 0 &\chi \end{smallmatrix} \right)$ under this basis then Galois must fix the basis element $P$ which happens if and only if $P$ is defined over $K$. This is not always true and hence we can't always put $\rho$ in this form. In fact, if we can write $\rho$ as $\left( \begin{smallmatrix} \psi_1 & \ast \\ 0 &\psi_2 \end{smallmatrix} \right)$ then we have a Galois stable subspace generated by the multiples of $P$ which implies we have an $l$-isogeny (with kernel $\langle P \rangle$) which is also not possible in general.

Now let $e(P,Q)$ denote the Weil pairing. Recall that the Weil pairing is bilinear, alternating, nondegenerate and Galois-equivariant. In particular, alternating implies that $e(T,T)=1$ and $e(S,T)=e(T,S)^{-1}$ for any $S,T \in E[l]$. Let $e(P,Q)=\zeta$, where $\zeta$ is a (necessarily as nondegenerate) primitive $l$-th root of unity.

Let $g \in G_K$ act as $\left( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \right)$ on $E[l]$, i.e. $gP=aP+cQ, gQ=bP+dQ$. Then

\begin{eqnarray*} g(\zeta) &=& g(e(P,Q)) \\ &=& e(gP,gQ) \quad \text{ as Galois equivariant} \\ &=& e(aP+cQ,bP+dQ) \quad \text{ by above} \\ &=& e(aP,bP)e(aP,dQ)e(cQ,bP)e(cQ,dQ) \quad \text{ as bilinear} \\ &=& e(P,P)^{ab}e(P,Q)^{ad}e(Q,P)^{cb}e(Q,Q)^{cd} \quad \text{ bilinearity again} \\ &=& e(P,Q)^{ad}e(Q,P)^{bc} \quad \text{ as alternating} \\ &=& e(P,Q)^{ad}e(P,Q)^{-bc} \quad \text{ as alternating} \\ &=& e(P,Q)^{ad-bc} \\ &=& \zeta^{ad-bc} \end{eqnarray*}

By definition of the cyclotomic character, $g(\zeta)=\zeta^{\chi(g)}$ and hence $\chi(g)=ad-bc$ which is precisely the determinant of $\rho(g)$.

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  • $\begingroup$ Thanks for your wonderful answer! And I think that for passing to inverse limit, we need to construct a basis $\{P_n, Q_n\}$of $Z/l^n Z$-module $E[l^n]$ for every n such that $lP_{n+1}=P_n$, $lQ_{n+1}=Q_n$ and ${e(P_{n+1}, Q_{n+1})}^l=e(P_n, Q_n)$. Does such a basis exist ? $\endgroup$ – Eiang Jul 10 at 12:16
  • $\begingroup$ That's right and is part of the construction. You can actually see this nicely through the complex uniformisation (assuming $K$ has characteristic $0$ so embeds into $\mathbb{C}$). Let $\Lambda=\mathbb{Z}+\mathbb{Z}\tau$ be the corresponding lattice, then the compatible system of points $P_n=\frac{1}{l^n}, Q_n=\frac{\tau}{l^n}$ does exactly what you want with the Weil pairing (see exercise 1.15 in Silverman's Advanced topics) $\endgroup$ – Matt B Jul 11 at 23:46
  • $\begingroup$ That’s nice! What about the $char=p$ case? In this case, we can’t embed $K$ into C. $\endgroup$ – Eiang Jul 12 at 12:40
  • $\begingroup$ And there is also a question. Not every field of $char=0$ can be embedded into $C$. See [this question](math.stackexchange.com/questions/1464736/…) ! $\endgroup$ – Eiang Jul 12 at 12:47
  • $\begingroup$ But your answer is enough for me! Thanks again! $\endgroup$ – Eiang Jul 12 at 12:50

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