I begin by defining the Weil pairing in general (as in Oda's 1969 paper). My question is about an explicit formula for this pairing in the case of an elliptic curve over complex numbers.

Let $\lambda\colon A \rightarrow B$ be an isogeny between abelian schemes over a base scheme $S$ and let $\lambda^*\colon \mathrm{Pic}^0_{B/S} \rightarrow \mathrm{Pic}^0_{A/S}$ be the dual isogeny. Define the Weil pairing $$< , >_{\lambda}\colon\mathrm{Ker}(\lambda) \times \mathrm{Ker}(\lambda^*) \rightarrow \mathbb{G}_m$$ as follows. For a line bundle $\mathscr{L} \in \mathrm{Ker}(\lambda^*)(S)$, choose an isomorphism $$ \alpha\colon \lambda^* \mathscr{L} \xrightarrow{\sim} \mathscr{O}_{A},$$ note that for an $S$-point $x \in \mathrm{Ker}(\lambda)$ one has $\lambda^* = T_x^* \lambda^*$ canonically, where $T_x$ is translation by $x$, and define $<x, \mathscr{L}>_{\lambda}$ to be the element of $\mathbb{G}_m(S)$ that gives rise to the composite isomorphism $$\mathscr{O}_{A} \xrightarrow{\alpha^{-1}} \lambda^* \mathscr{L} = T_x^* \lambda^* \mathscr{L} \xrightarrow{T_x^*\alpha} T_x^* \mathscr{O}_A = \mathscr{O}_A.$$

Now suppose that $S = \mathrm{Spec} (\mathbb{C})$ and $\lambda$ is the multiplication by $n$ isogeny on an elliptic curve. The above pairing then *is* (by definition) the Weil pairing
$$ E[n] \times E[n] \rightarrow (\mathbb{G}_m)_{\mathbb{C}}.$$
Suppose that $E$ is presented via a complex uniformization: $E = \mathbb{C}/(\mathbb{Z} + \mathbb{Z} \cdot (x + iy))$ with $y > 0$. My question is: why does the Weil pairing agree with the pairing
$$ \left(\frac{a}{n}, \frac{b}{n} \cdot (x + iy)\right) \mapsto e^{-\frac{2\pi i}{n} ab}$$
(say, as opposed to its complex conjugate or some other pairing).

P.S. I am not interested in suggestions to use a different "easier" definition of the Weil pairing that works only for elliptic curves over fields (unless, of course, the suggestion is accompanied by a proof of the agreement between the pairings). The point is to understand the correct definition, not to avoid using it.

Abelian varieties. $\endgroup$ – abx Feb 13 '15 at 16:36