TITLE: Deligne Pairing v.s. Weil Pairing on a Family of curves
QUESTION [7 upvotes]: We have the Deligne Pairing on a family of curve $\pi:X\to S$ by using
$$\langle L,M\rangle_{\mathrm{Pic}^0(X/S)}=\det R\pi_*(L\otimes M) \otimes (\det R\pi_*L)^{-1}\otimes (\det R\pi_*M)^{-1} \otimes \det R\pi_*\mathcal{O}_X $$
where $\det R\pi_*E:=\det R^0\pi_* E \otimes (\det R^1\pi_* E)^{-1} $ is the determinant of cohomology.
Using the fact that the Deligne's pairing is multiplicative: $\langle L_1\otimes L_2,M\rangle = \langle L_1,M\rangle\otimes \langle L_2,M\rangle$ as well as $\langle L,M_1\otimes M_2\rangle = \langle L,M_1\rangle\otimes \langle L,M_2\rangle$, for any $n$-torsion of $\mathrm{Pic}^0(X/S)$ (denoted by  $\mathrm{Pic}^0(X/S)[n]$) we can have two different trivialization.
$\langle L,M\rangle^{\otimes n}=\langle L^{\otimes n},M\rangle \cong \mathcal{O}_S$ or $\langle L,M\rangle^{\otimes n}=\langle L,M^{\otimes n}\rangle \cong \mathcal{O}_S$.
Here's the question:
I've seen here that the "difference of these" will be the Weil pairing, but I have a hard time proving that it actually is. I think this should be a standard fact, but I didn't find it anywhere. Any comment/reference suggestion is welcomed!

REPLY [7 votes]: Let $C$ be a smooth projective curve (say over $\mathbf{C}$ for simplicity). 
 Given $L,M \in \mathrm{Pic}^0(C)[n]$, recall that the Weil pairing $e_n(L,M)$ is defined as follows. First of all, for any $f \in \mathbf{C}(C)$ and any divisor 
$$D = \sum_{p \in C}n_pp$$ 
on $C$ with support disjoint from $\mathrm{div}(f)$, define
$$f(D) = \prod_{p \in C} f(p)^{n_p}.$$
Let $l$ and $m$ be nonzero meromorphic sections of $L$ and $M$ and let $D_l = \mathrm{div}(l)$ and $D_m = \mathrm{div}(m)$ with disjoint supports. Then the Weil pairing of $L$ and $M$ is
$$e_n(L,M) := \frac{l^n(D_m)}{m^n(D_l)} \in \mu_n.$$
The well-definedness of $e_n(L,M)$ is based on Weil's reciprocity law.
If $\pi: X \to S$ is a family of curves over a connected base and 
$$\mathcal{L}, \mathcal{M} \in \mathrm{Pic}(X/S)^0[n],$$
then  $e_n( \mathcal{L}_{|X_s}, \mathcal{M}_{|X_s})$ does not depend on $s \in S$ and will be denoted by $e_n( \mathcal{L}, \mathcal{M})$.
In Geometry of algebraic curves Vol.2 p. 366-379, there is an equivalent definition of Deligne's pairing $\langle L,M\rangle$ as follows. Let $V$ be the free vector space generated by the symbols $(l,m)$ where $l$ and $m$ run through all nonzero meromorphic sections of $L$ and $M$ with disjoint supports. If $\sim$ is the equivalence relation on $V$ generated by
$$(fl,m) \sim f(D_m)(l,m),$$
$$(l,fm) \sim f(D_l)(l,m)$$
for every $f \in \mathbf{C}(C)$, then Deligne's pairing of $L$ and $M$ is defined to be
$$\langle L,M\rangle := V/\sim,$$
which, again based on Weil's reciprocity law, is a 1-dimensional vector space. This construction of Deligne's pairing can be generalized to a family of curves $\pi: X \to S$ which gives a line bundle over $S$ and Theorem XIII.5.8 of [GAC2] shows that it coincides with Deligne's definition, namely the one in your question (denoted by $\langle \bullet, \bullet \rangle_\pi$ in what follows).
To compare Deligne's pairing with Weil's pairing, as before let $\pi: X \to S$ be a family of curves over a connected base and 
$$\mathcal{L}, \mathcal{M} \in \mathrm{Pic}(X/S)^0[n].$$ 
Assume for simplicity that there exist rational sections $l,m$ of $\mathcal{L}$ and $\mathcal{M}$ such that $l_s := l_{|X_s}$ and $m_s := m_{|X_s}$ are rational functions on $X_s$ with disjoint supports for every $s\in S$. Then $\langle \mathcal{L}^{\otimes n},\mathcal{M}\rangle_{\pi,s}$ is generated by $(l_s^n,m_s)$ and we have
$$(l_s^n,m_s) = l_s^n(D_{m_s}) \cdot (1,m_s).$$
Similarly, $\langle \mathcal{L},\mathcal{M}^{\otimes n}\rangle_{\pi,s}$ is generated by
$$(l_s,m_s^n) = m_s^n(D_{l_s}) \cdot (l_s,1).$$
As $\langle \mathcal{L},\mathcal{O}_X \rangle_\pi$ and $\langle \mathcal{O}_X,\mathcal{M} \rangle_\pi$ are canonically isomorphic to $\mathcal{O}_S$, the "difference of trivializations" between $\langle \mathcal{L}^{\otimes n},\mathcal{M} \rangle_\pi$ and $\langle \mathcal{L},\mathcal{M}^{\otimes n} \rangle_\pi$ is therefore ${l_s^n(D_{m_s})}/{m_s^n(D_{l_s})} = e_n(\mathcal{L},\mathcal{M})$.