We have the [Deligne Pairing](https://publications.ias.edu/sites/default/files/Number58.pdf) 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](https://mathoverflow.net/a/125438/31342) 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! 



Deligne Pairing v.s. Weil Pairing on a Family of curves
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](https://mathscinet.ams.org/mathscinet/search/publdoc.html?agg_author_51660=51660&arg3=&batch_title=Selected%20Matches%20for%3A%20Title%3D%28geometry%20of%20algebraic%20curves%29&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&fmt=doc&pg4=ICN&pg5=TI&pg6=PC&pg7=ALLF&pg8=ET&review_format=html&s4=&s5=geometry%20of%20algebraic%20curves&s6=&s7=&s8=All&searchin=&sort=newest&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq&r=1&mx-pid=2807457) 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 ${l\_s^n(D\_{m\_s})}/{m\_s^n(D\_{l\_s})} = e\_n(\mathcal{L},\mathcal{M})$.


