# Deligne Pairing v.s. Weil Pairing on a Family of curves

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!

• Maybe I am missing something subtle, but it appears to me that there is nothing "deep" behind the statement. I think the professor was simply trying to re-cast the definition using the language of algebraic group. If I recall correctly, Deligne pairing can be defined for any locally free sheaf over an algebraic curve. – Bombyx mori Feb 11 at 6:13
• @Bombyxmori That's right, what I'm missing is the correct way of defining Deligne pairing, as the determinant notation doesn't give me a nice way to see this standard fact. Thanks for your comment. – Misaka01034 Feb 11 at 15:35

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 $${l_s^n(D_{m_s})}/{m_s^n(D_{l_s})} = e_n(\mathcal{L},\mathcal{M})$$.