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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!

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  • $\begingroup$ 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. $\endgroup$ – Bombyx mori Feb 11 at 6:13
  • $\begingroup$ @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. $\endgroup$ – Misaka01034 Feb 11 at 15:35
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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 ${l_s^n(D_{m_s})}/{m_s^n(D_{l_s})} = e_n(\mathcal{L},\mathcal{M})$.

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  • $\begingroup$ Thanks! This helps clarifying my confusion a lot! $\endgroup$ – Misaka01034 Feb 11 at 15:35

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