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Suppose I have an abelian variety $A$ with split toric reduction over a non-archimedean field $K$, so that $A$ can be realized as $T / \Lambda$ where $\Lambda$ is a lattice in the torus $T$. Letting $A^\vee$ denote the dual of $A$, we have the $\ell$-adic Weil pairing $T_\ell(A^\vee) \times T_\ell(A) \to \lim_{\leftarrow n} \mu_{\ell^n}$. I want to see this pairing explicitly in terms of the non-archimedean uniformization $A \cong T / \Lambda$ and that of its dual.

It is known (and proved in an article of Bosch and Lütkebohmert) that $A^\vee$ can be realized as $T^\vee / \Lambda^\vee$ where $T^\vee = \mathrm{Hom}(\Lambda, K^\times)$ and $\Lambda^\vee = \mathrm{Hom}(T, K^\times)$. On page 137 of Mihran Papikian's article "Non-archimedean uniformization and monodromy pairing", in the proof of Theorem 5.8, Papikian construct a pairing of $T_\ell(T^\vee)$ with $\Lambda \otimes \mathbb{Z}_\ell$ given by evaluation: any $\ell^n$-torsion element of $T^\vee$ is a homomorphism $\Lambda \to \mu_\ell \subset K^\times$ factoring through $\Lambda / \ell \Lambda$; now pass this evaluation map to the inverse limit. Papikian says "it is not too hard to check" that this is in fact the Weil pairing (restricted to $T_\ell(T^\vee) \times T_\ell(A)$ and factoring on the right through the right kernel $T_\ell(A)^I \subset T_\ell(A)$ of inertia-fixed elements). Coleman in his article "The Monodromy Pairing" simply casually the same thing in the proof of his Theorem 2.1 (p. 319) without proof or even bothering to say "it is not too hard to check".

My question is, how do we actually see this? The Weil pairing can seem a bit mysterious. I can view the Weil pairing as equivalent to the identification of $\mathrm{Hom}(A[n], \mathbb{G}_m)$ with $A^\vee[n]$ via Cartier duality but it doesn't seem obvious that that identification is the same as the evaluation discussed in the above paragraph. I imagine that if I went through Bosch and Lütkebohmert's Section 2 of "Degenerating Abelian Varieties" carefully enough with the divisor / line bundle definition of Weil pairing in mind, I could prove it by unwinding the definitions the right way, but this doesn't seem immediate and I suspect (from Papikian's and Coleman's way of mentioning it without proof) that there is a more obvious explanation I may be missing here.

An essentially equivalent variation on this statement that I would like to see is that the pairing of $T^\vee[n]$ with $\Lambda / n\Lambda$ given by evaluation is the one induced by the mod-$n$ Weil pairing for any integer $n$. (And I don't see why $n$ should have to be relatively prime to the residue characteristic here; in the context where this comes up in Papikian's article, for instance, $\ell$ is assumed relatively prime to the residue characteristic. This is a very relevant point for my purposes!)

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  • $\begingroup$ I doubt that there is a way of seeing this without looking in detail at the identification of $A^\vee$ with $T^\vee / \Lambda^\vee$ since the statement depends on this identification. $\endgroup$
    – naf
    Commented Nov 1 at 23:40
  • $\begingroup$ This makes logical sense; I'm just very suspicious of missing something immediate given that Coleman casually uses the fact in his paper (proof of Theorem 2.1) without comment. $\endgroup$ Commented Nov 2 at 16:01
  • $\begingroup$ Well, if you change the identification by a sign then the statement would not be true. I suspect that Coleman does not say anything about the proof because it is a simple check (as Papikian says). In the Bosch-Lütkebohmert paper the key is the couple of paragraphs beginning with "Now the proof of 2.1 is quite easy" on p. 663 where the Poincare bundle is constructed (which gives rise to the identification). Using this and comparing with the construction of the Weil pairing in Mumford's book (in terms of line bundles) should give a proof. $\endgroup$
    – naf
    Commented Nov 3 at 1:35
  • $\begingroup$ Also, I agree that no assumption on the residue characteristic should be needed. $\endgroup$
    – naf
    Commented Nov 3 at 1:38

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