This question is a follow up to Why is the norm map dual to restriction under Tate local duality?

Let $A$ and $B$ be dual abelian schemes over a base scheme $S$. For an integer $n \ge 1$, consider the Cartier duality cup product pairing $$ H^1(S, B[n]) \times H^1(S, A[n]) \rightarrow H^2(S, \mathbb{G}_m) $$ (which agrees with the edge-Ext-pairing by interpreting $B[n]$ as $\mathscr{Hom}(A[n], \mathbb{G}_m)$ and using the edge map $H^1(S, B[n]) \rightarrow \mathrm{Ext}^1(A[n], \mathbb{G}_m)$ --- see 1970-71 papers of Gamst & Hoechsmann for this agreement). Consider also the "Tate duality pairing" $$ B(S) \times H^1(S, A) \rightarrow H^2(S, \mathbb{G}_m) $$ defined by using the identification $B(S) = \mathrm{Ext}^1(A, \mathbb{G}_m)$ and the Ext-product again. The two pairings are related by the connecting homomorphism $$ B(S) \rightarrow H^1(S, B[n]) $$ and by $$ H^1(S, A[n]) \rightarrow H^1(S, A). $$ Why are the two pairings compatible? I.e., why does the obvious diagram commute? If possible, it would also be interesting to discuss the relation of the second pairing with the biextension cup product pairing $$ H^0(S, B) \times H^1(S, A) \rightarrow H^1(S, B \otimes^{\mathbb{L}} A) \rightarrow H^2(S, \mathbb{G}_m). $$ If it makes a difference, feel free to assume that $S$ is the spectrum of a nonarchimedean local field.

The compatibility that I am asking about is used in III.7.8 of Milne's "Arithmetic Duality Theorems" in the proof of Tate local duality for abelian varieties. It is also taken for granted whenever needed in a number of other places in the literature.

  • $\begingroup$ The discussion of Tate's theorem as Theorem 4.1 of Chapter X of Lang's book "Topics in cohomology of groups" (notes that Lang took based on weekly letters he got from Tate in the late 1950's) treats the case of $p$-adic fields (i.e., char. 0). Right at the start occurs the commutative diagram being asked about in this question (with $S = {\rm{Spec}}(K)$ for a $p$-adic field $K$), where it is claimed to hold "by the general theory of the augmented cup product" (no reference provided). So over fields of characteristic 0 there should be a more down-to-earth proof using cocycle calculations. $\endgroup$
    – user27920
    Jul 29 '14 at 23:58

To avoid notational confusion between translation in the derived category (of abelian fppf sheaves on the category of lfp $S$-schemes) and torsion in abelian schemes, I'll denote the $n$-torsion in $A$ as $A_n$ rather than $A[n]$, and likewise for $B$.

The $n$-torsor Kummer sequence for $B:=A^t = \mathscr{Ext}^1_S(A,\mathbf{G}_m)$ gives rise to the mapping cone morphism $d:B \rightarrow B_n[1]$ that "computes" the associated connecting homomorphism (under various $\delta$-functors applied to the Kummer sequence) up to a sign. The crux of the matter is to describe this in terms of $A$ in a more "direct" manner. This rests on two vanishing theorems that we now recall.

First, we have $\mathscr{Hom}_S(A,\mathbf{G}_m)=1$ (which follows from rigidity for abelian schemes; see Corollary 6.2 in section 1 of Chapter 6 of Mumford's GIT book). This gives $$B = (\tau_{\le 1}\mathbf{R}\mathscr{Hom}_S(A,\mathbf{G}_m))[1].$$ Second, we have $\mathscr{Ext}^1_S(G,\mathbf{G}_m)=1$ for any commutative finite locally free $S$-group scheme $G$. (See Proposition 3.3.1 in Exp. VIII of SGA7 for a very nifty proof, using bi-additivity of Ext and Deligne's theorem on $G$ being killed by its order to reduce to replacing $\mathbf{G}_m$ with $\mu_N$ for the order $N$ of $G$, after which Cartier duality does the job; Deligne's theorem isn't needed for $G=A_n$). Thus, the Cartier dual $\mathbf{D}(G)$ satisfies $$\mathbf{D}(G) = \tau_{\le 1}\mathbf{R}\mathscr{Hom}_S(G,\mathbf{G}_m).$$ Recall also that the identification of $B_n$ with $\mathbf{D}(A_n)$ is defined by the sheafified Ext-sequence associated to the $n$-torsion Kummer sequence of $A$: this gives the exact sequence $$0 \rightarrow \mathbf{D}(A_n) \stackrel{\delta}{\rightarrow} \mathscr{Ext}^1_S(A,\mathbf{G}_m) \stackrel{n}{\rightarrow} \mathscr{Ext}^1_S(A,\mathbf{G}_m)\rightarrow 0$$ that is the $n$-torsion Kummer sequence of $B$ combined with the identification of $B_n$ with $\mathbf{D}(A_n)$. Of course, to avoid confusion one should go back to Oda's thesis and check whether this identification of the $n$-torsion group schemes is the same as what is done there, or is off from it by a universal sign (independent of $n$).

This final exact sequence gives rise to the mapping cone map $\mathscr{Ext}^1_S(A,\mathbf{G}_m) \rightarrow \mathbf{D}(A_n)[1]$ which is our old friend $d$ mentioned near the start. But the sheafified Hom-Ext description of the Kummer sequence for $B$ has a crucial advantage: via the above displayed formulas for $B$ and $\mathbf{D}(A_n)$ in derived-category terms, it implies (after some standard homological algebra considerations left to the interested reader) that $d$ is exactly the natural map $$(\tau_{\le 1}\mathbf{R}\mathscr{Hom}(A,\mathbf{G}_m))[1] \rightarrow (\tau_{\le 1}\mathbf{R}\mathscr{Hom}(A_n,\mathbf{G}_m))[1]$$ induced by $j_n:A_n \hookrightarrow A$, up to a universal sign (independent of $n$); sign confusion is due to the signs which intervene under the translation functor.

Using the natural evaluation-pairing morphism $A_n \otimes^{\mathbf{L}} \mathbf{D}(A_n) \rightarrow A_n \otimes \mathbf{D}(A_n) \rightarrow \mathbf{G}_m$, we get an associated composite mapping $$A_n[1] \otimes^{\mathbf{L}} B \stackrel{1 \otimes d}{\rightarrow} A_n[1] \otimes^{\mathbf{L}} B_n[1] = (A_n \otimes^{\mathbf{L}} B_n)[2] \rightarrow \mathbf{G}_m[2]$$ where the middle equality involves a universal sign ambiguity (independent of $n$) when pulling the translation functors outside the derived tensor product (first left and then right, or first right and then left, or...?). Our above description of $d$ in derived-category terms with $A$ immediately implies that this composite map coincides (up to universal signs independent of $n$) with the composite map $$A_n[1] \otimes^{\mathbf{L}} B \stackrel{j_n[1] \otimes 1}{\rightarrow} A[1] \otimes^{\mathbf{L}} (\tau_{\le 1}\mathbf{R}\mathscr{Hom}(A,\mathbf{G}_m))[1] = (A \otimes^{\mathbf{L}} \tau_{\le 1}(\mathbf{R}\mathscr{Hom}(A,\mathbf{G}_m)))[2] \rightarrow \mathbf{G}_m[2],$$ where the final step composes $\tau_{\le 1}(C) \rightarrow C$ (for any complex $C$) with the derived-evaluation morphism (defined using an injective resolution of $\mathbf{G}_m$).

Putting this all together, we get a diagram in the derived category $$\begin{matrix} A_n[1] \otimes^{\mathbf{L}} B & \stackrel{j_n\otimes 1}{\rightarrow} & A[1] \otimes^{\mathbf{L}} B \\ \downarrow & & \downarrow \\ A_n[1] \otimes^{\mathbf{L}} B_n[1] & \rightarrow & \mathbf{G}_m[2] \end{matrix}$$ that commutes up to a universal sign independent of $n$, where the left side is $1 \otimes d$, the bottom is the "evaluation map", and the right side is a "derived evaluation" map. Given $c \in {\rm{H}}^1(S,A_n) = {\rm{Hom}}(\mathbf{Z}_S,A_n[1])$ and $b \in B(S) = {\rm{Hom}}(\mathbf{Z}_S,B)$ (Hom's in the derived category), composing with the map $c \otimes^{\mathbf{L}} b: \mathbf{Z}_S = \mathbf{Z}_S \otimes^{\mathbf{L}} \mathbf{Z}_S \rightarrow A_n[1] \otimes^{\mathbf{L}} B$ into the upper-left corner of the diagram, going around both ways gives two equal maps $\mathbf{Z}_S \rightarrow \mathbf{G}_m[2]$ up to a universal sign independent of $n$.

If we go around the left and bottom sides or the top and right sides then we get the same "Brauer class" on $S$ either way, up to a universal sign independent of $n$, and this expresses the equality (up to a universal sign independent of $n$) for the two composite maps $${\rm{H}}^1(S,A_n) \times B(S) \stackrel{1 \times {\rm{Kum}}_n}{\rightarrow} {\rm{H}}^1(S,A_n) \times {\rm{H}}^1(S,B_n) \rightarrow {\rm{H}}^2(S,\mathbf{G}_m)$$ (final step being the pairing resting on $n$-torsion duality for $A$ and $B$) and $${\rm{H}}^1(S,A_n) \times B(S) \rightarrow {\rm{H}}^1(S,A) \times B(S) \rightarrow {\rm{H}}^2(S,\mathbf{G}_m)$$ (final step resting on the derived-evaluation morphism $A \otimes^{\mathbf{L}} B \rightarrow \mathbf{G}_m[1]$). But the equality of those maps is exactly the desired commutative diagram, where I am taking the "derived evaluation" map $A \otimes^{\mathbf{L}} B \rightarrow \mathbf{G}_m[1]$ as my definition of the "bi-extension map"; this avoids having to know what a bi-extension is, and in my comments to the previous question by the same OP (linked at the top) I explained why this makes the route across the top and right also compute the pairing defined by Ext-pairings as well. This also justifies the commutativity of the left square at the bottom of page 285 of (the 2nd edition of) Milne's book. Of course, to complete the proof of Tate's theorem in characteristic $p$ one has to use Neron models, and for that purpose one needs to work with bi-extensions. But when bi-extensions are developed then I am pretty sure it should be just a matter of staring at definitions to see that the map $A \otimes^{\mathbf{L}} B \rightarrow \mathbf{G}_m[1]$ provided by bi-extensions agrees with the derived-evaluation map used above.

  • $\begingroup$ Thanks for such a nice answer! Regarding the homological algebra considerations left to the reader, is the idea in a nutshell that for an exact triangle $X \rightarrow Y \xrightarrow{y} Z \xrightarrow{z} X[1]$, if one builds the cone triangle $Z \xrightarrow{z} X[1] \rightarrow Cone(z) \xrightarrow{w} Z[1]$, then $w$ identifies with $y[1]$ (up to a universal sign?), whereas $\mathbf{R}\mathscr{Hom}$ and $\tau_{\le 1}$ are triangulated, so their intervention is harmless? $\endgroup$ Jul 30 '14 at 1:19
  • 1
    $\begingroup$ @QuestionMark: That is certainly the idea, but one has to be careful to ensure that the identifications made are functorial and not plagued by the ambiguity of some unknown isomorphism (as happens whenever using the TR3 axiom for triangulated categories to make a construction as opposed to verifying a property). So I grinded it out by working concretely with an injective resolution of $\mathbf{G}_m$, avoiding constructions based on abstract axioms; I don't think one can make the argument via abstract cone stuff (well, unless you're an $\infty$-categorical person, which I am not). $\endgroup$
    – user27920
    Jul 30 '14 at 2:31

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