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Let $L/K$ be a finite Galois extension of nonarchimedean local fields, and let $A$ and $A^t$ be dual abelian varieties over $K$. Tate local duality tells us that $A^t(K)$ and $H^1(K, A)$ are Pontryagin dual locally compact Hausdorff abelian topological groups, and likewise over $L$. Under this duality, why does the dual of the restriction map $H^1(K, A) \rightarrow H^1(L, A)$ identify with the norm map $A^t(L) \rightarrow A^t(K)$?

I am aware that this is "well-known" and that there are references that treat special cases, but I am interested in a proof or a reference in the stated generality.

Edit: Since $K$ is any nonarchimedean local field, the Tate local duality pairing $A^t(K) \times H^1(K, A) \rightarrow H^2(K, \mathbb{G}_m)$ that I am using is defined using biextensions (as in III.7 of Milne's Arithmetic Duality Theorems).

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2 Answers 2

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Note first of all that the "norm" map you speak of does not make sense unless the field extension is separable. That is, for a separable field extension $k'/k$ of finite degree and a commutative $k$-group scheme $C$ there is a canonical "norm" homomorphism $C(k') \rightarrow C(k)$ induced by the passage to Galois-invariants on the homomorphism $C(k' \otimes_k k_s) \rightarrow C(k_s)$ arising from the canonical $k_s$-algebra decomposition of $k' \otimes_k k_s$ as a product of copies of $k_s$ indexed by ${\rm{Hom}}_k(k',k_s)$. The same method defines a "norm" homomorphism $C(k' \otimes_k S) \rightarrow C(S)$ for any $k$-algebra $S$, so it corresponds to a natural $k$-homomorphism ${\rm{R}}_{k'/k}(C_{k'}) \rightarrow C$ in terms of the Weil restriction when $C$ is finite type. This generalizes to when $k'/k$ is replaced with a finite etale cover. But beyond the separable setting there is no "reasonable" norm through finite flat maps whose formation commutes with base change.) So for this reason, we are going to assume that $L/K$ is separable.

First, by the relationship between cup products and connecting homomorphisms (and the identification of $A^t(K)$ with ${\rm{Ext}}^1_K(A, \mathbf{G}_m)$ functorially in $A$) we see that the pairing $$A^t(K) \times {\rm{H}}^1(K,A) \rightarrow {\rm{H}}^2(K, \mathbf{G}_m) = \mathbf{Q}/\mathbf{Z}$$ identifies covariant functoriality in degree-1 cohomology as adjoint to dual-functoriality of abelian varieties.

Let $B$ denote the Weil restriction of scalars ${\rm{R}}_{L/K}(A_L)$; this is an abelian variety precisely because $L/K$ is separable. (If $L/K$ is not separable then $B$ is a smooth connected commutative $K$-group of dimension $[L:K]\dim(A)$ but is always non-proper if $A \ne 0$.) Let $j:A \rightarrow B$ be the natural inclusion. Ultimately we are going to transform your question into the above functoriality of the Tate pairing over $K$ applied to the $K$-homomorphism $j$.

By Shapiro's Lemma considerations, we naturally identify ${\rm{H}}^i(L,A)$ with ${\rm{H}}^i(K,B)$, and (check!) this identifies the restriction map on ${\rm{H}}^1$'s with ${\rm{H}}^1(j)$. Likewise, by the compatibility of Weil restriction with the formation of dual abelian variety (using the "norm" of the Weil restriction of the Poincare bundle), the norm map $A^t(L)\rightarrow A^t(K)$ is identified with the map on $K$-points induced by the dual homomorphism $j^t:B^t \rightarrow A^t$. Also, and most crucially, by a bit of diagram chasing (using the role of "norm of Poincare bundle" above) we see that the Tate pairing for $B$ over $K$ is identified with the composition of the Tate pairing for $A_L$ over $L$ and the "norm" map on Brauer groups $${\rm{Br}}(L) = {\rm{H}}^2(L, \mathbf{G}_m) = {\rm{H}}^2(K, {\rm{R}}_{L/K}(\mathbf{G}_m))\rightarrow {\rm{H}}^2(K,\mathbf{G}_m)={\rm{Br}}(K).$$ But when these flanking Brauer groups are identified with $\mathbf{Q}/\mathbf{Z}$, this composite map is the identity, as we see by analyzing pre-composition with the surjective restriction ${\rm{Br}}(K) \rightarrow {\rm{Br}}(L)$ (that intertwines with $[L:K]$ on $\mathbf{Q}/\mathbf{Z}$ via local class field theory, and the composition of $\mathbf{G}_m \rightarrow {\rm{R}}_{L/K}(\mathbf{G}_m)$ with the "norm" map ${\rm{R}}_{L/K}(\mathbf{G}_m) \rightarrow \mathbf{G}_m$ is $t \mapsto t^{[L:K]}$).

So putting it all together, the diagram you want to commute for $L/K$ separable really does translate into the elementary functoriality of the Tate pairing over $K$, applied to the map $j$ between abelian varieties over $K$.


Remark: In case $L/K$ is inseparable there is still something one might wish to prove. Namely, focusing on the case when $L/K$ is purely inseparable of degree $p^n$ with $p = {\rm{char}}(K)>0$, so $L = K^{1/p^n}$ over $K$, we use the $p^n$-power isomorphism of local fields $L \simeq K$ to identify $A^t(L)$ with $(A^t)^{(p^n)}(K) = (A^{(p^n)})^t(K)$, so we could define the norm $A^t(L) \rightarrow A^t(K)$ to be the map on $K$-points induced by the $n$-fold relative Verschiebung homomorphism $V_{A^t/K,n}:(A^t)^{(p^n)} \rightarrow A^t$. (The Verscheibung homomorphism is defined rather generally for commutative flat group schemes in section 4.2 of Exp. VII$_{\rm{A}}$ of SGA3, and it is dual to the $n$-fold relative Frobenius isogeny by applying 4.3.3 of loc. cit. on $p^m$-torsion for all $m \ge 0$, or cheat and define it to be the dual of that Frobenius isogeny.) So you could ask if this "norm" is Tate-dual to the $L/K$-restriction map in degree-1 Galois cohomology.

The restriction ${\rm{H}}^1(K,A) \rightarrow {\rm{H}}^1(L,A) \simeq {\rm{H}}^1(K,A^{(p^n)})$ is likewise identified with ${\rm{H}}^1(F_{A/K,n})$. Thus, since $V_{A/K,n}$ is dual to $F_{A/K,n}$, the elementary functoriality of the Tate pairing applied to the $K$-homomorphism $F_{A/K,n}$ reduces the observation that that the Tate pairing for $A_{K^{1/p}}$ over $K^{1/p}$ is equal to that of $A^{(p)}$ over $K$ via the isomorphism $K^{1/p} \simeq K$.

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    $\begingroup$ Ah, sorry, I didn't see "Galois" in the question; anyway, you definitely don't want to restrict to just the Galois case in practice. The proof of compatibility of dual with Weil restriction cannot be checked after base change unless you first define over $K$ the compatibility map that you intend to prove is an isomorphism. That is why I described it in terms which make sense directly over the ground field; of course, once the map is defined, to actually prove it is an isomorphism we make base change to split things and pass to the product situation that you suggested to think about. $\endgroup$
    – user27920
    Commented Jul 28, 2014 at 5:13
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    $\begingroup$ @QuestionMark: Yes, I completely agree with your preceding comment. But the Tate pairing doesn't need biextensions for its definition, nor did I intend for it to be defined that way: given an extension of $K$-groups $1 \rightarrow \mathbf{G}_m \rightarrow E \rightarrow A \rightarrow 1$ we get a connecting map ${\rm{H}}^1(K,A) \rightarrow {\rm{H}}^2(K,\mathbf{G}_m)$ that I would say is the Tate pairing. This ties in well with another description via composition of maps in the derived category of abelian sheaves on the fppf or etale sites on the category of $K$-schemes locally of finite type. $\endgroup$
    – user27920
    Commented Jul 28, 2014 at 16:15
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    $\begingroup$ Concerning the first diagram in the proof of III.7.8 in Milne's book, I remember thinking about it a long time ago, but I'd have to refresh my memory about the details. The equality between my proposed definition and the bi-extension definition also requires some thought. I'll try to come back to this a bit later. $\endgroup$
    – user27920
    Commented Jul 28, 2014 at 16:40
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    $\begingroup$ @QuestionMark: In SGA7 there is a discussion about the relationship between the biextension map and Cartier duality at finite level, and I expect that if one unpacks what is going on in there then it will provide a proof that the first diagram in the proof of III.7.8 in Milne's book commutes and that my proposed definition of the Tate pairing agrees with the biextension definition. However, I have never read that part of SGA7 carefully, so I cannot say anything more definite concerning that possible resolution to the uncertainty. I am busy these days, but will try to think about it later. $\endgroup$
    – user27920
    Commented Jul 29, 2014 at 2:03
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    $\begingroup$ For any scheme $S$ and abelian $S$-scheme $A$ with dual $B$, the Tate pairing $B(S)\times{\rm{H}}^1(S,A) \rightarrow{\rm{H}}^2(S,\mathbf{G}_m)$ (fppf abelian sheaves on the category of lfp $S$-schemes) via my proposed method agrees (up to a universal sign) with the one via biextensions. Indeed, by design composing the bi-extension morphism $A \otimes^{\mathbf{L}} B \rightarrow \mathbf{G}_m[1]$ with $b:\mathbf{Z}_S \rightarrow B$ gives the mapping cone $f:A\rightarrow \mathbf{G}_m[1]$ for the $S$-group extension associated to $b\in B(S)={\rm{Ext}}^1_S(A,\mathbf{G}_m)$. Composing $f[1]$ with... $\endgroup$
    – user27920
    Commented Jul 29, 2014 at 5:46
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If $m$ is an integer coprime to the characteristic of $k$, then the pairing between $A^t(K)/m A^t(K)$ and the $m$-torsion part of $H^1(K,A)$ is compatible with the cup pairing $$H^1(K,A^t[m]) \times H^1(K,A[m]) \to H^2(K,\mu[m])\cong \mathbb{Z}/m\mathbb{Z}$$ through the short exact sequences from Kummer theory $$ 0\to A(K)/mA(K) \to H^1(K,A[m])\to H^1(K,A)[m]\to 0$$ and likewise for $A^t$. Now, for the local duality in $H^1$, the dual map to restriction is corestriction, because of the formula $\xi\cup\operatorname{cor}(\eta) = \operatorname{cor}(\operatorname{res}(\xi)\cup\eta)$ and the compatibility of the Hasse invariant with the corestriction in $H^2$.

That is a sketch how to prove it for the prime-to-characteristic part. The original paper by Tate already made use of this in the proof.

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  • $\begingroup$ The above before the mention of corestriction works as written even if $m > 0$ is divisible by the characteristic (using fppf cohomology and fppf Kummer sequence), so it seems the merit of Galois cohomology (which needs $m$ not divisible by the characteristic) vs. fppf cohomology is the availability of corestriction (which ultimately expresses exactness of finite etale pushforward for the etale topology, whereas finite flat pushforward is not fppf exact) and "norm". $\endgroup$
    – user27920
    Commented Jul 28, 2014 at 12:39
  • $\begingroup$ Thanks for your response, Chris! You base your argument on the compatibility that I have alluded to in my previous comment: it would be helpful if you could indicate how to prove this compatibility (I am aware that people take this for granted)? Note that in characteristic $p$ the "biextension definition" of the pairing has to be used (in place of augmented cup products), so checking the compatibility involves dealing with edge maps in some spectral sequences, which is what makes this checking difficult for me. Any comments you may have on this would be most helpful. $\endgroup$ Commented Jul 28, 2014 at 13:16
  • $\begingroup$ It was not clear to me what definition of the pairing you are using. I thought you refer to the original definition by Tate (which is away from the characteristic). The compatibility between the two pairings is given in Proposition 2.1 in dpmms.cam.ac.uk/~taf1000/papers/ctpair.html by Tom Fisher (and probably in many other places). $\endgroup$ Commented Jul 28, 2014 at 13:27
  • $\begingroup$ I wouldn't know a good reference for the comparison with the bi-extension pairing. Maybe Milne's Airhtmetic Dualities ? $\endgroup$ Commented Jul 28, 2014 at 13:29
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    $\begingroup$ @QuestionMark: Sure, just let $k'/k$ be a purely inseparable extension of degree $p$ in characteristic $p$, and apply ${\rm{R}}_{k'/k}$ to the $p$-torsion Kummer sequence over $k'$. Letting $f:{\rm{Spec}}(k') \rightarrow {\rm{Spec}}(k)$ be the finite flat structure map, you get that $f_{\ast}(\mathbf{G}_m)$ is "${k'}^{\times}$ as a $k$-group", and in particular $p$-power on this $p$-dimensional smooth commutative affine group is valued in its 1-dimensional maximal $k$-torus (modulo which it becomes $p$-torsion, hence unipotent). In particular, $f_{\ast}$ applied to $p$-power isn't onto. $\endgroup$
    – user27920
    Commented Jul 28, 2014 at 16:44

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