Timeline for Why is the norm map dual to restriction under Tate local duality?
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| Jul 29, 2014 at 23:53 | history | edited | user27920 | CC BY-SA 3.0 |
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| Jul 29, 2014 at 13:57 | comment | added | Question Mark | Thanks for very helpful comments! I have made a question about III.7.8 here: mathoverflow.net/questions/177361/… | |
| Jul 29, 2014 at 7:26 | comment | added | user27920 | My guess was right about the relevance of SGA7 (Exp. VIII) to prove commutativity of III.7.8 in Milne's book, and fortunately one needs only Prop. 3.3.1 (whose proof is very cool). It is a general compatibility of pairings in cohomology over any base scheme $S$ (pairings valued in ${\rm{H}}^2(S,\mathbf{G}_m)$, always using the result from section 11 of Brauer III that for coefficients in a smooth group scheme etale and fppf abelian cohomology coincide, so in char. $p$ we do not lose contact with Galois cohomology when using fppf methods). If you make it an MO question then I'll answer it. | |
| Jul 29, 2014 at 5:58 | comment | added | user27920 | So the upshot is that my answer does prove as well that the bi-extension method has the same functorial property which you wanted (up to sorting out the sign issues, which doesn't affect whether or not the pairing is perfect, so it is legitimate to now use this functorial property in a proof of perfectness, as Milne's book does for the $p$-part in characteristic $p$). | |
| Jul 29, 2014 at 5:53 | comment | added | user27920 | $\xi:\mathbf{Z}_S \rightarrow A[1]$ corresponding to a class in ${\rm{H}}^1(S,A)$ gives $\mathbf{Z}_S \rightarrow \mathbf{G}_m[2]$ that "is" the bi-extension method (up to a sign: careful with $f[1]$!). But connecting homomorphisms are induced by the negative of the mapping cone morphism, so my proposed method via connecting homomorphism applies to the extension of $A$ corresponding to $f$ really does give the same up to universal signs. Good luck sorting out the signs (also relevant if you compute another way by passing translation through left or right side of a derived tensor product). | |
| Jul 29, 2014 at 5:46 | comment | added | user27920 | 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... | |
| Jul 29, 2014 at 2:03 | comment | added | user27920 | @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. | |
| Jul 28, 2014 at 20:06 | comment | added | Question Mark | That is a nice way to define the pairing and works for every $K$. It is a pity though that there doesn't seem to be a reference that would prove Tate local duality for abelian varieties for all $K$ at once using this definition. I am very interested in any comments on the diagram in III.7.8. | |
| Jul 28, 2014 at 16:40 | comment | added | user27920 | 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. | |
| Jul 28, 2014 at 16:15 | comment | added | user27920 | @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. | |
| Jul 28, 2014 at 14:29 | vote | accept | Question Mark | ||
| Jul 28, 2014 at 14:28 | comment | added | Question Mark | I think that under interpreting $\mathrm{Pic}^0$ as those $\mathbb{G}_m$-torsors that come from group extensions, your definition of the map simply corresponds to taking $\mathrm{Ext}^1(B, \mathrm{norm})$ as in my second comment. Then, for the compatibility of the dual with restriction of scalars, the eventual input (after base change to $K^s$) is that the Poincare bundle of a product of abelian varieties is the external product of Poincare bundles of the factors. | |
| Jul 28, 2014 at 13:08 | comment | added | Question Mark | I agree about Galois vs. separable (I restricted to Galois for simplicity). Thanks for your insightful comment about duality and Weil restriction; I will clarify to myself your definition of the map. As for Milne's book, this is one of the many diagrams in his book whose commutativity is not proved. It seems to me that sometimes the commutativity of the diagrams that Milne is basing his arguments on are actually difficult to prove: e.g., why does the first one in the pf. of III.7.8 commute (it is supposed to follow from III.C.5 which is not proved either)? | |
| Jul 28, 2014 at 5:30 | comment | added | user27920 | It seems that the proof of the $p$-part of Tate's theorem in characteristic $p > 0$ in Ch. III of Milne's "Arithmetic duality theorems" rest on the compatibility that you asked about (in the special case $L/K$ Galois cyclic); look at the middle square in the displayed diagram in the proof of Lemma 7.9 of Ch. III (whose commutativity does not seem to be addressed there, but perhaps either Milne discusses it elsewhere in the book or there is a simpler argument in the Galois case than the one I give for the separable case). | |
| Jul 28, 2014 at 5:18 | comment | added | user27920 | To answer your question about the "bit of diagram chasing"...this is exactly due to the mechanism by which I defined (over $K$!!) how to identify ${\rm{R}}_{L/K}(A_L)^t$ with ${\rm{R}}_{L/K}(A^t_L)$. So I suggest making sure you agree with my proposed specific mechanism to define how duality commutes with Weil restriction (which I guess from your first comment you tried to ignore), and then you should see that this implies the affirmative answer to the question in your second comment. | |
| Jul 28, 2014 at 5:13 | comment | added | user27920 | 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. | |
| Jul 28, 2014 at 3:23 | comment | added | Question Mark | The bit of "diagram chasing" that you mention ultimately reduces to showing that under the $\mathrm{Ext}^1$ point of view on the dual, the bijection $A^t(L) = B^t(K)$ is given by the composition $\mathrm{Ext}^1_L(A, \mathbb{G}_m) \xrightarrow{\mathrm{R}_{L/K}} \mathrm{Ext}^1_K(B, \mathrm{R}_{L/K}\mathbb{G}_m) \xrightarrow{\mathrm{Ext}^1_K(B, \mathrm{norm})} \mathrm{Ext}^1_K(B, \mathbb{G}_m)$. Would you be willing to indicate why $A^t(L) = B^t(K)$ identifies like this? | |
| Jul 28, 2014 at 2:36 | comment | added | Question Mark | Thank you very much for your detailed answer, I'll check the details. I did assume though that the extension is separable by saying that $L/K$ is Galois. Also, couldn't one check the compatibility of Weil restriction with duals by exploiting the compatibility of Weil restriction with base change to reduce to the compatibility of duals and products (to me this seems simpler than using the "norm of the Poincare bundle")? | |
| Jul 28, 2014 at 1:20 | history | edited | user27920 | CC BY-SA 3.0 |
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| Jul 28, 2014 at 0:59 | history | edited | user27920 | CC BY-SA 3.0 |
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| Jul 28, 2014 at 0:31 | history | answered | user27920 | CC BY-SA 3.0 |