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Let $K$ be a local field (in fact, finite extension of $\mathbb{Q}_p$) and let $A$ and $B$ be abelian varieties over $K$.

Associated to $A$ and $B$ are the Tate-modules $T_p(A)$ and $T_p(B)$.

Both of these are canonically $G_K:=\mathrm{Gal}(\bar{K}/K)$-modules (or $K[G_K]$-modules if someone prefers this).

What I want to do is compute $\mathrm{Ext}^1_{K[G_K]}(T_l(A),T_l(B))$ for all $l$, even $l=p$ (and if possible $\mathrm{Ext}^2_{K[G_K]}(T_l(A),T_l(B))$).

Are there any results out there in this direction?

If not, does anyone have a clue how to compute a projective resolution of a Tate-module (as a $K[G_K]$-module$?

Also, would it matter if I instead considered the completed group algebra $K[[G_K]]$?

Aside Yesterday I attended the ceremony where the named mathematician collected his Abel-price, and today, the Abel-lectures by said mathematician and Richard Taylor. Unfortunately, I realized I needed to know the answer to this question when I got back to my office, otherwise I could've pestered recipient and Taylor. :P But hey, now you all get to think about this instead!

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  • $\begingroup$ Over a finite field, these Ext groups are studied in one of my 1968 Inventiones papers. The same techniques may work over local fields. $\endgroup$
    – JS Milne
    Commented May 26, 2010 at 18:20
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    $\begingroup$ Daniel, probably you mean to write $\mathbf{Z}_ {\ell}[G_K]$ rather than $K[G_K]$. Also, the Ext's are same as (profinite) Galois cohomology of the Tate module for $B$ tensored against the linear dual of the Tate module of $A$. This linear dual is Tate twist of Cartier dual, and Cartier dual is Tate module of dual abelian variety. So using Tate duality would convert the Ext^2 into a concrete thing. $\endgroup$
    – BCnrd
    Commented May 26, 2010 at 18:28
  • $\begingroup$ @Brian, I see your point with $\mathbb{Z}_l$. I was thinking $K$ but you're right: the way I phrased the question it should be the $l$-adic integers. However, I don't understand what you are saying for the rest. Care to elaborate? Maybe as a de facto answer? $\endgroup$
    – Daniel Larsson
    Commented May 26, 2010 at 20:42
  • $\begingroup$ @Jim: somewhere in the back of my mind I had a faint memory of reading something by you and Waterhouse (quite recently actually) where you refer to a paper of yours where this was proven. I haven't looked at it yet, but I will, and see if the same technique, or some version of it, goes through in the local case also. Thanks! $\endgroup$
    – Daniel Larsson
    Commented May 26, 2010 at 20:45
  • $\begingroup$ @Daniel: I'm just alluding to the principle whereby Hom from a "finite free" object is same as invariants in tensor product against its dual. Now pass to derived functors (first at torsion level, then pass to inverse limits) to relate an Ext to group cohomology. $\endgroup$
    – BCnrd
    Commented May 26, 2010 at 22:12

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