8
$\begingroup$

Let $G$ be the abolute Galois group of $\mathbb Q_p$, let $\delta_1, \delta_2: G\rightarrow L^{\times}$ be continuous characters, where $L$ is a finite extension of $\mathbb Q_p$. Assume that $\delta_1\delta_2^{-1}$ is neither trivial nor the cyclotomic character then $Ext^1_{G}(\delta_2, \delta_1)$ is one dimensional. Hence there exists a unique non-split extension:

$0\rightarrow \delta_1\rightarrow V\rightarrow \delta_2\rightarrow 0$.

When is $V$ de Rham?

I believe that the answer is if and only if both $\delta_1$ and $\delta_2$ are de Rham and the Hodge-Tate weight of $\delta_1\delta_2^{-1}$ is $\ge 1$ (at least if the Hodge-Tate weights of $\delta_1$ and $\delta_2$ are distinct) and I guess I could x it out by using Bloch-Kato's paper in Grothendieck Festschrift, bu the answer must be well known and maybe even written down somewhere.

Ideally, I would like to be able to quote a reference, where this has been worked out.

$\endgroup$

3 Answers 3

8
$\begingroup$

Dear Vytas,

lemma 6.5 of my 2002 inventiones paper says that if $V$ is any de Rham representation all of whose HT weights are at least 1, then any extension of $Q_p$ by $V$ is itself de Rham. This holds for reps of $G_K$ where the residue field $k$ of $K$ can be any perfect field (not merely finite).

If $k$ is finite, then this was well-known before and follows from the results of Bloch and Kato (which I think you should quote). See proposition 1.28 of Nekovar's "On $p$-adic height pairings" where this is stated explicitly and proved using BK's computations.

EDIT: see also the "Proposition" on page 196 of Perrin-Riou's "Représentations $p$-adiques ordinaires". It predates Nekovar's paper, and although the result is less strong, it's enough for what you need.

$\endgroup$
0
6
$\begingroup$

Dear Vytas,

Twisting by $\delta_1^{-1}$, we may restrict attention to non-split extensions $$0 \to 1 \to V \to \delta \to 0,$$ and of course we may assume that $\delta$ is de Rham. If the HT wt. of $\delta$ is $\geq 0$, such a non-trivial extension is never de Rham, while if the HT wt. of $\delta$ is $< 0$, such a non-trivial extension is always de Rham (assuming as you do that $\delta$ is neither trivial nor inverse cyclotomic).

A statement of these facts is incorporated into the statement of Prop. 4.4.4 of my Coates paper, but the statement is standard, and follows e.g. from a theorem of Hyodo, stating that a de Rham ext. of pst. reps. is in fact pst. (Of course this follows from the more general statement that de Rham implies pst., but is easier. I don't know the original reference, but see e.g. Thm. 4.5.4 and Cor. 4.5.9 of my never-quite-finished paper on crystalline extensions with Mark Kisin.)

In any event, knowing that one has only has to check whether the extension is pst. (either by Hyodo or by the more general equivalence of de Rham and pst), a simple and direct calculation with pst. Dieudonne modules gives what is required.

I'm sorry I can't just give you a clean reference.

Best wishes,

Matt

$\endgroup$
3
$\begingroup$

Take a look at section 16 of this document of Laurent Berger : http://www.umpa.ens-lyon.fr/~lberger/barcelone/BergerBarcelone.pdf

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.