Let $p$ be a prime number, $\psi:G_\mathbb{Q} \rightarrow \bar{\mathbb{Q}}_p$ be an odd character of conductor $N$ prime to $p$, with finite image and such that $\psi(p)=1$. Let $\mathcal{W}$ be the weight space representing the homomorphism $\mathbb{Z}_p^{\times} \rightarrow \mathbb{G}_m$ and $\epsilon_p$ denotes the $p$-adic cyclotomic character.

Let $k$ be a character of $\mathcal{W}$ corresponding to some integer at least equal to two. Can we vary a non trivial $1$-co cycle of $Ext^{1}_{G_\mathbb{Q}}(1,\psi \epsilon_p^k)$ $p$-adically with respect to $k$ (here $Ext^{1}_{G_\mathbb{Q}}(1,\psi \epsilon_p^k)=Ext^{1}_{f}(1,\psi \epsilon_p^k)$ and it is of dimension one since $k\geq 2$).


1 Answer 1


Yes, this can be done.

One can realise the universal character $\epsilon^{\mathrm{univ}}$ as an $R$-linear representation $G_{\mathbf{Q}} \to R^\times$, where $R$ is isomorphic to a direct product of finitely many copies of $\mathbf{Z}_p[[T]]$. Thus $R$ is Noetherian semilocal, and if $S$ is the finite set of primes dividing $Np$, then the groups $H^i(G_{\mathbf{Q}, S}, R(\psi \epsilon^{\mathrm{univ}}))$ are finitely-generated $R$-modules, zero in degrees $\ge 3$. (This is e.g. a consequence of the theory in Nekovar "Selmer Complexes", but this special case can be easily verified directly using Nakayama's lemma and Tate's finiteness theorems for finite modules.) The support of the torsion submodule of $H^2$ is a finite subset of $Spec R$, and away from this support, the formation of $H^1$ commutes with base change. Consider the set of integers $\ell \ge 2$ congruent to $k$ modulo $p-1$; this is dense in the component of $Spec R$ containing $k$, and for each such point, the rank of the speicalisation of $H^1$ is 1. Hence its generic rank is 1 and you can pick a non-torsion element, and if you rig things sensibly this will be non-vanishing under specialisation at $k$ and thus will give the deformation you seek.

Of course, the specialisation of the resulting extension at integers $\ell \le 0$ will never be in $H^1_f$, away from the finite (possibly empty) set where it is zero.

(PS: This kind of cohomology group -- cohomology of the universal cyclotomic twist of a a fixed $\mathbf{Z}_p$-linear representation -- is rather well-studied under the name of "Iwasawa cohomology"; somehow I failed to notice this initially, despite having spent a large chunk of my career thinking about Iwasawa cohomology, which goes to show that one should maybe not write MO answers at 10pm after a long day. Anyway, the above answer would work for any family over a Noetherian semilocal ring, it doesn't have to be the universal cyclotomic twist of a fixed representation.)

  • $\begingroup$ I have another question related to the question above : A small computation tells us that the dimension of $Ext^1_{G_\mathbb{Q}}(1,\psi)$ is one, do you think that the specialization of $H^1(G_{\mathbf{Q}, S}, R(\psi \epsilon^{\mathrm{univ}}))$ to $H^1(G_{\mathbb{Q},S},\psi)$ is not the zero morphism ? $\endgroup$ Dec 2, 2016 at 14:26
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    $\begingroup$ The specialisation map $H^1(G_{\mathbf{Q}, S}, V \otimes R(\epsilon^{\mathrm{univ}})) \to H^1(G_{\mathbf{Q}, S}, V)$ is injective for every $V$ (because the Iwasawa $H^0$ always vanishes). Does that answer your question? $\endgroup$ Dec 3, 2016 at 8:49

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