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Let $L/\mathbf{Q}_p$ be a finite extension and we consider a fixed $L$-linear representation $V$ of the absolute Galois group $G:=\operatorname{Gal}(\overline{\mathbf{Q}}_p/\mathbf{Q}_p)$. Assume that $V$ is crystalline with distinct Hodge-Tate weights $k_1 < k_2 < \dotsb < k_n$. We do not assume any condition on the irreducibility/semi-simplicity of $V$. In fact, we assume that we have a $G$-stable filtration $V^- \subset V$ such that $V^-$ contains all of the non-negative Hodge-Tate weights (the cyclotomic character has weight $-1$, by the way). Note: everything happening is in characteristic zero. Such a $V$ might be said to satisfy the Panchiskin Condition.

Let $A$ be a local Artin $L$-algebra with residue field $L$ (a morphism between two such rings should induce the identity on $L$). We consider the of formal deformation functor $X(A)$ of $V$ to $A$. Inside, we define a subfunctor $X^- \subset X$ with the property that a deformation $V_A$ of $V$ to $A$ is in $X^-(A)$ if and only if it has a Galois stable filtration $V_A^{-} \subset V_A$ (a free $G$-stable $A$-submodule with free quotient $V_A/V_A^{-}$) and the isomorphism $V_A\otimes_A L \simeq V$ induces an isomorphism $V_A^{-}\otimes_L A \simeq V^-$.

I believe under some reasonable hypotheses the functor $X^-$ is (relatively) representable (over $X$). Here representable means "pro-representable". For example, if $V^-$ and $V/V^-$ are each absolutely irreducible then I think $X^-$ is representable. In the case that $V^-$ is a line, e.g. $k_1 = 0$, then one can proceed as in the proof that ordinary deformations of an ordinary representation are relatively representable (regardless now of the irreducibility of $V/V^-$).

Is there a reference which discusses the representability of this functor and the relatively representability of $X^- \rightarrow X$?

The content of the statement is a descent one

If one has a deformation $V_A \in X(A)$ and $A \subset A'$ such that $V_A \otimes_A A'$ is in $X^-(A')$, then is $V_A \in X^-(A)$?

Again, I essentially know how to do this when $V^-$ is a line.

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You can probably extract what you're looking for from the discussion of "paraboline deformation functors" in Chenevier's paper on the U(3) infinite fern. – David Hansen May 30 '13 at 3:11
up vote 3 down vote accepted

Let $\pi_1, \dots, \pi_r$ be the irreducible representations appearing as a sub quotient in $V^-$. Note that none of these irreducible representation appear in $V/V^-$ because these rep. have negative HT weights and $V/V^-$ non-negative HT weights.

Now consider the functor $F$ from the category of $L$-representations of $G$ to itself which to a representation $V$ attaches the largest sub-representation $F(V)$ of $V$ whose all irreducible sub quotients are among $\pi_1,\dots,\pi_r$. Note that this definition makes sense because the sum of two sub-representations of $V$ whose only irreducible subquotients are among $\pi_1,\dots,\pi_r$ has the same property (so $F(V)$ is well-defined), and the image of such a sub representation by a morphism also has this property (so $F$ is a functor). Also it is easy to see that the functor $F$ is additive and left-exact. By the remark made in the first paragraph, $F(V)=V^-$.

Now let $V_A$ be a deformation of $V$ to an Artinian local $L$-algebra $A$ of residue field . $L$, $M$ a finite $A$-module. I claim that $\dim_L F( V_A \otimes_A M) \leq (\dim_L M) d'$. Indeed, $\dim_L M$ is the length of $M$ as an $A$-module, and $M$ has a filtration of length $\dim_L M$ whose graded pieces are isomorphic to $L$. Thus $V_A \otimes_A M$ is, as a representation of $G$ over $L$, a successive extension of $\dim_L M$ copies of $V$, and the result follows by left-exacness of $F$.

Now assuming your hypothesis, $F(V_{A'})$ has dimension $(\dim A') d'$. One has an exact sequence $$ 0 \rightarrow F(V_A) \rightarrow F(V_{A'}) \rightarrow F(V_{A'} \otimes_{A'} A'/A $$ The last term has dimension at most $d'(\dim A'/A)$ by the claim applied to $M=A'/A$, the middle term exactly $d' \dim A'$, so the first term $F(V_A)$ has dimension at least $d'(\dim A)$. But by the claim applied to $M=A$, it has dimension at most $d' \dim A$. Hence $$\dim F(V_A) = d' (\dim A).$$

Also, by functoriality, $F(V_A)$ is a sub $A$-module of $V_A$, and its image in $V$ is just $F(V)=V^-$, so we deduce easily that $F(V_A)$ is free, hence direct summand (over a local artinian ring, ant free sub-module of a free module is direct summand), hence $V_A /F(V_A)$ is projective, hence free as well. So $F(V_A)$ is the sub representation you were looking for.

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