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 $V/V^-$).



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> Is there a reference which discusses the representability of this functor and the relatively representability of $X^- \rightarrow X$?
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The content of the statement is a descent one 



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> 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 $V\_A \in X^-(A)$?
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Again, I essentially know how to do this when $V^-$ is a line.



Galois deformations with Panchiskin condition
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.


