Let $f$ be a normalized newform of weight $k\geq 2$, $p$ a prime, and $V$ the associated Galois representation (with coefficients in a finite extension $K$ of $\mathbf{Q}\_p$ with ring of integers $\mathscr{O}$). Assume the residual representation attached to $V$ is absolutely irreducible, so that the choice of a $G\_\mathbf{Q}$-stable $\mathscr{O}$-lattice $T$ in $V$ is unique up to $\mathscr{O}$-scaling. Let $A=V/T$ be the quotient, which is a discrete $G\_\mathbf{Q}$-module isomorphic as an $\mathscr{O}$-module to $(K/\mathscr{O})^2$. 


My question: is the Cartier dual $A^\*=\mathrm{Hom}\_\mathscr{O}(A,(K/\mathscr{O})(1))$, a free $\mathscr{O}$-module of rank $2$, necessarily (isomorphic to) a $G\_\mathbf{Q}$-lattice in a representation associated to a newform (of some weight and level)? Along these same lines, do the representations $V^\vee=\mathrm{Hom}\_K(V,K)$ and $V(1)$ also come from a newform? I believe a positive answer to the second question would also give a positive answer to my first question, but I might be mistaken. If the answer to the first question is yes, and the original form $f$ is $p$-ordinary, can I always find a $p$-ordinary form giving rise to $A^\*$?


The reason I ask is because I'm trying to prove some things that require me to know certain facts about the Galois cohomology of $A^\*$ which I know for $A$ (because $A$ comes from a modular form). So if I knew $A^\*$ also came from a modular form, I'd have what I need. 



Duals and Tate twists of Galois representations attached to modular forms.
There are several ways to normalize the Galois representation attached to a modular eigenform,
which differ by a Tate twist, but when you choose one normalization, no Tate twist of a 
Galois representation attached to a modular form can ever be attached to another modular form.
One way to see this is the following: the Galois representation attached to a form of weight 
$k$ has, for my preferred normalization, a determinant $\omega^{k-1} \epsilon$, where 
$\omega$ is the cyclotomic character and $\epsilon$ a finite-order character (the nebentypus).
It has also Hodge-Tate weights at $p$ equal to $0$ and $k-1$. If you twist by a power of the cyclotomic, you will change the power of $\omega$ in the determinant, but not the difference between the Hodge-Tate weights, so something will not be right.


Hence the answer to your second question is negative. For the first, it is positive, in the sense that if $V$ is attached to a modular form $f$, $V^\ast(1)$ is also up to a twist (or no twist, depending on the way you normalize $V$) attached to a modular form (which is $f$ itself
in the case of trivial nebentypus). That simply because dual of representations of dimension 2 are isomorphic to themselves, up to a twist by the inverse of their determinant.


But it seems that your real question has to do with Galois cohomology. I am surprised that you need to know that your representation is associated to a modular form, because I know of no
theorem in Galois cohomology which applies only to Galois rep. attached to modular forms.
What do you have in mind?


