Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
Regarding your second question, if I understand correctly, making a Tate twist does not change the associated modular form. Refer to the answers in mathoverflow.net/questions/72886/… –  Dror Speiser Sep 25 '12 at 0:14
Dear @Dror, I'm not sure I understand. Are you saying that both $V$ and $V(1)$ come from the same modular form? Tate twisting should, for example, shift the Hodge-Tate weights at $p$, so how I don't see how $V$ and $V(1)$ can both come from $f$. –  Keenan Kidwell Sep 25 '12 at 1:29
add comment

1 Answer

up vote 4 down vote accepted

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?

share|improve this answer
Dear @Joël, Thank you so much for your answer! I am doing stuff with Iwasawa invariants of Selmer groups attached to $p$-ordinary newforms over $\mathbf{Z}_p$-extensions of general number fields. One of the hypotheses I need is that the local invariants $H^0(F_v,A)$ and $H^0(F_v,A^*)$ are finite for primes $v$ of my number field $F$ which divide the level of $N$. They are definitely finite for $v$ not dividing the level, and I believe, under certain hypotheses where I can invoke theorems on the local structure of $V$ at the prime $\ell\mid v$, $\ell\mid N$, that $H^0(F_v,A)$ will be finite. –  Keenan Kidwell Sep 25 '12 at 2:40
So I wanted to know if $A^*$ also came from a modular form so that I could try to apply the same reasoning to $H^0(F_v,A^*)$. It's entirely possible these invariants are finite for some other reason, independent of whether or not they come from modular forms. Regarding your response, I'm definitely not wanting to assume a trivial character for my newform, so in that case, are you saying that $V^*(1)$ (in which $A^*$ is a lattice) will come from a twist of $f$? BTW, I'm using the normalization where char. poly. of arithmetic Frob at $\ell\nmid pN$ is $X^2-a_\ellX+\omega^{k-1}\epsilon$. –  Keenan Kidwell Sep 25 '12 at 2:46
Keenan, but is A your finite group $K/mathscr{O}^2$? Then also its dual is finite, as module, and it will definitely have trivial invariants...or? –  Filippo Alberto Edoardo Sep 25 '12 at 7:59
Dear Keenan, if this is what you are aiming for, then I guess you are in good shape. Indeed, the arguments I know to prove the results you want rely on the local Langlands correspondence and the purity of the local Galois representation. Both results are (essentially) invariant under twisting, so you'll be able to describe the local structure of the $G_{F_{v}}$-representation $A$ and/or $A^*$ by essentially the same techniques. –  Olivier Sep 25 '12 at 8:16
Dear @Olivier, Thanks. You've given me hope :) –  Keenan Kidwell Sep 25 '12 at 12:01
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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