Fix an odd prime $p$ and an integer $r\geq 1$. Fix a Dirichlet character $\chi$ of primitive conductor $p^r$. Let $f\in S_2(p^r,\chi)$ be a normalized newform of weight $2$, level $p^r$ and nebentype $\chi$. Assume $f$ is ordinary at $p$.

Let $K_{f,p}$ be the finite extension of $\mathbb{Q}_p$ generated by the fourier coefficients of $f$. Associated to $f$ there is a Galois representation $$ \varrho_f: G_{\mathbb{Q}} \longrightarrow \mathrm{GL}_2(K_{f,p}) $$ which is unramified at all primes different from $p$. By a theorem of Mazur and Wiles, the restriction of $\varrho_f$ to a decomposition subgroup $D_p$ of $G_{\mathbb Q}$ is of the form $$ \varrho_{f|D_p}: \begin{pmatrix} \alpha_p & \star \\ 0 & \beta_p \end{pmatrix} $$ on a suitable basis. Here $\alpha_p$ and $\beta_p$ are characters of $D_p$.

Pretty much is known about the behavior of these two characters, but a precise description of them depends on the way one has constructed $\varrho_f$. To be sure what are we talking about, let me sketch (and fix) one of the various possible such constructions (all them are the same up to twist):

Let $X_r:=X_1(p^r)/\mathbb{Q}$ denote the modular curve associated to the moduli problem of classifying pairs $(E,i)$ where $E$ is an elliptic curve and $i$ is an isomorphism between the group scheme $\mu_{p^r}$ and a closed finite flat subgroup scheme of $E$. This model of $X_1(p^r)$ over $\mathbb{Q}$ satisfies that the cusp at infinity is a rational point. Don't confuse this model with another natural model $X'_1(p^r)$ this curve has over $\mathbb{Q}$, which arises by replacing the above level structure $i$ by the level structure consisting of the choice of a point of exact order $p^r$ on $E$. Curves $X_1(p^r)$ and $X'_1(p^r)$ are not isomorphic over $\mathbb{Q}$, but one is the twist of the other over $\mathbb{Q}(\mu_{p^r})$.

Let $A_f/\mathbb{Q}$ denote the abelian variety associated by Eichler-Shimura to $f$. This is a simple abelian variety over $\mathbb{Q}$ equipped with a surjection $\mathrm{Jac}(X_r) \rightarrow A_f$. The endomorphism algebra $\mathrm{End}(A_f)\otimes \mathbb{Q}$ is isomorphic to a number field $K$ such that $d=[K:\mathbb{Q}]=\mathrm{dim}(A_f)$. Let $V_p(A_f):=H^1_{\mathrm{et}}(\bar{A}_f,\mathbb{Q}_p)(1)$ denote the $p$-adic Tate module of $A_f$. The Tate twist "$(1)$" by the cyclotomic character is taken here so that $V_p(A_f)$ is Kummer self-dual, that is to say, there is an isomorphism of Galois modules $V_p(A_f) \simeq \mathrm{Hom}(V_p(A_f),\mathbb{Q}_p(1))$. The Galois group $G_{\mathbb{Q}}$ acts on $V_p(A_f)$ in a natural way, giving rise to a representation $r: G_{\mathbb{Q}} \longrightarrow \mathrm{GL}(V_p(A_f))\simeq \mathrm{GL}_{2d}(\mathbb{Q}_p)$. Since the action of $G_{\mathbb{Q}}$ commutes with the endomorphisms of $A_f$, it follows that $r$ factors through $\prod_{\wp} \mathrm{GL}_{2}(K_{\wp})$, where $\wp$ runs over the prime ideals of $K$ over $p$ and $K_{\wp}$ denotes the completion of $K$ with respect to this prime. Projecting to one of these factors yields the sough-after representation, denoted above $$ \varrho_f: G_{\mathbb{Q}} \longrightarrow \mathrm{GL}_2(K_{f,p}). $$

The question I would like to ask is: with this so-defined $\varrho_f$, what is $\alpha_p$ and what $\beta_p$? One of the two is surely the unramified character $\eta_f$ sending $\mathrm{Frob}_p$ to $a_p(f)$ (or its inverse, $\eta_f^{-1}$), and the other encodes also the cyclotomic character and the finite character $\chi$.

There are many papers where one can find explicit formulae describing $\alpha_p$ and $\beta_p$, but most often they state something like "associated to an ordinary newform $f$ there is a Galois representation whose restriction to $D_p$ is ... with $\alpha_p = ...$ and $\beta_p= ...$". My question asks for something more precise: if $\varrho_f$ is the one constructed above, who is $\alpha_p$ and who is $\beta_p$?


1 Answer 1


The usual normalizations for Galois representation attached to an eigencusp form $f$ is either to take the étale cohomology or the usual Tate module of the abelian variety cut out in the Jacobian of the modular curve by the morphism $\lambda_{f}$ sending the Hecke operator $T(\ell)$ to $a_\ell$ such that $T(\ell)f=a_\ell f$. If I understood correctly your careful explanations, your choice of normalization is thus slightly unusual because you chose to take étale cohomology and then twisted by the cyclotomic character.

Under the usual étale normalization, the determinant sends the geometric Frobenius morphism $Fr(\ell)$ for $\ell\neq p$ to $\chi(\ell)\ell$ so under your normalization, the determinant of $V_{p}(A_{f})$ sends the geometric Frobenius $Fr(\ell)$ for $\ell\neq p$ to $\chi(\ell)\ell^{-1}$ (the $-1=2-1-2$ comes from the determinant of the étale representation, contributing a $2-1$ and the Tate twist, contributing a $-2$) so the determinant of $V_p(A_f)$ is $\chi_{cyc}\chi$.

In your situation, neither $\alpha$ nor $\beta$ is $\eta_{f}$. Rather, $\alpha$ is $\chi_{cyc}\eta_f$ and $\beta$ is $\chi\eta_f^{-1}$.

It is not clear to me whether you want a proof of this statement or not, but anyway, here follows a sketch of the simplest that I know (and it is already quite hard, in fact). First notice that it is enough to prove that $\alpha_{et}$, the character appearing in the étale normalization without twist, is $\eta_{f}$. In order to prove this latter claim, first assume that $\pi(f)$ is a principal series at $p$, then appeal to the existence of compatible system of Galois representations, then switch to an auxiliary prime $q$ and then use local-global compatibility of the Langlands correspondence (proved in this case by Deligne, Langlands and Carayol). This compatibility shows that the $p$-divisible group attached to $H^1_{et}$ has a subgroup on which $T(p)$ acts by its invertible eigenvalue (and hence on which $Fr(p)$ acts by the invertible eigenvalue of $T(p)$). In order to prove the claim when $\pi(f)_{p}$ is Steinberg and to complete the proof, use the density of principal series points in a Hida family to reduce to the principal series case.

  • $\begingroup$ Thanks a lot, Olivier, it's a very interesting way of proving that, pretty different to Mazur-Wiles'. Let's use the etale cohomology with no twist if you prefer, in which case you tell me $\alpha=\eta_f$ and $\beta=\chi \chi_{\mathrm{cyc}}^{-1} \eta_f^{-1}$. Where are you using the model of $X_r$ I fixed? If instead of that, we use $X'_r$, what is according to you $\alpha'$ and $\beta'$? Perhaps $\alpha'=\eta_f \chi^{-1}$ and $\beta'=\chi_{\mathrm{cyc}}^{-1}\eta_f^{-1}$? $\endgroup$
    – user37837
    Oct 22, 2013 at 21:58
  • $\begingroup$ Your choice of a model is used in my sketch of proof to choose an adelic compact open subgroup in the Langlands correspondence: with a 1 in the below right corner or in the upper left corner mod $p^r$. Your model corresponds to 1 in the below right corner. Choosing the other model, or the other compact open subgroup, amounts to the twisting involution on GL2 and thus to the Fricke involution on the modular curves. I don't know offhand what happens to $\alpha$ and $\beta$ then and anyway recommend you work out your own normalization carefully. Mistakes are too easy to make otherwise. $\endgroup$
    – Olivier
    Oct 22, 2013 at 22:17
  • $\begingroup$ For what it's worth, the impact of the Fricke involution on the Galois representation is worked out in one of my papers (in a slightly more general context), but I'm quite sure I don't recommend looking it up more than figuring everything out yourself, for the reasons mentioned at the end of my previous comment. $\endgroup$
    – Olivier
    Oct 22, 2013 at 22:19

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