Let $f=\sum_{n=1}^\infty a_nq^n$ be a $p$-ordinary newform of weight $k\geq 2$, level $N$, and character $\chi$, and let $\rho_f:G_\mathbf{Q}\rightarrow\mathrm{GL}_2(K_f)$ be the associated $p$-adic Galois representation, where $K_f$ is the finite extension of $\mathbf{Q}_p$ obtained by adjoining the Fourier coefficients of $f$. Let $\mathscr{O}_f$ be the ring of integers of $K_f$, and $A_f$ a cofree $\mathscr{O}_f$-module of corank $2$, i.e., $(K_f/\mathscr{O}_f)^2$, on which $G_\mathbf{Q}$ acts by $\rho_f$ (so we've chosen an integral model of $\rho_f$).

My question involves the local invariants of $A_f$. Specifically, let $F$ be a number field, and let $v$ be a finite prime of $F$ not dividing $p$ or the conductor of $\rho_f\vert_{G_F}$. Fix a decomposition group $G_v$ of $v$ in $G_F\leq G_\mathbf{Q}$. Is it true that $H^0(G_v,A_f)$ is finite?

I'm really interested in whether or not $\ker(H^1(G_v,A_f)\rightarrow H^1(I_v,A_f))$ vanishes ($I_v\leq G_v$ the inertia group), but with my hypotheses on $v$, the vanishing of this kernel is equivalent to the finiteness of $H^0(G_v,A_f)$ (because the kernel in question is divisible of the same $\mathscr{O}$-corank as $H^0(G_v,A_f)$). This vanishing seems to be implicit in a couple papers I've been looking at, and I'm not sure why it's true.


If the invariants were infinite, they would be divisible, and so they would correspond to an invariant line in $V_f$ (the representation on $K_f^2$ attached to $\rho$). Let $\ell$ be the rational prime lying undre $v$. The char. poly. of $\mathrm{Frob}_{\ell}$ acting on this rep'n is exactly the $\ell$th Hecke polynomial, so by Ramanujan--Petterson, the eigenvalues of $\mathrm{Frob}_{\ell}$ are Weil numbers of weight $(k-1)/2$. In particular, they are not roots of unity (provided the weight $k > 1$). The eigenvalues of $\mathrm{Frob}_v$ are powers of the eigenvalues of $\mathrm{Frob}_{\ell}$ (since $\mathrm{Frob}_v$ is a power of $\mathrm{Frob}_{\ell}$), and so they cannot be $1$. Consequently, $H^0(G_v,V_f) = 0$. QED

(If $k = 1$ this argument breaks down, and of course the statement is false.)

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    $\begingroup$ Thanks, Matt. Just so I understand correctly, you're saying that a copy of $K_f/\mathscr{O}_f$ in $A^{\mathrm{Frob}_v=1}$ would give me a copy of $K_f$ in $V_f$ on which $\mathrm{Frob}_v$ acts trivially, which can't happen because $\mathrm{Frob}_v$ does not have $1$ as an eigenvalue? $\endgroup$ – Keenan Kidwell Aug 7 '12 at 19:01
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    $\begingroup$ @Keenan: Dear Keenan, Yes, that's exactly what I'm saying (but less succinctly!). Cheers, $\endgroup$ – Matt E Aug 7 '12 at 19:11
  • $\begingroup$ Great. Thanks very much! $\endgroup$ – Keenan Kidwell Aug 7 '12 at 19:12
  • $\begingroup$ Dear @Matt, If $E/\mathbf{Q}$ is an elliptic curve with $F$-rational $p$-torsion and $f$ is the associated eigenform, then won't $A_f$ (or possibly it's Cartier dual, depending on normalizations) have non-zero $G_v$-invariants (but still only finitely many by your argument)? $\endgroup$ – Keenan Kidwell Aug 30 '12 at 1:59
  • $\begingroup$ Dear Keenan, Yes, I think so (if I've understood your set-up correctly), although you have to be a little careful, because (depending on what you mean by $A_f$), the curves $E$ and $A_f$ will only be isogenous a priori, not necessarily isomorphic, and the $F$-rational $p$-torsion you have in mind may be killed off in the isogeny. Regards, $\endgroup$ – Matt E Aug 30 '12 at 3:11

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