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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.

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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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