## Explicitly describing a two-dimensional reducible representation of G_{Q_p}

Let $j$ be some integer which is neither 0 nor 1. The Galois cohomology space $H^1({\mathbb Q}_p,{\mathbb Q}_p(j))$ is then 1-dimensional, and thus there is a unique non-split 2-dimensional local Galois representations $\rho_j$ with ${\mathbb Q}_p(j)$ as a sub and ${\mathbb Q}_p$ as a quotient.

I think that $\rho_j$ is crystalline if $j>1$, and not crystalline if $j<1$. (One can just write down all of the 2-dimensional reducible admissible modules, and this should follow...I hope.)

Now take $\rho_j$ and restrict it to the absolute Galois group of ${\mathbb Q}_p(\mu_p^\infty)$. This restricted representation then looks like $\begin{pmatrix} 1 & \psi \newline 0 & 1 \end{pmatrix}$ when $\psi$ is some additive character of $G_{{\mathbb Q}_p(\mu_p^\infty)}$.

My question: Should we be able to explicitly describe this character $\psi$? For instance, is $\psi$ an unramified character?

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 Can't this be done for most $j$ using Soule's cocycle, that is, for all $j$ such that $\zeta_p(j)\neq 0$? (Perhaps you don't need the non-vanishing for this, but only when you're looking at some other extension coming from the projective line minus three points. I'm too lazy now to look this up.) – Minhyong Kim Dec 15 2010 at 3:16 If $j \ne 0$, then all such extensions are ramified. Indeed, the inertia group has open image. The image of inertia is normal in the image of $\rho$, and the quotient of the image is cyclic. Yet, by inspection, the only cyclic unramified quotients that could occur are annihilated by $p^n$, where $n = v(j)$. I'm not sure what you mean by "explicit". By Kummer theory, the mod-$p^n$ extension comes from taking the $p^n$th root of a unit in the $\omega^{1-j}$-part of $\Q_p(\zeta_{p^n})^{\times}/\Q_p(\zeta_{p^n})^{\times p}$, which is essentially unique when $j \ne 0,1$. – Lavender Honey Dec 15 2010 at 4:56 Although it's not very explicit, you can say that $H^1(G_{Q_p(\mu_{p^\infty})},Qp)$ is a $\Lambda$-module and by the inflation-restriction map you character is in the part on which $\Gamma$ acts by $\chi^j$. If you're wondering what $H^1(G_{Q_p(\mu_{p^\infty})},Qp)$ looks like, the theory of $(\varphi,\Gamma)$-modules may help... – Laurent Berger Dec 15 2010 at 8:31