If it had Hodge-Tate weights all 0, wouldn’t the image of inertia be finite? So it would be de Rham.

@dragoboy I'm referring to Theorem 3.1 on p191. You're right that the result on p186 only applies to level $1$ forms.

@olivier Do you know of anywhere where this is discussed? How much is known? And what is actually expected?

Thanks again. I guess what I'm after is an intrinsic way to work out if $G/N$ is solvable, without actually knowing what $N$ is! I can actually reduce to ruling out the case where $G/N = A_5$, since, after enlarging $N$, $G/N\hookrightarrow\mathrm{Aut}(M_2(k)) = \mathrm{PGL}_2(k)$).

As you say, the situation is indeed simple in the finite case! In the infinite case, it comes down to whether $G/N$ has an index two subgroup. Is there any information that is intrinsic to the representation that I can make use of? I know nothing about $N$ a priori.

That's only true if $k =\mathbb C$ or a finite field. Typically, $k$ will be something like $\overline{\mathbb Q}_p$ for me, and the image won't be finite.

@JohannesHahn I really care about Galois representations, so $G = \mathrm{Gal}(\overline K/K)$ is profinite, and representations are continuous.

My motivation behind the question is this: I have a four dimensional representation $\rho$ which decomposes as $\sigma + \sigma$ after restriction to $N$. I only know that $N$ is a finite index subgroup, but nothing else. I'd like to know under what circumstances there exists an $H$ and a rep $\sigma'$ of $H$ such that $\rho =\mathrm{Ind}_H^G\sigma'$.

Thanks for your answer! I'm not sure the corollary helps: if $\mathrm{End}(\mathrm{Res}^G_H(V)) = k\times k$, then enlarging $H$ if necessary, $V=\mathrm{Ind}^G_H(U)$, where $U$ is one of the two distinct subreps. For any $g\in G$, wouldn't $\chi_V(g)$ usually (always?) be even?