Fix a prime number $p$. Can there exist a continuous irreducible representation $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})\to \mathrm{GL}\_3(\mathbb{Q}\_p)$ that is unramified at almost all primes, is de Rham at $p$ and whose Hodge-Tate weights at $p$ are 0, 1 and 3?



Irreducible global Galois representation with weights 0, 1, 3?
Here are two arguments for why such a representation $\rho$ cannot exist.


* Automorphic argument: Fontaine and Mazur have conjectured that any irreducible $n$-dimensional geometric representation $\rho$ of $Gal(\overline{\mathbf{Q}} / \mathbf{Q})$ comes from a cuspidal automorphic representation $\pi$ of $GL\_n(\mathbf{A}\_{\mathbf{Q}})$, and that "local-global compatibility at $\infty$" should hold, which amounts to saying that the Archimedean component $\pi\_\infty$ should be determined by the Hodge–Tate weights of $\rho$ – up to a certain explicit shift, the multiset of Hodge–Tate weights of $\rho$ is the Harish-Chandra parameter of $\pi\_\infty$. However, the possibilities for the representations $\pi\_\infty$ which can show up as Archimedean components of automorphic representations are pretty restricted, so $(0, 1, 3)$ isn't possible. (This is essentially the argument sketched in David Hansen's comment from 2010 that you linked to.)
* Motivic argument: Fontaine and Mazur also made a (separate) conjecture that any such $\rho$ is the $p$-adic realisation of a pure motive over $\mathbf{Q}$. By the comparison isomorphism of Faltings–Tsuji relating étale and de Rham cohomology, this implies that the Hodge–Tate weights of $\rho$ give the graded pieces of a pure Hodge structure. Since a pure Hodge structure has a weight $w$ and an action of complex conjugation which switches the $(p, w-p)$ and $(w-p, p)$ parts, this means the set of weights must be symmetric around $w/2$.


