# Irreducible global Galois representation with weights 0, 1, 3?

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?

• So I'm not a reputable source, eh? :) – David Hansen Feb 1 at 13:10

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 $$w/2$$.
• Fontaine and Mazur only stated the automorphic conjecture for $n = 2$. For the general version, see Conjecture 1.2.1 (4) of Patrikis' monograph "Variations on a theorem of Tate" (AMS Memoirs, 2019) -- you can find a pdf of it on Patrikis' website math.utah.edu/~patrikis/variationsrevision.pdf. – David Loeffler Feb 2 at 10:29
• Could you elaborate on your motivic argument? I don't quite see how to deduce the symmetry of weights because we might be given an object of the category of motives over $\mathbb{Q}$ with coefficients in a field $F$ that has no real embeddings, so we are not getting a real Hodge structure out of it. For instance, if $K$ is a quadratic imaginary field of class number $1$ and $E$ is an elliptic curve over $K$ with CM by $K$ then the motive $h^1(Res_{K/\mathbb{Q}}E)$ splits into 4 motives of weights $0,0,1,1$ in the category of motives with coefficients in $K$. – SashaP Feb 3 at 20:41
• I don't think I agree with your analysis of this CM elliptic curve example. It splits into 2 motives with $K$-coeffs, each of which has weight 1 and Hodge types \{ (0, 1), (1, 0) \}$. Both summands split further when you restrict them to motives over$K\$ with K-coefficients but that's a different thing. – David Loeffler Feb 3 at 22:13