7
$\begingroup$

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?

$\endgroup$
1
  • 3
    $\begingroup$ So I'm not a reputable source, eh? :) $\endgroup$ Feb 1, 2020 at 13:10

1 Answer 1

6
+250
$\begingroup$

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 $w/2$.

$\endgroup$
6
  • $\begingroup$ thank you for your incredible answer! would you happen to have a precise reference (book or article, page number) for the automorphic Fontaine-Mazur conjecture? I have trouble locating it in the literature. $\endgroup$
    – user145520
    Feb 2, 2020 at 1:29
  • 1
    $\begingroup$ 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. $\endgroup$ Feb 2, 2020 at 10:29
  • $\begingroup$ then do you know historically why is it attributed to these two guys? Did they state in a talk? $\endgroup$
    – user145520
    Feb 3, 2020 at 16:29
  • $\begingroup$ 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 $K$. $\endgroup$
    – SashaP
    Feb 3, 2020 at 20:41
  • $\begingroup$ 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. $\endgroup$ Feb 3, 2020 at 22:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.