Consider a representation $\rho: \operatorname{Gal} (\overline{\mathbb Q} | \mathbb Q ) \to GL_n ( \overline{\mathbb Q}_\ell)$ that is a subrepresentation of $H^i(X, \overline{\mathbb Q}_\ell (j))$ for $X$ a smooth projective variety over $\mathbb Q$. This is a Galois representation of weight $w = i-2j$.

When do we expect a smooth projective variety $Y$ over $\mathbb Q$ such that $\rho$ shows up in $H^w(Y, \overline{\mathbb Q}_\ell)$?

I can see two obvious necessary conditions. One is that the coefficients of the characteristic polynomial of every Frobenius element had better be algebraic integers. Another is that the Hodge-Tate weights of $Y$ had better be in the interval $[0,w]$.

Do there exist any (conjectural) relations between the three conditions

  1. Actually shows up in cohomology without Tate twisting.
  2. Integral characteristic polynomial of Frobenius
  3. Correct Hodge-Tate weights

other than 1 implying 2 and 3? Should they all be equivalent? Are there any counterexamples?

There are some partial results toward the claim that 2 and 3 imply 1.

Having an integral characteristic polynomial is a sufficient condition in the weight $0$ case. This would imply the eigenvalues of Frobenius are roots of unity. With compactness, they are all roots of unity in a bounded degree extension of $\mathbb Q_\ell$, hence of bounded order, so by Chebotarev all elements of the Zariski closure of the image of $\rho$ have eigenvalues roots of unity of bounded order, hence the image is finite, so it shows up in the cohomology of a $0$-dimensional scheme. I guess I'm counting non-geometrically connected schemes as varieties here.

Having the correct Hodge-Tate weights is a sufficient condition in the weight 0 and 1 case, at least assuming the Hodge and Tate conjectures. By the Tate conjecture we can associate a motive to the Galois representation, to which we can associate a Hodge structure. In the weight 0 case, this Hodge structure is supported in $H^{0,0}$ so all the classes are Hodge classes and hence are algebraic, hence the image of $\rho$ is finite again. In the weight 1 case, this Hodge structure is supported in $H^{1,0}$ and $H^{0,1}$, hence corresponds to an abelian variety. $H^1$ of that abelian variety is a motive with the same Hodge structure as the motive of $\rho$. Because we assumed the Hodge conjecture, it must be the same motive and thus have the same Galois representation. A more detailed version of that argument is in this paper.

I guess the Langlands program will also give some cases of this.


1 Answer 1


Do there exist any (conjectural) relations between the three conditions? Well Grothendieck essentially makes this conjecture, see $\S 2$ p.300 of the following paper:


Especially the second paragraph of p. 301. Most people I have spoken to are fairly agnostic about this conjecture, but it at least suggests that finding a counterexample will not be so easy.

For two dimensional representations, it comes down to the question of whether any modular form $f$ has at least one ordinary prime (if not, then $\rho_{f}(-1)$ satisfies $2$). I think every one expects this to be true and nobody has any idea how to prove it (for weights $\ge 4$). At least $3 \Rightarrow 1,2$ is OK here, and, as you noted, in some [but not all] other contexts where modularity is available)

One can make natural generalizations of the (infinitely many ordinary primes) condition to other motives as well, but since (trusting Grothendieck) we expect the conditions you listed to be equivalent, and since (considering ordinary primes for modular forms) we can't prove it even in the (almost) simplest case, that leaves us pretty much up the creek without a paddle with regard to proving anything.

  • $\begingroup$ Regarding the question of whether every modular form has an ordinary prime, I asked a related question a long time ago: mathoverflow.net/questions/8003/… $\endgroup$ Jun 19, 2015 at 13:22
  • $\begingroup$ Oh, good point! I had seen Grothendieck's generalized Hodge conjecture but I didn't realize that this follows from that, though of course it does. I started thinking about this question precisely because I was trying to understand ordinary primes better. $\endgroup$
    – Will Sawin
    Jun 19, 2015 at 14:21
  • 7
    $\begingroup$ "I was trying to understand ordinary primes better." Good luck to you, son. Make any real progress on this problem and Nick's Moustache will fall off in surprise. $\endgroup$
    – user75178
    Jun 20, 2015 at 4:40

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