# Finiteness or infiniteness for Galois representations with unusual Hodge numbers

Say a representation $$\operatorname{Gal}(\mathbb Q) \to GL_n(\overline{\mathbb Q}_\ell)$$ has big monodromy if the Zariski closure of the image of $$\operatorname{Gal}(\mathbb Q)$$ contains $$SO_n$$ or $$Sp_n$$.

Let $$a_1,\dots a_n$$ be natural numbers with $$a_{n+1-i} =w-a_i$$. Assume that

(1) some natural number that isn't $$w/2$$ appears twice among the $$a_i$$, and

(2) some gap appears among the $$a_i$$.

Do there necessarily exist infinitely many representations $$\rho: \operatorname{Gal}(\mathbb Q) \to GL_n(\overline{\mathbb Q}_\ell)$$, that are finitely ramified, de Rham at $$\ell$$ with Hodge-Tate weights $$a_1,\dots,a_n$$, and with big monodromy, up to twisting by one-dimensional characters?

Do there necessarily exist finitely many such representations?

The idea of the conditions is that (1) should prevent one from proving that there are infinitely many by automorphic methods, as the associated automorphic forms will not be discrete series (I think...) and (2) should prevent one from proving that there are infinitely many by geometric methods, as Griffiths transversality will prevent a family of such representations with big monodromy.

The "up to twisting by one-dimensional characters" prevents finding infintely many by trivial methods and the big monodromy prevents finding many automorphic forms on a different group and applying some representation with multiplicity in its weight spaces.

With these tools removed, are there still infinitely many representations? Or are there finitely many?

• Nice question. In (1), do you mean $w - a_i$, not $w-i$? Dec 14, 2018 at 22:34
• @DavidLoeffler Thanks! And you're of course right. (In fact if that were the condition it would be impossible for (1) or (2) to be satisfied.) Dec 14, 2018 at 22:40
• Another way of constructing such representations is by deformation theory. Although your conditions do make this hard to apply, if you have at least one such representation then it seems quite possible that one can get infinitely many such by deforming the mod $l$ reduction (if this is nice enough).
– naf
Dec 15, 2018 at 5:18

This does not answer the question, but it may help clarify an issue with using deformation theory. Expanding the comment made by Ulrich, one could start with a residual representation $$\bar{\rho}:G_{\mathbb{Q}}\rightarrow \text{GSp}_4(\mathbb{F}_p)$$ which is odd and irreducible and has repeated characters on the diagonal torus. The oddness criterion is defined for instance in the first condition of the main theorem in Lifting irreducible Galois Representations- Fakhruddin, Khare and Patrikis. The reason for choosing $$GSp_4(\mathbb{F}_p)$$ and not $$GL_4$$ is that there are no such odd (in this particular sense) residual representations to $$GL_n(\mathbb{F}_p)$$ for $$n>2$$. The reason for not choosing $$\text{GL}_2(\mathbb{F}_p)$$ is that in this case once a residual representation is odd it has to have distinct characters on the diagonal torus.

Now one must construct an example of such an irreducible representation which satisfies the other conditions in the main theorem of Lifting irreducible Galois Representations- Fakhruddin, Khare and Patrikis. One should take some care in making sure that the multiplicity freeness condition number 3 is satisfied.

Finally, the Theorem tells us that $$\bar{\rho}$$ lifts to a geometric Galois representation $$\rho:G_{\mathbb{Q}}\rightarrow \text{GSp}_4(\mathbb{Z}_p)\subset \text{GL}_4(\mathbb{Z}_p)$$. In fact one can produce infinitely many such lifts ( which is a consequence of the method used). There is however a caveat, which is that one does not have control over what the characters on the torus for the lift are going to be. One only has a choice for what the similitude character of the lift can be. This is described in the statement of Theorem 1.5. Hence as I understand it, these lifts can have distinct characters on the diagonal torus even though the residual representation does not.

• This does not actually answer the question, does it? One gets no control at all of the Hodge--Tate weights of the lift. Dec 15, 2018 at 9:09
• Yes that's right, I believe there is a very different strategy which could be used which is to lift a Galois representation $\bar{\rho}$ which is decomposable, for instance perhaps a Galois representation which is odd and globally a sum of characters.Such Galois representations (far from being irreducible) have not been shown to exhibit geometric lifts but I do think that it's possible to prove geometric lifting theorems for such representations by adapting some of the lifting methods for reducible and indecomposable representations. In this case one may control the characters on the diagonal.
– user130124
Dec 15, 2018 at 10:04
• In the language of deformation theory, when $\bar{\rho}$ is a sum of characters of the diagonal, the adjoint representation decomposes as a Galois module as a direct sum of the adjoint representation from the diagonal and a complement consisting of matrices with zeros on the diagonal.
– user130124
Dec 15, 2018 at 10:07
• Therefore, it is possible to in the case described above fix lifts of characters on the diagonal and study the deformation theory in that case, of course untill one proves that such an adaptation can be made to work the question remains unanswered by this particular approach.
– user130124
Dec 15, 2018 at 10:13