In arithmetic geometry one often encounters continuous representations $\rho:\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})\to \operatorname{GL}_n(\mathbb{Q}_l)$ for some $n\geq 1$ and some prime number $l$ such that no Tate twist of $\rho$ factors through $\operatorname{GL}_n(K)$ for any number field $K$. An example is the first $l$-adic cohomology of an elliptic curve over $\mathbb{Q}$: if $\rho \otimes \mathbb{Q}_l(i)$ were to factor then $\wedge^2 (\rho\otimes \mathbb{Q}_l(i))\approx \wedge^2 \rho \otimes \mathbb{Q}_l(2i)\approx \mathbb{Q}_l(2i-1)$ would factor as well which is a contradiction since it is infinitely ramified.

This determinant trick is the only argument I know for proving that a representation does not factor. Are there explicit examples of representations unramified outside a finite set of primes and de Rham at $l$ with a trivial determinant and trivial Hodge-Tate weights that do not factor? Do such representations arise naturally in number theory or arithmetic geometry?


1 Answer 1


Since you mention the term "de Rham" in your question, you are clearly aware of the existence of p-adic Hodge theory; so I am surprised that you do not realise that this theory allows you to write down examples without any effort at all.

The point is that when a representation arises naturally in geometry, then you can read off its Hodge--Tate weights (i.e. the jumps in the filtration of its de Rham space) from the Hodge numbers of the variety. On the other hand, any representation that factors through a finite quotient of the Galois group has to have all its HT weights zero.

So e.g. if $W$ is $H^1$ of an elliptic curve, and $V$ is the quotient of $W \otimes W^\vee$ by its obvious 1-dimensional trivial subrep, then $V$ has trivial determinant, but its Hodge-Tate weights are $\{-1, 0, 1\}$.

EDIT. The question has now been rather radically changed by adding the assumption that the representation has trivial Hodge--Tate weights. This is a very different matter. It is expected that no such representations exist, and this is equivalent to the "tame Fontaine-Mazur conjecture", a difficult open problem which has been intensively studied by Boston among others.

  • $\begingroup$ I now realize that I forgot to ask for irreducibility of the representation (otherwise one can take the direct sum of $H^1(E, \mathbb{Q}_l)$ and the cyclotomic character). Is your 3-dimensional representation irreducible? $\endgroup$
    – user145520
    Jan 22, 2020 at 9:13
  • $\begingroup$ If the elliptic curve does not have CM then $V$ is irreducible. (This is easily seen from Serre's open image theorem, although that is using a sledgehammer to crack a nut.) $\endgroup$ Jan 22, 2020 at 15:53

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