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?


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 '20 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.) $\endgroup$ – David Loeffler Jan 22 '20 at 15:53

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