# Galois representations with trivial determinant that do not factor through a number field

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

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.

• 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? – user145520 Jan 22 '20 at 9:13
• 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.) – David Loeffler Jan 22 '20 at 15:53