# 2-dimensional $\ell$-adic representations [closed]

In an assignment, I have to give an example of a 2-dimensional $\ell$-adic representation of the absolute Galois group of $\mathbb{Q}$, bu I am faced with the problem that I do not a lot of these. Or not enough to find the example I am looking for.

More precisely, let $G$ be the absolute Galois group in question, and let $K$ be a (fixed) finite extension of $\mathbb{Q}_\ell$, for some prime $\ell$. I am looking for a representation $$\rho : G \to GL_2(\bar{K}).$$

The examples I do know are the trivial representation and the ones arising by taking the direct sum or the tensor product of characters (that is, 1-dimensional representations) of $G$. The problem is that they do not seem to give what I am looking for.

Added: Most $\ell$-adic representations that arise from geometry have image lying in $GL_2(\mathcal{O}_K)$. Although these are very interesting representations, and that (in some sense) you can always reduce it to this case (see BR comments below), what I am looking for are representations that cannot be conjugated (by any element of $GL_2(\bar{K})$) in such a way that the image lies in $GL_2(\mathcal{O}_K)$.

The reason the original phrasing was ambiguous and unclear is that I was hoping to build a better repertoire of Galois representations, in the hope of eventually finding an example with the desired properties.

• Is there a reason why the $\ell$-adic Tate module of an elliptic curve would not suffice? – B R Nov 26 '11 at 22:15
• @ BR: Yes. Basically, I don't want the image to lie in $GL_2(\mathcal{O}_K)$. – M Turgeon Nov 26 '11 at 22:48
• Could you clarify what you are looking for in a representation? For example, since $G$ is compact, the image of the representation will be a compact subgroup of $GL_2(K)$, for some finite extension $K$ of $\mathbb Q_\ell$, so will lie in $GL_2(\mathcal O_K)$. – B R Nov 26 '11 at 23:24
• Well, I don't want to give to many details, because they would take to much space, but also I don't want to be given the answer. But as for your comment, I believe that what happens is that the image lies in some finite extension $L/K$, and so the image (after possibly conjugation) will lie in $\mathcal{O}_{L}$, but nothing forces $L$ to be equal to $K$. Or maybe I missed something... – M Turgeon Nov 26 '11 at 23:40
• If you want the representation to be continuous, then, as B R points out, the image will be conjugate to a subgroup of $\mathrm{GL}_2(K)$ for $K/\mathbf{Q}_\ell$ finite, and any continuous representation $G\rightarrow\mathrm{GL}_2(K)$ will stabilize an $\mathcal{O}_K$-lattice, hence will have image conjugate to a subgroup of $\mathrm{GL}_2(\mathcal{O}_K)$. – Keenan Kidwell Nov 27 '11 at 1:24

The question remains unclear. As BR and Keenan Kidwell have pointed out, any continuous Galois representations with values in $GL_2(K)$ (with $K$ a finite extension of $\mathbb Q_{\ell}$) can always be conjugated by an element of $GL_2(K)$ into $GL_2(\mathcal O_K)$.
Added in response to the comment below: If a continuous Galois rep. takes values in $GL_2(\overline{K})$, then it lies in $GL_2(L)$ for some finite extension $L$ of $K$, and so lies in $GL_2(\mathcal O_L)$ after conjugating by an element of $GL_2(L)$. So this changes nothing, except for replacing the label $K$ by the label $L$ (and since $K$ was arbitrary, this is not a real change).
• As added above (since it seems it wasn't clear in the original phrasing), the Galois representation lies in $GL_2(\bar{K})$, so with coefficients in an algebraic closure of $K$. Hence, we can't conclude that it will fall in $GL_2(\mathcal{O}_K)$. – M Turgeon Nov 27 '11 at 3:12