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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.

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    Is there a reason why the $\ell$-adic Tate module of an elliptic curve would not suffice?2011-11-26
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    @ BR: Yes. Basically, I don't want the image to lie in $GL_2(\mathcal{O}_K)$.2011-11-26
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    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 $GL_2(\mathcal O_K)$.2011-11-26
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    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...2011-11-26
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    I agree with your analysis of the situation, I just felt that the question was ambiguous enough to possibly allow that interpretation. Would, for example, conjugating a representation whose image is in $GL_2(\mathcal O_K)$ by an element of $GL_2(\bar K)$ (not in the center or in $GL_2(\mathcal O_K)$) work?2011-11-27
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    Ok, I see that the question is really ambiguous. I'll try to add some details.2011-11-27
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    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 $\mathrm{GL}_2(\mathcal{O}_K)$.2011-11-27
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    M Turgeon, it seems to me that you are looking for conditions or examples in which the representation genuinely lies in a finite extension of some fixed base field. Is that correct?2011-11-27
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    Unfortunately I can't edit my comment now, but $K$ should be called $L$ and $\mathbf{Q}_\ell$ should be called $K$.2011-11-27
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    M Turgeon, a potentially interesting (I don't know whether it is interesting or not) question somewhat near to what you are asking would be to ask for examples of "exotic" $\ell$-adic representations (as well as what qualifies as exotic). This may not net you the example you want, but it might indicate the range of outcomes that exist and are considered interesting. I'll leave it to you to decide if it is worth asking or not.2011-11-27
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    (I should note that, now that I understand the question, I know that I don't know an answer to it, though I'm sure other people do.)2011-11-27
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    As I pointed out in my answer below, I still don't really understand the question. The label $K$ seems to have been chosen arbitrarily, and so it seems we may just as well replace it by $L$. Do you have some fixed $K$ in mind? Regards,2011-11-27
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    @MattE I don't know why you think $K$ is arbitrary. In my question, it is clearly written that $K$ is a fixed finite extension. And it's been so since the first revision.2011-11-27
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    Dear M Turgeon, Then I am again confused by the question. Having fixed $K$, just choose a representation with values in $L$ (for $L$ a proper finite extension of $K$) which is not defined over $K$. Is this what you want? Regards,2011-11-27
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    @MattE I see now that the problem, as stated, is much easier than I first thought. Although a representation such as the one you propose would not answer the particular question I am being asked in my assignment, I feel like my question has confused too many people as it is, and so I will not reformulate the question another time to include the extra hypotheses. I will just go back to my homework. Thank you all for your time, and sorry for the confusion.2011-11-27

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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 $GL_2(\mathcal O_K)$.

What exactly is is that you want?

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).

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    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 $GL_2(\mathcal{O}_K)$.2011-11-27
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    @M Turgeon: Dear M Turgeon, See my additional remarks. Regards,2011-11-27