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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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closed as not a real question by Qiaochu Yuan Nov 28 '11 at 23:12

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question.

    
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
1  
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

1 Answer 1

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 $GL_2(\mathcal{O}_K)$. –  M Turgeon Nov 27 '11 at 3:12
    
@M Turgeon: Dear M Turgeon, See my additional remarks. Regards, –  Matt E Nov 27 '11 at 12:36

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