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Let $E$ be an elliptic curve over $\mathbb{Q}_\ell$ and consider the image of the $\ell$-adic representation. Is there a description of this image similar to Serre's description of the image of the mod $\ell$ representation? (I.e., which depends on whether $E$ has e.g. bad multiplicative/good ordinary/good supersingular reduction?)

I'm looking for a description similar to Serre's description of the mod $\ell$ image (see e.g. Lemma 3.4 of this).

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  • $\begingroup$ 1. Do you want the image up to conjugacy in $GL_2(\mathbb Z_\ell)$ or $GL_2(\mathbb Q_\ell)$? 2. For bad multiplicative and good ordinary reduction the image can be easily seen to be upper-triangular, hence an extension of $\chi_1$ by $\chi_2$ where $\chi_1\chi_2$ is the cyclotomic character, and in both cases one can compute one of the characters, so the only difficulty is understanding the extension. $\endgroup$
    – Will Sawin
    Mar 22, 2016 at 16:43
  • $\begingroup$ Understanding the extension: in the good ordinary case the ratio of the characters isn't cyclotomic because of e.g. the Hasse bound, so there's only one non-split extension, however I know of no simple way to figure out whether the extension is split or not -- am I missing something? In the bad multiplicative case you can use the Tate curve to see that the extension is controlled by the $j$-invariant (in a precise sense using Kummer theory). In the supersingular case if $a_\ell=0$ you're induced from a quadratic extension. $\endgroup$
    – znt
    Mar 22, 2016 at 20:42
  • $\begingroup$ If $a_\ell$ is non-zero, but zero mod $\ell$ (which can happen if $\ell=2$ or 3) then the Galois representation is irreducible and crystalline and I would bet my bottom dollar that it admitted no concrete description at all, although the associated Fontaine module is easy enough to write down and this is sometimes enough for applications. Finally in the additive case I guess I'd make a base extension until we were semistable and then try and run through the same arguments as above. In some sense the most impenetrable case then is the non-ordinary case with $a_\ell\not=0$. $\endgroup$
    – znt
    Mar 22, 2016 at 20:44
  • $\begingroup$ @WillSawin GL2(Z_l) (though, Q_l might be useful too). $\endgroup$ Mar 23, 2016 at 18:23
  • $\begingroup$ The image may be hard to compute, but the Lie algebra of the image is known. See the appendix to Serre's book "Abelian ell-adic representations and elliptic curves". $\endgroup$ Mar 30, 2016 at 7:28

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