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I am aware that classifying all $SL_2(\mathbb{Z})$ representations is more or less completely intractable, but I was wondering what is known about the following simpler question: How do I recognize when a representation $SL_2(\mathbb{Z}) \to GL_N(\mathbb{C})$ is the restriction of a representation of $SL_2(\mathbb{C})$?

Ideally I'd like the criteria to be some finite collection of properties of how finitely many specific elements of $SL_2(\mathbb{Z})$ act. Here's an example of what I have in mind: If we ask the analogous question for $SL_n(\mathbb{Z})$ with $n\ge3$, then a representation $SL_n(\mathbb{Z}) \to GL_N(\mathbb{C})$ comes from a representation of $SL_n(\mathbb{C})$ if and only if the elementary matrix $E_{1,2}(1)$ acts unipotently.

A few things I have found during a brief search of the literature:

  • There is a low dimensional classification of representations of Tuba and Wenzl in "Representations of the Braid Group $B_3$ and of $SL(2,\mathbb{Z})$". It seems that the unipotency condition is not sufficient when $N = 5$ (obviously it is still necessary), but there are only finitely many other representations with this property.
  • If there is a representation of $SL_2(\mathbb{Z})$ with a compatible action of the Borel $B_2(\mathbb{C})$ then a result of Demazure in "Groupes reductifs de rang semi-simple" says they extend to $SL_2(\mathbb{C})$. This is nice, but I'd still want criteria for when such an action of the Borel can be constructed.

Is this question addressed somewhere in the literature?

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  • $\begingroup$ What kind of representations are you looking at? finite-dimensional ones? $\endgroup$ – YCor Mar 20 '19 at 19:48
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    $\begingroup$ Yes, finite dimensional ones. $\endgroup$ – Nate Mar 20 '19 at 19:51
  • $\begingroup$ "iff an elementary matrix acts unipotently": you mean "every" elementary matrix? $\endgroup$ – YCor Mar 20 '19 at 19:53
  • $\begingroup$ Well by "elementary matrix" I meant the identity matrix with a single extra 1 somewhere off the diagonal. They are all conjugate, so there is no difference. I will try to clarify. $\endgroup$ – Nate Mar 20 '19 at 20:02
  • $\begingroup$ This is because usually $e_{ij}(x)$ is called elementary for arbitrary $x$, and then they're not conjugate in $SL_2(Z)$. For instance in $SL_2(Z)$, $e_{12}(1)$ and $e_{12}(2)$ are not conjugate. So the condition that the second one is mapped to a unipotent element doesn't imply that the first is mapped to a unipotent element. $\endgroup$ – YCor Mar 20 '19 at 21:48

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