# Recognizing a restriction from $SL_2(\mathbb{C})$ to $SL_2(\mathbb{Z})$

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?

• What kind of representations are you looking at? finite-dimensional ones? – YCor Mar 20 '19 at 19:48
• Yes, finite dimensional ones. – Nate Mar 20 '19 at 19:51
• "iff an elementary matrix acts unipotently": you mean "every" elementary matrix? – YCor Mar 20 '19 at 19:53
• 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. – Nate Mar 20 '19 at 20:02
• 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. – YCor Mar 20 '19 at 21:48