Representations of $PSL_2(Z)$ and $\Gamma(2)$.

I would like a finite dimensional irreducible representation of ${PSL}_2(\mathbb{Z})$ which contains a proper sub-representation when restricted to the congruence subgroup $\Gamma(2)$, the kernel of $\mathrm{mod}_2:{PSL}_2(\mathbb{Z})\to {PSL}_2(\mathbb{Z}_2)$.

This would be used in constructing exact sequences of differential calculi on group algebras, and examining what relations exist between the differential invariants and the group structure. A seriously infinite noncommutative example would be really useful.

I have read the paper on the irreps of $B_3$ and ${PSL}_2(\mathbb{Z})$ by Imre Tuba and Hans Wenzl, but after calculations on some (I have not tried all of them!) of their examples, I cannot get a reducible rep for $\Gamma(2)$ which was not already reducible for ${PSL}_2(\mathbb{Z})$.

• Since the quotient is non-abelian finite group of order 6, it has an irreducible 2-dimensional rep. The resulting rep of $SL_2(Z)$ is irreducible and trivial (hence non-irreducible) in restriction to the kernel.
– YCor
Oct 4 '16 at 14:21
• This was the first thing that occurred to me; I thought the OP had in mind the case when the restriction to the subgroup contains non-trivial irrep, and did not mention this Oct 4 '16 at 16:04

Take the standard representation $\rho$ of $PSL(2,\mathbb{Z}$ on $\mathbb{C}^3$ and tensor it with a representation $\theta$ (of dimenstion not one) of the finite group $PSL_2(\mathbb{Z}/2\mathbb{Z})$ (such a representation exists since the group $PSL_2(\mathbb{Z}/2\mathbb{Z})$ is not abelian). Take $\pi =\rho \otimes \theta$; restriced to $\Gamma (2)$, this is a sum of copies of $\rho$.
The connected component of the Zariski closure of $\Gamma$ and $\Gamma (2)$ are the same both for $\rho$ and for $\pi$; hence for $\rho$ the Zariski closure of $\Gamma (2)$ is $G=PSL(2,\mathbb{C})\simeq SO(3,\mathbb{C})$.Thus the Zariski closure for $\Gamma$ in the larger representation $\pi$ is $G\times PSL(2,\mathbb{Z}/2\mathbb{Z})$, which shows that $\pi$ is irreducible since the latter group acts irreducibly on $\rho \otimes \theta\simeq \pi$.