Natural rep of $ SL(2,\mathbb{Z})$ and decomposing tensor powers into irreps

Question

$\DeclareMathOperator\SL{SL}$ Consider the natural representation of $ \SL(2,\mathbb{Z}) $ on $ V=\mathbb{C}^2 $.

What is the decomposition of $ V^{\otimes n} $ into irreps of $ \SL(2,\mathbb{Z}) $?

I know about the decomposition of $ (\mathbb{C}^2)^{\otimes n} $ into irreps for the natural representation of $ \SL(2,\mathbb{C}) $.

The $ \SL(2,\mathbb{Z}) $ decomposition must be at least that much, but maybe there is further branching of some $ \SL(2,\mathbb{C}) $ irreps into direct sums of $ \SL(2,\mathbb{Z}) $ irreps?

Answer 1

It's just the same as for $SL_2(\mathbb{C})$. Restricting an algebraic representation of $SL_2(\mathbb{C})$ to $SL_2(\mathbb{Z})$ preserves irreducibility. The key fact here is that $SL_2(\mathbb{Z})$ is Zariski dense in $SL_2(\mathbb{C})$.