Congruence subgroups from representations of the three-strand braid group (Q2012032)

Let \(A\) be an \(n \times n\) matrix with eigenvalues \(\lambda_1, \lambda_2, \ldots, \lambda_n\). Define the projective order of \(A\) to be \[ \operatorname{po}(A) = \min \{t > 1\mid \lambda_1^t = \ldots = \lambda_n^t \} \] where this is allowed to be infinite. The projective order of a matrix is invariant under scaling, that is, \(\operatorname{po}(A) = \operatorname{po}(\theta A)\) for all \(\theta \in \mathbb{C}^*\). The three-strand braid group \(B_3 = \langle \sigma_1, \sigma_2\mid \sigma_1 \sigma_2 \sigma_1 = \sigma_2 \sigma_1 \sigma_2 \rangle\) is the central extension: \[ 1 \to \mathbb{Z} = \langle (\sigma_1 \sigma_2)^3 \rangle \to B_3 \overset{\pi}{\to} \mathrm{PSL}(2, \mathbb{Z}{)} 1. \] Suppose that \(d = 2\) or \(3\) and \(\rho : B_3 \to \mathrm{GL}(d, \mathbb{C})\) is a \(d\)-dimensional representation with finite image that factors through \(\pi\). The first main result of the present paper approves that if \(2 \leq \operatorname{po}(\rho(\sigma_1)) \leq 5\) then \(\pi(\operatorname{ker} \rho)\) is a congruence subgroup with level equal to the order of \(\rho(\sigma_1)\). The second main result points out that for three dimensional representations, once the projective order of \(\rho(\sigma_1)\) is greater than five, the projective order is not enough to determine the congruence properties of the induced kernel. More accurately, for any positive integer of the form \(2l > 2\) with \(l\) odd, there are two representations \(\rho_{l,\pm} : B_3 \to \mathrm{GL}(3, \mathbb{C})\) with finite image that factors through \(\pi\) for which \(\operatorname{po}(\rho_{l,\pm}(\sigma_1)) = 2l\) and each \(\pi(\operatorname{ker} \rho_{l,\pm})\) is not a congruence subgroup.