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Let $\Gamma = PSL(2,\mathbb{Z}) = \langle S,T \ | \ S^2=(ST)^3=1 \rangle$. Let $G$ be some mystery normal subgroup of $\Gamma$ that we happen to think may be congruence. Recall that a subgroup of $\Gamma$ is a congruence subgroup of level $N$ if it contains the principal congruence subgroup of level $N$.

Are there methods for establishing upper bounds on the possible levels of $G$, say given information about $\Gamma/G$? Establishing lower bounds is trivial: the minimal $N$ such that $T^N \in G$ (if such an $N$ exists) is the minimal possible level of $G$.

For example, suppose $\Gamma/G \cong (\mathbb{Z}_3 \times \mathbb{Z}_9) \rtimes S_3$ or more generally of the form (finite abelian group)-semidirect product-(symmetric group). I'm generally thinking about the case where $G$ is the kernel of some unitary representation with finite image.

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If you only look at upper bounds, you can use the result by Lubotzky: For every algebraic group G, there is some $c$, such that the level of a congruence subgroup $U$ is bounded by the $c(G:U)$.

If you want to determine the level algorithmically, you can look at the image of the subgroup in $PSL(\mathbb{Z}/p^n\mathbb{Z})$. Determine the cusp split of the subgroup, if there are two cusp width $c_1, c_2$, and $p$ a prime with $p\nmid c_1c_2$, then $U$ contains different parabolic elements of order coprime to $p$, hence $U\Gamma(p)/\Gamma(p)$ contains two different parabolic subgroups, hence $U\Gamma(p)/\Gamma(p)=PSL(\mathbb{F}_p)$, thus $p$ does not divide the level.

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  • $\begingroup$ Does "cusp split" mean the orbits of the rationals and infinity under the subgroup? $\endgroup$
    – Joseph Ricci
    Commented Nov 22, 2015 at 18:53
  • $\begingroup$ $\mathbb{H}/\Gamma$ has one point at infinity. If $\Delta<\Gamma$, then $\mathbb{H}/\Delta$ has several points $p_1, \ldots, p_k$. Each $p_i$ has a neighbourhood intersecting $c_i$ copies of $\mathbb{H}/\Gamma$, where $\sum_i c_i=(\Gamma:\Delta)$. The partition $c_1, \ldots, c_k$ of $(\Gamma:\Delta)$ is the cusp split of $\Delta$ and is easy to compute, see Millington, Subgroups of the classical modular group, J. LMS 1 (1969) 351-357. $\endgroup$
    – Jan-Christoph Schlage-Puchta
    Commented Nov 23, 2015 at 13:08

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