# Cusps of hyperbolic surfaces under finite covers

The following statement seems true, but I don't know a proof or a reference for it (and I would like one).

Let $$\Gamma< \operatorname{PSL}(2,\mathbb R)$$ be a nonuniform lattice with one cusp. We may conjugate $$\Gamma$$ so that an element $$\begin{pmatrix} 1 & s\\ 0 & 1 \end{pmatrix}$$ generates the cusp of $$\Gamma$$. Let $$\Gamma'$$ be such that $$\Gamma<\Gamma'$$ be an index $$2$$ subgroup. Then, the element $$\begin{pmatrix} 1 & s/2\\ 0 & 1 \end{pmatrix}$$ generates the cusp of $$\Gamma'$$.

I imagine a similar statement would be true if both $$\Gamma$$ and $$\Gamma'$$ had two cusps.

• I think you should have -1 on the diagonal. – Moishe Kohan Sep 9 '20 at 22:18
• Sorry I meant that $\Gamma< \operatorname{PSL}(2,\mathbb{R})$, and I edited the question to reflect this. So I don't think such things matter. – Chris Z Sep 11 '20 at 15:04

Assume just that $$\Gamma$$ has index $$k$$ in $$\Gamma'$$. Let $$C \subset \mathbb R \cup \{\infty\}$$ be the set of parabolic points for the action of $$\Gamma$$. Then $$C$$ is also the set of parabolic points for the action of $$\Gamma'$$, because if $$\gamma \in \Gamma'$$ is parabolic with fixed point $$x$$ then for some integer $$i \ge 1$$, $$\gamma^i \in \Gamma$$ is parabolic with the same fixed point $$x$$.
In general every $$\Gamma'$$ orbit of $$C$$ decomposes as a union of $$\Gamma$$ orbits, so $$\#\text{cusps}(\Gamma') \le \#\text{cusps}(\Gamma)$$ with equality if and only if the every $$\Gamma'$$ orbit of $$C$$ is just a single $$\Gamma$$-orbit. So assuming that $$\#\text{cusps}(\Gamma') = \#\text{cusps}(\Gamma)$$ it follows for each $$x \in C$$ the group $$\text{Stab}(x;\Gamma)$$ has index $$k$$ in $$\text{Stab}(x;\Gamma')$$.
For the case $$x=\infty \in C$$ it follows that if $$\begin{pmatrix} 1 & s\\ 0 & 1 \end{pmatrix}$$ generates $$\text{Stab}(\infty;\Gamma)$$ then $$\begin{pmatrix} 1 & s/k\\ 0 & 1 \end{pmatrix}$$ generates $$\text{Stab}(\infty;\Gamma')$$.