For which quadratic number field, the algebraic integers are cusps for some Coxeter group?

Question

1. Let $H^2=\{(x,y)\mid y>0\}$ be the hyperbolic upper-half plane.
2. Let $K=Q(\sqrt{d})$ be a quadratic number field, and $\mathcal{O}_K$ be the ring of algebraic integers in it.
3. Let $\Gamma=\Delta(p,q,r)$ be a triangle group of hyperbolic type, whose fundamental domain at least one ideal vertex on the boundary of $H^2$. That is, at least one of $p,q,r$ is $\infty$. $\Gamma$ gives a hyperbolic tiling of $H^2$ by non-compact triangles.

Here is an example: Let $\Gamma=(2,q,\infty)$ be a Hecke group where $q\geq3$ is an integer. The fundamental domain $D$ of $\Gamma$ has vertices at $i, e^{i\pi/q},\infty$. The triangles $\{w(D)\mid w\in\Gamma\}$ tile $H^2$. The cusps are the set $\{w(\infty)\mid w\in\Gamma\}$.

My question is this:

For which choices of $d$ does there exist a triangle group $\Gamma$ such that the cusps in the tiling are exactly $\mathcal{O}_K$? That is, $\mathcal{O}_K=\{w(\infty)\mid w\in\Gamma\}$?

Answer 1

This is a partial answer to say that this won't happen very often.

Let's work with the triangle group where $s_1 s_2$ has order $p$, $s_2 s_3$ has order $q$ and $s_1 s_3$ has order $\infty$. Let $\omega_2$ be the ideal vertex of the base triangle, so $\omega_2$ is fixed by $s_1$ and $s_3$. Consider the four points $\omega_2$, $s_2 \omega_2$, $s_1 s_2 \omega_2$, $s_3 s_2 \omega_2$ in $\partial H^2 \cong \mathbb{RP}^1$. (Note that, if $p=2$, then $s_1 s_2 \omega_2 = s_2 s_1 \omega_2 = s_2 \omega_2$, so we only have three distinct points.) In this answer, I'll compute the cross ratio of these points and get that it is $\tfrac{\cos(\pi/q)}{\cos(\pi/p)}$. In particular, if we want these points to lie in $\mathbb{P}^1(\mathbb{Q}(\sqrt{d}))$, then we must have $\tfrac{\cos(\pi/q)}{\cos(\pi/p)} \in \mathbb{Q}(\sqrt{d}) \cup \{ \infty \}$. At some point I might come back and carefully do the Galois theory to work out all the cases where this happens but, for typical $(p,q)$, it certainly doesn't.

The computation: Let $A$ be the Cartan matrix: $$A = \begin{bmatrix} 2 & -2 \cos(\pi/p) & -2 \\ -2 \cos(\pi/p) & 2 & - 2 \cos(\pi/q) \\ -2 & -2 \cos (\pi/q) & 2 \\ \end{bmatrix}.$$ Let $\alpha_1$, $\alpha_2$, $\alpha_3$ be the simple roots, so we have $\langle \alpha_i, \alpha_j \rangle = A_{ij}$ for the invariant dot product. Let $\omega_1$, $\omega_2$, $\omega_3$ be the dual basis, so $\langle \alpha_i, \omega_j \rangle = \delta_{ij}$. The action of the Coxeter group in the dual basis is $$s_i \omega_j = \begin{cases} - \omega_i - \sum_{k \neq i} A_{ik} \omega_k & i=j \\ \omega_j & i \neq j \\ \end{cases}.$$ I compute that $$\begin{array}{rcrcrcr} \omega_2 &=& && \omega_2 && \\ s_2 \omega_2 &=& -A_{12} \omega_1 &-& \omega_2 &-& A_{23} \omega_3 \\ s_1 s_2 \omega_2 &=& A_{12} \omega_1 &+& (A_{12}^2-1) \omega_2 &-& (2 A_{12} + A_{23}) \omega_3 \\ s_3 s_2 \omega_2 &=& - (A_{12} + 2 A_{23})\omega_1 &+& (A_{23}^2-1) \omega_2 &+& A_{23} \omega_3 \\ \end{array}$$

Now, these coordinates are in $\mathbb{R}^3$, or I can think of them as homogeneous coordinates in $\mathbb{P}^2$. I want to compute their images under the isomorphism $\partial(H^2) \to \mathbb{RP}^1$. To do this, I don't actually need to compute the equation of the conic $\partial H^2$ in $\mathbb{P}^2$: For any conic $C$ in $\mathbb{P}^2$, and any point $\omega$ in $C$, the isomorphism $C \to \mathbb{P}^1$ is given by linear projection from $\omega$.

However, there is one subtlety: To figure out where the point $\omega$ maps to, we need to figure out the equation of the tangent line to $C$ at $\omega$. The tangent line to $C$ at $\omega$ is $\beta^{\perp}$ where the vector $\beta$ is determined by the properties that $\langle \beta, \beta \rangle = \langle \beta, \omega \rangle = 0$. Then the image of $\omega$ under the map $C \to \mathbb{P}^1$ is given by applying the linear projection from $\omega$ to a point on $\beta^{\perp}$ other than $\omega$.

We take the projection from $\omega_2$. Note that the vector $\beta:= \alpha_1 + \alpha_3$ obeys the conditions $\langle \beta, \beta \rangle = \langle \beta, \omega_2 \rangle = 0$, so $\beta^{\perp}$ is the tangent line to $\partial(H^2)$ at $\omega_2$. Note also that $\omega_1 - \omega_3$ is a point on the line $\beta^{\perp}$.

So, we take the linear projection $\mathbb{R}^3 \to \mathbb{R}^2$ which projects away the $\omega_2$ coordinate. We apply this projection to $\omega_1 - \omega_3$, $s_2 \omega_2$, $s_1 s_2 \omega_2$ and $s_3 s_2 \omega_2$ in order to get the points of $\mathbb{P}^1$ corresponding to these points. Concretely, these points are $$(1:-1),\ (-A_{12}: -A_{23}),\ (A_{12}: - 2 A_{12} - A_{23}),\ (-A_{12} - 2 A_{23} : A_{23}).$$ I compute that the crossratio is $$\frac{(A_{12}+A_{23})(2 A_{12} A_{23} + 2 A_{23}^2)}{(A_{12}+A_{23})(2 A_{12}^2+2 A_{12} A_{23} )} = \frac{A_{23}}{A_{12}} = \frac{\cos(\pi/q)}{\cos (\pi/p)}.$$

Sorry for the partial answer, but this seems long enough to be worth writing down by itself. If I cared enough about this problem, I'd compute a bunch more ideal points and their cross ratios, to get a better idea of what is going on.