# Example of cocompact discrete subgroups of $\operatorname{SL}_2(\mathbb R)$

Let $$G = \operatorname{SL}_2(\mathbb R)$$, and let $$\Gamma$$ be a discrete subgroup of $$G$$. When studying automorphic forms, one might look at the quotient space $$\Gamma \backslash \mathbb H$$, where $$\mathbb H$$ is the upper half plane, or perhaps even $$L^2(\Gamma \backslash \mathbb H)$$, or $$L^2(\Gamma \backslash G)$$ (the elements of $$\Gamma \backslash \mathbb H$$ identify with right $$K$$-invariant functions on $$\Gamma \backslash G$$, where $$K = \operatorname{SO}(\mathbb R)$$).

Typically there will be a fundamental domain $$\mathcal F$$ for the action of $$\Gamma$$ on $$\mathbb H$$, through which one gets a measure $$d \dot \tau$$ on $$\Gamma \backslash \mathbb H$$, such that an "unwinding formula" holds:

$$\int\limits_{\mathbb H} f(\tau) \frac{dxdy}{y^2} = \sum\limits_{\gamma \in \Gamma} \int\limits_{\Gamma \backslash \mathbb H} f(\gamma. \tau) d \dot \tau$$

The classical case is when $$\Gamma$$ is a congruence subgroup of $$G$$. In this case, $$\Gamma \backslash \mathbb H$$ and $$\Gamma \backslash G$$ both have finite measure, but are not compact.

The compact case is nicer in a lot of ways. $$L^2(\Gamma \backslash G)$$ and $$L^2(\Gamma \backslash \mathbb H)$$ decompose as a Hilbert space direct sum of irreducible representations of $$G$$, for example.

What are some nice examples of subgroups $$\Gamma$$ such that $$\Gamma \backslash G$$ (equivalently, $$\Gamma \backslash \mathbb H$$) is compact? What does the fundamental domain look like?

• For nice examples over $\Bbb C$ see this MO-question. For $SL_2(\Bbb R)$ see Arithmetic Fuchsian groups, which gives many examples: "all orders in quaternion algebras (satisfying the above conditions) which are not $M_{2}(\mathbb {Q} )$ yield cocompact subgroups. – Dietrich Burde Dec 9 '19 at 20:02
• For lots of pretty pictures of this, see chapters 17 and 18 of "The Symmetries of Things" (Conway, Burgiel, and Goodman-Strauss). – Ted Dec 11 '19 at 2:02

Many examples arise by application of the Poincare polygon theorem, which you can find for example in Theorem 11.2.1 of Ratcliffe's book "Foundations of Hyperbolic Manifolds".

Here's a general description. Consider a finite polygon $$P$$ with $$2n$$ sides. Let $$f_1,...,f_n \in \text{Isom}(\mathbb H)$$ be isometries satisfying the following conditions:

1. For each $$i=1,...,n$$ there exist sides $$a_i,a'_i$$ such that $$P \cap f_i(P) = a_i = f_i(a'_i)$$
2. The list of sides of $$P$$ is $$a_1,a'_1,...,a_n,a'_n$$.
3. The total angle of each "vertex cycle" is $$2\pi$$ (more can be said here, but I'll refer to topology textbooks that cover the classification of surfaces).

The conclusion of the Poincare Polygon Theorem is that $$f_1,...,f_n$$ are generators of a discrete group $$\Gamma$$, $$P$$ is a fundamental polygon for $$\Gamma$$, the quotient map $$\mathbb H \mapsto \mathbb H / \Gamma$$ is a universal covering map, and the composition $$P \hookrightarrow \mathbb H \mapsto \mathbb H / \Gamma$$ is the quotient map obtained by gluing $$a_i$$ to $$a'_i$$ using $$f_i$$, for each $$i=1,...,n$$.

Let me just do one example to give the idea, the "standard gluing pattern" of an octagon. Let $$P \subset \mathbb H$$ be a regular octagon with angles $$2\pi/8$$, so all side lengths are equal. Starting from a vertex $$v_0$$ and going around in counterclockwise order, list the sides like this: $$a_1, a_2, a'_1, a'_2, a_3, a_4, a'_3, a'_4$$ Choose $$f_i$$ to be the unique orientation preserving isometry satisfying the condition above with respect to $$a_i$$ and $$a'_i$$. Now one has to check that all 8 vertices are identified to a single point on the quotient (this is what one does by going around the "vertex cycle"). Since each of these has angle $$2\pi/8$$, their sum is $$2\pi$$ as required.

• The defect implies for all $\alpha <3\pi/4$ there is a (regular) hyperbolic octogon whose interior angles are $\alpha$ ? – reuns Dec 9 '19 at 23:32
• That's right, just an intermediate value theorem argument, as the vertices go out to infinity. – Lee Mosher Dec 10 '19 at 0:57
• How is this a subgroup of $\operatorname{SL}_2(\mathbb R)$? – D_S Dec 10 '19 at 0:59
• $SL_2(\mathbb R) = \text{Isom}_+(\mathbb H)$. – Lee Mosher Dec 10 '19 at 0:59
• What is $\operatorname{Isom}_+(\mathbb H)$? – D_S Dec 10 '19 at 1:03