# Do general discrete subgroups of $\operatorname{SL}_2(\mathbb R)$ have fundamental domains in the upper half plane?

Let $$\Gamma$$ be a congruence subgroup of $$\operatorname{SL}_2(\mathbb Z)$$. The quotient $$\Gamma \backslash \mathbb H$$ has the structure of a one dimensional complex manifold, such that the quotient map $$\pi: \mathbb H \rightarrow \Gamma \backslash \mathbb H$$ is holomorphic. There is a nice fundamental domain $$D \subset \mathbb H$$ coming from the usual fundamental domain for $$\operatorname{SL}_2(\mathbb Z)$$ which, up to some boundary identification, gives us a copy of $$\Gamma \backslash \mathbb H$$ inside $$\mathbb H$$.

The Borel measure $$\mu = \frac{dx dy}{y^2}$$ on $$\mathbb H$$ descends to a Borel measure $$\overline{\mu}$$ on $$\Gamma \backslash \mathbb H$$ which we may define using the fundamental domain: if $$U \subset \Gamma \backslash \mathbb H$$ is Borel, then we set

$$\overline{\mu}(U) := \mu \bigg(\pi^{-1}(U) \cap D \bigg) \tag{1}$$

Now, assume $$\Gamma$$ is an arbitrary discrete subgroup of $$\operatorname{SL}_2(\mathbb R)$$.

1. Is there a canonical measure $$\bar{\mu}$$ on $$\Gamma \backslash \mathbb H$$ coming from $$\mu = \frac{dx dy}{y^2}$$?

2. Does there always exist a fundamental domain $$D$$ for $$\Gamma$$?

3. Can $$\overline{\mu}$$ arise from a differential form on $$\Gamma \backslash \mathbb H$$? That is, does $$\overline{\mu}$$ come from a (unique?) smooth differential $$2$$-form $$\overline{\omega}$$ on $$\Gamma \backslash \mathbb H$$ (thought of as a smooth manifold) which pulls back to the differential form on the smooth manifold $$\mathbb H$$ corresponding to $$\mu = \frac{dx dy}{y^2}$$?

For intuition, I'm thinking of $$\mathbb R$$ modulo the action of $$\mathbb Z$$. Up to boundary identification, $$[0,1]$$ is a fundamental domain for the action of $$\mathbb Z$$ on $$\mathbb R$$. For the Haar measure $$\bar{\mu}$$ on $$\mathbb R/\mathbb Z$$, we can get it in two ways. First, if $$\pi: \mathbb R \rightarrow \mathbb Z$$ is the quotient map, we can measure subsets of $$\mathbb R/\mathbb Z$$ by pulling them back to $$\mathbb R$$, intersecting them with $$[0,1]$$, then measuring. Second, $$\bar{\mu}$$ comes from the unique invariant nonvanishing $$1$$-form on $$\mathbb R/\mathbb Z$$ which pulls back to the top form $$dx$$ on $$\mathbb R$$ giving the Lebesgue measure on $$\mathbb R$$.

• If $\Gamma$ is a discrete subgroup of $G$, won't the way $K$ acts on the $3$-dimensional locally isomorphic real manifolds $G$ and $\Gamma \setminus G$ ($1$-dimensional continuous and compact right action of $K$) means that $\Gamma \setminus G/ K$ is a $2$-dimensional real manifold whose topology is locally isomorphic to that of $G/K\cong \mathbb{H}$, this gives the fundamental domain, and considering $G/K$ as a $1$-dimensional complex manifold (a Riemann surface) then so is $\Gamma \setminus G/ K$ ? – reuns Jan 7 '19 at 2:38
• Where is the fundamental domain? I don't understand – D_S Jan 7 '19 at 2:45
• I'm not immediately convinced..there is also the issue that generally $G/K \rightarrow \Gamma \backslash G/K$ is not a local homeomorphism quite everywhere because you might have some elliptic points. – D_S Jan 7 '19 at 3:00
• Elliptic points (that is $\gamma g K = g K, \gamma \not \in g Kg^{-1}$) are a good point and quite the main issue then. And it seems they are automatically removed from the Dirichlet domain. Anyway starting with the $G$ invariant metric on $G/K$ should make things easier. – reuns Jan 7 '19 at 4:40

I'd recommend not thinking that choice of a "nice fundamental domain" is of much importance for the basic development of things. Yes, the details of a fundamental domain tell something about generators of the discrete group $$\Gamma$$, but that's not universally necessary, nor intelligible.

So, for example, the key relationship between functions on $$\mathfrak H$$ and $$\Gamma\backslash \mathfrak H$$, or, equivalently, $$\Gamma\backslash G$$ and $$G$$, or $$\Gamma\backslash G/K$$, where $$G=SL_2(\mathbb R)$$ and $$K=SO(2,\mathbb R)$$, can be described very well without any mention or choice of "fundamental domain". This is fortunate. Namely, one proves the lemma that the averaging map $$f\to \sum_{\gamma\in \Gamma} f\circ \gamma$$ from $$C^o_c(G)$$ to $$C^o_c(\Gamma\backslash G)$$ is surjective. Then the uniqueness of invariant distributions... shows that there is a unique integral/measure on $$\Gamma\backslash G$$ such that "unwinding" is correct, namely, that $$\int_G f(g)\;dg \;=\; \int_{\Gamma\backslash G} \sum_{\gamma\in\Gamma} f(\gamma g)\;d\dot{g}$$ with $$d\dot{g}$$ denoting that measure on the quotient.

(Happily for us, the only things that this set-up depends upon are that $$G$$ be a unimodular topological group, and $$\Gamma$$ a discrete subgroup.)

• Does the passage of measures from $G$ to $\Gamma \backslash G$ induce a passage of measures from $G/K$ to $\Gamma \backslash G /K$? That seems to be the remaining issue here – D_S Jan 7 '19 at 0:45
• @D_S, yes, the averaging map from $G$ to $\Gamma\backslash G$ commutes with the right $K$ action, and $K$ is compact, so we can likewise show that $K$-invariant things surject to $K$-invariant, etc. No problem. – paul garrett Jan 7 '19 at 18:14
• Thanks for your answer. I wrote an answer of my own explaining in my own words what you wrote. Would you mind seeing if there is anything incorrect about what I wrote? I want to make sure I understand your answer completely. – D_S Jan 8 '19 at 3:49

To answer the question asked, yes, every discrete subgroup of $$SL_2(\mathbb R)$$ has a fundamental domain, called a Dirichlet domain.

Let $$G = \operatorname{SL}_2(\mathbb R), K = \operatorname{SO}_2(\mathbb R)$$, and $$\Gamma$$ a discrete subgroup of $$G$$. We identify the measures $$\mu, \dot \mu, \bar{\mu}$$ on $$G, \Gamma \backslash G$$ and $$G/K$$ respectively with positive linear functionals $$T, \dot T, \bar{T}$$ on $$\mathscr C_c(G), \mathscr C_c(\Gamma \backslash G), \mathscr C_c(G/K)$$ respectively. For example,

$$T(f) = \int\limits_G f(g)dg \tag{f \in \mathscr C_c(G)}$$

$$\bar{T}(f) = \int\limits_{G/K} f(gK)d\bar{\mu}(gK) \tag{f \in \mathscr C_c(G/K)}$$

The upper half plane $$\mathbb H$$ with the hyperbolic measure $$\frac{dxdy}{y^2}$$ are identified with $$G/K$$ and $$\bar{\mu}$$, respectively.

Since $$K$$ is compact, we have a natural identification of $$\mathscr C_c(G/K)$$ with those elements of $$\mathscr C_c(G)$$ which are right $$K$$-invariant, by precomposing with the quotient map $$G \rightarrow G/K$$. We may similarly identify $$\mathscr C_c(\Gamma\backslash G/K)$$ with the right $$K$$-invariant elements of $$\mathscr C_c(\Gamma \backslash G)$$ via the quotient map $$\Gamma \backslash G \rightarrow \Gamma \backslash G/K$$. The surjection

$$\delta: \mathscr C_c(G) \rightarrow \mathscr C_c(\Gamma \backslash G)$$

$$\delta(f)(\Gamma g) = \sum\limits_{\gamma \in \Gamma} f(\gamma g)$$

can be shown to restrict to a surjection

$$\mathscr C_c(G/K) \rightarrow \mathscr C_c(\Gamma \backslash G /K)$$

The answer to my question follows if I can prove the following proposition.

Proposition: There exists a unique linear positive functional $$\widetilde{T}$$ on $$\mathscr C_c(\Gamma \backslash G/K)$$, with corresponding measure $$\widetilde{\mu}$$, such that

$$\bar{T} = \widetilde{T} \circ \delta$$

In terms of measures, what this says is that if we define $$\widetilde{T}(f) =: \int\limits_{\Gamma \backslash G/K} f(\Gamma g K) d\widetilde{\mu}(\Gamma g K)$$, then

$$\int\limits_{G/K} f(gK) d \bar{\mu}(gK) = \int\limits_{\Gamma \backslash G/K} \space \space \space\bigg( \sum\limits_{\gamma \in \Gamma} \space \space f(\gamma g K) \bigg)\space d \widetilde{\mu}(\Gamma gK)$$

or

$$\int\limits_{\mathbb H} f(z)y^{-2} dxdy = \int\limits_{\Gamma \backslash \mathbb H} \sum\limits_{\gamma \in \Gamma} f(\gamma.z) d \widetilde{\mu}(\Gamma z)$$