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 $\mathbb R$.

  • $\begingroup$ 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 $\Gamma \setminus G/ K$ ? $\endgroup$ – reuns Jan 7 '19 at 2:38
  • $\begingroup$ Where is the fundamental domain? I don't understand $\endgroup$ – D_S Jan 7 '19 at 2:45
  • $\begingroup$ 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. $\endgroup$ – D_S Jan 7 '19 at 3:00
  • $\begingroup$ 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. $\endgroup$ – 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.)

  • $\begingroup$ 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 $\endgroup$ – D_S Jan 7 '19 at 0:45
  • $\begingroup$ @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. $\endgroup$ – paul garrett Jan 7 '19 at 18:14
  • $\begingroup$ 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. $\endgroup$ – 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

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


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


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