Questions about universal covering space and fundamental domain

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I recently learned the proof of Picard little theorem using universal holomorphic covering map from upper half plane $H$ to $C\backslash \{0, 1\}$. The proof constructed a fundamental domain using $ \{ z|0 < Re(z) < 1, |z - 1/2| > 1/2 \} $ and modular group $\Gamma(2)$. I tried another definition of universal covering map using set of paths quotient equivalence of homotopic paths. It seems that every closed path in base space will go across several fundamental domain in covering space and these two definitions are equivalent.

My question is that:

Does every Riemannian surface admit a universal covering space which has a fundamental domain acted by some discrete group, just like $C\backslash \{0, 1\}$?

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Let me first lay out a theorem which you might know, at least in bits and pieces. It's a theorem that combines a few things from covering space theory and from Riemann surface theory.

Theorem: If $S$ is a Riemann surface then

  1. There exists a covering map $p : U \to S$ such that $U = S^2$, $\mathbb C$, or $\mathbb H^2$, and such that $p$ is conformal.
  2. The deck transformation group of $p$, meaning the group $\mathcal D$ of all conformal automorphisms $h : U \to U$ such that $h \circ p = p$, is isomorphic to the fundamental group $\pi_1(S)$.
  3. If $p_i : U_i \to S$ are two conformal covering maps as in conclusion 1, with deck transformation groups $\mathcal D_i$ as in conclusion 2, then:
  • there exists a conformal homeomorphism $g : U_1 \to U_2$ such that $h_2 \circ g = h_1$;
  • there exists a group isomorphism $\mathcal G : \mathcal D_1 \to \mathcal D_2$ such that for all $h \in \mathcal D_1$ we have $g \circ h = \mathcal G(h) \circ g$.

I'll refer to any $p$ satisfying conclusion 1 as a conformal universal covering map of $S$.

What conclusion 3 says, in your context, is this: it does not matter which conformal universal covering map for $S$ you pick, because if one of them has a fundamental domain acted on by some discrete group just like $C \setminus \{0,1\}$ then so does every other conformal universal covering map for $S$.

This theorem answers part of your question, namely the existence of a conformal covering map of appropriate type. But through a further analysis of the deck transformation group $\mathcal D$ the theorem also contains the keys to answering the rest of your question, regarding fundamental domains. For that purpose, from conclusion 1 you can perhaps see that there are the three separate cases to consider. Let $S$ have universal covering map $p : U \to S$ as in conclusion $1$ above, with deck transformation group $\mathcal D$ as in conclusion $2$.

  • If $U=S^2$ then $\mathcal D$ is the trivial group. The entire sphere is a fundamental domain, and $S$ is conformally isomorphic to $S^2$.
  • If $U=\mathbb C$ then $\mathcal D$ is a discrete group of translations of $\mathbb C$, isomorphic to a free abelian group of rank $0$, $1$ or $2$. If $\mathcal D$ has rank $0$ then the entire plane is a fundamental domain, and $S$ is conformally isomorphic to $\mathbb C$. If $\mathcal D$ has rank $1$ then there is a fundamental domain which is a strip bounded by parallel lines, and $S$ is a cylinder. If $\mathcal D$ has rank $2$ then there is a fundamental domain which is a parallelogram, and $S$ is a torus.
  • If $U = \mathbb H$ then $\mathcal D$ is a discrete subgroup of the conformal group $\mathcal{Aut}(\mathbb H) \approx \text{PSL}(2,\mathbb R)$. There is a fundamental domain which is a convex polygon $P \subset \mathbb H^2$, and $P$ has finitely many sides if and only if the group $\mathcal D \approx \pi_1(S)$ is finitely generated.

As you might be able to see, the last case $U = \mathbb H$ is the most important (well, the most common, at least, namely all cases except $S^2$, $\mathbb C$, cylinders, and toruses). In that case, one useful construction of a fundamental domain can be found using the key words Dirichlet domain. This wikipedia page is perhaps a good starting point.