# Concerning (hyperbolic) compact Riemann surfaces of genus $g>1$

Let $X$ be a compact Riemann surface of genus $g>1$ with $G$ a strict open subset of $\mathbb{C}$ and $U=\bigcup_{j}U_j$ strictly a subset of $G$ for $U_j$ compact (for all $j$ such that $\bigcup_{j}U_j$ is compact). Then $\tilde X$ is the open unit disk $\Delta$ which is biholomorphic to $G$ (by the Riemann mapping theorem). Thus the first fundamental group $\pi_1(X)$ is nontrivial, for $X$ is multiply-connected. So $X$ is hyperbolic and we can write it as the quotient of the upper-half plane by a Fuchsian group, i.e. $X=\mathbb{H}/\Gamma.$

Is it true that $X\cong G$ for $\phi:X\to G\subset\mathbb{C}$ biholomorphic?

I am essentially asking if the Riemann surface $X$ of genus $g>1$ is diffeomorphic to an open subset of $\mathbb{C},$ $G\cong\Delta.$

• Did you mean something like that ? Let $X$ be a compact Riemann surface. Find a simply connected open $U \subset \mathbb{C}$ together with a covering $\phi : U \to X$. Let $G$ be the group of those biholomorphisms $\rho : U \to U$ such that $\phi \circ \rho = \phi$. Then $X = U/ G$. If you find a biholomorphism $\psi : U \to \mathbb{H}$ then $X = \psi^{-1}(\mathbb{H}/\Gamma)$ where $\Gamma = \psi(G)$ is a subgroup of $SL_2(\mathbb{R})/ \{\pm Id\}$ the group of biholomorphisms $\mathbb{H} \to \mathbb{H}$. Aug 16, 2017 at 8:03
• Sorry I am very new to the subject area-is there anything in particular that is difficult to understand? I am simply asking if it is true that a compact Riemann surface $X$ of genus $g>1$ is isomorphic to an open subset $G\cong\Delta.$ @reuns Aug 16, 2017 at 18:16
• An example of compact Riemann surface of genus $1$ is $\mathbb{C}/(\mathbb{Z}+i\mathbb{Z})$. Does it look isomorphic to an open subset of $\mathbb{C}$ ? Example of other genus is $\mathbb{H}/\Gamma$ for some discrete subgroup $\Gamma$ of $SL_2(\mathbb{R})/ \{\pm Id\}$ Aug 17, 2017 at 3:20
In a somewhat trivial, but perhaps significant, way, the answer is "no": compact connected Riemann surfaces of genus >1 are not diffeomorphic to open subsets of $\mathbb C$". Namely, open subsets of $\mathbb C$ are not compact. If we try to dodge this by taking closures, then the uniformization theorem will require that we identify some points on the boundaries, so the map (from a non-Euclidean polygon to the Riemann surface) will definitely not be one-to-one.
• Thanks for the answer! So $X$ of genus $g>$1 cannot be diffeomorphic to $U$ a compact subset of $\mathbb{C}?$@paulgarrett Aug 17, 2017 at 3:50
• Perhaps my question should be phrased as: To what subset of $\mathbb{C}$ can a Riemann surface of genus $g>1$ be diffeomorphic to? @paulgarrett Aug 17, 2017 at 3:54