Is a non-compact Riemann surface an open subset of a compact one ?

Question

Let $X$ be a non-compact holomorphic manifold of dimension $1$. Is there a compact Riemann surface $\bar{X}$ suc that $X$ is biholomorphic to an open subset of $\bar{X}$ ?

Edit: To rule out the case where $X$ has infinite genus, perhaps one could add the hypothesis that the topological space $X^{\mathrm{end}}$ (is it a topological surface?), obtained by adding the ends of $X$, has finitely generated $\pi_1$ (or $H_1$ ). Would the new question make sense and/or be of any interest?

Edit2: What happens if we require that $X$ has finite genus? (the genus of a non-compact surface, as suggested in a comment below, can be defined as the maximal $g$ for which a compact Riemann surface $\Sigma_g$ minus one point embeds into $X$)

Answer 1

You should probably check the following article: Migliorini, Luca, "On the compactification of Riemann surfaces". Here is the Mathscinet review about it: "In this paper the author studies some questions concerning the compactifications of Riemann surfaces. It is proved that if $X$ is an open connected Riemann surface then X has finite genus if and only if there exists a holomorphic injection $i: X \hookrightarrow \tilde{X}$ (with $\tilde{X}$ a compact Riemann surface), $i(X)$ being dense in $\tilde{X}$..."

Answer 2

No. Take a surface of infinite genus.

Answer 3

Useful references for your question are Robert Brooks' "Platonic surfaces" and Dan Mangoubi's "Conformal Extension of Metrics of Negative Curvature" (both on arxiv).

I emailed Luca Migliorini requesting his paper. He told me it was basically his undergraduate thesis, published in a defunct italian journal, and that no copy of it remains. In otherwords, utterly useless.

The basic fact on compactifying a riemann surface is this: if $S$ is a finite area riemann surface, then there exists a compact riemann surface $S^c$ and a finite set of points $p_1, \ldots, p_k$ on $S^c$ such that $S^c \setminus {{p_1, \ldots, p_k}}$ is conformally equivalent to $S$.

In Brooks' paper, he states that this riemann surface $S^c$ is unique. However I'll admit to not be convinced of this uniqueness. The expression he uses throughout is "conformally filling punctures" -- a phrase which I think deserves more explanation than is given.

Lemma 1.1 in Brooks is interesting, and justifies the above claim. Of course we know what cusps on riemann surfaces look like. A cuspidal neighborhood $C$ of a Riemann surface can be taken isometric to the quotient of $\{ z\in \mathbb{H}^2: \Im(z)\geq 1/y \}$ by the isometry $z\mapsto z+1$, for some $y>0$. The parameter $y$ gives a measure on the size of the cusp, i.e. gives a geodesic loop homotopic to the puncture with hyperbolic length $y$. So the cusp $C$ is really isometric to the punctured ball of euclidean radius proportional to $1/y$ via the mapping $z\mapsto e^{2\pi i z}$ on the punctured open unit disk $D^\ast$ equipped with the metric $ds^*=\frac{-1}{r \log r} |dz|$. However $ds^*$ blows-up as $r\to 0$ like $1/r$.

Brooks (and afterwords Mangoubi more explicitly) gives, for any $\epsilon>0$, smooth bump functions $\delta$ concentrated at the origin on $D$ such that $e^\delta ds^*$ extends to a smooth metric past the origin and whose curvature remains pinched $-1 \pm \epsilon$.

I am going to include the details of this construction, together with some remarks relating to Donaldson's compactification of algebraic curves (from his book) shortly.