Holomorphic maps from a Riemann surface of infinite genus

Question

Let $X$ be a Riemann surface of infinite genus and let $n$ be an arbitrarily large natural number. Do there always exist a closed Riemann surface $Y$ of genus greater than $n$ and a nonconstant holomorphic map $f:X\to Y$?

Answer 1

The following is not a real answer but an extensive comment on the OP.

First, recall that an (open) Riemann surface is said to have type $P_{AB}$ (resp. $O_{AB}$) if it admits (resp., does not admit) a nonconstant bounded holomorphic function. A detailed discussion of where the surfaces of this type fit in the classification of types of Riemann surfaces can be found in the book "Riemann surfaces" by Ahlfors and Sario. (Clearly, each surface of type $P_{AB}$ admits a bounded harmonic function, but the converse is false already for planar surfaces.) Since every Riemann surface contains a subdomain biholomorphic to the open init disk in ${\mathbb C}$, it follows that every Riemann surface of type $P_{AB}$ admits nonconstant holomorphic maps to compact Riemann surfaces of arbitrarily high genus. Hence, the OP is really about Riemann surfaces of type $O_{AB}$. A good example of a Riemann surface of infinite topological type (i.e. with infinite rank of $H_1$) which has the type $O_{AB}$ is the complex plane with an infinite discrete subset $E$ removed. (More generally, one can remove from the complex plane any subset of Hausdorff dimension $<1$; the result will have type $O_{AB}$.) Let $Z$ denote the surface in this example.

Now, suppose that $X, Y$ are hyperbolic Riemann surfaces (i.e. their universal covering spaces are biholomorphic to the unit disk). In what follows, I will always equip hyperbolic Riemann surfaces with the unique conformal Riemannian metrics of constant curvature $-1$ and the metric notions will be understood with respect to these metrics. Then every holomorphic map $X\to Y$ is distance non-increasing (Schwarz lemma). Hence, the fact that a compact Riemannian surface has positive injectivity radius implies:

Lemma 1. Suppose that $X, Y$ are hyperbolic Riemann surfaces, $Y$ is compact, and, with respect to its hyperbolic Riemann metric, $X$ contains a sequence of loops $c_n$ such that $\lim_{n\to\infty} Length(c_n)=0$, and $f: X\to Y$ is a holomorphic map. Then for all sufficiently large $n$, the loop $f(c_n)$ is null-homotopic in $Y$.

From this, it follows for instance, that the surface used in Sam's answer does not admit $\pi_1$-injective holomorphic maps to compact hyperbolic Riemann surfaces. One can do a bit more with this observation:

Lemma 2. There exists a Riemann surface of infinite topological type, namely, the surface $Z$ above, which does not admit a nonconstant holomorphic map to any compact hyperbolic Riemann surface $Y$.

Proof. Consider a holomorphic map $f: Z\to Y$. Observe that for each puncture in the surface $Z$ (corresponding to a point $p$ of $D$ removed from ${\mathbb C}$), the hyperbolic metric on $Z$ has a cusp at $p$. Hence, by Lemma 1, for a small simple loop $c$ around $p$, the loop $f(c)$ is null-homotopic in $Y$. But the loops $c$ as above generate $\pi_1(Z)$. Hence, the map $f_*: \pi_1(Z)\to \pi_1(Y)$ is trivial. In particular, $f$ lifts to a map $F$ from $Z$ to the universal covering space of $Y$, which is the unit disk. But $Y$ has type $O_{AB}$, hence, $F$ is constant. Thus, $f$ is constant as well. qed

A small variation on this argument yields an open Riemann surface $X$ of infinite type and genus $1$ such that $X$ admits no nonconstant holomorphic maps to compact hyperbolic Riemann surfaces. (Recall that the genus of a surface $M$ is the supremum of the numbers $n$ such that $M$ contains a 1-dimensional compact submanifold $C\subset M$ consisting of $n$ connected components and such that $M\setminus C$ is connected. This definition works regardless of orientability and compactness. Thus, the surface $Z$ above has genus $0$.)

This leads to, what I think, is the right version of the OP:

Question. Is there a Riemann surface $X$ of genus $\ge 2$ (necessarily of type $O_{AB}$) which admits no nonconstant holomorphic maps to compact hyperbolic Riemann surfaces?

Answer 2

Edit: as Moishe points out, my answer (just below) is for a different, and easier, question. I will leave the answer up, as it does feel “related”.

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The answer is no.

I find it conceptually easier to work in the category of hyperbolic surfaces (but these are Riemann surfaces too, so it is permitted!).

We build $X$ by gluing together an infinite collection of surfaces $X_k$ (for $k \in \mathbb{Z}$) with each $X_k$ a copy of the genus one surface with two boundary components. The boundary components all have length one (say) and $X_k$ glues to $X_{k \pm 1}$. We arrange matters so that there is a simple closed hyperbolic geodesic in $X_k$ with length less than $1/|k|$. It follows that $X$ does not cover any closed surface.