# Are concrete Riemann surfaces Riemann domains over $\mathbb C$?

While reading about Riemann surfaces I stumbled upon the following two definitions. A Riemann surface of a complete analytic function is an example of both definitions while Abstract Riemann Surface is a generalisation of both definitions. So it seems like every (non singular, concrete) Riemann surface can be considered as Riemann domain over $$\mathbb C$$ or maybe vice versa. But I am unable figure it out. I am also curious about the historical origins of these definitions.

Riemann surface:

A subset $$S \subseteq \mathbb C^2$$ is called a (non-singular, concrete) Riemann surface if for each point $$s \in S$$, there is a neighbourhood $$U$$ of $$s$$ and a holomorphic function $$F$$ on $$U$$ with non-zero gradient at $$s$$ such that $$S \cap U$$ is the zero set of $$F$$.

Riemann domain over $$\mathbb C$$:

A topological space (Hausdorff and connected) $$G$$ is called a Riemann domain over $$\mathbb C$$ if there exists a local homeomorphism map $$\pi:G \to \mathbb C$$.

• Hint: use the implicit function theorem – Wojowu Oct 18 '18 at 20:01
• Implicit function theorem is local theorem, how can I use it to construct global map $\pi$? – Mayuresh L Oct 18 '18 at 20:04
• Sorry, I misunderstood the definition. I thought we require there to be locally a homeomorphism between neighbourhoods (which is precisely what implicit function theorem would guarantee) – Wojowu Oct 18 '18 at 20:06
• What are the sources of these definitions? – Armando j18eos Oct 24 '18 at 10:55
• Riemann surface definition is from Riemann surface notes by C. Teleman available online and Riemann domain definition is from SCV book by Narasimman/Fritsche and Grauert – Mayuresh L Oct 24 '18 at 11:27

## 1 Answer

There are theorems of Gunning and Narasimhan (Gunning, R. C., and Raghavan Narasimhan. "Immersion of open Riemann surfaces." Mathematische Annalen 174.2 (1967): 103-108.) and a theorem of Narasimhan (Narasimhan, R. "Imbedding of open Riemann surfaces", Göttingen Nachrichten, No. 7 (1960), pp. 159-165; also see American Journal of Mathematics Vol. 82, No. 4 (Oct., 1960), pp. 917-934) that almost(!) demonstrate an equivalence.

The first paper proves (quoting from the paper):

Theorem: Any (connected) open Riemann surface $$X$$ admits a holomorphic immersion into the complex plane; that is, there is a holomorphic mapping $$F:X\to\mathbb{C}$$ which is a local homeomorphism.

In other words, given any surface $$S$$ as in the original post above there is a holomorphic mapping $$F:S\to\mathbb{C}$$ exhibiting it as a $$G$$ as in the original post.

In the second paper it is proved that (quoting from the third paper!) "any open Riemann surface has a non-singular imbedding in $$\mathbb{C}^3$$".

In general, going from $$\mathbb{C}^3$$ to $$\mathbb{C}^2$$ will introduce some singularities. So going back all the way to $$\mathbb{C}^2$$ may not be possible.

A possible reason for such a obstruction is given below, but it is not a counter-example until proven!

For example, a smooth compact Riemann surface of any genus has an embedding in $$\mathbb{CP}^3$$. However, a general smooth compact Riemann surface of genus 6 does not have an embedding in $$\mathbb{CP}^2$$.

One can produce curves of degree 6 in $$\mathbb{CP}^2$$ with 4 nodes whose desingularisations are compact Riemann surfaces of genus 6. The 4 nodes will (in general) not lie on a line, so one cannot get an open Riemann surface of the type above by removing a $$\mathbb{CP}^1$$. Note, however, that this is not a proof that it cannot be done since there are more holomorphic functions on an open Riemann surfaces than those obtained by restrictions of meromorphic algebraic functions on the compact one.

• Thanks for the answer. Does first theorem also hold for abstract Riemann surface (i.e. 1 dimensional complex manifold)? – Mayuresh L Oct 28 '18 at 15:50
• @MayureshL Yes. Any open connected 1 dimensional complex manifold. – Kapil Oct 29 '18 at 2:18
• Ohh! Means studying open connected 1 dimensional complex manifold (i.e. open connected Abstract Riemann surface) is same as studying Riemann domain over $\mathbb C$ – Mayuresh L Oct 30 '18 at 7:34
• The reverse direction question seems to be at a research level so I have asked it on MO as a new question. Hope that is OK with @MayureshL. – Kapil Oct 30 '18 at 9:04
• I understood your comment. Indeed, all abstract open Riemann surfaces are of type $G$. The question of whether a $G$-type is also expressible as an $S$-type is the Bell-Narasimhan Conjecture according to answer I got on Math Overflow! – Kapil Oct 30 '18 at 12:06