0

Is it automatic that a compact Riemann surface $X$ of genus $g>1$ is not homeomorphic to a compact subset of the complex projective line $\mathbb{CP}^1?$

Note: I apologize for the confusion-I am just a high school student with an interest in math. This question was an offspring of my earlier question. Since it seemed sufficiently distant in nature, I decided to ask a new question.

Thanks in advance!

Sergio Charles
  • 1,366
  • 1
    The first examples of compact Riemann surface are $\mathbb{P}^1(\mathbb{C})$ the Riemann sphere (genus $0$) and $\mathbb{C}/(\mathbb{Z}+i\mathbb{Z})$ a torus (a donuts, genus $1$). A compact Riemann surface of genus $n$ is in some sense $n$ tori glued together. – reuns Aug 19 '17 at 20:21
  • See the Weierstrass function $\wp$ which is the example of a meromorphic function $\mathbb{C}/(\mathbb{Z}+\tau\mathbb{Z}) \to \mathbb{C}$ (an analytic function $\mathbb{C}/(\mathbb{Z}+\tau\mathbb{Z})\to \mathbb{P}^1(\mathbb{C})$) – reuns Aug 19 '17 at 21:27
  • One last question, when you say that a compact Riemann surface of genus $n$ ($X_n$) is (in some sense) $n$ tori glued together, does this mean the disjoint union of $n$ tori $X_n=\coprod_n(S^1\otimes S^1),$ or something else?@reuns – Sergio Charles Aug 20 '17 at 20:39
  • Or more precisely $\coprod_{i=1}^n\mathbb{C}/(\mathbb{Z}+\tau_i\mathbb{Z})$ @reuns – Sergio Charles Aug 20 '17 at 21:09
  • Does it look like a direct product of two complex tori ? Did you construct the complex torus $\mathbb{C}/(\mathbb{Z}+i\mathbb{Z})$ yet ? – reuns Aug 20 '17 at 22:00
  • No, but maybe possibly under a deformation? Yes, I did. @reuns – Sergio Charles Aug 22 '17 at 17:22
  • 1
    A complex torus is the quotient of $\mathbb{C}$ by a lattice $\Lambda = \omega_1\mathbb{Z}+\omega_2\mathbb{Z}$ (whose elements act on $\mathbb{C}$ by translation). The direct product of two complex tori is $\mathbb{C}^2/(\Lambda \times \Lambda')$ the quotient of $\mathbb{C}^2$ by a lattice $\Lambda \times \Lambda'$ in $\mathbb{C}^2$. Locally it looks like an open of $\mathbb{C}^2$, it is not a (Riemann) surface. – reuns Aug 22 '17 at 18:17
  • Is there a standard construction of $X$ (of genus $g\ge 1$) from complex tori (I am using Forster's Lectures and I have yet to see this construction)? Thanks. @reuns – Sergio Charles Aug 23 '17 at 02:30
  • No. You need to stay in the genus $1$ setting and look at the complex elliptic curve $y^2 = 4 x^3-ax-b$ before looking at $g \ge 2$ and quotients $\mathbb{H}/\Gamma$ of the upper half-plane by congruence subgroups. Of course, before all this you need to know a lot of complex analysis (theory of holomorphic functions) – reuns Aug 23 '17 at 02:52

1 Answers1

2

The genus is a topological invariant, in particular a surface of genus $g \geq 1$ can't be homeomorphic to the Riemann sphere. Moreover, any subset of the Riemann sphere has boundary unlike a surface of genus $g$ so such an homeomorphism is also not possible.

On the other hand, any surface $S$ of genus $g \geq 2$ can be obtained as a quotient of the unit disk $D \subset \Bbb C$. A surface of genus 1 is more or less by definition $\Bbb C/ \Lambda$ where $\Lambda$ is a lattice.

  • +1 Thanks for the answer. In view of this, can a compact subset of $X$ (with $g\ge 2$) be homeomorphic to a compact subset of $\mathbb{CP}^1$ where the compact subsets (of $X$) cover $X?$ @N.H. – Sergio Charles Aug 19 '17 at 20:55
  • 1
    Sure, you can cover $X$ by closed disks which are subset of $\Bbb CP^1$ but it not so interesting. On the other hand, something in this direction would be a branched covering : it's a map $f : X \to \Bbb P^1$ which is a local diffeomorphism except at finitely many points. This is a very interesting class of map. –  Aug 19 '17 at 21:09
  • @Multivariablecalculus Why don't you work on $\mathbb{C}/(\mathbb{Z}+i\mathbb{Z})$ instead of asking about things you can't even define (compact Riemann surfaces) ? – reuns Aug 19 '17 at 21:15
  • Do you think you can provide me with a reference to this map (perhaps Forster's lectures)? Also, is there a way to show that $\infty$ is one of these (finitely many) points? @N.H. – Sergio Charles Aug 19 '17 at 21:16
  • Sorry, I am not quite sure what you mean-compact Riemann surfaces are defined in most introductory texts on complex geometry. @reuns – Sergio Charles Aug 19 '17 at 21:20
  • @Multivariablecalculus Obviously you don't understand this definition.. – reuns Aug 19 '17 at 21:21
  • This question was solely asked to help me make sure that I understood this object. As I said, I am still a novice in the subject area. @reuns – Sergio Charles Aug 19 '17 at 21:25
  • 1
    I think you should get a little more background in math before trying to understand Riemann surfaces, especially if you are a high-school student. Do you have a good understanding of complex analysis or topology ? I would advice you to read the book "Introduction to topological manifolds" by Lee which contains all the prerequisites in topology for studying Riemann surfaces. It is an excellent book. –  Aug 19 '17 at 21:32
  • Thanks, I have just begun an "Introduction to Smooth Manifolds" by Lee. And yes, I have taken complex analysis (and have only been introduced to elementary topology via analysis) @N.H. – Sergio Charles Aug 20 '17 at 17:37