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Inspired by a discussion with Neil Strickland I am very interested to hear of explicit examples (one per answer, please), as follows.


A compact Riemann surface can be presented in many different ways. For example:

  1. A smoothly embedded embedded surface in the three-sphere $S^3$.
  2. A smooth projective curve (say cut out of $\mathbb{CP}^2$ by a single equation).
  3. A quotient of the hyperbolic plane by some fuchsian group.

For each of these "ways" we can accept some minor modifications.

  1. We prefer embeddings into $S^3$ (or the three-torus $\mathbb{T}^3$) because we want to actually "see" the surface. However giving the surface as a level-set of a nice function, or via some other nice analytical construction (for example as a minimal surface with symmetries), is also welcome.

  2. I am not algebraic enough to deform the condition given in (2) - I hope some reader will suggest the correct modifications.

  3. We also accept quotients of $\mathbb{C}$ by a lattice $\mathbb{Z} + \mathbb{Z}\omega$. We also accept tilings of the upper half-plane as long as the tiling has no "moduli" (or has enough explicit side conditions) so that the fuchsian group can be deduced, with sufficient amount of hyperbolic trig). We may also modify (3) in another way - for example giving square-tiled surfaces or more generally surfaces given by gluing explicitly described polygons in $S^2$, $\mathbb{E}^2$, or $\mathbb{H}^2$.

Question: Give explicit examples of pairs (or preferably triples) of isomorphic Riemann surfaces of the above types.

We first dispose of the trivial example of the sphere. Here (1) and (3') are addressed by saying "the round sphere". (2) is addressed by saying (for example) "$x + y + z = 0$".

As an actual example, we have tori. For (3), we specify $\omega$ and thus the lattice $\mathbb{Z} + \mathbb{Z}\omega$. For (2), we have the Eisenstein series giving the modular invariants. For (1) we have the Hopf tori embedded in $S^3$ (and thus, after stereographic projection, embedded in three-space). Pinkall [Inventiones, 1985] defines these, and has explicit constructions (see Figure 4b (reproduced below) and Figure 6b). See also Figure 5 of Sullivan [Bridges, 2011].

I give further examples in the comments here but please see, in addition, Strickland's talk.

Hopf torus taken from Figure 4b of Pinkall's 1985 paper.

Sam Nead
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    How do you get a Riemann surface structure from (1) (a smoothly embedded surface in $S^3$)? – YCor May 03 '22 at 11:33
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    @YCor: the standard metric on $S^2$ restricted on the embedded surface defines a conformal structure on this surface. – Alexandre Eremenko May 03 '22 at 11:52
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    A given Riemann surface has very many embeddings to $S^3$, so it is unclear how to choose a canonical one among them. – Alexandre Eremenko May 03 '22 at 11:57
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    In the theory of Minimal surfaces, and more generally surfaces of constant mean curvature, there are some explicit examples of embedding of (punctured) Riemann surfaces to $R^3$ or $S^3$. – Alexandre Eremenko May 03 '22 at 12:12
  • There is no desire here to have a canonical embedding. All that is asked is for an explicit embedding in the correct conformal class. I will add to the question that the embedding should also be "pretty" if possible (so, a short description, extra symmetries, additional properties... add to its beauty). – Sam Nead May 03 '22 at 12:15
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    Given (2) it should be possible to get (1) relatively easily. Given an algebraic curve in $\mathbb{P}^2_{\mathbb{C}}$ we can choose a point $p$ that does not lie on it. Then $\mathbb{P}^2_{\mathbb{C}}-{p}$ is a line bundle on $\mathbb{P}^1_{\mathbb{C}}$ whose $S_1$ bundle is $S^3$.

    It should be relatively easy to "push" the curve into this $S_1$ bundle quite explicitly.

    – Kapil May 04 '22 at 02:24
  • @Kapil - That is a cool idea. I suppose (?) that the difficult bit is proving that “pushing” preserves the isomorphism type (that is, the conformal class) of the Riemann surface? – Sam Nead May 04 '22 at 06:24
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    @SamNead You are right. However, it appears to work in the hyperelliptic case. $y^2=f(x)$ for $f$ an odd degree polynomial. Higher degree covers of $\mathbb{P}^1$ would be complicated. – Kapil May 04 '22 at 16:42
  • @Kapil - That is very interesting... I would very much like to hear more details of this. – Sam Nead May 04 '22 at 19:27

3 Answers3

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A particular example is that of the Lawson surface $\xi_{g,1}$ of genus $g$. As defined here, it is a compact minimal surface in the 3-sphere, obtained by reflecting the solution of the Plateau problem for a specific geodesic 4-gon. The surface has many symmetries, so it is easy to deduce that it is given by the algebraic equation $$y^2=z^{2g+2}-1.$$ Finally, the Riemann surface has so many symmetries that you can write down its uniformisation oper (and its monodromy) explicitly: the former is given by the desingularisation of the pull-back (by the holomorphic map to the projective line obtained by quotienting out all symmetries) of the Fuchsian system $$\nabla\,=\,d+\begin{pmatrix}\frac{1}{8}&0\\0&-\frac{1}{8}\end{pmatrix}\frac{dz}{z}+ \begin{pmatrix}-4\rho^2&1\\ \rho^2-16\rho^4&4\rho^2\end{pmatrix}\frac{dz}{z-1}$$ where $\rho=\tfrac{g}{2g+2}.$ The Fuchsian group can be computed explicitely from the monodromy of the Fuchsian system, which is (conjugated to) the representation of the 3-punctured sphere with monodromies $$ M_0\,=\,\begin{pmatrix}\frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix}$$ $$M_1\,=\,\begin{pmatrix} -\cos{\frac{\pi}{k}}-\sqrt{(1+\cos{\frac{\pi}{k}}) \cos\frac{\pi}{k}}& 1 + \cos{\frac{\pi}{k}}+\sqrt{(1+\cos{\frac{\pi}{k}})\cos{\frac{\pi}{k}}}\\ -1 -\cos{\frac{\pi}{k}}+\sqrt{(1+\cos{\frac{\pi}{k}})\cos{\frac{\pi}{k}}}& -\cos{\frac{\pi}{k}}+\sqrt{(1+\cos{\frac{\pi}{k}}) \cos\frac{\pi}{k}} \end{pmatrix}$$ $$M_\infty\,=\,\begin{pmatrix} \frac{1}{\sqrt{2}}&\frac{-1- 2\cos{\frac{\pi}{k}}-2 \sqrt{(1+\cos{\frac{\pi}{k}})\cos{\frac{\pi}{k}}}}{\sqrt{2}}\\ \frac{1+ 2\cos{\frac{\pi}{k}}-2\sqrt{(1+\cos{\frac{\pi}{k}})\cos{\frac{\pi}{k}}}}{\sqrt{2}}& \frac{1}{\sqrt{2}} \end{pmatrix},$$ where $k=g+1.$

Photo, by Wjatscheslaw Kewlin, of 3D print of the genus two Lawson surface.

A 3D print of the (stereographic projection of the) genus two Lawson surface. The (conformal) parameterisation is due to Sebastian Heller and Nicholas Schmitt - the print is due to Nicholas Schmitt and Wjatscheslaw Kewlin, and the photo is due to Wjatscheslaw Kewlin.

Sam Nead
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Sebastian
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There are indeed very few pairs (except spheres with 3 or 4 singularities, or tori, and what can be obtained from them by finite coverings, where correspondence 2)-3) is completely explicit. See:

H. P. de Saint-Gervais, Uniformisation des surfaces de Riemann, ENS Editions, 2010 (there is an English translation), Chap. IX.

Concerning pairs 1)-3), many beautiful examples occur in the theory of minimal surfaces; they can be seen in the Bloomington Virtual Minimal Surface Museum.

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    One example per answer, please. :) Surely you have a favourite one? – Sam Nead May 03 '22 at 14:05
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    I do not see any reason for the arbitrary rule "one example per answer", and I am not going to follow it. If you find it useful, you can upvote it, or vise versa. – Alexandre Eremenko May 03 '22 at 14:38
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    “One answer per answer” is a typical rule for “big list” questions. I’ve downvoted your answer because you have given no examples, explicit or otherwise. – Sam Nead May 03 '22 at 17:38
  • Sure. When you are not satisfied with an answer, you may downvote it. If the number of downvotes exceeds the number of upvotes, I may consider removing it. P.S. As I hinted in my answer, the list is not going to be big:-) – Alexandre Eremenko May 04 '22 at 21:58
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A classical, wonderful example in which is possible to explicitly see all the three descriptions is the Klein quartic.

  • Where is the conformally correct embedding into $S^3$ given? – Sam Nead May 03 '22 at 12:17
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    Did you edit the question? In a previous version, you said that for (1) you also accept a visualization in $\mathbb{R}^3$ (which is provided by the Wikipedia article linked). – Francesco Polizzi May 03 '22 at 12:19
  • I did edit the question, but not in that way... Greg Egan's embedding is not smooth, and it is not conformally correct. It does show the combinatorics of the KQ, but it is not "isomorphic" as a Riemann surface to the KQ. The remarks hold for Helaman Ferguson's sculpture. Don't get me wrong: both are useful and lovely - they are just not examples of – Sam Nead May 03 '22 at 12:27
  • Ok, I understand what you mean. In fact, there is no smooth embedding of the Klein quartic in $\mathbb{R}^3$, so these combinatorial models are the best one can achieve, in a sense. – Francesco Polizzi May 03 '22 at 12:30
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    Every Riemann surface admits a conformal embedding into $\mathbb{R}^3$. See https://mathoverflow.net/questions/53999/ for references and a discussion. However, those "constructions" are not explicit (and I would argue, not pretty!). – Sam Nead May 03 '22 at 12:33
  • I am confused since the automorphism group of the Klein quartic is not a subgroup of $\operatorname{SO}(3)$, so this conformal model in $\mathbb{R}^3$ cannot have a rotational symmetry... – Francesco Polizzi May 03 '22 at 12:39
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    The symmetries of the embedded Riemann surface need not (and usually will not) be induced by isometries of the ambient three-manifold. This happens, for example, for the hexagonal Hopf torus embedded in the three-sphere. – Sam Nead May 03 '22 at 12:58