Talk:Riemann surface

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[edit] Early 2003

How about an explanation in layman's english, too? Theanthrope 06:46 Jan 23, 2003 (UTC)

I'll try. However, while it's possible to sort of talk about a manifold in layman's english, the distinction between Riemann surfaces and 2-manifolds is very subtle and difficult to describe without using any math. Loisel 07:08 Jan 23, 2003 (UTC)

Axel, I saw you improved the opening paragraph. However, I'm concerned that perhaps we could serve Theanthrope's request (see above) by moving some of the jargon to a second paragraph, and keeping the opening paragraph as light on jargon as possible. I think it would also help if someone could give an interesting example of a 2-manifold that can't be viewed as a Riemann surface. This way, we can somehow say "Riemann surfaces are mainly what everyone thinks of as surfaces, however some surfaces such as X aren't Riemann surface, see below for technical details." Unfortunately, I'm not sure which 2 surface does not admit a Riemann surface structure. What do you think? Loisel 06:18 Feb 12, 2003 (UTC)

I moved the term "holomorphic" into the second paragraph, even though I think it's the whole point of the enterprise. We certainly need examples: open sets of the complex plane, the Riemann sphere, the two-sheet thingy you get from the square root function etc.

Everything2 has a nice writeup at http://everything2.com/index.pl?node_id=1007348

All Riemann surfaces are orientable, so the Moebius strip can't be made into a Riemann surface. AxelBoldt 16:56 Feb 12, 2003 (UTC)

I didn't know that. I added an argument outline to the article. I didn't want to infect the article with a lengthy proof. Delete it if you don't like it. Loisel 03:27 Feb 13, 2003 (UTC)

We need to de-TeX this article a bit, it takes far too long to load. -- Tarquin 16:39 Feb 12, 2003 (UTC)

Yes, I agree. AxelBoldt 17:15 Feb 12, 2003 (UTC)

Me too - plus it's horrible to read text where the font size jumps up and down by a factor of 2. Chas zzz brown 12:04 Feb 13, 2003 (UTC)


Axel, I'm not sure I have the correct idea for the charts of concrete riemann surfaces. I'm not completely sure that what I have in mind is a chart independent of the a_k. Can you clarify that for me? Loisel 22:07 Feb 17, 2003 (UTC)

Actually, now that I think about it: it doesn't seem like all "concrete Riemann surfaces" as I defined them are Riemann surfaces. It seems the concrete ones may have singularities while our abstrace Riemann surfaces cannot. I don't know how to fix this right now; maybe we should just take the example of concrete surfaces out, even though historically it was the motivating one. AxelBoldt 00:21 Feb 18, 2003 (UTC)

I'm interested in the "only if" portion of what you just wrote. It says that if a real two-manifold X is orientable, then it has a Riemann surface structure. Is there an easy way to see this, or a book where I can look this stuff up? I've been sitting here trying to take some charts and making sure they're Riemann compatible, but try as I might, I can't figure it out. Take for instance R^2. In some open region, I use the identity map for a chart f(x)=x. In another open region, I use a function g(x) which isn't analytic at some point (like exp(-1/x^2) at 0 or something like that.) How do I turn these charts into Riemann charts? Do I need to know some facts about the ring C^\infty(X) to prove this? Loisel 02:32 Feb 18, 2003 (UTC)


I found the claim in http://www.jchl.co.uk/maths/Riemann%20Surfaces.pdf I don't know how to prove it. But I do know that every differentiable real manifold has an atlas whose elements are compatible as real analytic functions. You somehow need to use the orientability to get from real analytic to holomorphic. AxelBoldt 06:23 Feb 18, 2003 (UTC)


Can you point me to the page you're thinking of? Are you referring to the realization of Riemann surfaces as algebraic surfaces of the form you suggested ("concrete Riemann surfaces") or is it something else? Those are good notes, by the way. Loisel 07:34 Feb 18, 2003 (UTC)

On the top of page 6 of the above given PDF file, he makes the claim that every orientable real 2-manifold can be turned into a Riemann surface. I don't know if he proves it later in the notes. AxelBoldt 16:56 Feb 18, 2003 (UTC)

I see. Interesting. Loisel 17:50 Feb 18, 2003 (UTC)

[edit] sheaves

Has anyone else seen the definition of "sheaf" used in this article. I was under the impression a sheaf was a contravariant functor from the category of open sets and inclusions in a topological space to the category of sets, rings, etc. such that if all restrictions of certain sections line up a unique new element can be patched together from the data. The "sheaf" in the article seems to be a global analytic function constructed as the union of the functions given by local power series expansions; it seems like this object would be a section of some sheaf of holomorphic functions rather than a sheaf itself. Is this wrong? Vivacissamamente

The usage isn't exactly correct, perhaps; but it is rather close. That is, with the definition of sheaf space of a sheaf (an alternative to the functor approach, and earlier too), the projection of G really will be the kind of local homeomorphism that is to be identified with a sheaf. As if the Riemann surface required is really constructed as a quite small subsheaf of the sheaf of holomorphic functions, would say it another way.

Charles Matthews 14:32, 8 Jun 2004 (UTC)

I've never been able to parse the algebraic sheaves, although I realize they're much more talked about than the complex analysis sheaves today. However, I believe that the nomenclature originated in complex analysis, and I'm fairly certain that the definition here is the one for complex analysis.

Loisel 23:12, 9 Jun 2004 (UTC)

Historically, I believe, this goes back to the Poincaré-Volterra theorem? Charles Matthews 07:23, 10 Jun 2004 (UTC)


Which one is that? Loisel 21:01, 10 Jun 2004 (UTC)

Nevermind, got it. Loisel 16:06, 11 Jun 2004 (UTC)

[edit] Analytic continuation

Would it make sense to move the whole bit about analytic continuation and sheaves to the article on analytic continuation? Its not clear to me why this stuff about analytic continuation is even in this article... (other that that g composed with f-1 needs to be analytic, etc. Anyone up for that, or will my edits get reverted ??linas 20:03, 22 Jan 2005 (UTC)

Still planing on making good on the above threat...real sooner now. linas 14:59, 14 Apr 2005 (UTC)
The connection is because the 'maximum analytic continuation' is to a Riemann surface; and this is one of the historical roots of the topic. That said, it might be just as clear to place this material in the other article. Charles Matthews 15:44, 14 Apr 2005 (UTC)
Moved. I want to expand this article a bit, and the stuff on analytic continuation was taking up space. I also remember learning analysis for the first time, and being frustrated by the lack of a good definition of analytic continuation. For whatever reason, courses in begining complex analysis don't really use the words riemann surface very much. linas 17:14, 16 Apr 2005 (UTC)

[edit] sheaves

the article on sheaves contains a link to this article for an example of a sheaf. I removed this line from the text, because there is no such example here. If anyone does add an example at some point, then the link should be restored in the article on sheaves. -Lethe | Talk 19:53, May 31, 2005 (UTC)

Looks like the relevant material was moved in April to analytic continuation. Charles Matthews 20:48, 31 May 2005 (UTC)

[edit] Hyperbolic surfaces have nontrivial π1?

the article says the following things: the disk is among the simply connected complex surfaces, the disk is hyperbolic, hyperbolic surfaces have nontrivial fundamental group. One of these three sentences doesn't fit. I think it's probably the last one? -Lethe | Talk 13:29, Jun 16, 2005 (UTC)

thinking more on the matter, I think a surface of constant negative curvature must have negative euler characteristic, and so high genus, and so nontrivial pi1. So the first statement is in error? Also, can I infer that the Poincaré half plane is multiply connected? That seems surprising. -Lethe | Talk 14:19, Jun 16, 2005 (UTC)

The last statement is not correct, the disk is the unique (up to biholomorphism) simply connected hyperbolic surface. The hyperbolic surfaces which aren't simply connected have the disk as the universal cover. The upper half plane is biholomorphic to the disk and so also hyperbolic and simply connected. -- Fropuff 15:05, 16 Jun 2005 (UTC)

Maybe the author of that stuff meant compact hyperbolic Riemann surfaces? The relationship between curvature and genus is maybe not straightforward for noncompact surfaces. I find similar erroneous statements in several other related articles as well. -Lethe | Talk 22:28, Jun 16, 2005 (UTC)

That is entirely possible. Statements further down (regarding the area of the surface) certainly apply only in the compact case. These statements all need to be clarified -- Fropuff 04:59, 17 Jun 2005 (UTC)

[edit] corrections

Yes, certainly it should be changed. In Riemann surface, it seems to me that the sentence

"Hyperbolic Riemann surfaces are multiply connected and thus have a non-trivial fundamental group."

ought not be the first sentence in the paragraph. I mean, this isn't particular to hyperbolic geometry, right? C is simply connected noncompact and parabolic, and C mod some rank 2 discrete lattice group is closed parabolic, which is not simply connected (pi1 = lattice group). At this point, I don't see what's special about the hyperbolic case: H is simply connected noncompact and hyperbolic. while H mod Fuchsian group is closed hyperbolic (pi1 = Fuchsian group). It's the same, so why is the hyperbolic case getting singled out?

I propose we change it to something along these lines:


"For every closed parabolic Riemann surface, the fundamental group is isomorphic to a rank 2 lattice group, and thus the surface can be constructed as C/Γ, where C is the complex plane and Γ is the lattice group. The representatives of the cosets are called fundamental cells. The area of such a surface is arbitrary (is this true?).
OK I copied this paragraph into the article. Notice the red links. Sould we redirect fundamental cell to free regular set for now, until someone writes a "real" article, or is that a bad idea? Should we redirect lattice group to lattice (group)? Or maybe redirect to torus? As to area of surface being arbitrary, that sure sounds right to me, but as you notice I'm error-prone. I left the sentance out. linas 00:33, 18 Jun 2005 (UTC)
Actually, there's already an article fundamental domain. I guess fundamental cell should redirect to that. And yeah, I guess lattice group should redirect to lattice (group). As for the area of a torus, the complex structure depends on the ratio of the lengths of the sides, so you can scale the area to anything you want. You can also see that the Gauss-Bonnet theorem can't tell you the area of the torus of constant curvature, since the curvature is zero. -02:10, Jun 19, 2005 (UTC)
"Similarly for every hyperbolic Riemann surface, the fundamental group is isomorphic to a Fuchsian group, and thus the surface can be modelled by a Fuchsian model H/? where H is the upper half plane and ? is the Fuchsian group. The representatives of the cosets of H/? are called fundamental polygons. The total area of such a surface , where g is the genus of the surface; the area is obtained by applying the Gauss-Bonnet theorem to the area of the fundamental polygon."

Can we say some more about the elliptic case? Is the Riemann sphere the only closed elliptic surface (up to biholomorphism)? We should link also to some article on Lie groups that says pi1 of universal cover mod discrete subgroup = discrete subgroup.

Last I looked, Riemann sphere was unique. As to the modulo-group remark, I think that the article covering map states this, if rather densely. Also, by lattice group do you mean wallpaper group? Or is there another definition? linas 14:03, 17 Jun 2005 (UTC)
No, a lattice group is not the same thing as a wallpaper group, although I think the latter includes the former as a special case. see Lattice (group). -Lethe | Talk 23:20, Jun 17, 2005 (UTC)
OK, so let me see if I have it straight: there is only one elliptic Riemann surface, and it's simply connected. For the parabolic case, there is one simply connected surface C, one surface with pi1=Z (C mod lattice of rank 1 = C*), and an infinite number of tori with pi1=ZxZ (C mod lattice of rank 2). And in the hyperbolic case, I'm a bit fuzzy. Is there only one surface for each genus >1? each of them has H as universal cover? -Lethe | Talk 02:10, Jun 19, 2005 (UTC)
Hmm. You know, I am at best a novice w.r.t Riemann surfaces. I don't know the parabolic case. Saying "its a lattice" essentially means "its a torus"; I have no clue if there are any parabolic riemann surfaces that aren't toruses. Can't say I've heard of such beasts, but also can't say I've heard of a theorem stating otherwise. Maybe its trivial? linas 00:33, 18 Jun 2005 (UTC)
Well, if the lattice group is rank 1, you get a parabolic manifold that is not a torus. You get the cylinder, which is the same as the multiplicative group of nonzero complex numbers. C, C*, and a bunch of tori, those are the only parabolic surfaces. -Lethe | Talk 02:10, Jun 19, 2005 (UTC)

The article on hyperbolic geometry has the similar statement "Hyperbolic surfaces have a non-trivial fundamental group , known as the Fuchsian group. The quotient space H/? of the upper half-plane modulo the fundamental group is known as the Fuchsian model of the hyperbolic surface.", which also needs to me changed. I thought I saw something like this in a third article too, this morning, but now I can't find it. Maybe it was just the two. -Lethe | Talk 05:42, Jun 17, 2005 (UTC)

Several quick remarks: don't sacrifice overall clarity of an introduction to exactness. Most hyperbolic surfaces do have non-trivial genus; only one doesn't, the hyperbolic disk. I think that a minor edit could take care of the one-exception case. Other than that, the above proposed paragraphs look reasonable. linas 13:51, 17 Jun 2005 (UTC)
Sure, most manifolds in any isogeny class have nontrivial pi1, except for the one that doesn't. This is true for any isogeny class of Lie groups. But what does that have to do with hyperbolic geometry? The sentence makes it seem like somehow the hyperbolicity implies nontrivial pi1. That's true in the compact case, not true in the noncompact case. -Lethe | Talk 22:36, Jun 17, 2005 (UTC)
My apologies, I just realized you were complaining about sentences I had written back when. I am sorry they're misleading, this was not my intent, and perhaps shows my ignorance of the subject. The corrections you propose seem reasonable; I invite you to make them, or you can wait for me to do it. However, I have a backlog of unfinished projects, so waiting my prove to be irritating. linas 23:37, 17 Jun 2005 (UTC)
Never mind, I fixed by copying you wording. Please review for additional corrections. linas 00:33, 18 Jun 2005 (UTC)

Also, is it correct to say the representatives are the fundamental polygon? Isn't a representative of a coset simply a point? -Lethe | Talk 05:54, Jun 17, 2005 (UTC)

No, its not a point. Technically, the representatives are free regular sets which can be converted to fundamental polygons after a set of operations that re-arrange the area of the thing so that its bounded by geodesics, etc. linas 13:51, 17 Jun 2005 (UTC)
I think a certain set of representatives form the fundamental domain/polygon/free regular set, but the representatives themselves are simply points. I mean, I get what it's trying to say, but I think as it stands it is worded incorrectly. Actually, you can see the article on free regular sets to see the correct wording. -Lethe | Talk 22:36, Jun 17, 2005 (UTC)
Oh gosh, very sorry, yes, I misunderstood, you are absolutely right; the wording in the article is wrong. What can I say? Brain in neutral, keyboard in overdrive. I'll fix this in a few minutes. linas 23:37, 17 Jun 2005 (UTC)

[edit] Hyperbolic/parabolic/elliptic terminology

I am a little confused as to the meaning of hyperbolic/parabolic/elliptic in regards to Riemann surfaces. This article defines a Riemann surface to be hyperbolic/parabolic/elliptic according to whether the universal cover is the unit disk, the complex plane, or the Riemann sphere. Or, equivalently, whether the surface carries a Riemannian metric with constant curvature −1/0/+1. However, the book by Farkas and Kra (Riemann Surfaces 2nd ed. 1992, ISBN 0-387-97703-1) uses a different definition for these terms. In section IV.3.2 they define a Riemann surface to be

  • elliptic iff it is compact,
  • parabolic iff it is noncompact and does not carry a negative nonconstant subharmonic function,
  • hyperbolic iff it does carry a negative nonconstant subharmonic function (necessarily noncompact).

According to this classification the three simply connected surfaces are still hyperbolic/parabolic/elliptic as before, but the meaning changes for another surfaces. For example, all compact surfaces (of any genus) are elliptic according to Farkas and Kra but most (g ≥ 2) are hyperbolic according to this article.

Can anyone comment of the different definitions? Are either or both standard? Should we mention the Farkas and Kra meaning in this article? I note that the Farkas and Kra definition is potentially more interesting since nearly every surface is hyperbolic in the sense of this article (the sole exceptions being the Riemann sphere, the complex plane, the punctured plane, and complex tori).

-- Fropuff 8 July 2005 04:30 (UTC)

I don't know. Jost, Compact Riemann Surfaces doesn't appear to attempt such a naming convention. linas 15:29, 11 July 2005 (UTC)
No, I don't suppose Jost would, as by the title of his book he only treats the compact case, which are all elliptic according to Farkas and Kra. I'll have to see if I can find any other references. -- Fropuff 16:01, 11 July 2005 (UTC)

I've checked a few other sources. Both Ahlfors (Riemann Surfaces, 1960) and Beardon (A Primer on Riemann Surfaces, 1984) use the Farkas and Kra meaning for parabolic and hyperbolic. They don't define elliptic spaces, refering to those spaces simply as compact surfaces. These books also mention that hyperbolic surfaces (in their sense) are precisely those surfaces which admit Green's functions.

On the other hand, Forster's book (Lectures on Riemann Surfaces, 1981) uses the meaning described in this article. He notes however that no one really refers to the Riemann sphere as an elliptic Riemann surface, as it might cause confusion with elliptic curves (i.e. complex tori, compact surfaces of genus 1).

Anyone have any others? -- Fropuff 19:20, 11 July 2005 (UTC)

Gamelin's Complex Analysis has a chapter on Riemann surfaces, with no mention of elliptic, parabolic, or hyperbolic surfaces. Lang's Complex Analysis, doesn't do Riemann surfaces, nor does Wells' Differential Analysis on Complex Manifolds, nor does Kodaira's Complex Manifolds and Deformations of Complex Structures.
On the other hand, it is easy to find the words elliptic/hyperbolic/parabolic in a book on Riemannian geometry where they mean positive/negative/zero curvature. I guess that usage is the source of the terms in this article.
It would be nice if the two usages agreed, but I guess they don't. I guess they agree one which ones are the simply connected versions: the sphere is elliptic, the plane parabolic, the half-plane hyperbolic. But with the nontrivial surfaces, they seem to be in total disagreement. Farkas and Kra want the many-handled torus to be elliptic, while its Riemannian geometry is hyperbolic. Hrmm...
I guess I think the article should be brought in line with Farkas and Kra, but it's not enough to just shuffle the words, some stuff has to be said about subharmonic functions and such. Anyway, I always thought it was weird how, according to this article, there is only one elliptic surface. Hardly seems worth giving it a name, if there's only the one, right? -Lethe | Talk 22:35, July 11, 2005 (UTC)

Hi! I would stick with the curvature definition. I guess the different terminology comes from the different applications. Riemann surfaces play a role in many different areas. From the geometric viewpoint, you should name them after their curvature. sphere with curvature 1, torus curvature 0, the two special cases kind of, and then all higher genus surfaces with curvature -1 and the hyperbolic space as universal cover.

Note that if you care for any geo-"metric" aspects (maybe in contrast to purely analytic aspects) this seems very important, as the sphere is the prototype of the sphere (somewhat stupid) the plane a prototype for a neighborhood on the torus and a higher genus surface looks like hyperbolic space IH locally.

Hope that makes sense! Norman - nhilbert.de

[edit] formal definition

Hello,

I must say the definition given here looks the most accessible one I have found on the internet. However there is one thing I don't understand and perhaps someone could clarify this : when talking about compatible maps f and g, suppose f is a map from G (open subset in X) to H (subset of C) and g is analogously from I to J

we require that G and I are not disjoint( intersecting domains) but what kind of map is f after g^(-1) then

from J to H or are there more restrictions?

Yes, there's a restriction, its defined only on the intersection of H and I. The article Atlas (topology) is supposed to explain this in detail, but it doesn't  :( linas 22:09, 15 August 2005 (UTC)

I don't see why we need the appeal to Zorn's Lemma. To get a maximal atlas containing a given one can't we just grab all charts compatible with all the charts already in the atlas. Any pair of these new charts must be compatible because they're compatible with all the charts of the original atlas, which covers. Can someone back me up or smack me down? Amling 16:40, 22 January 2006 (UTC)

Sounds right. In the smooth category, your statement is Lemma 1.10(a) in Lee's Introduction to Smooth Manifolds, GTM. Melchoir 18:30, 22 January 2006 (UTC)

[edit] Brave New World

Just thought I'd note that in the book "Brave New World", they play tennis on what is described as Riemann Surfaces. :-) In case anyone wanted to know. —The preceding unsigned comment was added by 62.31.86.15 (talk • contribs) 23:25, 1 January 2006.

Dang, I don't have a copy of the book on me. Could someone else add it to the article? Melchoir 00:42, 2 January 2006 (UTC)
Never mind, I got it. Melchoir 00:48, 2 January 2006 (UTC)

[edit] Fundamental Error

The article makes this claim: " A two-dimensional real manifold can be turned into a Riemann surface (usually in several inequivalent ways) if and only if it is orientable."

Alas, this is simply not true. Perhaps it goes without saying that a Riemann surface must be Hausdorff, but this is not sufficient. It must also have a countable base to its topology. One counterexample is the square of the long line. Another quite surprising example is the Prüfer manifold: Start with an open half-plane H = {(x,y) in R^2  : y > 0}. For each (c,0) on the x-axis, we augment H in a strange way: Define H_c as the homeomorphic image of H by the (real analytic) homeomophism h_c: H -> H (which is strictly into):

                                            h_c(x,y)  =  (x,y) + (x-c,y)/sqrt((x-c)^2 + y^2).

(That is, we push each (x,y) directly away from (c,0) by exactly one unit.) Now take the union of H_c with the original H, which has the effect of adding a semicircle of radius 1 about (c,0) to what was H (which for each c we identify with H_c). Finally, perform this augmentation once for each (c,0) on the x-axis. The result is a real-analytic surface, the "Prüfer manifold", which obviously does not have a countable base to its topology (or a countable dense subset), since we have added uncountably many disjoint (2D) semicircles to what was H. Such a manifold cannot be made into a Riemann surface.Daqu 06:10, 16 May 2006 (UTC)

"Manifolds", in the sense of this article, are implicitly second-countable. See Manifold. Melchoir 06:31, 16 May 2006 (UTC)
Granted, but this only shows that Wikipedia's various definitions of a manifold need to be put on a sounder logical footing. E.g., the entry Topological manifold makes no such assumption of Hausdorff or separability -- which is exactly as it should be.) When topologists get serious, they specify a manifold as being topological with such-and-such precise additonal structure. In the present case, a significant episode in the history of Riemann surfaces was to determine whether or not all Hausdorff locally Euclidean connected 2-manifolds with holomorphic transition maps had to be separable (or equivalently are second countable). The Prüfer manifold especially called this into question. The fact that a 1-dimensional Hausdorff connected complex manifold is automatically separable is quite surprising.
Perhaps just as surprising is that higher-dimensional complex manifolds defined only as locally Euclidean, Hausdorff, connected and having an atlas with holomorphic transition functions need not be second countable! (Just take the complexification of the Prfer manifold.)
So while I don't advocate getting too technical too quickly, it's also a good idea not to make easily-misinterpreted statements either (like that every connected surface can be given a Riemann surface structure).Daqu 14:30, 16 May 2006 (UTC)
Actually, when topologists get serious, they do geometry ;) Sam nead 23:16, 23 October 2006 (UTC)

[edit] Attempting to understand

I've got a question which doesn't seem to be in the article. What are the conditions for a real 2-manifold to be equivalent to a Riemann surface? From what I gather they are that: if on any overlap between a (x,y) coordinate patch and a (X,Y) coordinate patch Z = X+iY\, is holomorphic in z=x+iy\,; and that the Cauchy-Riemann equations hold: \frac{\delta X}{\delta x} = \frac{\delta Y}{\delta y}, \frac{\delta X}{\delta y} = -\frac{\delta Y}{\delta x}. Are those all the conditions? —porges(talk) 10:52, 10 July 2006 (UTC)

Also, is this equivalent? "A complex function f(x + iy) = u + iv is holomorphic if and only if it satisfies the Cauchy-Riemann equations and u and v have continuous first partial derivatives with respect to x and y."

I guess the shorter version is "if a real surface has holomorphic transition functions and it satisfies the Cauchy-Riemann equations, is it the same as a Riemann surface?" —porges(talk) 00:24, 12 July 2006 (UTC)
The first part is correct. If a surface has holomorphic transition functions, then it is a Riemann surface. Holomorphicity means that the transition functions satisfy the Cauchy-Riemann equations, so you don't have to say that twice. -lethe talk + 00:58, 12 July 2006 (UTC)
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