Talk:Riemann surface

message: 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)
thread_title: sheaves

message: 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)
replies: text: Still planing on making good on the above threat...real sooner now. linas 14:59, 14 Apr 2005 (UTC)
replies: text: 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)
replies: text: 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)
thread_title: Analytic continuation

message: 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)
thread_title:  sheaves 

message: 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)
replies: text: 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)
thread_title:  formal definition 

message: 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. —Preceding unsigned comment 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)
replies: text: Never mind, I got it. Melchoir 00:48, 2 January 2006 (UTC)
thread_title:  Brave New World 

message: 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)
replies: text: 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)
replies: text: 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)
thread_title:  Attempting to understand 

message: I'd rather not correct this myself, since I'm neither a complex analysis expert, nor a Wikipedia expert. This is just a thought:
The image of the surface for \sqrt{z} depicted in the upper left corner is actually only depicting the real part of a multi-valued function f: C \rightarrow C. What about including also the imaginary part as a separate image? It might be confusing to beginners, that's for sure. Maybe it wouldn't be bad to also include a page about ways how to imagine such mappings and provide a link. I.e. functions R \rightarrow R^2 and R^2 \rightarrow R can be visualised simply in 3D, but that's about it. Even though 4D cannot be imagined, we can still get the general idea by emplying projections into 3D.
On the other hand, Wikipedia is supposed to be a reference rather than a textbook, so I don't know how appropriate this would be.
193.85.150.3 (talk) 11:27, 10 December 2007 (UTC)
thread_title:  The image 

message: These are depicted as having branch points at the origin . . . and without having poles there.
But the functions z-n for n = 2,3,4 have no branch points, and each has a pole at the origin.  Can it be that the images are in fact the Riemann surfaces for z1/n for n = 2,3,4 ?  In any case, something seems to need fixing here.Daqu (talk) 18:01, 9 February 2009 (UTC)
thread_title: Images of Riemann surfaces labeled as for for z<sup>-2</sup>, z<sup>-3</sup>, z<sup>-4</sup>

message: Do you think that it is clear,when you 'draw' the Riemann surface associated to the holomorphic function,for example,f(z)=\sqrt z,what does this drawing represent ? A function drawn in a 3-dimension space must represent,in the simplest interpretation,a real function z in the two real variables x and y,so that z=f(x,y).A holomorphic function f in the complex variable z,should at least be represented graphically by TWO surfaces in a 3-dimension space,that is one for the real part and one for the imaginary part of f.
Don't you think that this elementary issue should be cleared up,before talking of anything about the Riemann surfaces ?  —Preceding unsigned comment added by 90.0.205.206 (talk) 19:23, 22 March 2010 (UTC)
replies: text: I gave it a shot. Better now? Hanche (talk) 22:56, 23 March 2010 (UTC)
thread_title: Please, explain what the drawings of surface in the article represent 

message: The fourth paragraph reads in full:
"Geometrical facts about Riemann surfaces are as "nice" as possible, and they often provide the intuition and motivation for generalizations to other curves, manifolds or varieties."
I love Riemann surfaces, but:
To claim that certain geometrical facts are "nice as possible" -- with or without the quotation marks around the word "nice" -- is just meaningless, and hence just confusing.  If there is an underlying fact here, then why not just state it?
Certainly many Riemann surfaces have conformal automorphism groups that are much smaller than the largest possible symmetry group for a Riemannian surface of the same genus.  This certainly isn't "as nice as possible", however the phrase is defined.
Nor is the inability to measure length, a staple of most forms of geometry.
Also, every  compact Riemann surface is conformally equivalent to some (projective) algebraic curve, so it seems weird to compare them with "other" curves.Daqu (talk) 01:06, 12 May 2012 (UTC)
thread_title:  Confusing statement should be removed 

message: As I understand it (undergraduate with limited analysis/topology experience), manifolds proper don't self-intersect or local homeomorphisms would fail, right (an X is not locally homeomorphic to a line, for instance)? So perhaps a note that sqrt(z) does not properly self-intersect, only its immersion in 3D, would be worthwhile beneath the page image. 67.79.154.194 (talk) 00:45, 15 December 2012 (UTC)
replies: text: I think this highlights a serious problem with the diagram (and others in the article). The diagram illustrates a multi-valued function on the complex plane, and not a Riemann surface. This is likely to cause confusion for those that do not already understand what a Riemann surface is. A key point about a Riemann surface is that there is no vertical axis: it may be thought of as the domain of a function, not as the function itself. This misrepresentation implicit in the diagrams should really be addressed. — Quondum 06:22, 15 December 2012 (UTC)
replies: text: What he said. 76.106.149.108 (talk) 19:12, 30 June 2013 (UTC)
thread_title:  Self-intersection 

message: The Simple English Wikipedia article on this. 76.106.149.108 (talk) 19:11, 30 June 2013 (UTC)
thread_title:  Needed: 

message: The article begins with the statement:
"Riemann surfaces can be thought of as deformed versions of the complex plane: . . .."
No, they can't be thought of this way — not without deeply confusing the thinker.Daqu (talk) 20:50, 10 February 2015 (UTC)
thread_title:  As "deformed versions of the complex plane" ??? 

message: The article states that "Riemann surface [...] is a one-dimensional complex manifold". But later it states that "Every Riemann surface is a two-dimensional real analytic manifold (i.e., a surface)". So how many dimensions does it have? Since it's a surface, it should be 2, right? 
89.229.8.110 (talk) 20:45, 26 August 2017 (UTC)
replies: text: (I know this question is 2 years old but...) A dimension d complex manifold is a dimension 2d real manifold (with a complex structure added). Complex manifold are locally modelled on C^d, and C^d is R^2d if we forget the complex structure. --Arnaud Chéritat (talk) 23:35, 26 December 2019 (UTC)
thread_title:  Dimension 

message: The definitions 1 and 2 are not fully equivalent: one mentions connectedness and not the other. --Arnaud Chéritat (talk) 23:37, 26 December 2019 (UTC)
thread_title:  Connected or not 

message: Paragraph three of the introduction states a necessary condition for a 2d manifold to be a Riemann surface - it must be orientable and metrizable.  It then gives three counterexamples of non-orientable manifolds.  Seems to me that the article would be improved by providing a non-metrizable counterexample. Not crucial to the article but an improvement.
Suggestions for non-metrizable counterexamples cheerfully accepted.  Mr. Swordfish (talk) 23:45, 30 March 2021 (UTC)
replies: text: metrisability for locally compact separable (in T3 sense) spaces is more or less equivalent to being second countable so the non-metrisable connected surfaces are exactly these. For a explicit example you can take the long line times a circle. jraimbau (talk) 12:13, 31 March 2021 (UTC)
replies: text: Thanks.  Long line cross S2 seems like an obvious counterexample (obvious once you point it out, that is).  Should we mention it in the article? We'd need a cite, of course.  Mr. Swordfish (talk) 23:38, 31 March 2021 (UTC)
thread_title:  Non-metrizable counterexample? 

message: The description of the first image says "For this function, it is also the height after rotating the plot 180° around the vertical axis." ("it" refers to the color). 
I don't see how a rotation in the vertical axis would achieve this. Atoll (talk) 07:31, 2 March 2023 (UTC)
thread_title:   in the first image, after rotation colour = height, really?