# Talk:Riemann sphere

WikiProject Mathematics (Rated B-class, Top-priority)

## This article needs lots of work

Take a look at the short paragraph below:

(Another reader notes that afaik this article has the stereographic projection wrong - the sphere and the plane should not have the same origin. The sphere rests on the plane and the sphere point is defined as the 'other' intersection of the line with the sphere. 62.49.9.94

Stereographic projection can be performed onto many different planes. Choosing the plane to have the same origin as the sphere is particularly convenient for the Riemann sphere, because it makes the unit circles line up, etc. Joshua R. Davis 15:47, 3 September 2007 (UTC)

## Complex structure

The complex manifold structure on the Riemann sphere is specified by an atlas with two charts and coordinates z and w

${\displaystyle z:{\hat {\mathbb {C} }}\setminus \{\infty \}\to \mathbb {C} \,}$
${\displaystyle w:{\hat {\mathbb {C} }}\setminus \{0\}\to \mathbb {C} \,}$

The transition function between the two patches is w = 1/z, which is clearly holomorphic and so defines a complex structure. To see that these charts give the topology of the sphere note that we can give an atlas on S2 by stereographic projection onto the complex planes tangent to the north and south poles respectively. Labeling points in S2 by (x1, x2, x3) where ${\displaystyle x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=1}$, we have

${\displaystyle z={\frac {x_{1}+ix_{2}}{1+x_{3}}}}$
${\displaystyle w={\frac {x_{1}-ix_{2}}{1-x_{3}}}}$

which satisfies the equation w = 1/z. In terms of standard spherical coordinates (θ, φ)

${\displaystyle z=e^{i\phi }\tan {\frac {\theta }{2}}}$
${\displaystyle w=e^{-i\phi }\cot {\frac {\theta }{2}}}$

### Problems

• For the map:
${\displaystyle z:{\hat {\mathbb {C} }}\setminus \{\infty \}\to \mathbb {C} \,}$,
what is the formula?
• For the map :${\displaystyle w:{\hat {\mathbb {C} }}\setminus \{0\}\to \mathbb {C} \,}$,
what is the formula?
• The equations
${\displaystyle z={\frac {x_{1}+ix_{2}}{1+x_{3}}}}$
${\displaystyle w={\frac {x_{1}-ix_{2}}{1-x_{3}}}}$

are not in the proper context, they implicitely assume that the connection between the Riemann sphere and the normal sphere is already established, while it was not.

Also, in a paragraph below in text one says:

When the sphere is given the round metric...

Same problem. This makes no sense until a correspondence between the normal sphere and the Riemann sphere is explicitely stated, and that correspondence is fixed, so there is no ambiguity. Oleg Alexandrov 01:57, 15 Feb 2005 (UTC)

Perhaps the wording in the article is not the best. If you think things would be clearer, feel free to make it more rigorous. The idea is very simple though. You have two coordinate charts (both copies of C) with coordinates z and w. The transition function on the overlap (where both z and w are nonzero) is given by z = 1/w. That's all the information needed to specify the complex manifold. The equations relating z and w to the x's is defining the relationship between the geometrical unit 2-sphere and the Riemann sphere. -- Fropuff 04:23, 2005 Feb 15 (UTC)

From my knowlege of differential geometry, to define a manyfold you need (a) some charts covering it (b) a map from each chart to a subset of R^n (C^n) (c) the transition function (that follows from (b) actually). You did (a) and (c), but did not specify (b). I will do some changes in this article tomorrow. Oleg Alexandrov 04:40, 15 Feb 2005 (UTC)
My comments were exaggerated, sorry. One could have realized that by z you mean the identity map, z→z, and that by
${\displaystyle z={\frac {x_{1}+ix_{2}}{1+x_{3}}}}$
you mean the map which starts at a point (x_1, x_2, x_3), takes as output this fraction, which you denote by z, and which you further identify with the chart going from ${\displaystyle \mathbb {C} }$ to ${\displaystyle {\hat {\mathbb {C} }}}$ or the other way around. But making these explicit as I did, certainly does not hurt. Oleg Alexandrov 20:04, 15 Feb 2005 (UTC)

No worries. It is a little confusing with C serving as both part of the manifold and as coordinate charts. One can be clever with the notation but it is perhaps more confusing in the end. -- Fropuff 02:30, 2005 Feb 16 (UTC)

One certainly should not make notation stay in the way of understanding. Point taken. The only reasons I introduced the notation f and g was to make explicit the charts, as unless you work all your life in complex analysis and know that z is the default variable, one would not understand how z is defined. If you feel notation is getting in the way somewhere, please let me know, I don't quite see it myself.
You said that it is a little confusing with C serving as both part of the manifold and as coordinate charts. I don't know what we can do about it. After all, that's how the Riemann sphere is defined, as
${\displaystyle {\hat {\mathbb {C} }}=\mathbb {C} \cup \{\infty \}.}$
Or this is not what you meant? Oleg Alexandrov 02:51, 16 Feb 2005 (UTC)

That is what I meant. But as you say, there is little that can be done about it. I think what is there now works fine. -- Fropuff 03:01, 2005 Feb 16 (UTC)

## Complexity of the article (presentation angle)

(No puns intended on complexity or angle)

I have found several much simpler explanations for the complex sphere. Examples include the one in Penrose's The Road to Reality or even the more technical one in Needham's Visual Complex Analysis. Their common simplifying factor is presenting it in terms of geometry, with diagrams to boot.

Presenting the Riemann sphere in terms of a set theory and mappings seems unnecessarily abstract. Giving the transformations and set-theoretic definitions, IMO, should be secondary to giving the geometric derivations of mapping the complex plane (+ infinity) to points on a sphere.

Disclaimer: I am a high school senior learning these things on my own. I don't know how textbooks do it. I just look at what I'm given and judge. In the case of Penrose and Needham, I say "this makes sense and is really cool!" But for Wikipedia, I come out of it more like "huh? Why do it this way? What the heck does this even have to do with spheres?"

So basically, I don't pretend to be an authority on this. —Preceding unsigned comment added by DomenicDenicola (talkcontribs)

Well, if you wish, you could start by adding diagram and some text, and we may help. What you say makes sense. Oleg Alexandrov (talk) 21:55, 8 December 2005 (UTC)
The presentation in the article is geometrical; at least to a graduate student in mathematics. But, you are right, the Riemann sphere is often encountered by bright high school students long before graduate school. This page could do with some pictures and a gentler introduction. -- Fropuff 22:07, 8 December 2005 (UTC)

### Working draft of geometric rewrite

(Notes: this is meant to be standalone, so when it's done we would find overlap from other sections and remove it here or there.)

Define ${\displaystyle {\widehat {\mathbb {C} }}=\mathbb {C} \cup \{\infty \}}$ (i.e. the complex numbers joined with the point at infinity). The Riemann sphere is based on the transformation from ${\displaystyle {\widehat {\mathbb {C} }}}$ to ${\displaystyle {\widehat {\mathbb {C} }}}$ and is in the form

${\displaystyle w=f(z)={\frac {1}{z}}}$,

where ${\displaystyle w,z\in {\widehat {\mathbb {C} }}}$ and ${\displaystyle {\frac {1}{0}}=\infty }$.

We visualize the Riemman sphere as a sphere in 3-space, i.e. in ${\displaystyle \mathbb {R} ^{3}}$. Every point on the sphere has both a ${\displaystyle z}$ value and ${\displaystyle w}$ value, related by the above transformation. That is, ${\displaystyle f(z)}$ transforms the sphere onto itself.

#### Stereographic projection

To establish the correspondence between points in the extended complex plane and the Riemann sphere, we first place the ${\displaystyle z}$ plane across the sphere's equator. We then use stereographic projection from the south pole of the sphere. This is done by drawing a line from the south pole that intersects both the sphere and the complex plane; a unique, one-to-one correspondence is then established between points on the complex plane and points on the Riemann sphere. Note that points on the complex plane inside the unit circle will map to the upper hemisphere, and points outside will map to the lower hemisphere.

In order to complete this one-to-one correspondence for the extended complex plane, we define the south pole to be ${\displaystyle z=\infty }$. Note that the north pole is ${\displaystyle z=0}$.

The correspondence between the ${\displaystyle w}$ plane and the Riemann sphere is done in much the same way, simply "upside down." That is, the ${\displaystyle w}$ plane is an equitorial plane oriented oppositely to the ${\displaystyle z}$ plane, such that ${\displaystyle w=1,i,-1,-i}$ matches to ${\displaystyle z=1,-i,-1,i}$. We then perform the stereographic projection from the north pole, and similarly define the north pole to be ${\displaystyle w=\infty }$. Now, every point on the sphere has both a ${\displaystyle z}$ and ${\displaystyle w}$ coordinate, related by the transformation above.

#### Geometric features of note

The equator of the sphere is the unit circle in the complex plane; in a similar fashion, circles can be found for the imaginary line and real lines. Note that these are shared between the two projections, because the relation ${\displaystyle w={\frac {1}{z}}}$ is holomorphic.

This is a specific case of how stereographic projection maps all lines and circles in the complex plane to circles on the Riemann sphere. The reason that lines are mapped to circles is that a line with infinite length can simply be thought of as a circle that passes through the point at infinity.

#### Möbius transformations

Möbius transformations, which send ${\displaystyle {\widehat {\mathbb {C} }}}$ to ${\displaystyle {\widehat {\mathbb {C} }}}$, are often visualized as acting on the Riemann sphere. They are in the form

${\displaystyle t=f(z)={\frac {az+b}{cz+d}}}$,

where ${\displaystyle t,z\in {\widehat {\mathbb {C} }}}$, ${\displaystyle a,b,c,d\in \mathbb {C} }$, and ${\displaystyle ad-bc\neq 0}$. They map the sphere to itself, preserving important features such as angles and circles/lines. This is because they are only composed of dilations, translations and inversions.

#### Problems

I think I've confused myself with the whole Möbius transformation thing. I believe my explanation is only valid for the transformation ${\displaystyle w=1/z}$. So how do Möbius transformations fit in? I know they map Riemann spheres to Riemann spheres, but I don't think they do so in the manner I described (simple projection through the matched planes).

I think I fixed this, although I'm a little shaky on the need for/accuracy of the new Möbius transformation section. I just thought that since they were so important to the Riemann sphere the article should say something about them. Domenic Denicola 00:20, 10 December 2005 (UTC)

Does this sentence even make sense? I might not be saying what I'm trying to say... "Note that these are shared between the two projections, because the Möbius transformation is holomorphic."

Made irrelevant by the above fix; I certainly know that ${\displaystyle w={\frac {1}{z}}}$ is holomorphic! Domenic Denicola 00:20, 10 December 2005 (UTC)

Comments on explanation? I really like it, but hey, that's because it makes sense to me.

I really need diagrams, especially ones just showing the sphere superimposed with at least the ${\displaystyle z}$ plane and with ${\displaystyle 1,-1,i,-i}$ labelled.

This will depend on how we place the planes, but I could scan Penrose if we stay with equitorial planes. Domenic Denicola 00:20, 10 December 2005 (UTC)

Domenic Denicola 19:50, 9 December 2005 (UTC)

Scanning Penrose is bound to violate copyright laws. I really think we should stick with tangent planes anyway. You could put up a request on Wikipedia:Requested_pictures#Mathematics. -- Fropuff 00:28, 10 December 2005 (UTC)
OK; I'll do some searching around and then probably put in a request. I think I found why Penrose likes the equatorial version; it works better for Twistor theory. See this page with diagrams.

### Comments of above suggestion

• Mobius transformations are usually defined in terms of the Riemann sphere and not the other way around. One should leave them out of the definition to avoid circular logic. The only transformation you have to worry about is w = 1/z.
I'm a litte confused as to what exactly the point of having a ${\displaystyle w}$ coordinate even is then. What does it give you? Why not just map a ${\displaystyle z}$ plane onto a sphere?
(Speculation) Maybe it's because Riemann surfaces are always defined in terms of transformations or functions? I don't have much background on those either, but e.g. I know that ${\displaystyle w=\log(z)}$ has a twisting spiral plane surface; maybe you need the transformation ${\displaystyle w=1/z}$ to get a sphere surface. Domenic Denicola 00:32, 10 December 2005 (UTC)
• It is more common to stereographically project onto tangent planes rather than to planes through the equator. This has the added advantage of making the z-plane and the w-plane geometrically distinct.
After thinking about this statement for a while, I realized you meant planes tangent to the poles (right?). I guess it makes sense, but (probably just because I read the stacked planes version first) I don't like it as much. Could you find a web page that shows this in more detail, so I can hopefully be enlightened on why it's a nice way of doing things? I think we should change it too, if it's the conventional way of doing it; however, I'd like to understand the new version more thoroughly first. Domenic Denicola 00:32, 10 December 2005 (UTC)
• The section on Complex structure in the article needs to remain as it shows how the Riemann sphere takes on the structure of a 1-dimensional complex manifold. I would suggest adding your material (sufficiently revised) as an introductory/motivation section.

-- Fropuff 21:15, 9 December 2005 (UTC)

Agree with Fropuff. Now, the current article introduction, saying
In mathematics, the Riemann sphere, named after Bernhard Riemann, is the unique simply-connected, compact, Riemann surface.
is intimidating indeed. I would start the article by saying the Riemann sphere is obtained by imagining that all the rays emanating from the origin of the complex plane eventually meet again at a point called the infinity, in the same way that all the meridians from the north pole of a sphere get to meet each other at the south pole. I don't know. Ideas? Oleg Alexandrov (talk) 22:08, 9 December 2005 (UTC)
Yes, it would be good to have something like that as the introduction. Another good blurb is "a useful visualization of the extended complex plane, especially when doing Mobius transformations." Other ideas would be good too :). Domenic Denicola 00:32, 10 December 2005 (UTC)
Feel free to add this kind of stuff, looks good. We may edit later if need be. About pictures. I could make a picture if necessary, but please note that we already have a picture (the very top of the article), and putting a sphere and a plane in the same picture while showing how the correspondance goes might not be possible. Oleg Alexandrov (talk) 02:03, 10 December 2005 (UTC)
I think these are some good samples of what I am looking for: sample 1 sample 2 sample 3.
Also, I am beginning to wonder about Fropuff's statement that tangent to the poles is more common. At least in stereographic projection---if not in Riemann spheres, necessarily---it seems the equitorial approach crops up a lot more often in my (highly unscientific) Google image search.

I deleted these sentences because I wasn't sure they worked for tangent planes:

Note that points on the complex plane inside the unit circle will map to the upper hemisphere, and points outside will map to the lower hemisphere.
The equator of the sphere is the unit circle in the complex plane; in a similar fashion, circles can be found for the imaginary line and real lines. Note that these are shared between the two projections, because the relation ${\displaystyle w={\frac {1}{z}}}$ is holomorphic.

If they only fit in with the equitorial projection, I could add them in the section on that; alternatively, if they work in both, I'll restore them to the appropriate places. I need guidance, however.

The planes are still switched in orientation, right?

Hope everyone likes the results!

Domenic Denicola 06:32, 13 December 2005 (UTC)

The rewrite looks good, thanks. I have some comments though.
1. I don't understand the sentence
The Riemann sphere is based on the transformation from ::::${\displaystyle {\widehat {\mathbb {C} }}}$ to ${\displaystyle {\widehat {\mathbb {C} }}}$ in the form
${\displaystyle w=f(z)={\frac {1}{z}}}$,
where ${\displaystyle w,z\in {\widehat {\mathbb {C} }}}$ and ${\displaystyle {\frac {1}{0}}=\infty }$.
and I don't think it belongs in the elementary introduction. The article disucsses Mobius transformations later.
I didn't mean this as a Mobius transformation. Perhaps it is not necessary, but I think since we talk a lot about how w and z coordinates are related it should be there. I was under the impression that all Riemann surfaces were representations of some function (e.g. ${\displaystyle w=log(z)}$ is a spiraling plane), and the Riemann sphere was the representation of ${\displaystyle w=1/z}$. The appropriate way to introduce this to someone not familiar with Riemann surfaces (in order to keep this article free from "required reading") seems to say that it is "based on the transformation." Does that sound right? How could I phrase it to make it more understandable? Domenic Denicola 20:49, 13 December 2005 (UTC)
2. The sentence: "To establish the correspondence between points in the extended complex plane and the Riemann sphere," is not right. The Riemann sphere and the extended complex plane are the same thing. The correspondence is between the extended complex plane (also known as the Riemann sphere) and the ususual sphere in R^3, a correspondence which justifies the name.
Proposed new section (please check for accuracy):
To establish the correspondence between points in the extended complex plane and a 3-sphere, we first place the ${\displaystyle z}$ plane tangent to the sphere's north pole. We then use stereographic projection from the south pole of the sphere. This is done by drawing a line from the south pole that intersects both the sphere and the complex plane; a unique, one-to-one correspondence is then established between points on the complex plane and points on the 3-sphere. The "Riemann sphere" is thus a name for either of these structures, as they are mathematically equivalent. Domenic Denicola 20:49, 13 December 2005 (UTC)
I don't know. The point is that the Riemann sphere is not a sphere to start with, it is just the complex plane plus the point of infinity. That's be definition. Now, one can explain why it is called a Riemann sphere, but it does not chang ee the fact. Oleg Alexandrov (talk) 21:27, 13 December 2005 (UTC)
3. I am not sure I like the way the article is split in sections and subsections.
Overall, looks good, thanks, great job! These are all minor things. Oleg Alexandrov (talk) 16:09, 13 December 2005 (UTC)
Did Fropuff fix that for us? And thanks :). Domenic Denicola 20:49, 13 December 2005 (UTC)

## Baffled

This article leaves me utterly confused. I don't expect to grasp the formulas toward the bottom of the page in any math-related article; I don't have sufficient background. But none of this makes any sense at all, not from the first few words.

How is a Riemann sphere distinct from an ordinary sphere? Explain it so my daughter can understand it. She knows the difference between a sphere and a circle; is this the same thing? Circle, sphere, Riemann sphere? (Square, cube, hypercube.)

Is there any difference between Riemann sphere and Riemann space? If so, what? Say it without any numbers or special words. If I lived on the surface of a Riemann sphere, what would be different about my life? Would all my doughnuts turn into coffeecups? Are squares still square?

Is there more than one Riemann sphere? Can they come in different sizes? (I suspect the answers are no and no.) Could I even tell if I did live in a Riemann sphere? It looks to me, offhand, as if the thing is of infinite size. Wouldn't any finite zone or section (my local known universe) always appear Euclidean? Is any point on the Riemann sphere distinguished? If so, what would happen if I stood there? Would I blow up? Would my left and right hands disappear or get stuck together? Could I even tell?

What is the use of this thing? How can I apply it and to what? Automotive engineering? Faster-than-light spacecraft theory? Molecular biology? Game theory? Can I win a bar bet with it; if so, what's the bet?

I came here because graphics were requested; I'm willing to do graphics. But first, somebody will have to tell me what it is. John Reid 20:14, 14 April 2006 (UTC)

Good points, but you are expecting too much from an article on an abstract topological concept. So here we are:
The Riemann sphere is not a sphere, it is the complex plane plus an extra point (paragraph 1)
The Riemann sphere is topologically equivalent to the real sphere (paragraph 2). Sorry, but that topological homeomoprhisms can't be explained easily to people who don't know what topology is.
The question about different sizes of Riemman spheres does not make sense, again, a Riemann sphere is the complex plane with an extra point.
One of its uses is to conveniently deal with meromorphic functions, or otherwise, functions taking infinite values. I know that is not what you are looking for, but this is an abstract concept, used in other areas of math, and not directly in real life.
In short, your questions are appreciated, and you are right that there is always room for improvement for abstract math articles, but you should also not expect way too much, some things are abstract and complex because they must be so. Oleg Alexandrov (talk) 20:23, 14 April 2006 (UTC)

The fine points may be lost on a layman but please have the courtesy not to look down your nose at me. I have enough background in topology to understand why I can remove my vest without taking off my coat. I know that if I glue together all four corners of a checkerboard and glue (along each edge) four squares to four squares then I have created a bag-like thing which is not equivalent to a sphere; it is not even a manifold because there are distinguished points. I know why I can never comb a hairy ball. And I even have a pretty fair idea why the N-color problem has a different solution on a torus. If you can't explain the subject of this article to me, I think you don't have a clear grasp of the concept -- only what you have been told. No offense intended.

The question about different sizes makes a great deal of sense. One of the first things I do when trying to understand a new concept is to search for limits. Is there one only or more than one? Does it come in more than one size or variety? As I said, I suspect there is only one, of infinite size. You still have not told me if the point at infinity is distinguished. Indeed, you've failed to answer most of my questions. That's okay; but if you don't know the answers then please don't be so dismissive of the questions.

Nothing useful is purely abstract. Experts who work with a concept regularly may be comfortable with an abstract representation; but there is always some connection to reality -- otherwise the concept is nonsensical. You may have difficulty associating this topic to a practical aspect of life but I suggest that there is such a connection. If not, I'm tempted to say that however important it may be to a specialist, it holds no possible interest for the general reader. John Reid 22:20, 14 April 2006 (UTC)

I guess I found your question to be rather inflammatory, with wording like "utterly confused, baffled, etc". It was as if you are were saying "you people did an awfully bad job here" rather than "Hi all, I don't know a huge lot about topology, so would you be so kind to explain". But it seems that I answered in kind.
That said, I will take some time to think of a gentle answer to your question. Oleg Alexandrov (talk) 23:28, 14 April 2006 (UTC)
I am completely in favour of making all concepts as accessible to as many people as possible by describing them first in the most undemanding intuitive way possible, but it is wise to accept the limits of what is reasonable. It would be dishonest to claim that all mathematical topics can be made accessible to readers with all levels of knowledge. The first paragraph (if a reader ignores words they might not understand) should get the partial message across that the Riemann sphere is the complex plane plus one point called "infinity" which can be viewed as a 2-dimensional sphere (the stereographic projection section explains how this works). This is almost as much as is possible to get across without the concept of a complex manifold. The one thing I think is missing from the introduction is a mention of the fact that the point at infinity is used to extend the domain and the range of many complex functions in a useful way. Without a knowledge of complex analysis, the most honest thing to say would be that this would not be a useful article to read at all. Mathematics is hierarchical, and in many cases it is necessary to have a fair knowledge of certain subjects in order to get to grips with more advanced ones: I would not expect to understand an advanced text on any subject where I didn't understand the prerequisites, although I have often wished I could. This is simply being realistic. As an elementary analogy, in order to understand the purpose of the sieve of Eratosthenes (which I fondly remember executing by hand late in primary school) it is necessary to first study multiplication. Paraphrasing Euclid's honest words to king Ptolemy I, "there is no royal road to" differential "geometry" Elroch 01:03, 15 April 2006 (UTC)
Problems with the introduction at the moment are that there are a number of things mentioned which are really part-truths and distractions from the main idea. The Riemann sphere is not constructed using lines, nor is it constructed using the 1-point compactification of the plane. The first is a geometrical concept, the second a topological one. The Riemann sphere is actually constructed (as a Riemann surface) in one step using two charts, as explained in the article, and the topology derives incidentally from this. The message to get across is:
(1) that using the function 1/z identifies the extended complex plane with zero omitted to a second copy of the complex plane, and allows us to transfer the notions of differentiability we know about on the complex plane to the extended complex plane with zero omitted.
(2) fortunately, this notion of differentiability is identical with the usual one on the overlap with the complex plane of the space in (1) (i.e on the complex plane with zero omitted). Hence it gives us a well-defined notion of differentiability on the whole extended complex plane.
(3) this notion of differentiability on the whole extended complex plane allows us to talk about the differential properties of functions from the extended complex plane (or open subsets of it) to itself (essentially, complex functions that may also be evaluated at infinity and may take the value infinity). A key example is that meromorphic functions extend in a unique way.
(4) almost incidentally, the topology on the Riemann sphere that derives from the differential structure is that of a 2-dimensional sphere. Elroch 23:52, 15 April 2006 (UTC)

To reply to John Reid, from the top down, here it goes. The Riemann sphere is just the complex plane with an extra point added in, called the point at infinity. For analogy, look at the real line. There, when dealing with limits, it is convenient to pretend that there exist two points ∞ and -∞ which are endpoints of the real line. Then ∞+∞=∞, and all other formal rules makes it easier to deal with limits without worrying much about particular cases of infinite limit.

In the same way, one can pretend that all rays in the complex plane originating from 0 actually have an endpoint, and they all eventually meet at infinity, a point far-far away (not accurate as Elroch mentions above, but helpful in imagining things).

The Riemann sphere is not the same as the usual sphere, but they are topologically equivalent. Imagine a normal sphere, remove the north pole, and make the obtained hole there larger and larger (assume the sphere is made of very flexible rubber). Eventually, that sphere without a point can be flattened in a plane, the complex plane. The original north pole corresponds to the point at infinity in the complex plane.

There is only one Riemann sphere, as the point at infinity is just a symbol, its actual nature is not relevant. In the same way that there exists essentially one normal sphere. The radiuses may differ, but any sphere can be deformed gently into another sphere, without tearing the surface. In exactly the same way a sphere is the same as the surface of a cube, but not with the surfce of a donut.

You can't say if any portion of the Riemann sphere appears Euclidean, or whether it is infinite in size or not. That because there is no concept of distance and size on the Riemann sphere. Any portion of the sphere can be stretched/shrank in any way as long as the sphere does not burst or separate patches merge.

The Riemann sphere does not get applied directly much beyond math, or otherwise I never heard of it. It is a useful construct, but rather abstract.

I was under the impression that it was used in quantum theory as a way of visualizing CP1, which represents the spin state of an electron. At least, that is, by Roger Penrose. Domenic Denicola 03:46, 16 April 2006 (UTC)

Above I talked about the topology of the Riemann sphere, not its differential geometry . But that would be harder to explain.

I don't know how satisfactory you found the answers. Try to read them though, and let me know if you have questions. Oleg Alexandrov (talk) 02:37, 16 April 2006 (UTC)

Thank you. My understanding of this topic has been broadened. Here is what I have gleaned so far:
(1) This subject is not a geometric thing at all, but only topological. Size and distance are meaningless here, so there's no point discussing things like the sum of the angles of a triangle.
(2) The subject is a manifold; no point is distinguished and a bug at any point would perceive the same local universe as one at any other. Indeed it's just another genus-0 spheroid, topologically.
(3) This subject has no practical application. It is a pure abstraction of interest only to specialists. There are no concrete examples possible, no jokes, no bar bets, nothing that might reach beyond mathematics into the commonplace world.
(4) This subject is highly resistant to any treatment that might make it accessible or useful to a general audience. It doesn't have a hot or sexy handle sticking out of it.
Editors are welcome to enlarge on and correct this understanding but I'm content. It might be nice if there was an attempt, very early in the introduction, to somehow indicate to the general readership that the density of the article is intrinsic to the topic and not an artifact of poor or obscure writing. This is beyond my capabilities; I'm moving on. John Reid 14:46, 17 April 2006 (UTC)
I'm a year late, but I feel that these assertions deserve a response.
• The Riemann sphere is homeomorphic to the unit sphere in R3. So as far as topology is concerned the Riemann sphere is just a sphere. But topology is not the point...
• Complex analysis is the raison d'etre. The Riemann sphere is a complex 1-manifold, usually described by two charts from C, each of which covers all but one point of the sphere. If z and w are the two coordinates, then the transition map is z = 1 / w. This is equivalent to the view of it as the complex projective line --- which connects seriously to algebraic geometry.
• Geometrically, there is a "most popular" metric, namely the Fubini-Study metric, which ends up being the round metric. Under this metric the Riemann sphere is locally conformal to C. You can see this either directly from the Fubini-Study formula or by identifying the Riemann sphere isometrically with the unit sphere in R3 and stereographically projecting.
• The z = 1 / w business lets us build up an elegant theory of meromorphic functions, which are crucial to complex analysis and hence applicable to many things that use complex analysis, probably including minimal surfaces/soap films, quantum mechanics, maybe two-dimensional fluid mechanics via stream functions (but now I'm out of my expertise). My personal sense is that just about every professional mathematician is familiar with the Riemann sphere, probably.
• The average person does not have a strong appreciation of the distinction between a real 2-manifold and a complex 1-manifold; hence they will not find this article very compelling compared to, say, Sphere. The "sexy handle" (a vivid image, by the way) could be quantum mechanics string theory; a gentler, more pedagogical approach could come via rational functions.
Joshua R. Davis 20:40, 25 April 2007 (UTC)

## Recent changes

Elroch made a lot of good edits to this article. It is now more mathematically correct, but it is hard to understand for somebody not knowing math however. I believe the geometric viewpoint, which, if not entirely accurate, was helpful in illustrating what is going on. Wonder what you think. Oleg Alexandrov (talk) 15:09, 17 April 2006 (UTC)

Thank you for your kind comment, Oleg. However, I now realise I have been unnecessarily narrow in my viewpoint. The thing that has shifted my viewpoint is finally realising (as Hipparchus did a little earlier :-) ) that the stereographic projection, although of course not preserving distances, is conformal. This gives the geometrical viewpoint a fully justified role (with all non-trivial holomorphic maps, including Möbius transformations) being conformal on the geometrical sphere). I wish this had sunk in before I made my second "improvement" to the introduction. Oh well, at least this isn't on paper. Of course this justifies increasing the prominence of the geometrical viewpoint again, which should help to address your suggestion. Elroch 19:11, 17 April 2006 (UTC)
I have now returned the geometrical viewpoint to the introduction, and added a little more detail. Hopefully this will mean the net change has been positive, and anything that is still less than ideal will be flagged or fixed by those more perceptive than me. Elroch 19:48, 17 April 2006 (UTC)
Thanks, your edits improved the article. Oleg Alexandrov (talk) 03:51, 18 April 2006 (UTC)

## Stereo proj: equatorial vs. polar-tangent

This article (like the Stereographic projection article) prefers the projection onto a plane tangent at a pole over the projection onto a plane through the equator. It even asserts that the former is more popular. This contradicts my experience. Among the texts in front of me, the equatorial version is used books by Rudin, Bredon, Thorpe, Oprea, and Brown/Churchill; the pole-tangent version is used in Do Carmo; Spivak uses both versions.

More importantly, this article contradicts itself. Its claims about the transition maps being ${\displaystyle z=1/w}$, ${\displaystyle w=i}$ corresponding to ${\displaystyle z=-i}$, etc. are all assuming the equatorial version. (In the equatorial version, the unit circle in ${\displaystyle \mathbb {C} }$ is sent to the equator; in the pole-tangent version, the circle of radius 2 is sent there, so the unit circles from the two complex planes don't match up.)

In short, while the pole-tangent version of stereographic projection has uses in differential geometry, the equatorial version seems unequivocally better suited to complex analysis, which is where this article comes in. So unless there are objections I'm going to start making these changes. Joshua R. Davis 19:55, 25 April 2007 (UTC)

As with many articles I suspect the contradictions have crept in from various editors using different conventions and not being careful as to whether their edits were consistent or not. I agree that the equatorial version is much better suited to this article (despite whatever assertions I may have made earlier on this page) -- Fropuff 03:25, 26 April 2007 (UTC)
Actually, now that I look back at my earliest version of this page, I realize that the mistake lies partially with me. The equations I used were for an equatorial projection but I asserted that they were onto a polar-tangent plane. Clearly, I was confused. All due apologies for accusations of edit creep and editor sloppiness. I'll just go and flog myself. -- Fropuff 03:55, 26 April 2007 (UTC)

## Need citations

I appreciate that mathematical experts may be writing this article from their own minds, without referring to source material. I've done the same in articles where I have expertise. This is Wikipedia though, and at least a few citations are needed. At a minimum, cite a few books or papers that describe a Riemann sphere. Davidwr 15:10, 3 May 2007 (UTC)

Yes, this is a serious problem with the article. I am finishing up some revisions that do cite references. Perhaps they will appear later today. Joshua R. Davis 16:03, 3 May 2007 (UTC)

## Proposed major revision

Hello all. I just added applications and references to the article. But in fact I have rewritten the entire article. I think the new version is more precise/rigorous, clearer about what the metric is and what depends on the metric, better organized, less redundant, etc. (But then I would think that. :) I have already transplanted some of the purely geometric material to Stereographic projection (also newly rewritten).

Because there has been recent activity on this article and this talk page (moreover by editors whom I know and respect) I don't want to impose a complete rewrite on everyone out of the blue, so I'm posting it here for comments/incorporation. Joshua R. Davis 04:18, 4 May 2007 (UTC)

In mathematics, the Riemann sphere is a way of extending the plane of complex numbers with one additional point at infinity, in a way that makes expressions such as

${\displaystyle 1/0=\infty }$

well-behaved and useful, at least in certain contexts. It is named after 19th century mathematician Bernhard Riemann. It is also called the complex projective line, denoted ${\displaystyle \mathbb {CP} ^{1}}$.

On a purely algebraic level, the complex numbers with an extra infinity element constitute the extended complex numbers. Arithmetic with infinity does not obey all of the usual rules of algebra, and so the extended complex numbers do not form a field. However, the Riemann sphere is geometrically and analytically well-behaved, even near infinity; it is a one-dimensional complex manifold, also called a Riemann surface.

In complex analysis, the Riemann sphere facilitates an elegant theory of meromorphic functions. The Riemann sphere is ubiquitous in projective geometry and algebraic geometry as a fundamental example of a complex manifold, projective space, and algebraic variety. It also finds utility in other disciplines that depend on analysis and geometry, such as quantum mechanics and other branches of physics.

### As a complex manifold

As a one-dimensional complex manifold, the Riemann sphere can be described by two charts, both with domain equal to the complex number plane ${\displaystyle \mathbb {C} }$. Let ${\displaystyle \zeta }$ and ${\displaystyle \xi }$ be complex coordinates on ${\displaystyle \mathbb {C} }$. Identify the nonzero complex numbers ${\displaystyle \zeta }$ with the nonzero complex numbers ${\displaystyle \xi }$ using the transition maps

${\displaystyle \zeta =1/\xi ,}$
${\displaystyle \xi =1/\zeta .}$

Since the transition maps are holomorphic, they define a complex manifold, called the Riemann sphere.

Intuitively, the transition maps indicate how to glue two planes together to form the Riemann sphere. The planes are glued in an "inside-out" manner, so that they overlap almost everywhere, with each plane contributing just one point (its origin) missing from the other plane.

In other words, (almost) every point in the Riemann sphere has both a ${\displaystyle \zeta }$ value and a ${\displaystyle \xi }$ value, and the two values are related by ${\displaystyle \zeta =1/\xi }$. The point where ${\displaystyle \xi =0}$ should then have ${\displaystyle \zeta }$-value "${\displaystyle 1/0}$"; in this sense, the origin of the ${\displaystyle \xi }$-chart plays the role of "${\displaystyle \infty }$" in the ${\displaystyle \zeta }$-chart. Symmetrically, the origin of the ${\displaystyle \zeta }$-chart plays the role of ${\displaystyle \infty }$ with respect to the ${\displaystyle \xi }$-chart.

Topologically, the resulting space is the one-point compactification of a plane into the sphere. However, the Riemann sphere is not merely a topological sphere. It is a sphere with a well-defined complex structure, so that around every point on the sphere there is a neighborhood that can be biholomorphically identified with ${\displaystyle \mathbb {C} }$. On the other hand, the two-dimensional sphere admits a unique complex structure turning it into a one-dimensional complex manifold.

The uniformization theorem, a central result in the classification of Riemann surfaces, states that the only simply-connected one-dimensional complex manifolds are the complex plane, the hyperbolic plane, and the Riemann sphere. Of these, the Riemann sphere is the only one that is closed (compact and boundaryless).

### As the complex projective line

The Riemann sphere can also be defined as the complex projective line. This is the subset of ${\displaystyle \mathbb {C} ^{2}}$ consisting of all pairs ${\displaystyle (\alpha ,\beta )}$ of complex numbers, not both zero, modulo the equivalence relation

${\displaystyle (\alpha ,\beta )=(\lambda \alpha ,\lambda \beta )}$

for all nonzero complex numbers ${\displaystyle \lambda }$. The complex plane ${\displaystyle \mathbb {C} }$, with coordinate ${\displaystyle \zeta }$, can be mapped into the complex projective line by

${\displaystyle (\alpha ,\beta )=(\zeta ,1).}$

Another copy of ${\displaystyle \mathbb {C} }$ with coordinate ${\displaystyle \xi }$ can be mapped in by

${\displaystyle (\alpha ,\beta )=(1,\xi ).}$

These two complex charts cover the projective line. For nonzero ${\displaystyle \xi }$ the identifications

${\displaystyle (1,\xi )=(1/\xi ,1)=(\zeta ,1)}$

demonstrate that the transition maps are ${\displaystyle \zeta =1/\xi }$ and ${\displaystyle \xi =1/\zeta }$, as above.

This treatment of the Riemann sphere connects most readily to projective geometry. For example, any line or smooth conic in the complex projective plane is biholomorphic to the complex projective line. It is also convenient for studying the sphere's automorphisms, later in this article.

### As a sphere

The Riemann sphere can be visualized as the unit sphere ${\displaystyle x^{2}+y^{2}+z^{2}=1}$ in the three-dimensional real space ${\displaystyle \mathbb {R} ^{3}}$. To this end, consider the stereographic projection from the unit sphere minus the point ${\displaystyle (0,0,1)}$ onto the plane ${\displaystyle z=0}$, which we identify with the complex plane by ${\displaystyle \zeta =x+iy}$. In Cartesian coordinates ${\displaystyle (x,y,z)}$ and spherical coordinates ${\displaystyle (\phi ,\theta )}$ on the sphere (with ${\displaystyle \phi }$ the zenith and ${\displaystyle \theta }$ the azimuth), the projection is

${\displaystyle \zeta ={\frac {x+iy}{1-z}}=\cot(\phi /2)\;e^{i\theta }.}$

Similarly, stereographic projection from ${\displaystyle (0,0,-1)}$ onto the ${\displaystyle z=0}$ plane, identified with another copy of the complex plane by ${\displaystyle \xi =x-iy}$, is written

${\displaystyle \xi ={\frac {x-iy}{1+z}}=\tan(\phi /2)\;e^{-i\theta }.}$

(The two complex planes are identified differently with the plane ${\displaystyle z=0}$. An orientation-reversal is necessary to maintain consistent orientation on the sphere, and in particular complex conjugation causes the transition maps to be holomorphic.) The transition maps between ${\displaystyle \zeta }$-coordinates and ${\displaystyle \xi }$-coordinates are obtained by composing one projection with the inverse of the other. They turn out to be ${\displaystyle \zeta =1/\xi }$ and ${\displaystyle \xi =1/\zeta }$, as described above. Thus the unit sphere is diffeomorphic to the Riemann sphere.

Under this diffeomorphism, the unit circle in the ${\displaystyle \zeta }$-chart, the unit circle in the ${\displaystyle \xi }$-chart, and the equator of the unit sphere are all identified. The unit disk ${\displaystyle |\zeta |<1}$ is identified with the southern hemisphere ${\displaystyle z<0}$, while the unit disk ${\displaystyle |\xi |<1}$ is identified with the northern hemisphere ${\displaystyle z>0}$.

### Metric

A Riemann surface does not come equipped with any particular Riemannian metric. However, the complex structure of the Riemann surface does uniquely determine a metric up to conformal equivalence. (Two metrics are said to be conformally equivalent if they differ by multiplication by a positive smooth function.) Conversely, any metric on an oriented surface uniquely determines a complex structure, which depends on the metric only up to conformal equivalence. Complex structures on an oriented surface are therefore in one-to-one correspondence with conformal classes of metrics on that surface.

Within a given conformal class, one can use conformal symmetry to find a representative metric with convenient properties. In particular, there is always a complete metric with constant curvature in any given conformal class.

In the case of the Riemann sphere, the Gauss-Bonnet theorem implies that a constant-curvature metric must have positive curvature K. It follows that the metric must be isometric to the sphere of radius ${\displaystyle 1/{\sqrt {K}}}$ in ${\displaystyle \mathbb {R} ^{3}}$ via stereographic projection.

In the ${\displaystyle \zeta }$-chart on the Riemann sphere, the metric with ${\displaystyle K=1}$ is given by

${\displaystyle ds^{2}=\left({\frac {2}{1+|\zeta |^{2}}}\right)^{2}\,|d\zeta |^{2}={\frac {4}{\left(1+\zeta {\bar {\zeta }}\right)^{2}}}\,d\zeta d{\bar {\zeta }}.}$

In real coordinates ${\displaystyle \zeta =u+iv}$, the formula is

${\displaystyle ds^{2}={\frac {4}{\left(1+u^{2}+v^{2}\right)^{2}}}\left(du^{2}+dv^{2}\right).}$

Up to a constant factor, this metric agrees with the standard Fubini-Study metric on complex projective space (of which the Riemann sphere is an example).

Conversely, let S denote the sphere (as an abstract smooth or topological manifold). By the uniformization theorem there exists a unique complex structure on S. It follows that any metric on S is conformally equivalent to the round metric. All such metrics determine the same conformal geometry. The round metric is therefore not intrinsic to the Riemann sphere, since "roundness" is not an invariant of conformal geometry. The Riemann sphere is only a conformal manifold not a Riemannian manifold. However, if one needs to do Riemannian geometry on the Riemann sphere, the round metric is a natural choice.

### Automorphisms

The study of any mathematical object is aided by an understanding of its group of automorphisms, meaning the maps from the object to itself that preserve the essential structure of the object. In the case of the Riemann sphere, an automorphism is an invertible biholomorphic map from the Riemann sphere to itself. It turns out that the only such maps are the Möbius transformations. These are functions of the form

${\displaystyle f(\zeta )={\frac {a\zeta +b}{c\zeta +d}},}$

where ${\displaystyle a}$, ${\displaystyle b}$, ${\displaystyle c}$, and ${\displaystyle d}$ are complex numbers such that ${\displaystyle ad-bc\neq 0}$. Examples of Möbius transformations include dilations, rotations, translations, and complex inversion. In fact, any Möbius transformation can be written as a composition of these.

The Möbius transformations are profitably viewed as transformations on the complex projective line. In projective coordinates, the transformation ${\displaystyle f}$ is written

${\displaystyle f(\alpha ,\beta )=(a\alpha +b\beta ,c\alpha +d\beta )={\begin{pmatrix}a&b\\c&d\end{pmatrix}}{\begin{pmatrix}\alpha \\\beta \end{pmatrix}}.}$

Thus the Möbius transformations correspond to ${\displaystyle 2\times 2}$ complex matrices with nonzero determinant. These are the projective linear transformations ${\displaystyle \mathrm {PGL} _{2}(\mathbb {C} )}$.

If one endows the Riemann sphere with the Fubini-Study metric, then not all Möbius transformations are isometries; for example, the dilations and translations are not. The isometries form a proper subgroup of ${\displaystyle \mathrm {PGL} _{2}(\mathbb {C} )}$, namely ${\displaystyle \mathrm {PSU} _{2}}$. It is isomorphic to the rotation group ${\displaystyle \mathrm {SO} (3)}$, which is the isometry group of the unit sphere in ${\displaystyle \mathbb {R} ^{3}}$.

Overall, I think this is a big improvement. It's very nicely organized and well written. My main criticism is with the section "As a sphere". This section should really start out with a description of the stereographic projection rather than its inverse. The complex coordinates should be given by

${\displaystyle \zeta ={\frac {x+iy}{1-z}}\qquad \xi ={\frac {x-iy}{1+z}}}$

(which are incidentally much simpler and easier to grok than their inverses). This shows very geometrically how to put complex coordinates onto the sphere to make it into a complex manifold. One can then give the inverse transformations to complete the picture.

The inverses are parametrizations, which tell you how to "plot" the sphere, so I think they're helpful to non-experts. But you are right that the projections should also be included.

There is a subtlety that occurs in these equations which should be explained. When projecting from opposite poles, one is not quite projecting onto the same plane. The planes are complex conjugates of each other. This is necessary to make the transition functions holomorphic rather than antiholomorphic. I remember this confused me a good deal when I first encountered the Riemann sphere. I think it is also the reason I, at one time, preferred projections onto the polar tangent planes: as the planes were distinct it was less easy to confuse them.

You are quite right. I was not sure how to raise the point, but I guess we shouldn't ignore it, either.

Minor quips: I prefer using z and w for the coordinates rather than ξ and ζ (which, by the way, appear nearly identical with my fonts), but I can see how this might lead to confusion. Also, for reasons unbeknownst to me, it seems much more natural to complex conjugate the southern hemisphere chart (i.e. when projecting from the north pole) rather than the northern one so that

${\displaystyle \zeta ={\frac {x-iy}{1-z}}\qquad \xi ={\frac {x+iy}{1+z}}.}$

Maybe its because I live in the northern hemisphere and I'm biased. But again this is a personal preference and I'm not bound to it. -- Fropuff 08:11, 4 May 2007 (UTC)

The z and w are common, but I thought z should be used for the third coordinate of R3. The previous version used x3, which I think is intimidating to non-experts. Then again, so are Greek letters. By the way, the article consistently uses Greek for complex and Roman for real (except a, b, c, d). By another way, the notation here exactly matches with that at Stereographic projection, by design. So I have put some thought into it, but of course I'm willing to be outvoted. And I'm certainly be willing to switch to some other Greek combo, like ζ and η.
The issue of which hemisphere to use is arbitrary, as far as I can tell. If I'm outvoted or if someone can show the other usage is standard, then let's switch. But how about this: When projecting onto a plane tangent at a point, it is most natural to project from the north pole onto the plane at the south pole, since then the sphere projects down onto the plane, like sunlight. Given that, it is then most natural when projecting onto the equatorial plane to project from the north pole. So I'm trying to explain my own bias.

Another thought regarding the metric: the way you've written it makes it sound like the round metric is chosen arbitrarily, which is, of course, far from the truth. Firstly, the metric is the pullback of the round metric on the sphere by the inverse stereographic projection. Secondly, the complex structure on the sphere uniquely determines a conformal class of metrics on the sphere. Within this class there is a unique (or so I recall) metric with constant curvature. Normalize the curvature to be +1 and you get the round metric. Perhaps it would be best to put all the stuff about the metric in its own section. -- Fropuff 08:43, 4 May 2007 (UTC)

The Riemann sphere, being a complex manifold, does not come with any default choice of metric. To say that it's the pullback from the round sphere by stereographic projection is to make a choice --- albeit a popular one, exactly as popular as the Fubini-Study metric. However, it does come with a unique conformal class of metric, necessarily isometric to a round sphere of some radius, but the radius/scaling is unspecified. When I wrote that the Riemann sphere could be a non-round ellipsoid, that was indeed wrong. Thank you for pointing it out.
I'm busy right now, but later I'll start incorporating your comments (editing the version above in place, unless that is sinful). Joshua R. Davis 12:39, 4 May 2007 (UTC)
I have tried to incorporate your major suggestions, and I certainly think they were for the better. Thank you. The projections, the conjugation on the two complex planes, and the unique conformal structure are in there. The metric is in its own section. I have not put in the notation changes or the switch to conjugating the other hemisphere.
By the way, does the metric have to have positive curvature, as I wrote above? I need to think/read about this. Joshua R. Davis 17:21, 4 May 2007 (UTC)
Yes, if the curvature is constant then it must be positive, since its integral is positive, by the Gauss-Bonnet theorem. Sorry for so many little posts. Joshua R. Davis 18:01, 4 May 2007 (UTC)
Actually, you weren't wrong to say you could give the Riemann sphere a non-round metric. In fact, any metric on the sphere is conformally equivalent to the unit round metric. So the statement that the complex structure uniquely determines a conformal class of metrics is somewhat vacuous in the case of the sphere (not so for other surfaces which generally admit more than one complex structure). There is a unique complex structure on the sphere and any metric determines it. It is just "nice" to choose a metric with constant curvature. This can be done with any Riemann surface. And again its nicest to fix the normalization to be +1 (as you say the curvature must be positive by Gauss-Bonnet). More later, I'm in a bit of a rush right now. -- Fropuff 20:01, 4 May 2007 (UTC)
I think the damage is being contained to the Metric section. I've changed the text to something that's temporarily correct, until I change it again because it's incorrect. Wikipedia helps me learn! Joshua R. Davis 21:00, 4 May 2007 (UTC)
I had a stab at rewriting the section on the metric. It is a little more explicit now, but perhaps less accessible. I'm not sure I've worded things in the best possible manner. Let me know what you think. -- Fropuff 06:48, 5 May 2007 (UTC)
It looks much better. I've done a little cosmetic rewording, but nothing big that I can recall. Are you interested in moving this to the article any time soon? All the big points of the current article are covered here, but is there anything that the current article does better? And what should we do about the ζ and ξ notation? Joshua R. Davis 16:25, 6 May 2007 (UTC)

If we want to stick with Greek letters for complex variables then ζ and ξ are fine, and like it or not, your choice of which plane to conjugate does seem more common in the literature so I guess I'll have to live with it. The only thing I really miss from the present article is a description of the stereographic projection in spherical coordinates. In present notation we should have

${\displaystyle \zeta =e^{i\phi }\cot {\frac {\theta }{2}}\qquad \xi =e^{-i\phi }\tan {\frac {\theta }{2}}}$

where θ is the zenith angle and φ is the azimuth angle. I find these formulas extremely useful. They show clearly that circles of latitude on the sphere map to circles on the plane while lines of longitude map to radial lines. I realize these formulas are at stereographic projection but its nice to see them in complex notation. Other than that I am happy to replace the article with the new version. -- Fropuff 00:18, 7 May 2007 (UTC)

## Loxodromes on the Riemann sphere

Why do we have this image? What does it half to do with the Riemann sphere? It's relevence is never explained. As far as I can tell, only the title of the image has anything to do with the article. The Riemann sphere is topologically a sphere, but has no one intrinsic geometry (you can get a metric by pulling back a metric with stereographic projection, but even then there is no unique choice of projection). Loxodromes are intrinsically geometric objects, and so would seem to have little to do with the Riemann sphere. And even if you do give the Riemann sphere a metric, the obvious choice would be a pullback via stereographic projection, and even then a loxodrome wouldn't come from any natural curve on the complex plane. The curve that maps to a loxodrome would be a "spiral" starting out at the origin and winding around finitely many times before shooting out to infinity (as opposed to natural spirals, which wrap around infinitely often). No natural function does that. So the image strikes me as a complete nonsequiter, and even if it isn't, it needs to be tied into the article somehow. skeptical scientist (talk) 12:41, 1 July 2007 (UTC)

Hmm. A loxodrome makes a fixed angle with all meridians, so its image makes the same fixed angle with all radii in the plane, so its image is an equiangular spiral? Which, back on the sphere, wraps infinitely many times around both poles (unless it's a meridian)? If a loxodrome on the sphere starts out at a pole, then it must be a meridian?
But you are right that loxodromes have little to do with the Riemann sphere. I vote to remove the image. Joshua R. Davis 22:35, 4 July 2007 (UTC)
Agree. Remove the image. Though pretty, its relevance is minimal. Is there a more standard image, for instance of the stereographic projection, which would be better? Silly rabbit 22:49, 4 July 2007 (UTC)
So it does. My intuition has failed me. Anyways, stereographic projection would be a natural choice, but it appears later on in the article, so that might not be the best image. I don't know what a better one would be though. skeptical scientist (talk) 19:01, 13 July 2007 (UTC)

## Where is an explicit projection formula?

I cannot deduce from the formulaes in the article how to transform a point P=x+i·y of the Gaussian plane onto the x,y,z Cartesian coordinates (or longitude,latitude spherical coordinateS) of the Riemann sphere. :-/ Is it that complicated? --RokerHRO (talk) 09:50, 18 May 2010 (UTC)

The section "As a sphere" contains the formula ζ = (x + i y) / (1 - z). So if we write ζ = X + i Y (like your P = x + i y), we have X = x / (1 - z) and Y = y / (1 - z). This is the map from the sphere to the plane; you just have to invert it to get the map from the plane to the sphere. There are more details in the Stereographic projection article. Mgnbar (talk) 19:32, 18 May 2010 (UTC)

## ∞ + ∞ = ∞ is false (in general)

∞ + ∞ = ∞ is stated at the section Arithmetic operations. I think this operation must be moved to the undefined operations (together with ∞ - ∞ and 0·∞). You can find a counterexample at page 30 here. I'm not an expert, so I'm afraid I may not be correct and I didn't change it. So please, if someone agrees, change it.--Ssola (talk) 17:57, 7 March 2012 (UTC)

I'll change it (as I have no annswers).--Ssola (talk) 14:33, 19 March 2012 (UTC)
For the title of this section: there is no "in general" situation for addition with ∞, because ∞ is not a real number, so "in general" is a bit confusing. You are correct that ∞ + ∞ is undefined on the extended complex plane, however (as your source clearly indicates). TricksterWolf (talk) 03:18, 6 February 2014 (UTC)

## Riemann Sphere disproven

This article is to dispell the obvious mistake made in the Riemann sphere definition which states that "∞" infinity is near to very large numbers and that zero "0" is near to very small numbers. This is false for two reasons: 1) nothing and infinity are unlimited concepts and are therefore by their very nature completely incompatible mathematically and conceptually with limited numbers. 2) The concept that zero is close to small numbers and infinity is close to large numbers is based on the assumants inability to conceive of something infinite outside of space and time resulting in the assumants mind counting up or down forever caught in a limited process that given enough time would never reach its infinite goal for the simple fact that it is always limited.

Conclusion this theory is logically refuted as false. — Preceding unsigned comment added by 193.200.145.253 (talk) 10:52, 23 July 2013 (UTC)

When mathematicians use the Riemann sphere, they do not make broad claims about the philosophical nature of nothingness and infinity. Rather, they are talking about a specific mathematical space, with very precisely defined properties. Nothing about the Riemann sphere is controversial in mainstream mathematics. On the other hand, if you can find reliable sources to support your argument, then perhaps we can incorporate that material into the article. Mgnbar (talk) 18:06, 23 July 2013 (UTC)

## Dividing 0 by 0?

The article addresses the problem of division by zero and rules out the special cases ∞ + ∞, ∞ - ∞ and 0 ⋅ ∞ as undefined. But what about 0/0? Is it still left undefined, as it is with the real numbers? My intuition tells me it would... SBareSSomErMig (talk) 07:04, 25 September 2013 (UTC)

Yes, it is still undefined. By that I mean that the Riemann sphere does not solve the problem of giving "0/0" a definition.
This talk page is not for questions about mathematics; this talk page exists to discuss how to edit the Riemann sphere article. If you have further questions about math, try asking at Wikipedia:Reference desk/Mathematics. Cheers. Mgnbar (talk) 12:42, 25 September 2013 (UTC)

## Infinity plus infinity

Why does this article say infinity plus infinity is undefined?? It has always made sense to define it as infinity. Infinity plus negative infinity is undefined though. Georgia guy (talk) 22:00, 16 January 2014 (UTC)

Really more of a question for the refdesk, but I suppose it might come up again here, so OK. The ∞ of the Riemann sphere is unsigned (negative infinity equals infinity). Look at the figure of the sphere near the top of the article — you can approach ∞ from any direction. --Trovatore (talk) 23:02, 16 January 2014 (UTC)
Ssola provided this counterexample already. See chapter 3 to learn more about why these particular definitions are chosen for this model. (You must realize math is a tool for selecting certain parameters and seeing what comes up; not for discovering the One True Algebra. There are lots of One True Algebras. They're like reproductions of the One Ring: cheap and ubiquitous.) TricksterWolf (talk) 03:38, 6 February 2014 (UTC)

## Confused on "the" Riemann sphere vs. "a" Riemann sphere.

Some parts of this article seem to imply any stereographic projection of the complex plane to a sphere is "a Riemann sphere", whereas other parts seem to imply there is only one canonical projection which forms "the Riemann sphere". Some consistency would be nice here.

My intuition from elsewhere in mathematics is that the former is true, even though the wording of the latter is more likely (e.g. in computational complexity we may refer to "the first Turing machine enumerated in shortlex form as a member of {0, 1}*", ignoring the fact that which Turing machine this may be depends entirely upon the coding language being used to model the machine in its coded form--generally because it has been proven that the choice of coding method is immaterial as long as all possible machines can be enumerated). TricksterWolf (talk) 03:31, 6 February 2014 (UTC)

Hmm. Here's what's supposed to be happening. The Riemann "sphere" is a particular complex manifold. To visualize it, we use a diffeomorphism (a 1-1 correspondence that is smooth in both directions) between the Riemann "sphere" and the unit sphere in R^3. There is nothing canonical about this diffeomorphism. Does that make sense so far?
If I understand you correctly, your complaint arises principally from the "As a sphere" section, which favors one (actually two!) particular stereographic projection(s). So we should change the wording in this section, to make it clearer that this is just an illustration or visualization of the Riemann sphere? Mgnbar (talk) 06:43, 6 February 2014 (UTC)
Let me see if I understand. You're saying the projection is unimportant because the "sphere" is a manifold in the topological sense? That makes perfect sense to me now. I think others might be confused as to how the same "point" on two spheres could represent a different "point" on the plane, yet both of those are "the same sphere".
I guess I'm thinking about the sphere from an applied perspective, though. Does the projection not matter when the sphere is actually used? As in, wouldn't there be error if I used one projection in an initial stage of a proof, then used another later on? This is sort of where intuition suggests to me that while the definition doesn't matter topologically-speaking, the definition needs to be fixed if it's being used as a map back and forth to the plane. I may be misunderstanding how the sphere is applied (perhaps the exact mapping is never used in proofs). TricksterWolf (talk) 14:53, 6 February 2014 (UTC)

Any two stereographic projections give the same complex structure on the sphere (and also the same conformal structure). So if one thinks of the Riemann sphere as a complex or conformal manifold only, then there is no preferred projection. However, if one thinks of the Riemann sphere as the extended complex plane, then there is a preferred point at infinity and a preferred stereographic projection. Which of these is intended depends on the context. (Ironically, one of the participants in this very discussion has a username that illustrates both of these cases, with g=0 and different values of n.) Sławomir Biały (talk) 16:33, 6 August 2014 (UTC)

## All holomorphic functions are rational?

It is claimed in the Rational functions section that "the set of complex rational functions ... form all possible holomorphic functions from the Riemann sphere to itself". As far as I know, ${\displaystyle e^{x}}$ is not a rational function, yet it is holomorphic. So this claim appears to be false. The only way out I can see is if we consider ${\displaystyle e^{x}}$ to be a "polynomial" on the grounds that it has a power series, but this seems to be a stretch. Am I mistaken? Luqui (talk) 05:19, 8 July 2018 (UTC)

Hi. I'm not an expert, but I think that ${\displaystyle e^{x}}$ is not a counterexample. To make it a counterexample, you must extend ${\displaystyle e^{x}}$ from the complex plane to the Riemann sphere, by defining its value at ${\displaystyle \infty }$. Let ${\displaystyle y=1/x}$, so your function is ${\displaystyle e^{1/y}}$. Now try to define the value of the function at ${\displaystyle y=0}$. I think that you will run into problems. Mgnbar (talk) 11:26, 8 July 2018 (UTC)
Another tack: You're going to have a problem defining ${\displaystyle e^{\infty }}$ to make ${\displaystyle e^{x}}$ continuous (let alone holomorphic). Because in the real numbers ${\displaystyle \lim _{x\to -\infty }e^{x}=0}$ and ${\displaystyle \lim _{x\to \infty }e^{x}=\infty }$, but in the Riemann sphere ${\displaystyle -\infty }$ and ${\displaystyle \infty }$ show up as the same point. Mgnbar (talk) 13:31, 8 July 2018 (UTC)