Riemannian surfaces as one dimensional complex manifolds

[TrickyDicky]

- a metric does not induce a topology

From Wikipedia page about spaces:

"A hierarchy of mathematical spaces: The inner product induces a norm. The norm induces a metric. The metric induces a topology."

The extended complex plane - as a topological space - is a topological sphere.

See above

BTW: You sai that all metrics on the sphere are conformally equivalent. Without a reference, can you give a proof of this?

I don't remember saying exactly this. Can you quote it with the context?

 

[lavinia]

From Wikipedia page about spaces:

"A hierarchy of mathematical spaces: The inner product induces a norm. The norm induces a metric. The metric induces a topology."




See above



I don't remember saying exactly this. Can you quote it with the context?

Once a conformal structure exists - which is the case you were talking about, the topology already exists. the metric does not induce the topology. I do not understand why you refuse to listen. I will not respond to this thread again.

 

[TrickyDicky]

Once a conformal structure exists - which is the case you were talking about, the topology already exists. the metric does not induce the topology.

Fine, and it so happens that this topology is the one-point compactification of the complex plane and the round metric is not intrinsic to this topology, for instance when immersed in non-euclidean manifolds, other conformally equivalent metrics might be used.

 

[Ben Niehoff]

As Lavinia pointed out, the "extended complex plane" IS a topological 2-sphere, which is why I have interpreted your posts to be talking about topological 2-spheres.

I notice that when talking about submanifolds I've been using the term embedded when I meant immersed, several times and that might lead to confusion, sorry about that.

The horosphere is not an immersion of a topological 2-sphere (i.e. extended complex plane) into H^3 either, precisely because of the missing point at infinity. It is, however, an embedding (and hence an immersion) of the open disk (i.e. infinite plane).

I agree with Lavinia, it is quite frustrating to talk to you, because you say the opposite of what you mean and then blame the confusion on everyone else. Goodbye.

 

[TrickyDicky]

To be honest I'm also a bit fed up with you and Lavinia. So goodbye to you both too. What I write next is just for the benefit of others.
This is from the Handbook of Complex variables (page 83) by Steven George Krantz: available at Google books:

"Stereographic projection puts the extended complex plane into one-to-one correspondence with the two dimensional sphere S in R^3 in such a way that topology is preserved in both directions of the correspondence.[...]
Note that, under stereographic projection, the "point at infinity" in the plane corresponds to the north pole N of the sphere. For this reason, the extended complex plane is often thought of as "being" a sphere, and is then called, for historical reasons, the Riemann sphere. "
From this is obvious that the identification of the extended complex plane with the 2-sphere depends on a embedding in R^3, and works basically as a visual aid to better understand, in no case the extended complex plane IS the 2 sphere, that would be a very naive identification that is not valid if we are not doing the stereographic projection in R^3.
It is also obvious that the extended complex plane in H^3 can be given a flat metric basically because in the context of a negatively curved space a euclidean metric has constant positive curvature relative to the ambient space (just like the round metric wrt the euclidean ambient), so the geometric relations are conserved and the stereographic projection still has a one-to-one correspondence.

Whether stereographic projection can put the extended complex plane into one-to-one correspondence to a horosphere in H^3 is left as an open question.

 

[TrickyDicky]

It turns out my intuition in the final lines of post #33 was correct and the Riemann sphere IS (can be thought of, just like S^2 is diffeomorphic to the extended complex plane in R^3) the boundary of H^3 which is diffeomorphic to the space of horospheres, that has euclidean metric of course.

For reference:
Geometric group theory by Niblo and Roller, page 12.
Wikipedia: Mobius transformation page, under hyperbolic geometry heading.

 

[TrickyDicky]

What I still find a little puzzling is not that a couple of people who admittedly think they know about differential geometry seem to never have heard of second fundamental forms or of surface embeddings other than in R^3, but that no one in this forum has had anything to say about this (either to correct me or others). I'm thinking for instance of mathwonk or micromass to name some of those that I find knowledgeable.

 

[mathwonk]

the problem is one of communication. i.e. people with math training use words in a certain way that is not understood perfectly without that same training. so people misunderstand each other. to paraphrase s certain us president it depends on what the meaning of "is" is. when they say the extended complex plane "is" a 2 sphere, they mean it has a unique topology in which it is homeomorphic to the 2 sphere. you are right when you remark that stereographic projection is such a homeomorphism, but there are many others, not depending at all on an embedding.

now to get picky, you are also right in the sense that if one wants the word "is" to imply that there is a precise given way to identify points of the extended plane with points of the 2 sphere, one needs to choose a particular homeomorphism. so it could be argued that when lavinia says the ex plane is the 2 sphere, she means there is some unspecified homeomorphism, while when you say this following krantz, you mean they are identified via stereographic, or some other specified map.so you are to some extent arguing because you are using the same words in a different way. they agree with each other because, having the same training, they are using those words in the same way. you on the other hand might have understood them better had they said "can be viewed as" rather than "is".

and they are frustrated with you because they do know more about some aspects of the subject, whereas you seem not to grant them that. I.e. you give the impression of assuming because of some things you read, but may not fully understand as a mathematician would, that you are justified in insisting that they are wrong.

i have been hanging out here hoping to learn a few things, since i have sort of a gap in my own education when it comes to differential geometry and riemannian metrics. so i don't feel fully authoritative here although i do know a good deal about the complex side of things.

it also appears to me you seem not to realize that the word "metric" has more than one meaning also. in the wikipedia venn diagram you displayed, it apparently means the distance function in a metric space, which does then define a topology.

but the term metric in the sense of a riemannian metic, means something different, it means a dot product on the tangent bundle of a manifold, in which case that manifold already has a topology.

I suggest your problem would largely be solved if, when you think your reading has shown someone else is totally wrong, you asked instead somewhat as follows: "i must be missing something here, because your answer seems to me at odds with this paragraph of wikipedia. can you help me see what i am misunderstanding?"

best wishes,

and compliments on your intellectual curiosity.

 

[TrickyDicky]

Thanks for your comment mathwonk, I agree with you and think you are a real gentleman.
It certainly seems to be a problem of communication and there I have a big share of fault, not being a mathematician or having any formal mathematical training. Self-learning has certainly its drawbacks.
I have to say I also sensed an unwillingness to admit certain errors on the other side.

Now, back to bussiness, my point was just about embedding the Riemann sphere in H^3 instead of R^3 and what consequences that might have on the metric that can be used for this Riemannian surface in this particular embedding, my claim was that in contrast with the R^3 case where it only admits a round metric in H^3 it can admit a flat metric, taking into account that Gaussian curvature (that of course it can also be derived in terms of only the first fundamental form, but here we are talking about Riemannian surfaces as submanifolds) is equal to the determinant of the shape operator and this shape operator is an extrinsic curvature that will depend on the embedding.
Mathwonk, please let me know if you find this approximately correct. Thanks again.

 

[mathwonk]

wonderful response! now here is the acid test. i do not immediately know the answer to your question. maybe someone else will, if we ask them in a way that motivates them! good luck.

however i will try to the limit of my knowledge. i think the sphere cannot have a flat (riemannian) metric in any embedding, because the gauss bonnet theorem says that the integral of the curvature is the same under every embedding. and moreover that integral equals the euler characteristic.

since the euler characteristic of the sphere is not zero, it follows that it is impossible to give the sphere a "flat" (riemannian) metric, i.e.one with zero curvature, in any embedding. I.e. any surface that has an embedding with metric of zero curvature, must have euler characteristic zero. but this is not true of the sphere. so the sphere has no such embedding.

how does this sound?

 

[TrickyDicky]

i think the sphere cannot have a flat (riemannian) metric in any embedding, because the gauss bonnet theorem says that the integral of the curvature is the same under every embedding. and moreover that integral equals the euler characteristic.
since the euler characteristic of the sphere is not zero, it follows that it is impossible to give the sphere a "flat" (riemannian) metric, i.e.one with zero curvature, in any embedding. I.e. any surface that has an embedding with metric of zero curvature, must have euler characteristic zero. but this is not true of the sphere. so the sphere has no such embedding.

how does this sound?

I see what you mean and it is correct, my only point is that in the case of the extended complex plane (a conformal manifold, not a Riemannan manifold) in a negative constant curvature ambient space, it would no longer be a sphere (in this space), what I mean is that I understand the sphere representation of the extended complex plane as a particularity of its stereographic projection in R^3.
Let's take the gauss-bonnet theorem, it says the integral of K(gaussian curvature) gives us the euler characteristic, but first we have to introduce K, and I can see that for the extended complex plane as a manifold or embedded in R^3, K is a positive constant curvature +1, but in a negative curvature embedding K is calculated with reference to the ambient space and the shape operator is different since here the extrinsic curvature of the extended complex plane is zero, and a stereographic projection puts it in one-to-one correspondence with a R^2+∞ plane in H^3 (as it is known this extended euclidean plane can't be embedded in R^3).
Hope this is intelligible, if not I'm open to specific questions. Or anything leading me to see how my arguments might be flawed.

Also I've given references where it is said that the boundary of a hyperbolic 3-manifold with boundary (wich is flat) is the extended complex plane. In the Poincare ball model for instance, the conformal boundary of hyperbolic space which is a sphere in R^3, is actually flat in H^3.

 

[mathwonk]

well again there is a communication problem, as you are using the word "sphere" in a restricted sense that does not agree with mine it seems.

in modern mathematics, objects have am intrinsic structure that is independent of an y representation in euclidean space. so a topological sphere is any topological space homeomorphic to a sphere in R^3.

a riemannian sphere is any riemannian manifold which is diffeomorphic to the sphere in R^3, plus possibly some condition on the induced map on tangent bundles.

a conformal sphere is presumably any conformal manifold conformally equivalent to the sphere in R^3 with its riemannian structure. i.e. i never heard of them before this minute, but wikipedia says a conformal manifold is reprsented by a riemnnian m,anifold, but two riemannian sturctures on the same manifokld can be coformally equivalent even if not equivalent in the same riemannian sense.it is also not quite clear to me what a flat conformal manifold is since we are only measuring angles, not lengths. well i guess curvature does have a manifestation in etrms of angle sums of triangles, but it is not clear to me that we have geodesics, but we must if we are going to have anglkes.

anyway i do not klnow this subject and do not have time to learn it right now and explain it. but there may be others here that do, if they are patient enough to try to communicate un der these circumstances.

by the way, wiki states flatly that the sphere can have a local flat conformal structure but not a global one.

"A conformal metric is conformally flat if there is a metric representing it that is flat, in the usual sense that the Riemann tensor vanishes. It may only be possible to find a metric in the conformal class that is flat in an open neighborhood of each point. When it is necessary to distinguish these cases, the latter is called locally conformally flat, although often in the literature no distinction is maintained. The n-sphere is a locally conformally flat manifold that is not globally conformally flat in this sense, whereas a Euclidean space, a torus, or any conformal manifold that is covered by an open subset of Euclidean space is (globally) conformally flat in this sense. A locally conformally flat manifold is locally conformal to a Möbius geometry meaning that there exists an angle preserving local diffeomorphism from the manifold into a Möbius geometry. In two dimensions, every conformal metric is locally conformally flat. "

best wishes,

for now at least.

 

[mathwonk]

well i'll throw in one more remark and a guess.

the remark; again words are slippery little suckers in this subject, because even after stating that a sphere cannot have a globally flat conformal structure, the wiki article begins to speak of the flat conformal geometry on the sphere, but they don't mean that, they mean the locally flat conformal geometry. so they are redefining the word "flat" so as to fool me as to its meaning.

this causes another problem for us when you quote some sentence where they say "flat" which in my brain i hear as "globally flat", but the writer meant "locally flat".

you saved me from misunderstanding you that time by emphasizing that you were speaking not of a riemannian manifold but a conformal one, so i could look up what a conformal geometer means by "flat".

the guess: the role played by stereographic projection is to define the locally flat conformal structure on the open set in the sphere consisting of all but the north pole.i.e. this i a conformal map from that open set on the sphere to a flat plane.

so there is no metric preserving map from any open set on a sphere to a flat plane but there is an angle preserving one. even that cannot be done globally however.

so it is correct to say that stereographic projection is used to put a locally flat conformal structure on a sphere and hence does define the conformal structure of the LOCALLY conformally flat sphere.

but this isn't my game... it interests me though.

I am looking at a geometry book by gunter ewald who develops euclidean spherical and hyperbolic geometries at the same time, by using conformal ideas rather than metric ones.

i.e. he looks at reflections, which are apparently the "inversions" we are seeing in these wiki articles on conformal manifolds.

then he separates out the three kinds of geometry from a curvature standpoint, by the different axioms on parallels, but it could as well be axioms on angle sum of triangles,

or curvature.

 

[homeomorphic]

I didn't read very carefully, but I have a couple comments.

First of all, Tricky wondered why the Euler characteristic of an odd-dimensional closed manifold is zero. That comes from Poincare duality.

http://en.wikipedia.org/wiki/Poincaré_duality

You can calculate the Euler characteristic by the alternating sum of ranks of homology groups. For an odd-dimensional manifold, there are an even number of these. Poincare duality plus universal coefficients pairs the n-kth homology with the kth homology and says they have the same rank. But they have opposite sign in the sum, so they all cancel in pairs.

Secondly, some comments about the two different uses of the word metric. It's not quite true that the metric does not induce a topology--however, there is already a topology on the manifold because manifolds have a topology. The metric in the Riemannian sense determines a metric in the metric space sense. Since we have geodesics, we have distances between different points, given by the inf of the lengths of curves joining them. This distance is a metric in the metric space sense. Thus, it DOES determine a topology, but it agrees with the one the manifold already has, so it is superfluous.

Thirdly, you do have geodesics in conformal geometry. Because the metric is defined up to scale, lengths of curves are also defined up to scale. The geodesics are not going to depend on this rescaling because if you minimize arclength with respect to one representative metric, you also minimize it for a scalar multiple of the metric.

 

[TrickyDicky]

Interesting comments, thanks.

 

[mathwonk]

here are some more comments that confused me recently. what does it mean to have an isometric embedding of a manifold into euclidean space?

consider the sphere in euclidean space. it has two natural metrics (not riemannian metrics, but metric space metrics), both induced by the embedding, namely we can measure the distance between two points by using the distance between them as points of R^3, or we can use the distance measured along a great circle on the sphere.

since the second one is induced by the notion of lengths of curves on the sphere, it is the metric on the sphere, induced by the riemannian metric on the sphere. now suppose we think an isometric embedding is an embedding that preserves the metric, rather than the riemannian metric. if this is the case, and we ask for such an embedding which takes the second metric of the sphere to the intrinsic metric of euclidean space, then there is no such embedding.

i.e. there is no isometric embedding of the usual sphere in R^3, if by that you mean the embedding takes the metric measured along great circles, to the restriction of the euclidean metric. that is because that would require geodesics in the riemannian metric to map to geodesics in the embedded metric, which would mean all great circles would map to straight lines.

although that does happen for stereographic projection, that projection does not preserve lengths. i.e. under projection, finite great circles go to infinite straight lines.

so apparently an isometric embedding of riemannian manifolds means one such that the riemannian metric is preserved, but not the metric induced by the riemannian metric. i am a little over the line from dinner so could be wrong even here. but it is confusing hey?>

 

[Ben Niehoff]

so apparently an isometric embedding of riemannian manifolds means one such that the riemannian metric is preserved, but not the metric induced by the riemannian metric. i am a little over the line from dinner so could be wrong even here. but it is confusing hey?>

Yes, this is what I've understood "isometric embedding" to mean: that the pullback of the ambient (Riemannian) metric is equal to the (Riemannian) metric on the submanifold.

 

[TrickyDicky]

here are some more comments that confused me recently. what does it mean to have an isometric embedding of a manifold into euclidean space?

consider the sphere in euclidean space. it has two natural metrics (not riemannian metrics, but metric space metrics), both induced by the embedding, namely we can measure the distance between two points by using the distance between them as points of R^3, or we can use the distance measured along a great circle on the sphere.

since the second one is induced by the notion of lengths of curves on the sphere, it is the metric on the sphere, induced by the riemannian metric on the sphere. now suppose we think an isometric embedding is an embedding that preserves the metric, rather than the riemannian metric. if this is the case, and we ask for such an embedding which takes the second metric of the sphere to the intrinsic metric of euclidean space, then there is no such embedding.

i.e. there is no isometric embedding of the usual sphere in R^3, if by that you mean the embedding takes the metric measured along great circles, to the restriction of the euclidean metric. that is because that would require geodesics in the riemannian metric to map to geodesics in the embedded metric, which would mean all great circles would map to straight lines.

although that does happen for stereographic projection, that projection does not preserve lengths. i.e. under projection, finite great circles go to infinite straight lines.

so apparently an isometric embedding of riemannian manifolds means one such that the riemannian metric is preserved, but not the metric induced by the riemannian metric. i am a little over the line from dinner so could be wrong even here. but it is confusing hey?>

I think you are getting close to where I come from. So following your line of reasoning (when you realize that the metric induced by the Riemannian metric is not necessarily preserved(in this case the sphere metric) it might make sense that the ambient with negative constant curvature induces in the conformal structure of the extended complex plane a different metric than the the Euclidean ambient does.
It is important to remember again that the extended complex plane as an abstract object is only going to preserve in any embedding its infinitesimal shape (its angles, not the lengths) and that as you said in a previous post is expressed by saying it is "locally" conformally flat (actually all Riemann surfaces are conformally flat, and this can only be understood in "local" terms since conformal geometry is only concerned with preserving angles and infinitesimal shapes which are local notions), the sphere is conformally flat (locally), what this means is that it will accept any metric from the ambient space as long as it is a conformally flat one.
Here enters the significative fact that stereographic projection preserves angles, not lenghts: "it is conformal, meaning that it preserves angles. It is neither isometric nor area-preserving: that is, it preserves neither distances nor the areas of figures" (from WP).
The angles are preserved then by the stereographic conformal projection but lengths and areas are left to be determined by the ambient's Riemannian metric, that in the case of being Euclidean determine a sphere metric (only locally flat) and in the case of hyperbolic (negatively curved) space (but you need to think about it in terms of a hyperbolic manifold with boundary, that is, points on the boundary are considered to be part of the manifold) determine a flat metric (globally flat) by stereographic projection of the extended complex plane.

 

[mathwonk]

I haven't read your post yet tricky, but here are some more comments on confusing use of words.

homeomorphism said:

":Thirdly, you do have geodesics in conformal geometry. Because the metric is defined up to scale, lengths of curves are also defined up to scale. The geodesics are not going to depend on this rescaling because if you minimize arclength with respect to one representative metric, you also minimize it for a scalar multiple of the metric."this comment led me to think that he meant a conformal map takes geodesics to geodesics. it seems it does do so but only locally. e.g. a stereographic projection from the north pole does take geodesics on the sphere, i.e. great circles, which pass through the origin, to geodesics in the plane, i.e. to straight lines through the origin.

But it does NIOT take all geodesics on the sphere to geodesics in the plane. i.e. other great circles, not passing through the south pole go to circles in the plane which are not geodesics in the plane. Indeed it would not be possible for all geodesics to go to geodesics, since triangles would go to triangles, and their angles would be preserved, and thus the angle sum would be preserved.

But on a sphere triangles have angle sum more than 180 and in the plane they have angle sum exactly 180. so the maps we are looking at that give "flat" conformal structure to the sphere locally, do not take triangles to triangles, hence do not take any three non collinear geodesics to geodesics.the argument for it that homeomorphisms gave used "scaling" but what does that mean? apparently it means a scaling oriented at a single point. I.e. in the plane, multiplying vectors by 2 scales them homogeneously centered at the origin, but does not scale uniformly from other centers.

 

[mathwonk]

tricky says:

"Here enters the significative fact that stereographic projection preserves angles, not lenghts: "it is conformal, meaning that it preserves angles. It is neither isometric nor area-preserving: that is, it preserves neither distances nor the areas of figures" (from WP)."
This brings out the fact that conformal geometry does not have a notion of "curvature" since that notion depends on comparing lengths and areas, e.g. comparing the circumference of a circle to its area. this makes us naive customers wonder how they have the nerve to speak of "flat" conformal manifolds, when curvature makes no sense. Such things cause great confusion amongst the uninitiated.We seem to have understood the meaning now of isometric embedding, as one inducing isomorphism of riemannian metrics but not the actual metrics derived from them.

e.g. an embedding preserving riemannian metrics preserves the length of tangent vectors, but even that does not end geodesics to geodesics. I.e. does not send geodesics to straight lines, as would be required if the actual metrics were preserved, i.e. if the metric derived from the riemannian metric were to become equal to the embedded euclidean distance function.e.g. i thought an "isometric" embedding of the hyperbolic plane in euclidean space would send geodesics to straight lines, but this is not even true for the usual isometric embedding of the sphere. i.e. the term "isometric" is a misnomer since it means literally "unchanged metric", whereas all that is unchanged is the riemannian or infinitesimal metric. but as usual, to save words, people adopt special meanings for them, that cannot be understood by outsiders.

 

[mathwonk]

tricky:

"I think you are getting close to where I come from. So following your line of reasoning (when you realize that the metric induced by the Riemannian metric is not necessarily preserved(in this case the sphere metric) it might make sense that the ambient with negative constant curvature induces in the conformal structure of the extended complex plane a different metric than the the Euclidean ambient does."this fact had failed to interest me much since the statements that were being debated were thought by me to be global ones. i.e. no matter what metric is induced by the negative ambient, it induces the same topology on the extended plane, and this topology determines the global properties of the metric. I.e. there cannot be any globally flat metric on the extended plane that induces the usual sphere topology.

I think this point was at the heart of several misunderstandings earlier.

Topologically, the extended plane "is" just (i.e. is homeomorphic to) a sphere, no matter what riemannian or conformal metric induces its topology.another matter that bothers me is this: in WP there is the statement that 2 dimensional conformal geometry "is just the geometry of riemann surfaces".

To me this is a patently false statement, since riemann surfaces are oriented and conformal manifolds are not. I.e. riemann surface geometry is induced by a complex analytic coordinate cover, which is a proper subset of the associated conformal cover.

so it seems to me there may be two riemann surfaces conformally equivalent to each conformal surface?

at any rate their geometry, to me, admittedly a rookie, is not "the same". so this is another example of blunt statements in WP that are confusing at best, false at worst.

 

[mathwonk]

my experience with WP is that I have learned a lot from reading articles there that I knew little about. it is different in respect to articles that I know a lot about. In those cases I have tried to edit them to improve them, but afterwards someone later re edited them to remove my contribution. this is the basic problem with an openly editable article.

I only had the nerve to edit articles concerning subjects I had studied for some 30 years or so, including reading all the relevant historical sources, and doing research in the area for decades. The article I looked at was already very very good and there were only a tiny number of changes I thought could make.

that was years ago and those articles continue to be modified frequently, one as recently as last month, so that by now it does not even resemble the one I modified. When this happens the article that remains tends to be a list of facts resembling the introduction to a textbook.

I suspect a lot of the information there is put there by very smart fast readers who synopsize what they themselves have read elsewhere but sometimes without fully understanding it.

This causes me to wonder just how much I am learning from the articles that I must trust because of my ignorance.

mathoverflow is an excellent place to get superb and authoritative answers to math questions from experts, but they answer only research level questions. math stackexchange is a similar place for lower level questions, things that could be learned in books or wikipedia.

 

[homeomorphic]

homeomorphism said:

":Thirdly, you do have geodesics in conformal geometry. Because the metric is defined up to scale, lengths of curves are also defined up to scale. The geodesics are not going to depend on this rescaling because if you minimize arclength with respect to one representative metric, you also minimize it for a scalar multiple of the metric."


this comment led me to think that he meant a conformal map takes geodesics to geodesics. it seems it does do so but only locally. e.g. a stereographic projection from the north pole does take geodesics on the sphere, i.e. great circles, which pass through the origin, to geodesics in the plane, i.e. to straight lines through the origin.

No, I didn't mean anything beyond what I said. Just that there is a notion of geodesic in conformal geometry. I didn't read this thread carefully and was just making a couple casual comments in passing, so I wouldn't read anything into the significance of what I said.

 

[mathwonk]

is it your impression that a conformal surface is always orientable? if the definition is a surface given an equivalence class of riemannian metrics at each point under scaling, it seems any riemannian surface determines a conformal surface. that would seem to include non orientable ones, another argument that the statement in WP that their geometry is precisely that of riemann surfaces is false. i.e. it would seem the klein bottle has a conformal structure.

 

[TrickyDicky]

tricky says:

"Here enters the significative fact that stereographic projection preserves angles, not lenghts: "it is conformal, meaning that it preserves angles. It is neither isometric nor area-preserving: that is, it preserves neither distances nor the areas of figures" (from WP)."



This brings out the fact that conformal geometry does not have a notion of "curvature" since that notion depends on comparing lengths and areas, e.g. comparing the circumference of a circle to its area. this makes us naive customers wonder how they have the nerve to speak of "flat" conformal manifolds, when curvature makes no sense. Such things cause great confusion amongst the uninitiated.

Well, actually for higher dimensional manifolds than Riemann surfaces it makes more sense to speak of conformally flat manifolds and has implications on the curvature tensor. But yes, in the specific case we are concerned with, it does not affect curvature since as previously stated all riemann surfaces are conformally flat.

 

[homeomorphic]

is it your impression that a conformal surface is always orientable? if the definition is a surface given an equivalence class of riemannian metrics at each point under scaling, it seems any riemannian surface determines a conformal surface. that would seem to include non orientable ones, another argument that the statement in WP that their geometry is precisely that of riemann surfaces is false. i.e. it would seem the klein bottle has a conformal structure.

I don't think so. Why not consider a conformal structure on the Klein bottle? It's probably false, but all you need to do to fix it is say that the geometry of Riemann surfaces is the oriented case. I guess you can't get a complex structure if you're non-orientable (since a complex structure determines an orientation), so you don't have the same connection with complex analysis anymore, but I don't see anything wrong with considering metrics up to scale, which was their definition.

All manifolds are compact, oriented, connected...whatever you want them to be. The definition sometimes shifts depending on the writer or speaker's willingness to keep saying all the words that they need.

 

[mathwonk]

i agree it may be true to say that oriented conformal geometry agrees with the geometry of riemann surfaces, but that is not what they said - they said all conformal geometry is the same as the geometry of riemann surfaces. i agree that is an easy fix, but we are very picky in math, if there i an easy fix then make that fix before speaking. there is true and there is false, this is not horseshoes. or else do as i do, and admit one is not an expert.

 

[homeomorphic]

I was being sarcastic.

It's okay to fudge things a little if everyone involved understands what's going on (a big if) or if you are clear on the fact that you are fudging. Which isn't really the case, here. I agree it should be fixed. I was just pointing out that the error was a small one. Wikipedia works pretty well most of the time, but I guess we found some cracks.

Personally, it's hard for me to say I'm an expert on anything, except maybe a couple very very specialized topics.

 

[TrickyDicky]

This is a bit OT but I'll give it a shot.

is it your impression that a conformal surface is always orientable? if the definition is a surface given an equivalence class of riemannian metrics at each point under scaling, it seems any riemannian surface determines a conformal surface. that would seem to include non orientable ones, another argument that the statement in WP that their geometry is precisely that of riemann surfaces is false. i.e. it would seem the klein bottle has a conformal structure.

Can you give any example of a non orientable conformal surface? I don't think the klein bottle has a conformal structure.
Wikipedia says:
"A complex structure gives rise to a conformal structure by choosing the standard Euclidean metric given on the complex plane and transporting it to X by means of the charts. Showing that a conformal structure determines a complex structure is more difficult"

"Complex manifolds are canonically oriented, not just orientable".

Taking all this into account, I think the Wikipedia claim is basically correct if it is restricted, as it is, to conformal geometry in two real dimensions if we agree that riemann surfaces are complex manifolds.
But if you were able to provide counterexamples like showing how the klein bottle has a conformal structure that would be interesting.

Going back to topic for a minute, I'd like to say again that the topology of the extended complex plane (positive curvature) is preserved by a flat metric if we think about it in terms of relative curvature wrt the ambient space so it keeps being homeomorphic to the topology of the sphere, this means IMO there can be a globally flat metric on the extended plane that induces the usual sphere topology if the ambient space negative curvature allows the euclidean metric to be considered positively curved wrt to it.

 

[lavinia]

i agree it may be true to say that oriented conformal geometry agrees with the geometry of riemann surfaces, but that is not what they said - they said all conformal geometry is the same as the geometry of riemann surfaces. i agree that is an easy fix, but we are very picky in math, if there i an easy fix then make that fix before speaking. there is true and there is false, this is not horseshoes. or else do as i do, and admit one is not an expert.

Since I am still reading this thread I would like to rejoin if that is OK. My apologies to Tricky and everyone else for getting frustrated.

There are many conformal structures on the Klein bottle (by which I mean a smooth surface that is homeomorphic to a topological Klein bottle).

One may obtain different conformal flat Klein bottles from different conformal flat tori. Given a flat torus mod out by an isometry that is a fixed point free involution to get a flat Klein bottle. For example, if the flat torus is obtained from the standard lattice in the plane then add the transformation (x,y) -> (x+1/2,-y). Note that the square of this transformation is a lattice point so it projects to an involution of the flat torus. In fact it projects to a fixed point free isometry of the flat torus so the quotient is a flat Klein bottle. For other lattices the procedure is the same.

Conversely, starting with a flat Klein bottle there is a flat torus that covers it by a local isometry. So conformal classes of flat Klein bottles correspond to conformal classes of flat tori.

 

[lavinia]

here are some more comments that confused me recently. what does it mean to have an isometric embedding of a manifold into euclidean space?

consider the sphere in euclidean space. it has two natural metrics (not riemannian metrics, but metric space metrics), both induced by the embedding, namely we can measure the distance between two points by using the distance between them as points of R^3, or we can use the distance measured along a great circle on the sphere.

since the second one is induced by the notion of lengths of curves on the sphere, it is the metric on the sphere, induced by the riemannian metric on the sphere. now suppose we think an isometric embedding is an embedding that preserves the metric, rather than the riemannian metric. if this is the case, and we ask for such an embedding which takes the second metric of the sphere to the intrinsic metric of euclidean space, then there is no such embedding.

i.e. there is no isometric embedding of the usual sphere in R^3, if by that you mean the embedding takes the metric measured along great circles, to the restriction of the euclidean metric. that is because that would require geodesics in the riemannian metric to map to geodesics in the embedded metric, which would mean all great circles would map to straight lines.

although that does happen for stereographic projection, that projection does not preserve lengths. i.e. under projection, finite great circles go to infinite straight lines.

so apparently an isometric embedding of riemannian manifolds means one such that the riemannian metric is preserved, but not the metric induced by the riemannian metric. i am a little over the line from dinner so could be wrong even here. but it is confusing hey?>

For a smooth Riemannian manifold, an isometric embedding means that the inner product on the tangent space is preserved. Another way of saying this is that curves on the manifold have the same length after the embedding. Isometry means that the distance measure on the manifolds is preserved.

- If a manifold is already embedded in a Riemannian manifold of higher dimension then it inherits a Riemannian metric on its tangent space from the higher dimensional manifold. With this metric it is embedded isometrically in the bigger manifold. I am used to the terminology which says that this inherited metric is "induced".

If the map is merely an immersion, then the manifold also inherits a metric because the differential of an immersion is injective on each tangent plane.

- This idea of induced metrics generalizes to Riemannian vector bundles. A vector bundle,E, over a manifold is Riemannian if it has an inner product on each fiber that varies smoothly across the manifold.

An arbitrary vector bundle with a Riemannian metric still has a curvature tensor and still has an idea of parallel translation. But there is in general no way to translate this into a distance measure on the manifold. Therefore, a Riemannian metric on a vector bundle over a manifold in general does not induce a topology. But the topology already exists - otherwise you could not define the vector bundle in the first place.

In the case of the tangent bundle one does get a distance measure - the distance measure determined by geodesics. One can prove that it determines the same topology that the manifold already has. This is a basic theorem of Riemannian geometry.

So any Riemannian metric on a manifold determines the same topology.- When one says that a manifold is embedded in another manifold one is already assuming that the manifold has a topology. Embedding literally means that the map is smooth and injective, and that its image is surrounded by a tubular neighborhood. This means – prove it for yourself – that the image set of the embedding in the subspace topology is diffeomorphic to the original manifold. Its topology therefore will be the same as the subspace topology it has with respect to any metric that determines the topology of the ambient manifold.. The point is that embedding is an idea of differential topology. There is absolutely no need for any metric.

 

[lavinia]

This is a bit OT but I'll give it a shot.

Going back to topic for a minute, I'd like to say again that the topology of the extended complex plane (positive curvature) is preserved by a flat metric if we think about it in terms of relative curvature wrt the ambient space so it keeps being homeomorphic to the topology of the sphere, this means IMO there can be a globally flat metric on the extended plane that induces the usual sphere topology if the ambient space negative curvature allows the euclidean metric to be considered positively curved wrt to it.

There can not be a globally flat metric on the extended plane if by metric you mean a Riemannian metric on its tangent space and by flat you mean that the Gauss curvature is zero.

In hyperbolic 3 space, as Ben demonstrated by direct calculation, the horosphere has a flat Riemannian metric. Therefore this metric can not be extended to the extended complex plane.

 

[homeomorphic]

Can you give any example of a non orientable conformal surface? I don't think the klein bottle has a conformal structure.

The definition of conformal structure that they give is a metric up to scale. That means, yes, the Klein bottle has a conformal structure on it, since the Klein bottle has a metric on it (as does any manifold, by a partition of unity argument or by embedding it in R^n). And this makes perfect sense. Conformal mappings are just about angles. If you have a metric up to scale, you have angles, so you can talk about conformal mappings.

Wikipedia says:
"A complex structure gives rise to a conformal structure by choosing the standard Euclidean metric given on the complex plane and transporting it to X by means of the charts. Showing that a conformal structure determines a complex structure is more difficult"

There's something fishy about that because a conformal structure is supposed to be a globally defined metric, up to scale, but they are trying to define it using a local, coordinate-dependent procedure. A conformal structure does not determine a complex structure if the surface is not orientable, and, as I have argued above, by their definition, the Klein bottle most certainly has a conformal structure. It may be that a conformal structure determines a complex structure if the surface is oriented.

"Complex manifolds are canonically oriented, not just orientable".

Yes, as I have mentioned.

Taking all this into account, I think the Wikipedia claim is basically correct if it is restricted, as it is, to conformal geometry in two real dimensions if we agree that riemann surfaces are complex manifolds.
But if you were able to provide counterexamples like showing how the klein bottle has a conformal structure that would be interesting.

I just did. Embed it in R^5 by Whitney's embedding theorem. It then inherits a metric from R^5. Consider it up to scale. Done.

Going back to topic for a minute, I'd like to say again that the topology of the extended complex plane (positive curvature)

You're using the word topology in a very problematic way. Of course, you're doing so in reference to Gauss-Bonnet. You shouldn't say that the topology has positive curvature. What could that possibly mean? Curvature is defined in terms of structures other than the topology. Plus, the sphere can have different metrics, and they don't have to be positively curved. It's just that the integral of the curvature over the sphere is positive. You could embed a sphere so it has lots of saddle points and negative curvature, locally. But, of course, the metrics induced by these embeddings aren't the standard metric on the sphere. The usual one is the one inherited from R^3 when you consider it as the unit sphere. That one has constant positive curvature. There's only one topological 2-sphere. But there are many different Riemannian 2-spheres.

is preserved by a flat metric

How can topology be preserved by a metric?

if we think about it in terms of relative curvature wrt the ambient space so it keeps being homeomorphic to the topology of the sphere,

Any metric will induce the same topology, the standard one on the sphere. Maybe that's what you are trying to say?

this means IMO there can be a globally flat metric on the extended plane that induces the usual sphere topology

As we've been trying to say, the topology is already there on the sphere. Any metric on the sphere induces the topology that it already has. The topology comes PRIOR to the metric. So, what you are saying is a tautology (the globally flat part doesn't sound right, though, but I think we don't know what you mean).

if the ambient space negative curvature allows the euclidean metric to be considered positively curved wrt to it.

 

[TrickyDicky]

Since I am still reading this thread I would like to rejoin if that is OK. My apologies to Tricky and everyone else for getting frustrated.

Welcome back Lavinia.

There can not be a globally flat metric on the extended plane if by metric you mean a Riemannian metric on its tangent space and by flat you mean that the Gauss curvature is zero.

Ok, this is understood. Let's change flat by euclidean and let's think about a situation where a euclidean surface could have a extrinsic positive curvature, for instance a negatively curved ambient space, what would be wrong in this picture in your opinion?

The definition of conformal structure that they give is a metric up to scale. That means, yes, the Klein bottle has a conformal structure on it, since the Klein bottle has a metric on it (as does any manifold, by a partition of unity argument or by embedding it in R^n). And this makes perfect sense. Conformal mappings are just about angles. If you have a metric up to scale, you have angles, so you can talk about conformal mappings.

Ok, let me try to ask again: Can you give any example of a non orientable conformal surface in R^3? I'm not trying to cheat by adding this but I think the WP page is referring to euclidean ambient space because it defines orientability as:" a property of surfaces in Euclidean space". Not trying to defend wikipedia here, just pointing out that in this case there seems to be a contextual ambiguity.

There's something fishy about that because a conformal structure is supposed to be a globally defined metric, up to scale, but they are trying to define it using a local, coordinate-dependent procedure. A conformal structure does not determine a complex structure if the surface is not orientable, and, as I have argued above, by their definition, the Klein bottle most certainly has a conformal structure. It may be that a conformal structure determines a complex structure if the surface is oriented.

Ok,I think this is related with what I write above on the context of conformal surfaces in Euclidean space and the klein bottle impossibility to be embedded in euclidean space (R^3). It seems in R^3 all surfaces conformal structures are orientable and therefore determine complex structures.

I just did. Embed it in R^5 by Whitney's embedding theorem. It then inherits a metric from R^5. Consider it up to scale. Done.

See above.

You're using the word topology in a very problematic way. Of course, you're doing so in reference to Gauss-Bonnet. You shouldn't say that the topology has positive curvature. What could that possibly mean? Curvature is defined in terms of structures other than the topology. Plus, the sphere can have different metrics, and they don't have to be positively curved. It's just that the integral of the curvature over the sphere is positive. You could embed a sphere so it has lots of saddle points and negative curvature, locally. But, of course, the metrics induced by these embeddings aren't the standard metric on the sphere. The usual one is the one inherited from R^3 when you consider it as the unit sphere. That one has constant positive curvature. There's only one topological 2-sphere. But there are many different Riemannian 2-spheres.

Yes, I'm using this problematic way of dealing with topology in reference to Gauss-Bonnet, rather than to the "Riemannian metric" use of curvature. I see you caught my drift here.

How can topology be preserved by a metric?

Never mind this, bad choice of words again. What I meant was the topology is of course not changed by embedding in a hyperbolic manifold, but the metric induced on that topology by the Negatively curved ambient Riemannian metric could change and admit a euclidean metric on the topology.

As we've been trying to say, the topology is already there on the sphere. Any metric on the sphere induces the topology that it already has. The topology comes PRIOR to the metric. So, what you are saying is a tautology (the globally flat part doesn't sound right, though, but I think we don't know what you mean).

I see this, that's why I keep talking about the "one-point compactification of the complex plane" topology, but of course my expressions lack mathematical rigor.

I know the globally flat part doesn't sound right, I'm not sure about it either but all the time I've had the feeling that I'm not being able to get across what I mean, I don't mind being wrong but it bothers me not to be able to express what i mean in a sound mathematical way, so I guess I have to study this harder.

 

[lavinia]

The Mobius band is a non-orientable surface in 3 space. It naturally has a flat metric. There are no closed surfaces without boundary in 3 space that are unorientable. For instance the Klein bottle can not be embedded in 3 space.