Riemannian surfaces as one dimensional complex manifolds

[Ben Niehoff]

It arose in several places like math journals written by people that like writing complete nonsense. For instance:
http://www.igt.uni-stuttgart.de/LstDiffgeo/Kuehnel/preprints/totalcurv.pdf
jus take a look at the introduction.
Or: http://www.crm.es/Publications/08/Pr805.pdf
page 18
just to mention a couple that are freely available.

OK, I see that the Gauss-Bonnet formula can be re-written in terms of various extrinsic curvatures, that's kinda neat! But the point remains that what it is calculating (i.e. the Euler characteristic) is an invariant, intrinsic property (a fact restated many times in those very papers!). This really shouldn't be any surprise; as I've already pointed out, the Gauss curvature itself can be written in terms of extrinsic quantities, and yet it is intrinsic.

In any case, the Euler characteristic of a 2-surface M can always be calculated via

2 \pi \, \chi(M) = \int_M K \, dV + \int_{\partial M} k_g \, ds
and every quantity appearing here is intrinsic to M; i.e., does not depend on whatever space M might be embedded in.

If a horosphere is a topological sphere then it must have \chi = 2, so obviously this will come by paying careful attention to the boundary term (since the Gauss curvature of a horosphere is zero).

This is nonsense (clearly hyperbolic geometry is not your field of expertise) and unrelated to what I wrote.

It is unfortunate that you don't see the relevance, but without pointing out any specific issue, I can't help you much.

You have a habit of misusing mathematical terminology and neglecting to give examples of exactly what you're talking about. Perhaps this is why you get responses that you think are irrelevant.

Earlier you said

But anyway my point is that a submanifold surface metric is determined by the embedding 3-manifold so it makes little sense to calculate the Ricci tensor of the submanifold, the Ricci curvature in this case is referred to the 3D manifold.

which leads me to believe you don't quite understand what's going on. The Euler characteristic is an intrinsic property of the surface, so why do you keep harping on about ambient 3-manifolds? If you have some embedding into an ambient 3-manifold, then it serves one purpose: it induces a metric on the 2-surface. Then this metric can be used to compute the Ricci scalar on the 2-surface, which can be integrated to give the Euler characteristic.

These papers you're reading show that you can also calculate the Euler characteristic via some other routes. The answer will still be the same, though. In fact I would say that's the whole point of those papers.

 

[Ben Niehoff]

So, because I'm bored, let's actually calculate the Euler characteristic of a horosphere in H^3. Using the Poincare ball model and spherical coordinates, the metric of hyperbolic space is

ds^2 = \frac{4}{(1-r^2)^2} (dr^2 + r^2 \, d\theta^2 + r^2 \, \sin^2 \theta \, d\phi^2)
Now, choose a horosphere tangent to the boundary sphere at (x,y,z) = (0,0,1) having the equation

r = \cos \theta
Hence dr^2 = \sin^2 \theta \, d\theta^2, and the induced metric can be written

ds_{\mathrm{ind}}^2 = \frac{4}{\sin^4 \theta} (d\theta^2 + \cos^2 \theta \, \sin^2 \theta \, d\phi^2)
Now make the change of coordinates \rho = 2 \cot \theta, d\rho^2 = (4/\sin^4 \theta) \, d\theta^2 to get

ds_{\mathrm{ind}}^2 = d\rho^2 + \rho^2 \, d\phi^2
The range of \theta was (0, \pi/2], and so the range of \rho is [0, \infty).

This is the standard flat metric on R^2. Therefore the Gauss curvature is zero, and the Euler characteristic can be calculated from the boundary term. We can assume a circular boundary that grows to infinite radius:

\begin{align}<br /> 2 \pi \, \chi(M) &amp;= \int_M K \, dV + \int_{\partial M} k_g \, ds \\<br /> &amp;= 0 + \lim_{\rho \to \infty} \int_0^{2\pi} \frac{1}{\rho} \, \rho \, d\theta \\<br /> &amp;= 2 \pi \\<br /> \chi(M) &amp;= 1<br /> \end{align}<br />
And so, as a submanifold of hyperbolic space, a horosphere is NOT a topological sphere, but rather an infinite R^2, just as I said earlier.

The reason for this is precisely the reason I gave earlier: the "point at infinity" that would have closed the sphere is not a point belonging to hyperbolic space. Hence no submanifold of hyperbolic space can contain that point!

The catch here is that the boundary sphere of the Poincare ball model is not actually a boundary of H^3. After all, H^3 is infinite in extent and has no boundary. Rather, the boundary sphere of the Poincare ball model is what's called a "conformal boundary" of H^3. This means the boundary only shows up after using a conformal transformation to map H^3 into a region of finite volume. (The Poincare ball coordinates can be considered such a map.)

The overall lesson is don't be fooled by appearances. A horosphere "looks like" a sphere in Poincare ball coordinates, but in fact it is an infinite flat plane, and is neither compact nor closed!

 

[TrickyDicky]

Hi, Ben

OK, I see that the Gauss-Bonnet formula can be re-written in terms of various extrinsic curvatures, that's kinda neat!

Glad you finally see this. Remember you previously deemed it as "total nonsense".

But the point remains that what it is calculating (i.e. the Euler characteristic) is an invariant, intrinsic property (a fact restated many times in those very papers!). This really shouldn't be any surprise; as I've already pointed out, the Gauss curvature itself can be written in terms of extrinsic quantities, and yet it is intrinsic.

In any case, the Euler characteristic of a 2-surface M can always be calculated via

2 \pi \, \chi(M) = \int_M K \, dV + \int_{\partial M} k_g \, ds
and every quantity appearing here is intrinsic to M; i.e., does not depend on whatever space M might be embedded in.

You are saying this as if you thought I believe otherwise, this leads me to think that I haven't been able yet to get my point thru to you.
Please remember that quantity K depends on the metric and the metric when we are talking of submanifold structures that admit more than one metric can give rise to K=1 but also K=0 or K=-1.

If a horosphere is a topological sphere then it must have \chi = 2, so obviously this will come by paying careful attention to the boundary term (since the Gauss curvature of a horosphere is zero).

This confirms you haven't understood what my point is, I never implied a horosphere is a topological sphere, on the contrary my claim was that it had euclidean metric.

You have a habit of misusing mathematical terminology and neglecting to give examples of exactly what you're talking about.

This I admit freely, I'll try to improve on it.

The Euler characteristic is an intrinsic property of the surface, so why do you keep harping on about ambient 3-manifolds?

I have insisted several times that I think the metric induces the topology on the conformal complex structure, so if we are talking about complex structures immersed in a 3-space that imposes a metric on the complex structure up to conformal equivalence (that is, not any metric can be imposed in complex structure of ℂ\cup∞, only constant curvature metrics).

These papers you're reading show that you can also calculate the Euler characteristic via some other routes. The answer will still be the same, though. In fact I would say that's the whole point of those papers.

Those papers also claim that makes a difference to calculate it in euclidean or hyperbolic ambient space, but it would help me to know exactly how you interpret what is written in page 18 of the second reference I linked.

So, because I'm bored, let's actually calculate the Euler characteristic of a horosphere in H^3...
This is the standard flat metric on R^2. Therefore the Gauss curvature is zero, and the Euler characteristic can be calculated from the boundary term. We can assume a circular boundary that grows to infinite radius:

And so, as a submanifold of hyperbolic space, a horosphere is NOT a topological sphere, but rather an infinite R^2, just as I said earlier.

The overall lesson is don't be fooled by appearances. A horosphere "looks like" a sphere in Poincare ball coordinates, but in fact it is an infinite flat plane, and is neither compact nor closed!

First, I'd like to say that I used the horosphere as an example only based on intuition, and I said that obviously I'm not sure if it is a valid example of the extended complex plane in hyperbolic space.
But the key point is that I didn't use it as an example of a topological sphere as you keep saying. Your calculation confirms what I said about horospheres having euclidean metric.
I'd like to call on mathematicians that might be lurking to help me on this and say something about whether an abstract extended complex plane in hyperbolic space can be identified with a horosphere. (Not with a topological sphere)

The reason for this is precisely the reason I gave earlier: the "point at infinity" that would have closed the sphere is not a point belonging to hyperbolic space. Hence no submanifold of hyperbolic space can contain that point!

You are right it doesn't belong to the hyperbolic space, but it can belong to the hyperbolic 3-manifold which is a quotient space of H^3/\Gamma, that is hyperbolic 3-space over the conformal transformations of the Riemann sphere (Kleinian group), and it is equipped with a complete Riemannian metric, being complete is analogous to a closed set:contains all its limit point including the point at infinity.
Thinking about this perhaps the horosphere is not the object that is equivalent to the extended complex plane but the very boundary of the hyperbolic 3-manifold would be a good candidate. Certainly it wouldn't belong to hyperbolic space but to the hyperbolic manifold.

 

[lavinia]

Please remember that quantity K depends on the metric and the metric when we are talking of submanifold structures that admit more than one metric can give rise to K=1 but also K=0 or K=-1.

By K do you mean the Gauss curvature of the surface? If the sumanifold is homeomorphic to an open disk as I guess the horosphere is, then it can have metrics of constant positive, negative of zero Gauss curvature.

I have insisted several times that I think the metric induces the topology on the conformal complex structure, so if we are talking about complex structures immersed in a 3-space that imposes a metric on the complex structure up to conformal equivalence (that is, not any metric can be imposed in complex structure of ℂ\cup∞, only constant curvature metrics).

- a metric does not induce a topology
- on a surface a conformal structure is the same as a complex structure
- the extended complex plane can have metrics of non-constant Gauss curvature that it inherits from a manifold that it is embedded in.
- if the sphere is immersed rather than embedded I would like to see a proof that it can inherit a metric of constant Gauss curvature.
[/QUOTE]

But the key point is that I didn't use it as an example of a topological sphere as you keep saying. Your calculation confirms what I said about horospheres having euclidean metric.
I'd like to call on mathematicians that might be lurking to help me on this and say something about whether an abstract extended complex plane in hyperbolic space can be identified with a horosphere. (Not with a topological sphere)

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

Thinking about this perhaps the horosphere is not the object that is equivalent to the extended complex plane but the very boundary of the hyperbolic 3-manifold would be a good candidate. Certainly it wouldn't belong to hyperbolic space but to the hyperbolic manifold.

I don't know anything about these manifolds but if the boundary is covered by the bounding plane of H^3 then it can not be a sphere and therefore can not be the extended complex plane unless you mean something different than what is usually meant when you say extended complex plane.BTW: You sai that all metrics on the sphere are conformally equivalent. Without a reference, can you give a proof of this?

Maybe it is not so hard. Choose any conformal diffeomorphism of the 2 spheres. Then maybe in isothermal coordinates the diffeomorphism is actually a conformal mapping of the metrics.

 

[TrickyDicky]

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.
(Though, I think when referring to Riemannian geometry and Riemannian manifolds like in this case an isometric embedding is an immersion between Riemannian manifolds which preserves the Riemannian metrics)

 

[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.