2-sphere manifold intrinsic definition

[pasmith]

Sorry, I believe your equivalence relation $\sim$ first glues the vertical edges together to form a vertically laid down cylinder.

Yes, I meant the vertical sides.

Then, starting from this intrinsic definition of the 2-sphere as quotient, one can define an embedding in $\mathbb R^3$.

By the obvious $(\theta, \phi) = (\pi(1 - y),2\pi x)$.

Btw, in your definition of equivalence relation $\sim$, why there is the last term $x_1=x_2 \text { and } y_1=y_2$ ?

An equivalence relation is by definition reflexive, so $(x,y) \sim (x,y)$ must always hold.

 

[fresh_42]

From 1821 until the mid-1840s, Gauß was in charge of the surveying and later the land surveying in the Kingdom of Hanover. At the end of this project, over 3,000 triangulation points had been determined, almost all of which had been corrected by Gauß or later by people he trusted. The overview map shows the triangular networks in northern Germany.

The impetus to survey the Kingdom of Hanover came from its northern neighbor. From 1816 onwards, Gauß's friend and colleague Heinrich Christian Schumacher had carried out a survey of the Kingdom of Denmark. At that time, Denmark also included the duchies of Schleswig and Holstein.

The Danish triangulation thus extended to the gates of Hamburg, which was to be followed by the Hanover survey. King George IV entrusted Carl Friedrich Gauß with this task in 1820. Like Schumacher, Gauß also used the triangulation method, in which the country is covered with a network of triangles that meet at their sides.

Source.

This brings up a fun fact:

Maybe it was Georg IV who finitely was responsible for the beginning of differential geometry! Or was it Frederick VI of Denmark who initiated the measurement in Denmark and Holstein?

 

[cianfa72]

By the obvious $(\theta, \phi) = (\pi(1 - y),2\pi x)$.

Sorry, the embedding is in $\mathbb R^3$ so the embedding map should go to $x,y,z$.

An equivalence relation is by definition reflexive, so $(x,y) \sim (x,y)$ must always hold.

Ok yes. $(x,y)$ and $(x,y)$ are always related by $\sim$ and belong to the same equivalence class.

 

[cianfa72]

Consider spherical polar coordinates $(r, \theta, \phi)$ with $r = 1$.

Ah ok, so $\pi$ in your map in post #36 is not the projection map, it is actually the irrational 3,14159... and $x,y,z \in \mathbb R^3$.

 

[mathwonk]

just a remark about embedding abstract manifolds. this is not at all difficult, and not as sophisticated as Whitney's theorems, which precisely bound the embedding dimension.
As suggested above, if f1,...,fj, is a covering set of charts, and if g1,...,gj is a corresponding partition of unity, then (f1g1,...,fjgj, g1,...,gj) embeds the (compact) manifold. qed.

Riemann may not have known exactly about partitions of unity, or even topological spaces, but I believe he would have understood at least intuitively, that one can extend a chart to a global function. This makes me wonder if there was ever actually any confusion as to whether embedded and abstract manifolds are the same, as it is hard to believe any expert could have missed this simple argument.

Or maybe it was believed, but an actual proof awaited the development of the necessary language? Or maybe it was the non -compact case that was more challenging? I note that Whitney also embeds his manifolds as analytic sub manifolds, but makes no mention in his introduction (to his 1936 Annals paper) that any instance of the general embedding result was previously known. I have not read the later sections, which require some privileged access.

According to wikipedia, Poincare' already defined local overlapping atlases on manifolds.

Aha: My search for the origin of partitions of unity turns up Whitney, in 1934. maybe that technical tool for extending functions was the key? Hard to believe Riemann would not have known this sort of thing, if only in his intuitive way.

well, the topological case seems covered by the tietze urysohn extension lemma and urysohn seems to have died in 1924, so I would think everyone knew a compact topological manifold embeds in Euclidean space well before Whitney.

here is a lecture note going through the easy construction of a partition of unity from urysohn's lemma, applied to embedding compact manifolds, but he makes it look as if there is possibly a little technical supplement needed to apply urysohn. It is really hard though for me to believe that the topological embedding theorem could have been considered news in 1936. Maybe the "smooth urysohn lemma", i.e. smooth PofU, was the new wrinkle, and the consequent smooth embedding theorem?

http://staff.ustc.edu.cn/~wangzuoq/Courses/21S-Topology/Notes/Lec15.pdf

 

[martinbn]

Speaking of Riemann, the sphere is genus zero Riemann surface, no embedding used.

 

[cianfa72]

By the obvious $(\theta, \phi) = (\pi(1 - y),2\pi x)$.

Thinking again about it, I believe $x,y$ in your map are actually coordinates for the square $[0,1]^2$ that is quotiented according $\sim$ to "intrinsically" define the 2-sphere. So the above is a map from it to the $\theta, \phi$ region and its composition with the following map into $\mathbb R^3$
$$\begin {align}
x & = \sin \theta \cos \phi \nonumber \\
y & = \sin \theta \sin \phi \nonumber \\
z & = \cos \theta \nonumber
\end {align}$$ define an embedding for the 2-sphere (intrinsically defined as quotient space).

 

[A.T.]

About the 2-sphere manifold intrinsic definition without looking at its embedding in $\mathbb R^3$

- Constant intrinsic curvature
- All geodesics closed and of the same length

Does this uniquely define a sphere?

 

[mathwonk]

I am not expert, but it seems already the only Riemannian surfaces on which all geodesics can be closed are the sphere and the projective plane. Moreover, for the sphere, having all geodesics closed implies they are all the same length. I do not know about the projective plane. However there are apparently Riemannian structures on the sphere where not all geodesics are closed, and where it is only known that at least 3 geodesics are closed. I do not know what is added by requiring the constancy of the curvature. I would guess the projective plane may have these properties as well as the sphere, (presumably a geodesic on the projective plane is doubly covered by a geodesic on the sphere), but I don't know.

https://www.imaginary.org/sites/default/files/snapshots/snapshots-2017-005.pdf

 

[A.T.]

I do not know what is added by requiring the constancy of the curvature. I would guess the projective plane may have these properties as well as the sphere,

For a sphere it would be positive constant curvature. And the geodesics would be required to have finite equal length.