2-sphere intrinsic definition by gluing disks' boundaries

[cianfa72]

TL;DR

On the equivalence of the definition of 2-sphere from intrinsic vs extrinsic point of view

A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from $\mathbb R^2$ and then taking the quotient topology obtained by gluing their boundaries.

Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of $\mathbb R^3$ in the subspace topology.

 

[cianfa72]

I was thinking as follows: call $D_{+}$ the disk of radius 1 centered at the origin of $\mathbb R^2$ and $D_{-}$ the disk of radius 1 centered at $(5,5)$. Define $f$ as: $$f(x,y) = \begin{cases} (x,y,+\sqrt{1 - x^2 - y^2}) , & x,y \in D_{+} \\ (x,y,-\sqrt{1 - (x-5)^2 - (y-5)^2}), & x,y \in D_{-} \end{cases}$$
Let's check it is homemorphism onto the image on the quotient defined by gluing disks' boundaries. Take the subspace topology from $\mathbb R^2$ on both (closed) disks. $f(x,y)$ as function $f: D = D_{+} \cup D_{-} \to \mathbb R^2$ is continuous and defines a bijective continuous function from the quotient onto the image equipped with the subspace topology from $\mathbb R^3$.

In the end, we have a bijective continuos function between a compact and the image that is Hausdorff (since $\mathbb R^3$ is) therefore $f$ is homeomorphism, i.e it defines a (topological) embedding into $\mathbb R^3$.

 

[fresh_42]

Why do you make it so complicated? You can choose two identical copies $D_1=D_2=\{(x,y)\,|\,x^2+y^2\leq 1\}$ and map them to $D_1'=\{(x,y,z)\,|\,(x,y)\in D_1\, , \,z=x^2+y^2\}$ and $D_2'=\{(x,y,z)\,|\,(x,y)\in D_2\, , \,z=-x^2-y^2\}$ and then glue them. If you choose $D_2$ without boundary, you won't even need the quotient.

 

[cianfa72]

Ok, coming back to the OP, just to be clear, what is a bijective continuos map is the function $\tilde f$ resulting by passing $f$ to the quotient.

 

[fresh_42]

Ok, coming back to the OP, just to be clear, what is a bijective continuos map is the function $\tilde f$ resulting by passing $f$ to the quotient.

The quotient building is the identification of certain points in $D_+$ with certain points in $D_-.$ You have to show that these points have the same value under $f.$ If so, then $\tilde{f}$ is simply the same as $f$ without mentioning these points twice.

 

[cianfa72]

The quotient building is the identification of certain points in $D_+$ with certain points in $D_-.$ You have to show that these points have the same value under $f.$

Yes of course. From the definition of $f$ given in #2 it takes the same value ($f=0$) on the pair of points in $D_+$ and $D_-$ identified by the quotient building (i.e. gluing pair of "corresponding" points on the boundaries of the disks).

If so, then $\tilde{f}$ is simply the same as $f$ without mentioning these points twice.

Yes, definitely.

P.s. pls note that in #2 the definition of $f$ on $D_-$ is actually $(x - 5, y -5, - \sqrt{1 - (x-5)^2 - (y-5)^2})$

 

[lavinia]

@cianfa72

Your thoughts brought to mind a few considerations that I thought would be of interest to you.

You start out by saying that the definition of the 2-sphere is the quotient space of two disks with their boundary circles glued together. To me this definition is incomplete because it does not specify what gluing map is used.

In general,when two manifolds with boundaries are glued along their boundaries, the gluing map matters. For instance the 3 sphere and the tangent circle bundle to the 2 sphere in Euclidean space are both solid tori glued together along their boundaries.

If one takes as the definition of the 2-sphere, any manifold that is homeomorphic to the points of distance one to the origin in R^3, then your proof shows that such a topological manifold is the quotient of two disks glued together along their boundary circles. ( It seems BTW that this thinking would lead one to start with the sphere in R^3 and then slice it along its equator. Both hemispheres then project onto the disk of radius one in the x,y-plane.) This seems different than first defining a topological sphere as a quotient space of two disks because then one needs to show that all such quotient spaces are homeomorphic.

In a similar vein, one might define a topological 2-sphere as any closed 2-manifold that can be covered by only two coordinate charts, two open sets on the manifold that are mapped homeomorphically onto R^2. And more generally a topological n-sphere as any closed n-manifold made from two n-dimensionlal coordinate charts. A priori there might be many such manifolds. I would guess that these are all homeomorphic to spheres in Euclidean space.

 

[cianfa72]

You start out by saying that the definition of the 2-sphere is the quotient space of two disks with their boundary circles glued together. To me this definition is incomplete because it does not specify what gluing map is used.

Of course the relevant gluing map glues "corresponding points" on disks
' boundary (e.g. point $(1,0)$ of $D_{+}$ is glued with point $(6,5)$ on $D_{-}$).

If one takes as the definition of the 2-sphere, any manifold that is homeomorphic to the points of distance one to the origin in R^3, then your proof shows that such a topological manifold is the quotient of two disks glued together along their boundary circles. (It seems BTW that this thinking would lead one to start with the sphere in R^3 and then slice it along its equator. Both hemispheres then project onto the disk of radius one in the x,y-plane.) This seems different than first defining a topological sphere as a quotient space of two disks because then one needs to show that all such quotient spaces are homeomorphic.

By the same logic, one could take as definition of topological 2-sphere, any manifold homeomorphic to a specific pair of closed disks glued on their boundaries as prescribed by the gluing map above.

 

[lavinia]

Of course the relevant gluing map glues "corresponding points" on disks
' boundary (e.g. point $(1,0)$ of $D_{+}$ is glued with point $(6,5)$ on $D_{-}$).

In the category of topological spaces and homeomorphisms, one can define those manifolds which are made from gluing two disks along their boundaries. It turns out that they are all homeomorphic to the standard sphere. Once this is proved, the topological sphere can be defined as two disks pasted together along their boundaries.

 

[lavinia]

@cianfa72

This thought came up while thinking about your thread. A topological 7 spere is two 7 disks that are glued together along their 6 sphere boundaries. Any such gluing creates a topological 7 sphere. But the gluing map still matters since the topological 7 sphere can have more than one differential structure yet the 6 sphere and the 7 disk both have only one differentiable structure.

I wonder if anyone has described these gluings explicitly.

 

[mathwonk]

apparently (by Smale's h-cobordism theorem) the crux is the pseudo-isotopy class of the (orientation preserving) gluing map:
https://math.stackexchange.com/ques...ffeomorphisms-of-the-sphere-modulo-isotopy-an
apparently the equivalence classes of these gluings form a group which is sometimes computable via cobordism and/or homotopy methods. E.g. there is only one class of gluing of the 60-dimensional sphere, and consequently only one class of smooth structures on the 61-dimensional sphere.
In particular there are 28 classes of gluings for the 6-sphere. I don't how explicitly they have been described, but it seems not very. Every source I found just says there is an exotic smooth structure on S^7, hence it is given by a gluing of S^6 which is not isotopic to the identity.

 

[lavinia]

apparently (by Smale's h-cobordism theorem) the crux is the pseudo-isotopy class of the (orientation preserving) gluing map:
https://math.stackexchange.com/ques...ffeomorphisms-of-the-sphere-modulo-isotopy-an
apparently the equivalence classes of these gluings form a group which is sometimes computable via cobordism and/or homotopy methods. E.g. there is only one class of gluing of the 60-dimensional sphere, and consequently only one class of smooth structures on the 61-dimensional sphere.
In particular there are 28 classes of gluings for the 6-sphere. I don't how explicitly they have been described, but it seems not very. Every source I found just says there is an exotic smooth structure on S^7, hence it is given by a gluing of S^6 which is not isotopic to the identity.

@mathwonk
Thank you for this wonderful reply. It inspires me to enter into this incrediblle world.