Sphere to donut mathematically

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In summary: S^{2}→R^{3}## with ##f_{0}=f## and ##f_{1}## as the embedding of the torus I the sphere. The idea is to have a continuous path of embeddings connecting the two surfaces. In this restricted sense one can not continuously deform a sphere into a torus. However, this idea is not necessary for the intuitive idea of continuous deformation to be valid and it does not go against the idea of bringing opposite poles of a sphere together to form a torus.In summary, the conversation is about the possibility of continuously deforming a sphere into a torus and the argument of a mathematician about the use of the term "dough
  • #1
Dopamine
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I just read through this thread* where a mathematician tries to shutdown another guys use of the "sphere to donut" analogy relative to ripping spacetime because a sphere is a "2-dimensional manifold". That is so completely and entirely irrelevant when both are still just 3 dimensional shapes, and shapes that are easily changed to and from one another both mathematically and via a CGI representation. I felt the need to comment on the thread but then found out I have insufficient privelages. :*(
 
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  • #2
Which thread is it? It is probably too old to accept new replies.
 
  • #4
Dopamine said:
I just read through this thread* where a mathematician tries to shutdown another guys use of the "sphere to donut" analogy relative to ripping spacetime because a sphere is a "2-dimensional manifold".
A sphere and a torus (doughnut) or topologically different, with the sphere being a genus-0 surface and the torus being a genus-1 surface. It is not possible to continuously deforn a sphere into a doughnut. That's probably what the mathematician was talking about regarding ripping spacetime.
Dopamine said:
That is so completely and entirely irrelevant when both are still just 3 dimensional shapes, and shapes that are easily changed to and from one another both mathematically and via a CGI representation. I felt the need to comment on the thread but then found out I have insufficient privelages. :*(
Maybe so using CGI, but not so mathematically.
 
  • #5
Your technical jargon I'm sure is accurate, but bringing the highest and lowest point of the sphere on a given axis inwards is only possible via cgi, because it is possible mathematically. Are you implying it's not possible because the surface area of the sphere wouldn't be equivalent? or...

If you are so bright as to understand these concepts, surely you can explain why it's not possible to mathematically "take your finger and poke a hole in a solid donut", to an idiot like me. Let's say it is a strawberry jelly donut.
 
  • #6
Dopamine said:
If you are so bright as to understand these concepts, surely you can explain why it's not possible to mathematically "take your finger and poke a hole in a solid donut", to an idiot like me. Let's say it is a strawberry jelly donut.
Try doing it without the donut ripping. You can't.
 
  • #7
Greg Bernhardt said:
Try doing it without the donut ripping. You can't.
You cannot do it continuously:
Mark44 said:
It is not possible to continuously deforn a sphere into a doughnut.
You could look up a mathematical statement called the "no-retraction theorem". On Google Scholar this yields, for example:

Y. Kannai, An elementary proof of the no-retraction theorem, The American Mathematical Monthly, 1981.

which may be worth a look. (Note: I do not know this reference myself, but the title suggests it is readable without too much advanced knowledge.)

The intuitive idea is that near the spot where you start the hole there will always be two points that, no matter how close they are on the initial, solid doughnut, they will end up on opposite sides of the hole.

Dopamine said:
Lets say it is a strawberry jelly donut.
Why not a chocolate doughnut? I want a chocolate doughnut instead. In fact, I would like two. :-p
 
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  • #8
Dopamine said:
Your technical jargon I'm sure is accurate, but bringing the highest and lowest point of the sphere on a given axis inwards is only possible via cgi, because it is possible mathematically. Are you implying it's not possible because the surface area of the sphere wouldn't be equivalent? or...

If you are so bright as to understand these concepts, surely you can explain why it's not possible to mathematically "take your finger and poke a hole in a solid donut", to an idiot like me. Let's say it is a strawberry jelly donut.

The mathematician in the thread is making a semantic argument about what "doughnut" means in Mathematics. He says it means a solid torus rather than a torus and then says that it is "absurd" to continuously deform a three dimensional manifold into a two dimensional manifold. This is just nit picking about terminology and has nothing to do with whether one can deform a sphere into a torus or not.

You are right that a change in topology can occur through continuous deformation and your example of bringing opposite poles of a sphere together works. Once the poles touch one no longer has a sphere. The resulting topological space is called a "pinched torus" The reason is that the same space can be obtained by pinching a transversal circle on a torus to a single point. This can be done by continuous deformation of the torus.

From this one sees that just because two topological spaces have different topology does not mean that they can not be continuously deformed into each other. So it is false to say that the sphere can not be deformed into a torus just because the two are topologically distinct.

If one thinks of topological spaces as made of infinitely flexible material, then a single point may be viewed as a circle with zero radius and as such can be expanded into a circle of positive radius by continuous deformation. In the case of the pinched torus this means that the pinch point is a circle of zero radius that can be expanded continuously into a circle. The result after this radius expansion is a torus back again Another way to say this is that the pinching can be reversed. Puttting together the two deformations: the pinching of the sphere followed by the reverse pinching of the torus in some sense continuously transforms a sphere into a torus.


- There is a more restricted idea of continuous deformation which perhaps may underlie some of the things said here. So I thought at least to mention it. If one imagines an abstract topological sphere then its "realization" as a surface in 3 space can be thought of as an embedding ##f:S^{2}→R^{3}##. A continuous deformation can be thought of as a 1 parameter family of continuous mappings ##F:S^{2}×I→R^{3}## where at time zero ##F## is an
embedding. Such a map transforms the sphere into whatever space one has at time 1. This may be another embedded sphere or some other topological space. One can ask whether the space at time 1 can be a torus.
 
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  • #9
Greg Bernhardt- the article listed is about the ripping of the "Fabric" of space time, so your statement is supporting the layman who's argument was semantically and, honestly, falsely disputed.

Lavinia states it can be done continuously; Krylov states that it cannot, and given my experiential data from this morning.. I postulate that it rips, and some stuff falls out.
 
  • #10
Dopamine said:
Lavinia states it can be done continuously; Krylov states that it cannot, and given my experiential data from this morning.. I postulate that it rips, and some stuff falls out.
More precisely, I referred you to an article that discusses a mathematical statement that I felt relates to your question. If you want to talk about (im)possibilities, you first need to be mathematically precise about what exactly is supposed to be (im)possible.
 
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  • #11
We are analogizing the fabric of time and space to donuts. I'd suggest re-calibrating your humor sensor.I will add, just for the sake of some vainglorious psychological analysis, that you are stronger in maths than physics. Thank you for the statement though, I will check it out a bit later after I clean up this jelly stain.
 
  • #12
Dopamine said:
We are analogizing the fabric of time and space to donuts. I'd suggest re-calibrating your humor sensor.I will add, just for the sake of some vainglorious psychological analysis, that you are stronger in maths than physics. Thank you for the statement though, I will check it out a bit later after I clean up this jelly stain.
Right... Shall we (try to) talk about mathematics?

You could attempt to understand Lavinia's post #8 above and/or look at the no-retraction theorem. She knows a lot more about geometry and topology than I do. Then you could ask questions here.
 
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  • #13
Please keep in mind any attempts to discuss mathematics with me will have to be at the level of d(potato)/dy = 2potato.

I understood her post though, appreciate the added subtext, and will definitely be reading your posted statement asap.

Again, all comments and discourse are appreciated, genuinely.
 
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  • #14
https://www.math.washington.edu/~morrow/336_11/papers/jack.pdf

Ok, since this is the only public paper I can find easily, I've attempted to delve into it to better understand your perspective(s). If at any time my questions are overwhelming, obtuse, irrelevant or irritating I encourage you just to ignore my pedantic attempts at grasping things far beyond my level of education.

lavinia mentions "3 space". I'm assuming that's 3d space? When referring to an "embedding", you are just talking about the function f: s^2 -> r^3 being embedded into the 3d space you're discussing?

I see those constants, S^2 -> R^3 used throughout the paper in reference to different things. As some constant squared approaches another constant cubed?

You list that at time 0, F is an embedding. I do not follow this logic through since I do not understand the relationship of the function listed, to time. I guess it wouldn't really matter? But At time 0, F would just be equivalent to the starting embedded function/shape? whether pinched torus or sphere?
 
  • #15
Now I'm also a little lost as to what relevance this non-retraction theorem has to two 3 dimensional shapes morphing into each other either. It states that

A homeomorphism from one topological space to another is a
bicontinuous (and thus invertible) mapping. Homeomorphisms preserve a wide variety of topological properties. If there
exists a homeomorphism between two spaces, they are said to be homeomorphic to each other.

So given lavinia's post we can conclude that the sphere and pinched torus are homeomorphic to each other?

Thanks again

And sorry, little addendum here, but if we can conclude that the shapes are homeomorphic, and then giving them some assumed values typical of basic physics (time/space being warped by mass), wouldn't that even imply it is possible to warp a sphere of time/space given enough mass to cause the very rip we are postulating about?
^^ Please excuse the **huge** leap in logic here.
 
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  • #16
The torus and the sphere are not homeomorpic. As explained, the transformations change the topology. There is no homeomorphism.

If you want to deform the sphere by a 1 parameter family of homeomorphisms then you will never get anything except another topological sphere.

There are many ways to prove that the torus is not homeomorphic to the sphere. Here is a short list:

1) The torus can be given a flat Riemannian metric. The sphere cannot.
2) The fundamental group of the torus is not zero. The fundamental group of the sphere is zero. The fundamental group is a homeomorphism invariant.
3) The Euler characteristic of the torus is zero. The Euler characteristic of the sphere is 2.The Euler characteristic is a homeomorphism invariant.
4) The universal covering space of the torus is the plane. The universal covering space of the sphere is itself.
5) The torus can be given the structure of a Lie group. The sphere cannot.
 
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  • #17
Ok, so we are changing the topology. Either way, it's mainly a physics discussion and the mathematics put forth weren't relevant to what the poster was trying to portray.

The Einstein-Rosen Bridge
^^ would have been what the poster was getting at
 

1. What is the mathematical formula for converting a sphere to a donut?

The mathematical formula for converting a sphere to a donut, also known as a torus, is (x^2 + y^2 + z^2 + R^2 - r^2)^2 - 4R^2(x^2 + y^2) = 0, where R is the radius of the sphere and r is the radius of the donut.

2. How does the shape of a sphere change when it is converted to a donut mathematically?

When a sphere is converted to a donut mathematically, its overall shape remains the same but it gains a hole in the center. The resulting shape is called a torus and it is a three-dimensional shape with a continuous surface.

3. Can a sphere be converted to a donut with any size?

Yes, a sphere can be converted to a donut with any size. The radius of the sphere and the radius of the donut can be adjusted to create a torus of any desired size.

4. What are the real-life applications of converting a sphere to a donut mathematically?

Converting a sphere to a donut mathematically has various real-life applications, including in the fields of computer graphics, architecture, and engineering. It is also used in mathematical modeling and simulations.

5. Is there a way to visualize the conversion of a sphere to a donut mathematically?

Yes, there are various ways to visualize the conversion of a sphere to a donut mathematically. One way is to use computer software to create a 3D model of the torus. Another way is to use physical models or diagrams to understand the transformation visually.

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