A space (shape) is contractible if you can “contract” (shrink) it to a point without cutting, pinching or punching through holes. For example, a mattress is contractible, since you can shrink it to the center - each point can follow the line to the center, continuously. Meanwhile, a doughnut, a circle or a hollow sphere are not contractible, you can never remove the inner “hole” to shrink to a point without cutting.
In general, any dimensional sphere is not contractible… Until it is - infinite dimensional sphere is contractible. Somehow, it loses the “hollow space” inside.
For each finite dimension n (1, or 2, or 3, etc…), the sphere in dimension n can’t be contracted because of that empty n-dimensional space it surrounds. But that same sphere is the “equator” of the sphere in the next higher dimension, n+1. There, the n-dimensional equator can contract along one of the hemispheres, to a pole. But then that whole (n+1)-dimensional sphere still isn’t contractible, because of the (n+1)-dimensional space it surrounds.
BUT the (n+1)-dimensional sphere can contract along one of the hemispheres in the (n+2)-dimensional sphere. And so on.
For any particular finite dimension n, there is an n-dimensional obstruction to contracting the sphere in that dimension. But if you go all the way to infinitely-many dimensions, there is no obstruction that ever stops contractibility of the infinite-dimensional sphere.
Explanation?
I am not a topologist, but I can try…
A space (shape) is contractible if you can “contract” (shrink) it to a point without cutting, pinching or punching through holes. For example, a mattress is contractible, since you can shrink it to the center - each point can follow the line to the center, continuously. Meanwhile, a doughnut, a circle or a hollow sphere are not contractible, you can never remove the inner “hole” to shrink to a point without cutting.
In general, any dimensional sphere is not contractible… Until it is - infinite dimensional sphere is contractible. Somehow, it loses the “hollow space” inside.
For each finite dimension n (1, or 2, or 3, etc…), the sphere in dimension n can’t be contracted because of that empty n-dimensional space it surrounds. But that same sphere is the “equator” of the sphere in the next higher dimension, n+1. There, the n-dimensional equator can contract along one of the hemispheres, to a pole. But then that whole (n+1)-dimensional sphere still isn’t contractible, because of the (n+1)-dimensional space it surrounds.
BUT the (n+1)-dimensional sphere can contract along one of the hemispheres in the (n+2)-dimensional sphere. And so on.
For any particular finite dimension n, there is an n-dimensional obstruction to contracting the sphere in that dimension. But if you go all the way to infinitely-many dimensions, there is no obstruction that ever stops contractibility of the infinite-dimensional sphere.
Ty
Sphere refers to the surface area, correct? Ball would be the volume inside?
Indeed.
On what level? A proof? Or just the meaning of the words?
What is contractible in mathematics? Wikipedia had article on contractible spaces but that was way beyond be.
https://en.wikipedia.org/wiki/Contractibility_of_unit_sphere_in_Hilbert_space
I can offer no ELI5 but here’s the context