How to make a 2-sphere in 4 dimensions

[jk22]

I want a closed two dimensional manifold embedded in four dimensions.

Is the following way a good one :

Parametrize a 3 sphere with $$\theta,\phi,\chi$$

Put $$\chi=f(\theta,\phi)$$

Does it make the manifold closed ?

 

[fzero]

In Euclidean 4-space, the solution to any quadratic in 3 variables will define a topological 2-sphere (times a line). If you want to obtain the 2-sphere as a submanifold of a 3-sphere, then the natural construction is called the Hopf fibration.

 

[mathwonk]

any closed 2 manifold embedded in 3 dimensions is already also wmbedded in 4 dimensions, so the usual 2 sphere in R^3 works, where R^3 is embedded in R^4 via (x1,x2,x3) --> (x1,x2,x3,0). So i don't know what you really mean.

the hopf fibration by the way is not an embedding of the 2 sohere into the 3 sphere but a fibration map from the 3 sphere onto the 2 sphere, with circles as fibers. The 2 sphere is easily embedded in the 3 sphere as its "equator".

 

[jk22]

any closed 2 manifold embedded in 3 dimensions is already also wmbedded in 4 dimensions, so the usual 2 sphere in R^3 works, where R^3 is embedded in R^4 via (x1,x2,x3) --> (x1,x2,x3,0). So i don't know what you really mean.

I imagined that there could be more possibilities in 4d like kind of 'knotted' spheres that we could not make in 3d.

 

[mathwonk]

like maybe take a knotted circle in 3 space and cone it off in two different directions in 4 space, joining its points to a single point in R^+ x 3 space, and also to a single point in R^- x 3 space? would that work?

 

[disregardthat]

If you didn't want any "trivial" cases of this, you always have the classic example of the Klein bottle, which is a compact 2-dimensional manifold embeddable in $\mathbb{R}^4$ but not $\mathbb{R}^3$. Otherwise, mathwonks example is pretty neat.

 

[mathwonk]

i thought he wanted one homeomorphic to a 2 sphere, but i like the klein bottle.

 

[disregardthat]

Ah, well now I see the title.

 

[jk22]

For example a 2sphere in 4d could be given by the following embedding :

$$\phi=2\pi sin(u)sin(v)\\ \psi=\pi sin(u)cos(v) \\ \chi=\pi Cos(u) \\ w=sin(\phi)sin(\psi)sin(\chi) \\ x=sin(\phi)sin(\psi)cos(\chi)\\
Y=sin(\phi)cos(\psi)\\
Z=cos(\phi)$$

But i cannot visualize what it represents.

Btw i would like to write a c++ program to project 4d objects maybe a collaborative online work is possible. If any interested.

 

[sgd37]

Do you mean the Hopf map

http://mathworld.wolfram.com/HopfMap.html

 

[lavinia]

Another example, similar to the Klein bottle, is the projective plane. It can not be realized in 3 space but can in 4. What it and the Klein bottle have in common is that they are not orientable surfaces. It is a theorem that a compact surface without boundary( called a "closed surface") in 3 space must be orientable.

Orientability is a topological property of a surface. But there are also geometrical restrictions on a surface that can be realized in 3 space. For instance, a theorem of Hilbert says that a closed surface must have a point of positive Gauss curvature. So any closed surface of non-positive curvature e.g. the flat torus or a closed surface of constant negative curvature can not be realized in 3 space. However if you remove the restriction that the surface has no boundary then there are examples. For instance, the cylinder and the Mobius band are flat and the pseudosphere has constant negative curvature. Other examples can be found in Struik's book on classical differential geometry.

I wonder if one can put a geometry on a disk so that no open subset of it can be realized in 3 space.

 

[WWGD]

Another example, similar to the Klein bottle, is the projective plane. It can not be realized in 3 space but can in 4. What it and the Klein bottle have in common is that they are not orientable surfaces. It is a theorem that a compact surface without boundary( called a "closed surface") in 3 space must be orientable.

Orientability is a topological property of a surface. But there are also geometrical restrictions on a surface that can be realized in 3 space. For instance, a theorem of Hilbert says that a closed surface must have a point of positive curvature. So any closed surface of non-positive curvature e.g. the flat torus or a closed surface of constant negative curvature can not be realized in 3 space. However if you remove the restriction that the surface has no boundary then there are examples. For instance, the cylinder and the Mobius band are flat and the pseudosphere has constant negative curvature. Other examples can be found in Struik's book on classical differential geometry.

I wonder if one can put a geometry on a disk so that no open subset of it can be realized in 3 space.

By a geometry you mean a choice of Riemannian metric?

 

[lavinia]

By a geometry you mean a choice of Riemannian metric?

yes

 

[lavinia]

like maybe take a knotted circle in 3 space and cone it off in two different directions in 4 space, joining its points to a single point in R^+ x 3 space, and also to a single point in R^- x 3 space? would that work?

Mathwonk's example answers your question by showing how a topological 2 sphere can be knotted in 4 space. By knotted one means much the same thing as a knotted string in 3 space. Clearly a sphere can not be knotted in 3 space. His example is called a suspension knot since it is an embedding of the suspension of a knotted 1 sphere.

There are other ways to start with knotted strings to get knotted spheres in 4 space. For instance, tie a knot in an open string and place it in the upper half space of R3 with its end points touching the boundary 2 plane. In R4, rotate the half space around the bounding plane. The result is a knotted 2 sphere in R4.

To picture this think of spinning a half plane in R3 around its edge line. A point in the half plane will spin to become a circle. Similarly each point in the knotted string will spin into a circle in R4 except the two end points since they are anchored to the edge plane which is stationary. This shows that the spun knot is a 2 sphere. These knotted spheres are called spin knots.

A third method is to allow the knot to twist around as it spins.

- One may ask the question of when are two knots are essentially the same. For instance, two unknotted strings with different shapes could be thought of as equivalent but an unknotted string and a knotted one are not. Similarly there is an idea of an unknotted 2 sphere in 4 space. There is a rich theory that deals with this question. It applies not only to two spheres in 4 space but to all dimensions. There are amazing examples of non-trivially knotted spheres in 4 space.

- Here is a weird example. Take a knot that has infinitely many knotted loops and then apply Mathwonk's construction.

 

[WWGD]

I imagined that there could be more possibilities in 4d like kind of 'knotted' spheres that we could not make in 3d.

I think in general knotedness (for any object, not just for spheres) is necessarily a codimension-2 issue.

 

[WWGD]

I want a closed two dimensional manifold embedded in four dimensions.

Is the following way a good one :

Parametrize a 3 sphere with $$\theta,\phi,\chi$$

Put $$\chi=f(\theta,\phi)$$

Does it make the manifold closed ?

I see, so you are trying to use a version of implicit/inverse function theorem? Then, yes, if you find a regular point , then the inverse is a submanifold.

 

[zinq]

In Euclidean 4-space, the solution to any quadratic in 3 variables will define a topological 2-sphere (times a line). If you want to obtain the 2-sphere as a submanifold of a 3-sphere, then the natural construction is called the Hopf fibration.

This isn't right. The Hopf fibration does not involve the 2-sphere as a submanifold of the 3-sphere. The 2-sphere in the Hopf fibration is what becomes of the 3-sphere when each fibre (which is a circle) is collapsed to a point. Which makes the 2-sphere the base space of this fibre bundle.

 

[WWGD]

And I think Hopf Fibration also gives an example of a non-trivial class of higher $\pi_n$ for general spheres, i.e., $S^m$s, specifically, the map is a non-trivial element of $\pi_3 (S^2)$.

 

[zinq]

. . .
. . .

I wonder if one can put a geometry on a disk so that no open subset of it can be realized in 3 space.

Great question. After a bit of web searching . . . it apparently turns out the answer is not always if the metric is smooth (by which I mean C^∞; see http://arxiv.org/pdf/math/0208127.pdf). But always if the metric is real-analytic (Lewy H (1938) On the existence of a closed convex surface realizing a given Riemannian metric. Proc Natl Acad Sci 24:104–106).

 

[lavinia]

Great question. After a bit of web searching . . . it apparently turns out the answer is not always if the metric is smooth (by which I mean C^∞; see http://arxiv.org/pdf/math/0208127.pdf). But always if the metric is real-analytic (Lewy H (1938) On the existence of a closed convex surface realizing a given Riemannian metric. Proc Natl Acad Sci 24:104–106).

very cool. i haven't read the paper yet but the theorems seem amazing.

 

[zinq]

The papers require a great deal of knowledge (especially about PDEs) that I don't have.

On second thought, it looks as if that 1938 paper I cited covers only the case of a convex surface. But I read somewhere else that it's always true in the real-analytic case.

 

[lavinia]

... Then if M and N are topologically equivalent manifolds, there must be a homeomorphism h: M —> N. This implies that for any point p of M, the relative homology groups H_k(M,M-p) and H_k(N,N-h(p)) are isomorphic.

If the dimensions dim(M) and dim(N) were not equal, let's assume, without loss of generality, that d = dim(M) > dim(N). It can be shown that for any manifold M of dimension d, the relative group H_d(M,M-p) is isomorphic to the group ℤ of integers. It can also be shown that for any manifold N with d greater than the dimension of N, the homology group H_d(N,N-q) is the trivial group {0}, where q is any point of N.

But if M and N were homeomorphic they would have isomorphic relative groups. Since they do not, this shows that any two manifolds of unequal dimensions cannot be homeomorphic.

It is worth elaborating this method of proof since it shows the power of homology theory.

If M is any manifold then by definition any point,p, in M has a neighborhood homeomorphic to an open ball,A, in Rn. By the excision axiom of homology

Hq(M,M-p) = Hq(M-(M-A), (M-p) - (M-A)) = Hq(A,A-p) for all dimensions,q.

A is contractible so its homology is zero in all dimensions except dimension 0.
A-p deforms onto a n-1 sphere so its homology is zero in all dimensions except 0 and n-1. In dimension n-1 its homology(with Z coefficients) is Z.

By the exact sequence of this pair one gets

... Hq(A) -> Hq(A,A-p) -> Hq-1(A-p) -> Hq-1(A) ...

Since A is contractible this is

0 -> Hq(A,A-p) -> Hq-1(A-p) -> 0 for q>1

For q = n, the dimension of the manifold, the sequence is

0 -> Hn(A,A-p) -> Z ->0 so Hn(A,A-p) = Z.

In the other dimensions (except 1) the sequence is

0 -> Hq(A,A-p) -> 0 ->0 so Hq(A,A-p) = 0.

In dimension 1 the sequence is

0 -> H1(A,A-p) -> H0(A-p) -> H0(A) and the last arrow on the right is an isomorphism.This is an example of the power of homology theory. Difficult theorems are reduced to simple calculations. Another example is the homology proof of Brouwer's Fixed Point Theorem. A difficult analytical proof was discovered much later by Milnor.