# How to make a 2-sphere in 4 dimensions

1. Jul 30, 2015

### 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 ?

2. Jul 30, 2015

### 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.

3. Jul 30, 2015

### 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".

4. Jul 31, 2015

### jk22

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

5. Jul 31, 2015

### 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?

6. Jul 31, 2015

### 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.

Last edited: Aug 1, 2015
7. Jul 31, 2015

### mathwonk

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

8. Aug 1, 2015

### disregardthat

Ah, well now I see the title.

9. Aug 8, 2015

### 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.

10. Aug 11, 2015

### sgd37

11. Sep 28, 2015

### 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.

Last edited: Sep 28, 2015
12. Sep 28, 2015

### WWGD

By a geometry you mean a choice of Riemannian metric?

13. Sep 28, 2015

### lavinia

yes

14. Sep 28, 2015

### lavinia

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.

Last edited: Sep 29, 2015
15. Oct 1, 2015

### WWGD

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

16. Oct 1, 2015

### WWGD

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.

17. Oct 5, 2015

### zinq

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.

18. Oct 5, 2015

### 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 $\pi_3 (S^2)$.

19. Oct 6, 2015

### zinq

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).

20. Oct 6, 2015

### lavinia

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