# Curved spacetime: embedded or not?

1. Oct 17, 2009

### Dmitry67

Can we think about 2 different "interpretations" of GR:

1. Without embedding: space is curved, that's all
2. With embedding: curved non-euclidean space can be embedded in higher-demensional euclidean space.

In another words, spacetime is just curved (1), or it is curved is something (2)

Or in another words, curved spacetime in GR can be embedded in euclidean space, but it is just a pure mathematical result (1) or alternatively there might be some physical meaning of that super-space? (2)

Note that some topological solutions, like helf-spehere, is possible in (1) but not in (2)

2. Oct 17, 2009

### Nabeshin

I'm fairly certain the embedding in a higher-dimensional space is simply for our convenience in visualizing something. I.e it's pretty much impossible to think about the surface of a sphere without actually picturing a sphere embedded in R^3. I don't think any of the equations or derivations from GR require the existence of a higher dimensional space that spacetime be embedded in, at least not any I've encountered so far.

3. Oct 17, 2009

### Bob_for_short

Logunov's RTG is built in the plane Minkowski space-time. The Riemann geometry metric is an effective metric for matter but the gravitational filed equations contain the Minkowski metric and the gravitational field separately, so the physical background is plane (R=0). The gravitational field is then as physical as the other fields. One can say that the curvature of space-time is embedded in the plane space time of the same dimension (3+1) for matter but not for gravitational field.

In GR it is not so. Splitting the entire metric is not possible since R ≠ 0.

Last edited: Oct 17, 2009
4. Oct 17, 2009

### Spinnor

If number 2 how many dimensions would we need for the curved space-time we live in?

5. Oct 17, 2009

### JesseM

I bookmarked a thread on this long ago--here it is, post #3 on that thread mentions that the maximum number of dimensions needed for a flat embedding spacetime would be 90, 87 spacelike dimensions and 3 timelike, though that's just an upper limit, so for any given curved spacetime (including the one we live in) the answer could be lower.

This result was probably just derived out of mathematical interest though, there is no need for an embedding spacetime in GR, curvature can be defined "intrinsically" using differential geometry and adding an embedding spacetime shouldn't change any physical predictions about what will be observed within curved 4D spacetime.

Last edited: Oct 17, 2009
6. Oct 18, 2009

### Dmitry67

I can give you an example where predictions can be different.

Klein bottle

If space is NOT embedded, you can do it.
If space IS embedded, then you will intersect space with itself in attempt to make it.

7. Oct 18, 2009

### Hurkyl

Staff Emeritus
I can't tell precisely what you mean.

But I think it's relevant to point that:
. any 4-dimensional manifold contains Klein bottles
. Euclidean 3-space does not contain Klein bottles

8. Oct 18, 2009

### Dmitry67

yes, but is Klein bottle a 'bottle' in 4D space?
In 3D it is not realistic because it cant be made without intersections to itself
In 4D you can build it, but 'water' would be able to escape from it (like a ring is a 'bottle' in 2D but not in 3D)

9. Oct 18, 2009

### Hurkyl

Staff Emeritus
What does that have to do with anything? I would have assumed you're just playing a dumb word game, but you sound like you really meant what you said....

10. Oct 18, 2009

### Phrak

I've heard that spacetime can be embedded in an R5 Euclidean space. Is there any truth to this rumor?

11. Oct 18, 2009

### JesseM

It can't be embedded in 3 dimensions, but in general there's no guarantee that a curved N-dimensional surface (with the curvature at every point defined using differential geometry) can be embedded in a Euclidean space of N+1 dimensions, you may need more dimensions to do it. The proof I mentioned earlier shows that every curved 4D spacetime allowed by GR (which would presumably include spacetimes where space has the topology of a Klein bottle) can be embedded in a flat spacetime of at most 90 dimensions.

12. Oct 18, 2009

### JesseM

Did you see my post #5 above? The proof mentioned there, that every 4D spacetime allowed by GR can be embedded in a flat spacetime of at most 90 dimensions, would be pretty pointless if there was also a proof it could be embedded in a flat spacetime of only 5 dimensions, so I'm pretty confident the rumor is wrong.

13. Oct 18, 2009

### Dmitry67

Another interesting question: is it possible that these "extra" dimensions can also include additional 26-4, 11-4 or 10-4 dimensions from the superstring theory?

14. Oct 18, 2009

### Phrak

I found something on it. The wiki claims that a pseudo-Riemann manifold can be embedded in an n(n+1)/2 pseudo-Euclidian space. This would be 10 for spacetime.

More, a 4 dimensional Riemann can be embedded in a 5 dimensional manifold of zero Ricci curvature. It's a little unclear whether this includes pseudo-Riemann manifolds.

My numerologist's handbook doesn't come up anything special about the number 90, other than [n(n+1)/2][-1+n(n+1)/2]. Do you know where it comes from?

15. Oct 18, 2009

### JesseM

What wiki page says that? I didn't see a wiki link on that thread.
I don't know how it was derived, the paper the result comes from is Chris Clarke's "On the global isometric embedding of pseudo-Riemannian manifolds" which doesn't seem to be available online (I'm sure I wouldn't understand it even if it was), though I did find this paper on arxiv.org which references Clarke's paper and seems to be discussing similar issues (but again the math is beyond me).

16. Oct 18, 2009

### Phrak

The "[URL [Broken] article[/URL], is on Campbell's theorem.

thanks, that the one I also came up with in a search.

Last edited by a moderator: May 4, 2017
17. Oct 18, 2009

### JesseM

Interesting. The Wolfram MathWorld page on this says that Campbell's theorem says "Any n-dimensional Riemannian manifold can be locally embedded into an (n+1)-dimensional manifold with Ricci curvature Rab = 0" and that "any n-dimensional pseudo-Riemannian manifold can be locally and isometrically embedded in an n(n+1)/2-dimensional pseudo-Euclidean space" (italics mine)--I wonder about that use of the word "locally", maybe this is different in some way than an embedding of the entire manifold? I also found this paper on arxiv.org which says on p. 5:

Last edited by a moderator: May 4, 2017
18. Oct 18, 2009

### Phrak

It does get involved, doesn't it? I can only guess that 'local' embedding is far less stringent.

Then there is the matter of differentiability. Ideally we'd like an embedding that is one-to-one and onto. So how do we embed a spacetime that contains singularities?

19. Oct 18, 2009

### cesiumfrog

Any (3+1 dimensional) spacetime can be embedded in 8 dimensional Euclidean space. (This was mentioned in Jesse's first source.) It's only if you want to embed it in Minkowski space (in a manner which completely determines the metric) that you may need an order of magnitude more dimensions. A local embedding wouldn't really make sense, you'd only need 4 dimensions for that. I'm not aware if there is any useful physics yet that follows from the existence of embeddings?

20. Oct 18, 2009

### Phrak

Thanks for that, cesiumfrog. Jesse's previous reference actually has something to say about this, although these are probably not the sort of embedding you have in mind, such as Kaluza-Klein theory.