Flat Embedding of Our Universe: Questions & Answers

[gerald V]

TL;DR

Can it make sense to embed the universe in 8 flat dimensions?

It is mathematically proven that any intrinsically curved manifold can be derived from embedding in a flat space of sufficient number of dimensions. I heard somewhere, but lost that reference, that for our universe 10 flat dimensions are needed in the most general case.. May I ask

- Is the number 10 correct? If so, what about the metric signature of the embedding space (9+1, 8+2, 5+5, ...)?

- Is there literature on this topic (preferrably with not so much mathematical jargon)?

- If the embedding space only had 8 dimensions, what aspects of our universe would get lost? (as far as I know, for the special case of a homogeneous and isotropic universe, even 5 dimensions would be sufficient).Thank you very much in advance.

 

[George Jones]

Summary:: Can it make sense to embed the universe in 8 flat dimensions?

It is mathematically proven that any intrinsically curved manifold can be derived from embedding in a flat space of sufficient number of dimensions. I heard somewhere, but lost that reference, that for our universe 10 flat dimensions are needed in the most general case..

If you want to embed the universe isometrically, then the general result is much greater than 8 or 10. According to

Clarke, C. J. S., "On the global isometric embedding of pseudo-Riemannian
manifolds," Proc. Roy. Soc. A314 (1970) 417-428

every 4-dimensional spacetime can be embedded isometrically in higher dimensional flat space, and that 90 dimensions suffices - 87 spacelike and 3 timelike. A particular spacetime may be embeddable in a flat space that has dimension less than 90, but 90 guarantees the result for all possible spacetimes.

 

[gerald V]

Many thanks, George Jones.

Maths can surprise me again and again. In differential geometry - Isometric embedding - Mathematics Stack Exchange it is discussed that the notions used can refer to quite different mathematical structures.

I mean embedding in the very pedestrian sense of a submanifold with the metric on it given by Gauss formula for an induced metric. The embedding only shall conserve distances locally, as any nonsingular metric can be brought to Minkowski form locally. The embedding should cover the universe as a whole including its singularities at which points an induced metric simply does not exist, like (the circumference of) a square can be embedded in a plane.

As I understand from the quoted thread as well as from wagt `s answers in the thread Embedding curved spacetime in higher-d flat spacetime | Physics Forums , in this case Nash’s theorem holds and the number of necessary dimensions of the flat embedding space is 10 in the most general case. I shall look up the references given there.

However, Amanheis says in the same thread that 8 dimensions are sufficient in the most general case. Wikipedia (which is not a scientific source, I know, but usually is correct) Nash embedding theorem - Wikipedia says that for C^1 embedding, twice the dimensions of the embedded manifold - that would be 8 - are sufficient. This wikipedia article as well as Amanheis refer to Whitney embedding, which seems to be another animal.

Thanks in advance for any possible clarification.

 

[martinbn]

What's the motivation behind this question?

 

[gerald V]

Now, in a texbook I found an explanation, which is embarrasingly simple. To embed n-dimensional space parametrized by coordinates x into N-dimensional space parametrized by coordinates X, there are exactly N embedding equations. This is because for any point on the embedded manifold it must be well defined where it is situated in the embedding space.

On the other hand, Gauss‘ formula reads $g_{\mu \mu} = \frac {G_{ab}\partial X^a \partial X^b}{\partial x^\mu \partial x^\nu}$, where $g$ and $G$ are the metrices of the embedded space and of the embedding space, respectively.

As a symmetric tensor, $g$ has $\frac{n(n+1)}{2}$ independent entrances in the general case, and there must be as many embedding equations. Hence, $N \ge \frac{n(n+1)}{2}$. For our universe this means $N \ge 10$ if no further symmetries are around.

The freedom of coordinate transformations on the embedded manifold can be used to identify all the coordinates on it with a subset of $n$ coordinates on the embedding space. So $n$ embedding equations can be made trivial.

 

[PAllen]

If you want to embed the universe isometrically, then the general result is much greater than 8 or 10. According to

Clarke, C. J. S., "On the global isometric embedding of pseudo-Riemannian
manifolds," Proc. Roy. Soc. A314 (1970) 417-428

every 4-dimensional spacetime can be embedded isometrically in higher dimensional flat space, and that 90 dimensions suffices - 87 spacelike and 3 timelike. A particular spacetime may be embeddable in a flat space that has dimension less than 90, but 90 guarantees the result for all possible spacetimes.

That link doesn’t work for me. I assume (can you verify?) they mean a smooth isometric embedding. Otherwise, I would think way fewer dimensions are required.

Then I would think part of the issue is the generality of topology possible in principle. I took the OP to mean what is required for a spacetime plausible for our universe. That is, a wrinkly FLRW, with embedded BH, none of which have white hole parts, and for which one assumes perturbative stability removes the baroque interior topology of Kerr solutions. Perhaps no one has investigated this, but my guess is that far less than 90 dimensions are required for a smooth isometric embedding.