1. Sep 11, 2014

### stevendaryl

Staff Emeritus
According to a quote from a past Physics Forums article, 4D spacetime can be embedded isometrically (preserving the metric) in 90-dimensional flat spacetime:
My intuition is that smaller regions of spacetime will require many fewer dimensions. By analogy: a 2D torus requires 3D for an embedding, but any small enough section can be embedded in 2D. My question is whether there is some kind of limit theorem of the form:

For any 4D spacetime, every point belongs to an open set that can be isometrically embedded in flat spacetime of N dimensions or fewer.

where N is some number much smaller than 90. (If we drop the "isometrically", then the answer is clearly N=4, because every 4D manifold is by definition made up of 4D sections stitched.) together.

2. Sep 11, 2014

### A.T.

I don't understand the qualifiers "spacelike" and "timelike" for embedding dimensions.

When I think about embedding a curved 2D-spacetime in a flat 3D embedding space, I don't see the 3 embedding dimensions as having any physical meaning. They are neither representing space nor time, and the orientation of the 2D-spacetime-surface relative to those 3 dimensions is completely arbitrary.

3. Sep 11, 2014

### Staff: Mentor

In this context I would interpret "timelike" and "spacelike" as referring to the signs of the metric coefficients.

4. Sep 11, 2014

### cosmik debris

There is something called the Whiteleigh (spelling???) embedding theorem. From my rapidly fading memory I thought Minkowski space/time would embed in (88, 2).

5. Sep 11, 2014

### PAllen

On the torus: a torus whose curvature is zero needs 4-d to embed in. Thus, if its patches embed in flat 2-d, the whole thing needs 4-d. If it has the right curvature, then both its patches and the whole thing embed in 3-d.