Does de Sitter Spacetime Have Flat Foliation?

[andrewkirk]

TL;DR Summary

de Sitter spacetime is curved despite containing no mass-energy, because of a positive cosmological constant. Does it have a foliation into spatially flat hypersurfaces though?

I was just reading about de Sitter space and the following question occurred to me:

de Sitter spacetime is curved despite containing no mass-energy, because it has a positive cosmological constant. Does it have a foliation into spatially flat, constant-time hypersurfaces though?

Maybe it's just me but I find interesting the question of whether space is curved as well as spacetime. I looked at a few articles on de Sitter spacetime but could not readily find the answer. I was hoping somebody who knows lots about de Sitter spacetimes could tell me.

Thank you.

 

[pervect]

Wiki gives a conformally flat metric,. Which isn't quite what you're looking for, but might be of interest.

https://en.wikipedia.org/w/index.php?title=De_Sitter_space&oldid=961657095#Flat_slicing

If I'm reading the article correctly, the metric is something like

$$ds^2 = a(t) (dx^2 + dy^2 + dz^2 - dt^2)$$

One non-general possibility for a(t) is 1/(1-t)^2.

I mashed this into GrTensor. Using the orthonormal basis dt/(1-t), dx/(1-t),dy/(1-t),dz/(1-t) , the computer calculates the Einstein tensor G in this basis as:

$$G_{\hat{a}\hat{b}} = \begin{bmatrix} 3&0&0&0\\0&-3&0&0\\0&0&-3&0\\0&0&0&-3 \end{bmatrix}$$

which looks right to me, a positive constant density and a negative diagonal pressure with the same absolute value. There is a glaring coordinate singularity at t=1.

 

[PeterDonis]

Does it have a foliation into spatially flat, constant-time hypersurfaces

Yes:

https://en.wikipedia.org/wiki/De_Sitter_space#Flat_slicing

 

[PeterDonis]

Wiki gives a conformally flat metric

This slicing isn't just conformally flat, it's flat. The constant time hypersurfaces are flat Euclidean 3-spaces.

 

[PeterDonis]

If I'm reading the article correctly

The line element for the flat slicing is

$$
ds^2 = - dt^2 + a^2(t) \left( dx^2 + dy^2 + dz^2 \right)
$$

Exactly as you'd expect for a flat slicing. The scale factor is $a(t) = e^{t / \alpha}$, where $\alpha$ is related to the cosmological constant by $\Lambda = 3 / \alpha^2$.

The article also mentions a conformally flat form of the line element, but that is not the flat slicing.