# Is de-Sitter spacetime homogeneous and isotropic?

• A
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The question is in the title. I believe the answer is yes.

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The spatial slices are, yes. But not only that, the de Sitter space is maximally symmetric.

vanhees71
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I would have thought that the existence of a flat slicing for de Sitter space, as described here, implies homogeneity and isotropy.

But then I got confused by this paper, which appears to state that Schwarzschild spacetime, which is neither isotropic nor homogeneous, also has a flat foliation. Although they claim that can be observed from the metric, which I can't see, since the transformed metric they present in support of that claim still depends on radius ##r##, which I would have thought implies non-homogeneity.

Staff Emeritus
Would it be correct to say that special relativity (SR) requires the local constancy of the speed of light, isotropy, homogoneity, and flatness? I don't recall ever seeing flatness ever spelled out as a requirement for SR, but I wouldn't regard de-Sitter space as being part of SR - and it has the other requirements I mentioned.

dextercioby
romsofia
If you have access to "introduction to General relativity, black holes, and cosmology" by the great Yvonne Choquet-Bruhat, on page 169 she does the math for "Isotropic and homogenous Riemannian manifolds" in problem VII 9.1, might be helpful

dextercioby
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I believe the answer is yes.
I agree.

Mentor
I would have thought that the existence of a flat slicing for de Sitter space, as described here, implies homogeneity and isotropy.
No, a flat slicing is not sufficient for homogeneity and isotropy, as the counterexample of Schwarzschild spacetime with Painleve coordinates shows. Nor is a flat slicing even a necessary condition for homogeneity and isotropy, since a closed FRW universe, for which spatial slices are 3-spheres, is homogeneous and isotropic and has no flat spatial slicing.

they claim that can be observed from the metric, which I can't see, since the transformed metric they present in support of that claim still depends on radius ##r##, which I would have thought implies non-homogeneity.
The paper you cited isn't claiming that Schwarzschild spacetime is homogeneous. They are only claiming that it has a flat spatial slicing. The way to see that from the Painleve metric is to note that the spatial part of the metric (i.e., the part you get when you set ##dt = 0## so all the terms with a ##dt## in them vanish) is the metric for flat Euclidean 3-space.

Mentor
I don't recall ever seeing flatness ever spelled out as a requirement for SR
It isn't spelled out explicitly in any treatment I'm aware of, but the existence of global Lorentz transformations (i.e., Lorentz transformations preserving the metric globally, instead of just locally) implies zero geodesic deviation (any two geodesics that start out parallel will always remain parallel), which is equivalent to flatness.

dextercioby
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Would it be correct to say that special relativity (SR) requires the local constancy of the speed of light, isotropy, homogoneity, and flatness? I don't recall ever seeing flatness ever spelled out as a requirement for SR, but I wouldn't regard de-Sitter space as being part of SR - and it has the other requirements I mentioned.
SR simply assumes a flat spacetime manifold (the pseudo-Euclidean affine space called Minkowski space), which is fixed once and for all. In SR there are global inertial frames and any observer at rest wrt. an inertial frame by assumption describes space as a Euclidean affine space (with all its symmetries of homogeneity and isotropy).

dextercioby
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No, a flat slicing is not sufficient for homogeneity and isotropy, as the counterexample of Schwarzschild spacetime with Painleve coordinates shows. Nor is a flat slicing even a necessary condition for homogeneity and isotropy, since a closed FRW universe, for which spatial slices are 3-spheres, is homogeneous and isotropic and has no flat spatial slicing.

The paper you cited isn't claiming that Schwarzschild spacetime is homogeneous. They are only claiming that it has a flat spatial slicing. The way to see that from the Painleve metric is to note that the spatial part of the metric (i.e., the part you get when you set ##dt = 0## so all the terms with a ##dt## in them vanish) is the metric for flat Euclidean 3-space.
From what I read in the paper, if you set ##dt=0## you do not get Euclidean space as it does not have a ##dr^2## term. The ##dr## only appears together with a ##dt##.

Gold Member
From what I read in the paper, if you set ##dt=0## you do not get Euclidean space as it does not have a ##dr^2## term. The ##dr## only appears together with a ##dt##.
I'm not familiar with Gullstrand-Painleve coordinates, but a Google search suggests that the formula used in the paper referred to above is wrong: they seem to have omitted a ##dr^2## term.

vanhees71
I would have thought that the existence of a flat slicing for de Sitter space, as described here, implies homogeneity and isotropy.

But then I got confused by this paper, which appears to state that Schwarzschild spacetime, which is neither isotropic nor homogeneous, also has a flat foliation. Although they claim that can be observed from the metric, which I can't see, since the transformed metric they present in support of that claim still depends on radius ##r##, which I would have thought implies non-homogeneity.
Just consider setting t constant in the metric from that paper. Then you have Euclidean flat metric left, in polar coordinates. The r of polar coordinates doesn’t make it non flat. In fact, within such a t constant slice, you can trivially transform to Cartesian coordinates, getting the Euclidean flat Cartesian metric.

vanhees71
I'm not familiar with Gullstrand-Painleve coordinates, but a Google search suggests that the formula used in the paper referred to above is wrong: they seem to have omitted a ##dr^2## term.
Oops, yes, but the conclusion is still correct. I’ve seen this many times before, so my mind filled in the missing dr2 term.

vanhees71 and DrGreg
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I don't recall ever seeing flatness ever spelled out as a requirement for SR
From page 60 of Wald's "General Relativity": "Thus, the theory of special relativity asserts that spacetime is the manifold ##\mathbb{R}^4## with a flat metric of Lorentz signature defined on it. Conversely, the entire content of special relativity as we have presented it thus far is contained in this statement, since, given ##\mathbb{R}^4## with a flat Lorentz metric, we can use the geodesics of this metric to construct global inertial coordinates, etc."

(Wald's italics, not mine.)

vanhees71, dextercioby and robphy
Mentor
From what I read in the paper, if you set ##dt=0## you do not get Euclidean space as it does not have a ##dr^2## term. The ##dr## only appears together with a ##dt##.
I think the paper is using unusual notation; by ##d\Omega^2## I think they mean the metric for Euclidean 3-space in spherical coordinates, not the metric of a 2-sphere in angular coordinates.

As others have pointed out, it is easily confirmed by checking other references that the correct metric for Schwarzschild spacetime in Painleve coordinates has flat (Euclidean 3-space) spatial slices of constant coordinate time.

vanhees71
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I think the paper is using unusual notation; by dΩ2 I think they mean the metric for Euclidean 3-space in spherical coordinates, not the metric of a 2-sphere in angular coordinates.
I think it is a typo. They do write ##r^2 d\Omega^2## so it would be a very strange notation to have the ##r^2## in front when there is no ##r^2## in the ##dr^2## term.

vanhees71
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It isn't spelled out explicitly in any treatment I'm aware of, but the existence of global Lorentz transformations (i.e., Lorentz transformations preserving the metric globally, instead of just locally) implies zero geodesic deviation (any two geodesics that start out parallel will always remain parallel), which is equivalent to flatness.
To be fair, the textbook on Relativity (Ray D'Inverno's Introducing Einstein's Relativity, Clarendon Press, 1992, p. 113) which I like most says:
Axiom I:
”Space and time are represented by a four-dimensional manifold endowed with a symmetric affine connection ##\Gamma^{a}_{~bc}## and a metric tensor ##g_{ab}## satisfying the following:
(i) ##g_{ab}## is non-singular with signature ##-2##;
(ii)## \nabla_c g_{ab} = 0##;
(iii) ##R^{a}_{~b\vert cd} = 0.##"

Quoting just under Axiom II on the same page: "The first axiom defines the geometry of the theory [...]. Thus the first axiom states that ##\Gamma^{a}_{~bc}## is the metric connection (by I(i)) and that the metric is flat (by I(iii)) [...]".

So the flatness in explicitly defining SR is a matter of reading (the good) books.

vanhees71 and pervect
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I think it is a typo. They do write ##r^2 d\Omega^2## so it would be a very strange notation to have the ##r^2## in front when there is no ##r^2## in the ##dr^2## term.
Hm, yes, good point.

vanhees71