On wikipedia I can find that a de Sitter-space has maximal symmetry and a constant curvature. Recently the interest of de Sitter spaces has increased as it could serve as a model for the universe, an universe which has (almost) no matter density, however a sensible non-zero value of the cosmological constant (large amount of dark energy compared to much smaller amound of dark & visible matter). So how can the de Sitter-space serve a model for the universe if the latter is supposed to be flat whereas the de Sitter space has constant curvature (I guess it is meant non-zero constant curvature)?
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asked Aug 22, 2017 at 13:17
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1$\begingroup$ The universe has zero spatial curvature but nonzero spacetime curvature. $\endgroup$– JavierAug 22, 2017 at 13:58
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$\begingroup$ @Javier If I look a the graphical plot of the de Sitter space in wikipedia, with dimensional reduction, then I see a hyperbolical cylinder surface with a spatial $\phi$-coordinate. So spatially I can walk around from e.g. $\phi=0$ to $\phi=2\pi$ and so come back to the same place. (The other coordinate, time $t$, however, is hyperbolical.) For me this is not compatible with a flat universe. $\endgroup$– Frederic ThomasAug 22, 2017 at 14:05
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$\begingroup$ @Javier no we don't know that. Measurements are consistent with zero; that's the most we can say. They are also consistent with non-zero. There is an infinite ratio between non-zero and zero. $\endgroup$– Andrew SteaneMay 28, 2020 at 21:55
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2 Answers
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Those are two different curvatures you are talking about.
First, you can talk about curvature of the spacetime i.e. treating one temporal and three spatial coordinates on equal footing. Then de Sitter spacetime has constant spacetime curvature, it's basically 4d hyperboloid. Realistic cosmological solutions also all have some spacetime curvature that however is not constant.
On the other hand, in cosmology it's common to consider a slice of constant time getting some 3d space. The time in question is chosen in such a way that everything on this 3d space is to a high degree homogeneous. The resulting 3d space can be of various topology and has some 3d curvature that is completely different from the spacetime curvature. Now the observed cosmology corresponds to zero 3d curvature but non-zero spacetime curvature.
You may ask what would be the 3d curvature for the de Sitter spacetime? The curious thing is how do you define the slice of constant time. The de Sitter spacetime is highly symmetric and you can actually slice it in many ways obtaining homogeneous space. Those possibilities fall into three categories that can be illustrated by this picture (which I made from this)
So while de Sitter spacetime has some positive spacetime curvature it can be viewed as having arbitrary constant 3d curvature.
UPD:
Let me be more technical now. The spacetime is characterized by its metric $g_{\mu\nu}$ determined by the Einstein equations. You can however consider various coordinates on the same spacetime. The definition of the space (and "now") as a slice of constant time $t\equiv x^0=\mathrm{const}$ is totally artificial. You may consider any coordinate system you like with all kinds of non-equivalent "spaces".
However in cosmology we are interested in the particular form of the metric, namely the Friedmann-Robertson-Walker (FRW) metric, \begin{equation} ds^2=g_{\mu\nu}dx^\mu dx^\nu=dt^2-a^2(t)d\vec{\Sigma}_k^2 \end{equation} where $d\vec{\Sigma}_k^2$ is a metric of 3d homogeneous space, either Euclidean space ($k=0$), unit sphere ($k=1$) or unit hyperboloid ($k=-1$). The slice of constant time $t=\mathrm{const}$ defines the 3d space with the metric $a^2(t)d\vec{\Sigma}_k^2$.
If you try to consider different coordinates $g_{\mu\nu}$ transforms as any tensor does. If the transformation is not a time shift, spatial rotation or spatial translation (or its analog in case of sphere and hyperboloid) you will lose the FRW form of the metric. That's why our desire to study the particular form of the metric prefers certain coordinate system and therefore certain notion of 3d space (which of course exists only to the extent that we can approximate actual spacetime with FRW metric)
The de Sitter spacetime however is exception. First, as I've already mentioned you can find three classes of the coordinate systems that will give you all three versions of the FRW metrics. So all three metrics, \begin{aligned} &ds^2=dt^2-e^{2t}d\vec{\Sigma}_0^2,\\ &ds^2=dt^2-\cosh^2(t)d\vec{\Sigma}_{+1}^2,\\ &ds^2=dt^2-\sinh^2(t)d\vec{\Sigma}_{-1}^2 \end{aligned} all describe the same de Sitter spacetime in different coordinates while being in the FRW form.
Second, the de Sitter spacetime has $SO(1,4)$ symmetry group coming from the Lorentz symmetry of its possible embedding as a hyperboloid into the $(1+4)$ Minkowski spacetime. Because of that all three metrics actually admit continuous symmetries that you can consider as generalizations of the Lorentz boosts.
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$\begingroup$ Thank you very much for this answer. I was already suspecting something like this. However, now I've even more questions. The de Sitter space is a solution of the Einstein's equations. So everything is already fixed. By which condition is the spacelike cut fixed, by more than was is already given by the field equations? Can you comment on this more ? Thanks $\endgroup$ Aug 24, 2017 at 22:15
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$\begingroup$ @FredericThomas I tried to answer your question in the new version. Comment if something still is not understood $\endgroup$– OONAug 25, 2017 at 21:03
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$\begingroup$ Thank you for the update. Actually, it still does not hit the heart of my main question, so let me be a bit provocative: It seems to be (at least for me) that changing the coordinates I can change the metrics from e.g. k=-1, to k=0 or k=+1. That's seems to be rather strange. Also I understood that the LFRW metric is based on some non-zero matter density which is apparently not the case for the de Sitter space. $\endgroup$ Aug 28, 2017 at 7:54
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2$\begingroup$ @FredericThomas Well, "seems to be rather strange" is not really a good argument, isn't it? You talk about how 3d curvature change under 4d coordinate transformations. 3d curvature is not a 4d tensor. Probably what is so "strange" about all that - it's just an example that in the general relativity our beloved 3d is just an artificial auxiliary construct, just some imaginary slicing of the spacetime. Concerning matter density, you can always include cosmological constant term into the matter part like dark energy. So de Sitter spacetime is not really empty in a certain sense. $\endgroup$– OONAug 28, 2017 at 19:21
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2$\begingroup$ Thank you very much! Isn't it amazing ? This is the reason I've been asking so insistingly. When I posted my principal question I would never thought that. Still I can hardly believe it. $\endgroup$ Sep 11, 2017 at 16:16
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Hi if you pick 3d euclidean space and you foliate it using nested spheres, you get these spheres have a certain induced metric and non 0 curvature. If you foliate it using a stack of parallel planes, you get a different induced metric and 0 curvature of these plane slices. Yet the manifold of which these slicings are taken has a fixed metric, and curvature (0 in this case). Now the diffrent foliations in deSitter space represent different physical models of the galaxies etc because they have different energy momentum tensors. Also the behavior of the galaxies is different in each model. So basically what we have here is that in the maximal desitter space we can sit several setups (matter distribution + geometry) that are solutions to einsteins equations. whatitis at youtube has some videos on this.
