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)?

# [Physics] Is de Sitter space with non-zero curvature an acceptable model for the universe

cosmologycurvaturede-sitter-spacetimespacetime

#### Related Solutions

The reason why the AdS/CFT correspondence is interesting is not that AdS space is supposed to describe our universe, which, as you have correctly pointed out, would lead to conflicts with experiments. In the context of the correspondence, a four-dimensional (conformal) field theory is mapped to a string theory living in an $AdS_5\times S^5$ space, although there exist generalizations in which the AdS part is of higher or lower dimension than five.

This duality in principle allows one to carry calculations from one side to the other, making it possible to choose the framework in which the solution to the problem at hand can be found conveniently. One key observation in this context is that the duality can map a strongly coupled theory to a weakly coupled one, circumventing the failure of perturbation series. This is especially interesting with respect to QCD, where a a conventional perturbative low energy description is not possible. Even though an exact holographic dual of QCD is yet to be found, there are theories (for example the Sakai-Sugimoto model) that capture important features of QCD surprisingly well.

One may now ask what is so special about AdS space that allows for such a duality? One way to approach this is to point out the rich symmetry content of this kind of spacetime. The isometry group of Anti-de Sitter space is given by $SO(4,2)$, which is precisely the conformal group in four dimensions.

Regarding de Sitter space: the nature of this spacetime makes it difficult to formulate a correspondence analogous to its positively curved counterpart. See this article for more information.

it isn't obvious to me how the FRW spacetime is related to the de Sitter spacetime.

De Sitter space is the FRW solution in which there is no baryonic or dark matter, only dark energy.

Maybe this comment refers only to a very early universe dominated by Λ that looked like de Sitter.

You have this the wrong way around. In a cosmology with nonzero $\Lambda$, i.e., dark energy, dark energy always dominates at *late* times. This is because the contribution of dark energy to the stress energy stays the same as expansion continues, whereas contributions from other matter fields fall off like some negative power of the scale factor $a$. The early universe was radiation-dominated, because radiation has an exponent of $-4$, which is the biggest.

Our universe is currently quite well approximated by de Sitter space.

## Best Answer

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 curvaturebut non-zerospacetime 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 curvatureit can be viewed as having arbitrary constant3d 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.