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The present experimental value for the cosmological constant is tiny, but nonzero: $\Lambda \approx 1.19·10^{−52}$ $1/m^2$.

The Poincare group is the contraction of the deSitter group in the limit $\Lambda \rightarrow 0$, analogous to how the Galilean group is the contraction of the Poincare group in the limit $c\rightarrow \infty$.

Doesn't a non-zero $\Lambda$ mean that the exact spacetime symmetry group is the deSitter group and the Poincare group is only a good approximation because $\Lambda$ is so small?

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  • $\begingroup$ The answer to the question is "Yes.", which is too short to even submit as an answer. Can you be more specific what you want to know? $\endgroup$ – ACuriousMind Dec 19 '16 at 14:32
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    $\begingroup$ @ACuriousMind Ah okay, perfect. Does this has any implications? The representation theory of the Poincare group is essential for particle physics. The deSitter group is a quite different group (for example, it is a simple group) and thus I suppose the representation theory is quite different. I was just wondering, because this is never mentioned in particle physics textbooks. $\endgroup$ – jak Dec 19 '16 at 14:50
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Yes, if our universe is indeed deSitter, then the correct group to seek unitary representations of in quantum field theory is indeed the deSitter group $\mathrm{O}(1,n)$ instead of the Poincaré group $\mathrm{SO}(1,n-1)\ltimes\mathbb{R}^n$.

As shown in "Contractions of representations of de Sitter groups" by Mickelsson and Niederle (see also their references to Wigner and Inonu), it is fortunately the case that at least all the massive representations of the Poincaré group can be obtained by contracting representations of the deSitter group, meaning the approximation is also consistent on this level. What exactly the mass parameter itself means in deSitter space is a somewhat open question.

A longer overview over the Wigner classification for $\mathrm{O}(1,4)$, i.e. 4D deSitter space, and the physical meaning of the different possible representations is given in "Group theory and de Sitter QFT" by Boers.

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  • $\begingroup$ For the entire universe (cosmologically) to be deSitter, would it suffice to say that the cosmological constant is non-zero? But anyway, for the Poincare group to be replaced with the deSitter group, I think the non-zero cosmological constant should suffice. Aren't we sufficiently sure that the cosmological constant is non-zero? If we are then is it a consensus that using the deSitter group is the right way rather than using the Poincare group? $\endgroup$ – Dvij Mankad May 14 '17 at 12:00

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