# The deSitter group vs. the Poincare group for a non-zero cosmological constant?

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?

• 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? – ACuriousMind Dec 19 '16 at 14:32
• @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. – jak Dec 19 '16 at 14:50

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 $\mathrm{SO}(1,n-1)\ltimes\mathbb{R}^n$.
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.