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Wigner classfied elementary particle as unitary irreducible representations (UIR) of the Poincaré group.

Suppose the spacetime is curved with symmetries $G$. Should the elementary particles in this spacetime always be UIR of $G$?

Specifically: Our universe has a positive cosmological constant, so shouldn't we look for unitary irreducible representations of the de Sitter group (as the next best approximation)? Or given a Schwarzschild spacetime: Should the elementary particles in this spacetime be UIR of its symmetries?

(It is clear to me that in general spacetimes have no symmetries at all. I'm wondering why it is the Poincaré group that has physical significance for our elementary particles and not, say, the de Sitter group. Or why do we even believe that elementary particles are UIR of Poincaré even though realistic spacetime of our universe do not have Poincaré symmetries.)

ungerade
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1 Answers1

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I'm at around the same point as you, so I'm not sure how well I'll be able to answer, but I'll try. I don't know if you've come across the Einstein-Cartan formulism for GR yet, but it's basically treating gravity as a gauge theory, in an analogous way to Yang-Mills. Gravity is the gauge theory of the Poincare group so spacetime does have a Poincare symmetry, but it's a local one. I suppose, you could maybe also think about it like any particle is so small, that it is in an inertial reference frame and so just has the minkowski metric.

The symmetries of the spacetime I think you're thinking about are connected more to conserved quantities in the spacetime. As you're probably aware, in Schwarzschild, we have spherical symmetry so the angular momentum is conserved. I'm unsure as to why it is we wouldn't get extra particles, not just conserved quantities, from those UIR from the extra spacetime symmetries. Although, the stuff I covered in Wigner's classification, not all irreps actually show up as physical particles (e.g the continuous spin ones for massless particles) and I was given no real reason for this. Hopefully someone can else can answer that. Also, remember the FRW-metric is only a 0th order approximation too, so not accurate once you come zoom in to some normal distance.

baker_man
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