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The classification of particles in general spacetime? Is it still meaningful to say spin-0, 1/2 ,1 field in general spacetime?

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In 3+1 dim Minkovski spacetime, the classification of particle, that is spin-0, 1/2 , 1..., depends on the representation of the universal covering group of $SO(1,3)$, that is $SL(2,C)$. When we study the particle in $d+1$ dim Minkovski spacetime, is it still meaningful to say spin0, 1/2, 1...?

My question is:

(1) In $d+1$ dimensional Minkovski spacetime, in principle we need to study the irreducible representation of $Spin(1,d)$ group to do the classification of particle. Therefore is it still meaningful to say a spin-1/2 partilcle in higher dimensional Minkovski spacetime, becasue spin-1/2 is a representation of $Spin(1,3)=SL(2,C)$.

(2) In general 4 dimensional curved spacetime with no isometry, why is it still meaningful to say particle with spin-0,1/2,1... in curved spacetime? 

(3) In particular, in 4-dimensional de-Sitter spacetime with isometry group $SO(1,4)$, why we don't do irreducible representation of $SO(1,4)$ to classify the particles in de-Sitter spacetime? Why do we still say spin-0,1/2,1... in de-Sitter spacetime?

Firstly which books or papers will handle above problems? Secondly I really want to know the irreducible representation of $Spin(1,d)$ group, and where can I find the answers?

asked Aug 7, 2015 in Theoretical Physics by Alienware (185 points) [ revision history ]

1 Answer

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Yes. Tensor fields exist on any manifold ad define integer spin representations. On manifolds that have a spin structure (which is a nontrivial restriction) one can also define spinor fields, and then as tensor products fields of arbitrary half-integral spin.

In general relativity in the Palatini form, one has beyond the manifold structure a local Lorentz gauge structure, which giveses representations of the Poincare group on the tangent space. This is the relevant classification - not Spin(,3)! Note that scattering happens on the tangent space level since it is essentially a local process. This is why particle physics can work without problems in Minkowski space.

answered Aug 10, 2015 by Arnold Neumaier (12,355 points) [ no revision ]

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