Let SO(1,d−1)+ be the restricted Lorentz Group in d dimensions. Are there projective irreducible representations of this group that do not descend from a representation of Cℓ1,d−1?
In other words, it is known that any representation of the Clifford algebra induces a representation of the corresponding Spin group; is the converse true, i.e., does any representation of the Spin group correspond to some representation of the corresponding Clifford algebra?
Any set of matrices {γμ} satisfying γ(μγν)=ημν
My question is: is it true that for any set of matrices {Sμν} satisfying (2) we will have a set of matrices {γμ} satisfying (1)?
Note: when considering projective representations of this group, only two phases are possible, ±1. Needless to say, here I am asking about those corresponding to −1. For the other sign the answer is obvious.
Answer
Any irreducible complex representation of a Clifford algebra in d dimensions has dimension 2⌊d/2⌋. A proof of this claim can be found e.g. in this post by Qmechanic.
As the question already says, any representation of a Clifford algebra induces a representation of its corresponding Lorentz algebra.
So, let us take an arbitrary irreducible representation of the Lorentz algebra in four dimensions, labeled by (s1,s2)∈(12Z)2 of dimension D=(2s1+1)(2s2+1). There are three cases:
D<4: This is only the case for one si=1/2,1 and the other being zero. The (1/2,0)-representations are the Weyl spinors and are subrepresentations of the single irreducible representations of the Clifford algebra in four dimensions, the Dirac spinors. The (1,0)-representations are those of (anti-)self-dual 2-forms and do not descend from the Clifford algebra, however these also have "phase +1" as a projective representation, so this falls into the case where the question considers the answer "obvious".1
D=4: The only four-dimensional irrep of the Lorentz group is (1/2,1/2), the ordinary 4-vectors, which do not carry a representation of the Clifford algebra.
D>4: No irrep of the Lorentz algebra of dimension larger than 4 can be compatible with a representation of the Clifford algebra by points 1. and 2. above: If there were a compatible representation of the Clifford algebra, it would have to be reducible by point 1, i.e. have a proper subrepresentation. But by point 2. this would also induce a proper subrepresentation of the Lorentz algebra, meaning the irrep was not irreducible, which yields a contradiction.
Therefore, in particular, any representation of the Lorentz algebra with D>4 and s1+s2 non-integer is a projective representation that does not come from a representation of the Clifford algebra.
1A linear representation of the Lorentz algebra integrates to a linear (and not merely projective) representation of the proper orthochronous Lorentz group if and only if s1+s2 is integer.
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