mathematical physics - Are there projective representations of the Lorentz Group NOT coming from a Clifford algebra?

Let $\mathrm{SO}(1,d-1)_{+}$ be the restricted Lorentz Group in $d$ dimensions. Are there projective irreducible representations of this group that do not descend from a representation of $\mathrm{C}\ell_{1,d-1}$?

In other words, it is known that any representation of the Clifford algebra induces a representation of the corresponding $\mathrm{Spin}$ group; is the converse true, i.e., does any representation of the $\mathrm{Spin}$ group correspond to some representation of the corresponding Clifford algebra?

Any set of matrices $\{\gamma^\mu\}$ satisfying $$ \gamma^{(\mu}\gamma^{\nu)}=\eta^{\mu\nu}\tag1 $$ leads to a set of matrices $S^{\mu\nu}:=\frac i2\gamma^{[\mu}\gamma^{\nu]}$ satisfying $$ [S^{\mu\nu},S^{\rho\sigma}]=\eta^{\mu\rho}S^{\nu\sigma}+\text{perm.}\tag2 $$

My question is: is it true that for any set of matrices $\{S^{\mu\nu}\}$ satisfying $(2)$ we will have a set of matrices $\{\gamma^\mu\}$ satisfying $(1)$?

Note: when considering projective representations of this group, only two phases are possible, $\pm1$. Needless to say, here I am asking about those corresponding to $-1$. For the other sign the answer is obvious.

Answer

1. 
   

   Any irreducible complex representation of a Clifford algebra in $d$ dimensions has dimension $2^{\lfloor d/2\rfloor}$. A proof of this claim can be found e.g. in this post by Qmechanic.

   
   


2. 
   

   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 $(s_1,s_2) \in (\frac {1}{2}\mathbb{Z})^2$ of dimension $D = (2s_1 + 1)(2s_2 + 1)$. There are three cases:

• 
  

  $D < 4$: This is only the case for one $s_i = 1/2,1$ 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 $s_1 + s_2$ non-integer is a projective representation that does not come from a representation of the Clifford algebra.

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^{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 $s_1 + s_2$ is integer.}