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.

  • $\begingroup$ What definition do you use for the $\text{Spin}$ group? $\endgroup$
    – DanielC
    Commented Oct 1, 2017 at 20:16
  • $\begingroup$ @DanielC for our purposes, it is the double cover of $\mathrm{SO}$ $\endgroup$ Commented Oct 1, 2017 at 20:19
  • $\begingroup$ @AccidentalFourierTransform, can you, perhaps, take a look here? arxiv.org/abs/math-ph/0509040v1 It should start/contain the path to finding an answer to your question. $\endgroup$
    – DanielC
    Commented Oct 1, 2017 at 22:36

2 Answers 2


This answer is based on the seminal paper by Berg, DeWitt-Morette, Gwo and Kramer (BDGK) about the physics of the double covers of the Lorentz groups.

Although, the article treats the general case of multiple space and time dimensions; in the following answer, only the $(3,1)$ case will be considered. Also, only the finite dimensional (nonunitary) spinor representations will be considered, as we know how to promote them into infinite dimensional unitary representations of the Poincaré group.

First, the physically interesting groups are the double covers of $O(3,1)$ rather than $SO(3,1)$ because they contain distinct parity and time reversal operators rather than their product only (BDGK: Figure-1 page 17 and Figure-2 page 19).

The double covers of $O(3,1)$ are called Pin groups (Their representative vectors are called Pinors). $O(3,1)$ possesses 8 types of double covers called $\mathrm{Pin}^{abc}(3,1)$ ($a,b.c\in \mathbb{Z_2}$ corresponding to (BDGK-appendix C): $$\Lambda_P^2 = a$$ $$\Lambda_T^2 = b$$ $$\Lambda_{PT}^2 = c$$ ($P=$ parity, $T=$ time reversal, $\Lambda$ is the representation matrix)

Only two of the above double covers can be obtained from a Clifford algebra they correspond to: $\Lambda_P^2 = -\Lambda_T^2 = -\Lambda_{PT}^2 = 1$ and $\Lambda_T^2 = -\Lambda_P^2 = -\Lambda_{PT}^2 = 1$ respectively. These groups are called Cliffordian and usually denoted by: $\mathrm{Pin}(3,1)$ and $\mathrm{Pin}(1,3)$ respectively.


  1. I am not aware of any physical application of the non-Cliffordian Pin groups.

  2. Basically, we do not know the Pin groups of the elementary particles except for the neutrino in the neutrinoless double beta decay. BDGK suggest some experiments which can distinguish the type of the Pin group.

  • $\begingroup$ Hi D.B.M., thank you for your answer. Unfortunately, I'm not entirely sure I actually understand it. I've edited my question to be as explicit as I can. Perhaps you could give it a second look just to be sure we are on the same page here. $\endgroup$ Commented Oct 16, 2017 at 14:32
  • $\begingroup$ Hi AccidentalFourierTransform I dealt with representations of the double covers because the true representations of the double covers are the projective representations of the group itself. Secondly I took the liberty of treating the (double covers of the) groups $O(3,1)$: $Pin(3,1)$ instead of (those of ) $SO(3,1)$: $Spin(3,1)$ for of two reasons: $\endgroup$ Commented Oct 17, 2017 at 7:28
  • $\begingroup$ cont. 1) Only the Pin groups contain the parity and time reversal transformation as a part of the groups, the Spin groups contain only their product. I think that this is important physically (the physical fermions must be represented by Pinors rather than by Spinors) $\endgroup$ Commented Oct 17, 2017 at 7:28
  • $\begingroup$ cont. 2) Only Pin groups provide a nontrivial answer to your question. The spinor representations of $Spin(p,q)$ always stem from representations of the corresponding Clifford algebra as follows: in the case of $p+q=\mathrm{odd}$. The generators of the Clifford algebra are a part of the generators of the Lie algebra of the group, and the relations $S^{\mu \nu} = \gamma^{[\mu}\gamma^{\nu]}$ are just a part of the commutation relations generating the Lie algebra corresponding to of the $Spin(p,q-1)$ subgroup of $Spin(p,q)$. $\endgroup$ Commented Oct 17, 2017 at 7:30
  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.

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.


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