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It seems to me that spinors (pinors) are loosely defined as representations of the spin (pin) group $Spin(p,q)$ ($Pin(p,q)$), which double covers the spacetime symmetry group $SO(p,q)$ ($O(p,q)$). $\gamma$ matrices is a matrix representation of the Clifford algebra that generates the spin (pin) group, and spinors are vectors that transform under the matrix representation of the spin (pin) group constructed from $\gamma$ matrices. I have two questions though.

  1. One of the main purposes of using spinors is the spin group, $Spin(3,1)$ for Minkowski space is simply connected as opposed to $SO(3,1)$, so a projective representation of $SO(3,1)$ can be promoted to a non-projective representation of $Spin(3,1)$, see Weinberg's QFT. However, not all spin groups are simply connected and it depends on the dimension of spacetime and the signature. I am wondering what's the physics motivation of considering spinors in that case?

  2. $\gamma$ matrices for odd dimension of spacetime is not a faithful matrix representation of the Clifford algebra. Is the representation of the spin group faithful in this case? If not, is there any faithful representation of the spin group (maybe better candidate for the name spinor)?

Qmechanic
  • 201,751

1 Answers1

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  1. The spin group ${\rm Spin}(p,q)\cong {\rm Spin}(q,p)$ is connected if $\max(p,q)\geq 2$. If we exclude multiple temporal dimensions, i.e. if we consider only Minkowskian and Euclidean signatures, then the component of ${\rm Spin}(p,q)$ that is connected to the identity is simply connected; except in the cases 2+0D and 2+1D where the fundamental group is $\pi_1\cong\mathbb{Z}$. Spinors are needed to describe Dirac fermions, which have half-spin. Therefore if a spinor representation happens to be projective, it seems we'll have to live with that.

  2. For OP's 2nd question, let's for simplicity consider the complexification. Then the signature is irrelevant. For even (resp. odd) spacetime dimension $d$, the full (resp. even) Clifford algebra ${\rm Cl}(d,\mathbb{C})$ (resp. ${\rm Cl}(d,\mathbb{C})_{\rm even}$) is faithfully represented by spinors. Since the spin group $\rm Spin(d,\mathbb{C})\subseteq {\rm Cl}(d,\mathbb{C})_{\rm even}$, it is also faithfully represented.

Qmechanic
  • 201,751