Bispinor - Derivation of A Bispinor Representation - An Example

An Example

Let X=2πM12 so that X generates a rotation around the z-axis by an angle of 2π. Then Λ = eiX = I ∈ SO(3;1)+ but eiπ(X) = -I ∈ GL(U). Here, I denotes the identity element. If X = 0 is chosen instead, then still Λ = eiX = I ∈ SO(3;1)+, but now eiπ(X) = I ∈ GL(U).

This illustrates the double valued nature of a spin representation. The identity in SO(3;1)+ gets mapped into either -I ∈ GL(U) or I ∈ GL(U) depending on the choice of Lie algebra element to represent it. In the first case, one can speculate that a rotation of an angle 2π will turn a bispinor into minus itself, and that it requires a 4π rotation to rotate a bispinor back into itself. What really happens is that the identity in SO(3;1)+ is mapped to -I in GL(U) with an unfortunate choice of X.

It is impossible to continuously choose X for all g ∈ SO(3;1)+ so that S is a continuous representation. Suppose that one defines S along a loop in SO(3;1) such that X(t)=2πtM12, 0 ≤ t ≤ 1. This is a closed loop in SO(3;1), i.e. rotations ranging from 0 to 2π around the z-axis under the exponential mapping, but it is only "half"" a loop in GL(U), ending at -I. In addition, the value of I ∈ SO(3;1) is ambiguous, since t = 0 and t = 2π gives different values for I ∈ SO(3;1).

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