**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).

Read more about this topic: Bispinor, Derivation of A Bispinor Representation

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“Our intellect is not the most subtle, the most powerful, the most appropriate, instrument for revealing the truth. It is life that, little by little, *example* by *example*, permits us to see that what is most important to our heart, or to our mind, is learned not by reasoning but through other agencies. Then it is that the intellect, observing their superiority, abdicates its control to them upon reasoned grounds and agrees to become their collaborator and lackey.”

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