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D=3 null translations have two signs and independent central phases

Example

For D=3, use the universal Lorentz cover K∞=L~. Its null little group is the direct product R×Z, with deck generator z=(0,1). For t∈R and α∈R/Z, define

θt,α(u,j)=eitue2πiαj.

These are one-dimensional irreducible null little-group representations. For every ρ>0, the two signs t=ρ and t=−ρ are inequivalent, and z acts by the independently chosen phase e2πiα. Thus all phases occur on K∞; on a finite cover KN, the allowed phases are the Nth roots of unity, Nα∈Z, and either sign remains available.

Facts & Assumptions

Given: D=3, a nonzero future null momentum, and the universal or finite connected Lorentz cover.

[F1]

Global Lorentz stabilizers and their connected-cover preimages: the D=3 null stabilizer is R×Z on K∞, with z=(0,1); on KN it is R×(Z/NZ).

[F2]

Null little-group unitary duals, including low dimensions: every irreducible strongly continuous unitary representation of a D=3 null little group is θt,α(u,j)=eitue2πiαj; the real parameter has both signs, and finite N requires Nα∈Z.

[F3]

Central kernel characters control descent of Poincaré inductions: the character descends from K∞ to KN exactly when it is trivial on zN, which is the condition Nα∈Z independent of t.

[F4]

Positive-energy Wigner–Mackey classification for connected covers: inducing any such irreducible null little-group character over a future null momentum gives an irreducible positive-energy representation.

[A1]

AC is assumed through [F2], [F3], and [F4]. The Axiom of Choice

Proof

Given: The groups and parameters in the Example.

1.1F1givenalgebra

The displayed formula is multiplicative because R×Z is a direct product: θt,α((u,j)+(v,k))=eit(u+v)e2πiα(j+k)=θt,α(u,j)θt,α(v,k). It is continuous in u, and Z is discrete. A one-dimensional unitary representation is irreducible.

2.1F1F2step 1.1algebra

For ρ>0, evaluating the characters with t=ρ and t=−ρ at (u,j)=(π/(2ρ),0) gives respectively i and −i, so they are inequivalent. This is consistent with the orbit classification because the connected rotation factor is SO(1)={1} and cannot reverse the real translation character.

2.2F1F2step 1.1algebra

The deck generator is (0,1), so θt,α(z)=e2πiα and the value is independent of t. As α ranges over R/Z, these are all of T. On KN, the relation zN=1 forces and is forced by e2πiNα=1, equivalently Nα∈Z, giving precisely the N possible phases.

3.1F2F3F4A1step 2.1step 2.2∎

For every t=±ρ and allowed α, [F4] gives a positive-energy irreducible Poincaré induction. The independent central-phase calculation in step 2.2 and the sign calculation in step 2.1 show that neither the finite-cover descent condition [F3] nor the phase parameter identifies the two signs. This proves the stated D=3 distinction.

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