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An irrational D=3 massive spin descends to no finite cover

Example

In $D = 3$ , take the universal Lorentz cover $K_{\infty} = \tilde{L}$ and a future timelike momentum $p_{m} = m e_{0}$ , $m > 0$ . Its massive little group is $R$ , with coordinate $r$ normalized so the deck generator $z$ is $r = 2 π$ . The character

θ_{\sqrt{2}} (r) = e^{i \sqrt{2} r}

defines an irreducible positive-energy induced Poincaré representation. It descends to no finite connected Lorentz cover $K_{N} = \tilde{L} / ⟨ z^{N} ⟩$ , $N \geq 1$ .

Facts & Assumptions

Given: $D = 3$ , $m > 0$ , $p_{m} = m e_{0}$ , and the universal Lorentz cover.

[F1]

Global Lorentz stabilizers and their connected-cover preimages: the $D = 3$ universal-cover massive stabilizer is $R$ , its deck generator has coordinate $2 π$ , and the finite connected covers are $K_{N} = \tilde{L} / ⟨ z^{N} ⟩$ .

[F2]

Positive-energy Wigner–Mackey classification for connected covers: a character of the massive little group over $p_{m} \in C_{+}$ induces an irreducible positive-energy representation.

[F3]

Central kernel characters control descent of Poincaré inductions: descent from the universal cover to $K_{N}$ occurs exactly when the little-group character is trivial on $z^{N}$ .

[A1]

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

Proof

Given: $m > 0$ and the groups and character in the Example.

1.1F1F2A1given

The map $r \mapsto e^{i \sqrt{2} r}$ is a continuous one-dimensional unitary character of the additive group $R$ , so it is irreducible. The momentum $p_{m}$ is future timelike and lies in $C_{+}$ . Applying [F2] gives an irreducible positive-energy induced representation on the universal cover.

2.1F1F3step 1.1

By [F1], $z^{N}$ has little-group coordinate $2 π N$ . Its character value is $θ_{\sqrt{2}} (z^{N}) = e^{2 π i N \sqrt{2}}$ . By [F3], descent to $K_{N}$ would require $N \sqrt{2} \in Z$ .

3.1step 2.1algebra∎

No positive integer $N$ satisfies this condition. If $N \sqrt{2} = k \in Z$ , then $\sqrt{2} = k / N$ is rational. To recall the elementary contradiction, write $\sqrt{2} = a / b$ in lowest positive terms; $a^{2} = 2 b^{2}$ makes $a$ even, then makes $b$ even, contradicting coprimality. Thus the representation fails the descent test for every finite $N$ .

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Sources

Xavier Bekaert and Nicolas Boulanger, The unitary representations of the Poincaré group in any spacetime dimension, v2 (2021) (standard reference, not scraped)
Niklas Beisert, Quantum Field Theory I, Lecture Notes, ETH Zürich, 2025 HS (standard reference, not scraped)