Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedaudited 2026-09-22
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Character lattices of SU(2) and SO(3)

Example

Assume the Axiom of Choice. For type A1 with fundamental weight ω one has P=Zω and Q=2Zω. The character lattice of the maximal torus of SU(2)={g∈U(2):det⁡g=1} is P, while that of the maximal torus of SO(3)≅SU(2)/{±I} is Q; consequently precisely the even SU(2) highest weights descend to SO(3).

Facts & Assumptions

Given: Assume the Axiom of Choice; SU(2) with maximal torus TSU(2)={diag⁡(z,z−1):∣z∣=1} and the standard central quotient identification SO(3)≅SU(2)/{±I}.

[L1]

For a compact connected semisimple group Q⊆X∗(T)⊆P, the simply connected form has character lattice P and the adjoint form has character lattice Q; central quotients correspond to intermediate lattices (Root and weight lattice sandwich, Central quotients and intermediate character lattices).

[L2]

For type A1, Q=Zα=2Zω and P=Zω, and characters of TSU(2) are the maps diag⁡(z,z−1)↦zn with weight nω (Characters are the integral weights). [L1]

Verification

technique · direct
1.1L1L2

SU(2) is simply connected and semisimple with root system A1, so X∗(TSU(2))=P=Zω by [L1]; the characters are diag⁡(z,z−1)↦zn, with weight nω.

2.1L2step 1.1

The kernel of SU(2)→SO(3) is {±I}. Since −I=diag⁡(−1,−1) corresponds to z=−1 in TSU(2), the character of weight nω takes the value (−1)n there, so it is trivial on the kernel exactly when n is even.

3.1L1step 2.1

By [L1] the intermediate lattice of SO(3) is X∗(TSO(3))={characters trivial on {±I}}=2Zω=Q, and the adjoint-form computation of [L1] gives the same answer.

4.1L1step 2.1step 3.1∎

Hence a dominant SU(2) highest weight nω descends to SO(3) exactly when n is even, i.e. exactly when nω∈Q; this is the explicit form of the finite central quotient obstruction.

Depends on

Used by

Dependency tree · two levels

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Sources