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Weyl integration for SU(2)

Example

Assume the Axiom of Choice. For a continuous class function f on SU(2), ∫SU(2)f(g) dg=12∫S1f(diag⁡(z,z−1)) ∣1−z−2∣2 dz, with dz the normalized Haar measure of the circle.

Facts & Assumptions

Given: Assume the Axiom of Choice; SU(2) with its diagonal maximal torus T={diag⁡(z,z−1):∣z∣=1}, Weyl group of order two, and normalized Haar measures.

[L1]

The diagonal matrices form a maximal torus of SU(2) with Weyl group W≅S2 of order two acting by z↦z−1 (Maximal tori and Weyl groups of U(n) and SU(n)).

[L2]

Weyl integration: for a class function f on a compact connected G with maximal torus T, ∫Gf dg=∣W∣−1∫Tf(t)J(t) dt with J(t)=∏α>0∣1−α(t)−1∣2 (Weyl integration formula, Weyl Jacobian).

[L3]

In sl2(C), the adjoint action of t=diag⁡(z,z−1) sends E12 to z2E12 and E21 to z−2E21, while it fixes the diagonal trace-zero line. Thus the roots of (SU(2),T) are the two characters z↦z±2, and one may choose α(t)=z2 as the positive root. [algebra]

Verification

technique · direct
1.1L1L2

Substituting ∣W∣=2 into [L2] and using that the class function is constant on Weyl orbits gives ∫SU(2)f dg=12∫S1f(diag⁡(z,z−1))J(diag⁡(z,z−1)) dz.

2.1L3step 1.1

By [L3] the single positive root satisfies α(diag⁡(z,z−1))=z2, so J(diag⁡(z,z−1))=∣1−z−2∣2, which is the displayed factor.

3.1step 2.1∎

As a check, f≡1 gives 12∫S1∣1−z−2∣2 dz=12∫S1(2−z2−z−2) dz=12⋅2=1, consistent with the normalization of Haar measure.

Depends on

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