Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-08
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A limit of discrete series is not square-integrable

Statement refuted

The claim that every limit of discrete series is square-integrable in the all-matrix-coefficients sense of The limits of discrete series are not square-integrable is false. In the fixed Haar normalization of Left Haar integral and left Haar measure and KAK integration formula for K-bi-invariant functions on SL2(R), the limit D1+ has a unit matrix coefficient whose squared modulus has infinite integral. By contrast, for each genuine discrete-series parameter n≥2, the corresponding normalized extremal coefficient has finite squared integral.

Facts & Assumptions

Given: AC, G=SL2(R), its fixed left Haar measure, the odd compact-picture model I1,0, and the holomorphic discrete-series models for n≥2.

[F1]

The unit vector f1(kθ)=eiθ lies in the closed limit summand D1+ of I1,0, and D1+ is an irreducible strongly continuous unitary representation (The two limits of discrete series, Unitarity and irreducibility of the limits of discrete series).

[F2]

Matrix coefficients are cv,w(g)=⟨π(g)v,w⟩ with pairing linear in the first variable, and they are continuous for strongly continuous unitary representations (Matrix coefficient of a unitary representation).

[F3]

In the compact picture, c(g)=⟨Π0(g)f1,f1⟩ satisfies c(aτ)=sech⁡(τ/2) for every τ∈R (Matrix-coefficient formulas and decay for the discrete and principal series(b)).

[F4]

For a continuous function with unitary-character left and right K-transformation, its modulus is K-bi-invariant, and the fixed Haar measure gives ∫G∣c(g)∣2 dg=2π∫0∞∣c(aτ)∣2sinh⁡τ dτ (KAK integration formula for K-bi-invariant functions on SL2(R)).

[F5]

For each n≥2, the normalized extremal vector un in a genuine discrete-series model has coefficient ⟨πn(aτ)un,un⟩=cosh⁡(τ/2)−n (Matrix-coefficient formulas and decay for the discrete and principal series(a)).

[F6]

The square-integrability condition used here requires every matrix coefficient of an irreducible unitary representation to lie in L2(G) (The limits of discrete series are not square-integrable).

[F7]

If 0≤gm↑g pointwise, then ∫gm dμ↑∫g dμ (Monotone convergence for the integral).

[A1]

AC is assumed and inherited through the normalized principal-series and fixed-Haar model interfaces (The Axiom of Choice).

Counterexample

technique · direct

Given: The assumptions and notation above.

1.1F1F2F3F4A1

Let π be the restriction of I1,0 to D1+ and set c(g)=⟨π(g)f1,f1⟩. By [F1], this is a matrix coefficient of an irreducible unitary limit representation; by [F2] it is continuous, and [F3] gives c(aτ)=sech⁡(τ/2). If χ(kθ)=eiθ, unitarity and π(k)f1=χ(k)f1 give c(k1gk2)=χ(k1)χ(k2)c(g). Thus [F4] applies to ∣c∣2.

2.1F1F3F4F7step 1.1algebra

The KAK formula and sinh⁡τ=2sinh⁡(τ/2)cosh⁡(τ/2) give ∫G∣c(g)∣2 dg=2π∫0∞sech⁡2(τ/2)sinh⁡τ dτ=2π∫0∞2tanh⁡(τ/2) dτ. For τ≥log⁡3, 2tanh⁡(τ/2)≥1. The indicators 1[log⁡3,log⁡3+m] increase to 1[log⁡3,∞) and their integrals are m, so [F7] shows the last nonnegative integral is infinite.

3.1F4F5F6step 2.1algebra∎

By [F6], the irreducible unitary representation D1+ is not square-integrable because the coefficient in step 2.1 is not in L2(G). For n≥2, [F4] and [F5] give the corresponding extremal coefficient integral 2π∫0∞cosh⁡−2n(τ/2)sinh⁡τ dτ. With u=tanh⁡(τ/2) this equals 8π∫01u(1−u2)n−2 du=4π/(n−1)<∞, confirming the endpoint contrast and refuting the claim in the Statement.

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