False statementConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedaudited 2026-09-22

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A compact group can have infinite-dimensional unitary representations

Statement

Assume the Axiom of Choice. Every unitary representation of a compact Lie group is finite-dimensional.

Facts & Assumptions

Given: Assume the Axiom of Choice; the group $G = S^{1}$ with its normalized Haar measure and the circle characters $z \mapsto z^{n}$ , $n \in Z$ .

[L1]

On $L^{2} (G)$ the left regular representation $(L_{x} f) (y) = f (x^{- 1} y)$ is a well-defined unitary representation of $G$ on a Hilbert space (Left and right regular representations on L2(G)).

[L2]

The characters of the torus are the maps $z \mapsto z^{n}$ , $n \in Z$ , and distinct characters are pairwise orthonormal in $L^{2} (S^{1})$ ; in particular ${z \mapsto z^{n} : n \in Z}$ is an infinite orthonormal family (Peter–Weyl theorem, Irreducible characters are orthonormal class functions).

Refutation

technique · direct

1.1

The family $(z \mapsto z^{n})_{n \in Z}$ is orthonormal in $L^{2} (S^{1})$ by [L2]; an orthonormal family of infinitely many nonzero vectors has no finite spanning set, because vectors in a finite-dimensional space are subject to the finite bound on the cardinality of linearly independent families; hence $L^{2} (S^{1})$ is infinite-dimensional.

2.1

By [L1] the left regular representation makes $L^{2} (S^{1})$ a unitary representation of the compact Lie group $S^{1}$ ; it is infinite-dimensional by step 1.1.

L1step 1.1

3.1

Hence there exists a unitary representation of a compact Lie group that is not finite-dimensional, so the statement of this item is false; Peter–Weyl decomposes the regular representation into finite-dimensional pieces but does not make the whole Hilbert space finite-dimensional.

L1step 2.1∎

Depends on

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Sources

Anthony W. Knapp, Lie Groups Beyond an Introduction, 2nd ed. (standard reference, not scraped)