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TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-09-27
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The regular representations are unitary, strongly continuous, and the left one is faithful

Statement

Assume AC. Let G be an LCH group with a fixed left Haar measure μ, and let λ,ρ be the left and right regular representations on L2(G) (Left and right regular unitary representations of an LCH group). Then λ and ρ are strongly continuous unitary representations of G on the Hilbert space L2(G), and λ is faithful: λ(g)=id implies g=e.

Facts & Assumptions

Given: An LCH group G with fixed left Haar measure μ, its modular function ΔG, the Hilbert space L2(G), the maps λ,ρ, and AC.

[F1]

λ(g)ξ(x)=ξ(g−1x) and ρ(g)ξ(x)=ΔG(g)1/2ξ(xg) define complex-linear isometries of L2(G), with λ(g)−1=λ(g−1), ρ(g)−1=ρ(g−1), and group laws λ(gh)=λ(g)λ(h), ρ(gh)=ρ(g)ρ(h) (Left and right regular unitary representations of an LCH group).

[F2]

Strong continuity: for ξ∈L2(G) and ϵ>0 there is a neighbourhood V of e with ∥λ(g)ξ−ξ∥2<ϵ and ∥ρ(g)ξ−ξ∥2<ϵ for all g∈V; and at every point g0 of G (Strong continuity of left and modular right translations on L1 and L2).

[F3]

L2(G) is a Hilbert space with inner product ⟨ξ,η⟩=∫Gξη‾ dμ, whose induced norm is ∥⋅∥2, and Cc(G)=Cc(G;C) is dense in it (Hilbert space, Complex Haar L^p spaces and compactly supported functions, Completeness of the complex Haar L1 and L2 spaces and density of Cc, Compact support, Cc(X), and C0(X)).

[F4]

In the terminology of the cited definition a representation is unitary when it takes values in the group of unitary operators, and strongly continuous when g↦π(g)ξ is continuous for every ξ (Continuous and unitary representations).

[F5]

If h∈L2(G) vanishes a.e. then h=0: the set where a continuous representative is nonzero is open and thus has positive measure when nonempty, and a class vanishing a.e. has the zero class as its only continuous representative (Haar measure is positive on nonempty open sets and finite on compact sets).

[F6]

Under Dependent Choice, if K is compact and contained in an open U, there is f∈Cc(G) with 1K≤f≤1U; AC implies Dependent Choice. Taking K={e} gives a nonzero f which vanishes off U (LCH Urysohn cutoff, The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain).

[A1]

AC is assumed, in the choice-function form of the cited definition; it is used in step 2.2 through the cutoff function of [F6] (The Axiom of Choice).

Proof

technique · direct
1.1

The group laws already hold by [F1]: λ(gh)=λ(g)λ(h), ρ(gh)=ρ(g)ρ(h) and λ(e)=ρ(e)=id, and the inverses are λ(g−1), ρ(g−1).

F1
1.2

Each λ(g) and ρ(g) is a bijective complex-linear isometry of L2(G): it is complex-linear and isometric by [F1], and surjective because the displayed inverse is a two-sided inverse. In the Hilbert-space setting this makes each of them a unitary operator, in the sense recalled in [F4] and in the definition [F1].

F1F4
1.3

Strong continuity. By [F2] the maps g↦λ(g)ξ and g↦ρ(g)ξ are continuous at e for every ξ∈L2(G); continuity at an arbitrary g0∈G follows because ∥λ(g)ξ−λ(g0)ξ∥2=∥λ(g0−1g)ξ−ξ∥2→0 as g→g0 and likewise for ρ, using the group laws and isometry of [F1]. Hence λ and ρ are strongly continuous in the sense of [F4].

F1F2F4
2.1

Unitarity of the two homomorphisms. By steps 1.1 and 1.2, λ and ρ are group homomorphisms G→U(L2(G)); by step 1.3 they are strongly continuous. Thus both are strongly continuous unitary representations of G on the Hilbert space L2(G).

F4step 1.1step 1.2step 1.3
2.2

Faithfulness of λ. Suppose g≠e and λ(g)=id. Since G is Hausdorff, choose an open neighbourhood W of e with g∉W. The map q(x,y)=xy−1 is continuous and q(e,e)=e, so there are open neighbourhoods V1,V2 of e with V1V2−1⊆W; their intersection V0=V1∩V2 satisfies V0V0−1⊆W. By local compactness choose a compact neighbourhood C of e and an open neighbourhood O with e∈O⊆C. The set C is closed because G is Hausdorff, so U:=V0∩O is an open neighbourhood of e with compact closure contained in C. Since UU−1⊆W and g∉W, one has U∩gU=∅. By [F6], under the Dependent Choice derived from [A1], choose f∈Cc(G) with 1{e}≤f≤1U. Then f(e)≥1, f vanishes off U, and f∈L2(G) by [F3]. The class λ(g)f−f is zero since λ(g)=id. But {x:f(x)>0} is a nonempty open set; for each such x we have x∈U and x∉gU, so g−1x∉U and (λ(g)f)(x)=f(g−1x)=0. Thus λ(g)f−f=−f on this nonempty open set, so it is not the zero class by [F5], a contradiction. Hence λ is faithful.

A1F3F5F6step 1.1
3.1

Combining the steps: λ and ρ are strongly continuous unitary representations by step 2.1, and λ is faithful by step 2.2. ∎

step 2.1step 2.2

Remarks

  • Why the left representation is the faithful one. Both λ and ρ are faithful as well whenever G is such that ρ(g)=id forces g=e; the statement records faithfulness only for λ, because that is the case the proof above establishes directly using left translates.
  • Choice cost. [A1] is used only in step 2.2, for the cutoff function supported in a neighbourhood disjoint from its translate; the unitarity and strong continuity statements use only the fixed measure and its modular function.

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