Normed versus topological groups: Dichotomy and duality

Normed groups 33(ii) For α ∈ H unif (X) and given ε > 0, choose δ > 0, so that (u-cont) holds. Then, forβ, γ with ˆd(β, γ) < δ, we have d(β(t), γ(t)) < δ for each t, and henceˆd(αβ, αγ) = sup d(α(β(t)), α(γ(t))) ≤ ε.(iii) Again, for α ∈ H unif (X) and given ε > 0, choose δ > 0, so that (u-cont) holds.Thus, for β, η with ˆd(β, η) < δ, we have d(β(t), η(t)) < δ for each t. Hence for ξ withˆd(α, ξ) < ε we obtainConsequently, we haved(α(β(t)), ξ(η(t))) ≤ d(α(β(t)), α(η(t))) + d(α(η(t)), ξ(η(t)))≤ ε + ˆd(α, ξ) ≤ ε + ε.ˆd(αβ, ξη) = sup d(α(β(t)), ξ(η(t))) ≤ 2ε.Comment. See also [AdC] for a discussion of the connection between choice of metricand uniform continuity. The following result is of interest.Proposition 3.11 (deGroot-McDowell Lemma, [dGMc, Lemma 2.2]). Given Φ, a countablefamily of self-homeomorphism of X closed under composition (i.e. a semigroup inAuth(X)), the metric on X may be replaced by a topologically equivalent one such thateach α ∈ Φ is uniformly continuous.Definition. Say that a homeomorphism h is bi-uniformly continuous if both h and h −1are uniformly continuous. WriteH u = {h ∈ H unif : h −1 ∈ H unif }.Proposition 3.12 (Group of left-shifts). For a normed topological group X with rightinvariantmetric d X , the group T r L (X) of left-shifts is (under composition) a subgroupof H u (X) that is isometric to X.Proof. As X is a topological group, we have T r L (X) ⊆ H u (X) by Cor. 3.6; T r L (X) is asubgroup and λ : X → T r L (X) is an isomorphism, becauseλ x ◦ λ y (z) = λ x (λ y (z)) = x(λ y (z) = xyz = λ xy (z).Moreover, λ is an isometry, as d X is right-invariant; indeed, we haved T (λ x , λ y ) = sup z d X (xz, yz) = d X (x, y).We now offer a generalization which motivates the duality considerations of Section12.