Normed versus topological groups: Dichotomy and duality

Normed groups 127(t, x) ✛✻Φ T✲ (t, tx)✻❄(x, t)✛Φ X❄✲ (t, xt)Here the two vertical maps may, and will, be used as identifications, since (t, tx) →(t, x) → (t, xt) are bijections (more in fact is true, see [Ost-knit]).Definitions. Let X be a topological group with right-invariant metric d X . We definefor x ∈ X a map ξ x : H(X) → H(X) by puttingWe setξ x (s)(z) = s(λ −1x (z)) = s(x −1 z), for s ∈ H u (X), z ∈ X.Ξ := {ξ x : x ∈ X}.By restriction we may also write ξ x : H u (X) → H u (X).Proposition 12.1. Under composition Ξ is a group of isometries of H u (X) isomorphicto X.Proof. The identity is given by e Ξ = ξ e , where e = e X . Note thatξ x (e S )(e X ) = x −1 ,so the mapping x → ξ x from X to Ξ is bijective. Also, for s ∈ H(X),(ξ x ◦ ξ y (s))(z) = ξ x (ξ y (s))(z) = (ξ y (s))(x −1 z)= s(y −1 x −1 z) = s((xy) −1 z) = ξ xy (s)(z),so ξ is an isomorphism from X to Ξ and so ξ −1x = ξ x −1.For x fixed and s ∈ H u (X), note that by Lemma 3.8 and Cor. 3.6 the map z → s(x −1 z)is in H u (X). Furthermored H (ξ x (s), ξ x (t)) = sup z d X (s(x −1 z), t(x −1 z)) = sup y d X (s(y), t(y)) = d H (s, t),so ξ x is an isometry, and hence is continuous. ξ x is indeed a self-homeomorphism ofH u (X), as ξ x −1 is the continuous inverse of ξ x .Remark. The definition above lifts the isomorphism λ : X → T r L (X) to H u (X). IfT ⊆ H u (X) is λ-invariant, we may of course restrict λ to operate on T. Indeed, ifT = T r L (X), we then have ξ x (λ y )(z) = λ y λ −1x (z), so ξ x (λ y ) = λ yx −1.In general it will not be the case that ξ x ∈ H u (H u (X)), unless d X is bounded. Recallthat‖x‖ ∞ := sup s∈H(X) ‖x‖ s = sup s∈H(X) d X s (x, e) = sup s∈H(X) d X (s(x), s(e)).