Normed versus topological groups: Dichotomy and duality

Normed groups 129Fix s. Since s is uniformly continuous, δ = δ(s) is well-defined andd(s(z ′ ), s(z)) ≤ 1,for z, z ′ such that d(z, z ′ ) < δ. In the definition of the word-net take ε < 1. Now supposethat w(x) = w 1 ...w n(x) with ‖z i ‖ = 1 2 δ(1 + ε i) and |ε i | < ε, where n(x) = n(x, δ) satisfiesPut z 0 = z, for 0 < i < n(x)1 − ε ≤ n(x)δ‖x‖ ≤ 1 + ε.z i+1 = z i w i ,and z n(x)+1 = x; the latter is within δ of x. Aswe haveHenced(z i , z i+1 ) = d(e, w i ) = ‖w i ‖ < δ,d(s(z i ), s(z i+1 )) ≤ 1.d(s(z), s(xz)) ≤ n(x) + 1 < 2‖x‖/δ.The final assertion follows from the subadditivity of the Lipschitz norm (cf. Theorem3.27).If {δ(s) : s ∈ H u (X)} is unbounded (i.e. the inverse modulus of continuity is unbounded),we cannot develop a duality theory. However, a comparison with the normedvector space context and the metrization of the translations x → t(z + x) of a linear mapt(z) suggests that, in order to metrize Ξ by reference to ξ x (t), we need to take accountof ‖t‖. Thus a natural metric here is, for any ε ≥ 0, the magnification metricd ε T (ξ x , ξ y ) := sup ‖t‖≤ε d T (ξ x (t), ξ y (t)).(mag-eps)By Proposition 2.14 this is a metric; indeed with t = e H(X) = id X we have ‖t‖ =0 and, since d X is assumed right-invariant, for x ≠ y, we have with z xy = e thatd X (x −1 z, y −1 z) = d X (x −1 , y −1 ) > 0. The presence of the case ε = 0 is not fortuitous; see[Ost-knit] for an explanation via an isomorphism theorem. We trace the dependence on‖t‖ in Proposition 12.5 below. We refer to Gromov’s notion [Gr1], [Gr2] of quasi-isometryunder π, in which π is a mapping between spaces. In a first application we take π to bea self-homeomorphism, in particular a left-translation; in the second π(x) = ξ x (t) with tfixed is an evaluation map appropriate to a dual embedding. We begin with a theorempromised in Section 3.Theorem 12.4 (Uniformity Theorem for Conjugation). Let Γ : G 2 → G be the conjugationΓ(g, x) := g −1 xg.Under a bi-invariant Klee metric, for all a, b, g, h ,d G (a, b) − 2d G (g, h) ≤ d G (gag −1 , hbh −1 ) ≤ 2d G (g, h) + d G (a, b),and hence conjugation is uniformly continuous.