Normed versus topological groups: Dichotomy and duality

38 N. H. Bingham and A. J. OstaszewskiUsing compactness, we may pass to a convergent subsequence, g m → g (in the norm‖.‖ G ). Since multiplication is jointly continuous in G we obtain the contradiction that‖geg −1 ‖ = ‖e‖ = 0 > ε.Remarks. 1. Suppose as usual that d R is a right-invariant metric on a group G. Theright-shift ρ g (x) = xg is uniformly continuous, asHowever, it is not necessarily bounded, asd R (xg, yg) = d R (x, y).‖ρ g ‖ H = sup x d R (xg, x) = sup x ‖g‖ x = ‖g‖ ∞ .But on the subgroup {ρ g : ‖g‖ ∞ < ∞}, the norm ‖ρ g ‖ is bi-invariant, since ‖g‖ ∞ isbi-invariant.2. The condition (n-adm) used in Theorem 3.17 to check admissibility of the supremumnorm may be reformulated, without reference to the group-norm, topologically thus:g n z n g −1n → e for z n → e,with g n arbitrary. In a first-countable topological group this condition is equivalent tothe existence of a bi-invariant metric (see Proposition 2.15; cf. Theorem 3.3.4 in [vM2,p. 101]). We will see several related conditions later: (ne) in Th.3.30, and (W-adm) and(C-adm) ahead of Lemma 3.33 below; we recall here the condition (H-adm) of Prop. 2.14.3. Note that SL(2, R), the set of 2 × 2 real matrices with determinant 1, under matrixmultiplication and with the subspace topology of R 4 forms a (locally compact) topologicalgroup with no equivalent bi-invariant metric; for details see e.g. [HR, 4.24], or [vM2]Example 3.3.6 (p.103), where matrices a n , g n are exhibited with z n := a n g n → e andg n a n ↛ e, so that g n (a n g n )gn−1 ↛ e. (See also [HJ, p.354] for a further example.)We now apply the last theorem and earlier results to an example of our greatestinterest.Example. Let X be a normed group with right-invariant metric d X . Give the groupG = H(X) the usual group-norm‖f‖ H := sup x d X (f(x), x).Finally, for f, g ∈ G recall that the g-conjugate norm and the conjugacy refinement normare‖f‖ g := ‖gfg −1 ‖ H , and ‖f‖ ∞ := sup{‖f‖ g : g ∈ G}.Thus‖f‖ ∞ = sup x sup g d X g (f(x), x).