Normed versus topological groups: Dichotomy and duality

32 N. H. Bingham and A. J. OstaszewskiTheorem 3.9 (Ambidextrous Refinement). For X a normed group with norm ‖.‖, putd X S (x, y) := max{‖xy −1 ‖, ‖x −1 y‖} = max{d X R (x, y), d X L (x, y)}.Then X is a topological group under the right (or left) norm topology iff X is a topologicalgroup under the symmetrization refinement metric d X S iff the topologies of dX S and of dX Rare identical.Proof. Suppose that under the right-norm topology X is a topological group. Then d X L isd X R -continuous, by Th. 3.4 (continuity of inversion), and hence dX S is also dX R -continuous.Thus if x n → x under d X R , then also, by continuity of dX L , one has x n → x under d X S .Now if x n → x under d X S , then also x n → x under d X R , as dX R ≤ dX S . Thus dX S generatesthe topology and so X is a topological group under d X S .Conversely, suppose that X is a topological group under d X S . As X is a topological group,its topology is generated by the neighbourhoods of the identity. But as already noted,d X S (x, e) := ‖x‖,so the d X S -neighbourhoods of the identity are also generated by the norm; in particularany left-open set aB(ε) is d X S -open (as left shifts are homeomorphisms) and so right-open(being a union of right shifts of neighbourhoods of the identity). Hence by Th. 3.4 (orTh. 3.3) X is a topological group under either norm topology.As for the final assertion, if the d X S topology is identical with the dX R topology theninversion is d X R -continuous and so X is a topological group by Th. 3.4. The argument ofthe first paragraph shows that if d X R makes X into a topological group then dX R and dX Sgenerate the same topology.Thus, according to the Ambidextrous Refinement Theorem, a symmetrization thatcreates a topological group structure from a norm structure is in fact redundant. We areabout to see such an example in the next theorem.Given a metric space (X, d), we let H unif (X) denote the subgroup of uniformly continuoushomeomorphisms (relative to d), i.e. homeomorphisms α satisfying the conditionthat, for each ε > 0, there is δ > 0 such thatd(α(x), α(x ′ )) < ε, for d(x, x ′ ) < δ.(u-cont)Lemma 3.10 (Compare [dGMc, Cor. 2.13]). (i) For fixed ξ ∈ H(X), the mapping ρ ξ :α → αξ is continuous.(ii) For fixed α ∈ H unif (X), the mapping λ α : β → αβ is in H unif (X) – i.e. is uniformlycontinuous.(iii) The mapping (α, β) → αβ is continuous from H unif (X) × H unif (X) to H(X) underthe supremum norm.Proof. (i) We haveˆd(αξ, βξ) = sup d(α(ξ(t)), β(ξ(t))) = sup d(α(s), β(s)) = ˆd(α, β).