Normed versus topological groups: Dichotomy and duality

Normed groups 31argument as again proving that γ g is continuous; indeed we may now read the earlieridentity as asserting thatγ g (z n ) := gz n g −1 (zn−1 z n ) = [g −1 , zn −1 ] R z n ,for which the earlier argument continues to hold.3. For T a normed group with right-invariant metric d R one is led to study theassociated supremum metric on the group of bounded homeomorphisms h from T to T(i.e. having sup T d(h(t), t) < ∞) with composition ◦ as group operation:d A (h, h ′ ) = sup T d(h(t), h ′ (t)).This is a right-invariant metric which generates the norm‖h‖ A := d A (h, e A ) = sup T d(h(t), t).It is of interest from the perspective of topological flows, in view of the following observation.Lemma 3.8 ([Dug, XII.8.3, p. 271]). Under d A on A = Auth(T ) and d T on T, theevaluation map (h, t) → h(t) from A×T to T is continuous.Proof. Fix h 0 and t 0 . The result follows from continuity of h 0 at t 0 viad T (h 0 (t 0 ), h(t)) ≤ d T (h 0 (t 0 ), h 0 (t)) + d T (h 0 (t), h(t))≤ d T (h 0 (t 0 ), h 0 (t)) + d A (h, h 0 ).4. Since the conjugate metric of a right-invariant metric need not be continuous, oneis led to consider the earlier defined symmetrization refinement of a metric d, which werecall is given byd S (g, h) = max{d(g, h), d(g −1 , h −1 )}.(sym)This metric need not be translation invariant on either side (cf. [vM2, Example 1.4.8] );however, it is inversion-invariant:d S (g, h) = d S (g −1 , h −1 ),so one expects to induce topological group structure with it, as we do in Th. 3.13 below.When d = d X R is right-invariant and so induces the group-norm ‖x‖ := d(x, e) andd(x −1 , y −1 ) = d X L (x, y), we may use (sym) to defineThen‖x‖ S := d X S (x, e).‖x‖ S = max{d X R (x, e), d X R (x −1 , e)} = ‖x‖,which is a group-norm, even though d X S need not be either left- or right-invariant. Thismotivates the following result, which follows from the Equivalence Theorem (Th. 3.4)and Example A4 (Topological permutations), given towards the start of Section 2.