Normed versus topological groups: Dichotomy and duality

Normed groups 9The following result is simple; we make use of it in the Definition which follows Lemma3.23.Proposition 2.2 (Symmetrization refinement). If ‖x‖ 0 is a group pre-norm, then thesymmetrization refinementis a group-norm.‖x‖ := max{‖x‖ 0 , ‖x −1 ‖ 0 }Proof. Positivity is clear, likewise symmetry. Noting that, for any A, B,a + b ≤ max{a, A} + max{b, B},and supposing without loss of generality thatmax{‖x‖ 0 + ‖y‖ 0 , ‖y −1 ‖ 0 + ‖x −1 ‖ 0 } = ‖x‖ 0 + ‖y‖ 0 ,we have‖xy‖ = max{‖xy‖ 0 , ‖y −1 x −1 ‖ 0 } ≤ max{‖x‖ 0 + ‖y‖ 0 , ‖y −1 ‖ 0 + ‖x −1 ‖ 0 }= ‖x‖ + ‖y‖ 0 ≤ max{‖x‖ 0 , ‖x −1 ‖ 0 } + max{‖y‖ 0 , ‖y −1 ‖ 0 }= ‖x‖ + ‖y‖.Remark. One can use summation and take ‖x‖ := ‖x‖ 0 + ‖x −1 ‖ 0 , as‖xy‖ = ‖xy‖ 0 + ‖y −1 x −1 ‖ 0 ≤ ‖x‖ 0 + ‖y‖ 0 + ‖y −1 ‖ 0 + ‖x −1 ‖ 0 = ‖x‖ + ‖y‖.However, here and below, we prefer the more general use of a supremum or maximum,because it corresponds directly to the intersection formula (Str) which defines the refinementtopology. We shall shortly see a further cogent reason (in terms of the refinementnorm).Proposition 2.3. If ‖ · ‖ is a group-norm, then d(x, y) := ‖xy −1 ‖ is a right-invariantmetric; equivalently, ˜d(x, y) := d(x −1 , y −1 ) = ‖x −1 y‖ is the conjugate left-invariant metricon the group.Conversely, if d is a right-invariant metric, then ‖x‖ := d(e, x) = ˜d(e, x) is a group-norm.Thus the metric d is bi-invariant iff ‖xy −1 ‖ = ‖x −1 y‖ = ‖y −1 x‖, i.e. iff the group-normis abelian.Furthermore, for (X, ‖ · ‖) a normed group, the inversion mapping x → x −1 from (X, d)to (X, ˜d) is an isometry and hence a homeomorphism.Proof. Given a group-norm put d(x, y) = ‖xy −1 ‖. Then ‖xy −1 ‖ = 0 iff xy −1 = e, i.e.iff x = y. Symmetry follows from inversion as d(x, y) = ‖(xy −1 ) −1 ‖ = ‖yx −1 ‖ = d(y, x).Finally, d obeys the triangle inequality, since‖xy −1 ‖ = ‖xz −1 zy −1 ‖ ≤ ‖xz −1 ‖ + ‖zy −1 ‖.