Normed versus topological groups: Dichotomy and duality

Normed groups 99Taking h(x) := ‖π(x)‖ Y , Cor. 2.9, Th. 7.10 and Th. 7.11 together immediately implythe following.Corollary 7.12. If X is a Baire normed group and π : X → Y a group homomorphism,where ‖.‖ Y is (µ-γ)-quasi-isometric to ‖.‖ X under the mapping π, then there existconstants K, L, M such that‖π(ux)‖ Y / ‖π(x)‖ Y < ‖u‖ K X (u ≥ L, ‖x‖ X ≥ M).8. The Subgroup TheoremIn this section G is a normed locally compact topological group with left-invariant Haarmeasure. We shall be concerned with two topologies on G : the norm topology and thedensity topology. Under the latter the binary group operation need not be jointly continuous(see Heath and Poerio [HePo]); nevertheless a right-shift x → xa, for a constant,is continuous, and so we may say that the density topology is right-invariant. We notethat if S is measurable and non-null then S −1 is measurable and non-null under the correspondingright-invariant Haar and hence also under the original left-invariant measure.We may thus say that both the norm and the density topologies are inversion-invariant.Likewise the First and Second Verification Theorems (Theorems 6.2 and 7.5) assert thatunder both these topologies shift homeomorphisms satisfy (wcc). This motivates a theoremthat embraces both topologies as two instances.Theorem 8.1 (Topological, or Category, Interior Point Theorem). Let G be given aright-invariant and inversion-invariant topology τ, under which it is a Baire space andsuppose that the shift homeomorphisms h n (x) = xz n satisfy (wcc) for any null sequence{z n } → e (in the norm topology). For S Baire and non-meagre in τ, the difference setS −1 S, and likewise SS −1 , is an open neighbourhood of e in the norm topology.Proof. Suppose otherwise. Then for each positive integer n we may selectz n ∈ B 1/n (e)\(S −1 S).Since {z n } → e (in the norm topology), the Category Embedding Theorem (Th. 6.1)applies, and gives an s ∈ S and an infinite M s such thatThen for any m ∈ M s ,{h m (s) : m ∈ M s } ⊆ S.sz m ∈ S , i.e. z m ∈ S −1 S,a contradiction. Replacing S by S −1 we obtain the corresponding result for SS −1 .One thus has again.