Normed versus topological groups: Dichotomy and duality

Normed groups 19(iii) The ¯d H -topology on H is equivalent to the d X -topology on H iff d X satisfies (Hadm),the metric admissibility condition on H.(iv) In particular, if d X is right-invariant, then H = X and ¯d H = d X .(v) If X is a compact topological group under d X , then ¯d H is equivalent to d X .Proof. (i) The argument relies implicitly on the natural embedding of X in Auth(X) asT r L (X) (made explicit in the next section). For x ∈ X we write‖λ x ‖ H := sup z d X (xz, z).For x ≠ e, we have 0 < ‖λ x ‖ H ≤ ∞. By Proposition 2.12, H(X) = H(X, T r L (X)) ={λ x : ‖λ x ‖ H < ∞} is a subgroup of H(X, Auth(X)) on which ‖ · ‖ H is thus a norm.Identifiying H(X) with the subset H = {x ∈ X : ‖λ x ‖ < ∞} of X, we see that on H¯d H (x, y) := sup z d X (xz, yz) = ˆd H (λ x , λ y )defines a right-invariant metric, as¯d H (xv, yv) = sup z d X (xvz, yvz) = sup z d X (xz, yz) = ¯d H (x, y).Hence withby Proposition 2.11‖x‖ H = ¯d H (x, e) = ‖λ x ‖ H ,‖λ x λ −1y ‖ H = ¯d H (x, y) = ‖xy −1 ‖ H ,as asserted.If d X is left-invariant, then¯d H (vx, vy) = sup z d X L (vxz, vyz) = sup z d X L (xz, yz) = ¯d H (x, y),and so ¯d H is both left-invariant and right-invariant.Note that‖x‖ H = ¯d H (x, e) = sup z d X L (xz, z) = sup z d X L (z −1 xz, e) = sup z ‖x‖ z = ‖x‖ ∞ .(ii) We note thatd X (z n , e) ≤ sup y d X (z n y, y).Thus if z n → e in the sense of d H , then also z n → e in the sense of d X . Suppose that themetric admissibility condition holds but the metric d H is not equivalent to d X . Thus forsome z n → e (in H and under d X ) and ε > 0,Thus there are y n withsup y d X (z n y, y) ≥ ε.d X (z n y n , y n ) ≥ ε/2,which contradicts the admissibility condition.For the converse, if the metric d H is equivalent to d X , and z n → e in H and under d X ,then z n → e also in the sense of d H ; hence for y n given and any ε > 0, there is N suchthat for n ≥ N,ε > ¯d H (z n , e) = sup y d X (z n y, y) ≥ d X (z n y n , y n ).