Normed versus topological groups: Dichotomy and duality

Normed groups 25shows that [z n , y] L → e iff [y n , y] L → e, i.e. (i)⇔(ii).Turning to the second equivalence in the chain, we see from continuity of inversion at e(or inversion invariance) that for any y[z n , y] L = z n yzn−1 y −1 → e iff [y, z n ] L = yz n y −1 zn −1 → e,giving (ii)⇔(iii). Finally, with the notation y n = z n y, the identity[y, y n ] L = yy n y −1 y −1nshows that [y, z n ] L → e iff [y, y n ] L → e, i.e. (iii)⇔(iv).= y(z n y)y −1 (y −1 zn−1 ) = yz n y −1 zn−1 = [y, z n ] LProposition 2.22. For a normed group X, with the right norm topology, and for g, h ∈X, the commutator map x → [x, h] L is continuous at x = g provided the map x →[x, hgh −1 ] L is continuous at x = e.Hence if all the commutator maps x → [x, y] L for y ∈ X are continuous at x = e, thenthey are all continuous everywhere.Proof. For fixed g, h and with h n = z n h we have the identity[h n , g] L = [z n , hgh −1 ] L [h, g] L= (z n hgh −1 z −1n hg −1 h −1 )(hgh −1 g −1 ).Suppose x → [x, hgh −1 ] L is continuous at x = e. The identity above now yields [h n , g] L →[h, g] L as h n → R h; indeed z n = h n h −1 → e X so w n := [z n , hgh −1 ] L → e X , and thuswith a := [h, g] L we have ρ a (w n ) = w n a → a.Remarks. 1. If the group-norm is abelian, then we have the left-right commutator inequality‖[x, y] L ‖ ≤ 2‖xy −1 ‖ = 2d R (x, y),because‖[x, y] L ‖ = ‖xyx −1 y −1 ‖ ≤ ‖xy −1 ‖ + ‖yx −1 ‖ = 2‖xy −1 ‖.The commutator inequality thus implies separate continuity of the commutator by Lemma2.21.2. If the group-norm is arbitrary, this inequality may be stated via the symmetrizedmetric:‖[x, y −1 ] R ‖ ≤ ‖xy −1 ‖ + ‖x −1 y‖ = d R (x, y) + d L (x, y)≤ 2 max{d R (x, y), d L (x, y)} := 2d S (x, y).3. Take u = f(tx), v = f(x) −1 etc.; then, assuming the Klee Property, we have‖f(tx)g(tx)[f(x)g(x)] −1 ‖ = ‖f(tx)g(tx)g(x) −1 f(x) −1 ‖≤ ‖f(tx)f(x) −1 ‖ + ‖g(tx)g(x) −1 ‖,