Normed versus topological groups: Dichotomy and duality

24 N. H. Bingham and A. J. Ostaszewskiwe call the maps x → [x, y] and y → [x, y] commutator maps and in the context of aspecified norm topology (either!), we say that the commutator [·, ·] is:(i) left continuous if for all y the map x → [x, y] is continuous at each x;(ii) right continuous if for all x the map y → [x, y] is continuous at each y,(iii) separately continuous if it is left and right continuous.We show that the commutators are like homomorphisms, in that their continuity may beimplied by continuity at the identity e X , but this does require that all the commutatormaps be continuous at the identity.Theorem 2.20. In a normed group an either-sided commutator is left continuous iff itis right continuous and so iff it is separately continuous.We deduce the above theorem from the following two more detailed results; see alsoTheorem 3.4 for further insights on this result.Proposition 2.21. In a normed group under either norm topology the following areequivalent for y ∈ X:(i) the commutator map x → [x, y] L is (left) continuous at x = y,(ii) the commutator map x → [x, y] L is (left) continuous at e, i.e. [z n , y] L → e, asz n → e,(iii) the commutator map x → [y, x] L is (right) continuous at e, i.e. [y, z n ] L → e, asz n → e,(iv) the commutator map z → [y, z] L is (right) continuous at z = y,(v) the commutator map z → [y −1 , z] R is (right) continuous at z = y −1 ,(vi) the commutator map x → [x, y −1 ] R is (left) continuous at x = y −1 .Proof. As the conclusions are symmetric without loss of generality we work in the rightnorm topology generated by the right-invariant metric d R and write → R to show that theconvergence is in d R . Note that y n → R x iff y n x −1 → e; there is no need for a subscriptfor convergence to e, as the ball B ε (e X ) is the same under either of the conjugate metrics(cf. Prop. 2.15). Indeed, writing y n = z n y, we have d R (z n y, y) = d R (z n , e) → 0.We first prove the chain of equivalences: (i)⇔(ii)⇔(iii)⇔(iv). The remaining equivalencesfollow from the observation that[z n , y] L = [zn−1 , y −1 ] Rand zn−1 is a null sequence iff z n is.In regard to the first equivalence, employing the notation y n = z n y, the identityi.e.[z n , y] L = z n yz −1ny −1 = (z n y)y(y −1 zn−1 )y −1 = y n yyn −1 y −1 = [y n , y] L ,[y n , y] L = [y n y −1 , y] L ,