Normed versus topological groups: Dichotomy and duality

130 N. H. Bingham and A. J. OstaszewskiProof. Referring to the Klee property, via the cyclic property we haved G (gag −1 , hbh −1 ) = ‖gag −1 hb −1 h −1 ‖ = ‖h −1 gag −1 hb −1 ‖≤ ‖h −1 g‖ + ‖ag −1 hb −1 ‖≤ ‖h −1 g‖ + ‖ab −1 ‖ + ‖g −1 h‖,for all a, b, yielding the right-hand side inequality. Then substitute g −1 ag for a etc., g −1for g etc., to obtaind G (a, b) ≤ 2d G (g −1 , h −1 ) + d G (gag −1 , hbh −1 ).This yields the left-hand side inequality, as d G is bi-invariant and sod G (g −1 , h −1 ) = ˜d G (g, h) = d G (g, h).Proposition 12.5 (Permutation metric). For π ∈ H(X), let d π (x, y) := d X (π(x), π(y)).Then d π is a metric, andd X (x, y) − 2‖π‖ ≤ d π (x, y) ≤ d X (x, y) + 2‖π‖.In particular, if d X is right-invariant and π(x) is the left-translation λ z (x) = zx, thend X (x, y) − 2‖z‖ ≤ d X z (x, y) = d X (zx, zy) ≤ d X (x, y) + 2‖z‖.Proof. By the triangle inequality,d X (π(x), π(y)) ≤ d X (π(x), x) + d X (x, y) + d X (y, π(y)) ≤ 2‖π‖ + d X (x, y).Likewise,d X (x, y) ≤ d X (x, π(x)) + d X (π(x), π(y)) + d X (π(y), y)≤ 2‖π‖ + d X (π(x), π(y)).If π(x) := zx, then ‖π‖ = sup d(zx, x) = ‖z‖ and the result follows.Recall from Proposition 2.2 that for d a metric on a group X, we write ˜d(x, y) =d(x −1 , y −1 ) for the (inversion) conjugate metric. The conjugate metric ˜d is left-invariantiff the metric d is right-invariant. Under such circumstances both metrics induce thesame norm (since d(e, x) = d(x −1 , e), as we have seen above). In what follows note thatξ −1x = ξ x −1.Theorem 12.6 (Quasi-isometric duality). If the metric d X on X is right-invariant andt ∈ T ⊂ H(X) is a subgroup, then˜d X (x, y) − 2‖t‖ H(X) ≤ d T (ξ x (t), ξ y (t)) ≤ ˜d X (x, y) + 2‖t‖ H(X) ,and hence, for each ε ≥ 0, the magnification metric (mag-eps) satisfies˜d X (x, y) − 2ε ≤ d ε T (ξ x , ξ y ) ≤ ˜d X (x, y) + 2ε.Equivalently, in terms of conjugate metrics,d X (x, y) − 2ε ≤ ˜d ε T (ξ x , ξ y ) ≤ d X (x, y) + 2ε.