Dehn Twists Have Linear Growth

In this note we give a simple proof that Dehn twists have linear growth with respect to the word metric on the mapping class group of an oriented surface of nite type (Theorem 1). This answers a question of Ivanov in 3]. A proof in the case of punctured surfaces was given by Mosher 6] using his techniques for constructing an automatic structure on the mapping class group. This result implies in particular that the embedding of the mapping class group as an orbit in the Teichm uller space is not a quasi isometry (Theorem 3), and this should be viewed in comparison with the opposite conclusion for higher-rank lattices, by Lubotzky-Mozes-Raghunathan 4]. Fix an oriented surface S of nite type. We assume throughout that S is not trivial, i.e. not the sphere, annulus, or thrice-punctured sphere. We also usually consider S to be hyperbolic, that is not the regular torus. Let Mod(S) denote the mapping class group of S, or the group of all homeomorphisms modulo those isotopic to the identity. If fg 1 ; : : : ; g N g is a set of generators for Mod(S), let j j W denote minimal word length with respect to these generators. Theorem 1. Let S be a nontrivial surface of nite type. Let t be a Dehn twist in Mod(S) and x a nite generating set. There exists a constant c > 0 so that jt n j W cjnj For all n. Equivlently, one can say that t has positive translation distance in the sense of Gersten and Short 2], that is d(t) = lim inf n!1 jt n j W n > 0: More generally, it follows by previously known results that Theorem 2. Every element of innnite order in Mod(S) has linear growth. The case of Dehn twists around a collection of disjoint curves is done in the same way; the case of an element h which is pseudo-Anosov or is reducible and has a power which is pseudo-Anosov on some subsurface is already known { the argument uses the exponential growth of lengths of curves under h (see Mosher 6]). By Thurston's classiication of elements of Mod(S) 1], these are all possible cases.