Linear growth of infinite order mapping class group elements

Fix a set of generators of the mapping class group of $ S$ (the group is finitely generated; there are many proofs of this fact, already known to Dehn [14] or [15] for an English translation; see also [31] and [24]) say $ \mathfrak{T}$. We assume that the identity is not in $ \mathfrak{T}$ and that if $ f$ is in $ \mathfrak{T}$ then so is $ f^{-1}$. The length of an element $ f$ of $ Mod(S)$, $ \vert\vert f\vert\vert$, is defined as the minimum integer $ n$ such that $ f$ can be expressed as a product of $ n$ elements of $ \mathfrak{T}$ (with the length of the identity equal to 0).

Theorem 4.2 (Farb, Lubotzky and Minsky [18])   Every Dehn twist $ t$ has linear growth in $ Mod(S)$; that is, there exists a positive constant $ c$ such that $ \vert\vert t^n\vert\vert \geq c \vert n\vert$, for all $ n\in \mathbb{Z}$.

Combining this result with the work of L. Mosher [38] one gets that every element of infinite order in $ Mod(S)$ has linear growth (theorem 1.2 in the cited paper).

If we now consider the Teichmüller metric on $ T(S)$, and fix a point $ X_0 \in
T(S)$, we have an embedding of the modular group on Teichmüller space by $ g
\mapsto g(X_0)$. It is known that if a group acts with certain properties on a space then this mapping is a quasi-isometry (see [8, proposition 8.19]). However we have that this is not true in the case under consideration.

Theorem 4.3   The word metric on the modular group and the metric induced by inclusion as an orbit in $ T(X)$ with the Teichmüller metric are not Lipschitz equivalent.

The proof is based on the fact that given Dehn twist $ t$, there exists a totally geodesic copy of the upper half plane (Teichmüller disc), invariant under $ t$, that contains $ X_0$ and on which $ t$ acts as a parabolic transformation ( $ z \mapsto z + 1$ on $ \mathbb{H}$).

Pablo Ares Gastesi 2005-08-31