Does there exist a quasiisometric embedding $$MCG(S) \to (\mathrm{Teich}(S), d)$$ for $d$ any "known" distance on the Teichmuller space (i.e. Teichmuller, WeilPetersson, Thurston...) ?

$\begingroup$ This is not exactly the answer you want but you can find some related results here <arxiv.org/abs/math/0701719> $\endgroup$– CuspOct 13, 2013 at 3:52

1$\begingroup$ I believe the answer is no for Teichmuller and WeilPetersson: I think this follows from work of Minsky, Rafi, and Brock, who found combinatorial models for measuring distance in the mapping class group, Teichmuller metric, and WeilPetersson metric respectively. $\endgroup$– Ian AgolOct 13, 2013 at 4:14

$\begingroup$ @Ian: I am not sure, since a qi embedding might be unrelated to the standard action of the mcg. $\endgroup$– MishaOct 13, 2013 at 14:29

$\begingroup$ Ok, good point Misha! I implicitly assumed he was looking for an equivariant quasiisometry, so whether the orbit of a point under the mcg is a q.i. embedding. $\endgroup$– Ian AgolOct 13, 2013 at 16:48
1 Answer
A result of Behrstock and Minsky (cf. Hamenstadt too) implies that the rank of mapping class groups is the maximal rank of abelian subgroups, which is $3g+p3$ for a connected hyperbolic surface of genus $g$ with $p$ boundary components. The rank of Teichmuller space with respect to the WeilPetersson metric they show is $\lfloor \frac{3g+p2}{2}\rfloor < 3g3+p$, so there is no q.i. embedding of the mapping class group to Teichmuller space with the WeilPetersson metric.
On the other hand, EskinMasurRafi show that the rank of the Teichmuller metric is equal to that of the mapping class group. They point out that the orbit of a point under the mapping class group is not a quasiisometric embedding. But this leaves open whether there might be a nonequivariant q.i. embedding.