# Complexification or 'real'ization of Mapping Class group.

So is there a complexification or 'real'ization of the mapping class group or can it be realised as a lattice in some lie group. like $PSL(2, \mathbb Z)$ in $PSL(2, \mathbb R)$. for g=1 this certainly seems to be the case. So I was thinking about the g > 1 case and thought of something very naive.

So the mapping class group acts properly discontinously on the Teichmuller Space. Namely it is generated by dehn twists., i.e. twists by 2$\pi$ around closed loops, but I was wondering why isnt this action extended continuously to a larger group action on the Teichmuller space, i.e. say by a slight fractional twist or a deformation of a surface with a pair of pants decomposition. More specifically say, we look at teichmuller space in fenchel nielsen coordinates of $(l_1, l_2, l_3.... l_{3g-3}, \tau_1, \tau_2, .... , \tau_{3g-3})$ here a dehn twist about a loop defining the pair of pants decomposition on a hyperbolic surface would just change the \tau_k parameter by 2$\pi$, but why cant we just add a parameter t(instead of $2\pi$) to it and continously extend the action of MCG. I am just trying to generalise from $PSL(2, \mathbb Z)$ as a subset of $PSL(2, \mathbb R)$. Please tell me If i am speaking complete nonsense. Again I am a novice in this area. So please pardon my ignorance.

• What you are describing is certainly studied very extensively. Reading the canonical sources is wise (e.g. Thurston's works on surfaces). Dec 7, 2012 at 1:44
• Mapping class group is not isomorphic to a lattice in a Lie group, except for the mapping class groups of the torus with at most one puncture and the sphere with at most 4 punctures. Mapping class group does embed in the group of Hamiltonian symplectomorpisms of the Teichmuller space, but the latter is not a Lie group in the traditional sense. Dec 7, 2012 at 4:29

Royden's theorem implies that the only isometries of the Teichmuller metric (in genus $>1$) is the mapping class group. In fact, since the Teichmuller metric is the Kobayashi metric on Teichmuller space, this is also the group of complex automorphisms.
So if you want to embed the mapping class group into a larger group, it will have to act on some larger space than Teichmuller space, or not preserve the complex structure. One can get larger groups acting e.g. by taking the cotangent bundle to Teichmuller space, which also has an action of $SL_2(\mathbb{R})$ on it.
As Misha mentions, his paper with Leeb, "Actions of discrete groups on nonpositively curved spaces" MR1411351, gives severe restrictions on CAT(0) structures on $MCG(S)$, which covers the case of lattice embeddings of $MCG(S)$ into a semisimple Lie group $G$, and more general Lie groups are then easily handled (@Misha I think this is what your comment refers to, please correct if not). This generalizes in a couple of directions. Kida's theorem, in Measure equivalence rigidity of the mapping class group'' MR2680399, puts severe restrictions on any embedding of the mapping class group $MCG(S)$ of a surface $S$ as a lattice in a locally compact, second countable group $G$. And as a consequence of the quasi-isometric rigidity theorem for $MCG(S)$ of Behrstock, Kleiner, Minsky and myself, MR2928983 "Geometry and rigidity of mapping class groups", one gets severe restrictions on embeddings of $MCG(S)$ as a cocompact discrete subgroup of a compactly generated group $G$. In each case, the limitations are similar: roughly speaking, $G$ differs from $MCG(S)$ itself only by compact or bounded normal subgroups and finite index stuff. The arguments apply as well to embeddings of finite index subgroups of $MCG(S)$.
Let me give a careful statement and proof for the case I know best, which uses our QI-rigidity theorem cited above, namely the case where $MCG(S)$ embeds as a cocompact discrete subgroup of a compactly generated group $G$. The conclusion is that there is a retraction homomorphism $\rho : G \to MCG(S)$ with bounded kernel (I think the kernel must also be compact open, which is true when $G$ is a Lie group, but I am not sure in general). I'll restrict to the case that $S$ is not a 2-holed torus because our QI-rigidity theorem is slightly convoluted in that case. Using the compactly generated word metric on $G$, cocompactness implies that there is a "closest point retraction" $p : G \to MCG(S)$ which moves points a bounded distance $\le A$, and so $p$ is a $1,2A$ quasi-isometry. For each $g \in G$ consider the left multiplication map $L_g : G \to G$ which is an isometry with respect to the compactly generated word metric. Precompose $L_g$ with the inclusion $i : MCG(S) \hookrightarrow G$ which is a $K,C$ quasi-isometry for some $K,C$, and postcompose with the closest point retraction $p$, and the resulting map $p \circ L_g \circ i : MCG(S) \to MCG(S)$ is a $K,C'$ quasi-isometry for some $K,C'$ independent of $g$. Write this map as $\phi \mapsto p(g \cdot \phi)$. By the QI-rigidity theorem cited above, there exists a constant $B$ independent of $g$ and there exists a unique $\rho(g) \in MCG(S)$, such that $d(p(g \cdot \phi),\rho(g) \cdot \phi) \le B$ for all $\phi \in MCG(S)$. This function $\rho$ is a retraction because if $g$ is already in $MCG(S)$ then the inequality $d(g \cdot \phi, \rho(g) \cdot \phi) < A$ for all $\phi$ implies $g=\rho(g)$, by uniqueness. A similar line of thought using uniqueness shows that $\rho$ is a homomorphism. The kernel of $\rho$ is bounded because, by the triangle inequality, if $\rho(g) = Id$ then the quantity $d(Id,g) = d(\rho(g),g) = d(\rho(g) \cdot Id,g \cdot Id)$ differs from the quantity $d(p(g \cdot Id),\rho(g) \cdot Id) \le B$ by at most $d(g \cdot Id,p(g \cdot Id)) = d(g,p(g)) \le A$.