# Classifying map for a surface bundle

Let $E\longrightarrow X$ be a surface (with holes) bundle. The structure group is then $M_{g, s}$, the mapping class group of the fiber. It follows from the famous work of Penner that the classifying space of $M_{g, s}$ is homotopy equivalent to the geometrical realization of the category of fatgraphs, i.e., $$BM_{g,s} \simeq |\mathit{Fat}_{g,s}|.$$ My question is as follows: how can one construct the classifying map $X\longrightarrow |\mathit{Fat}_{g,s}|$ explicitly given the initial bundle in these terms? This seems pretty intuitive, but I can not get to the formal description.

My only thought was to try to find something similar the natural map to the Milnor join construction of the classifying space.

Edit: the natural restrictions for the genus and number of holes are $s\geqslant 1$ and $s+2g \geqslant 3$.

• The triangulation of the moduli space of curves in terms of fatgraphs only works for surfaces with nonempty boundary; indeed, it is clear from the definition that the surface associated to a fatgraph has a nonempty boundary. – Andy Putman Jul 22 '16 at 15:46
• @AndyPutman yes, indeed, thank you. I will edit the question to avoid misunderstandings, the isomorphism is valid only for $s\geqslant 1$ and $s+2g\geqslant 3$. – Denis Gorodkov Jul 22 '16 at 16:03
• One of the things that makes this slightly unnatural is that your bundle is really a principal $\text{BDiff}(\Sigma_{g,s})$-bundle, not a principal $M_{g,s}$-bundle. It's a deep theorem of Earle-Eells that these are the same thing (i.e. that the identity component of $\text{Diff}(\Sigma_{g,s})$ is contractible). – Andy Putman Jul 27 '16 at 15:38
• Another difficulty is that there isn't a "canonical" identification of the classifying space of the mapping class group with the geometrical realization of the category of fatgraphs (and indeed, there are several such in the literature). I do know nice maps between standard models for the classifying space of the diffeomorphism group of a surface and the moduli space of Riemann surfaces, but they all use geometry in some essential way. One then has to map this to the fatgraph complex. – Andy Putman Jul 27 '16 at 15:42
• (actually, I'll add an answer with one such map; it doesn't answer the question as stated, but it might be interesting) – Andy Putman Jul 27 '16 at 15:44

For simplicity, I'm going to work with closed surfaces and with fiber bundles whose bases are smooth manifolds. Consider a fiber bundle $\pi:E \rightarrow X$ whose fibers are closed genus $g$ surfaces $\Sigma_g$ and whose base $X$ is a smooth manifold (not necessarily compact). Our goal is to construct a classifying map $\phi:X \rightarrow \mathcal{M}_g$, where $\mathcal{M}_g$ is the moduli space of Riemann surfaces. I'm going to think of $\mathcal{M}_g$ as the space of isomorphism classes of conformal structures on $\Sigma_g$.
We can find a smooth embedding $f:E \rightarrow \mathbb{R}^k$ for some large $k$. For each $x \in X$, the image $A_x:=f(\pi^{-1}(x))$ is a smooth submanifold of $\mathbb{R}^k$ which is diffeomorphic to $\Sigma_g$. Restrict the usual Riemannian metric on $\mathbb{R}^k$ to $A_x$ to make $A_x$ into a Riemannian manifold. The image $\phi(x) \in \mathcal{M}_g$ is then the conformal structure on $A_x$ given by this Riemannian metric.
Of course, this depends on the choice of embedding $f$. However, stabilizing $f$ by embedding $\mathbb{R}^k$ into $\mathbb{R}^{k+1}$ changes nothing. By making $k$ large enough, we can ensure that any two choices of embeddings are isotopic, which gives us a homotopy between the two classifying maps.