2
$\begingroup$

If $M_{g}$ is the moduli space of Riemann surface of genus $g$, $M^1_{g}$ is the moduli space of Riemann surface of genus $g$ with one boundary, how can we show that the natural map:

$M^1_{g} \rightarrow M_{g}$

the fiber over a point $S \in M_{g}$ is the sphere bundle associated to the tangent bundle of Riemann surface $S$?

I don't have appropriate coordinates that is probably why I can't show it.

I am asking this question because I want to show that if $M^{m,b}_{g,n}$ is the moduli space of Riemann surfacw of genus $g$ and $b$ boundary components, and with $n$ interior marked points, $m=(m_1,m_2,\ldots,m_{b})$ marked points on the boundary, $m_i$ marked points on the $i$th boundary component, then the homology dimension

$H_i(M^{m,b}_{g,n},Q)=0,$ for $i\ge 6g-7+2n+3b+m$ possibly excludes some bad low dimensional cases.

I learned one way is tp assoicate it to mapping class group (especially we got a virtual cohomological dimension of mapping class group by Harer"on virtual cohomological dimension of mapping class group of oriented surface") and homology of mapping class group is the same as homology of the above moduli space, but I still don't understand how it works. Thank you for help.

Improve this question
$\endgroup$
6
  • $\begingroup$ HYYY -- if you are considering coarse moduli spaces, then the forgetful map exists, but is is not a fibration: the fiber over a point is the quotient of the spherized tangent bundle by the automorphism group of the curve. One way to remedy this is to consider the classifying spaces of mapping class groups. Every automorphism of the closed surface can be isotoped (in many ways) to an automorphism that preserves the boundary pointwise. We have a surjection of the mapping class groups, which gives an honest fibration of the classifying spaces. $\endgroup$
    – algori
    Commented Aug 8, 2010 at 5:19
  • 1
    $\begingroup$ To relate my answer to algori's : short exact sequences of groups correspond to fibrations of classifying spaces. $\endgroup$
    – Andy Putman
    Commented Aug 8, 2010 at 5:48
  • $\begingroup$ Andy -- absolutely. To elaborate a bit more on that, here is how one can identify the kernel: suppose we have an isomorphism of the surface with boundary that is isotopic to the identity when extended to the closed surface. The isotopy however does not necessarily fix the boundary pointwise: the boundary can travel; its path can be viewed as a path in the spherized tangent bundle, which is a $K(\pi,1)$; from this one can easily deduce that the kernel is the fundamental group of the spherized bundle. $\endgroup$
    – algori
    Commented Aug 8, 2010 at 6:02
  • $\begingroup$ Thank you,Andy and algori, is the classifying space of mapping class group the same as the corresponding moduli space?Thanks! $\endgroup$
    – HYYY
    Commented Aug 8, 2010 at 8:43
  • 1
    $\begingroup$ HYYY -- whenever a group $G$ acts on a space $X$, there is a map from $(EG\times X)/G$to $X/G$; the fibers of this map are of the form $EG/Stab(x)=BStab(x)$; note that $(EG\times X)/G$ also maps to $EG/G=BG$ with fiber $X$ and so if $X$ is contractible, then $(EG\times X)/G$ is homotopy equivalent to $BG$. Apply this when $G$ is a mapping class group and $X$ is the Teichmueller space; we get a map from something homotopy equivalent to the classifying space to the moduli space; since the stabilizers of the MCG on the Teichmueller space are finite, the rational cohomology of the fibers vanishes. $\endgroup$
    – algori
    Commented Aug 8, 2010 at 11:38

1 Answer 1

Reset to default
4
$\begingroup$

I'll give you references for the appropriate fact about the mapping class group. Let $Mod_{g,b}^p$ be the mapping class group of a genus $g$ surface with $b$ boundary components and $p$ punctures $\Sigma_{g,b}^p$. There are really two fact. The first is the Birman exact sequence (by the way, I believe this was Joan Birman's thesis!). It says that there is a short exact sequence $$1 \longrightarrow \pi_1(\Sigma_{g,b}^{p-1}) \longrightarrow Mod_{g,b}^p \longrightarrow Mod_{g,b}^{p-1} \longrightarrow 1$$ except in the degenerate cases where either $g=0$ and $b+p$ is at most $3$, or $g=1$ and $b+p$ is at most $1$.

This was originally proven in her paper

J. S. Birman, Mapping class groups and their relationship to braid groups, Comm. Pure Appl. Math. {\bf 22} (1969), 213--238.

Two other nice sources for it are Birman's book "Braids, Links, and Mapping Class Groups" and Farb-Margalit's book "A Primer on Mapping Class Groups", the latter of which is available here.

The second is a sort of variant on it, which says that (except in the degenerate cases above) we have a short exact sequence $$1 \longrightarrow \pi_1(U\Sigma_{g,b-1}^p) \longrightarrow Mod_{g,b}^p \longrightarrow Mod_{g,b-1}^p \longrightarrow 1.$$ Here $U\Sigma_{g,b-1}^p$ is the unit tangent bundle of $\Sigma_{g,b-1}^p$.

It's a little harder to give a reference for this; the only place I know where it is proven is in section 3 of the following paper:

D. Johnson, The structure of the Torelli group. I. A finite set of generators for ${\cal I}$, Ann. of Math. (2) {\bf 118} (1983), no.~3, 423--442.

By the way, this also happens to be the earliest appearance of this in the literature that I know of, though I've seen it many times subsequently with no reference to Johnson's work and no proof.

The proof makes use of the Birman exact sequence above.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Dear Andy, the mapping class group in Birman exact sequence should be pure mapping class group,i.e.,it fixes each puncture individually, then the corresponding moduli space should be the moduli space of Riemann surface with ordered puncture right? $\endgroup$
    – HYYY
    Commented Aug 9, 2010 at 22:35
  • $\begingroup$ Yes. To get from there to the unordered situation, just mod out by the obvious S_n action. $\endgroup$
    – Andy Putman
    Commented Aug 9, 2010 at 22:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .