Timeline for Question related to the moduli space of Riemann surfaces and a fibration
Current License: CC BY-SA 2.5
12 events
| when toggle format | what | by | license | comment | |
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| Aug 8, 2010 at 12:35 | vote | accept | HYYY | ||
| Aug 8, 2010 at 11:39 | comment | added | algori | The space $(EG\times X)/G$ is called the Borel quotient or the Borel construction. | |
| Aug 8, 2010 at 11:38 | comment | added | algori | 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. | |
| Aug 8, 2010 at 8:43 | comment | added | HYYY | Thank you,Andy and algori, is the classifying space of mapping class group the same as the corresponding moduli space?Thanks! | |
| Aug 8, 2010 at 6:25 | history | edited | HYYY | CC BY-SA 2.5 |
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| Aug 8, 2010 at 6:24 | history | edited | Charles Matthews | CC BY-SA 2.5 |
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| Aug 8, 2010 at 6:02 | comment | added | algori | 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. | |
| Aug 8, 2010 at 5:48 | comment | added | Andy Putman | To relate my answer to algori's : short exact sequences of groups correspond to fibrations of classifying spaces. | |
| Aug 8, 2010 at 5:19 | comment | added | algori | 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. | |
| Aug 8, 2010 at 5:03 | history | edited | HYYY | CC BY-SA 2.5 |
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| Aug 8, 2010 at 4:37 | answer | added | Andy Putman | timeline score: 4 | |
| Aug 8, 2010 at 4:15 | history | asked | HYYY | CC BY-SA 2.5 |