Timeline for Why is the mapping class group of hyperbolic manifolds finite?
Current License: CC BY-SA 3.0
7 events
when toggle format | what | by | license | comment | |
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Oct 12, 2012 at 10:58 | vote | accept | Lor | ||
May 30, 2011 at 19:36 | comment | added | Ryan Budney | @Lor: notice the difference between Agol's response and my own. Depending on how you define "mapping class group" the question you ask is quite different. My response was assuming "mapping class group" meant $\pi_0 HomotopyEquivalences(M)$. Ian's response is for $\pi_0 Homeo(M)$. I don't believe Ian's response applies to $\pi_0 Diff(M)$ provided the dimension of $M$ is large. | |
May 30, 2011 at 19:31 | comment | added | Dave Futer | ``I know there is Dehn-Nielsen Theorem which states that $Out(\pi_1(M))$ is isomorphic to $MCG(M)$, but I know this to be true only in dimension 2...what can I say in dimension (at least) 3?'' Taking away the hypothesis that $M$ is hyperbolic turns this into a much more interesting question (in my humble opinion). An attempt at rephrasing it is here: mathoverflow.net/questions/66484/… | |
May 30, 2011 at 19:27 | answer | added | Ian Agol | timeline score: 10 | |
May 30, 2011 at 18:21 | answer | added | Ryan Budney | timeline score: 6 | |
May 30, 2011 at 15:09 | comment | added | Andy Putman | Hyperbolic manifolds have contractible universal covers (ie hyperbolic space), so they are $K(\pi,1)$'s. It is standard that there is a bijection between self-homotopy equivalences of $K(\pi,1)$'s and outer automorphisms of $\pi$. | |
May 30, 2011 at 14:57 | history | asked | Lor | CC BY-SA 3.0 |