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Oct 6, 2013 at 17:03 comment added Ian Agol This is essentially true because $Diff^+(S^1)\simeq U(1)$, so $BDiff^+(S^1)\simeq BU(1)$ from the fibration $Diff^+(S^1)/U(1)\to EDiff^+(S^1)/U(1) \to EDiff^+(S^1)/Diff^+(S^1)$ with contractible fibers.
Oct 6, 2013 at 13:55 vote accept user79530
Oct 6, 2013 at 13:43 answer added sara timeline score: 18
Oct 6, 2013 at 11:36 review First posts
Oct 6, 2013 at 11:37
Oct 6, 2013 at 11:33 history edited user79530 CC BY-SA 3.0
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Oct 6, 2013 at 11:16 history asked user79530 CC BY-SA 3.0