Skip to main content
14 events
when toggle format what by license comment
Jun 9, 2019 at 18:54 comment added Aleksandar Milivojević @IgorBelegradek In case you're still interested: It is not; by Corollary 11 in Kotschick's original paper arxiv.org/abs/math/0004009, a closed oriented four-manifold with $b_1=1$ and $b_2=0$ is geometrically formal iff it fibers over a circle. Now, $S^3\times S^1$ is geometrically formal, but if you connect sum that with a non-trivial rational homology four-sphere, then you get a manifold with the same rational homotopy type, but which does not fiber over a circle. (This last part, about not fibering over a circle, was shown to me by Corey Bregman; I can provide details if you're interested.)
Mar 24, 2018 at 0:40 history edited Adam P. Goucher CC BY-SA 3.0
Misplaced apostrophe
Nov 27, 2017 at 3:00 answer added Aleksandar Milivojević timeline score: 9
Mar 16, 2014 at 11:32 history edited archipelago CC BY-SA 3.0
added 408 characters in body
Mar 12, 2014 at 15:19 history edited archipelago CC BY-SA 3.0
edited body
Mar 12, 2014 at 14:23 history edited John Pardon CC BY-SA 3.0
replaced e.g. with i.e. (e.g. means "for example", you mean the latter)
Mar 12, 2014 at 13:58 answer added Greg Lupton timeline score: 5
Mar 12, 2014 at 12:50 history edited archipelago CC BY-SA 3.0
added 8 characters in body
Mar 12, 2014 at 3:13 comment added Igor Belegradek Is geometric formality preserved under rational homotopy equivalence?
Mar 12, 2014 at 0:34 history edited archipelago CC BY-SA 3.0
added 5 characters in body
Mar 11, 2014 at 20:43 comment added archipelago Note that examples of these kind cannot be homogeneous spaces or even biquotients, as they are rationally elliptic.
Mar 11, 2014 at 20:38 history edited archipelago CC BY-SA 3.0
deleted 2 characters in body
Mar 11, 2014 at 19:30 history edited Ben McKay CC BY-SA 3.0
spelling
Mar 11, 2014 at 18:42 history asked archipelago CC BY-SA 3.0