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Jun 5, 2013 at 22:14 comment added Fernando Muro All stable rational cohomology operations are trivial, i.e. $H\mathbb Q^{\ast}H\mathbb Q$ is $\mathbb Q$ concentrated in degree $0$.
Jun 5, 2013 at 15:09 vote accept Sean Tilson
Jun 5, 2013 at 15:07 vote accept Sean Tilson
Jun 5, 2013 at 15:08
Jun 5, 2013 at 11:34 answer added John Rognes timeline score: 21
Aug 20, 2012 at 23:27 answer added Martin Frankland timeline score: 12
Dec 28, 2010 at 4:32 answer added Charles Rezk timeline score: 11
Dec 28, 2010 at 4:31 answer added Tom Goodwillie timeline score: 32
Dec 27, 2010 at 23:02 history asked Sean Tilson CC BY-SA 2.5