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Apr 22, 2013 at 0:19 vote accept Clay Cordova
Apr 18, 2013 at 4:00 answer added Theo Johnson-Freyd timeline score: 2
Apr 18, 2013 at 3:50 comment added Theo Johnson-Freyd "if G˜ is the simply connected universal cover of G then all representations of g can be integrated to representations of G". You should include "finite-dimensional" somewhere in that sentence. The Lie algebra $\mathbb R$ acts on $\mathcal C^\infty(I)$, where $I$ denotes the open unit interval $I = (0,1)$, by sending the basis vector to $\frac{\partial}{\partial x}$, but this representation is not integrable to a representation of $\mathbb R$ on $\mathcal C^\infty(I)$.
Apr 17, 2013 at 11:53 answer added Allen Knutson timeline score: 8
Apr 17, 2013 at 3:04 answer added Sam Gunningham timeline score: 16
Apr 17, 2013 at 2:18 history asked Clay Cordova CC BY-SA 3.0