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Jun 3, 2013 at 10:46 comment added Dima Sustretov Thanks, Ralph. This seems to be the closest to what I have been hoping for. It's a pity that there is no natural pairing $H^{ab} \times H^1(H, A) \to A$, but alas, this how the things are.
Jun 3, 2013 at 8:54 vote accept Dima Sustretov
Jun 1, 2013 at 12:08 comment added Mark Grant @Ralph: I fixed a typo by changing an $H$ to a $Q$, I hope it's OK.
Jun 1, 2013 at 12:06 history edited Mark Grant CC BY-SA 3.0
fixed typo
May 31, 2013 at 18:09 history answered Ralph CC BY-SA 3.0