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Jun 5, 2017 at 14:20 vote accept Cepu
Jun 5, 2017 at 12:38 answer added Daniel Robert-Nicoud timeline score: 1
Jun 5, 2017 at 7:33 history edited Cepu CC BY-SA 3.0
some typo corrected
Jun 5, 2017 at 7:32 comment added Cepu Ups is a typo. it is exaktaly what I mean. I agree but $A\widehat{\otimes}\mathfrak{g}$ may be not pronilpotent since the Lie algebras $A\widehat{\otimes}\mathfrak{g}_{n}$ are not finite dimensional.
Jun 5, 2017 at 0:12 comment added Pavel Safronov I guess you mean $A$ is a commutative algebra rather than a Lie algebra. What do you mean by $\hat{\otimes}$? Is it just $A\hat{\otimes} \lim \mathfrak{g}_n = \lim (A\otimes \mathfrak{g}_n)$? Given any cdga $A$ and a nilpotent Lie algebra $\mathfrak{g}_n$, $A\otimes \mathfrak{g}_n$ is nilpotent.
Jun 4, 2017 at 15:36 history edited Cepu CC BY-SA 3.0
added 437 characters in body
Jun 4, 2017 at 15:18 history asked Cepu CC BY-SA 3.0