Timeline for Morphism from the Koszul associative cooperad into the Koszul Lie cooperad?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 7, 2018 at 5:03 | review | Close votes | |||
| Jan 7, 2018 at 11:50 | |||||
| Jan 7, 2018 at 4:44 | comment | added | Mark.Neuhaus | It might not be appropriate, right. But I wasn't sure if this is appropriate for math.stackexchange either. I'm fine with a vote for close or a transit to math.stackexchange. Don't want to delete it though, since I think its interesting nontheless. | |
| Jan 7, 2018 at 4:42 | comment | added | Gabriel C. Drummond-Cole | Your argument is essentially correct (modulo the 0 morphism from $Com$ to $Ass$). Rather than worrying about automorphisms of $S$ you could just tensor by $S^{-1}$ to get $\theta$. But I'm not sure that this question is mathoverflow appropriate. | |
| Jan 7, 2018 at 4:41 | comment | added | Mark.Neuhaus | I mean the Koszul dual of the Lie operad (in contrast to the linear dual). It's personal taste, I justthink it sounds better. | |
| Jan 7, 2018 at 4:38 | comment | added | Gabriel C. Drummond-Cole | It seems nonstandard to call $Lie^i$ the "Koszul Lie cooperad" There is a cooperad governing Lie coalgebras and it's not that one. | |
| Jan 7, 2018 at 4:36 | comment | added | Mark.Neuhaus | Because of the universality of the tensor product. Maybe there are other automorphism on $\mathcal{S}$ then $1$, but that wouldn't change the argument I think. | |
| Jan 7, 2018 at 4:02 | history | asked | Mark.Neuhaus | CC BY-SA 3.0 |