Timeline for Homology of solvable (nilpotent) Lie algebras
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| May 8, 2019 at 8:24 | comment | added | Simon Wadsley | Agreed. In fact I think that when $P=I$, the augmentation ideal which is the important case here, the generators of $L(P)$ are precisely the weights of adjoint representation. I suppose it remains possible that the bad $\lambda$ are all a linear combination of at most $\dim \mathfrak{g}$ of these weights with repeats allowed (although the converse is not true). Perhaps even $H^i(\mathfrak{g},\mathbb{C}_\lambda)$ is zero unless $\lambda$ is a linear combination of $i$ of the weights (but not conversely). | |
| May 7, 2019 at 21:16 | comment | added | YCor | I doubt this conjecture would match my example in (e) of my long post (for which the "bad" $\lambda$ are precisely $0$, $A$, $B$, $A+2B$, $2A+B$, and $2A+2B$ for suitable $A,B$, but not $A+B$). | |
| May 7, 2019 at 20:38 | comment | added | Simon Wadsley | If I were in the mood for making wild conjectures I might speculate that all homology groups of $\mathbb{C}_\lambda$ are trivial precisely if $\lambda$ is a sum of at most $\dim \mathfrak{g}/[\mathfrak{g,g}]$ of the generators of $L(I)$ given by Brown (if the sign was wrong before it is also wrong here). However I have no evidence for this. | |
| May 7, 2019 at 20:25 | comment | added | Simon Wadsley | It is possible that I've made sign errors near the end and that I mean that $(\mathbb{C}_{-\lambda})_S=0$ unless $\lambda\in L(P)\cup \{0\}$ and then $L(P)$ is the negation of what I claimed in the specific example. | |
| May 7, 2019 at 20:22 | history | answered | Simon Wadsley | CC BY-SA 4.0 |