Timeline for History of Koszul complex
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Jan 13, 2017 at 3:20 | history | edited | Gabriel C. Drummond-Cole | CC BY-SA 3.0 |
fixed a few typos
|
| S Jan 13, 2017 at 3:20 | history | suggested | New learner | CC BY-SA 3.0 |
fixed a few typos
|
| Jan 13, 2017 at 3:02 | review | Suggested edits | |||
| S Jan 13, 2017 at 3:20 | |||||
| Jul 5, 2016 at 4:53 | comment | added | Jonathan Gleason | I am confused as to what is meant here. The Chevalley-Eilenberg complex I know is a cochain complex, whereas the Koszul complex I know is a chain complex (that is, sans "co"). It seems that maybe application of $\mathrm{Hom}_R(-,R)$ to the Koszul complex I have in mind would yield the Chevalley-Eilenberg complex you mention. I suppose for $R$ a field and in finite-dimensions no information is lost by doing so, but in general it seems there would be. Could you clarify what you are referring to when you say CE complex and Koszul complex? | |
| Nov 1, 2013 at 14:20 | comment | added | YCor | @Sasha Pavlov: there are two incompatible definitions of metabelian in the air. In group theory it essentially always mean that the derived subgroup is abelian. Possibly Lie algebra theory, I think both definitions are still in use (see arxiv.org/pdf/1302.0825v1.pdf, the first given by google for "metabelian lie algebra", for an example where it means 2-step solvable). | |
| Oct 31, 2013 at 12:45 | comment | added | Sasha Pavlov | Marino, quantifier is missing. Koszul complex for any linear form is a Chevalley-Eilenberg complex for Lie algebra of that form. | |
| Oct 31, 2013 at 12:41 | comment | added | Sasha Pavlov | Yves, thanks for remark on affine homotheties, it did not occur to me. But I think definition for metabelian is different: metabelian is 2-step nilpotent, not 2-step solvable. | |
| Oct 31, 2013 at 6:16 | comment | added | Mariano Suárez-Álvarez | The Koszul complex is the special case of the Chevalley-Eilenberg complex for a less weird example: the abelian Lie algebra! | |
| Oct 30, 2013 at 15:51 | vote | accept | Sasha Pavlov | ||
| Oct 30, 2013 at 13:19 | comment | added | YCor | For your last question: note that the Lie algebras you get are pretty special since they are metabelian in the sense that the derived subalgebra is abelian. In case $R=K$ is a field and $f\neq 0$, the computation of this Lie algebra is rather straightforward: it is the Lie algebra of the group of affine homotheties of the hyperplane Ker($f$). | |
| Oct 30, 2013 at 7:24 | answer | added | Faisal | timeline score: 24 | |
| Oct 30, 2013 at 1:55 | history | asked | Sasha Pavlov | CC BY-SA 3.0 |