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what is Koszul resolution? what is its role played in the computation of spectral sequence?

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closed as not a real question by Yemon Choi, S. Carnahan Jun 23 '10 at 14:52

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

You should tell us what you want to know, exactly... that a Google search will not answer. – Mariano Suárez-Alvarez May 26 '10 at 15:00

The Koszul complex is defined at . In certain cases, one of which is explained there, the Koszul complex is a resolution (typically a free resolution, see

There is a complicated history of the Koszul complex, but really it began with Lie algebra cohomology, before becoming (also) a tool generally used in commutative algebra. Any free or projective resolution might appear in a spectral sequence argument: that query is not very specific. Maybe you want some standard argument from Lie algebra cohomology?

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More globally the Koszul complex gives you a locally free resolution of $\mathcal{O}Z$ where $Z \subset X$ is the zero-set of a section of a vector bundle. This is explained, e.g. in the Appendix of Fultons Book on Interseciton theory. – Heinrich Hartmann May 26 '10 at 16:09

When doing things related to free resolutions you may want the actual computation or you may want something highly structured. For the former you use the koszul resolution, it is nice and little and small. If you want a lot of structure you use the Bar resolution. This is sort of a philosophical things, so when you actually want to compute something you use the Koszul resolution since it is pretty small and you know that making this choice wont affect your answer.

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