Timeline for "Lie algebra" for a general group ?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 31, 2014 at 16:00 | comment | added | Zuriel | @Ralph, would you please give some reference to these constructions? | |
| Jun 22, 2012 at 1:31 | comment | added | Christopher Drupieski | In your answer above, I believe Quillen's isomorphism involves the restricted enveloping algebra of the (restricted) Lie algebra $\text{gr}^p(G)$. This is a finite-dimensional algebra, whereas the usual universal enveloping algebra is infinite-dimensional. The statement about cohomology then refers to the cohomology of $\text{gr}^p(G)$ as a restricted Lie algebra. | |
| Apr 26, 2012 at 17:53 | comment | added | Ralph | I checked it. It's an isomorphism of $k$-vector spaces. More specifically, there is a filtration on the group cohomology such that $H^i(\operatorname{gr}^p(G);k)\cong \operatorname{gr}H^i(G;k)$, but I didn't check if the latter preserves products. | |
| Apr 25, 2012 at 8:33 | comment | added | Ralph | The isomorphism above is the cohomological analogue of Theorem 4.3 ii) in Grünenfelder: On the homology of filtered and graded rings, J. Pure and Appl. Algebra, 14(1979), 21-37. As homology and cohomology are dual to each other (over a field) the isomorphism is true at least additively. But I guess it is also an isomorphism of rings. I'll have a look later and will also look into the Friedlander-Parshall paper. | |
| Apr 25, 2012 at 7:56 | comment | added | M T | There is a spectral sequence computing $H^*(G,k)$ from $H^*(\operatorname{gr}^p(G),k)$, but they won't be isomorphic as rings in general: the latter is finitely generated over its subring generated by degree two elements (e.g. Friedlander and Parshall "Geometry of p-unipotent Lie algebras" J.Alg. 109) but $H^*(G,k)$ may not have this property. If they did all periodic $kG$-modules would have period at most two, but examples of larger period are known. | |
| Apr 24, 2012 at 22:39 | history | answered | Ralph | CC BY-SA 3.0 |