All Questions
6 questions
4
votes
1
answer
133
views
Second cohomology group of the contact Lie algebra $K_3$
Let $F$ be a field of characteristic zero and, for all $n>0$, consider the contact Lie algebra $K_{2n+1}$. It follows from Corollary 3 of the paper [V. Guillemin - S. Shnider: Some stable results ...
4
votes
0
answers
75
views
Lie algebra "semi" coinvariants
In the process of my research, I've come across the need to understand the following construction:
Let $\mathfrak{g}$ be a (finite-dimensional) complex Lie algebra, $\beta\in \mathfrak{g}^*$ a Lie ...
1
vote
1
answer
129
views
Lie algebra cohomology: $H^i(R,V)=H^i(R,V^R)$ with $R$ reductive and $V$ an $R$-module
Let $R$ be a reductive, finite-dimensional Lie algebra over a field of characteristic 0, and let $V$ be a semisimple $R$-module (also finite dimensional). I have seen a reference to the fact that $H^i(...
9
votes
3
answers
1k
views
Poincaré duality for (co)homology of Lie algebras?
Let $R$ be a commutative ring and $\mathfrak{g}$ a Lie $R$-algebra that has an $R$-module basis with $n$ elements.
In Algebra, Geometry, and Software Systems by Joswig & Takayama on p.200, it ...
0
votes
1
answer
128
views
Lie algebra cohomology with values in injective module
I am looking for a proof of the following result: Let $\mathfrak{g}$ be a Lie algebra and $I$ an injective $\mathfrak{g}$-module. Then $\mathrm{H}^q(\mathfrak{g},I)=0$ $\forall q>0$. More precisely,...
6
votes
0
answers
465
views
Lagrangian (classical) BRST cohomology groups
I'm trying to understand BRST complex in its Lagrangian incarnation i.e. in the form mostly closed to original Faddeev-Popov formulation. It looks like the most important part of that construction (...