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Yes, it is true. In fact, it is true that $H^i(K_{2n+1},F)=0$ for $0<i\le 2n$. This can be deduced from the theorem of Feigin sketched in Feigin, B.L. Cohomology of contact Lie algebras. (Russian) C. …

The answer to this question may be well-known, but I failed to locate it in any obvious source. From the results of Bott (Ann. Math. 66 (1957), 203-248) and Kostant (Ann. Math. 74 (1961), 329-387), it …

A very useful source to learn about cohomology of infinite-dimensional Lie algebras is the book by D.B.Fuks "Cohomology of Infinite-Dimensional Lie Algebras" (shocking, I know). This source discusses …

To offer a slightly more geometric viewpoint on the same, the space $\bigoplus_q \mathop{\mathrm{Hom}}_K(\Lambda^q(\mathfrak{p}),\mathbb{C})$, which is the direct sum of all spaces you are considering …

The way the question is formulated, it is trivial. If $V\ne\mathfrak{g}$ ($\mathfrak{g}$ here being the adjoint module, in which case you already know everything), the corresponding sum is clearly d …

First of all, you might want to look at the MO question How to define the equivalence of Maurer-Cartan elements in an $L_\infty$-algebra?, since of course this is effectively what you need (morphisms …

First, let me expand on the reply of Dietrich Burde: I got hold of the paper of Hazewinkel, and can now be more precise about what is and what is not there (last time I saw it was some years ago). …

There is one silly answer to your question, and you are probably aware of it: if you use as coefficients $S(g)$, the symmetric algebra of $g$, and not just $g$, everything will work wonderfully. Alas …

Cohomology does not change under field extensions, so just extend everything to the algebraic closure to prove your result in char 0 in general. Of course, field extensions do not preserve irreducibil …

The other comment revealed most of the story. Let me add some points. First, the Chevalley-Eilenberg complex is defined for the most general case of $H^\bullet(g,M)$ --- cohomology with coefficients i …