Can you please help me out to prove the following (rather easy (but not for me obviously)) exercise which I didn't manage to solve (Exercise 7.4.2, page 229)?
Assume $\mathfrak{g}$ is the free $\mathbb{k}$-module on basis $\{ e_1, e_2,..., e_n \}$, made it into abelian Lie algebra. Show that $H_p(\mathfrak{g}, \mathbb{k}) = \bigwedge^k (\mathfrak{g})$, the $p^{th}$ exterior power of the $\mathbb{k}$-module $\mathfrak{g}$.
Thank you!
On
let $L$ be a Lie algebra over $K$. The complex $(C_{\bullet},\partial)$ is given by $C_p(L,V)=\Lambda^pL\otimes_KV$, where $V$ is an $L$-module, and $\partial_p$ is a boundary operator. For the trivial module $V=K$ these space of $p$-chains is just $\Lambda^pL$, and since $L$ is abelian, the boundary map $\partial_p$ is zero, so that we have $$ H_p(L,K)=\ker(\partial_p)/im (\partial_p)=\ker(\partial_p)=C_p(L,K)=\Lambda^pL. $$
The $a$ be a finite dimensional abelian Lie algebra, let $\{x_1,\dots,x_n\}$ be a basis for $a$ and let $U$ be its enveloping algebra. It is easy to see that $U$ is a polinomial ring with generators $x_1,\dots,x_n$. You can check that $x_1,\dots,x_n$ is a regular sequence in $U$, so using the result of Chapter 4 in the book that refer to th Koszul complex, you get a free resolution of the module $U/(x_1,\dots,x_n)$, which is just th trivial module over $a$, as a let $U$-module. Use that resolution to compute $Ext_U(k,k)$, which is the Lie cohomology $H^*(a,k)$.