Let $\mathfrak{g}$ be a solvable Lie algebra over $\mathbb{C}$ and $\lambda\in(\mathfrak{g}/[\mathfrak{g},\mathfrak{g}])^\*$ be a character of $\mathfrak{g}$. I'm interested in calculating homology for $\mathbb{C}\_\lambda$, the one-dimensional $\mathfrak{g}$-module with character $\lambda$.


I have calculated homology manually for the 2-dimensional algebra $\mathfrak{g}=\langle x,y \rangle$ with relation $[x,y]=y$. The thing that surprised me is that homology is nontrivial only for $\lambda(x)=0$ or $1$. 


In general, I conjecture is that the homology is nontrivial, iff $\lambda$ is a weight of the adjoint representation. I can't prove it (or find counterexample). I've tried to find the answer in books, but there is a lack on literature in homology theory for solvable lie algebras.



Homology of solvable (nilpotent) Lie algebras
Let me provide an elementary proof that for $\mathfrak{g}$ nilpotent finite-dimensional and $\lambda\neq 0$ we have $H\_\*(\mathfrak{g},V\_\lambda)=0$. (In general it seems to be particular case of results of Delorme in the 70s, and possibly known earlier.)


Recall that the homology of $V\_\lambda$ is the homology of the Chevalley-Eilenberg complex, with $C\_p=\Lambda^p\mathfrak{g}$, and $d\_p:C\_p\to C\_{p-1}$ is given by 
$$x\_1\wedge\dots\wedge x\_p\mapsto \sum\_{i=1}^p(-1)^{i+1}\lambda(x\_i)x\_1\wedge\dots\wedge \hat{x\_i}\wedge\dots\wedge x\_p+$$ $$+\sum\_{1\le i<j\le p}[x\_i,x\_j]\wedge x\_1\wedge\dots\wedge \hat{x\_i}\wedge\dots\wedge \hat{x\_j}\wedge\dots\wedge x\_p.$$ 


Let $\mathfrak{z}(\mathfrak{g})$ be the center of $\mathfrak{g}$. Then for all $z\in\mathfrak{z}(\mathfrak{g})\cap [\mathfrak{g},\mathfrak{g}]$ and $u\in\Lambda^{p-1}\mathfrak{g}$ we have $d\_p(u\wedge z)=d\_{p-1}(u)\wedge z$ (because all remaining terms include $[x\_i,z]$ or $\lambda(z)$ which are zero). The induction step is the following:



> 
> **Lemma.** Fix $p\ge 1$. Let $z$ be an element of $\mathfrak{z}(\mathfrak{g})\cap [\mathfrak{g},\mathfrak{g}]$. 
>  Suppose that $H\_i(\mathfrak{g}/Kz,V\_\lambda)$ is zero for $i\in\{p-1,p\}$. Then $H\_p(\mathfrak{g},V\_\lambda)=0$.
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Let $f$ be a cycle in degree $p$. Modulo $z$ it is a $p$-boundary. This means that there exists $g\in\Lambda^{p+1}\mathfrak{g}$ such that $d\_{p+1}(g)=f$ in $\lambda^p\mathfrak{g}/Kz$, which means that $d\_{p+1}(g)=f+\mu\wedge z$ for some $\mu\in\Lambda^{p-1}\mathfrak{g}$. Since $d\_pf=0$, we moreover have $d\_{p-1}(\mu)\wedge z=d\_p(\mu\wedge z)=0$. This means that $d\_{p-1}(\mu)=0$ in $\Lambda^{p-2}(\mathfrak{g}/Kz)$. Hence, there exists $\xi\in\Lambda^{p}\mathfrak{g}$ such that $d\_{p}\xi=\mu$ in $\Lambda^{p-1}(\mathfrak{g}/Kz)$. This means that $d\_p\xi=\mu+\eta\wedge z$ for some $\eta\in\Lambda^{p-2}\mathfrak{g}$. So $d\_{p+1}(\xi\wedge z)=d\_p(\xi)\wedge z=\mu\wedge z$ and hence $f$ is a boundary. (If $p=1$, we directly have $d\_0\mu=0$ since $d\_0=0$ so no need to introduce $\eta$.)



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> **Corollary.** $H\_\*(\mathfrak{g},V\_\lambda)=0$ for $\mathfrak{g}$ finite-dimensional nilpotent and $\lambda\neq 0$.
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Given that $H\_0(\mathfrak{g},V\_\lambda)=0$ for every $\lambda\neq 0$, the lemma reduces, by induction, to the case when $\mathfrak{g}$ is abelian. In this case, we choose the basis $(e\_0,\dots,e\_n)$ with $\lambda(e\_0)=1$, $\lambda(e\_i)=1$ for $i\ge 1$, and view the Lie algebra as graded in $\mathbf{Z}^n$ with $e\_i$ of grade $u\_i$ (the canonical basis of $\mathbf{Z}^n$) and $e\_0$ of grade $0$, noting that the boundary map preserves the grading. In $\Lambda^i\mathfrak{g}$, we have the grades $u\_I=\sum\_{i\in I}u\_i$ when $I$ is a subset of $\{1,\dots,n\}$ of cardinal $i$ or $i-1$. If $I$ has cardinal $i$, $(\Lambda^i\mathfrak{g})\_{u\_I}$ has dimension $1$, generated by the $x\_I=\wedge\_{i\in I}e\_i$, which is a $i$-cycle, and is also a boundary, namely of $e\_0\wedge x\_I$ (up to sign). If $I$ has cardinal $i-1$, $(\Lambda^i\mathfrak{g})\_{u\_I}$ has dimension $1$, generated by $y\_I=e\_0\wedge x\_I$, which has a nonzero image by $d\_i$. This proves the vanishing of all the homology.


