Homology of solvable (nilpotent) Lie algebras

Question

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.

Answer 1

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 $H_p(\mathfrak{g},V_\lambda)=0$.

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 $\eta$.)

Corollary. $H_*(\mathfrak{g},V_\lambda)=0$ for $\mathfrak{g}$ finite-dimensional nilpotent and $\lambda\neq 0$.

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.

Answer 2

Let $K$ be a field. Let $\mathfrak{g}$ be a Lie algebra over $K$. Let $\lambda:\mathfrak{g}\to K$ be a homomorphism. Let $V_\lambda$ be $K$ endowed with the structure of right $\mathfrak{g}$-module given by $\lambda$.

For a representation $V$ of $\mathfrak{g}$, let us say that $\lambda$ is a weight of $V$ is $V_\lambda$ is isomorphic to a subquotient of $V$ as $\mathfrak{g}$-module. This means that for some block-triangulation of the representation, $\lambda$ appears as a diagonal block. Say that $\lambda$ is a strong weight if $V_\lambda$ is isomorphic to a submodule of $V$, i.e., if there exists $y\in V\smallsetminus\{0\}$ such that $x\dot y=\lambda(x)y$ for all $x\in\mathfrak{g}$. For $\mathfrak{g}$ nilpotent and $\dim(V)<\infty$, weight and strong weight are equivalent, but not in general: for instance, for the adjoint representation of the 2-dimensional Lie algebra $\mathfrak{b}$ with basis $u,v$, $[u,v]=v$ and $\lambda(u)=0$ and $\lambda(v)=0$, the weights are $0$ and $\lambda$, but only $\lambda$ is a strong weight.

By definition, we have $H_1(\mathfrak{g},V_\lambda)=\mathrm{Ker}(\lambda)/I_\lambda$, where $I_\lambda\subset\mathfrak{g}$ is the image of $d:\Lambda^2\mathfrak{g}\to\mathfrak{g}$, $x\wedge y\mapsto \lambda(x)y-\lambda(y)x-[x,y]$.

If $\lambda=0$, then $H_1(\mathfrak{g},V_0)$ is the abelianization, and hence is zero iff $\mathfrak{g}$ is perfect (which is the same as $0$ being a strong weight of the coadjoint representation. Note that for $\mathfrak{g}$ finite-dimensional, $0$ is a weight of the adjoint/coadjoint representation iff it is not semisimple (for $\mathfrak{g}$ solvable this means iff $\mathfrak{g}\neq 0$). It is a strong weight of the adjoint representation iff $\mathfrak{g}$ has nontrivial center.

Next I assume $\lambda\neq 0$.

The non-vanishing of $H_1(\mathfrak{g},V_\lambda)$ means that there exists a linear form $f$ on $\mathfrak{g}$, not proportional to $\lambda$, such that $f([x,y])=\lambda(x)y-\lambda(y)x$ for all $x,y$. The latter condition means that $x\mapsto \lambda(x)u+f(x)v$ is a homomorphism into the 2-dimensional Lie algebra $\mathfrak{b}$, and the non-proportionality means that it is nonzero. In other words, the vanishing of $H_1(\mathfrak{g},V_\lambda)$ means that every lift $\mathfrak{g}\to\mathfrak{b}$ of $\lambda$ has a 1-dimensional image.

It is clear that the latter condition only depends on the metabelianization $\mathfrak{g}/\mathfrak{g}^{(2)}$.

Also it's clear that if $\mathfrak{g}$ is nilpotent then $\mathfrak{b}$ is not quotient of $\mathfrak{g}$, so in this case $H_1(\mathfrak{g},V_\lambda)=0$ for all $\lambda\neq 0$.

For $\mathfrak{g}$ finite-dimensional metabelian and $\lambda\neq 0$, let me check that $H_1(\mathfrak{g},V_\lambda)\neq 0$ iff $\lambda$ is weight of the adjoint representation, if and only if $\lambda$ is a strong weight of the adjoint representation.

Proof: the latter equivalence is true for metabelian Lie algebras and $\lambda\neq 0$. Indeed, if $\mathfrak{a}$ is a Cartan subalgebra and $\mathfrak{n}$ is the derived subalgebra, let $\lambda$ be a nonzero weight of $\mathfrak{g}$. So it is a strong weight of $\mathfrak{n}$ on $\mathfrak{n}$. If $y\neq 0$ and $[x,y]=\lambda(x)y$ for all $\in\mathfrak{a}$, then this is also true for $x\in\mathfrak{n}$ (because $\mathfrak{n}$ is abelian), and hence for all $x$ (since $\mathfrak{a}+\mathfrak{n}=\mathfrak{g}$).

If the $H_1(\mathfrak{g},V_\lambda)$ is nonzero, we have a surjection $\mathfrak{g}\to\mathfrak{b}$ projecting to $\lambda$ and by pull-back we deduce that $\lambda$ is a weight of $\mathfrak{g}$.

Conversely, $\lambda$ is a weight of $\mathfrak{g}$, by the above it is a strong weight. So there exists $y$ such that for all $x\in\mathfrak{g}$ we have $[x,y]=\lambda(x)y$. Since $\lambda\neq 0$, necessarily $y\in\mathfrak{g}^{(1)}$, the derived subalgebra, and this can be viewed as a condition on the adjoint representation of $\mathfrak{g}/\mathfrak{g}^{(1)}$ on $\mathfrak{g}^{(1)}$. We can decompose this representation into isotypic components, and kill all other components (for other eigenvalues as $\lambda$), and also kill an invariant hyperplane in the isotypic component of $\lambda$. After doing this, we preserve the property of $\lambda$ being a weight (maybe $y$ is not longer the same), and in addition the derived subalgebra is 1-dimensional. Using a Cartan subalgebra, we see that $\mathfrak{g}$ is then semidirect product of an abelian subalgebra $\mathfrak{a}$ with the 1-dimensional $\mathfrak{g}$, and killing the kernel of $\lambda$ in $\mathfrak{a}$ results in a 2-dimensional algebra. This quotient is precisely a 2-dimensional surjective lift of $\lambda$ onto $\mathfrak{b}$.

Corollary: for $\mathfrak{g}$ finite-dimensional and $\lambda\neq0$
$H_1(\mathfrak{g},V_\lambda)\neq 0$ iff $\lambda$ is a (strong) weight of the adjoint representation of $\mathfrak{g}$ on the metabelianization $\mathfrak{g}/\mathfrak{g}^{(2)}$.

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(a) Now here's an example for which $\lambda$ is a weight of the adjoint representation of $\mathfrak{g}$, but not of its metabelianization (so $H_1(\mathfrak{g},V_\lambda)=0$ anyway). Namely, consider a Lie algebra with basis $(s,x,y,z)$ and nonzero brackets $[s,x]=x$, $[s,y]=y$, $[s,z]=2z$, $[x,y]=z$. (This appears a parabolic subalgebra in $\mathfrak{su}(2,1)$ over the reals.) Define $\lambda$ mapping $s$ to 1 and other basis elements to 0. Then $2\lambda$ is a (strong) weight of the adjoint representation (with eigenspace generated by $z$), but not a weight of the 3-dimensional metabelianization (which only has the weights $0$ and $\lambda$), so $H_1(\mathfrak{g},V_{2\lambda})=0$. [Edit: actually $H_*(\mathfrak{g},V_{2\lambda})=0$, see (e) below.]

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Edits: here are remarks about $H_0$ and $H_d$ for $d=\dim(\mathfrak{g})$. First, we have $H_0(\mathfrak{g},V_\lambda)=0$ iff $\lambda\neq 0$.

(b) For $\mathfrak{g}$ finite-dimensional, say of dimension $d$, define $\tau_{\mathfrak{g}}(x)=\mathrm{Trace}(\mathrm{ad}(x))$ (for instance, $\mathfrak{g}$ is unimodular iff $\tau_{\mathfrak{g}}= 0$). Then $H_d(\mathfrak{g},V_\lambda)=0$ if and only if $\lambda=\tau_{\mathfrak{g}}$. However, $\tau_{\mathfrak{g}}$ is often not a weight.

(c) For instance, if $\mathfrak{g}$ is 3-dimensional with 1-dimensional abelianization and weights of the adjoint representation $0$, $\lambda$ and $t\lambda$ with $t\notin\{0,-1\}$, then $\tau_{\mathfrak{g}}=(1+t)\lambda$ is not a weight of the adjoint representation although $H_*(\mathfrak{g},V_{\tau_{\mathfrak{g}}})\neq 0$.

(d) Looking at $H_2$ also provides unimodular counterexamples. Namely, consider an $(n+1)$-dimensional Lie algebra with basis $(s,v_1,\dots,v_n)$ with $[s,v_i]=a_iv_i$. (It is unimodular iff $\sum a_i=0$.) Define $\lambda(s)=1$, $\lambda(v_i)=0$, so the weights are $0$ and the $a_i\lambda$ (which are strong weights). For $i\neq j$ we have $H_2(\mathfrak{g},V_{(a_i+a_j)\lambda})\neq 0$, although usually $(a_i+a_j)\lambda$ is not a weight; this gives unimodular counterexamples to the conjecture in dimension $n+1\ge 4$.

(e) Variant of (a): consider the Lie algebra with basis $(s,x,y,z)$ and nonzero brackets $[s,x]=ax$, $[s,y]=by$, $[s,z]=(a+b)z$, $[x,y]=z$, with $a,b,c$ scalars with $a,b,a+b\neq 0$. Let $\lambda$ map $s$ to $1$ and other basis elements to $0$. So the weights are the $t\lambda$ for $t\in\{0,a,b,a+b\}$.

Then a direct computation shows that the homology of $V_{t\lambda}$ is nonzero exactly for $t\in\{0,a,b,2a+b,a+2b,2a+2b\}$ (for $0$ the non-vanishing occurs in degree $0, 1$; for $a,b$ it occurs in degree $1$ and $2$; for $2a+b$ and $a+2b$ it occurs in degree $2$ and $3$ and for $2a+2b$ it occurs in degree $4$).

Thus in this example both implications of the conjecture fail: $(a+b)\lambda$ is a weight of the adjoint representation but $V_{(a+b)\lambda}$ has zero homology, while $t\lambda$ for $t\in\{2a+b,a+2b,2a+2b\}$ are not weights of the adjoint representation but for these values $V_{t\lambda}$ has homology.

Answer 3

To fill out my comment with a partial answer to the question.

Note that given any (left) $\mathfrak{g}$-module $V$ one can compute $H_i(\mathfrak{g},V)$ as $\mathrm{Tor}_i^{U(\mathfrak{g})}(\mathbb{C},V)$. If $S$ is any left Ore set that does not meet the augmentation ideal $I=\ker (U(\mathfrak{g})\to \mathbb{C})$ then the ring homomorphism $U(\mathfrak{g})\to \mathbb{C}$ factors through $U(\mathfrak{g})_S$. The functors $\mathbb{C}\otimes_{U(\mathfrak{g})} -$ and $\mathbb{C}\otimes_{U(\mathfrak{g})_S} (U(\mathfrak{g})_S \otimes_{U(\mathfrak{g})}-)$ are isomorphic and so have the same derived functors. Since $U(\mathfrak{g})_S$ is a flat right $U(\mathfrak{g})$-module (since $S$ was assumed a left Ore set) it follows that $$H_i(\mathfrak{g},V) = \mathrm{Tor}_i^{U(\mathfrak{g})_S}(\mathbb{C}, V_S)$$ where as usual $V_S$ denotes the localisation $U(\mathfrak{g})_S\otimes_{U(\mathfrak{g})}V$ of $V$ at $S$. In particular if $V_S=0$ then all homology groups of $V$ are trivial.

Now we come to the paper of Ken Brown numdam.org/article/CM_1984__53_3_347_0.pdf I mentioned in the comments. Theorem 5.3 tells us that the set Brown calls $\mathscr{S}'(I)$ is a left Ore set (here we use that $\mathfrak{g}$ is soluble). Chasing back through the paper we see that this set is the intersection $\mathscr{C}(I)\cap \bigcap_{\lambda\in L(I)} \mathscr{C}(\tau_\lambda(I))$. Here for a prime ideal $P$, $\mathscr{C}(P)$ denotes the elements of $U(\mathfrak{g})$ that are not zero divisors mod $P$, $L(I)$ is a sub-semigroup of $(\mathfrak{g}/[\mathfrak{g},\mathfrak{g}])^\ast$ generated by a certain explicit finite subset (containing at most $\dim \mathfrak{g}$ elements) of $\mathfrak{g}^{\ast}$ determined by the adjoint action of $\mathfrak{g}$ and $\tau_\lambda$ is the 'winding automorphism' of $U(\mathfrak{g})$ defined by $x\mapsto x+\lambda(x)$ for $x\in \mathfrak{g}$.

Now by Theorem 6.1 of the Brown paper if $S=\mathscr{S}'(I)$ then the localisation $U(\mathfrak{g})_S$ has countably many maximal one-sided ideals (which are all two sided) namely the ideals generated by the $\tau_\lambda(I)$ for $\lambda\in L(P)\cup\{0\}$. In follows that $(\mathbb{C}_\lambda)_S=0$ unless $\lambda\in L(P)\cup\{0\}$.

I believe that in the example in the question $L(P)$ consists of the elements of $(\mathfrak{g}/[\mathfrak{g},\mathfrak{g}])^\ast$ such that $x$ gets sent to a positive integer. So we see that this sufficient condition for vanishing homology is not necessary but gets you somewhere.

Perhaps I should close by noting that if $\mathfrak{g}$ is nilpotent then $L(P)=0$ and the fact that all homology groups $H_i(\mathfrak{g},V_\lambda)$ are trivial for any $\lambda\neq 0$ can be recovered.

Answer 4

This is an attempt to prove the (refined) conjecture I made in the comments of my previous answer. Let $\mathfrak{g}$ be a f.d soluble Lie algebra over $\mathbb{C}$. Let $\mathfrak{n}$ be its derived subalgebra which is nilpotent by Lie's theorem. I claim that for $\lambda\in ( \mathfrak{g}/\mathfrak{n})^\ast$, $H_i(\mathfrak{g}, \mathbb{C}_\lambda)\neq 0$ only if $\lambda$ is a sum of $i$ weights in the adjoint action.

To see this we pick a one-dimensional ideal $\mathfrak{z}=\mathbb{C}z$ in $ \mathfrak{g}$ contained in $\mathfrak{n}$. [If $\mathfrak{n}=0$ the result is easy by the now commutative methods in my previous answer.] Let's say z has weight $\mu$.

The Hochschild-Serre spectral sequence says that $H_i(\mathfrak{g}/\mathfrak{z} , H_j(\mathfrak{z} ,\mathbb{C}_\lambda))$ converges to $H_{i+j}(\mathfrak{g},\mathbb{C}_\lambda)$.

So we first understand $H_j(\mathfrak{z},\mathbb{C}_\lambda)$ as a $\mathfrak{g}$-module. This can be computed as $\mathrm{Tor}_j^{U(\mathfrak{g})}(U(\mathfrak{g})/(z), \mathbb{C}_\lambda)$ which is the $j$th homology group of $$0\to U(\mathfrak{g})\otimes_{U(\mathfrak{g})} \mathbb{C}_\lambda \to U(\mathfrak{g})\otimes_{U(\mathfrak{g})} \mathbb{C}_\lambda\to 0$$ where the non-zero map is multiplication by $z$. It follows that $$H_0(\mathfrak{z},\mathbb{C}_\lambda)=\mathbb{C}_\lambda$$ and $$H_1(\mathfrak{z}, \mathbb{C}_\lambda)=\mathbb{C}_{\lambda-\mu}.$$ By induction on $\dim\mathfrak{g}$ we're done.