Lie Algebra Cohomology and Kostants Nilradical Theorem

By [F4] fix a $k$-basis $B$ of $\mathfrak{a}$, well ordered by a total order $\le$. For $q\ge 0$ let $T_q$ be the set of strictly increasing $q$-tuples $i_1<\cdots <i_q$ in $B$, and for $i\in B$ write $e_i$ for the corresponding basis vector. The wedges $e_I=e_{i_1}\wedge \cdots \wedge e_{i_q}$, $I\in T_q$, form a $k$-basis of ${\mathrm{\Lambda }}^q\mathfrak{a}$: they span because ${\mathrm{\Lambda }}^q\mathfrak{a}$ is spanned by decomposables $x_1\wedge \cdots \wedge x_q$ and expanding each $x_j$ in $B$ multilinearly writes every such wedge as a finite sum of wedges of basis vectors, each of which is $\pm$ a wedge $e_I$ or zero; and they are independent because for each fixed $J\in T_q$ the alternating $q$-linear form ${\phi }_J(x_1,\dots ,x_q)=\det ({\lambda }_{j_r}(x_s){)}_{r,s}$, built from the coordinate functionals ${\lambda }_j(e_i)={\delta }_{ij}$, factors through ${\mathrm{\Lambda }}^q\mathfrak{a}$ and satisfies ${\phi }_J(e_I)={\delta }_{IJ}$, so a linear relation ${\sum }_I{c}_I{e}_I=0$ gives $c_J={\phi }_J({\sum }_I{c}_I{e}_I)=0$ for every $J$.

Put $P_q=U(\mathfrak{a}){\otimes }_k{\mathrm{\Lambda }}^q\mathfrak{a}$. Since $U(\mathfrak{a})$ is a $(U(\mathfrak{a}),k)$-bimodule, [F6] makes $P_q$ a left $U(\mathfrak{a})$-module with $s(u\otimes \omega )=(su)\otimes \omega$, and since the wedges $e_I$ form a $k$-basis of ${\mathrm{\Lambda }}^q\mathfrak{a}$, the map $U(\mathfrak{a}{)}^{(T_q)}\to P_q$ sending the standard basis vector at $I$ to $1\otimes e_I$ is an isomorphism of left $U(\mathfrak{a})$-modules: it is $U(\mathfrak{a})$-linear and surjective, and the balanced map $(u,\omega )\mapsto (c_{I}u{)}_{I\in T_q}\in U(\mathfrak{a}{)}^{(T_q)}$, where $\omega ={\sum }_I{c}_I{e}_I$, induces an inverse by the universal property of the tensor product. Hence every $P_q$ is a free $U(\mathfrak{a})$-module. Define ${\mathrm{\partial }}_q:P_q\to P_{q-1}$ for $q\ge 1$ on decomposable arguments by ${\mathrm{\partial }}_q(u\otimes x_1\wedge \cdots \wedge x_q)={\sum }_{r=1}^q(-1{)}^{r-1}u{x}_r\otimes x_1\wedge \cdots \widehat{x_r}\cdots \wedge x_q+{\sum }_{1\le r<s\le q}(-1{)}^{r+s}u\otimes [x_r,x_s]\wedge x_1\wedge \cdots \widehat{x_r}\cdots \widehat{x_s}\cdots \wedge x_q,$ and set ${\mathrm{\partial }}_0=\epsilon :U(\mathfrak{a})\to k$ for the algebra augmentation supplied by the universal property of $U(\mathfrak{a})$ from the zero map on $\mathfrak{a}$ ([F6], Universal enveloping algebra); the formula is $k$-multilinear and alternating in $x_1,\dots ,x_q$ and balanced in $u$, so it defines a $U(\mathfrak{a})$-linear map $P_q\to P_{q-1}$ on the tensor product of the free module with the exterior power.

For a left $U(\mathfrak{a})$-module $W$, evaluation on $1\otimes \omega$ is a natural bijection ${\mathrm{Hom}}_{U(\mathfrak{a})}(P_q,W)\to C^q(\mathfrak{a},W)$, $\phi \mapsto (x_1\wedge \cdots \wedge x_q\mapsto \phi (1\otimes x_1\wedge \cdots \wedge x_q))$, with inverse $f\mapsto (u\otimes \omega \mapsto u\cdot f(\omega ))$; it is the Hom-tensor adjunction for the free module on the basis $\{e_I\}$. Under this identification the precomposition $\phi \mapsto \phi \circ {\mathrm{\partial }}_{q+1}$ corresponds to the Chevalley–Eilenberg differential: for $f\in C^q(\mathfrak{a},W)$ and $x_0,\dots ,x_q\in \mathfrak{a}$, $\phi ({\mathrm{\partial }}_{q+1}(1\otimes x_0\wedge \cdots \wedge x_q))$ equals ${\sum }_{i=0}^q(-1{)}^i{x}_i\cdot f(x_0,\dots ,\widehat{x_i},\dots ,x_q)+{\sum }_{0\le i<j\le q}(-1{)}^{i+j}f([x_i,x_j],x_0,\dots ,\widehat{x_i},\dots ,\widehat{x_j},\dots ,x_q)$, which is exactly $(df)(x_0,\dots ,x_q)$ for the zero-based formula of [F1]. Hence ${\mathrm{Hom}}_{U(\mathfrak{a})}(P_{\bullet },W)$ is the Chevalley–Eilenberg complex $C^{\bullet }(\mathfrak{a},W)$.

Use the fixed ordered basis $B$ of the arbitrary Lie algebra $\mathfrak{a}$, and identify $S(\mathfrak{a})$ with the polynomial ring $k[x_i:i\in B]$ by [F5]. Put $F_N{P}_q=F_{N-q}U(\mathfrak{a}){\otimes }_k{\mathrm{\Lambda }}^q\mathfrak{a}$, with $F_{p}U=0$ for $p<0$. This filtration is increasing, exhaustive and bounded below in each degree. The action term of $\mathrm{\partial }$ raises PBW degree by one and lowers exterior degree by one, preserving total degree $N$; the bracket term lowers total degree by one. Therefore the associated graded differential is the Koszul differential on $S(\mathfrak{a}){\otimes }_k{\mathrm{\Lambda }}^{\bullet }\mathfrak{a}$: on each basis wedge it is the finite sum $\delta (f\otimes e_{i_1}\wedge \cdots \wedge e_{i_q})={\sum }_{r=1}^q(-1{)}^{r-1}f{x}_{i_r}\otimes e_{i_1}\wedge \cdots \widehat{e_{i_r}}\cdots \wedge e_{i_q}$. Its augmentation evaluates all variables at zero, giving $k$.

The chain identity is ${\mathrm{\partial }}_q{\mathrm{\partial }}_{q+1}=0$ for $q\ge 1$. In step 2.1 take $W=P_{q-1}$: the composite of the precomposition maps from $\mathrm{Hom}(P_{q-1},W)$ to $\mathrm{Hom}(P_{q+1},W)$ is the square of the Chevalley–Eilenberg differential, hence zero by [F1]. Applying it to ${\mathrm{id}}_{P_{q-1}}$ gives ${\mathrm{\partial }}_q{\mathrm{\partial }}_{q+1}=0$. Also $\epsilon {\mathrm{\partial }}_1=0$ because $\epsilon (ux)=0$ for $x\in \mathfrak{a}$. Thus the augmented $P_{\bullet }$ is a chain complex of free modules.

For every finite $S\subseteq B$, let $R_S=k[x_i:i\in S]$ and use the induced order on $S$. The ordered variables form an $R_S$-regular sequence: multiplication by each variable is injective on the polynomial ring in the variables not yet removed, as it shifts the corresponding monomial exponent by one; the successive quotients remove that variable. By [F7], the augmented Koszul complex $R_S{\otimes }_k{\mathrm{\Lambda }}^{\bullet }({\mathrm{span}}_k\{e_i:i\in S\})\to k$ is exact. Inclusions $S\subseteq T$ give inclusions of these complexes compatible with their augmentations. Every polynomial and exterior tensor has finite support in $B$, so their filtered colimit is exactly the augmented graded complex of step 2.2. Exactness of filtered colimits [F8] proves this graded complex exact, including its degree-zero augmentation. This argument uses finite polynomial-variable subcomplexes, not finite-dimensional Lie subalgebras. The differential preserves total polynomial-plus-exterior degree, so each homogeneous total-degree component is also exact.

The augmented complex $P_{\bullet }\to k$ is exact, including at $P_0$. The zero element is already a boundary. Let $z\mathrm{\ne }0$ in $P_q$ be a cycle with $q\ge 1$, or let $z\in \ker \epsilon \subseteq P_0$; let $N$ be minimal with $z\in F_N{P}_q$ and let $\bar{z}\in {\mathrm{gr}}_N{P}_q$ be its symbol. Since $\mathrm{\partial }z=0$ and the induced graded differential sends the symbol of an element to the symbol of its image, $\delta \bar{z}=0$; by exactness of the augmented associated graded complex from step 3.2, $\bar{z}=\delta \bar{w}$ for some $\bar{w}\in {\mathrm{gr}}_N{P}_{q+1}$ in the cycle case, or for $q=0$, $N\ge 1$ and $\bar{z}\in \ker ({\mathrm{gr}}_N{P}_0\to k)=\mathrm{im}\delta$ (and if $q=0$ and $N=0$ then $z\in F_0{P}_0=k\cdot 1$ with $\epsilon (z)=0$, so $z=0$). Choose a lift $w\in F_N{P}_{q+1}$ of $\bar{w}$; then $z-\mathrm{\partial }w\in F_{N-1}{P}_q$ is again a cycle and $\epsilon (z-\mathrm{\partial }w)=0$ in degree zero. Iterating this reduction finitely many times (each iteration lowers $N$ by at least one, and only finitely many choices of lifts are made) arrives at an element of $F_{q-1}{P}_q$, which is zero for $q\ge 1$, or at an element of $F_0{P}_0\cap \ker \epsilon =0$ for $q=0$. Hence $z\in \mathrm{im}{\mathrm{\partial }}_{q+1}$, and since $\mathrm{im}{\mathrm{\partial }}_{q+1}\subseteq \ker {\mathrm{\partial }}_q$ by step 3.1, this proves exactness at every degree.

Steps 2.2 and 3.2 apply to arbitrary $\mathfrak{a}$ without a finite-dimensionality hypothesis. The filtration reduction of step 4.1 terminates because each individual tensor has finite PBW degree. Consequently the augmented $P_{\bullet }\to k$ is exact for every Lie algebra: $\ker {\mathrm{\partial }}_q=\mathrm{im}{\mathrm{\partial }}_{q+1}$ for $q\ge 1$, $\ker \epsilon =\mathrm{im}{\mathrm{\partial }}_1$, and $\epsilon$ is surjective since $\epsilon (1)=1$.

By steps 1.2, 2.1 and 5.1, $P_{\bullet }\to k$ is a resolution of $k$ by free $U(\mathfrak{a})$-modules, hence by [F6] and the Axiom of Choice a projective resolution. Taking $W=V$ in step 2.1, the complex ${\mathrm{Hom}}_{U(\mathfrak{a})}(P_{\bullet },V)$ is the Chevalley–Eilenberg complex $C^{\bullet }(\mathfrak{a},V)$, so $H^n{\mathrm{Hom}}_{U(\mathfrak{a})}(P_{\bullet },V)\cong H^n(\mathfrak{a},V)$ for every $n\ge 0$. Iterating the free module on the underlying set of each kernel gives a projective resolution of every module, since these free modules are projective by [F6]. Injective resolutions exist by Every module admits an injective resolution under the assumed Axiom of Choice. Thus the module category has enough projectives and injectives, as required by the balanced Ext convention [F9]. Fix supplied resolution data on the objects being compared, with $Q_{\bullet }(k)=P_{\bullet }$; by [F9] and Dependent Choice, available by [F3], every supplied projective resolution datum on the same class computes Ext groups naturally isomorphic to those of $Q$, so ${\mathrm{Ext}}^n(k,V)\cong H^n\mathrm{Hom}(P_{\bullet },V)\cong H^n(\mathfrak{a},V)$ for every $n$. The isomorphisms are natural in $V$ because the identification of step 2.1 is natural in the coefficient module and the comparison maps of [F9] are natural. This proves the statement; no finite-dimensionality, field characteristic or coefficient hypothesis was used beyond the displayed ones.

By [L1] an element of $C^0(\mathfrak{a},V)$ is the same as a vector $v\in V$, and $d^{0}v=0$ means exactly that the linear map $x\mapsto x\cdot v$ vanishes, that is $x\cdot v=0$ for every $x\in \mathfrak{a}$. Hence $\ker d^0=\{v\in V:x\cdot v=0\text{ for every }x\in \mathfrak{a}\}$.

The space $C^{-1}(\mathfrak{a},V)$ is zero by [L1], so $\mathrm{im}{d}^{-1}=0$; substituting into [L3] gives $H^0(\mathfrak{a},V)=\ker d^0$, and step 1.1 identifies this with the invariant subspace $V^{\mathfrak{a}}$.

For the trivial module the action on $k$ is zero by definition, so every vector is invariant and $H^0(\mathfrak{a},k)=k$; this is the special case $M=k$ of [L4]. If $\mathfrak{a}=0$ the condition $x\cdot v=0$ is vacuous, so $H^0(0,V)=V$; if $V=0$ both sides are zero.

For fixed $x\in \mathfrak{p}$ the formula of the Statement defines a $k$-linear map ${\theta }_x:C^q(\mathfrak{a},V)\to C^q(\mathfrak{a},V)$: both displayed terms are $k$-linear in $\omega$, and they are alternating in the arguments because $\omega$ is and because the substitution $x_j\mapsto [x,x_j]$ is linear in the slot; the bracket lies in $\mathfrak{a}$ by [F4], so the expression is a legitimate cochain. Moreover $x\mapsto {\theta }_x$ is $k$-linear, since both the action on $V$ and the bracket are.

Write $A_x\omega =x\cdot \omega$ for action on values and $B_x\omega ={\sum }_j\omega (\dots ,[x,x_j],\dots )$ for action on argument slots, so ${\theta }_x=A_x-B_x$. Value and slot operators commute. By [F2], $[A_x,A_y]=A_{[x,y]}$. In $[B_x,B_y]$, substitutions in distinct slots cancel pairwise; the same-slot difference is $\omega (\dots ,[y,[x,x_j]]-[x,[y,x_j]],\dots )=-\omega (\dots ,[[x,y],x_j],\dots )$ by [F3]. Thus $[B_x,B_y]=-B_{[x,y]}$, and $[{\theta }_x,{\theta }_y]=A_{[x,y]}-B_{[x,y]}={\theta }_{[x,y]}$.

Each ${\theta }_x$ commutes with $d$. Evaluate ${\theta }_{x}d\omega -d{\theta }_x\omega$ on $y_0,\dots ,y_q\in \mathfrak{a}$, using the zero-based formula [F1]. Write $\rho (x)$ for the action on $V$. The action terms, after cancelling substitutions in the unacted slots, are ${\sum }_i(-1{)}^i([\rho (x),\rho (y_i)]-\rho ([x,y_i]))\omega (y_0,\dots ,\widehat{y_i},\dots ,y_q)$, which vanish by [F2]. The remaining bracket terms are ${\sum }_{i<j}(-1{)}^{i+j}\omega ([x,[y_i,y_j]]-[[x,y_i],y_j]-[y_i,[x,y_j]],y_0,\dots ,\widehat{y_i},\dots ,\widehat{y_j},\dots ,y_q)$, which vanish by Jacobi [F3]. Thus ${\theta }_{x}d=d{\theta }_x$.

For $x\in \mathfrak{a}$ the Cartan formula ${\theta }_x=d{i}_x+i_{x}d$ holds. Evaluate at $(x_1,\dots ,x_q)$. In $i_{x}d$, the terms of $d\omega$ in which the leading argument $x$ acts give $x\omega (x_1,\dots ,x_q)$, and the terms in which $x$ is the bracket argument give ${\sum }_j(-1{)}^j\omega ([x,x_j],x_1,\dots ,\widehat{x_j},\dots ,x_q)$; all terms in which one of the $x_i$ acts, and all bracket terms with two entries among the $x_i$, cancel against the corresponding terms of $d{i}_x$, because $\omega$ is alternating: $\omega (x,[x_i,x_j],\dots )=-\omega ([x_i,x_j],x,\dots )$ and the two appear with the same coefficient $(-1{)}^{i+j}$, while $x_i\omega (x,\dots )$ and $x_i\omega (x,\dots )$ appear with opposite coefficients $(-1{)}^i$ and $(-1{)}^{i-1}$. Since moving the bracket $[x,x_j]$ from the first slot into the $j$-th slot costs the sign $(-1{)}^{j-1}$, the surviving sum equals $-{\sum }_j\omega (x_1,\dots ,[x,x_j],\dots ,x_q)$, so $d{i}_x\omega +i_{x}d\omega =x\omega -{\sum }_j\omega (\dots ,[x,x_j],\dots )={\theta }_x\omega$.

For $x\in \mathfrak{a}$, ${\theta }_x$ induces the zero map on every cohomology group. If $\omega$ is a cocycle, then by step 2.3 ${\theta }_x\omega =d(i_x\omega )+i_x(d\omega )=d(i_x\omega )$ is a coboundary; and by step 2.2 ${\theta }_x$ maps coboundaries to coboundaries, since ${\theta }_x(d\alpha )=d({\theta }_x\alpha )$. Hence the induced endomorphism of $H^q(\mathfrak{a},V)=\ker d^q\mathrm{/}\mathrm{im}{d}^{q-1}$ is zero for every $q$.

The action descends. By steps 2.1 and 2.2, $\theta :\mathfrak{p}\to {\mathrm{End}}_k(C^{\bullet }(\mathfrak{a},V))$ is a representation preserving the differential, so it induces a representation of $\mathfrak{p}$ on each cohomology space $H^q(\mathfrak{a},V)$; by step 3.1 this induced representation kills $\mathfrak{a}$. Since $\mathfrak{a}$ is an ideal [F4] and the quotient map $q:\mathfrak{p}\twoheadrightarrow \mathfrak{p}\mathrm{/}\mathfrak{a}$ is a surjective Lie algebra homomorphism, there is a unique Lie algebra homomorphism $\bar{\theta }:\mathfrak{p}\mathrm{/}\mathfrak{a}\to {\mathrm{End}}_k(H^q)$ with $\bar{\theta }\circ q=\theta$: it is well defined because $\theta$ vanishes on $\mathfrak{a}$, and it is a homomorphism because $\theta$ is and $q$ is surjective. By [F4] the representation $\bar{\theta }$ makes $H^q(\mathfrak{a},V)$ a unital $U(\mathfrak{p}\mathrm{/}\mathfrak{a})$-module. For $\mathfrak{a}={\mathfrak{n}}^{+}◃\mathfrak{b}$ the quotient is $\mathfrak{h}$ by [F5], so $H^{\bullet }({\mathfrak{n}}^{+},V)$ is an $\mathfrak{h}$-module for every $\mathfrak{b}$-module $V$; and every $\mathfrak{g}$-module is a $\mathfrak{b}$-module by restriction. The same conclusion holds for $\mathfrak{p}=N_{\mathfrak{g}}(\mathfrak{a})$, which makes $\mathfrak{a}$ an ideal by [F4] applied to the normalizer definition, so the statement covers every subalgebra whose normalizer acts.

Multiplication by a central element is a cochain map: if $\omega \in C^q({\mathfrak{n}}^{+},V)$ and $(z\cdot \omega )(x_1,\dots ,x_q)=z\cdot \omega (x_1,\dots ,x_q)$, then $z\cdot (d\omega )=d(z\cdot \omega )$, because in the two sums of the differential the operator $z$ commutes with every $x_i$ and with every bracket on $V$; hence $z$ acts on each $H^q({\mathfrak{n}}^{+},V)$.

An injective $U(\mathfrak{g})$-module is injective over $U({\mathfrak{n}}^{+})$. Let $u:A\rightarrowtail B$ be an embedding of $U({\mathfrak{n}}^{+})$-modules and $f:A\to I$ a $U({\mathfrak{n}}^{+})$-linear map. By [F3], $U(\mathfrak{g}){\otimes }_{U({\mathfrak{n}}^{+})}-$ preserves monomorphisms, so $U(\mathfrak{g})\otimes u$ is injective, and the map $F:U(\mathfrak{g}){\otimes }_{U({\mathfrak{n}}^{+})}A\to I$, $s\otimes a\mapsto s\cdot f(a)$, is a well-defined $U(\mathfrak{g})$-linear map by the extension-of-scalars adjunction ${\mathrm{Hom}}_{U({\mathfrak{n}}^{+})}(A,I)\cong {\mathrm{Hom}}_{U(\mathfrak{g})}(U(\mathfrak{g}){\otimes }_{U({\mathfrak{n}}^{+})}A,I)$, which is the universal property of the tensor product over $U({\mathfrak{n}}^{+})$. Since $I$ is injective over $U(\mathfrak{g})$, $F$ extends along $U(\mathfrak{g})\otimes u$ to a $U(\mathfrak{g})$-linear $\tilde{F}$; the formula $g(b)=\tilde{F}(1\otimes b)$ defines a $U({\mathfrak{n}}^{+})$-linear map $B\to I$ extending $f$, because $1\otimes u(a)$ is the image of $1\otimes a$ and $\tilde{F}$ extends $F$. Hence $I$ has the extension property over $U({\mathfrak{n}}^{+})$.

Let $0\to V\to I\to I\mathrm{/}V\to 0$ be the short exact sequence obtained from an injective embedding $V\hookrightarrow I$ of [F4]. The connecting maps of its long exact sequence commute with multiplication by $z$ and with the $\mathfrak{h}$-action of [F7]: multiplication by $z$ is a $U({\mathfrak{n}}^{+})$-linear endomorphism of each of $V$, $I$, $I\mathrm{/}V$ commuting with the inclusion and the quotient, so naturality of the long exact sequence of [F6] gives $z\cdot \delta =\delta \cdot z$. For the Cartan action, represent a class by a cocycle $c$ in $C^{p-1}({\mathfrak{n}}^{+},I\mathrm{/}V)$ and lift it to $b$ in $C^{p-1}({\mathfrak{n}}^{+},I)$; then $\delta [c]=[db]$, with $db$ valued in $V$. The coefficient inclusion and quotient intertwine the cochain operators ${\theta }_h$, so ${\theta }_{h}b$ lifts ${\theta }_{h}c$; since ${\theta }_h$ commutes with $d$ by [F7], $\delta [{\theta }_{h}c]=[d{\theta }_{h}b]=[{\theta }_{h}db]={\theta }_h\delta [c]$. This proves Cartan equivariance directly, and therefore equivariance for $\mathrm{pr}(z)\in U(\mathfrak{h})$.

The identity holds in degree zero. Let $v\in H^0({\mathfrak{n}}^{+},V)=V^{{\mathfrak{n}}^{+}}$. By [F1] and [F2], $z-\mathrm{pr}(z)\in U(\mathfrak{g}){\mathfrak{n}}^{+}$, so $(z-\mathrm{pr}(z))v\in U(\mathfrak{g})({\mathfrak{n}}^{+}v)=0$; hence $z\cdot v=\mathrm{pr}(z)\cdot v$, where $\mathrm{pr}(z)\in U(\mathfrak{h})$ acts through the $\mathfrak{h}$-action of [F7] and the identification $U(\mathfrak{h})=S(\mathfrak{h})$.

Positive cohomology with injective coefficients vanishes: $H^q({\mathfrak{n}}^{+},I)=0$ for every $q>0$ and every injective $U(\mathfrak{g})$-module $I$. By step 1.2, $I$ is injective over $U({\mathfrak{n}}^{+})$, so by [F5] $H^q({\mathfrak{n}}^{+},I)\cong {\mathrm{Ext}}_{U({\mathfrak{n}}^{+})}^q(k,I)=0$ for $q>0$.