Chevalley–Eilenberg cohomology computes Ext of the trivial module

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.