Semisimple Lie Algebras, Cohomology, and Levi Theory

Linearity of $\rho$, composition, and trace makes $B_{\rho }$ bilinear. By [L2], $B_{\rho }(x,y)=\mathrm{tr}(\rho (x)\rho (y))=\mathrm{tr}(\rho (y)\rho (x))=B_{\rho }(y,x)$, so it is symmetric.

Put $X=\rho (x)$, $Y=\rho (y)$, and $Z=\rho (z)$. Since $\rho$ preserves brackets, the left side of the invariance identity is $\mathrm{tr}((ZX-XZ)Y)+\mathrm{tr}(X(ZY-YZ))$. After expansion, the middle terms cancel directly and [L2] gives $\mathrm{tr}(ZXY)=\mathrm{tr}(XYZ)$, so the remaining terms cancel as well.

The Killing form is the trace form for $\rho =\mathrm{ad}$, so steps 1.1–1.2 apply verbatim. The zero representation and zero-dimensional space cause no exception: every displayed trace is then zero.

The equations defining ${\mathfrak{i}}^{\perp }$ are linear in $u$, so it is a linear subspace. If $u\in {\mathfrak{i}}^{\perp }$, $z\in \mathfrak{g}$, and $v\in \mathfrak{i}$, invariance gives $B([z,u],v)=-B(u,[z,v])$. The second argument on the right belongs to $\mathfrak{i}$ by [L1], so the right side is zero. Hence $[z,u]\in {\mathfrak{i}}^{\perp }$, proving ideality.

Taking $\mathfrak{i}=\mathfrak{g}$ in step 1.1 gives that ${\mathfrak{g}}^{\perp }$ is an ideal. This includes the zero algebra, the zero form, and the nondegenerate case ${\mathfrak{g}}^{\perp }=0$ without any separate choice.

Suppose first that $\mathfrak{h}$ is solvable. Choose bases of $\mathfrak{h}$ and $V$, and let $k_0\subseteq k$ be the subfield generated over $\mathbb{Q}$ by the finitely many structure constants and matrix entries. The resulting $k_0$-form ${\mathfrak{h}}_0$ is solvable because its derived series becomes that of $\mathfrak{h}$ after the faithful scalar extension $k_0\subseteq k$. The finitely generated field $k_0$ embeds in $\mathbb{C}$. Over $\mathbb{C}$, [L1] gives a basis in which ${\mathfrak{h}}_0$ is upper triangular and its derived algebra is strictly upper triangular. Thus $\mathrm{tr}(yx)=0$ for all $x\in [{\mathfrak{h}}_0,{\mathfrak{h}}_0]$ and $y\in {\mathfrak{h}}_0$. These finitely many bilinear identities hold over $k_0$ and hence after extension to $k$.

We prove the converse first under the stronger hypothesis $\mathrm{tr}(xy)=0$ for every $x,y\in \mathfrak{h}$. Choose bases and let $k_0$ be the characteristic-zero subfield generated by the finitely many matrix entries of a basis of $\mathfrak{h}$ and its structure constants. The strong trace hypothesis is determined by the finitely many pairs of basis vectors, so it persists after extending the resulting $k_0$-form to $\mathbb{C}$. Solvability over $\mathbb{C}$ descends because every derived term commutes with scalar extension. We may therefore carry out the strong converse over $\mathbb{C}$.

Over $\mathbb{C}$, fix $x\in [\mathfrak{h},\mathfrak{h}]$. By [L6], choose a Jordan basis on $V$ and write $x=x_s+x_n$, where $x_s$ is diagonal, $x_n$ is block-nilpotent, and they commute. The commuting left and right multiplication operators on $\mathrm{End}(V)$ show that the semisimple part of ${\mathrm{ad}}_x=L_x-R_x$ is ${\mathrm{ad}}_{x_s}$. Write the diagonal entries of $x_s$ as $a_1,\dots ,a_n$, and let $\overline{x_s}$ have diagonal entries $\overline{a_1},\dots ,\overline{a_n}$. On the generalized eigenspaces of ${\mathrm{ad}}_x$, ${\mathrm{ad}}_{\overline{x_s}}$ has eigenvalues $\overline{a_i-a_j}$. Finite Hermite interpolation on the finitely many differences $a_i-a_j$, taking zero to zero, expresses ${\mathrm{ad}}_{\overline{x_s}}$ as a polynomial without constant term in ${\mathrm{ad}}_x$. Since $\mathfrak{h}$ is stable under ${\mathrm{ad}}_x$, it follows that $[\overline{x_s},\mathfrak{h}]\subseteq \mathfrak{h}$.

Write $x$ as a finite sum of commutators. By [L5], $\mathrm{tr}(A[B,C])=\mathrm{tr}([A,B]C)$. Step 1.3 and the strong trace hypothesis therefore give $\mathrm{tr}(\overline{x_s}x)=0$. In the Jordan basis this trace is ${\sum }_i\overline{a_i}{a}_i$, so every $a_i=0$ and $x$ is nilpotent. If $x^m=0$ on $V$, then the commuting operators $L_x,R_x$ on $\mathrm{End}(V)$ satisfy $(L_x-R_x{)}^{2m-1}=0$, because every binomial term contains either $L_x^m$ or $R_x^m$. Restriction to the invariant derived algebra shows that each of its adjoint operators is nilpotent. Hence [L2] makes $[\mathfrak{h},\mathfrak{h}]$ nilpotent, and [L3] makes $\mathfrak{h}$ solvable.

Under the stated weaker hypothesis, apply steps 1.2–2.1 to the linear Lie algebra $[\mathfrak{h},\mathfrak{h}]$: the required strong trace vanishing holds for every pair in it. Its solvability implies solvability of $\mathfrak{h}$ by the derived-series definition. Together with step 1.1 this proves the linear criterion.

Apply the linear criterion to $\mathrm{ad}(\mathfrak{g})$. Its derived algebra is $\mathrm{ad}([\mathfrak{g},\mathfrak{g}])$, and [L7] says that the trace hypothesis is precisely the asserted Killing-form condition. Hence $\mathrm{ad}(\mathfrak{g})$ is solvable. The exact sequence $0\to Z(\mathfrak{g})\to \mathfrak{g}\to \mathrm{ad}(\mathfrak{g})\to 0$ has abelian kernel, so [L4] makes $\mathfrak{g}$ solvable. The reverse direction is step 1.1 for the adjoint representation. The zero algebra makes both conditions vacuous and is solvable.

Suppose $\mathfrak{g}$ is semisimple and let $\mathfrak{a}={\mathfrak{g}}^{\perp }$ be the radical of $K$. It is an ideal by [L1]. For $x,y\in \mathfrak{a}$, the operator ${\mathrm{ad}}_x{\mathrm{ad}}_y$ maps $\mathfrak{g}$ into $\mathfrak{a}$ and induces zero on $\mathfrak{g}\mathrm{/}\mathfrak{a}$; its trace on $\mathfrak{g}$ therefore equals the trace of its restriction to $\mathfrak{a}$. Thus the intrinsic Killing form of $\mathfrak{a}$ is the restriction of $K$, hence zero.

Conversely suppose $K$ is nondegenerate and let $\mathfrak{a}$ be an abelian ideal. For $a\in \mathfrak{a}$ and $x\in \mathfrak{g}$, put $T={\mathrm{ad}}_a{\mathrm{ad}}_x$. Its image lies in $\mathfrak{a}$, while $T$ vanishes on $\mathfrak{a}$ because $[x,\mathfrak{a}]\subseteq \mathfrak{a}$ and $\mathfrak{a}$ is abelian. Hence $T^2=0$, so $K(a,x)=\mathrm{tr}(T)=0$. Nondegeneracy forces $a=0$; thus $\mathfrak{g}$ has no nonzero abelian ideal.

By [L2], step 1.1 makes $\mathfrak{a}$ solvable. It is a solvable ideal of the semisimple algebra $\mathfrak{g}$, so [L3] gives $\mathfrak{a}=0$. Therefore $K$ is nondegenerate.

If $\mathfrak{g}$ had a nonzero solvable ideal $\mathfrak{r}$, let ${\mathfrak{r}}^{(m)}$ be the last nonzero term of its derived series: it is nonzero and abelian, and it is an ideal of $\mathfrak{g}$ because derived terms of an ideal are ideals of the ambient algebra by Jacobi. This contradicts step 1.2. Thus the radical is zero and [L3] makes $\mathfrak{g}$ semisimple. For $\mathfrak{g}=0$, the unique bilinear form has zero radical and is nondegenerate in the standard vacuous sense, so both directions still hold.

If $z\in Z(\mathfrak{g})$, then ${\mathrm{ad}}_z=0$, so [L2] gives $K(z,x)=0$ for every $x$. Nondegeneracy in [L1] yields $z=0$.

Let $D=[\mathfrak{g},\mathfrak{g}]$. Its orthogonal complement consists exactly of the elements $z$ with $K([x,y],z)=K(x,[y,z])=0$ for all $x,y$, hence $[y,z]=0$ by [L1]. Thus $D^{\perp }=Z(\mathfrak{g})=0$ by step 1.1, and finite-dimensional nondegeneracy gives $D=\mathfrak{g}$.

When $\mathfrak{g}=0$, both conclusions read $0=0$; no nonempty choice was used.

The assertion for $\mathfrak{g}=0$ is the empty direct sum. For nonzero $\mathfrak{g}$, if $\mathfrak{a}$ is any ideal, then $\mathfrak{a}\cap {\mathfrak{a}}^{\perp }$ is abelian. Indeed, for $u,v$ in that intersection and $z\in \mathfrak{g}$, invariance gives $K([u,v],z)=K(u,[v,z])=0$, because $[v,z]\in \mathfrak{a}$. By [L1], $[u,v]=0$. Semisimplicity makes this abelian ideal zero.

If $\mathfrak{g}\mathrm{\ne }0$, choose a nonzero ideal $\mathfrak{a}$ of least positive dimension; finite dimension makes this a choice from a finite set of integers. By step 1.1 and dimension, $\mathfrak{g}=\mathfrak{a}\oplus {\mathfrak{a}}^{\perp }$. Both summands are ideals, and their bracket lies in their intersection, hence is zero.

The minimal ideal $\mathfrak{a}$ is not abelian, because a nonzero abelian ideal is solvable. If $0\mathrm{\ne }\mathfrak{j}⊲\mathfrak{a}$, then $[{\mathfrak{a}}^{\perp },\mathfrak{j}]=0$ and $[\mathfrak{a},\mathfrak{j}]\subseteq \mathfrak{j}$, so $\mathfrak{j}$ is an ideal of $\mathfrak{g}$; minimality gives $\mathfrak{j}=\mathfrak{a}$. Thus $\mathfrak{a}$ is simple. Assume inductively that every semisimple algebra of smaller dimension has the asserted decomposition.

The complement ${\mathfrak{a}}^{\perp }$ is semisimple: any solvable ideal in it is, because the two summands commute, also a solvable ideal of $\mathfrak{g}$, and hence zero. Its dimension is smaller, so the induction hypothesis in step 3.1 decomposes it into finitely many simple ideals. Adjoining $\mathfrak{a}$ proves the result; a simple $\mathfrak{g}$ is the one-summand case.

Let $p_i$ be projection to ${\mathfrak{g}}_i$. If $p_i(\mathfrak{a})\mathrm{\ne }0$, simplicity and nonabelianness give $[{\mathfrak{g}}_i,p_i(\mathfrak{a})]={\mathfrak{g}}_i$. But bracketing an element supported in the $i$th factor with an element of $\mathfrak{a}$ produces an element of $\mathfrak{a}$ supported only in that factor. Hence ${\mathfrak{g}}_i\subseteq \mathfrak{a}$. If the projection is zero, the factor does not occur. Therefore $\mathfrak{a}$ is exactly the direct sum of the factors for which its projection is nonzero.

The sum of the remaining factors is an ideal complement $\mathfrak{b}$, so $\mathfrak{g}=\mathfrak{a}\oplus \mathfrak{b}$. Both $\mathfrak{a}$ and $\mathfrak{b}$ are direct sums of simple ideals and hence semisimple. By [L2], projection restricts to an isomorphism $\mathfrak{b}\cong \mathfrak{g}\mathrm{/}\mathfrak{a}$, so the quotient is semisimple as well.

For $\mathfrak{g}=0$, the indexing family and both subfamilies are empty; the ideal and quotient are zero. The cases $\mathfrak{a}=0$ and $\mathfrak{a}=\mathfrak{g}$ correspond to the empty and full subfamilies and are included in step 2.1.

The form isomorphism $b:\mathfrak{g}\to {\mathfrak{g}}^{\ast }$ sends $x\mapsto B(x,-)$. Under $\mathfrak{g}\otimes {\mathfrak{g}}^{\ast }\cong \mathrm{End}(\mathfrak{g})$, the identity corresponds to ${\sum }_i{x}_i\otimes b(x^i)$. Hence ${\sum }_i{x}_i\otimes x^i$ is the inverse tensor of $B$ and is independent of the basis. Multiplication into $U(\mathfrak{g})$ proves the same for ${\mathrm{\Omega }}_B$.

For $z\in \mathfrak{g}$, the diagonal adjoint action on the inverse tensor is ${\sum }_i([z,x_i]\otimes x^i+x_i\otimes [z,x^i])$. Pairing its second factor with an arbitrary vector and using [L2] shows that this tensor is zero. Multiplication sends it to $[z,{\mathrm{\Omega }}_B]$, hence that commutator is zero. Since $\mathfrak{g}$ generates $U(\mathfrak{g})$, ${\mathrm{\Omega }}_B$ is central.

By [L3], any module action extends to $U(\mathfrak{g})$. Centrality gives $\rho (x)\rho ({\mathrm{\Omega }}_B)=\rho ({\mathrm{\Omega }}_B)\rho (x)$ for every $x$, precisely the intertwining condition. For $\mathfrak{g}=0$, the inverse tensor and Casimir are empty sums and all assertions reduce to $0=0$.

We first work over an algebraically closed field. Every one-dimensional $\mathfrak{g}$-module is trivial: its representation kills $[\mathfrak{g},\mathfrak{g}]$, which equals $\mathfrak{g}$ by [L5].

Suppose $V\mathrm{/}W$ is one-dimensional and $W$ is simple. Replace $\mathfrak{g}$ by its image $\mathfrak{h}$ in $\mathfrak{g}\mathfrak{l}(V)$; the kernel is an ideal and [L8] makes the quotient $\mathfrak{h}$ semisimple. If $\mathfrak{h}=0$ the action is trivial and every line complementary to $W$ is invariant, so assume $\mathfrak{h}\mathrm{\ne }0$. By [L9] its trace form on $V$ is invariant and symmetric, so [L10] makes its radical an ideal; [L6] makes that radical solvable, hence zero. Form its Casimir $C$ as in [L7]. It kills the trivial quotient $V\mathrm{/}W$, while [L2] and [L3] say that it acts on $W$ by a scalar $a$. Moreover ${\mathrm{tr}}_V(C)=\dim \mathfrak{h}$, by summing the dual-basis identities. Characteristic zero makes this nonzero, so $a\mathrm{\ne }0$. Consequently $\ker C$ is an invariant line complementary to $W$. This, together with $W=0$, is the induction base.

Retain $\dim (V\mathrm{/}W)=1$ but allow arbitrary $W$, and assume the assertion for submodules of smaller dimension. If $W$ is not simple, choose a nonzero proper submodule $W^{\mathrm{\prime }}$. By the induction hypothesis, $W\mathrm{/}{W}^{\mathrm{\prime }}$ has a complement $V^{\mathrm{\prime }}\mathrm{/}{W}^{\mathrm{\prime }}$ in $V\mathrm{/}{W}^{\mathrm{\prime }}$. Now $V^{\mathrm{\prime }}\mathrm{/}{W}^{\mathrm{\prime }}$ is one-dimensional, and another induction application splits $V^{\mathrm{\prime }}=W^{\mathrm{\prime }}\oplus L$. The line $L$ is disjoint from $W$ and complements it in $V$.

For general $W\subseteq V$, give ${\mathrm{Hom}}_k(V,W)$ the action $(xf)(v)=xf(v)-f(xv)$. Let $V_1$ consist of maps whose restriction to $W$ is a scalar multiple of the identity, and $W_1$ of maps vanishing on $W$. They are submodules and $V_1\mathrm{/}{W}_1$ is one-dimensional. Step 3.1 gives an invariant line $L=kf$ complementary to $W_1$. By step 1.1, $L$ is trivial, so $f$ intertwines the action. Its restriction to $W$ is a nonzero scalar—otherwise $f\in W_1$—and after rescaling it is the identity. Therefore $\ker f$ is an invariant complement to $W$.

For an arbitrary characteristic-zero ground field, choose bases of $\mathfrak{g}$ and $V$ adapted to $W$, and let $k_0$ be the subfield generated over $\mathbb{Q}$ by their finitely many structure and action coefficients. Nondegeneracy of the Killing matrix shows that the resulting $k_0$-form of $\mathfrak{g}$ is semisimple. Embed the finitely generated field $k_0$ in $\mathbb{C}$. Steps 1.1–4.1 over $\mathbb{C}$ produce an equivariant projection $P:V_{\mathbb{C}}\to W_{\mathbb{C}}$ restricting to the identity. The conditions $P{\mathrm{\mid }}_W=1$ and $P\rho (x)=\rho (x)P$ for the chosen finite bases form a finite linear system over $k_0$. Row reduction shows that consistency over $\mathbb{C}$ already gives a $k_0$-solution; extending it to the original field gives an invariant kernel complementary to $W$. Since $W$ was arbitrary, every submodule has an invariant complement. Induction on $\dim V$ now gives the direct-sum condition in [L1]: for $V\mathrm{\ne }0$, choose a nonzero submodule $S$ of least dimension, which is irreducible; if $S\mathrm{\ne }V$, split $V=S\oplus T$ and decompose the smaller module $T$ by induction. For $V=0$ the empty direct sum is [L1]'s convention. Every descent, embedding, and row reduction uses only finite data, so no choice principle is used.

Assume condition 1 and write $\mathfrak{z}=Z(\mathfrak{g})$. By [L1], $\mathfrak{q}=\mathfrak{g}\mathrm{/}\mathfrak{z}$ is semisimple. It acts on $\mathfrak{g}$ through adjoints, and $\mathfrak{z}$ is a trivial submodule. By [L2] there is a $\mathfrak{q}$-submodule $\mathfrak{s}$ with $\mathfrak{g}=\mathfrak{z}\oplus \mathfrak{s}$. Invariance says $[\mathfrak{g},\mathfrak{s}]\subseteq \mathfrak{s}$, so $\mathfrak{s}$ is an ideal, and projection identifies it with $\mathfrak{q}$. Hence it is semisimple by [L3]. Since $\mathfrak{z}$ is central and $\mathfrak{s}$ is perfect by [L4], $[\mathfrak{g},\mathfrak{g}]=\mathfrak{s}$. This is condition 2.

Assume condition 2. The adjoint module is the direct sum of the trivial module $Z(\mathfrak{g})$ and the adjoint module of the semisimple ideal $[\mathfrak{g},\mathfrak{g}]$. The latter is completely reducible by [L2], so the whole adjoint module is completely reducible. This proves condition 3.

Assume condition 3. By [L5], the adjoint module is a finite direct sum of irreducible submodules. Every submodule $W$ therefore has an invariant complement: starting with $C=0$, if $W\oplus C\mathrm{\ne }\mathfrak{g}$, some irreducible summand in the displayed finite decomposition is not contained in $W+C$; its intersection with $W+C$ is then zero by irreducibility, so adjoining it strictly enlarges $C$ while preserving $W\cap C=0$. Finite dimensionality makes this process terminate with $\mathfrak{g}=W\oplus C$. Write $\mathfrak{r}=\mathrm{rad}(\mathfrak{g})$ and use this observation to choose an invariant complement $\mathfrak{s}$. Both are ideals, so $[\mathfrak{r},\mathfrak{s}]\subseteq \mathfrak{r}\cap \mathfrak{s}=0$. Apply the observation again to the characteristic ideal ${\mathfrak{r}}^{\mathrm{\prime }}=[\mathfrak{r},\mathfrak{r}]$, choosing an invariant complement $\mathfrak{c}$ in $\mathfrak{g}$. If $\mathfrak{u}=\mathfrak{r}\cap \mathfrak{c}$, then the decomposition $\mathfrak{g}={\mathfrak{r}}^{\mathrm{\prime }}\oplus \mathfrak{c}$ restricts to $\mathfrak{r}={\mathfrak{r}}^{\mathrm{\prime }}\oplus \mathfrak{u}$. The summands ${\mathfrak{r}}^{\mathrm{\prime }}$ and $\mathfrak{u}$ are ideals and commute.

From step 1.3, $\mathfrak{r}={\mathfrak{r}}^{\mathrm{\prime }}\oplus \mathfrak{u}$ and $[{\mathfrak{r}}^{\mathrm{\prime }},\mathfrak{u}]=0$. Hence ${\mathfrak{r}}^{\mathrm{\prime }}=[\mathfrak{r},\mathfrak{r}]=[{\mathfrak{r}}^{\mathrm{\prime }},{\mathfrak{r}}^{\mathrm{\prime }}]+[\mathfrak{u},\mathfrak{u}]$. The second summand lies in both $\mathfrak{u}$ and ${\mathfrak{r}}^{\mathrm{\prime }}$, so it is zero; therefore ${\mathfrak{r}}^{\mathrm{\prime }}=[{\mathfrak{r}}^{\mathrm{\prime }},{\mathfrak{r}}^{\mathrm{\prime }}]$. But ${\mathfrak{r}}^{\mathrm{\prime }}$ is solvable, and a nonzero perfect algebra cannot be solvable, so ${\mathfrak{r}}^{\mathrm{\prime }}=0$. Thus $\mathfrak{r}$ is abelian and, because it also commutes with $\mathfrak{s}$, is central in $\mathfrak{g}$. Conversely the center is an abelian ideal and lies in the radical. This proves condition 1 and closes the equivalence. For $\mathfrak{g}=0$, all three conditions hold.

A subspace $U\subseteq \mathfrak{g}$ is stable under the adjoint action exactly when $[\mathfrak{g},U]\subseteq U$, which is exactly the ideal condition. Therefore [L1] says every ideal has an ideal complement.

An irreducible adjoint summand is a nonzero ideal with no nonzero proper ideal of $\mathfrak{g}$. It cannot be abelian, since that would be a solvable ideal of a semisimple algebra. Its ideal complement commutes with it, so an ideal inside the summand is also an ideal of $\mathfrak{g}$; hence the summand is simple. Conversely, each simple factor from [L2] has no proper adjoint submodule and is irreducible. The zero algebra has the empty decomposition.

On $\mathrm{Der}(\mathfrak{g})$ define $B(D,E)={\mathrm{tr}}_{\mathfrak{g}}(DE)$. By [L2] this trace form is invariant. Its restriction to the ideal $\mathrm{ad}(\mathfrak{g})$ is the Killing form, which is nondegenerate by [L3]. Finite-dimensional linear algebra therefore gives $\mathrm{Der}(\mathfrak{g})=\mathrm{ad}(\mathfrak{g})\oplus \mathrm{ad}(\mathfrak{g}{)}^{\perp }$, and [L5] makes the orthogonal complement an ideal.

The two ideals in step 1.1 commute: their bracket lies in their intersection, which is zero. If $D$ is in the orthogonal complement, then $0=[D,{\mathrm{ad}}_x]={\mathrm{ad}}_{D(x)}$ for every $x$; the last equality is the derivation identity. By [L4], $D(x)=0$ for every $x$, so $D=0$. Thus every derivation is inner.

If ${\mathrm{ad}}_x={\mathrm{ad}}_y$, then $x-y$ is central and [L4] gives $x=y$. For $\mathfrak{g}=0$, both derivation algebras are zero and uniqueness is vacuous.

Choose a basis of $\mathfrak{g}$. The equations $A[x,y]=[Ax,Ay]$ for basis pairs are finitely many polynomial equations in the matrix entries of $A$. Their common zero set inside $\mathrm{GL}(\mathfrak{g})$ is exactly $\mathrm{Aut}(\mathfrak{g})$ and is closed. By [L1] it is an embedded Lie subgroup; for a complex algebra, apply [L1] first to the underlying real group.

A tangent vector at the identity is represented by $A(t)=I+tD+o(t)$. Differentiating the bracket equation gives $D[x,y]=[Dx,y]+[x,Dy]$, so the tangent algebra is contained in $\mathrm{Der}(\mathfrak{g})$. Conversely, for a derivation $D$, the linear vector field $A\mapsto DA$ is tangent to the defining equations, or equivalently its local flow $\exp (tD)$ preserves the bracket by differentiating $\exp (tD)[x,y]-[\exp (tD)x,\exp (tD)y]$; hence every derivation is tangent. When $\mathfrak{g}$ is complex, the ambient group consists of complex-linear maps and the resulting space of complex-linear derivations is stable under multiplication by $i$; exponential charts therefore give the embedded subgroup its corresponding complex Lie-subgroup structure.

Step 1.2 identifies the Lie algebra with $\mathrm{Der}(\mathfrak{g})$, and [L2] identifies that with $\mathrm{ad}(\mathfrak{g})$. If $\mathfrak{g}=0$, the automorphism group is the one-point group and all tangent algebras are zero. Countable choice is used only through [L1], not in the polynomial or differentiation steps.

Expand $d(df)$ using [L1], and let $X^{ij}$ denote the ordered list obtained by omitting $x_i,x_j$. For fixed $i<j$, the outer action by $x_i$ followed by the action of $x_j$ has coefficient $(-1{)}^{i+j-1}$, whereas the reverse order has coefficient $(-1{)}^{i+j}$. Their sum is $(-1{)}^{i+j-1}[\rho (x_i),\rho (x_j)]f(X^{ij})$. There is exactly one term in which the outer differential forms $[x_i,x_j]$ and the inner differential lets that new first argument act; its coefficient is $(-1{)}^{i+j}$, so it contributes $(-1{)}^{i+j}\rho ([x_i,x_j])f(X^{ij})$. These three terms cancel by [L2].

It remains to account for action--bracket terms on three distinct indices. Fix $i<j$ and $k\mathrm{\notin }\{i,j\}$. Let $a$ be the number of $i,j$ that are less than $k$, and $b$ the number greater than $k$, so $a+b=2$. Acting first by $x_k$ and then forming $[x_i,x_j]$ has coefficient $(-1{)}^{i+j+k-b}$. Forming $[x_i,x_j]$ first and then letting $x_k$ act has coefficient $(-1{)}^{i+j+k+1-a}$. The exponents differ by $1-a+b=3-2a$, which is odd, while both terms have the same value $\rho (x_k)f([x_i,x_j],X^{ijk})$. Thus they cancel. This covers every mixed term with three distinct original indices.

Consider the terms in which both differentials use their bracket sums. For two disjoint pairs, the two possible orders have the same scalar sign: the numbers of cross-pair index shifts in the two orders add to $4$, so their sign exponents differ by an even integer. Their cochain values are $f([x_p,x_q],[x_i,x_j],X^{ijpq})$ and $f([x_i,x_j],[x_p,x_q],X^{ijpq})$, which cancel because $f$ is alternating. For $i<j<k$, the three terms in which the second bracket uses the bracket created by the first have common coefficient $(-1{)}^{i+j+k-1}$ and bracket sum $[[x_i,x_j],x_k]+[[x_j,x_k],x_i]-[[x_i,x_k],x_j]$. Since $-[[x_i,x_k],x_j]=[[x_k,x_i],x_j]$, this is zero by Jacobi [L3]. The expansion has now been partitioned into action--action plus created-bracket action (step 1.1), mixed action--bracket terms (step 1.2), disjoint double brackets, and nested double brackets; hence $d^{2}f=0$. For $n<0$ or a zero cochain space the assertion is the unique zero composite, and for $n=0$ step 1.1 is exactly the representation identity.

By [L1], $m$ lies in the degree-zero kernel exactly when every $x$ kills it, so $\ker d^0=M^{\mathfrak{g}}$.

The degree-minus-one cochain space is zero, hence $\mathrm{im}{d}^{-1}=0$. Substitution in [L2] proves the displayed equality. If $M=0$ both sides are zero; if $\mathfrak{g}=0$, the condition is vacuous and both sides are all of $M$.

For a $1$-cochain $\delta$, [L1] gives $(d\delta )(x,y)=x\delta (y)-y\delta (x)-\delta ([x,y])$. Thus its kernel is precisely the space of module-valued derivations in the displayed sense.

For $m\in C^0=M$, [L1] gives $(dm)(x)=xm$, so the image consists exactly of the inner module-valued derivations. Taking the quotient in [L2] proves the first identification.

If $M=\mathfrak{g}$ with the adjoint action, step 1.1 becomes $\delta ([x,y])=[\delta x,y]+[x,\delta y]$, the derivation law in [L3], while step 1.2 gives $x\mapsto [x,m]={\mathrm{ad}}_{-m}(x)$; its span is the same inner-derivation space. Zero algebras and zero modules satisfy the same formulas.

Given an extension, lift a finite basis of $\mathfrak{g}$ and extend linearly to a section $s:\mathfrak{g}\to \mathfrak{e}$. The formula $x\cdot m=[s(x),m]$ is independent of the lift because $M$ is abelian, and it is the fixed action. Define ${\omega }_s(x,y)=[s(x),s(y)]-s([x,y])\in M$. It is alternating. Expanding Jacobi for the three lifts and collecting their $M$-components gives exactly $d{\omega }_s=0$ with [L1]'s signs.