Cartan's criterion for solvability and semisimplicity

Intuition [Beginner]

Every Lie algebra has a built-in measuring device called the Killing form. It takes two elements of the algebra and produces a single number, computed by tracking how each element moves the algebra around and multiplying the resulting movements together. The Killing form detects whether the algebra is "complicated enough" to have interesting representation theory, or whether it collapses into a pattern of shrinking commutators.

A Lie algebra is solvable when repeated commutators shrink to zero. A Lie algebra is semisimple when it contains no solvable ideal larger than zero. The Killing form tells the two cases apart: it vanishes on the solvable part and is non-degenerate (never zero for a nonzero input paired with some partner) on the semisimple part. Cartan's criterion is the precise statement of this dichotomy.

The reason this matters is that semisimple Lie algebras are the ones with rich, classifiable representation theory: root systems, weight lattices, character formulas. Without Cartan's criterion, there is no way to identify which algebras admit this structure and which do not.

Visual [Beginner]

A diagram showing a three-dimensional Lie algebra as a cloud of arrows. Each arrow represents how one element moves the others via the bracket. The Killing form pairs two arrows by counting how much their movements overlap. In the solvable case, the cloud is flat: every pair of arrows gives a zero measurement, and the diagram is greyed out. In the semisimple case, the cloud is full-dimensional: pairing the right arrows recovers a non-zero number, and the diagram is coloured to show the non-degenerate pairing.

The picture captures the core mechanism: the Killing form is a bilinear measuring device, and its degeneracy pattern detects the solvable versus semisimple split.

Worked example [Beginner]

Consider the Lie algebra ${\mathfrak{sl}}_2\mathbb{C}$ of traceless $2\times 2$ matrices, with basis $H,E,F$ satisfying $[H,E]=2E$, $[H,F]=-2F$, $[E,F]=H$.

Step 1. Compute the adjoint matrices. The element $H$ acts on the basis $\{H,E,F\}$ as $[H,H]=0$, $[H,E]=2E$, $[H,F]=-2F$. So ${\mathrm{ad}}_H$ has matrix $\left(\begin{array}{ccc}0& 0& 0\\ 0& 2& 0\\ 0& 0& -2\end{array}\right)$.

Step 2. Compute $B(H,H)=\mathrm{tr}({\mathrm{ad}}_H\circ {\mathrm{ad}}_H)$. The matrix of ${\mathrm{ad}}_H$ squared is $\left(\begin{array}{ccc}0& 0& 0\\ 0& 4& 0\\ 0& 0& 4\end{array}\right)$, with trace $8$.

Step 3. Check the full Killing matrix. By similar computations: $B(E,E)=0$, $B(F,F)=0$, $B(E,F)=4$, $B(H,E)=0$, $B(H,F)=0$. The Killing matrix in the basis $\{H,E,F\}$ is $\left(\begin{array}{ccc}8& 0& 0\\ 0& 0& 4\\ 0& 4& 0\end{array}\right)$, with determinant $-128$, which is nonzero.

What this tells us: the Killing form is non-degenerate on ${\mathfrak{sl}}_2\mathbb{C}$, confirming by Cartan's criterion that ${\mathfrak{sl}}_2\mathbb{C}$ is semisimple (in fact simple).

Check your understanding [Beginner]

Formal definition [Intermediate+]

Let $\mathfrak{g}$ be a finite-dimensional Lie algebra over a field of characteristic zero. The Killing form is the symmetric bilinear form $$ B(X, Y) = \mathrm{tr}(\mathrm{ad}_X \circ \mathrm{ad}_Y), \qquad X, Y \in \mathfrak{g}. $$

The Killing form is invariant: $B([X,Y],Z)=B(X,[Y,Z])$ for all $X,Y,Z\in \mathfrak{g}$. This follows from the identity ${\mathrm{ad}}_{[X,Y]}=[{\mathrm{ad}}_X,{\mathrm{ad}}_Y]$ and the cyclic property of the trace.

The radical of the Killing form is $\mathrm{rad}(B)=\{X\in \mathfrak{g}:B(X,Y)=0\text{ for all }Y\in \mathfrak{g}\}$. The Killing form is non-degenerate if $\mathrm{rad}(B)=0$.

The radical of $\mathfrak{g}$ (not to be confused with $\mathrm{rad}(B)$) is the largest solvable ideal of $\mathfrak{g}$, denoted $\mathrm{Rad}(\mathfrak{g})$. A Lie algebra is semisimple if $\mathrm{Rad}(\mathfrak{g})=0$.

Counterexamples to common slips

• The radical of $B$ is not the same as the radical of $\mathfrak{g}$. The radical of the Killing form is $\mathrm{rad}(B)=\{X:B(X,\mathfrak{g})=0\}$, which is always an ideal. For solvable $\mathfrak{g}$, $\mathrm{rad}(B)=\mathfrak{g}$ (the whole Killing form is zero). For semisimple $\mathfrak{g}$, $\mathrm{rad}(B)=0$. The radical of $\mathfrak{g}$ is a solvable ideal that may be smaller than $\mathrm{rad}(B)$ in general, though Cartan's criterion identifies the two via $[\mathrm{Rad}(\mathfrak{g}),\mathrm{Rad}(\mathfrak{g})]\subseteq \mathrm{rad}(B)$.
• The Killing form can be degenerate on a non-solvable algebra. The radical $\mathrm{rad}(B)$ is a solvable ideal, so if $\mathfrak{g}$ is not semisimple then $\mathrm{rad}(B)\ne 0$. But $\mathfrak{g}$ may still not be solvable (e.g., a semisimple algebra plus a one-dimensional centre: the Killing form is degenerate because the centre contributes zero rows, but the algebra is not solvable).

Key theorem with proof [Intermediate+]

Theorem (Cartan's criterion for solvability). Let $\mathfrak{g}$ be a finite-dimensional Lie algebra over $\mathbb{C}$. Then $\mathfrak{g}$ is solvable if and only if $B(X,Y)=0$ for all $X\in [\mathfrak{g},\mathfrak{g}]$ and all $Y\in \mathfrak{g}$.

Proof. ($\Rightarrow$) Assume $\mathfrak{g}$ is solvable. By Lie's theorem (unit 07.06.14), there exists a basis of $\mathfrak{g}$ in which every ${\mathrm{ad}}_X$ is upper triangular. The commutators ${\mathrm{ad}}_{[X,Y]}={\mathrm{ad}}_X{\mathrm{ad}}_Y-{\mathrm{ad}}_Y{\mathrm{ad}}_X$ are then strictly upper triangular (diagonal entries of ${\mathrm{ad}}_X{\mathrm{ad}}_Y$ and ${\mathrm{ad}}_Y{\mathrm{ad}}_X$ agree and cancel). So ${\mathrm{ad}}_Z$ for $Z\in [\mathfrak{g},\mathfrak{g}]$ is strictly upper triangular. The composition ${\mathrm{ad}}_Z\circ {\mathrm{ad}}_Y$ is then strictly upper triangular times upper triangular, which is strictly upper triangular (zero diagonal). The trace of a strictly upper triangular matrix is zero, so $B(Z,Y)=0$.

($\Leftarrow$) Assume $B([\mathfrak{g},\mathfrak{g}],\mathfrak{g})=0$. It suffices to show that $[\mathfrak{g},\mathfrak{g}]$ is nilpotent, which implies $\mathfrak{g}$ is solvable (since ${\mathfrak{g}}^{(1)}=[\mathfrak{g},\mathfrak{g}]$ nilpotent implies solvable).

Let $\rho :\mathfrak{g}\to \mathfrak{gl}(V)$ be the adjoint representation restricted to $[\mathfrak{g},\mathfrak{g}]$, and let $A$ be the associative subalgebra of $\mathrm{End}(\mathfrak{g})$ generated by $\{{\mathrm{ad}}_X:X\in \mathfrak{g}\}$. For $Z\in [\mathfrak{g},\mathfrak{g}]$ and $Y\in \mathfrak{g}$, the hypothesis gives $\mathrm{tr}({\mathrm{ad}}_Z\circ {\mathrm{ad}}_Y)=0$. Since the trace is linear and $[\mathfrak{g},\mathfrak{g}]$ is spanned by commutators, $\mathrm{tr}(\rho ([X_1,X_2])\cdot T)=0$ for all $X_1,X_2\in \mathfrak{g}$ and all $T\in A$.

The algebraic argument now uses the following trace identity. Let $S$ be any subspace of $\mathrm{End}(V)$, and let $S^{\perp }=\{T\in \mathrm{End}(V):\mathrm{tr}(ST)=0\text{ for all }S\in S\}$. If $S$ is a linear subspace with $[S,S]\subseteq S^{\perp }$, then every element of $[S,S]$ is nilpotent. The proof: by Cartan's trace criterion (a linear-algebra lemma), $\mathrm{tr}(ST)=0$ for all $T$ in the associative algebra generated by $S$ implies that $S$ is simultaneously nilpotent when $S$ consists of commutators.

Concretely, for $Z=[X_1,X_2]\in [\mathfrak{g},\mathfrak{g}]$, the element ${\mathrm{ad}}_Z=[{\mathrm{ad}}_{X_1},{\mathrm{ad}}_{X_2}]$ is a commutator in $\mathrm{End}(\mathfrak{g})$. The trace-hypothesis $\mathrm{tr}({\mathrm{ad}}_Z\cdot T)=0$ for all $T\in A$ forces ${\mathrm{ad}}_Z$ to be nilpotent by the following lemma.

Lemma (Cartan's trace lemma). Let $V$ be a finite-dimensional vector space and $S\in \mathrm{End}(V)$ a linear operator of the form $S=[A,B]$ with $A,B\in \mathrm{End}(V)$. If $\mathrm{tr}(S\cdot T)=0$ for every $T$ in the associative subalgebra generated by $A$ and $B$, then $S$ is nilpotent.

Proof of lemma. Let ${\lambda }_1,\dots ,{\lambda }_n$ be the eigenvalues of $S$ (over $\mathbb{C}$). We show all ${\lambda }_i=0$. Choose $T=S^k$ for each $k=0,1,\dots$. Then $\mathrm{tr}(S^{k+1})={\lambda }_1^{k+1}+\cdots +{\lambda }_n^{k+1}=0$ for all $k\ge 0$. Newton's identities (power sums determine elementary symmetric polynomials) then force all ${\lambda }_i=0$. Hence $S$ is nilpotent.

Applying this to ${\mathrm{ad}}_Z$ for every $Z\in [\mathfrak{g},\mathfrak{g}]$: every ${\mathrm{ad}}_Z$ is nilpotent. By Engel's theorem 07.06.14, $[\mathfrak{g},\mathfrak{g}]$ acts nilpotently on $\mathfrak{g}$ via the adjoint representation. A Lie algebra whose adjoint representation consists of nilpotent operators is nilpotent (Engel's theorem applied to the adjoint representation). So $[\mathfrak{g},\mathfrak{g}]$ is nilpotent, hence $\mathfrak{g}$ is solvable. $\square$

Corollary (Cartan's criterion for semisimplicity). A finite-dimensional Lie algebra $\mathfrak{g}$ over $\mathbb{C}$ is semisimple if and only if its Killing form $B$ is non-degenerate.

Proof. ($\Rightarrow$) If $\mathfrak{g}$ is semisimple, then $\mathrm{Rad}(\mathfrak{g})=0$. The radical $\mathrm{rad}(B)$ of the Killing form is a solvable ideal (one verifies that $\mathrm{rad}(B)$ is an ideal using the invariance of $B$, and Cartan's solvability criterion applied to $\mathrm{rad}(B)$ shows it is solvable). Since $\mathrm{Rad}(\mathfrak{g})=0$ and $\mathrm{rad}(B)\subseteq \mathrm{Rad}(\mathfrak{g})$ (as a solvable ideal), we get $\mathrm{rad}(B)=0$, i.e., $B$ is non-degenerate.

($\Leftarrow$) If $B$ is non-degenerate, then $\mathrm{rad}(B)=0$. Suppose $\mathfrak{g}$ has a non-zero solvable ideal $\mathfrak{r}$. By Cartan's solvability criterion, $B(\mathfrak{r},[\mathfrak{r},\mathfrak{r}])=0$, and since $\mathfrak{r}$ is an ideal, $[\mathfrak{r},\mathfrak{g}]\subseteq \mathfrak{r}$, so $B([\mathfrak{r},\mathfrak{r}],\mathfrak{g})=B(\mathfrak{r},[\mathfrak{r},\mathfrak{g}])\subseteq B(\mathfrak{r},\mathfrak{r})$. But if $\mathfrak{r}\ne 0$ is solvable, the derived series gives $[\mathfrak{r},\mathfrak{r}]⊊\mathfrak{r}$. By induction on derived length, $B([\mathfrak{r},\mathfrak{r}],\mathfrak{g})=0$, meaning $[\mathfrak{r},\mathfrak{r}]\subseteq \mathrm{rad}(B)=0$. More directly: the largest solvable ideal $\mathrm{Rad}(\mathfrak{g})$ satisfies $[\mathrm{Rad}(\mathfrak{g}),\mathrm{Rad}(\mathfrak{g})]\subseteq \mathrm{rad}(B)$ by the solvability criterion applied to $\mathrm{Rad}(\mathfrak{g})$. Since $\mathrm{rad}(B)=0$, the bracket $[\mathrm{Rad}(\mathfrak{g}),\mathrm{Rad}(\mathfrak{g})]=0$. If $\mathrm{Rad}(\mathfrak{g})\ne 0$, pick $X\in \mathrm{Rad}(\mathfrak{g})$, $X\ne 0$. Then ${\mathrm{ad}}_X$ kills $\mathrm{Rad}(\mathfrak{g})$ (since the bracket is zero) and ${\mathrm{ad}}_X(\mathfrak{g})\subseteq \mathrm{Rad}(\mathfrak{g})$ (since $\mathrm{Rad}(\mathfrak{g})$ is an ideal). So ${\mathrm{ad}}_X^2=0$, making $X\in \mathrm{rad}(B)=0$, a contradiction. So $\mathrm{Rad}(\mathfrak{g})=0$ and $\mathfrak{g}$ is semisimple. $\square$

Bridge. Cartan's criterion builds toward 07.06.17 where the Cartan subalgebra exists and is conjugate because the Killing form provides the non-degenerate pairing that identifies semisimple elements. The foundational reason the criterion works is that Lie's theorem forces the commutator subalgebra into strictly upper-triangular form, killing all diagonal entries and hence all Killing-form pairings. This is exactly the mechanism appearing again in 07.06.14 where the vanishing of $B([\mathfrak{g},\mathfrak{g}],\mathfrak{g})$ is the trace-content of solvability. The central insight is that the Killing form is an intrinsic detector: its degeneracy locus is the radical of the Lie algebra, and its non-degeneracy identifies the semisimple piece. The bridge is between the structural decomposition of a Lie algebra into radical and Levi factor, and the algebraic invariant (the Killing form) that computes both pieces.

Exercises [Intermediate+]

Advanced results [Master]

Theorem 1 (Killing form on a direct sum). Let $\mathfrak{g}={\mathfrak{g}}_1\oplus \cdots \oplus {\mathfrak{g}}_k$ be a direct sum of ideals. Then $B_{\mathfrak{g}}=B_{{\mathfrak{g}}_1}\oplus \cdots \oplus B_{{\mathfrak{g}}_k}$: the Killing form is block diagonal, with $B({\mathfrak{g}}_i,{\mathfrak{g}}_j)=0$ for $i\ne j$.

This makes the Killing form of a semisimple algebra completely determined by the Killing forms of its simple summands.

Theorem 2 (Simple ideals are Killing-orthogonal). If $\mathfrak{g}$ is a semisimple Lie algebra with decomposition $\mathfrak{g}={\mathfrak{g}}_1\oplus \cdots \oplus {\mathfrak{g}}_k$ into simple ideals, then the ${\mathfrak{g}}_i$ are pairwise orthogonal under $B$: $B({\mathfrak{g}}_i,{\mathfrak{g}}_j)=0$ for $i\ne j$, and $B{|}_{{\mathfrak{g}}_i}$ is non-degenerate for each $i$.

Theorem 3 (Radical and Levi decomposition). Every finite-dimensional Lie algebra $\mathfrak{g}$ over $\mathbb{C}$ has a unique maximal solvable ideal $\mathrm{Rad}(\mathfrak{g})$, called the radical. The quotient $\mathfrak{g}/\mathrm{Rad}(\mathfrak{g})$ is semisimple, and there exists a subalgebra $\mathfrak{s}\le \mathfrak{g}$ (a Levi subalgebra) with $\mathfrak{g}=\mathrm{Rad}(\mathfrak{g})⋊\mathfrak{s}$. Any two Levi subalgebras are conjugate under $\exp ({\mathrm{ad}}_X)$ for $X$ in the nilradical.

The Levi decomposition reduces the structure theory of arbitrary Lie algebras to the solvable case (handled by Lie's theorem 07.06.14) and the semisimple case (handled by root systems 07.06.03).

Theorem 4 (Characterisation of the radical via $B$). The radical of $\mathfrak{g}$ satisfies $\mathrm{Rad}(\mathfrak{g})\cap [\mathfrak{g},\mathfrak{g}]=[\mathfrak{g},\mathfrak{g}]\cap \mathrm{rad}(B)$, and $\mathrm{Rad}(\mathfrak{g})$ is the largest ideal of $\mathfrak{g}$ contained in the orthogonal complement of $[\mathfrak{g},\mathfrak{g}]$ under $B$. In particular, $\mathrm{Rad}(\mathfrak{g})=\{0\}$ if and only if $B$ is non-degenerate.

Theorem 5 (Uniqueness of the Killing form on simple algebras). On a simple Lie algebra $\mathfrak{g}$ over $\mathbb{C}$, every invariant symmetric bilinear form is a scalar multiple of the Killing form. The space of invariant symmetric bilinear forms on a semisimple Lie algebra $\mathfrak{g}={\mathfrak{g}}_1\oplus \cdots \oplus {\mathfrak{g}}_k$ is $k$-dimensional, parametrised by choosing one scalar for each simple summand.

This rigidity is the reason the Killing form is canonical: up to a positive scalar on each simple factor, there is only one natural invariant form.

Theorem 6 (Whitehead's lemma). Let $\mathfrak{g}$ be a semisimple Lie algebra over $\mathbb{C}$ and $V$ a finite-dimensional $\mathfrak{g}$-module. Then $H^1(\mathfrak{g},V)=0$: every derivation of $\mathfrak{g}$ into $V$ is inner. This is the cohomological shadow of the non-degeneracy of the Killing form and Cartan's criterion, and it underpins the Levi decomposition's conjugacy theorem.

Theorem 7 (Negative-definiteness on compact forms). If $\mathfrak{k}$ is a compact real form of a semisimple complex Lie algebra $\mathfrak{g}$, then $B_{\mathfrak{g}}{|}_{\mathfrak{k}}$ is negative definite. This is the analytic counterpart of Cartan's criterion: the Killing form detects the compact structure just as it detects the semisimple structure.

Synthesis. The foundational reason Cartan's criterion works is that the Killing form is an invariant bilinear form whose degeneracy locus coincides with the radical of the Lie algebra. The central insight is that solvability is detected by the vanishing of $B$ on $[\mathfrak{g},\mathfrak{g}]\times \mathfrak{g}$, and semisimplicity is detected by the non-degeneracy of $B$ on all of $\mathfrak{g}\times \mathfrak{g}$. Putting these together, the Killing form computes both pieces of the Levi decomposition: the radical is the kernel of $B$ restricted to the largest solvable ideal, and the Levi factor is the Killing-orthogonal complement.

This is exactly the structure that identifies the radical with the degeneracy of $B$ and the semisimple quotient with the non-degenerate remainder. The bridge is between the abstract decomposition theorem (every Lie algebra is a semidirect product of its radical and a semisimple subalgebra) and the computable invariant (the Killing form matrix, whose rank equals the dimension of the semisimple part). The pattern recurs in 07.06.10 where the Casimir element is built from the Killing form's dual basis and acts as a scalar on each irreducible, and in 07.06.17 where the Cartan subalgebra is identified using the Killing form's restriction to semisimple elements. The generalises direction runs toward the full Cartan-Killing classification: Cartan's criterion identifies the semisimple algebras, and the root-space decomposition 07.06.18 classifies them.

Full proof set [Master]

Proposition 1 (Simple ideals are Killing-orthogonal). Let $\mathfrak{g}$ be a semisimple Lie algebra with simple ideal decomposition $\mathfrak{g}={\mathfrak{g}}_1\oplus \cdots \oplus {\mathfrak{g}}_k$. Then $B({\mathfrak{g}}_i,{\mathfrak{g}}_j)=0$ for $i\ne j$.

Proof. For $X\in {\mathfrak{g}}_i$ and $Y\in {\mathfrak{g}}_j$ with $i\ne j$: the bracket $[X,Y]=0$ (since ${\mathfrak{g}}_i$ and ${\mathfrak{g}}_j$ are distinct ideals, $[{\mathfrak{g}}_i,{\mathfrak{g}}_j]\subseteq {\mathfrak{g}}_i\cap {\mathfrak{g}}_j=0$). So ${\mathrm{ad}}_X$ maps ${\mathfrak{g}}_j$ to zero and ${\mathrm{ad}}_Y$ maps ${\mathfrak{g}}_i$ to zero. In the decomposition $\mathfrak{g}={\mathfrak{g}}_i\oplus {\mathfrak{g}}_j\oplus \mathfrak{r}$ (where $\mathfrak{r}$ is the remaining summands), ${\mathrm{ad}}_X$ is block diagonal with zero block on ${\mathfrak{g}}_j$, and ${\mathrm{ad}}_Y$ is block diagonal with zero block on ${\mathfrak{g}}_i$. Their composition ${\mathrm{ad}}_X\circ {\mathrm{ad}}_Y$ is block diagonal with zero on ${\mathfrak{g}}_i$ (from ${\mathrm{ad}}_Y$), zero on ${\mathfrak{g}}_j$ (from ${\mathrm{ad}}_X$), and zero on $\mathfrak{r}$ (from $[{\mathfrak{g}}_i,\mathfrak{r}]\cap [{\mathfrak{g}}_j,\mathfrak{r}]$ orthogonality). So $\mathrm{tr}({\mathrm{ad}}_X\circ {\mathrm{ad}}_Y)=0$. $\square$

Proposition 2 (Killing form rigidity on simple algebras). On a simple Lie algebra $\mathfrak{g}$ over $\mathbb{C}$, every invariant symmetric bilinear form is a scalar multiple of the Killing form.

Proof. Let $\beta$ be an invariant symmetric bilinear form on $\mathfrak{g}$. Since $B$ is non-degenerate (semisimplicity), the map $f:\mathfrak{g}\to \mathfrak{g}$ defined by $\beta (X,Y)=B(f(X),Y)$ for all $Y$ is well-defined: $f(X)$ is the unique element with $B(f(X),\cdot )=\beta (X,\cdot )$. Invariance of $\beta$ gives $\beta ([Z,X],Y)=\beta (X,[Z,Y])$ for all $Z$. Rewriting via $f$: $B(f([Z,X]),Y)=B([Z,f(X)],Y)$ for all $Y$, using the invariance of $B$. So $f([Z,X])=[Z,f(X)]$, i.e., $f$ commutes with ${\mathrm{ad}}_Z$ for all $Z$. By Schur's lemma (the adjoint representation is irreducible when $\mathfrak{g}$ is simple), $f=\lambda \cdot \mathrm{id}$ for some $\lambda \in \mathbb{C}$. Then $\beta (X,Y)=B(\lambda X,Y)=\lambda B(X,Y)$. $\square$

Proposition 3 (The radical of $B$ is solvable). The radical $\mathrm{rad}(B)=\{X\in \mathfrak{g}:B(X,\mathfrak{g})=0\}$ of the Killing form is a solvable ideal.

Proof. That $\mathrm{rad}(B)$ is an ideal was Exercise 5. For solvability: the subalgebra $\mathfrak{r}=\mathrm{rad}(B)$ has its own Killing form $B_{\mathfrak{r}}$. By definition, $B_{\mathfrak{r}}(X,Y)=\mathrm{tr}({\mathrm{ad}}_X^{\mathfrak{r}}\circ {\mathrm{ad}}_Y^{\mathfrak{r}})$ where ${\mathrm{ad}}^{\mathfrak{r}}$ is the adjoint in $\mathfrak{r}$. Since $\mathfrak{r}$ is an ideal, ${\mathrm{ad}}_X^{\mathfrak{r}}={\mathrm{ad}}_X{|}_{\mathfrak{r}}$ for $X\in \mathfrak{r}$, and the Killing form of $\mathfrak{r}$ is a restriction of the Killing form of $\mathfrak{g}$: $B_{\mathfrak{r}}(X,Y)=B(X,Y)$ for $X,Y\in \mathfrak{r}$ (when the trace is computed in the same ambient space). Since $\mathfrak{r}=\mathrm{rad}(B)$, we have $B(X,Y)=0$ for all $X,Y\in \mathfrak{r}$. In particular $B_{\mathfrak{r}}([\mathfrak{r},\mathfrak{r}],\mathfrak{r})\subseteq B(\mathfrak{r},\mathfrak{r})=0$. By Cartan's solvability criterion (the converse direction, applied to the restriction), $\mathfrak{r}$ is solvable. $\square$

Connections [Master]

• Engel's theorem and Lie's theorem 07.06.14. The proof of Cartan's solvability criterion relies on Lie's theorem to produce the upper-triangular form that forces $B([\mathfrak{g},\mathfrak{g}],\mathfrak{g})=0$ for solvable $\mathfrak{g}$, and on Engel's theorem to conclude that nilpotency of every ${\mathrm{ad}}_Z$ for $Z\in [\mathfrak{g},\mathfrak{g}]$ implies $[\mathfrak{g},\mathfrak{g}]$ is nilpotent. The two theorems of 07.06.14 are the engine inside Cartan's criterion.

• Casimir element 07.06.10. The Casimir element $C={\sum }_a{X}_a{X}^a$ is built from the Killing form's dual basis, which exists precisely when $B$ is non-degenerate — the condition Cartan's criterion identifies with semisimplicity. On a non-semisimple algebra the Killing form is degenerate and the Casimir is undefined. The complementarity is exact: Cartan's criterion detects where the Casimir construction works.

• Root system 07.06.03. Root-space decomposition 07.06.18 requires a semisimple Lie algebra, and Cartan's criterion is the test for semisimplicity. The Killing form induces the inner product on the root lattice, and the non-degeneracy of $B$ on the Cartan subalgebra is what makes the root system Euclidean rather than degenerate. The entire root-system apparatus presupposes the non-degeneracy that Cartan's criterion guarantees.

Historical & philosophical context [Master]

Wilhelm Killing introduced the bilinear form $B(X,Y)=\mathrm{tr}({\mathrm{ad}}_X\circ {\mathrm{ad}}_Y)$ in his four-part paper Die Zusammensetzung der stetigen endlichen Transformationsgruppen published in Mathematische Annalen (volumes 31, 33, 34, 36, 1888–1890) ^{[Killing1888]}. Killing's goal was the classification of continuous transformation groups; the form that bears his name arose as a computational tool for identifying the structure of the root system, though Killing did not isolate the non-degeneracy criterion as a theorem.

Élie Cartan, in his 1894 Paris thesis Sur la structure des groupes de transformations finis et continus ^[Cartan1894], gave the first rigorous proof of the criterion distinguishing solvable from semisimple Lie algebras via the trace form. Cartan's thesis reformulated and corrected Killing's classification, and the solvability criterion (now called Cartan's criterion) was the key structural input that made the classification of simple Lie algebras possible: one first identifies the semisimple algebras by non-degeneracy of $B$, then decomposes into simple ideals by the orthogonality of the Killing form on the direct summands.

The Levi decomposition (every Lie algebra is a semidirect product of its radical and a semisimple subalgebra) was conjectured by Cartan and proved by Eugenio Elia Levi in 1905 ^[Levi1905]. The conjugacy of Levi subalgebras was proved by Malcev in 1942. The cohomological interpretation via Whitehead's lemma and the vanishing $H^1(\mathfrak{g},V)=0$ for semisimple $\mathfrak{g}$ crystallised with the work of Chevalley and Eilenberg in 1948.

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