The Adjoint Representation and Killing Form

⚡ Key Takeaways

The adjoint representation $\mathrm{ad}: \mathfrak{g} \to \mathfrak{gl}(\mathfrak{g})$ sends $X \mapsto [X,-]$ , realizing a Lie algebra as operators on itself; the Killing form $B(X,Y) = \mathrm{tr}(\mathrm{ad}\,X\,\mathrm{ad}\,Y)$ is the canonical invariant symmetric bilinear form built from it. Key result: Cartan's criteria — $\mathfrak{g}$ is solvable iff $B(\mathfrak{g},[\mathfrak{g},\mathfrak{g}])=0$ , and semisimple iff $B$ is nondegenerate. Applies in: structure theory of Lie algebras, representation theory, differential geometry of Lie groups, gauge theory. Who/when: Wilhelm Killing (1888–1890), Élie Cartan (1894 thesis).

💡 The Core Insight

A Lie algebra has no preferred way to "measure" its elements — but the bracket gives one for free. By letting each element act as a linear operator (via $\mathrm{ad}$ ) and then taking the trace of products of these operators, we manufacture a canonical inner-product-like object out of nothing but the bracket itself. The miracle is that this object's degeneracy locus is exactly the obstruction to semisimplicity, turning a subtle structural question into a determinant computation.

DefinitionAdjoint Representation

Let $\mathfrak{g}$ be a Lie algebra over a field $k$ . For each $X \in \mathfrak{g}$ define the linear operator $\mathrm{ad}_X : \mathfrak{g} \to \mathfrak{g}$ by

1

\quad \mathrm{ad}_X(Y) = [X,Y], \qquad X,Y \in \mathfrak{g}.

Derivation hint: This is just the bracket with the first slot fixed; linearity in $Y$ is bilinearity of the bracket. The map $\mathrm{ad} : \mathfrak{g} \to \mathfrak{gl}(\mathfrak{g})$ , $X \mapsto \mathrm{ad}_X$ , is a Lie algebra homomorphism:

2

\quad \mathrm{ad}_{[X,Y]} = [\mathrm{ad}_X, \mathrm{ad}_Y] = \mathrm{ad}_X \circ \mathrm{ad}_Y - \mathrm{ad}_Y \circ \mathrm{ad}_X.

Derivation hint: Apply both sides to arbitrary $Z$ and expand using the Jacobi identity $[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0$ . The kernel $\ker(\mathrm{ad})$ is the center $\mathfrak{z}(\mathfrak{g}) = \{X : [X,\mathfrak{g}]=0\}$ ; thus $\mathrm{ad}$ is faithful iff $\mathfrak{z}(\mathfrak{g})=0$ .

DefinitionKilling Form

The Killing form of $\mathfrak{g}$ (finite-dimensional) is the symmetric bilinear form

3

\quad B(X,Y) = \mathrm{tr}\!\left(\mathrm{ad}_X \circ \mathrm{ad}_Y\right), \qquad X,Y \in \mathfrak{g}.

Derivation hint: $\mathrm{ad}_X\mathrm{ad}_Y \in \mathrm{End}(\mathfrak{g})$ is a finite-rank operator; its trace is well-defined and symmetric since $\mathrm{tr}(AB)=\mathrm{tr}(BA)$ . In a basis $\{e_a\}$ with structure constants $[e_a,e_b] = f^c_{ab} e_c$ , the Gram matrix is

4

\quad B_{ab} = B(e_a,e_b) = f^d_{ac}\, f^c_{bd}.

Derivation hint: $(\mathrm{ad}_{e_a})^c{}_d = f^c_{ad}$ ; compose two such matrices and take the trace by setting the outer indices equal and summing. The form is invariant (associative): for all $X,Y,Z$ ,

5

\quad B([X,Y],Z) = B(X,[Y,Z]).

Derivation hint: Use $[3]$ , the homomorphism property $[2]$ , cyclicity of trace, and $\mathrm{tr}([A,B]C)=\mathrm{tr}(A[B,C])$ .

TheoremCartan's Solvability Criterion

A finite-dimensional Lie algebra $\mathfrak{g}$ over a field of characteristic $0$ is solvable if and only if

6

\quad B(X,Y)=0 \quad\text{for all } X\in\mathfrak{g},\ Y\in[\mathfrak{g},\mathfrak{g}].

Derivation hint: One direction follows from Lie's theorem (solvable ⇒ simultaneously upper-triangularizable ⇒ $\mathrm{ad}$ of a commutator is strictly upper triangular, hence traceless products); the converse uses Cartan's trace lemma on $\mathfrak{gl}(V)$ .

TheoremCartan's Semisimplicity Criterion

A finite-dimensional Lie algebra $\mathfrak{g}$ over a field of characteristic $0$ is semisimple (zero radical) if and only if its Killing form is nondegenerate:

7

\quad \mathfrak{g}^{\perp} := \{X : B(X,Y)=0\ \forall Y\} = 0 \iff \det(B_{ab}) \neq 0.

Derivation hint: The radical of $B$ is a solvable ideal (by $[6]$ applied internally), so semisimple ⇒ $\mathrm{rad}\,B=0$ ; conversely $\mathrm{rad}\,B$ contains the solvable radical.

ProofThe adjoint map is a homomorphism (Equation 2)

Step 1. Fix $X,Y,Z \in \mathfrak{g}$ . Compute the right-hand side applied to $Z$ :

[\mathrm{ad}_X,\mathrm{ad}_Y](Z) = \mathrm{ad}_X\mathrm{ad}_Y(Z) - \mathrm{ad}_Y\mathrm{ad}_X(Z) = [X,[Y,Z]] - [Y,[X,Z]].

Step 2. The Jacobi identity in the form $[X,[Y,Z]] = [[X,Y],Z] + [Y,[X,Z]]$ gives

[X,[Y,Z]] - [Y,[X,Z]] = [[X,Y],Z].

Step 3. But $[[X,Y],Z] = \mathrm{ad}_{[X,Y]}(Z)$ . Since $Z$ was arbitrary, $[\mathrm{ad}_X,\mathrm{ad}_Y] = \mathrm{ad}_{[X,Y]}$ . $\blacksquare$

∎

ProofInvariance of the Killing form (Equation 5)

Step 1. Expand using the definition and the homomorphism property:

B([X,Y],Z) = \mathrm{tr}\big(\mathrm{ad}_{[X,Y]}\,\mathrm{ad}_Z\big) = \mathrm{tr}\big([\mathrm{ad}_X,\mathrm{ad}_Y]\,\mathrm{ad}_Z\big).

Step 2. Write $A=\mathrm{ad}_X$ , $\,C=\mathrm{ad}_Y$ , $\,D=\mathrm{ad}_Z$ . Then

\mathrm{tr}([A,C]D) = \mathrm{tr}(ACD - CAD) = \mathrm{tr}(ACD) - \mathrm{tr}(CAD).

Step 3. By cyclicity $\mathrm{tr}(CAD)=\mathrm{tr}(ADC)$ , so

\mathrm{tr}([A,C]D) = \mathrm{tr}(ACD - ADC) = \mathrm{tr}(A[C,D]).

Step 4. Translate back: $\mathrm{tr}(A[C,D]) = \mathrm{tr}(\mathrm{ad}_X\,\mathrm{ad}_{[Y,Z]}) = B(X,[Y,Z])$ . Hence $B([X,Y],Z)=B(X,[Y,Z])$ . $\blacksquare$

∎

ProofThe radical of

B

is an ideal

Step 1. Let $\mathfrak{r} = \mathfrak{g}^\perp = \{X : B(X,Y)=0\ \forall Y\}$ . Take $X\in\mathfrak{r}$ , $W\in\mathfrak{g}$ , $Y\in\mathfrak{g}$ .

Step 2. By invariance $[5]$ , $B([W,X],Y) = B(W,[X,Y])$ . We want to show $[W,X]\in\mathfrak{r}$ , i.e. $B([W,X],Y)=0$ .

Step 3. Note $B(W,[X,Y]) = -B(W,[Y,X])$ and apply invariance again: $B(W,[X,Y]) = -B([Y,W]... )$ . More directly, rewrite $B([W,X],Y)=B(X,[Y,W])$ using $[5]$ twice; since $X\in\mathfrak{r}$ and $[Y,W]\in\mathfrak{g}$ , this vanishes.

Step 4. Therefore $[W,X]\in\mathfrak{r}$ for all $W$ , so $\mathfrak{r}$ is an ideal. Combined with Cartan's criterion $[6]$ , $\mathfrak{r}$ is a solvable ideal, hence contained in the radical. $\blacksquare$

∎

mathematician · Wilhelm Killing | 1847–1923

A German mathematician and gymnasium teacher, Killing introduced the eponymous form and the root-system machinery between 1888 and 1890 in his series "Die Zusammensetzung der stetigen endlichen Transformationsgruppen," seeking to classify all continuous transformation groups. Working largely in isolation and burdened with heavy teaching duties, he discovered Cartan subalgebras, root systems, the Cartan matrix, and essentially the full list of simple Lie algebras — including the exceptional ones $G_2, F_4, E_6, E_7, E_8$ . His proofs contained gaps and errors that Élie Cartan later repaired and completed in his 1894 doctoral thesis; Cartan generously credited Killing for the central ideas. The phrase "Killing form" was coined much later, but the object is genuinely his.

🧠 Intuition

Think of $\mathrm{ad}_X$ as the infinitesimal rotation/shear that $X$ induces on the algebra when you "stir" with the bracket. The Killing form $B(X,Y)$ asks: how much do the stirring motions of $X$ and $Y$ reinforce each other on average? — averaged via the trace over all directions in $\mathfrak{g}$ . If $X$ is central ( $\mathrm{ad}_X=0$ , no stirring), it contributes nothing to $B$ , so it lands in the radical of the form — degeneracy. More generally, the parts of $\mathfrak{g}$ that act "nilpotently / triangularly" on the whole algebra produce traceless products and slip into the kernel of $B$ . Solvable algebras are exactly those that can be simultaneously triangularized, which is why their commutators always pair to zero. Semisimple algebras have no such hiding places: every nonzero direction couples to something, so $B$ is nondegenerate — a perfect pairing that lets you raise and lower indices, define Casimir elements, and identify $\mathfrak{g} \cong \mathfrak{g}^*$ as a representation.

Figure

The Killing form as a pairing: nondegenerate (semisimple) vs. degenerate (with radical)

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ExampleThe Killing form of

\mathfrak{sl}_2(\mathbb{C})

Given. Basis $H,E,F$ with relations $[H,E]=2E$ , $[H,F]=-2F$ , $[E,F]=H$ .

Step 1 — compute $\mathrm{ad}$ matrices in the ordered basis $(H,E,F)$ . $\mathrm{ad}_H$ : $[H,H]=0$ , $[H,E]=2E$ , $[H,F]=-2F$ , so $\mathrm{ad}_H = \mathrm{diag}(0,2,-2)$ . $\mathrm{ad}_E$ : $[E,H]=-2E$ , $[E,E]=0$ , $[E,F]=H$ , giving columns:

\mathrm{ad}_E = \begin{pmatrix} 0&0&1\\ -2&0&0\\ 0&0&0\end{pmatrix}.

$\mathrm{ad}_F$ : $[F,H]=2F$ , $[F,E]=-H$ , $[F,F]=0$ :

\mathrm{ad}_F = \begin{pmatrix} 0&-1&0\\ 0&0&0\\ 2&0&0\end{pmatrix}.

Step 2 — diagonal entries. $B(H,H)=\mathrm{tr}(\mathrm{ad}_H^2)=0^2+2^2+(-2)^2=8.$ $B(E,F)=\mathrm{tr}(\mathrm{ad}_E\,\mathrm{ad}_F)$ . Multiplying: $(\mathrm{ad}_E\mathrm{ad}_F)$ has diagonal $(2,0,2)$ ... compute: trace $=4$ . By symmetry $B(F,E)=4$ .

Step 3 — cross terms. $B(H,E)=\mathrm{tr}(\mathrm{ad}_H\mathrm{ad}_E)=0$ , $B(H,F)=0$ , $B(E,E)=B(F,F)=0$ (nilpotent products are traceless here).

Boxed answer.

\boxed{\,B = \begin{pmatrix} 8&0&0\\ 0&0&4\\ 0&4&0\end{pmatrix},\quad \det B = -8\cdot 16 = -128 \neq 0\,}

Nondegenerate ⇒ $\mathfrak{sl}_2$ is semisimple (indeed simple). Note $B = 4\,\mathrm{tr}(XY)$ for the defining $2\times2$ representation, illustrating $B_{\mathfrak{g}} = 2h^\vee\,\langle\cdot,\cdot\rangle$ with dual Coxeter number $h^\vee=2$ .

ExampleA solvable algebra: the

2

-dimensional non-abelian Lie algebra

Given. Basis $X,Y$ with the single nontrivial relation $[X,Y]=Y$ (so $[\mathfrak{g},\mathfrak{g}]=\mathbb{R}Y$ ).

Step 1 — $\mathrm{ad}$ matrices in basis $(X,Y)$ . $\mathrm{ad}_X$ : $[X,X]=0$ , $[X,Y]=Y$ , so $\mathrm{ad}_X = \begin{pmatrix}0&0\\0&1\end{pmatrix}$ . $\mathrm{ad}_Y$ : $[Y,X]=-Y$ , $[Y,Y]=0$ , so $\mathrm{ad}_Y = \begin{pmatrix}0&0\\-1&0\end{pmatrix}$ .

Step 2 — Killing entries. $B(X,X)=\mathrm{tr}(\mathrm{ad}_X^2)=0+1=1.$ $B(X,Y)=\mathrm{tr}(\mathrm{ad}_X\mathrm{ad}_Y)=\mathrm{tr}\begin{pmatrix}0&0\\0&1\end{pmatrix}\begin{pmatrix}0&0\\-1&0\end{pmatrix}=\mathrm{tr}\begin{pmatrix}0&0\\-1&0\end{pmatrix}=0.$ $B(Y,Y)=\mathrm{tr}(\mathrm{ad}_Y^2)=\mathrm{tr}\begin{pmatrix}0&0\\0&0\end{pmatrix}=0.$

Step 3 — check solvability criterion. For $Z\in[\mathfrak{g},\mathfrak{g}]=\mathbb{R}Y$ : $B(X,Y)=0$ and $B(Y,Y)=0$ . So $B(\mathfrak{g},[\mathfrak{g},\mathfrak{g}])=0$ — consistent with Cartan's criterion $[6]$ .

Boxed answer.

\boxed{\,B=\begin{pmatrix}1&0\\0&0\end{pmatrix},\quad \det B = 0,\quad \mathrm{rad}\,B = \mathbb{R}Y\,}

The form is degenerate; the radical equals the derived subalgebra. The algebra is solvable, not semisimple.

⚠️ Common mistake

The Killing form is NOT always an inner product, and not always nondegenerate. Three frequent errors: 1. " $B$ is positive-definite." False even for semisimple algebras: $B$ on $\mathfrak{sl}_2(\mathbb{R})$ has signature $(2,1)$ . Definiteness (negative-definite) holds only for compact real forms. 2. " $B$ nondegenerate $\Leftrightarrow$ $\mathfrak{g}$ simple." No — nondegeneracy means semisimple (a direct sum of simples). Simplicity is stronger. 3. "Cartan's criteria hold in any characteristic." They require characteristic $0$ (or characteristic $>$ some bound). In positive characteristic, the trace machinery fails and there exist semisimple algebras with degenerate Killing form.

Counterexample

Abelian algebra has identically zero Killing form. Let $\mathfrak{g}=\mathbb{R}^n$ with all brackets zero. Then $\mathrm{ad}_X=0$ for every $X$ , so $B\equiv 0$ — maximally degenerate, $\mathrm{rad}\,B=\mathfrak{g}$ . This shows $B$ cannot detect the dimension or fine structure of solvable parts; it only "sees" semisimple structure.

A subtler counterexample — degeneracy in characteristic $p$ . Over $\mathbb{F}_p$ , the algebra $\mathfrak{sl}_p(\mathbb{F}_p)$ contains the scalar identity (since $\mathrm{tr}(I_p)=p=0$ ), giving a nonzero center; the Killing form becomes degenerate even though the algebra is "morally semisimple." Cartan's criterion genuinely needs $\mathrm{char}\,k=0$ .

connection

The adjoint representation and Killing form sit at a crossroads of algebra, geometry, and physics:

Root Systems and Cartan Subalgebras — restricting $B$ to a Cartan subalgebra $\mathfrak{h}$ and dualizing gives the inner product on the weight lattice that defines root lengths and angles; the entire classification of Dynkin Diagrams is geometry of $B|_{\mathfrak{h}^*}$ .
Casimir Element — when $B$ is nondegenerate, choosing dual bases $\{e_a\},\{e^a\}$ with $B(e_a,e^b)=\delta_a^b$ yields the quadratic Casimir $\Omega = \sum_a e_a e^a$ in the universal enveloping algebra, central by invariance $[5]$ .
Compact Lie Groups and Bi-invariant Metrics — on a compact semisimple group, $-B$ is the canonical bi-invariant Riemannian metric; geodesics are one-parameter subgroups.
Semisimple Lie Algebras — Cartan's criterion $[7]$ is the definition-in-practice, and nondegeneracy drives Weyl's complete reducibility theorem.
Levi Decomposition — the radical of $B$ probes the solvable radical, the obstruction split off by Levi's theorem.

🔬 Where this appears

⏳ History

- 1873–1880s — Sophus Lie develops the theory of continuous transformation groups and their infinitesimal generators, implicitly the adjoint action.

- 1888 — Wilhelm Killing publishes the first installment of "Die Zusammensetzung der stetigen endlichen Transformationsgruppen," introducing what is now the Killing form and the characteristic equation of

\mathrm{ad}_X

.

- 1888–1890 — Killing completes four papers, discovering Cartan subalgebras, roots, the Cartan integers, and the classification list including the exceptional algebras.

- 1894 — Élie Cartan's doctoral thesis "Sur la structure des groupes de transformations finis et continus" rigorously establishes the solvability and semisimplicity criteria (Equations

[6]

–

[7]

) and corrects Killing's gaps.

- 1913–1925 — Cartan and Hermann Weyl develop the representation theory; Weyl proves complete reducibility (1925) using the Casimir built from

B

.

- 1925 — Weyl's "unitarian trick" exploits the negative-definiteness of

B

on compact real forms to transfer results between compact and complex groups.

- 1934 — Hans Zassenhaus and others systematize the structure theory; the name "Killing form" gains currency through Dynkin and Bourbaki later.

- 1947–1952 — Eugene Dynkin introduces Dynkin diagrams, recasting the

B

-geometry of roots into combinatorial form.

- 1955–1960s — Chevalley constructs the algebras over arbitrary fields, exposing the characteristic-

p

failures of the Killing-form criteria.

open-problem

Positive characteristic and modular Lie algebras. Cartan's criteria collapse when $\mathrm{char}\,k = p > 0$ : the trace form may be degenerate on simple algebras, and the classification of simple modular Lie algebras (restricted and non-restricted, including Cartan-type and Melikyan algebras) was only completed for $p>3$ by Block, Wilson, Strade, and Premet in a decades-long program finished around 2004–2009. The case $p=2,3$ remains incompletely understood, and the very notion of "Killing form" must be supplemented by other invariant bilinear forms.

Infinite-dimensional analogues. For Kac–Moody algebras and Lie algebras of vector fields, the naive trace defining $[3]$ diverges. One replaces $B$ by a symmetrizable invariant form defined axiomatically (the Killing form of an affine algebra is degenerate along the center–derivation axis), and questions about which infinite-dimensional Lie algebras admit nondegenerate invariant forms remain active.

Optimal normalization conventions. Across physics literature there is persistent ambiguity in normalizing $B$ relative to the trace form of a chosen representation (factors of $2h^\vee$ ); pinning down convention-independent statements (anomaly cancellation, level quantization) is a recurring technical subtlety rather than a solved formula.

The Adjoint Representation and Killing Form

Formal Statement

Why It Holds / Derivation

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also