research32 min studyLesson 502 in Mathematics

The Killing form and Cartan's criteria

Use the adjoint trace pairing to distinguish nondegeneracy, solvability, and positivity.

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What you’ll understand

  • Compute a Killing matrix in a declared Lie-algebra basis.
  • Apply Cartan's semisimplicity and solvability criteria under characteristic-zero hypotheses.
  • Explain why an isotropic vector does not imply a degenerate bilinear form.

The bracket gives an adjoint action on the Lie algebra itself. Taking traces of products of those actions produces the Killing form, a bilinear pairing whose degeneracy detects semisimplicity in characteristic zero. It is not a Euclidean norm and should not be tested by positivity. In this lesson all Lie algebras are finite dimensional over ℂ, and all bilinear forms are complex bilinear rather than Hermitian unless explicitly stated otherwise.

The adjoint trace pairing is intrinsic and invariant

Define B(x,y) as the trace of ad(x)ad(y). Cyclicity of trace makes it symmetric, and the Jacobi identity gives ad([x,y])=[ad(x),ad(y)]. Combining these identities proves invariance: B([x,y],z)=B(x,[y,z]). Thus the form is independent of a chosen matrix realization of g. A defining-representation trace form can be proportional to the Killing form in important examples, but it is not automatically the same normalization or even nondegenerate for an arbitrary representation.

The radical of a bilinear form consists of vectors pairing to zero with every vector. Invariance makes the Killing radical an ideal. A matrix representing B is invertible exactly when this radical is zero. One vector with B(x,x)=0 is merely isotropic; it need not pair to zero with other vectors. Over ℂ, even a nondegenerate symmetric form has isotropic vectors in many dimensions, so positivity is the wrong diagnostic. Hermitian positivity belongs to a different type of form.

B(x,y)=tr⁡(ad⁡x ad⁡y),B([x,y],z)=B(x,[y,z]),rad⁡B={x:B(x,g)=0}B(x,y)=\operatorname{tr}(\operatorname{ad}x\,\operatorname{ad}y),\qquad B([x,y],z)=B(x,[y,z]),\qquad \operatorname{rad}B=\{x:B(x,\mathfrak g)=0\}
The trace is taken on the finite-dimensional adjoint module. The form radical is defined by pairing with every element, not by a single self-pairing.↔ Scroll wide equations horizontally.

Two Cartan criteria have different conclusions

Cartan's semisimplicity criterion states that g is semisimple if and only if its Killing form is nondegenerate, for finite-dimensional Lie algebras over a characteristic-zero field; here the field is ℂ. Cartan's solvability criterion instead states that g is solvable if and only if B(g,[g,g])=0. The second condition does not mean the whole Killing form is zero. A solvable algebra can have a nonzero trace pairing on elements outside its derived algebra, as the two-dimensional affine example shows.

These are structural theorems, not consequences of matrix symmetry alone. A singular Killing matrix rules out semisimplicity but does not by itself prove that the whole algebra is solvable; an algebra with a semisimple factor and nonzero center supplies a counterexample. To prove solvability by the criterion, check orthogonality of the entire derived algebra to all of g. Likewise a nonzero determinant proves nondegeneracy in a chosen basis, but the value of that determinant changes under basis rescaling while the structural conclusion does not.

g semisimple  ⟺  rad⁡B=0,g solvable  ⟺  B(g,[g,g])=0\mathfrak g\text{ semisimple}\iff\operatorname{rad}B=0,\qquad \mathfrak g\text{ solvable}\iff B(\mathfrak g,[\mathfrak g,\mathfrak g])=0
Both equivalences use finite dimensionality and characteristic zero. The solvability condition is an orthogonality statement for the derived algebra, not a claim that B vanishes everywhere.↔ Scroll wide equations horizontally.

A form's normalization matters to later constructions

For sl(n,C), the Killing form is 2n times the ordinary trace of a product in the defining n-dimensional representation. This formula permits efficient calculations once its normalization is known. On a simple complex Lie algebra every invariant symmetric bilinear form is a scalar multiple of the Killing form. On a semisimple direct sum, separate simple factors may be scaled independently. Nondegeneracy survives multiplication by a nonzero scalar, but a dual-basis construction such as a Casimir element changes inversely with that scale.

Restriction to a compact real form gives a negative-definite Killing form, so its negative defines a useful positive real inner product there. The same assertion is false on arbitrary noncompact real forms, and it is not a positivity assertion on the complex vector space. Keeping the field and type of pairing explicit prevents a common transfer error between compact-group geometry and complex representation theory. This lesson uses the complex bilinear form to analyze algebraic structure, leaving positive Hermitian forms on representations to their separate context.

The Killing matrix of sl(2,C)

Use h=diag(1,−1), e=E₁₂, and f=E₂₁, with [h,e]=2e, [h,f]=−2f, and [e,f]=h. Compute B in the ordered basis h,e,f.

  1. The adjoint action of h has eigenvalues 0,2,−2, so B(h,h)=0²+2²+(−2)²=8. The defining trace formula is B(x,y)=4tr(xy).

  2. It gives B(e,f)=B(f,e)=4 and all other entries except B(h,h) equal zero. Thus the matrix is [[8,0,0],[0,0,4],[0,4,0]].

  3. Its determinant is 8·(−16)=−128, so the form is nondegenerate and sl(2,C) is semisimple. Although B(e,e)=0, the value B(e,f)=4 shows that e is not in its radical.

The result

The form is nondegenerate with an isotropic nonzero vector e. Nondegeneracy and positive definiteness are different properties.

Explore the model

Interactive lab

Interactive lab · vectors

Test pairing against more than one vector

Reset, set angle 90° and magnitude 1. The vector's pairing with the first coordinate vector is its x projection, zero; its pairing with the second is its y projection, one. Compare a fresh run with Reset, angle 0° and magnitude 1, reversing those roles. This Euclidean illustration tests the quantifier 'with every vector'; it does not display the complex Killing form or its isotropic self-pairings.

Test pairing against more than one vectorPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA zero pairing with one test vector is insufficient to place a vector in a bilinear-form radical.

Check your understanding

Which observation proves that a vector x lies in the Killing-form radical?

Try it yourself

For the affine algebra with basis h,e and [h,e]=e, compute its Killing matrix and check Cartan's solvability criterion.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

ad(h) sends h to zero and e to e; ad(e) sends h to −e and e to zero.

Compare with the solution

In the basis h,e, ad(h)=[[0,0],[0,1]] and ad(e)=[[0,0],[−1,0]]. Their trace products give Killing matrix [[1,0],[0,0]]. The derived algebra is Ce, orthogonal to all of g, so Cartan's solvability criterion holds. The whole form is not zero, but its radical is Ce and it is degenerate, ruling out semisimplicity as expected.

Take it with you

  • The Killing form is an invariant complex bilinear trace pairing.
  • Cartan's two criteria test different orthogonality conditions.
  • Field, positivity, and scale conventions must remain explicit.

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