Cartan's Criterion for Semisimplicity

Derives Lemma A31.3. (1) If \(\mathfrak{b} \subseteq \mathfrak{a}\) then \(\mathfrak{b}^{(j)} \subseteq \mathfrak{a}^{(j)}\) by induction; if \(\phi\) is a homomorphism then \(\phi\left(\mathfrak{a}^{(j)}\right) = \phi(\mathfrak{a})^{(j)}\), again by induction, since \(\phi\) preserves brackets.

(2) If \(\left(\mathfrak{a}/\mathfrak{b}\right)^{(m)} = 0\) then \(\mathfrak{a}^{(m)} \subseteq \mathfrak{b}\) by (1) applied to the quotient map, and if \(\mathfrak{b}^{(l)} = 0\) then \(\mathfrak{a}^{(m+l)} \subseteq \mathfrak{b}^{(l)} = 0\).

(3) By induction on \(j\); the case \(j = 0\) is the hypothesis. Let \(\mathfrak{b} = \mathfrak{a}^{(j)}\) be an ideal. For \(Z \in \mathfrak{g}\) and \(X,Y \in \mathfrak{b}\) the Jacobi identity gives

and both terms on the right lie in \(\comm{\mathfrak{b}}{\mathfrak{b}} = \mathfrak{a}^{(j+1)}\) because \(\comm{Z}{X}\) and \(\comm{Z}{Y}\) lie in \(\mathfrak{b}\).

(4) Let \(\mathfrak{s} \neq 0\) be a solvable ideal and let \(m\) be largest with \(\mathfrak{s}^{(m)} \neq 0\). Then \(\comm{\mathfrak{s}^{(m)}}{\mathfrak{s}^{(m)}} = \mathfrak{s}^{(m+1)} = 0\), so \(\mathfrak{s}^{(m)}\) is abelian, and it is an ideal of \(\mathfrak{g}\) by (3).

Derives Proposition A31.5. Let \(X \in \mathfrak{a}\) and \(Y \in \mathfrak{g}\), and put \(N = \ad_{X}\ad_{Y}\). Since \(\mathfrak{a}\) is an ideal, \(\ad_{X}\mathfrak{g} = \comm{X}{\mathfrak{g}} \subseteq \mathfrak{a}\), so \(N\mathfrak{g} \subseteq \mathfrak{a}\); and since \(\mathfrak{a}\) is an ideal, \(\ad_{Y}\mathfrak{a} \subseteq \mathfrak{a}\), while \(\ad_{X}\mathfrak{a} = \comm{\mathfrak{a}}{\mathfrak{a}} = 0\) because \(\mathfrak{a}\) is abelian. Hence

so \(N^{2} = 0\). A nilpotent endomorphism has all eigenvalues zero and therefore vanishing trace, so \(\kappa(X,Y) = \tr N = 0\) for every \(Y\), that is, \(X \in \mathfrak{g}^{\perp}\).

The same computation in components, with the index conventions of Equation (18.15), is worth displaying because it is the form in which the statement is usually met. Choose a basis adapted to \(\mathfrak{a}\): indices \(i,j,k\) label a basis of \(\mathfrak{a}\) and indices \(\mu,\nu\) complete it. That \(\mathfrak{a}\) is an ideal says \(C^{\mu}{}_{a i} = 0\) for every \(a\); that it is abelian says \(C^{c}{}_{ij} = 0\). Then, from Equation (18.15),

the second equality because \(C^{c}{}_{id}\) vanishes unless the upper index lies in \(\mathfrak{a}\), the third because \(C^{d}{}_{bj}\) vanishes unless \(d\) does, and the last because \(C^{j}{}_{ik} = 0\).

Derives Theorem A31.6. Let \(a_{1},\ldots,a_{r}\) be the distinct eigenvalues of \(x\) and \(\prod_{i}(t-a_{i})^{m_{i}}\) its characteristic polynomial, which splits because the field is algebraically closed. Put \(\mathbb{V}_{i} = \ker\left(x-a_{i}\right)^{m_{i}}\). The polynomials \((t-a_{i})^{m_{i}}\) are pairwise coprime, so by the Chinese remainder theorem in \(\mathbb{F}[t]\) — applied modulo the characteristic polynomial, which annihilates \(x\) by Cayley–Hamilton — the \(\mathbb{V}_{i}\) are the images of projections that are polynomials in \(x\), and \(\mathbb{V} = \bigoplus_{i}\mathbb{V}_{i}\).

By the same theorem choose \(p \in \mathbb{F}[t]\) with

the last congruence being imposed only when \(0\) is not among the \(a_{i}\) — when it is, say \(a_{1} = 0\), the congruence \(p \equiv 0 \pmod{t^{m_{1}}}\) already forces \(p(0) = 0\), and the moduli stay pairwise coprime in either case. Set \(x_{s} = p(x)\) and \(x_{n} = x - x_{s}\). On \(\mathbb{V}_{i}\) the first congruence gives \(x_{s} = a_{i}\identity\), so \(x_{s}\) is diagonalizable, and \(x_{n} = x - a_{i}\identity\) there, which is nilpotent on \(\mathbb{V}_{i}\) by definition of \(\mathbb{V}_{i}\); being nilpotent on each summand of a finite direct sum it is nilpotent. Both are polynomials in \(x\) with \(p(0) = 0\), hence commute with \(x\) and with each other.

Uniqueness. Let \(x = s' + n'\) be another such decomposition. Since \(s'\) and \(n'\) commute with each other they commute with \(x\), hence with every polynomial in \(x\), hence with \(x_{s}\) and \(x_{n}\). Then \(x_{s} - s' = n' - x_{n}\) is at once diagonalizable — a difference of two commuting diagonalizable maps, which are simultaneously diagonalizable — and nilpotent, being a difference of two commuting nilpotent maps. A map that is both is zero.

Derives Lemma A31.7. Let \(e_{1},\ldots,e_{m}\) be a basis of eigenvectors of \(x_{s}\), \(x_{s}e_{i} = a_{i}e_{i}\), and let \(E_{ij}\) be the corresponding matrix units. Then \(\ad_{x_{s}}E_{ij} = (a_{i}-a_{j})E_{ij}\), so \(\ad_{x_{s}}\) is diagonalizable. Writing \(L(y) = x_{n}y\) and \(R(y) = yx_{n}\), the maps \(L\) and \(R\) commute and are nilpotent, so \(\ad_{x_{n}} = L - R\) is nilpotent. And \(\comm{\ad_{x_{s}}}{\ad_{x_{n}}} = \ad_{\comm{x_{s}}{x_{n}}} = 0\). Hence \(\ad_{x} = \ad_{x_{s}} + \ad_{x_{n}}\) satisfies the three conditions of Equation (A31.6), and uniqueness identifies the two summands. The last sentence is the final statement of Theorem A31.6 applied to the endomorphism \(\ad_{x}\) of \(\operatorname{End}(\mathbb{V})\).

Derives Lemma A31.8. With \(L(y) = xy\) and \(R(y) = yx\) as above, \(L\) and \(R\) commute and each is nilpotent, since \(x^{N} = 0\) gives \(L^{N} = R^{N} = 0\). The binomial theorem for commuting maps gives \(\left(L-R\right)^{2N} = 0\).

Derives Theorem A31.9. Induct on \(\dim\mathfrak{h}\). If \(\mathfrak{h} = 0\) any nonzero \(v\) will do. Let \(\dim\mathfrak{h} \ge 1\) and choose a maximal proper subalgebra \(\mathfrak{m} \subsetneq \mathfrak{h}\), which exists because \(0\) is a proper subalgebra and dimensions are finite.

\(\mathfrak{m}\) is an ideal of codimension one. The adjoint action of \(\mathfrak{m}\) on \(\mathfrak{h}\) preserves \(\mathfrak{m}\), so it descends to an action on \(\mathfrak{h}/\mathfrak{m} \neq 0\); each \(\ad_{X}\), \(X \in \mathfrak{m}\), is nilpotent by Lemma A31.8, hence so is the induced map. The image of \(\mathfrak{m}\) in \(\mathfrak{gl}\left(\mathfrak{h}/\mathfrak{m}\right)\) is a subalgebra of dimension at most \(\dim\mathfrak{m} < \dim\mathfrak{h}\) consisting of nilpotent maps, so the induction hypothesis supplies a nonzero \(z + \mathfrak{m}\) annihilated by it: there is \(z \in \mathfrak{h} \setminus \mathfrak{m}\) with \(\comm{\mathfrak{m}}{z} \subseteq \mathfrak{m}\). Then \(\mathfrak{m} + \mathbb{K}z\) is a subalgebra strictly containing \(\mathfrak{m}\), so it is \(\mathfrak{h}\) by maximality, and \(\mathfrak{m}\) is an ideal of codimension one.

The common kernel. Let \(\mathbb{W} = \set{v \in \mathbb{V} \mid \mathfrak{m}v = 0}\), which is nonzero by the induction hypothesis applied to \(\mathfrak{m}\). It is stable under \(\mathfrak{h}\): for \(X \in \mathfrak{m}\), \(Y \in \mathfrak{h}\) and \(v \in \mathbb{W}\),

since \(\comm{X}{Y} \in \mathfrak{m}\). In particular \(z\) maps \(\mathbb{W}\) to itself, and \(z\) is nilpotent, so it has a nonzero kernel vector \(v \in \mathbb{W}\). That \(v\) is annihilated by \(\mathfrak{m}\) and by \(z\), hence by \(\mathfrak{h} = \mathfrak{m} + \mathbb{K}z\).

The flag. Induct on \(\dim\mathbb{V}\): take \(v\) as above as the first basis vector and apply the statement to the induced action on \(\mathbb{V}/\mathbb{K}v\), whose elements are again nilpotent. In the resulting basis every \(X \in \mathfrak{h}\) is strictly upper triangular.

Solvability. Let \(\mathfrak{n}_{r}\) be the space of matrices whose entries vanish except at least \(r\) places above the diagonal, so that \(\mathfrak{h} \subseteq \mathfrak{n}_{1}\) and \(\comm{\mathfrak{n}_{r}}{\mathfrak{n}_{s}} \subseteq \mathfrak{n}_{r+s}\) by matrix multiplication. Then \(\mathfrak{h}^{(j)} \subseteq \mathfrak{n}_{2^{j}}\) by induction, and \(\mathfrak{n}_{r} = 0\) once \(r \ge \dim\mathbb{V}\).

Derives Lemma A31.10. Let \(x = x_{s} + x_{n}\) be the Jordan decomposition (Theorem A31.6), let \(e_{1},\ldots,e_{m}\) be a basis of eigenvectors of \(x_{s}\) with \(x_{s}e_{i} = a_{i}e_{i}\), and let \(E \subseteq \mathbb{F}\) be the \(\Q\)-linear span of \(a_{1},\ldots,a_{m}\), a finite-dimensional vector space over \(\Q\). The claim is that \(E = 0\); then every \(a_{i} = 0\), so \(x_{s} = 0\) and \(x = x_{n}\) is nilpotent.

Since \(E\) is finite-dimensional over \(\Q\), it is enough to show that every \(\Q\)-linear functional \(f : E \to \Q\) vanishes. Fix such an \(f\) and let \(y \in \operatorname{End}(\mathbb{V})\) be defined by \(ye_{i} = f(a_{i})e_{i}\).

Step 1: \(\ad_{y}\) is a polynomial in \(\ad_{x}\) with zero constant term. In the basis of matrix units \(E_{ij}\) built from the \(e_{i}\),

Choose \(r \in \mathbb{F}[t]\) with \(r(0) = 0\) and \(r\left(a_{i}-a_{j}\right) = f(a_{i})-f(a_{j})\) for all \(i,j\). This is a consistent finite set of interpolation conditions: if \(a_{i}-a_{j} = a_{p}-a_{q}\) then, applying the \(\Q\)-linear \(f\), \(f(a_{i})-f(a_{j}) = f(a_{p})-f(a_{q})\); and if \(a_{i}-a_{j} = 0\) the prescribed value is \(0\), which is what \(r(0) = 0\) demands. Lagrange interpolation over the finitely many distinct values then produces \(r\). By Equation (A31.11), \(\ad_{y} = r\left(\ad_{x_{s}}\right)\), and by Lemma A31.7 \(\ad_{x_{s}} = q\left(\ad_{x} \right)\) for some \(q\) with \(q(0) = 0\). Hence \(\ad_{y} = r\left(q\left(\ad_{x}\right)\right)\), a polynomial in \(\ad_{x}\) without constant term.

Step 2: \(y \in M\). Since \(x \in M\) we have \(\ad_{x}(B) \subseteq A \subseteq B\), so \(\left(\ad_{x}\right)^{j}(B) \subseteq A\) for every \(j \ge 1\). A polynomial in \(\ad_{x}\) with zero constant term is a combination of such powers, so \(\ad_{y}(B) \subseteq A\), that is, \(y \in M\).

Step 3: the trace. By hypothesis \(\tr(xy) = 0\). Compute it. Since \(x_{n}\) commutes with \(x_{s}\) it preserves each eigenspace of \(x_{s}\), and inside each eigenspace a basis may be chosen making the nilpotent \(x_{n}\) strictly upper triangular; taking the \(e_{i}\) so adapted, \(x\) is upper triangular with diagonal entries \(a_{1},\ldots,a_{m}\) while \(y\) is diagonal with entries \(f(a_{1}),\ldots,f(a_{m})\). Then \(xy\) is upper triangular with diagonal entries \(a_{i}f(a_{i})\), so

The left-hand side lies in \(E\), since each \(f(a_{i}) \in \Q\) and \(E\) is a \(\Q\)-subspace containing the \(a_{i}\). Apply \(f\) to Equation (A31.12) and use \(\Q\)-linearity:

A sum of squares of rational numbers vanishes only if every term does, so \(f(a_{i}) = 0\) for all \(i\); as the \(a_{i}\) span \(E\) over \(\Q\), \(f = 0\). Every \(\Q\)-linear functional on \(E\) vanishes, so \(E = 0\).

Derives Theorem A31.11. Assume first that \(\mathbb{K}\) is algebraically closed. Apply Lemma A31.10 with \(A = \comm{\mathfrak{h}}{\mathfrak{h}}\), \(B = \mathfrak{h}\) and \(M\) as in Equation (A31.10); note \(\mathfrak{h} \subseteq M\). Let \(x \in \comm{\mathfrak{h}}{\mathfrak{h}}\) and \(y \in M\). By bilinearity it suffices to treat \(x = \comm{u}{v}\) with \(u,v \in \mathfrak{h}\), and then

using cyclicity of the trace twice. Now \(y \in M\) gives \(\comm{v}{y} \in \comm{\mathfrak{h}}{\mathfrak{h}}\), and \(u \in \mathfrak{h}\), so the right-hand side vanishes by Equation (A31.14). Lemma A31.10 therefore makes every element of \(\comm{\mathfrak{h}}{\mathfrak{h}}\) a nilpotent endomorphism, and Theorem A31.9 makes \(\comm{\mathfrak{h}}{\mathfrak{h}}\) solvable. Since \(\mathfrak{h}/\comm{\mathfrak{h}}{\mathfrak{h}}\) is abelian, Lemma A31.3(2) makes \(\mathfrak{h}\) solvable.

For general \(\mathbb{K}\), extend scalars to \(\overline{\mathbb{K}}\). The hypothesis Equation (A31.14) is bilinear and \(\comm{\mathfrak{h}}{\mathfrak{h}}\) spans \(\comm{\mathfrak{h}_{\overline{\mathbb{K}}}} {\mathfrak{h}_{\overline{\mathbb{K}}}}\), so the hypothesis holds for \(\mathfrak{h}_{\overline{\mathbb{K}}} \subseteq \mathfrak{gl}\left(\mathbb{V}_{\overline{\mathbb{K}}}\right)\), which is therefore solvable. Since \(\left(\mathfrak{h}^{(j)}\right)_{\overline{\mathbb{K}}} = \left(\mathfrak{h}_{\overline{\mathbb{K}}}\right)^{(j)}\) — brackets of spanning sets span — and a subspace vanishes if and only if its extension does, \(\mathfrak{h}\) is solvable.

Proof of Theorem A31.1. Derives Theorem A31.1. One direction is Proposition A31.5. For the other, assume \(\mathfrak{g}\) has no nonzero abelian ideal and let \(\mathfrak{s} = \mathfrak{g}^{\perp}\) be the radical of the Killing form, Equation (A31.1).

\(\mathfrak{s}\) is an ideal. Let \(X \in \mathfrak{s}\) and \(Z,Y \in \mathfrak{g}\). By the invariance of the Killing form, Equation (18.16),

so \(\comm{Z}{X} \in \mathfrak{s}\).

\(\ad(\mathfrak{s})\) is solvable. Consider \(\ad(\mathfrak{s}) \subseteq \mathfrak{gl}(\mathfrak{g})\), a Lie subalgebra because \(\ad\) is a homomorphism. Its derived algebra is \(\ad\left(\comm{\mathfrak{s}}{\mathfrak{s}}\right)\). Take \(x = \ad_{W}\) with \(W \in \comm{\mathfrak{s}}{\mathfrak{s}} \subseteq \mathfrak{s}\) and \(y = \ad_{Y}\) with \(Y \in \mathfrak{s}\); then

because \(W \in \mathfrak{s} = \mathfrak{g}^{\perp}\). So Theorem A31.11 applies and \(\ad(\mathfrak{s})\) is solvable.

\(\mathfrak{s}\) is solvable. The kernel of \(\ad\) restricted to \(\mathfrak{s}\) is \(\mathfrak{s} \cap Z(\mathfrak{g})\), where \(Z(\mathfrak{g}) = \set{X \in \mathfrak{g} \mid \comm{X}{\mathfrak{g}} = 0}\) is the centre; it is abelian, hence solvable; the image \(\ad(\mathfrak{s})\) is solvable; so Lemma A31.3(2) applies to the ideal \(\ker\left(\ad|_{\mathfrak{s}}\right)\) of \(\mathfrak{s}\).

Conclusion. If \(\mathfrak{s} \neq 0\) then, being a nonzero solvable ideal of \(\mathfrak{g}\), it would contain a nonzero abelian ideal of \(\mathfrak{g}\) by Lemma A31.3(4), contrary to hypothesis. Hence \(\mathfrak{s} = 0\) and \(\kappa\) is nondegenerate.

Derives Corollary A31.13. \(\mathfrak{a}^{\perp}\) is an ideal by the computation Equation (A31.16). Put \(\mathfrak{b} = \mathfrak{a} \cap \mathfrak{a}^{\perp}\), an ideal. For \(W \in \comm{\mathfrak{b}}{\mathfrak{b}} \subseteq \mathfrak{a}\) and \(Y \in \mathfrak{b} \subseteq \mathfrak{a}^{\perp}\) we have \(\tr\left(\ad_{W}\ad_{Y}\right) = \kappa(W,Y) = 0\), so Theorem A31.11 makes \(\ad(\mathfrak{b})\) solvable and, exactly as in the proof of Theorem A31.1, \(\mathfrak{b}\) solvable. A semisimple algebra has no nonzero solvable ideal (Remark A31.4), so \(\mathfrak{b} = 0\). Nondegeneracy of \(\kappa\) gives \(\dim\mathfrak{a}^{\perp} = \dim\mathfrak{g} - \dim\mathfrak{a}\), whence the direct sum. Then \(\comm{\mathfrak{a}}{\mathfrak{a}^{\perp}} \subseteq \mathfrak{a}\cap\mathfrak{a}^{\perp} = 0\), both being ideals.

An abelian ideal of \(\mathfrak{a}\) is an ideal of \(\mathfrak{g}\), since \(\mathfrak{a}^{\perp}\) brackets to zero with it; so it vanishes, and \(\mathfrak{a}\) is semisimple, as is \(\mathfrak{a}^{\perp}\). The projection \(\mathfrak{g} \to \mathfrak{a}^{\perp}\) along \(\mathfrak{a}\) is a homomorphism with kernel \(\mathfrak{a}\).

Iterating the splitting on a proper nonzero ideal, and stopping when none exists, writes \(\mathfrak{g} = \mathfrak{g}_{1}\oplus\cdots\oplus \mathfrak{g}_{s}\) with each \(\mathfrak{g}_{j}\) having no proper nonzero ideal; each is nonabelian, since an abelian \(\mathfrak{g}_{j}\) would be an abelian ideal of \(\mathfrak{g}\), so each is simple. The centre of \(\mathfrak{g}\) is an abelian ideal, hence zero. Finally \(\comm{\mathfrak{g}_{j}}{\mathfrak{g}_{j}}\) is a nonzero ideal of the simple \(\mathfrak{g}_{j}\) — nonzero because \(\mathfrak{g}_{j}\) is not abelian — so it is all of \(\mathfrak{g}_{j}\), and summing over \(j\) gives \(\comm{\mathfrak{g}}{\mathfrak{g}} = \mathfrak{g}\).

Derives Corollary A31.14. Let \(X \in \mathfrak{g}_{i}\), \(Y \in \mathfrak{g}_{j}\), \(i \neq j\). Then \(\ad_{Y}\) annihilates every summand but \(\mathfrak{g}_{j}\) and maps \(\mathfrak{g}_{j}\) into itself, while \(\ad_{X}\) annihilates \(\mathfrak{g}_{j}\); so \(\ad_{X}\ad_{Y} = 0\) and \(\kappa(X,Y) = 0\). For \(X,Y \in \mathfrak{g}_{j}\) the map \(\ad_{X}\ad_{Y}\) annihilates the other summands and preserves \(\mathfrak{g}_{j}\), so its trace over \(\mathfrak{g}\) equals its trace over \(\mathfrak{g}_{j}\).

Derives Corollary A31.15. That the \(J_{AB}\) with \(A<B\) are a basis is part (ii) of Proposition 18.59, and \(\eta\) is diagonal by Notation 18.1.

Off-diagonal entries vanish. Fix an index \(A_{0}\) and let \(\varsigma\) be the reflection \(x^{A_{0}} \mapsto -x^{A_{0}}\) of \(\R^{D}\), the other coordinates unchanged. It preserves \(\eta\), so its pushforward on vector fields maps Killing fields to Killing fields and, being induced by a diffeomorphism, preserves Lie brackets: it is an automorphism of \(\mathfrak{so}(p,q)\). On the generators Equation (18.86) it acts by \(\varsigma_{*}J_{AB} = \epsilon_{A}\epsilon_{B}J_{AB}\), where \(\epsilon_{A} = -1\) if \(A = A_{0}\) and \(+1\) otherwise, because both \(x_{A}\) and \(\pp_{B}\) pick up their own sign. The Killing form is invariant under any automorphism \(\phi\), since \(\ad_{\phi X} = \phi\,\ad_{X}\phi^{-1}\) and the trace is conjugation invariant; hence

If \(\set{A,B} \neq \set{C,D}\), choose \(A_{0}\) in the symmetric difference of the two pairs; then exactly one of the four factors is \(-1\) — the pairs have distinct entries — so the sign is \(-1\) and \(\kappa\left(J_{AB},J_{CD}\right) = 0\).

Diagonal entries. Take \(H = J_{12}\); the general case is the same computation with the indices renamed. From Equation (18.90), \(\comm{J_{12}}{J_{CD}} = \eta_{1C}J_{D2} + \eta_{2D}J_{C1} + \eta_{1D}J_{2C} + \eta_{2C}J_{1D}\), and since \(\eta\) is diagonal only the terms whose two indices coincide survive. For \(k \ge 3\),

while \(\ad_{H}J_{12} = 0\) and \(\ad_{H}J_{kl} = 0\) for \(k,l \ge 3\). Hence \(\ad_{H}^{2}\) is diagonal in the basis, acting as \(-\eta_{11}\eta_{22}\) on each of the \(2(D-2)\) generators \(J_{1k}\), \(J_{2k}\) with \(k \ge 3\), and as zero on the rest, so

which is Equation (A31.19) for \(A=C=1\), \(B=D=2\); the general case of Equation (A31.19) follows because both sides vanish off the diagonal and both are antisymmetric in \(AB\) and in \(CD\). Since \(\eta_{AA}\eta_{BB} = \pm1\), the diagonal entries are nonzero exactly when \(D \neq 2\), and Theorem A31.1 converts nondegeneracy into semisimplicity.