Jacobson-Morozov triple for a nilpotent element

We first prove the missing centralizer reduction in any finite-dimensional complex semisimple algebra $\mathfrak{a}$. For a semisimple $s\in \mathfrak{a}$, extend the toral subalgebra $\mathbb{C}s$ to a maximal toral subalgebra $\mathfrak{t}$: adjoining commuting semisimple elements preserves simultaneous diagonalizability, and dimensions bound the finite extension. Use [F3]'s root decomposition relative to this particular $\mathfrak{t}$. Set $\mathrm{\Psi }=\{\alpha :\alpha (s)=0\}$, $T={\mathrm{span}}_{\mathbb{C}}\{H_{\alpha }:\alpha \in \mathrm{\Psi }\}$ and $Z=\{u\in \mathfrak{t}:\alpha (u)=0\text{ for every }\alpha \in \mathrm{\Psi }\}$. The real span of these $H_{\alpha }$ lies in [F3]'s positive-definite real form, since each $H_{\alpha }$ is a nonzero real multiple of its coroot. Thus $B{\mathrm{\mid }}_T$ is nondegenerate, $Z=T^{\perp }\cap \mathfrak{t}$, and $\mathfrak{t}=T\oplus Z$ with $B{\mathrm{\mid }}_Z$ nondegenerate. Empty $\mathrm{\Psi }$ gives $T=0$, $Z=\mathfrak{t}$.

Every finite-dimensional complex semisimple algebra is a direct sum of simple ideals, as follows without a reductivity assumption. In a nonzero such algebra $\mathfrak{a}$, choose a nonzero ideal $I$ of minimal dimension. The radical of $B_{\mathfrak{a}}{\mathrm{\mid }}_I$ is an ideal of $\mathfrak{a}$ by invariance, hence is $0$ or $I$. If it is $I$, then $B_I=0$: for $x,y\in I$, their adjoint actions on $\mathfrak{a}\mathrm{/}I$ vanish, so the trace on $\mathfrak{a}$ equals the trace on $I$. The trace criterion [F4] makes ${\mathrm{ad}}_I(I)$ solvable; its central kernel then makes $I$ solvable by the derived series, contradicting semisimplicity. Thus $B_{\mathfrak{a}}{\mathrm{\mid }}_I$ is nondegenerate. Its orthogonal complement is an ideal, and $\mathfrak{a}=I\oplus I^{\perp }$ is a sum of commuting ideals. Every ideal of $I$ is now an ideal of $\mathfrak{a}$, so minimality makes $I$ simple and nonabelian. The complement has no nonzero solvable ideal, since any such ideal would be an ideal of $\mathfrak{a}$. Repeating in decreasing dimension proves the claim, including the empty decomposition of zero.

The centralizer in step 1.1 is $\mathfrak{l}={\mathfrak{a}}^s=\mathfrak{t}\oplus {⨁}_{\alpha \in \mathrm{\Psi }}{\mathfrak{a}}_{\alpha }$. Put $\mathfrak{k}=T\oplus {⨁}_{\alpha \in \mathrm{\Psi }}{\mathfrak{a}}_{\alpha }$. Root brackets show that $\mathfrak{k}$ is a subalgebra: sums of roots vanishing on $s$ again vanish on $s$, and opposite-root brackets lie in $\mathbb{C}{H}_{\alpha }\subseteq T$. Also $Z$ commutes with $\mathfrak{l}$, while each root vector belongs to $[\mathfrak{k},\mathfrak{k}]$ by its nonzero bracket with its own coroot, and $T$ is spanned by opposite-root brackets. Consequently $\mathfrak{l}=Z\oplus \mathfrak{k}$ is a sum of commuting ideals, $[\mathfrak{l},\mathfrak{l}]=\mathfrak{k}$, and $Z$ is exactly its center: any central element has zero root components and annihilates every root in $\mathrm{\Psi }$. The decomposition is $B$-orthogonal.

The algebra $\mathfrak{k}$ is semisimple. Indeed let $R$ be any solvable ideal in it. The commuting diagonalizable actions of $T$ preserve $R$, so $R$ is a sum of their weight intersections; projections onto the finitely many joint weights are interpolation polynomials in a finite basis of $T$. The restricted weights $\alpha {\mathrm{\mid }}_T$ for $\alpha \in \mathrm{\Psi }$ are nonzero and pairwise distinct: if $\alpha {\mathrm{\mid }}_T=\beta {\mathrm{\mid }}_T$, then $H_{\alpha }-H_{\beta }\in T$ is orthogonal to $T$, hence zero, and thus $\alpha =\beta$; the same argument compares a root to zero. If $R$ has a nonzero root component, the root space is one-dimensional, so it contains a nonzero $e_{\alpha }$. Bracketing with $f_{\alpha }\in \mathfrak{k}$ puts $h_{\alpha }$ in $R$, and then bracketing with $f_{\alpha }$ puts $f_{\alpha }$ in $R$. This embeds the perfect three-dimensional root-triple algebra in the solvable algebra $R$, impossible since a subalgebra of a solvable algebra is solvable. Thus $R\subseteq T$. Ideality gives $[R,{\mathfrak{a}}_{\alpha }]\subseteq R$, forcing every $\alpha \in \mathrm{\Psi }$ to vanish on $R$. Nondegeneracy on $T$ gives $R=0$. This proves semisimplicity; by step 1.2, $\mathfrak{k}$ is a direct sum of simple ideals. Thus $\mathfrak{l}$ is reductive in the explicit sense of a central abelian summand and a semisimple derived summand.

If $f\in \mathfrak{l}$ has nilpotent adjoint action on $\mathfrak{a}$, then $f\in \mathfrak{k}$. For $z\in Z$, the operators $\mathrm{ad}f$ and $\mathrm{ad}z$ commute. Their product is nilpotent because $(\mathrm{ad}f\mathrm{ad}z{)}^N=(\mathrm{ad}f{)}^N(\mathrm{ad}z{)}^N=0$. Hence $B(f,z)=0$. Since $B(\mathfrak{k},Z)=0$ and $B{\mathrm{\mid }}_Z$ is nondegenerate, the $Z$ component of $f$ is zero. If $s\mathrm{\ne }0$, its centralizer is proper: otherwise $s$ is central in the semisimple algebra, whose center is a solvable ideal and hence zero. Every simple ideal of $\mathfrak{k}$ then has dimension strictly less than $\dim \mathfrak{a}$.

We now prove existence of the triple by strong induction on dimension for all finite-dimensional complex semisimple algebras. At dimension $d$, assume the assertion for every such algebra of dimension less than $d$. For $f=0$, use $e=h=0$, including the zero algebra. If the algebra has more than one simple ideal, step 1.2 and induction give triples in each smaller ideal; the projections of $f$ are nilpotent because the restricted adjoint actions are nilpotent. Their sums give the required relations, since distinct ideals commute. It therefore suffices at each dimension to treat a simple $\mathfrak{g}$ and $f\mathrm{\ne }0$. If ${\mathfrak{g}}^f$ contains a nonnilpotent $x$, [F3]'s Jordan construction gives $x=s+n$ with nonzero semisimple $s$ commuting with $f$. Steps 1.1–4.1 put $f$ in the proper semisimple derived centralizer $\mathfrak{k}$. Its adjoint action on $\mathfrak{k}$ is the restriction of its nilpotent action on $\mathfrak{g}$. Induction in this smaller semisimple algebra supplies the triple in $\mathfrak{k}\subseteq \mathfrak{g}$.

In the remaining case every $x\in {\mathfrak{g}}^f$ is adjoint-nilpotent. Since $\mathrm{ad}f$ and $\mathrm{ad}x$ commute, their product is nilpotent, giving $B(f,x)=0$. Invariance and nondegeneracy give $\mathrm{im}(\mathrm{ad}f)=(\ker \mathrm{ad}f{)}^{\perp }=({\mathfrak{g}}^f{)}^{\perp }$, so $f\in \mathrm{im}(\mathrm{ad}f)$. Rescaling a preimage gives $h_0$ with $[h_0,f]=-2f$. Write $h_0=s+n$ for its commuting Jordan parts. On each eigenspace of $\mathrm{ad}s$, $\mathrm{ad}{h}_0$ equals the eigenvalue plus a commuting nilpotent operator. The equation $(\mathrm{ad}{h}_0+2)f=0$ therefore has no component in any eigenspace other than eigenvalue $-2$, because a nonzero scalar plus a nilpotent operator is invertible. Hence $[s,f]=-2f$. Set $h=s$, which is semisimple.

Put $h^{\mathrm{\prime }}=-h$, so $[h^{\mathrm{\prime }},f]=2f$. Jacobi makes ${\mathfrak{g}}^f$ an ideal of the Lie algebra $\mathbb{C}{h}^{\mathrm{\prime }}+{\mathfrak{g}}^f$. By [F4], the adjoint image of ${\mathfrak{g}}^f$ is strictly upper triangular and solvable, and its central kernel is abelian, so ${\mathfrak{g}}^f$ is solvable. Its one-dimensional extension is solvable as well. By [F5], this extension acts upper triangularly on $\mathfrak{g}$. Every $\mathrm{ad}x$ for $x\in {\mathfrak{g}}^f$ is nilpotent and thus has zero diagonal in this basis. Therefore $B(h^{\mathrm{\prime }},x)=\mathrm{tr}(\mathrm{ad}{h}^{\mathrm{\prime }}\mathrm{ad}x)=0$. Nondegeneracy and invariance again yield $h^{\mathrm{\prime }}\in ({\mathfrak{g}}^f{)}^{\perp }=\mathrm{im}(\mathrm{ad}f)$.

Choose $y$ with $[f,y]=h^{\mathrm{\prime }}$. Decompose $y$ under the diagonalizable $\mathrm{ad}{h}^{\mathrm{\prime }}$. The element $f$ has weight $2$, so only the weight-$-2$ component contributes to the weight-zero bracket $[f,y]=h^{\mathrm{\prime }}$. Call that component $e$; then $[f,e]=h^{\mathrm{\prime }}$ and $[h^{\mathrm{\prime }},e]=-2e$. As $h=-h^{\mathrm{\prime }}$, these equations give $[h,e]=2e$, $[h,f]=-2f$, and $[e,f]=-h^{\mathrm{\prime }}=h$. This closes the remaining induction case and proves the original assertion for every nilpotent $f$. Repeated $\mathrm{ad}e$ shifts the finite set of $h$-weights upward by $2$, so it is nilpotent as well.