Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 2 / 5

Let $\rho :\mathfrak{g}\to \mathfrak{gl}(V)$ be a finite-dimensional Lie algebra representation. The form denoted $B_V$ isIt is the Trace form of a Lie algebra representation; the Killing form $B$ without a subscript is specifically the case $V=\mathfrak{g}$, $\rho =\mathrm{ad}$. The distinction matters: the form of the trivial representation cannot detect whether the algebra is semisimple.

Linearity of $\rho$ and trace proves bilinearity, and cyclicity of the trace gives symmetry. With $A=\rho (X)$, $D=\rho (Y)$ and $C=\rho (Z)$, the representation identity givesThus it is an invariant bilinear form on a Lie algebra, and in particular the Adjoint representation preserves the Killing form.

The Cartan solvability criterion has two useful formulations. For a finite-dimensional complex Lie algebra $\mathfrak{g}$,Its matrix version says that a Lie subalgebra $L\subseteq \mathfrak{gl}(V)$ is solvable precisely when $\mathrm{tr}(xy)=0$ for all $x\in [L,L]$ and $y\in L$. We prove the matrix version first, with ordinary trace in $V$.

If $L$ is solvable, the Lie theorem makes all its matrices upper triangular. Their commutators are strictly upper triangular, so multiplying such a matrix by an upper triangular one still has zero diagonal and hence zero trace. This proves the easy direction.

Conversely assume the trace-orthogonality condition and fix $x\in [L,L]$. We show that all eigenvalues of $x$ vanish. Use its Additive Jordan decomposition $x=x_s+x_n$ with $[x_s,x_n]=0$. On the generalized eigenspace $V_{\lambda }$ of $x$, define an auxiliary endomorphism $y$ to be $\overline{\lambda }I$. This $y$ need not be in $L$; we only need control of its commutator with $L$.

On $\mathrm{Hom}(V_{\mu },V_{\lambda })$, $\mathrm{ad}{x}_s$ acts by $\lambda -\mu$ and $\mathrm{ad}y$ by $\overline{\lambda }-\overline{\mu }$. Choose a polynomial $q$ with $q(0)=0$ and $q(\lambda -\mu )=\overline{\lambda -\mu }$ at the finitely many distinct differences. Polynomial interpolation supplies it because equal differences have equal conjugates. Therefore $\mathrm{ad}y=q(\mathrm{ad}{x}_s)$.

The adjoint compatibility of additive Jordan decomposition identifies $\mathrm{ad}{x}_s$ as the semisimple part of $\mathrm{ad}x$. Elementary Jordan–Chevalley decomposition gives $\mathrm{ad}{x}_s=p(\mathrm{ad}x)$ for a polynomial $p$ with $p(0)=0$. Hence $\mathrm{ad}y$ is a polynomial in $\mathrm{ad}x$ with zero constant term. Since $\mathrm{ad}x$ maps $L$ into $[L,L]$ and preserves that Lie algebra ideal, we obtainNo assumption that $x_s$, $x_n$ or $y$ belongs to $L$ was made.

Write $x={\sum }_i[a_i,b_i]$ with $a_i,b_i\in L$. Cyclicity and the assumed orthogonality now giveOn the other hand the nilpotent parts have zero trace on each generalized eigenspace, soThus all $\lambda$ are zero and $x$ is a nilpotent endomorphism. This is the Conjugate-spectrum proof of Cartan solvability.

The permitted Engel theorem, in the form needed here, states: a finite-dimensional Lie subalgebra of endomorphisms in which every element is nilpotent has a nonzero vector annihilated by all its elements, and iteration on quotients makes every element simultaneously strictly upper triangular. Apply it to $[L,L]$. The strictly upper triangular algebra is nilpotent, so $[L,L]$ is a Nilpotent Lie algebra and therefore solvable. Since $L/[L,L]$ is abelian, the derived series of a Lie algebra of $L$ terminates too. This proves the matrix criterion.

Apply it to $L=\mathrm{ad}\mathfrak{g}$. The condition on $B$ is exactly the matrix condition on $L$. Thus $\mathrm{ad}\mathfrak{g}$ is solvable. The kernel of $\mathrm{ad}$ is the center of a Lie algebra, which is abelian, so $\mathfrak{g}$ is solvable as well: once the derived series maps to zero it is central, and its next term vanishes. Conversely a solvable $\mathfrak{g}$ has solvable adjoint image, proving the abstract Cartan solvability criterion in both directions.

A finite-dimensional Lie algebra is a semisimple Lie algebra when it has no nonzero solvable Lie algebra ideal, equivalently its solvable radical is zero. Let $R=\mathrm{rad}B=\{x:B(x,\mathfrak{g})=0\}$. Invariance makes $R$ an ideal. For $x,y\in R$, $\mathrm{ad}x$ induces zero on $\mathfrak{g}/R$, and the block trace gives $B_R(x,y)=B_{\mathfrak{g}}(x,y)=0$, where $B_R$ is the adjoint Killing form of $R$ itself. The Cartan solvability criterion therefore makes $R$ solvable. If $\mathfrak{g}$ is semisimple, $R=0$.

Conversely suppose $B$ is a nondegenerate bilinear form. If a nonzero solvable ideal existed, its last nonzero derived term $A$ would be an abelian ideal of $\mathfrak{g}$. For $a\in A$ and $x\in \mathfrak{g}$, the operator $\mathrm{ad}a\mathrm{ad}x$ has image in $A$ and is zero on $A$, so its square and its trace are zero. Thus $B(A,\mathfrak{g})=0$, contradicting nondegeneracy. This is Abelian ideals lie in the radical of the Killing form. We conclude the Cartan criterion for semisimplicity: