Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 1 / 2

Here the Cartan subalgebra of a finite-dimensional complex semisimple Lie algebra is a maximal toral subalgebra: it is abelian, its adjoint operators are simultaneously diagonalizable, and its centralizer is itself. These standard properties give the root-space decompositionThe roots of a root system $\alpha$ are the nonzero linear functionals in ${\mathfrak{h}}^{\ast }$ for which ${\mathfrak{g}}_{\alpha }\ne 0$; $\Phi$ is their finite set. Each root space is one-dimensional. The zero weight space is the centralizer of $\mathfrak{h}$, hence exactly $\mathfrak{h}$. Thus the displayed sum is a direct sum of simultaneous eigenspaces of the Adjoint representation. It is the Cartan root-space decomposition meant here.

The bracket respects weight addition: $[{\mathfrak{g}}_{\alpha },{\mathfrak{g}}_{\beta }]\subseteq {\mathfrak{g}}_{\alpha +\beta }$, where ${\mathfrak{g}}_0=\mathfrak{h}$ and a missing weight means the zero space. This follows by applying the Jacobi identity with $H\in \mathfrak{h}$ to the bracket of two root vectors. It explains how the vector-space decomposition encodes the Lie algebra structure.

The Killing form is the bilinear formThe Adjoint representation is linear, giving bilinearity, and the cyclic property of trace gives symmetry. To prove invariance, abbreviate the adjoint matrices of $X,Y,Z$ by $A_X,A_Y,A_Z$. The Jacobi identity gives $A_{[X,Y]}=[A_X,A_Y]$, and cyclic trace then givesThus $\boxed{B([X,Y],Z)=B(X,[Y,Z])}$, equivalently $B([X,Y],Z)+B(Y,[X,Z])=0$. This is infinitesimal invariance of the Killing form, with no choice of basis affecting the trace.

First show that the Cartan subalgebra is orthogonal, for the Killing form, to every nonzero root space. For $H\in \mathfrak{h}$ and $X\in {\mathfrak{g}}_{\alpha }$, choose $H_0\in \mathfrak{h}$ with $\alpha (H_0)\ne 0$. Invariance and the abelian nature of $\mathfrak{h}$ giveHence $B(H,X)=0$.

Now let $H$ be in the radical of the restriction to $\mathfrak{h}$, so $B(H,H^{\prime })=0$ for every $H^{\prime }\in \mathfrak{h}$. The preceding argument also gives $B(H,{\mathfrak{g}}_{\alpha })=0$ for every root. By the root-space decomposition, $H$ is orthogonal to all of $\mathfrak{g}$. The assumed nondegeneracy on $\mathfrak{g}$ forces $H=0$. Therefore the Killing form restricted to the Cartan subalgebra is nondegenerate.

Take the basis $(H_1,H_2,E_{12},E_{21},E_{13},E_{31},E_{23},E_{32})$ of ${\mathfrak{sl}}_3$, where $H_1=\mathrm{diag}(1,-1,0)$, $H_2=\mathrm{diag}(0,1,-1)$ and $E_{ij}$ is a matrix unit. For $D=\mathrm{diag}(1,-2,1)$, $[D,H_r]=0$ and $[D,E_{ij}]=(d_i-d_j)E_{ij}$. Thus the Adjoint representation matrix isSquaring and taking the trace computes the Killing form directly:The independent identity $B(X,Y)=6\mathrm{tr}(XY)$ for the special linear Lie algebra ${\mathfrak{sl}}_3$ also gives $6(1+4+1)=36$.