Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 2 / 5

The Killing form is the symmetric bilinear formThe facts needed here are its invariance, $B([A,C],D)=B(A,[C,D])$, and the root space identity $[H,X]=\alpha (H)X$. Invariance follows from $[\mathrm{ad}A,\mathrm{ad}C]=\mathrm{ad}[A,C]$ and cyclicity of the trace:Taking $A=H$, $C=X$, $D=Y$ provesNo normalization of $X,Y$ is needed for this identity. Semisimplicity gives nondegeneracy of the Killing form, but that fact is not needed for this particular equality.

Use the standard coroot normalization $\alpha (H_{\alpha })=2$. The symbols $H_{\alpha }$ are vectors in the Cartan subalgebra, not the root-space vectors $X_{\alpha }$. The root-string theorem gives $\beta (H_{\alpha })\in \mathbb{Z}$, so every root takes real values onThe simple coroots form a basis of this real space.

In the root-space decomposition, $\mathrm{ad}H$ is zero on $\mathfrak{h}$ and scalar multiplication by $\beta (H)$ on the one-dimensional space ${\mathfrak{g}}_{\beta }$. Therefore, for $H,K\in \mathfrak{h}$,For $H,K\in {\mathfrak{h}}_{\mathbb{R}}$ every term is real, so the restricted form is real. Moreover,If it is zero, all $\beta (H)$ vanish. Then $H$ commutes with every root space and with $\mathfrak{h}$, hence is central in $\mathfrak{g}$. The center of a Lie algebra vanishes for a semisimple Lie algebra, so $H=0$. Thus the Euclidean subspace of a Cartan subalgebra hasEquivalently, positivity follows because the roots span ${\mathfrak{h}}^{\ast }$.

Put $H_{12}=E_{11}-E_{22}=[E_{12},E_{21}]$. The six roots of ${\mathfrak{sl}}_3$ take values $2,-2,1,-1,-1,1$ on $H_{12}$, so the root-space decomposition givesUsing the preceding invariant-form identity with $\alpha =L_1-L_2$,HenceThis agrees with the Killing form of the special linear Lie algebra, $B(A,C)=2n\mathrm{tr}(AC)$, at $n=3$.

The weight lattice iswhere the fundamental weights satisfy ${\omega }_i(H_{{\alpha }_j})={\delta }_{ij}$. The coroots use the normalization fixed above.

The Killing form defines the complex-linear mapBy nondegeneracy of the Killing form on a Cartan subalgebra, this is an isomorphism. For $H\in {\mathfrak{h}}_{\mathbb{R}}$, expand its image in the dual fundamental-weight basis:The coefficients are real by the preceding positivity calculation, and the ${\omega }_i$ are in ${\Lambda }_W$. ConsequentlyIn fact both spaces have real dimension $r$, so this inclusion is equality.

Evaluate the Killing form on the root-space decomposition. For arbitrary $K\in \mathfrak{h}$,Thus the Killing dual of a coroot is the root expansionThe factor two in the positive-root expression occurs because the contributions of $\beta$ and $-\beta$ are equal.

For ${\mathfrak{sl}}_3$, let $a=L_1-L_2$, $b=L_2-L_3$; the positive roots are $a,b,a+b$. On $H_{12}$ their values are $2,-1,1$. ThereforeThis uses the unscaled Killing form $\mathrm{tr}(\mathrm{ad}A\mathrm{ad}C)$. Changing that normalization would change the scalar six.