Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 1 / 4

For a finite-dimensional Lie algebra, define its Killing form byThe trace identity $\mathrm{tr}(AB)=\mathrm{tr}(BA)$ proves symmetry, and linearity of trace proves bilinearity. The Jacobi identity gives $\mathrm{ad}[x,y]=[\mathrm{ad}x,\mathrm{ad}y]$. Consequently, by cyclic invariance of trace,This is the invariance identity. Equivalently $B([h,x],y)+B(x,[h,y])=0$ for all three elements.

If $x\in {\mathfrak{g}}_{\alpha }$, $y\in {\mathfrak{g}}_{\beta }$ and $h\in \mathfrak{h}$, invariance of the Killing form givesUnless $\alpha +\beta =0$, some $h$ makes its coefficient nonzero, so the two root spaces are orthogonal. Similarly $B(\mathfrak{h},{\mathfrak{g}}_{\alpha })=0$ for every root: choose $h$ with $\alpha (h)\ne 0$ and use that $\mathfrak{h}$ is abelian.

Now take nonzero $x\in {\mathfrak{g}}_{\alpha }$. Nondegeneracy on $\mathfrak{g}$ supplies some element pairing nontrivially with $x$. Its root-space decomposition shows that only its ${\mathfrak{g}}_{-\alpha }$ component can contribute. Thus the pairing between opposite root spaces is nonzero, and indeed nondegenerate. ThereforeHere nonzero means the bilinear pairing is not identically zero, not that every pair of vectors has a nonzero value.

If $h_0\in \mathfrak{h}$ is orthogonal to all of $\mathfrak{h}$, it is also orthogonal to every root space, by the preceding orthogonality. It is therefore orthogonal to all of $\mathfrak{g}$, forcing $h_0=0$. Hence $B{|}_{\mathfrak{h}}$ is nondegenerate.

Put $H_{12}=\mathrm{diag}(1,-1,0)$ and $H_{23}=\mathrm{diag}(0,1,-1)$. On the root space spanned by $E_{ij}$, their adjoint eigenvalues are $h_i-h_j$ and $k_i-k_j$; both act as zero on the Cartan subalgebra. Thus the Killing form is the sum of their products over the six roots. The pair of roots $\pm (L_1-L_2)$ contributes $-4$, the pair $\pm (L_2-L_3)$ contributes $-4$, and $\pm (L_1-L_3)$ contributes $2$. HenceThis also agrees with the Killing form of the special linear Lie algebra, $B(A,B)=6\mathrm{tr}(AB)$ for ${\mathfrak{sl}}_3$.

The displayed identity is valid for a highest-weight vector $v$, satisfying $Xv=0$ and $Hv=kv$, not for arbitrary $v\in V$. For example, in the defining module with $k=1$, take $v=y$: then $XYv=0$ but the proposed right side for $n=1$ is $v\ne 0$.

For a highest-weight vector, the sl2 triple relations give $H{Y}^{j}v=(k-2j)Y^{j}v$. The operator identitycomes from expanding a commutator with a product. Apply it to $v$, using $Xv=0$. Each term becomes $(k-2(n-1-j))Y^{n-1}v$, so the sum isThis proves the intended sl2 highest-weight lowering formula. The identity valid for an arbitrary vector is instead

On the basis $v,Yv,\dots ,Y^{k}v$ of the irreducible module, $XY$ has eigenvalue $(j+1)(k-j)$ on $Y^{j}v$, for $0\le j\le k$. Summing gives the trace of a raising-lowering product in an irreducible sl2 module:The last equality follows by the formulas for the sums of the first $k$ integers and their squares.

For a root $\alpha$, choose normalized root vectors $e_{\alpha }\in {\mathfrak{g}}_{\alpha }$, $f_{\alpha }\in {\mathfrak{g}}_{-\alpha }$ and the coroot element $h_{\alpha }\in \mathfrak{h}$ withThe requested sl2 subalgebra associated with a root isThese bracket relations identify its basis with the standard sl2 triple. The notation here denotes this subalgebra, not the root reflection $s_{\alpha }$ used for the Weyl group.

The weight diagram of the Adjoint representation of ${\mathfrak{sl}}_3$ has its six roots $L_i-L_j$, $i\ne j$, each of multiplicity one, and zero of multiplicity two, from the Cartan subalgebra.

Restrict to the root subalgebra generated by $X=E_{12}$, $Y=E_{21}$ and $H=H_{12}$. The explicit decomposition into irreducible modules isThe first summand has $H$ weights $2,0,-2$. The second has weights $1,-1$, since $[Y,E_{13}]=E_{23}$. The third likewise has weights $1,-1$, since $[Y,E_{32}]=-E_{31}$. The last commutes with all three generators and is trivial. These subspaces are stable under the sl2 triple, are independent and have dimensions totaling eight, so they exhaust the Adjoint representation.

The Killing form value is the trace of $\mathrm{ad}{E}_{12}\mathrm{ad}{E}_{21}$. By the formula just proved, the four summands contribute $4,1,1,0$. Therefore