= Solution
For a <root> $\alpha$, choose normalized <root> vectors $e_\alpha\in\mathfrak g_\alpha$, $f_\alpha\in\mathfrak g_{-\alpha}$ and the coroot element $h_\alpha\in\mathfrak h$ with
$$
[h_\alpha,e_\alpha]=2e_\alpha,\qquad [h_\alpha,f_\alpha]=-2f_\alpha,\qquad [e_\alpha,f_\alpha]=h_\alpha.
$$
The requested <sl2 subalgebra associated with a root> is
$$
\boxed{\mathfrak s_\alpha=\mathfrak g_\alpha\oplus\mathbb Ch_\alpha\oplus\mathfrak g_{-\alpha}\cong\mathfrak{sl}_2.}
$$
These 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 is
$$
\mathfrak{sl}_3=\langle E_{12},H_{12},E_{21}\rangle
\oplus\langle E_{13},E_{23}\rangle
\oplus\langle E_{32},-E_{31}\rangle
\oplus\langle\operatorname{diag}(1,1,-2)\rangle
\cong\Gamma_2\oplus\Gamma_1\oplus\Gamma_1\oplus\Gamma_0.
$$
The 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>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-1-adjoint.png]
{title=Adjoint sl3 weights coloured by their irreducible summands under the root sl2 subalgebra, with the two zero-weight vectors distinguished}
The <Killing form> value is the trace of $\operatorname{ad}E_{12}\operatorname{ad}E_{21}$. By the formula just proved, the four summands contribute $4,1,1,0$. Therefore
$$
\boxed{B(E_{12},E_{21})=6.}
$$
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