= Solution
The drawn simple roots have $\|\alpha\|^2=1$, $\|\beta\|^2=2$ and $\alpha\cdot\beta=-1$. Their <Cartan integers> are therefore $\langle\beta,\alpha^\vee\rangle=-2$ and $\langle\alpha,\beta^\vee\rangle=-1$, identifying a <B2 root system>. The corresponding <root reflections> act on coordinates by
$$
s_\alpha(u,v)=(-u,v),\qquad s_\beta(u,v)=(v,u).
$$
In particular $s_\beta(\alpha)=\alpha+\beta=(0,1)$ and $s_\alpha(\beta)=2\alpha+\beta=(1,1)$, so both new positive roots are forced. Reflection in any root also forces its negative. Thus
$$
\boxed{\Phi^+=\{\alpha,\beta,\alpha+\beta,2\alpha+\beta\},\qquad\Phi=\{\pm(1,0),\pm(0,1),\pm(-1,1),\pm(1,1)\}.}
$$
These eight roots are drawn below, with the given simple pair highlighted.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2003/iii/paper-1-b2-roots.png]
{title=All roots forced by the given simple roots of the B2 root system}
For completeness, no further roots can occur. In a finite reduced root system, every root is a <Weyl group> translate of a simple root: reflect a nonsimple positive root in a simple root with positive inner product to decrease its height, and iterate. This standard simple-root descent may also be quoted here. The eight-vector set just listed is stable under both generating reflections, and the orbits of the two simple roots already fill it. Hence no simple-root orbit, and therefore no root, lies outside that set. This establishes both the necessity of every drawn root and exhaustion of the root system.
Back to article page
