Past exam of the Lie algebras and representation theory of symmetric groups course of the University of Oxford / 2009 / A / 3 / b / Solution

Use the column-coroot convention $C_{ij}=2({\alpha }_i,{\alpha }_j)/({\alpha }_j,{\alpha }_j)$, consistent with the printed matrix and the stated shorter first root. It givesThe root string through $\beta$ in the $\alpha$ direction contains $\beta ,\beta +\alpha ,\beta +2\alpha ,\beta +3\alpha$. Reflecting $3\alpha +\beta$ in $\beta$ gives $3\alpha +2\beta$. Together with $\alpha$ and their negatives these giveThere are no further roots. Indeed the simple reflections act on coefficient pairs by $(a,b)\mapsto (-a+3b,b)$ and $(a,b)\mapsto (a,a-b)$; the displayed twelve-element set is closed under both. Every positive nonsimple root can be reflected to a positive root of smaller height: choose a simple root with positive inner product as in (a), and its positive integral Cartan pairing decreases height. Repetition reaches a simple root, so every root lies in an orbit already contained in the list. This is the G2 root system.

With $\alpha =e_1$, the inner product and length give the two reflected possibilities