Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 102 / 5

Let $C(\Delta )$ be the fundamental chamber of a root system determined by $\Delta$. Every positive root is a nonnegative linear combination of the simple roots, so every point in the interior of $C(\Delta )$ pairs strictly positively with every positive root. In particular, the interior meets none of the reflecting hyperplanes belonging to the subsystem ${\Phi }_0$ of roots of maximal length.

The connected set $C(\Delta )$ therefore lies in one chamber of the long-root subsystem. Let ${\Delta }_0$ be the unique fundamental system of a root system defining that chamber. Taking closures gives $C(\Delta )\subseteq C({\Delta }_0)$. Uniqueness follows because the interiors of two distinct chambers are disjoint. Thus there is a unique such ${\Delta }_0$.

Write the simple roots of the G2 root system as $\alpha$ short and $\beta$ long. The Long-root A2 subsystem of G2 has simple rootsIf ${\omega }_1,{\omega }_2$ are its fundamental weights, then the $A_2$ root-weight relations giveFor the $A_2$-dominant weight $\lambda =a{\omega }_1+b{\omega }_2$, therefore,The dominant weight inequalities for $G_2$ are thus equivalent tobecause $A_2$-dominance already gives $a,b\ge 0$.

The weights of the seven-dimensional irreducible $G_2$ representation are zero and the six short roots. In the $A_2$ weight lattice, the three positive short roots areTogether with their negatives, they split into the weight sets of the two dual three-dimensional Fundamental representations of sl3; the zero weight supplies a trivial representation. By the Restriction of the seven-dimensional G2 representation to long-root A2,

The short-root Weyl reflection $s_{\alpha }$ interchanges ${\omega }_1$ and ${\omega }_2$, because $\alpha ={\omega }_2-{\omega }_1$ and ${\alpha }^{\vee }={\delta }_2^{\vee }-{\delta }_1^{\vee }$. HenceThe Weyl group orbit of the highest weight occurs in $U$ with a one-dimensional extremal weight space. Let $v$ be a nonzero vector of weight $s_{\alpha }\lambda$. The same reflection exchanges the long simple roots: $s_{\alpha }{\delta }_1={\delta }_2$ and $s_{\alpha }{\delta }_2={\delta }_1$. If an $A_2$ raising operator did not annihilate $v$, then $s_{\alpha }\lambda +{\delta }_i$ would be a weight of $U$; applying $s_{\alpha }$ would make $\lambda +{\delta }_j$ a weight, contradicting the fact that $\lambda$ is the highest weight.

Thus $v$ is an $A_2$ highest-weight vector of weight $b{\omega }_1+a{\omega }_2$. The Complete reducibility of semisimple Lie algebra representations then supplies the corresponding irreducible summand, so