Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 102 / 5

Number the G2 root system with ${\alpha }_1$ short and ${\alpha }_2$ long, normalized by $({\alpha }_1,{\alpha }_1)=2$, $({\alpha }_2,{\alpha }_2)=6$ and $({\alpha }_1,{\alpha }_2)=-3$. Its positive roots are ${\alpha }_1,{\alpha }_2,{\alpha }_1+{\alpha }_2,2{\alpha }_1+{\alpha }_2,3{\alpha }_1+{\alpha }_2,3{\alpha }_1+2{\alpha }_2$, and the negatives complete the twelve-root system. The short roots form one hexagon, and the long roots form a rotated hexagon whose radius is $\sqrt{3}$ times larger. The picture also supplies the crystal requested in the next part.

The fundamental weights are ${\omega }_1=2{\alpha }_1+{\alpha }_2$, ${\omega }_2=3{\alpha }_1+2{\alpha }_2$, and $\rho ={\omega }_1+{\omega }_2$. The six positive coroots, in coordinates relative to ${\alpha }_1^{\vee },{\alpha }_2^{\vee }$, are $(1,0),(0,1),(1,3),(2,3),(1,1),(1,2)$. Applying the Weyl dimension formula to $\lambda =n_1{\omega }_1+n_2{\omega }_2$ therefore gives the G2 dimension polynomialThe denominator is the product $1\cdot 1\cdot 2\cdot 3\cdot 4\cdot 5$ of the corresponding pairings with $\rho$. As checks, the weights ${\omega }_1,{\omega }_2,2{\omega }_1$ give dimensions $7,14,27$ respectively.

The G2 dimension polynomial gives dimension seven at the short-root fundamental weight, soThe crystal of the seven-dimensional G2 representation is a chain whose weights, in order, areIts successive arrow colors are $1,2,1,1,2,1$, as drawn above. Each arrow subtracts its indicated simple root; in particular the central three vertices form a length-two string of color one. The six nonzero weights are exactly the six short roots, each with multiplicity one, and the zero weight also has multiplicity one.

Choose the chain basis $v_0,\dots ,v_6$ from the preceding part, normalized by $v_1=f_1{v}_0$, $v_2=f_2{v}_1$, $v_3=f_1{v}_2$. The associated sl2 Lie algebra strings give $e_1{v}_1=v_0$, $e_1{v}_3=2v_2$ and $e_2{v}_2=v_1$; other raising actions used below vanish because their proposed weights do not occur.

In the exterior square, $v_0\wedge v_1$ is a highest-weight vector of weight $2{\omega }_1-{\alpha }_1={\omega }_2$: $e_1$ produces $v_0\wedge v_0=0$, and $e_2$ kills both factors. Alsois a nonzero highest-weight vector of weight ${\omega }_1$. Its two $e_1$ terms cancel, and its $e_2$ image is zero. The Weyl complete reducibility theorem supplies irreducible summands with those weights; their dimensions $14+7=21$ exhaust $\dim {\Lambda }^{2}V$. ThusFor the symmetric square, $v_0\otimes v_0$ is a highest-weight vector of weight $2{\omega }_1$, giving a 27-dimensional summand. The longest Weyl-group element of $G_2$ acts as minus the identity, so the highest weight of a dual representation is $-w_0{\omega }_1={\omega }_1$: $V$ is self-dual. A nonzero intertwiner $V\to V^{\ast }$ gives a nondegenerate Lie-invariant bilinear form. It is symmetric, since an alternating bilinear form cannot be nondegenerate in odd dimension. Its inverse is a nonzero invariant vector in $S^{2}V$, supplying the scalar summand. Since $27+1=28=\dim S^{2}V$,These are the exterior and symmetric squares of the seven-dimensional G2 representation; all highest-weight and dimension claims are specified rather than relying on an unstated cross-product identity.