= Solution
Number the <simple roots> of the <G2 root system> so that $\alpha_1$ is short and $\alpha_2$ is long. In the Euclidean plane choose
$$
\alpha_1=(1,0),\qquad\alpha_2=(-3/2,\sqrt3/2).
$$
Their squared lengths are one and three, and their angle is $150$ degrees. The six <positive roots> are
$$
\alpha_1,\quad\alpha_2,\quad\alpha_1+\alpha_2,\quad2\alpha_1+\alpha_2,
\quad3\alpha_1+\alpha_2,\quad3\alpha_1+2\alpha_2.
$$
Together with their negatives these form the twelve-root diagram: two regular hexagons of radii one and $\sqrt3$, rotated by $30$ degrees relative to one another. Solving the <fundamental weight> equations gives
$$
\boxed{\omega_1=2\alpha_1+\alpha_2=(1/2,\sqrt3/2),\qquad
\omega_2=3\alpha_1+2\alpha_2=(0,\sqrt3).}
$$
In particular $\omega_1$ is the highest short root and $\omega_2$ the highest long root. The root diagram below marks both <weights>.
To evaluate the <Weyl dimension formula>, put $a=n_1$, $b=n_2$ and $\lambda=a\omega_1+b\omega_2$. The <Weyl vector> is $\rho=\omega_1+\omega_2$. In the order of <positive roots> listed above, their <coroots> have simple-coroot coefficient pairs
$$
(1,0),\quad(0,1),\quad(1,3),\quad(2,3),\quad(1,1),\quad(1,2).
$$
For example the short root $\alpha_1+\alpha_2$ has <coroot> $\alpha_1^\vee+3\alpha_2^\vee$, while the long root $3\alpha_1+2\alpha_2$ has <coroot> $\alpha_1^\vee+2\alpha_2^\vee$. Pairing with $\lambda+\rho$ and dividing by the same factors at $a=b=0$ gives the <G2 dimension polynomial>
$$
\boxed{\dim L(a\omega_1+b\omega_2)=
\frac{(a+1)(b+1)(a+b+2)(a+2b+3)(a+3b+4)(2a+3b+5)}{120}.}
$$
The denominator is $1\cdot1\cdot2\cdot3\cdot4\cdot5$. As checks, the fundamental representations have <dimensions> seven and fourteen, and $a=b=0$ gives the trivial representation of <dimension> one.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-4-g2-roots.png]
{title=The twelve G2 roots and the short-first fundamental weights omega one and omega two}
Back to article page
