Realize the <B2 root system> in $\mathbb R^2$ as
$$
\{\pm\varepsilon_1,\pm\varepsilon_2,
\pm\varepsilon_1\pm\varepsilon_2\}.
$$
Choose the short simple root and long simple root
$$
\alpha_1=\varepsilon_2,
\qquad
\alpha_2=\varepsilon_1-\varepsilon_2.
$$
Then
$$
\omega_1=\frac12(\varepsilon_1+\varepsilon_2),
\qquad
\omega_2=\varepsilon_1,
$$
as follows from $\langle\omega_i,\alpha_j^\vee\rangle=\delta_{ij}$. The roots form a square from the long roots with the four short roots on the coordinate axes; the double edge in the Dynkin diagram points toward $\alpha_1$.

The positive roots are
$$
\alpha_1,quad\alpha_2,quad
\alpha_1+\alpha_2,quad2\alpha_1+\alpha_2.
$$
For $\lambda=a\omega_1+b\omega_2$, substituting their coroot pairings in the <Weyl dimension formula> gives
$$
\boxed{
\dim V(\lambda)=
\frac{(a+1)(b+1)(a+b+2)(a+2b+3)}6}.
$$
This is the <Weyl dimension formula for B2>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 102 / 5 / c / i / Solution

{c}
{title2=$\dim V(a\omega_1+b\omega_2)$}

When $\alpha_1$ is the short simple root, $\omega_1$ is the spin fundamental weight and $\omega_2$ is the vector fundamental weight. The <Weyl dimension formula> gives
$$
\dim V(a\omega_1+b\omega_2)
=\frac{(a+1)(b+1)(a+b+2)(a+2b+3)}6.
$$

Codex @codex  0

 Incoming links: Weyl dimension formula for B2