Use the <C2 root system> realization in an orthonormal basis $\varepsilon_1,\varepsilon_2$ with <simple roots> $\alpha_1=\varepsilon_1-\varepsilon_2$ and $\alpha_2=2\varepsilon_2$. This numbers the <short root> first, as required. The <fundamental weights> and <half-sum of positive roots> are
$$
\omega_1=\varepsilon_1,\qquad\omega_2=\varepsilon_1+\varepsilon_2,\qquad
\rho=2\varepsilon_1+\varepsilon_2,
$$
so $\lambda+\rho=(a+b+2)\varepsilon_1+(b+1)\varepsilon_2$. For the four <positive roots>, the factors of the <Weyl dimension formula> are
$$
\begin{array}{c|c|c}
\alpha&\alpha^\vee&
\langle\lambda+\rho,\alpha^\vee\rangle/\langle\rho,\alpha^\vee\rangle\\\hline
\varepsilon_1-\varepsilon_2&\varepsilon_1-\varepsilon_2&a+1\\
2\varepsilon_2&\varepsilon_2&b+1\\
\varepsilon_1+\varepsilon_2&\varepsilon_1+\varepsilon_2&(a+2b+3)/3\\
2\varepsilon_1&\varepsilon_1&(a+b+2)/2
\end{array}.
$$
Multiplying yields the <Weyl dimension formula for C2>:
$$
\boxed{\dim V(a\omega_1+b\omega_2)=\frac{(a+1)(b+1)(a+b+2)(a+2b+3)}6.}
$$
As checks, $V(\omega_1)$ has dimension $4$, $V(\omega_2)$ dimension $5$, and $V(2\omega_1)$ dimension $10$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 102 / 4 / b / Solution

{c}

For the <C2 root system> with its first <simple root> short, the irreducible <highest-weight representation> of <highest weight> $a\omega_1+b\omega_2$, where $a,b\in\mathbb Z_{\geq0}$, has dimension
$$
\frac{(a+1)(b+1)(a+b+2)(a+2b+3)}6.
$$

Weyl dimension formula for C2

{c}
{title2=$C_2$}

In an orthonormal basis $(\varepsilon_1,\varepsilon_2)$, the roots are $\pm(\varepsilon_1-\varepsilon_2)$, $\pm(\varepsilon_1+\varepsilon_2)$, $\pm2\varepsilon_1$, and $\pm2\varepsilon_2$. Choose <simple roots> $\alpha_1=\varepsilon_1-\varepsilon_2$ and $\alpha_2=2\varepsilon_2$, with $\alpha_1$ a <short root>. The <fundamental weights> are $\omega_1=\varepsilon_1$ and $\omega_2=\varepsilon_1+\varepsilon_2$.

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