= Solution
Give the real root span an <inner product> with $\varepsilon_1,\varepsilon_2$ orthonormal. The <B2 root system> consists of four short roots on the coordinate axes and four long roots on the diagonals. Choose the long-first <simple roots>
$$
\boxed{\alpha_1=\varepsilon_1-\varepsilon_2,\qquad\alpha_2=\varepsilon_2.}
$$
The <positive roots> are $\alpha_1,\alpha_2,\alpha_1+\alpha_2=\varepsilon_1$, and $\alpha_1+2\alpha_2=\varepsilon_1+\varepsilon_2$. The corresponding simple reflections are
$$
s_1(x,y)=(y,x),\qquad s_2(x,y)=(x,-y).
$$
They generate every signed permutation of the two coordinates. Thus the <Weyl group> has order eight and is the symmetry group of a square; every element sends $(x,y)$ to $(\pm x,\pm y)$ or $(\pm y,\pm x)$, with independent signs.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2001/iii/paper-5-b2-roots.png]
{title=The B2 root system with long-first simple roots and all four positive roots}
Since $\alpha_1,\alpha_2$ form an integral basis of $\mathbb Z^2$, the <root lattice> is $Q=\mathbb Z\varepsilon_1\oplus\mathbb Z\varepsilon_2$. The <simple coroots> are $\alpha_1^\vee=\varepsilon_1-\varepsilon_2$ and $\alpha_2^\vee=2\varepsilon_2$. Therefore the <weight lattice> is
$$
\boxed{P=\{(x,y):x-y\in\mathbb Z,\ 2y\in\mathbb Z\}
=\mathbb Z^2\ \cup\ \big((\mathbb Z+\tfrac12)\times(\mathbb Z+\tfrac12)\big).}
$$
Solving $\langle\Lambda_i,\alpha_j^\vee\rangle=\delta_{ij}$ gives the <fundamental weights>
$$
\boxed{\Lambda_1=\varepsilon_1,\qquad\Lambda_2=\tfrac12(\varepsilon_1+\varepsilon_2).}
$$
The cone of <dominant weights> is
$$
\boxed{P^+=\{a\Lambda_1+b\Lambda_2:a,b\in\mathbb Z_{\ge0}\}
=\{(x,y)\in P:x\ge y\ge0\}.}
$$
Indeed $a=x-y$ and $b=2y$ are exactly the two simple-coroot pairings. This describes representations of the Lie algebra, or of its simply connected <Spin group>; half-integral weights are allowed. For representations descending to $\operatorname{SO}(5)$ itself, the weight must lie in $\mathbb Z^2$, equivalently $b$ must be even.
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