 Solution

Give the real root span an inner product with ${\epsilon }_1,{\epsilon }_2$ 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 rootsThe positive roots are ${\alpha }_1,{\alpha }_2,{\alpha }_1+{\alpha }_2={\epsilon }_1$, and ${\alpha }_1+2{\alpha }_2={\epsilon }_1+{\epsilon }_2$. The corresponding simple reflections areThey 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.

Since ${\alpha }_1,{\alpha }_2$ form an integral basis of ${\mathbb{Z}}^2$, the root lattice is $Q=\mathbb{Z}{\epsilon }_1\oplus \mathbb{Z}{\epsilon }_2$. The simple coroots are ${\alpha }_1^{\vee }={\epsilon }_1-{\epsilon }_2$ and ${\alpha }_2^{\vee }=2{\epsilon }_2$. Therefore the weight lattice isSolving $⟨{\Lambda }_i,{\alpha }_j^{\vee }⟩={\delta }_{ij}$ gives the fundamental weightsThe cone of dominant weights isIndeed $a=x-y$ 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 $\mathrm{SO}(5)$ itself, the weight must lie in ${\mathbb{Z}}^2$, equivalently $b$ must be even.