= Solution
Use the convention $A_{ij}=\langle\alpha_i,\alpha_j^\vee\rangle$, so the rows of the supplied <Cartan matrix> are the <simple roots> in <Dynkin label> coordinates. Realize the <B2 root system> in an orthonormal plane by
$$
\alpha_1=e_1-e_2,\qquad\alpha_2=e_2,\qquad
\alpha_1^\vee=e_1-e_2,\qquad\alpha_2^\vee=2e_2.
$$
Thus the first <simple root> is long and the second is short. The <Dynkin diagram> has two nodes joined by a double bond, with its arrow pointing to the short node:
$$
\boxed{\underset{\alpha_1\ \mathrm{long}}{\circ}\;\Longrightarrow\;\underset{\alpha_2\ \mathrm{short}}{\circ}.}
$$
A Euclidean <weight> $w=xe_1+ye_2$ has <Dynkin labels>
$$
(a,b)=(x-y,2y),\qquad (x,y)=(a+b/2,b/2).
$$
In particular the <fundamental weights> are $\omega_1=e_1$ and $\omega_2=(e_1+e_2)/2$. This long-first ordering is essential: sources that put the short <simple root> first interchange the two <Dynkin labels>.
The <highest weight> $(0,1)$ is the <spinor representation>. Its Euclidean <weights> are $(\pm\tfrac12,\pm\tfrac12)$, whose <Dynkin labels> are
$$
\boxed{\mathbf4=(0,1):\quad(0,1),\ (1,-1),\ (-1,1),\ (0,-1).}
$$
A lowering chain subtracts $\alpha_2=(-1,2)$, then $\alpha_1=(2,-2)$, then $\alpha_2$ again, passing through these four <weights> in the displayed order. All <weight multiplicities> are one.
The <highest weight> $(1,0)$ is the defining <vector representation>, with Euclidean <weights> $e_1,e_2,0,-e_2,-e_1$. Thus
$$
\boxed{\mathbf5=(1,0):\quad(1,0),\ (-1,2),\ (0,0),\ (1,-2),\ (-1,0).}
$$
Here a lowering chain subtracts $\alpha_1,\alpha_2,\alpha_2,\alpha_1$. Each <weight multiplicity> is again one.
Finally $(0,2)$ is the <highest root> $e_1+e_2=\alpha_1+2\alpha_2$, giving the <Adjoint representation>. Its nonzero <weights> are the eight <roots> $\pm e_1,\pm e_2,\pm(e_1-e_2),\pm(e_1+e_2)$. The <Cartan subalgebra> supplies zero with <weight multiplicity> two, so
$$
\boxed{\mathbf{10}=(0,2):\quad
\pm(1,0),\quad\pm(-1,2),\quad\pm(2,-2),\quad\pm(0,2),\quad(0,0)^{[2]}.}
$$
The <dimensions> are $4,5,10$. Independently the long-first <Weyl dimension formula> is
$$
\dim V(a,b)=\frac{(a+1)(b+1)(a+b+2)(2a+b+3)}6,
$$
which gives those same values. These are <Lie algebra> <representations>; the four-dimensional <spinor representation> integrates to the <Spin group> $\operatorname{Spin}(5)$, and does not descend to the <special orthogonal group> $SO(5)$. A concrete obstruction is a $2\pi$ rotation: it is the identity in $SO(5)$ but acts as $-I_4$ on the <spinor representation>, as verified in <4/d/solution>.
\Image[/past-exam-of-the-groups-and-representations-course-of-the-university-of-oxford/2019-b2-weight-systems.png]
{title=Three B2 weight systems in the long-first convention; point labels are Dynkin coordinates, and geometric positions are Euclidean weight coordinates. The adjoint zero weight has multiplicity two.}
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