= Solution
The <special orthogonal group> in five dimensions is
$$
SO(5)=\{R\in M_5(\mathbb R):R^TR=I,\ \det R=1\}.
$$
Its <Lie algebra> consists of real <skew-symmetric matrices>. There are $5\cdot4/2=10$ independent entries above the diagonal, so \b[$\dim SO(5)=10$]. Equivalently, the orthogonality equations impose fifteen independent constraints on twenty-five matrix entries; the <determinant> condition chooses a component without changing the dimension.
Fix the fifth coordinate. The matrices $\operatorname{diag}(S,1)$, $S\in SO(4)$, form an explicit subgroup. Under this subgroup the defining vector space splits as $\mathbb R^4\oplus\mathbb R\mathbf e_5$, so its <branching rule> is $\mathbf5\downarrow SO(4)=\mathbf4\oplus\mathbf1$. An element of the <so5 Lie algebra> can be written uniquely as
$$
X=\begin{pmatrix}B&w\\-w^T&0\end{pmatrix},\qquad B\in\mathfrak{so}(4),\quad w\in\mathbb R^4.
$$
Conjugation by $\operatorname{diag}(S,1)$ sends $B$ to $SBS^{-1}$ and $w$ to $Sw$. The first summand is the six-dimensional <Adjoint representation of a Lie algebra> of $\mathfrak{so}(4)$, and the second is its four-dimensional vector representation. Hence the <SO5 to SO4 branching> gives
$$
\boxed{\mathbf5\to\mathbf4\oplus\mathbf1,\qquad\mathbf{10}\to\mathbf6\oplus\mathbf4.}
$$
For the <left SU(2) subgroup of SO(4)>, identify $\mathbb R^4$ with the <quaternions>. Left multiplication by a unit <quaternion> is a real orthogonal transformation and gives an embedded <SU(2) group>. More generally, $q\mapsto a q b^{-1}$ gives the double cover $SU(2)_L\times SU(2)_R\to SO(4)$ with kernel $\{(1,1),(-1,-1)\}$. After complexifying, the vector representation is $(\mathbf2,\mathbf2)$ and the <Adjoint representation> is $(\mathbf3,\mathbf1)\oplus(\mathbf1,\mathbf3)$. Restricting to the left factor turns the right factor into a multiplicity space. Therefore
$$
\boxed{\mathbf4\to\mathbf2\oplus\mathbf2,\qquad\mathbf6\to\mathbf3\oplus\mathbf1\oplus\mathbf1\oplus\mathbf1,\qquad\mathbf1\to\mathbf1.}
$$
These are decompositions into complex <irreducible representations>; the real $\mathbf4$ is the underlying real representation of a quaternionic doublet. Combining the <branching rules> gives
$$
\mathbf5\to2\mathbf2\oplus\mathbf1,\qquad\mathbf{10}\to\mathbf3\oplus2\mathbf2\oplus3\mathbf1.
$$
Write a <weight> as $\lambda=(x,y)$. The integrality conditions for the <B2 root system> give $2x\in\mathbb Z$ and $y-x\in\mathbb Z$. Hence $x=m/2$, $y=m/2+n$, with $m,n\in\mathbb Z$. Thus
$$
\boxed{P=\mathbb Z^2\ \cup\ \left(\mathbb Z+\frac12\right)^2.}
$$
This is the <B2 weight lattice>, with an integer square lattice and a second square lattice shifted by $(1/2,1/2)$. The eight <roots of a root system> are
$$
\pm(1,0),\quad\pm(0,1),\quad\pm(1,1),\quad\pm(1,-1).
$$
The short roots lie on the coordinate axes and the long roots on the diagonals. The positive roots for the given simple-root choice are $\alpha$, $\beta$, $\alpha+\beta$, $2\alpha+\beta$.
The integrality conditions determine the <weight lattice> of the <Lie algebra>, equivalently of the simply connected <Spin group> $\operatorname{Spin}(5)$. For the global <special orthogonal group> $SO(5)$, a $2\pi$ rotation in either coordinate plane is the identity, so a genuine <group representation> requires integer $x,y$. Thus \b[the half-integer coset contains spin representations that do not descend to $SO(5)$]. Both representations requested here have integer weights, so their diagrams are unaffected by this distinction.
The <root system> of the displayed $SO(4)$ subgroup is $\{\pm(1,1),\pm(1,-1)\}$. The two orthogonal pairs give its two commuting $\mathfrak{su}(2)$ factors. Choose the left factor to have root $\delta=(1,1)$; exchanging the two diagonal pairs exchanges left and right. Its <coroot> pairs with a <weight> as
$$
\langle\lambda,\delta^\vee\rangle=\frac{2\lambda\cdot\delta}{\delta\cdot\delta}=x+y.
$$
This demonstrates the <diagonal-root SU(2) embedding in SO(5)> directly. The short-axis root $\alpha$ would instead give $2x$, and therefore a different subgroup: on the vector representation it would produce a triplet and two singlets rather than two doublets and a singlet.
For the vector representation, simultaneously rotate the first and second coordinate planes. Over $\mathbb C$, the two planes give opposite pairs of <weights>, while the fifth coordinate is fixed. Hence its <weight diagram> is
$$
\boxed{\operatorname{Wt}(\mathbf5)=\{(1,0),(-1,0),(0,1),(0,-1),(0,0)\},}
$$
with every <weight multiplicity> equal to one. Evaluating $x+y$ gives $+1$ twice, $-1$ twice and $0$ once, exactly two <SU(2) representations> of dimension two and one singlet.
For the <Adjoint representation>, the <root-space decomposition> has one one-dimensional space for each of the eight roots and a two-dimensional zero-weight <Cartan subalgebra>. Thus
$$
\boxed{\operatorname{Wt}(\mathbf{10})=\{\text{eight roots of }B_2\}\cup\{(0,0)\text{ with multiplicity }2\}.}
$$
The <coroot> values $x+y$ have multiplicities $1,2,4,2,1$ at $-2,-1,0,1,2$. One zero-weight state joins the $\pm2$ states to make a triplet; the $\pm1$ states form two doublets, leaving three zero-weight singlets. This verifies the earlier <branching rule> and accounts for all ten dimensions.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/iii/paper-44-b2-weights.png]
{title=B2 weight lattice and the weight diagrams of the vector and adjoint representations}
{height=480}
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