Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 302 / 4

The special orthogonal group isIts Lie algebra consists of antisymmetric $5\times 5$ matrices, determined by the entries above the diagonal. Therefore

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For every $R\in SO(4)$,is orthogonal with determinant one. Moreover $\iota (R_1{R}_2)=\iota (R_1)\iota (R_2)$ and $\iota$ is injective, so these block-diagonal matrices form an $SO(4)$ subgroup of $SO(5)$.

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Under the subgroup in part (b), $(v,s)\in {\mathbb{R}}^4\oplus \mathbb{R}$ transforms as $(Rv,s)$, henceAn element of ${\mathfrak{so}}_5$ has the unique block formConjugation by $\iota (R)$ sends $(A,v)$ to $(RA{R}^{-1},Rv)$. Thusand the dimensions check as $5=4+1$ and $10=6+4$. These are the SO5 to SO4 branching rules.

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The fundamental weights satisfy $2({\omega }_i,{\alpha }_j)/({\alpha }_j,{\alpha }_j)={\delta }_{ij}$. Since $|{\alpha }_1{|}^2=1$ and $|{\alpha }_2{|}^2=2$, solving givesThe full B2 root system isThe weight lattice is generated by ${\omega }_1,omeg{a}_2$; geometrically it consists of the integer lattice together with the translate in which both coordinates are half-integers.

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The Fundamental representations of B2 have weight setsso $\dim d_{[1,0]}=4$, andso $\dim d_{[0,1]}=5$. The Adjoint representation has all eight roots as nonzero weights and zero with multiplicity two. Its highest root isso its highest-weight label is $[2,0]$ and its dimension is $8+2=10$.

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Using ${\mathfrak{so}}_4\cong {\mathfrak{su}}_2\oplus {\mathfrak{su}}_2$, the five-dimensional vector representation restricts aswhich is the $4\oplus 1$ decomposition from part (c). The four-dimensional spin representation is naturally a representation of ${\mathfrak{so}}_5$ or $Spin(5)$ rather than an honest representation of $SO(5)$; it restricts asthe two chiral spin representations of ${\mathfrak{so}}_4$. Finally,namely the two three-dimensional summands of the $SO(4)$ adjoint plus its four-dimensional vector, agreeing with $10\to 6\oplus 4$ from the SO5 to SO4 branching calculation.

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