Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 302 (Symmetries, Particles, and Fields)

Put $\Lambda (t)=I+tX+O(t^2)$ in ${\Lambda }^T\eta \Lambda =\eta$. The coefficient of $t$ is $X^T\eta +\eta X$, so it must vanish. For $X=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ this condition isHence $a=d=0$ and $b=c$. Thus the Lie algebra is one-dimensional with basis

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Write $X(\phi )=\phi K$. Since $K^2=I$, the Exponential map of a Lie group givesThe hyperbolic addition formulas give $\Lambda (\phi )\Lambda (\psi )=\Lambda (\phi +\psi )$ and $\Lambda (\phi {)}^{-1}=\Lambda (-\phi )$, so these matrices form a subgroup of the One-dimensional Lorentz group. It is Abelian because addition in $\mathbb{R}$ is commutative. It is noncompact because $\cosh \phi$ is unbounded, equivalently because the subgroup is homeomorphic to $\mathbb{R}$.

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For the null coordinates in two-dimensional Minkowski spacetime, direct substitution givesThus a boost dilates one null direction and contracts the other by the reciprocal factor. It preservesThe invariant curves are therefore the level sets $x^2-y^2=c$: the branches of hyperbolas for $c\ne 0$, together with the two null lines when $c=0$. Each connected branch is preserved by the identity component.

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Solving ${\Lambda }^T\eta \Lambda =\eta$ with $\det \Lambda =1$ shows that every element is either $\Lambda (\phi )$ or $-\Lambda (\phi )$. HenceEach set is connected, but the sign of the $(1,1)$ entry cannot change continuously because its absolute value is at least one. Thus $SO(1,1)$ has two connected components. The matrix $-I$ lies in the component disjoint from the identity.

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For $\phi \ne \pm 1$, the Cayley transform isWriting its diagonal and off-diagonal entries as $a,b$, one has $a^2-b^2=1$, so $C^T\eta C=\eta$ and $\det C=1$. For $|\phi |<1$, put $\phi =\tanh (t/2)$ to obtain $C(\phi K)=\Lambda (t)$, so this interval covers the identity component. For $|\phi |>1$ the image lies in the other component and covers it except for $-I$, approached only as $|\phi |\to \infty$. The Exponential map of a Lie group reaches only the identity component, whereas the Cayley transform also reaches nonidentity-component elements but omits $-I$ and is undefined at $\phi =\pm 1$.

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Differentiating ${\Lambda }^T\eta \Lambda =\eta$ at the identity gives $X^T\eta +\eta X=0$. The six matricesobey this condition and form a basis of the Lorentz algebra. With cyclic indices,The $J_i$ mix the two spatial coordinates perpendicular to $x^i$ and therefore generate rotations about that axis; $K_i$ mixes $x^0$ with $x^i$ and generates a Lorentz boost in the $x^i$ direction.

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Matrix multiplication givesFor example, $[J_1,J_2]=J_3$, $[K_1,K_2]=-J_3$, and $[J_1,K_2]=K_3$. Over $\mathbb{C}$, defineThen $[A_i,A_j]={ϵ}_{ijk}{A}_k$, $[B_i,B_j]={ϵ}_{ijk}{B}_k$, and $[A_i,B_j]=0$. This proves the chiral decomposition of the complex Lorentz algebra. Its finite-dimensional irreducible representations are tensor products of irreducible representations of the two factors and are labelled by pairs $(j_L,j_R)$ of nonnegative half-integers.

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A coordinate transformation $x\mapsto Px$ conjugates an infinitesimal generator, hence $M^{\mu \nu }\mapsto P{M}^{\mu \nu }P$. It fixes the rotation generators $J_i$ and negates the boosts $K_i$. Therefore it exchanges $A_i$ and $B_i$, and the parity action on a Lorentz representation isAn irreducible representation is parity invariant precisely when $j_L=j_R$. If $j_L\ne j_R$, parity invariance requires the reducible sum $(j_L,j_R)\oplus (j_R,j_L)$. Thus the two Weyl-spinor representations $(1/2,0)$ and $(0,1/2)$ are exchanged, while their direct sum is the parity-invariant Dirac spinor.

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Parity preserves the Minkowski metric but reverses orientation. Hence ordinary tensor contraction makesparity even, while the Levi-Civita pseudotensor changes sign and makesparity odd. Equivalently, $C_1$ is proportional to ${\mathbf{J}}^2-{\mathbf{K}}^2$ and $C_2$ to $\mathbf{J}\cdot \mathbf{K}$. The combinations $C_1\pm i{C}_2$ are proportional to the two quadratic Lorentz Casimir invariants, and parity exchanges them exactly as it exchanges the two factors in the complexified algebra.

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The Adjoint representation of a Lie algebra isIt is linear. The Jacobi identity givesso it is a Lie algebra representation.

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The Killing form is $\kappa (X,Y)=\mathrm{Tr}({\mathrm{ad}}_X{\mathrm{ad}}_Y)$. From part (a),where the middle equality uses cyclicity of the trace.

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Each structure constant of a Lie algebra satisfies $f_{ab}{}^d=-f_{ba}{}^d$, so $f_{abc}=f_{ab}{}^d{\kappa }_{dc}$ is antisymmetric in $a,b$. Invariance of the Killing form givesThis cyclic symmetry together with antisymmetry in the first pair implies antisymmetry under every transposition. Thus $f_{abc}$ is totally antisymmetric.

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Let $W\subseteq \mathfrak{g}$ be invariant under the Adjoint representation of a Lie algebra. Then $[\mathfrak{g},W]\subseteq W$, so $W$ is an ideal. If $\mathfrak{g}$ is a Simple Lie algebra, its only ideals are $0$ and $\mathfrak{g}$. Hence the adjoint representation is irreducible. Compact type is compatible with the anti-Hermitian realization used below, although simplicity alone proves this adjoint irreducibility of a simple Lie algebra.

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The Trace form of a Lie algebra representation is invariant becauseChoose a basis orthonormal for the positive-definite form $-\kappa$ of the compact simple algebra. Any invariant bilinear form determines an endomorphism commuting with the irreducible adjoint action; Schur lemma makes it scalar. Thus $H(T_a,T_b)=c{\delta }_{ab}$. Since $d(T_a)$ is anti-Hermitian,The inequality is strict because the kernel of the nontrivial irreducible representation is an ideal and hence zero. Therefore $c=-\mu$ with $\mu >0$.

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For the trace trilinear form of a Lie algebra representation, the trace of a commutator vanishes:Taking $X=T_b$, $Y=T_a$, $Z=T_c$, $W=T_d$ and expanding each Lie bracket givesSince $f_a{}^{bc}$ is antisymmetric in $b,c$,Raising indices with the inverse Killing form gives $f_a{}^{bc}{f}_{bc}{}^e={\delta }_a{}^e$ in the stated normalization. Using $H(T_e,T_d)=-\mu {\delta }_{ed}$ proves

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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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