Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 302 (Symmetries, Fields and Particles)

A Lie group is a smooth manifold $G$ equipped with smooth maps $m:G\times G\to G$, $m(g,h)=gh$, and $\iota :G\to G$, $\iota (g)=g^{-1}$, for which multiplication is associative, there is an identity $e$, and every $g$ has the inverse $g^{-1}$. In equations,Thus the group axioms and the smooth structure are compatible.

In a coordinate chart centred at the identity, smooth multiplication has a local group lawThe identity gives $F(0,x)=x$ and $F(y,0)=y$, while associativity gives the functional equation $F(u,F(y,x))=F(F(u,y),x)$. Smooth inversion determines coordinates $\bar{x}$ with $F(\bar{x},x)=F(x,\bar{x})=0$. Changes of local coordinates alter the symmetric part of $B^k{}_{ij}$, whereas its antisymmetric part supplies the coordinate-independent Lie bracket on the tangent space at the identity.

Write $g(x)=I+x^i{T}_i+O(x^2)$ in a matrix representation and $[T_i,T_j]=c^k{}_{ij}{T}_k$. Direct multiplication givessoThe cancellation of the constant and linear terms explains why the group commutator first detects the Lie bracket at quadratic order.

The matrix is invertible precisely when $a\ne 0$, andThe largest group is thereforethe real affine group $\mathbb{R}⋊{\mathbb{R}}^{\ast }$ under $(a,b)(a^{\prime },b^{\prime })=(a{a}^{\prime },b+a{b}^{\prime })$.

As a smooth manifold, $G\cong {\mathbb{R}}^{\ast }\times \mathbb{R}$. It has two connected components, distinguished by the sign of $a$: each of $a>0$ and $a<0$ is connected, and a continuous path in $G$ cannot cross $a=0$.

The defining two-dimensional group representation is reducible, because the line $\mathbb{R}{e}_1$ is an invariant subspace:For $b\ne 0$, the complementary line $\mathbb{R}{e}_2$ is not invariant, so this representation need not split as a direct sum of one-dimensional representations.

The map $(a,b)\mapsto \mathrm{sgn}(a)$ is a surjective group homomorphism with kernel $H_0$, so $H_0$ is a normal subgroup and $G/H_0\cong C_2$. The subgroup $H_1$ is not normal: for a translation $t=(1,c)$,which leaves $H_1$ when $a\ne 1$ and $c\ne 0$. Finally, $(a,b)\mapsto a$ is a surjective homomorphism onto ${\mathbb{R}}^{\ast }$ with kernel $H_2$, so $H_2◃G$ and the quotient group $G/H_2\cong {\mathbb{R}}^{\ast }$.

With $\eta =\mathrm{diag}(1,-1,-1)$, the defining condition for the $(1+2)$-dimensional Lorentz group isTaking determinants gives $(\det \Lambda {)}^2=1$, hence $\det \Lambda =\pm 1$. The norm of the zeroth column givesso $|{\Lambda }^0{}_0|\ge 1$. The Proper orthochronous Lorentz group is the component with $\det \Lambda =1$ and ${\Lambda }^0{}_0\ge 1$.

Identify $x=(x^0,x^1,x^2)$ with the real symmetric matrixFor $A\in SL(2,\mathbb{R})$, define $X\mapsto AX{A}^T$. This linear action preserves $\det X$, so it defines $\rho (A)\in S{O}^{+}(1,2)$, and $\rho (AB)=\rho (A)\rho (B)$. Its kernel is $\{\pm I\}$ and it is onto, hence it is a two-to-one covering homomorphism rather than an isomorphism. Consequently $PSL(2,\mathbb{R})=SL(2,\mathbb{R})/\{\pm I\}\cong S{O}^{+}(1,2)$.

The connected $(1+2)$-dimensional Poincare group isThe semidirect product records that Lorentz transformations act nontrivially on translations, which is what organizes momentum orbits. For a representative momentum $p$, its little group is the stabilizer $G_p=\{\Lambda \in S{O}^{+}(1,2):\Lambda p=p\}$. Wigner's classification constructs irreducible particle representations by inducing a unitary irreducible representation of $G_p$ along the Lorentz orbit of $p$.

For a massive positive-energy orbit, choose $p_{\ast }=(m,0,0)$ with $m>0$. Its little group is $SO(2)$, whose unitary irreducible representations are the characters $e^{is\theta }$. Induction gives one-particle states $|p,s⟩$ on the mass shell $p^2=m^2$, $p^0>0$, labeled by mass and spin. For the Poincare group itself, single-valuedness gives $s\in \mathbb{Z}$; its double cover permits half-integers, and its universal cover permits any real $s$, producing anyonic spin in $(2+1)$ dimensions.

For the positive-energy massless orbit choose $k=(E,E,0)$. Its connected little group is the one-parameter group of null rotations, isomorphic to $(\mathbb{R},+)$. Its unitary irreducible representations are the characters $t\mapsto e^{i\rho t}$, $\rho \in \mathbb{R}$, and induction gives the massless one-particle representations. The physically usual finite-component representation has trivial little-group action, $\rho =0$; there is no $SO(2)$ helicity subgroup for a null momentum in $(2+1)$ dimensions.

The special unitary group is $SU(2)=\{U\in M_2(\mathbb{C}):U^{†}U=I,\det U=1\}$. Differentiating these equations at the identity shows that its Lie algebra isFor $T_i=-i{\sigma }_i/2$, the Pauli-matrix identity $[{\sigma }_i,{\sigma }_j]=2i{ϵ}_{ijk}{\sigma }_k$ givesso the structure constants in this basis are $c^k{}_{ij}={ϵ}_{ijk}$.

The Cartan-Weyl basis obeysFor a highest-weight vector $|\Lambda ⟩$, $E_{+}|\Lambda ⟩=0$ and $H|\Lambda ⟩=\Lambda |\Lambda ⟩$. Successive nonzero lowerings give the weight basisFinite dimensionality requires the highest weight $\Lambda$ to be a nonnegative integer. The weights are $\Lambda ,\Lambda -2,\dots ,-\Lambda$, each with multiplicity one, so $\dim d_{\Lambda }=\Lambda +1$.

In the defining representation,Multiplying the three factors in the disentangling identity and comparing with $e^{i\theta {\sigma }_1/2}$ givesThis factorization is the coordinate patch where $\cos (\theta /2)\ne 0$.

Since $-\theta T_1=i\theta {\sigma }_1/2$, the same disentangling identity applies in every Lie algebra representation. The rightmost $E_{+}$ kills $|\Lambda ⟩$, the leftmost $E_{-}$ kills $⟨\Lambda |$, and the middle factor acts by its highest weight. Therefore

Because $d(T_1)$ commutes with itself and is anti-Hermitian,Applying the preceding highest-weight matrix coefficient yields

For simple roots ${\alpha }_i$, the Cartan matrix isIts diagonal entries are two, while its off-diagonal entries encode the angles and relative lengths in the Dynkin diagram.

The fundamental weights ${\omega }_i$ are the basis dual to the simple coroots:

The Dynkin labels of a weight $\lambda$ areso $\lambda ={\sum }_i{\lambda }^i{\omega }_i$.

A dominant weight has nonnegative Dynkin labels. A dominant integral weight additionally lies in the weight lattice, so all those labels are nonnegative integers.

The highest weight of a representation is the weight $\Lambda$ of a nonzero weight vector annihilated by every positive-root operator. Every other weight is obtained from $\Lambda$ by subtracting a nonnegative integer combination of simple roots. A finite-dimensional irreducible representation of a complex semisimple Lie algebra is determined by its dominant integral highest weight.

Pairing ${\alpha }_i={\sum }_j{c}_{ij}{\omega }_j$ with ${\alpha }_k^{\vee }$ gives $c_{ik}=A_{ik}$. Hence the relation between the two bases is

With $A_{ij}=2({\alpha }_i,{\alpha }_j)/({\alpha }_j,{\alpha }_j)$ and roots labeled from left to right, the three Cartan matrices are

All roots of $A_3$ have the same length, so $|{\alpha }_2|/|{\alpha }_3|=1$. In $B_3$, ${\alpha }_2$ is long and ${\alpha }_3$ short, giving $\sqrt{2}$. In $C_3$ their roles are reversed, giving $1/\sqrt{2}$. These ratios also follow from $A_{23}/A_{32}=|{\alpha }_2{|}^2/|{\alpha }_3{|}^2$.

Disconnected nodes in a Dynkin diagram represent orthogonal roots, so $\angle ({\alpha }_1,{\alpha }_3)=90^{\circ }$ in all three cases. A single edge gives $120^{\circ }$, hence $\angle ({\alpha }_1,{\alpha }_2)=120^{\circ }$ and, for $A_3$, also $\angle ({\alpha }_2,{\alpha }_3)=120^{\circ }$. The double edge in both $B_3$ and $C_3$ gives $\angle ({\alpha }_2,{\alpha }_3)=135^{\circ }$.

Starting from $[1,0,0]$ and applying the simple-root lowerings gives the four weightsEach has weight multiplicity one, so $d_1$ is the four-dimensional defining representation of $A_3\cong {\mathfrak{sl}}_4(\mathbb{C})$.

The representation with highest weight $[0,0,1]$ is the dual representation of $d_1$. Its weights are the negatives of those of $d_1$:Thus $\dim d_2=4$.

For $B_3\cong {\mathfrak{so}}_7(\mathbb{C})$, the highest weight $[0,0,1]$ is the spinor representation. Its eight weights, all of multiplicity one, areEquivalently, in an orthonormal basis they are $(\pm e_1\pm e_2\pm e_3)/2$ with independent signs. Hence $\dim d_3=8$.

The tensor product of the defining representation and its dual is the endomorphism representation. Splitting an endomorphism into its matrix trace and traceless part givesIn Dynkin labels, the two irreducible summands have highest weights $[0,0,0]$ and $[1,0,1]$.

Applying $P[a,b,c]=[b,a,b+c]$ to the eight weights of $d_3$ givesThis is exactly the union of the weight sets of $d_1$ and $d_2$. It expresses the branching rule for the standard inclusion $D_3\cong A_3\subset B_3$: the eight-dimensional spinor representation of ${\mathfrak{so}}_7$ restricts to the two four-dimensional chiral spinor representations of ${\mathfrak{so}}_6\cong {\mathfrak{sl}}_4$,