Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 302 (Symmetries, Fields and Particles)

The integers are closed under addition, addition is associative, zero is an identity, and $-n$ is the additive inverse of $n$. Thus $(\mathbb{Z},+)$ is an abelian group.

The nonzero integers are closed under multiplication and contain the identity, but most elements have no inverse in the set: for example, the multiplicative inverse of $2$ is $1/2\notin {\mathbb{Z}}^{\ast }$. Hence $({\mathbb{Z}}^{\ast },\times )$ is not a group.

Composition is associative, andagain has nonzero slope. The identity is $(1,0)$ andThus these maps form the real affine group of the line, a nonabelian group.

In coordinates the Real Heisenberg group law isCommuting this with every $(q^{\prime },r^{\prime },s^{\prime })$ requires $q{r}^{\prime }-q^{\prime }r=0$ for all $q^{\prime },r^{\prime }$, hence $q=r=0$. Therefore

The Lie algebra consists of strictly upper-triangular matricesThe only nonzero basis bracket is $[Q,R]=S$. Thus the independent nonzero structure constant of a Lie algebra values are $f_{QR}{}^S=1$ and $f_{RQ}{}^S=-1$.

Since $X^3=0$,Every group element has the unique preimage $t=s-qr/2$, so the Exponential map of a Lie group is bijective. It is a diffeomorphism from ${\mathbb{R}}^3$ onto $G$, proving that $G$ is connected and simply connected.

The line $\mathbb{R}S$ is a nonzero central ideal, so the Heisenberg Lie algebra is not simple. It is two-step nilpotent and hence solvable; a nonzero solvable Lie algebra is not semisimple. It is therefore neither simple nor semisimple.

Strictly, the normalized functions form the unit sphere rather than a vector space; let $V=L^2(\mathbb{R})$ and restrict to normalized states when interpreting wavefunctions. The group law givesTranslation preserves Lebesgue measure and both exponential factors have unit modulus, so $D(g)$ preserves the inner product and is a unitary representation. This is the Schrödinger representation of the Heisenberg group: $q$ translates position, $r$ translates momentum, and $s$ contributes the physically irrelevant overall phase.

A Lie bracket is bilinear, antisymmetric, and satisfies the Jacobi identity. For a basis $T_a$,The structure constant of a Lie algebra coefficients satisfy $f_{ab}{}^c=-f_{ba}{}^c$ and $f_{ab}{}^d{f}_{dc}{}^e+f_{bc}{}^d{f}_{da}{}^e+f_{ca}{}^d{f}_{db}{}^e=0$.

Conjugation differentiates to the Adjoint representationBecause conjugations compose, ${\mathrm{Ad}}_{gh}={\mathrm{Ad}}_g{\mathrm{Ad}}_h$ and ${\mathrm{Ad}}_e=I$, so this is a group representation.

Differentiating ${\mathrm{Ad}}_{\exp (tX)}Y$ at zero gives the Adjoint representation of a Lie algebraThe Jacobi identity implies $[{\mathrm{ad}}_X,{\mathrm{ad}}_Y]={\mathrm{ad}}_{[X,Y]}$, proving that it is a Lie-algebra representation.

Since $({\mathrm{ad}}_{T_a}{)}^c{}_b=f_{ab}{}^c$, the Killing form has componentsThis is the matrix trace of ${\mathrm{ad}}_{T_a}{\mathrm{ad}}_{T_b}$.

The displayed basis obeys $[{\tau }_a,{\tau }_b]={ϵ}_{abc}{\tau }_c$. ThereforeThe basis is adapted to the Killing form in the usual orthogonal-basis sense, since its Gram matrix is diagonal; rescaling by $1/\sqrt{2}$ makes it orthonormal for $-\kappa$.

The complexification of a Lie algebra is $\mathfrak{su}(2{)}_{\mathbb{C}}=\mathfrak{su}(2){\otimes }_{\mathbb{R}}\mathbb{C}\cong {\mathfrak{sl}}_2(\mathbb{C})$. With $[H,E_{\pm }]=\pm 2E_{\pm }$ and $[E_{+},E_{-}]=H$, the Killing matrix in the ordered basis $(E_{+},E_{-},H)$ isThis Cartan-Weyl basis is adapted to the root decomposition, but it is not Killing-orthogonal because $\kappa (E_{+},E_{-})=4$.

Using the matrix-unit commutator and summing the adjoint indices givesEquivalently, in components this is $2n{X}^i{}_j{Y}^j{}_i-2X^i{}_i{Y}^j{}_j$. It vanishes on the scalar center, as expected because ${\mathfrak{gl}}_n$ is not semisimple.

The Lorentz group consists of linear maps $x\mapsto \Lambda x$ satisfying ${\Lambda }^T\eta \Lambda =\eta$. The Poincare group consists of affine isometries $x\mapsto \Lambda x+a$ and has multiplicationThus it is the semidirect product ${\mathbb{R}}^{1,3}⋊O(1,3)$, with the Lorentz group as the subgroup fixing the spacetime origin.

With cyclic spatial indices,Substitution in the given Poincare algebra brackets yields

Because the translations commute, $P^{\sigma }{P}^{\mu }$ is symmetric in $(\sigma ,\mu )$, whereas the Levi-Civita tensor in the Pauli-Lubanski pseudovector is antisymmetric, so $W_{\mu }{P}^{\mu }=0$. Moreover,because the two terms cancel after relabeling and each remaining momentum product is symmetric.

In the rest frame $P^{\mu }=(m,0,0,0)$, antisymmetry gives $W_0=0$. Taking ${ϵ}_{0123}=+1$ and $J_3=M^{12}$ givesHenceThe sign of the second eigenvalue reverses if the opposite Levi-Civita convention is chosen.

A Cartan subalgebra of a complex semisimple Lie algebra is a maximal abelian subalgebra consisting of semisimple elements.

For a Cartan subalgebra $\mathfrak{h}$, the root set consists of the nonzero linear functionals $\alpha \in {\mathfrak{h}}^{\ast }$ for which ${\mathfrak{g}}_{\alpha }=\{X:[H,X]=\alpha (H)X\}$ is nonzero. It is the root system of the Lie algebra.

A root string through $\beta$ in the $\alpha$ direction is the uninterrupted sequence $\beta -p\alpha ,\dots ,\beta +q\alpha$ of roots, with $p-q=⟨\beta ,{\alpha }^{\vee }⟩$.

For ordered simple roots, the Cartan matrix is $A_{ij}=⟨{\alpha }_i,{\alpha }_j^{\vee }⟩=2({\alpha }_i,{\alpha }_j)/({\alpha }_j,{\alpha }_j)$. It determines their relative lengths and angles.

The first diagram is $D_4$, with central node ${\alpha }_2$. Its Cartan matrix is

The second diagram is $F_4$; the arrow points from the long root ${\alpha }_2$ toward the short root ${\alpha }_3$. Thus

The classification of rank-two root systems gives $A_1\times A_1$, $A_2$, $B_2$, and $G_2$. The first is reducible and corresponds to a semisimple but nonsimple algebra. Hence the complex simple rank-two algebras are $A_2$, $B_2=C_2$, and $G_2$, so $N=3$.

This is the G2 root system, with $\alpha$ long and $\beta$ short. Its positive roots areand the full root set includes their negatives. Thus $\dim \mathfrak{g}=12+2=14$. Finallyand simple roots have obtuse angle, so $\theta =5\pi /6$.

For $x\alpha +y\beta$, the simple-coroot coordinates are $2x-y$ and $-3x+2y$. Solving for the fundamental weights givesHence $C_1=2$, $C_2=3$, and $C_3=2$. The weight $\omega$ is the fundamental weight of the short root, and its highest-weight representation is the seven-dimensional fundamental representation of $G_2$. Its weights are zero and the six short roots.

Under $\mathfrak{sl}(2{)}_{\alpha }$, a root $x\alpha +y\beta$ has weight $2x-y$. Counting the twelve root spaces and the two-dimensional Cartan subalgebra gives multiplicities one at weights $\pm 2$, four at $\pm 1$, and four at zero. Therefore

Under $\mathfrak{sl}(2{)}_{\beta }$, the weight is $-3x+2y$. The multiplicities are two at $\pm 3$, one at $\pm 2$, two at $\pm 1$, and four at zero. HenceThe dimensions are respectively $3+4\cdot 2+3=14$ and $2\cdot 4+3+3=14$, verifying both direct-sum decompositions of the Adjoint representation.