Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 302 (Symmetries, Fields and Particles)

The squared root lengths and inner product areWith $A_{ij}=2{\alpha }_i\cdot {\alpha }_j/|{\alpha }_j{|}^2$, the Cartan matrix isThis is the Cartan matrix of the G2 root system, with ${\alpha }_1$ short and ${\alpha }_2$ long.

The root string through ${\alpha }_1$ in the ${\alpha }_2$ direction contains ${\alpha }_1$ and ${\alpha }_1+{\alpha }_2$, so ${\alpha }_1$ can be raised once by ${\alpha }_2$. The string through ${\alpha }_2$ in the ${\alpha }_1$ direction containsso ${\alpha }_2$ can be raised three times by ${\alpha }_1$.

The Weyl reflection in the hyperplane perpendicular to $\alpha$ sends a weight $\mu$ to

The positive roots generated by the two simple roots areThe Adjoint representation of a Lie algebra has one weight space for each positive and negative root and a rank-two zero-weight space. Its nonzero weights are thereforetogether with their negatives. Thus the diagram consists of a hexagon of six long roots, a hexagon of six short roots, and the origin with multiplicity two. The representation has dimension $12+2=14$.

The fundamental weights satisfySolving these equations gives

Starting from the highest weight ${\mu }_1=2{\alpha }_1+{\alpha }_2$, lowering with the simple-root operators and closing under the Weyl group givesEvery weight has multiplicity one. The diagram is the hexagon of short roots with one central weight, so this is the seven-dimensional fundamental irreducible representation of $G_2$.

The identity and inverse axioms give

Setting either argument to zero in the expansion and comparing coefficients givesThus, through quadratic order,

Write ${\bar{x}}^r=-x^r+q^r{}_{st}{x}^s{x}^t+O(x^3)$. Substitution into $\phi (\bar{x},x)=0$ givesand henceOnly the symmetric part of $C^r{}_{st}$ contributes to this expression.

Insert the expansions from parts ii and iii successively into
$g(w)=g(x{)}^{-1}g(y{)}^{-1}g(x)g(y)$. All constant and linear terms cancel, leavingThe antisymmetric part of the local multiplication law is therefore the infinitesimal group commutator.

Since $z+dz=\phi (z,\theta )$,Thus

The infinitesimal change $dy$ can be writtenRight multiplication of $g(y)$ by $g(\theta )$ is also right multiplication of $g(z)=g(x)g(y)$ by the same element. Part v therefore givesso

Applying $T_a(y)={\mu }_a{}^s(y)\partial /\partial y^s$ to a function of $z$ and using part vi,HenceThese are the same left-invariant vector fields in different coordinates.

Differentiate the identity in part vi with respect to $y^t$, multiply by ${\mu }_a{}^s(y){\mu }_b{}^t(y)$, and use $\lambda \mu =I$. The product rule gives

Differentiate ${\lambda }_s{}^c{\mu }_a{}^s={\delta }_a{}^c$ to obtainMultiplying part viii by ${\lambda }_r{}^c(z)$ consequently gives

Antisymmetrize part ix in $a,b$. The left side vanishes because the second derivative is symmetric in $s,t$. ThereforeSince multiplication by an arbitrary $g(x)$ moves $y$ to any $z$ in the connected coordinate neighborhood, the $f^c{}_{ab}$ are constants. They are the structure constants.

Directly commuting the vector fields givesUsing the definition in part x and $\lambda \mu =I$ yields

In exponential coordinates, the Baker--Campbell--Hausdorff formula givesComparison with ${\phi }^c(x,y)=x^c+y^c+C^c{}_{ab}{x}^a{y}^b+O(3)$ gives

On the branch through the identity,Direct calculation gives $M^{†}M=I$ and $\det M=1$. Thus these matrices lie inproviding the three requested matrix groups.

A vector transforms as $x^{\prime a}=R^a{}_b{x}^b$. The matrix fixes $n$, preserves lengths and orientation, and rotates the perpendicular plane through $\theta$. It is a rotation by angle $\theta$ about the axis $n$, with orientation fixed by the epsilon-term convention.

Sinceconjugation by $A\in SU(2)$ preserves Hermiticity, tracelessness, and the determinant. Hence $A(x\cdot \sigma )A^{†}=x^{\prime }\cdot \sigma$ with $|x^{\prime }|=|x|$. Thus $\boxed{G=SU(2)}$.

From $A{\sigma }_b{A}^{†}=R^a{}_b{\sigma }_a$ and
$\mathrm{tr}({\sigma }_a{\sigma }_b)=2{\delta }_{ab}$,

The matrix isIts image is the rotation through $\theta$ about

The other element is $\boxed{-M}$. Conjugation by $M$ and $-M$ is identical, so the homomorphism $SU(2)\to SO(3)$ has kernel $\{\pm I\}$ and is the Spin group double cover.

Use the conventionfor the group commutator.

Invariance for every $x$ requiresThus $G=O(1,3)$, the Lorentz group; its identity component is $S{O}^{+}(1,3)$.

At first order, ${\Lambda }^T\eta \Lambda =\eta$ gives
${\omega }^T\eta +\eta \omega =0$. Therefore

Using ${\Lambda }_a^{-1}=I-{\omega }_a+{\omega }_a^2+O({\omega }_a^3)$,Thus $({\omega }_c{)}^{\mu }{}_{\nu }=({\omega }_2{)}^{\mu }{}_{\rho }({\omega }_1{)}^{\rho }{}_{\nu }-({\omega }_1{)}^{\mu }{}_{\rho }({\omega }_2{)}^{\rho }{}_{\nu }$.

Comparison of $U[{\Lambda }_c]$ with the operator group commutator gives the Lorentz algebraThere is no factor of $i$ because the generators are anti-Hermitian.

Substituting $J_m={ϵ}_{mij}{M}^{ij}/2$ and $K_i=M^{0i}$ yields

For $J_i^{\pm }=(J_i\pm i{K}_i)/2$,

The two commuting triples each generate a copy of the complexified SU(2) Lie algebra. Hence the chiral decomposition of the complex Lorentz algebra isIn physicists' compact notation this is
$\mathfrak{su}(2{)}_{+}\oplus \mathfrak{su}(2{)}_{-}$; the real Lorentz algebra relates the two factors by complex conjugation.