Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 302 (Symmetries, Fields and Particles)

A Lie group is a group that is also a smooth manifold, with differentiable multiplication and inversion. A Lie algebra is a vector space with a bilinear alternating Lie bracket satisfying the Jacobi identity.

For a Matrix Lie group $G$, put $\mathfrak{g}=T_{I}G$. If $X,Y\in \mathfrak{g}$, the matrix commutator $[X,Y]=XY-YX$ again lies in $\mathfrak{g}$: the group commutator $e^{tX}{e}^{sY}{e}^{-tX}{e}^{-sY}$ lies in $G$, and the coefficient of $ts$ in its matrix logarithm is $[X,Y]$. Bilinearity and antisymmetry are immediate, while associativity of matrix multiplication gives the Jacobi identity. Thus $T_{I}G$, with the commutator bracket, is the Lie algebra of a matrix Lie group $\mathcal{L}(G)$.

The special linear group $SL(2,\mathbb{R})$ is the inverse image of the regular value $1$ under the smooth determinant map, and multiplication and inversion are smooth. Differentiating $\det (I+tA)=1+t\mathrm{tr}A+O(t^2)$ shows thatThe Cayley-Hamilton theorem applied to a trace-zero two-by-two matrix gives

The exponential map of a matrix Lie group is the matrix exponentialSince $\det (e^A)=e^{\mathrm{tr}A}=1$, its image lies in $SL(2,\mathbb{R})$. Put $d=\det A$. The identity $A^2=-dI$ sums the series explicitly. If $d>0$, with $r=\sqrt{d}$,If $d=0$, the trace is $2$, while if $d<0$, with $r=\sqrt{-d}$,HenceBut $\mathrm{diag}(-2,-1/2)\in SL(2,\mathbb{R})$ has trace $-5/2$. It is therefore outside the image, so the exponential map is not surjective.

A Cartan subalgebra $\mathfrak{h}$ of a finite-dimensional complex Semisimple Lie algebra is a maximal abelian subalgebra consisting of semisimple elements. The root-space decomposition isand the nonzero functionals $\alpha$ are the roots. A Cartan-Weyl basis consists of a basis $H_i$ of $\mathfrak{h}$ and root vectors $E_{\alpha }\in {\mathfrak{g}}_{\alpha }$. Its brackets have the formand $[E_{\alpha },E_{\beta }]=N_{\alpha \beta }{E}_{\alpha +\beta }$ when $\alpha +\beta$ is a root, and zero when $\alpha +\beta$ is neither a root nor zero.

For the complexified so4 Lie algebra, take $H^1=T^{(12)}$ and $H^2=T^{(34)}$. WriteDirect use of the stated commutation relations givesThe simultaneous eigenvectors, hence the step generators, may be chosen asThus the roots relative to $(H^1,H^2)$ are $(\pm i,\pm i)$. Replacing $H^a$ by $-i{H}^a$ gives the usual real coordinates $(\pm 1,\pm 1)$. The only nonzero brackets between step generators, apart from those obtained by antisymmetry, are

An isomorphism of Lie algebras is a bijective linear map preserving the Lie bracket. DefineThenThe two spans are commuting copies of the complexified ${\mathfrak{su}}_2$, and together contain all six basis elements of ${\mathfrak{so}}_4$. Chiral decomposition of the complexified so4 Lie algebra therefore givesUnder this isomorphism the Adjoint representation is the direct sum of the adjoint representations of the two factors:

The root lattice is $Q={\sum }_i\mathbb{Z}{\alpha }^{(i)}$. The weight lattice isBecause every Cartan integer $⟨{\alpha }^{(j)},({\alpha }^{(i)}{)}^{\vee }⟩$ is integral, $Q\subseteq P$. The Dynkin labels of $\lambda$ are

There are three isomorphism classes of complex simple rank-three Lie algebras: types A3 root system$A_3$, B3 root system$B_3$, and C3 root system$C_3$. Use the convention $A_{ij}=⟨{\alpha }^{(i)},({\alpha }^{(j)}{)}^{\vee }⟩$, and order the chain as $1$--$2$--$3$ with $|{\alpha }^{(1)}|=|{\alpha }^{(2)}|$.

The Dynkin labels of a finite-dimensional irreducible representation are those of its highest weight. Thus $[1,0,0]$ means the fundamental representation $V({\omega }_1)$. The weights, written in Dynkin labels and in a lowering order, areFor $A_3$ these are the weights of the defining representation of ${\mathfrak{sl}}_4$; for $B_3$ they are $\{\pm e_1,\pm e_2,\pm e_3,0\}$ in the vector representation of ${\mathfrak{so}}_7$; for $C_3$ they are $\{\pm e_1,\pm e_2,\pm e_3\}$ in the defining representation of ${\mathfrak{sp}}_6$. Hence the requested dimensions are $4$, $7$, and $6$, respectively.

Let $G$ be a simple Lie group, let $T^a$ be Hermitian matrices for a finite-dimensional unitary representation $R$, and normalizeA matter field $\psi$ transforms locally as ${\psi }^{\prime }(x)=U(x)\psi (x)$, where $U(x)=e^{ig{ϵ}^a(x)T^a}$. An ordinary derivative of $\psi$ does not transform covariantly because it differentiates $U$. Introduce a gauge field $A_{\mu }=A_{\mu }^a{T}^a$ and the gauge covariant derivativeDemanding $D_{\mu }^{\prime }{\psi }^{\prime }=U{D}_{\mu }\psi$ determines the Yang-Mills gauge transformationTo first order in $ϵ$,

The gauge field strength is defined by $[D_{\mu },D_{\nu }]=-ig{F}_{\mu \nu }$:or, in components,Covariance of the commutator givesThe commutator term distinguishes Yang-Mills theory from an Abelian gauge theory and produces cubic and quartic gauge-boson interactions.

For a Dirac field of mass $m$ in $R$, the Lagrangian isEquivalently, the gauge term is proportional to $-\mathrm{tr}(F_{\mu \nu }{F}^{\mu \nu })$. The cyclic property of the trace and $F_{\mu \nu }^{\prime }=U{F}_{\mu \nu }{U}^{-1}$ make it invariant. Unitarity gives ${\bar{\psi }}^{\prime }=\bar{\psi }{U}^{-1}$, while $D_{\mu }^{\prime }{\psi }^{\prime }=U{D}_{\mu }\psi$, so both the matter kinetic term and mass term are invariant. A complex scalar $\varphi$ in a unitary representation may instead be coupled throughprovided the scalar potential $V$ is $G$-invariant.

The simplicity assumption means that the Lie algebra $\mathfrak{g}$ is nonabelian and has no proper nonzero Ideal of a Lie algebra. Its Adjoint representation is therefore irreducible, and every invariant symmetric bilinear form is proportional to the Killing form. Consequently the pure gauge kinetic term has one overall gauge coupling for a simple factor. The theory has no independent Abelian gauge direction; if the gauge algebra were a direct sum of simple and Abelian ideals, each factor could instead carry its own coupling. A simple group may still have a discrete center, but this does not add a gauge boson because gauge bosons are indexed by the Lie algebra.