Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 42 (Symmetries, Fields and Particles)

The SU(2) group consists of complex $2\times 2$ matrices $U$ with $U^{†}U=I$ and $\det U=1$. The SO(3) group consists of real $3\times 3$ matrices $R$ with $R^{T}R=I$ and $\det R=1$. Differentiate these equations along a path through the identity. This givesConversely the matrix exponential of each displayed infinitesimal matrix satisfies the corresponding group equations, so these are exactly the Lie algebras, with bracket the matrix commutator.

Using the Pauli matrices, put $t_a=-i{\sigma }_a/2$. They form a real basis of the SU(2) Lie algebra. The Pauli matrix commutator identity gives $[t_a,t_b]={ϵ}_{abc}{t}_c$. Define $J_a$ on ${\mathbb{R}}^3$ by $J_{a}v=e_a\times v$. These are a basis of the SO(3) Lie algebra; the vector triple-product identity gives $[J_a,J_b]={ϵ}_{abc}{J}_c$. Thusis a real linear bijection preserving the bracket. This is the SU(2)-SO(3) Lie algebra isomorphism. It is a Lie-algebra isomorphism, not a group isomorphism: the Adjoint double cover from SU(2) to SO(3) has kernel $\{I,-I\}$.

The SU(3) group is defined similarly by $U^{†}U=I$ and $\det U=1$ on complex $3\times 3$ matrices. One SU(2) subgroup is $\{\mathrm{diag}(V,1):V\in SU(2)\}$. The real orthogonal matrices with determinant one form an SO(3) group subgroup, since real orthogonality is also complex unitarity.

Under the defining group action, the group orbit of $e_1$ is exactly the unit sphere in ${\mathbb{C}}^3$:Unitarity proves containment. Conversely, extend a unit vector $z$ to an orthonormal basis and use it as the first column of a unitary matrix. Multiplying the last column by the inverse of its determinant makes that determinant one without changing the first column. The isotropy group of $e_1$ must also preserve its orthogonal complement, and therefore isThis proves the unit sphere orbit of the defining special unitary action, including $S^5\cong SU(3)/SU(2)$.

For the complex quadric, write $z=x+iy$. Its defining relation separates intoThe Hermitian norm is then $z^{†}z=1+2|y{|}^2$, which is not constant on this set. For example, $e_1$ and $\sqrt{2}{e}_1+i{e}_2$ both satisfy the complex bilinear relation but have Hermitian norms one and three. Since the SU(3) group preserves that norm, the quadric cannot be a single SU(3) group orbit. In fact it is not even invariant under the whole group: $\mathrm{diag}(e^{it},e^{-it},1)e_1$ fails the bilinear relation when $e^{2it}\ne 1$.

The real SO(3) group does preserve the quadric, acting simultaneously on $x$ and $y$. Its group orbits are classified completely by $r=|y|\ge 0$. For $r=0$, $y=0$ and $x$ is a real unit vector, giving the group orbit $S^2$ and stabilizer subgroup $SO(2)$. For $r>0$, the vectors $x/\sqrt{1+r^2}$ and $y/r$ are orthonormal. Adjoining their cross product gives an oriented orthonormal frame. There is a unique rotation taking the standard frame to this one; hence the action is transitive at fixed $r$ and the stabilizer subgroup is trivial. Thus the real rotation orbits on a complex unit quadric areIn particular the real subgroup does not act transitively on the whole quadric. The parametrization $x=\sqrt{1+|y{|}^2}n$ with $n\in S^2$ and $y\perp n$ also identifies the quadric, as a real manifold, with the tangent bundle of $S^2$.

Use anti-Hermitian gauge potentials and the convention $D_i={\partial }_i+A_i$ for the gauge covariant derivative. Commuting these operators defines the gauge field strength:It is antisymmetric in its two spacetime indices, so $F_{xx}=F_{yy}=0$ and $F_{yx}=-F_{xy}$. In two dimensions there is therefore only one independent component, although that component is itself Lie-algebra valued.

Put $B_x=({\partial }_{x}g)g^{-1}$ and $B_y=({\partial }_{y}g)g^{-1}$. Differentiating $g^{-1}$ gives ${\partial }_i{g}^{-1}=-g^{-1}({\partial }_{i}g)g^{-1}$. Equality of mixed derivatives then gives the right Maurer-Cartan equationConsequently the scaled right Maurer-Cartan gauge potential has curvatureThus $\alpha =0$ and $\alpha =-1$ give zero curvature for every smooth $g$. At zero the potential is zero. At minus one it is a pure gauge potential: transforming the zero connection by $g$ gives $A=-dg{g}^{-1}$. If the gauge Lie algebra is abelian, the commutator vanishes for every $\alpha$. If it contains $X,Y$ with $[X,Y]\ne 0$, choose $g(x,y)=e^{xX}{e}^{yY}$; at the origin $B_x=X$ and $B_y=Y$. Thus in that case the two displayed values are the only choices flat for every $g$. If the potential is instead defined using $D=\partial -A$ and field components ${\partial }_x{A}_y-{\partial }_y{A}_x-[A_x,A_y]$, the same calculation reads $\alpha (1-\alpha )[B_x,B_y]$ and the nonzero pure-gauge value is $+1$. The sign convention must be specified.

For the specified SU(2) exponential, let $r=\sqrt{x^2+y^2}$ and, away from the origin, $n=(x/r,y/r,0)$. The Pauli matrix multiplication law gives $(n\cdot \sigma {)}^2=I$, so summing the exponential series yieldsThe continuous extension at the origin is $I$. For $r>0$, the second term is zero precisely when $\sin (r/2)=0$, and henceThese are infinitely many distinct circles.

At the origin, differentiating the exponential at zero gives $B_x=t_1$ and $B_y=t_2$. Since $[t_1,t_2]=t_3=-i{\sigma }_3/2$, the gauge field strength there isTo evaluate it on the circles, use polar coordinates. On $r=2\pi k$ the angular derivative of $g$ vanishes, while $B_r=-i(n\cdot \sigma )/2$. Therefore $B_x=(x/r)B_r$ and $B_y=(y/r)B_r$ commute, and $F_{xy}=0$ everywhere on every such circle, for every $\alpha$.

One can also see these circular curvature zeros for a planar SU2 exponential from a formula valid away from the origin. Write $n=(\cos \vartheta ,\sin \vartheta ,0)$, $m=(-\sin \vartheta ,\cos \vartheta ,0)$, and $t_n=-i(n\cdot \sigma )/2$, $t_m=-i(m\cdot \sigma )/2$. Direct differentiation givesand henceBoth coefficients vanish at the positive circle radii, and the expression tends to $t_3$ at the origin, agreeing with the direct calculation.

A weight of a representation of $SU(3)$ is a joint eigenvalue of two commuting generators spanning a Cartan subalgebra. Equivalently it describes the character by which the diagonal maximal torus acts on a weight vector. Use the Hermitian generatorsTheir multiples by $i$ are in the SU(3) Lie algebra. A vector with weight $(a,b)$ acquires the phase $e^{i(a\theta +b\phi )}$ under $\exp (i\theta I_3+i\phi Y)$. The conventional generator $H_8$ is related by $Y=2H_8/\sqrt{3}$; using $H_8$ instead just rescales the vertical weight coordinate.

In the defining group representation, the coordinate vectors $e_u,e_d,e_s$ are joint eigenvectors. The weight diagram of the defining SU(3) representation therefore hasComplex conjugation reverses all torus phases, so the weights of $\overline{3}$ are $-{\mu }_u,-{\mu }_d,-{\mu }_s$, each with multiplicity one. The conjugation bar is essential: the tensor product here is $3\otimes \overline{3}$.

Weights add in a tensor product of group representations. Thus the nine weights of this product are ${\mu }_i-{\mu }_j$. There are three zero weights from $i=j$, and the remaining six areTo turn this weight calculation into a direct-sum decomposition, identify the tensor product with $\mathrm{End}({\mathbb{C}}^3)$ by $v\otimes \overline{w}\mapsto v{w}^{†}$. The group acts by $M\mapsto UM{U}^{-1}$. This gives two invariant spaces,The scalar line has weight zero and is the trivial group representation $1$. The traceless space has the six nonzero weights just computed, represented by the off-diagonal matrix units $E_{ij}$, and two zero weights, represented by traceless diagonal matrices. Its dimension is eight and it is the complex Adjoint representation $8$. ThereforeThis proves the octet and singlet in a fundamental SU3 tensor product with the correct central weight multiplicities.

The singlet is irreducible because it is one-dimensional. For the octet, an invariant complex subspace of ${\mathfrak{sl}}_3$ is stable under commutators with the complexified Lie algebra. Commuting Cartan generators project it into weight spaces. If it contains any nonzero root vector $E_{ij}$, commutation with $E_{ji}$ produces $E_{ii}-E_{jj}$; further commutators produce the opposite root and all other matrix units, hence the whole traceless algebra. If it contains only a nonzero diagonal traceless matrix $D$, two diagonal entries differ, so $[E_{ij},D]=(D_{jj}-D_{ii})E_{ij}$ supplies a root vector and reduces to the preceding case. There is no nonzero proper invariant subspace. The octet is irreducible, not a sum of six one-dimensional weight spaces and two singlets: the nondiagonal generators connect those spaces.

In the quark model, take $u,d,s$ as the defining flavour triplet and their antiquarks as the conjugate triplet. A colour-singlet quark-antiquark state with relative orbital angular momentum $L=0$ and total spin $S=0$ has $J=0$ and parity $(-1{)}^{L+1}=-1$. Thus this flavour decomposition classifies a pseudoscalar meson nonet: a meson octet and a singlet. The use of approximate flavor symmetry is important; the strange quark is heavier, so these states need not have equal masses.

In the $(I_3,Y)$ diagram, $Y$ is flavor hypercharge, not electroweak hypercharge. Mesons have baryon number zero, so $Y$ equals their strangeness, and electric charge is $Q=I_3+Y/2$. The outer six weight states areThe pions form an isospin triplet, completed at the centre by ${\pi }^0=(u\overline{u}-d\overline{d})/\sqrt{2}$. The four kaons occupy the two hypercharge-one and two hypercharge-minus-one weights. In particular electrically neutral kaons are not at the centre of the weight diagram.

An orthonormal basis of the three central flavour combinations isThe Eta octet state is the second zero weight of the octet; the eta singlet state is the separate invariant scalar. Both have isospin zero, whereas the neutral pion has isospin one. All have $I_3=Y=Q=0$, so the location of a weight alone does not determine either isospin or irreducible multiplet. Their neutral flavour-diagonal $L=S=0$ states have charge conjugation $C=(-1{)}^{L+S}=+1$ and hence $J^{PC}=0^{-+}$.

The physical eta and eta prime mesons are mixtures of the octet and singlet combinations, conventionally described at this level byFlavour breaking and the singlet axial anomaly affect their masses and mixing. The neutral pion is much lighter, about $135\mathrm{MeV}$, and decays predominantly to two photons. The eta has mass about $548\mathrm{MeV}$ and important two-photon and three-pion decay modes; the eta prime has mass about $958\mathrm{MeV}$ and important decays to $\eta \pi \pi$. The singlet axial anomaly explains why the eta prime is not an additional light Goldstone boson merely because the flavour tensor product contains a singlet. The centre contains two octet directions and one singlet direction; physical eta mixing combines the latter two, leaving the isospin-one neutral pion separate to a good approximation.

Start with the classification of finite-dimensional representations of SU2. Its irreducible complex group representations are the spin-$j$ spacesThe central element $-I$ acts on $V_j$ as $(-1{)}^{2j}$. These facts follow also by realizing $V_j$ as homogeneous polynomials of degree $2j$ in two variables: the raising and lowering operators connect all of their one-dimensional weight spaces, and the highest-weight classification supplies every irreducible.

Identify Euclidean four-space with the quaternions. The unit quaternions, each a copy of $SU(2)$, act byThe norm is multiplicative, so this is an orthogonal action. It preserves orientation because the acting group is connected. If it fixes every $q$, setting $q=1$ first gives $a=b$, and then this quaternion must commute with every quaternion; a real unit quaternion is $\pm 1$. The kernel is therefore $\{(1,1),(-1,-1)\}$.

For completeness, the differential is injective: if imaginary quaternions $u,v$ satisfy $uq-qv=0$ for every $q$, then $u=v$ is central and imaginary, hence zero. Both Lie algebras have dimension six. Thus the image contains a neighbourhood of the identity and is an open subgroup of the connected SO(4) group, so it is the whole group. This proves the Spin(4) double coverThe covering group is simply connected since each $SU(2)$ is a three-sphere.

The irreducible representations of a product of compact groups are tensor products of irreducibles of its two factors. One way to see this is to decompose an irreducible space into isotypic components for the first factor; the second commutes with the first, so only one isotypic component can occur. The multiplicity space must then be irreducible for the second factor. Hence the covering-group irreducibles are $V_{j_L}\otimes V_{j_R}$. By central parity on SU2 tensor products, the kernel element acts as $(-1{)}^{2j_L+2j_R}$. The group representation descends to $SO(4)$ precisely when that sign is positive. The representations of SO(4) from two SU2 spins are thereforeEvery finite-dimensional irreducible complex group representation of $SO(4)$ is obtained this way. Since the group is compact, these are also all its continuous irreducible unitary group representations, up to equivalence.

The Lie-algebra version makes the two spin labels visible locally. Choose rotation generators $J_i$ and generators $K_i$ mixing the fourth direction with the first three, normalized so thatThen $A_i=(J_i+K_i)/2$ and $B_i=(J_i-K_i)/2$ obey two commuting copies of $\mathfrak{su}(2)$. The quaternion quotient determines which Lie algebra representations integrate to the actual group, rather than only to its cover.

For example, $(0,0)$ is the scalar, $(1/2,1/2)$ is the four-vector, and $(1,0)$ and $(0,1)$ are the three-dimensional self-dual and anti-self-dual two-form group representations. The half-spin spaces $(1/2,0)$ and $(0,1/2)$ belong to the cover and do not descend to $SO(4)$. Restricting to rotations fixing the real quaternion axis gives the diagonal $SU(2)$, and the Clebsch-Gordan decomposition for SU2 yieldswith steps of one. For a descended group representation these diagonal spins are integers, as required for the spatial $SO(3)$ subgroup.

The Proper orthochronous Lorentz group is the connected Lorentz group in the question. Its double cover is $SL(2,\mathbb{C})$, viewed as a real Lie group. Identify a spacetime vector with the Hermitian matrixThe action $X\mapsto SX{S}^{†}$ for $S\in SL(2,\mathbb{C})$ preserves this determinant and hence the Minkowski metric. The group is connected, so its image is proper and orthochronous. Its kernel consists of $\pm I$: a matrix in the kernel first preserves $I$, hence is unitary, and then commutes with every Hermitian matrix, so is scalar; determinant one forces the two signs. Matrices in $SU(2)$ generate spatial rotations, and positive Hermitian determinant-one matrices generate boosts. Rotations and boosts generate the connected Lorentz group, so the action is onto. Polar decomposition also gives $SL(2,\mathbb{C})\cong SU(2)\times {\mathbb{R}}^3$ as a manifold, proving that it is simply connected. This establishes the Lorentz spinor double cover.

In an anti-Hermitian rotation-generator convention, the Lorentz algebra brackets areThe negative sign in the last bracket distinguishes boosts from Euclidean four-dimensional rotations. After complexification, setA direct bracket calculation gives $[A_i,A_j]={ϵ}_{ijk}{A}_k$, $[B_i,B_j]={ϵ}_{ijk}{B}_k$ and $[A_i,B_j]=0$. Thus the chiral decomposition of the complex Lorentz algebra isComplexification matters: the real Lorentz algebra is not the compact real algebra $\mathfrak{su}(2)\oplus \mathfrak{su}(2)$.

Each spin-$j$ homogeneous polynomial representation of SU2 extends from $SU(2)$ to $SL(2,\mathbb{C})$ as $D_j(S)={\mathrm{Sym}}^{2j}S$. Its complex-conjugate extension uses $\overline{S}$. The finite-dimensional irreducible complex group representations of the covering group are thereforeThe two separate complexified Lie-algebra factors act irreducibly on the two spin spaces, so their tensor product is irreducible. Conversely an invariant complex subspace for the real group is invariant under its complexified Lie algebra; the highest-weight classification for the two ${\mathfrak{sl}}_2$ factors gives exactly these tensor products. This constructs all finite-dimensional complex Lorentz representations.

Again $-I\in SL(2,\mathbb{C})$ acts by $(-1{)}^{2j_L+2j_R}$. Therefore the irreducible representations of the connected Lorentz group itself, in this finite-dimensional complex category, areIf the sum is a half-integer, the group representation is a group representation of the spin cover, or a projective group representation of the Lorentz group, and is not an ordinary single-valued group representation of the group named in the question.

The scalar $(0,0)$ and four-vector $(1/2,1/2)$ descend. The left and right Weyl spinors, $(1/2,0)$ and $(0,1/2)$, do not. Their direct sum is a Dirac spinor, reducible under the connected group; parity exchanges its two chiral summands. The group representations $(1,0)$ and $(0,1)$ describe the two complex chiral parts of an antisymmetric tensor. On the rotation subgroup, self-duality of $SU(2)$ irreducibles identifies the conjugate spin space with the usual spin space, so the Clebsch-Gordan decomposition for SU2 gives the same spin range $|j_L-j_R|,\dots ,j_L+j_R$ as in part (i).

These are group representations used for fields, and they need not be unitary for a positive-definite inner product. Indeed no nontrivial finite-dimensional group representation of this group is unitary: if it were, its differential would embed the simple real Lorentz algebra into an algebra of skew-Hermitian matrices. The trace form $-\mathrm{tr}(XY)$ would give an invariant positive-definite form $b$ on that algebra. Invariance and the boost brackets would then force $-b(J_3,J_3)=b([K_1,K_2],J_3)=b(K_1,[K_2,J_3])=b(K_1,K_1)$, impossible for positive-definite $b$. Equivalently nontrivial boosts in these polynomial group representations have real exponential rather than phase eigenvalues.

The qualification about dimension is necessary because a noncompact group also has infinite-dimensional unitary group representations. They too can be built using $SU(2)$ spin spaces, but not by a single finite pair. For example, the rotation content of induced Lorentz representations is obtained from normalized induced representations of the upper triangular subgroup of $SL(2,\mathbb{C})$, with $n\in \mathbb{Z}$, $\rho \in \mathbb{R}$ and unitary characters $(a/|a|{)}^n|a{|}^{2i\rho }$ on its diagonal $(a,a^{-1})$, trivial on its unipotent part. In the compact picture this uses functions on $SU(2)$ satisfyingExpanding functions into $SU(2)$ matrix coefficients selects a single right-torus weight from each spin space, giving the rotation contentNormalized induction supplies the boost action, coupling these infinitely many rotation spaces. The central sign is $(-1{)}^n$, so the even-$n$ family descends to the connected Lorentz group and has integer rotation spins. This explains both the finite-dimensional field construction and why a classification of unitary group representations cannot simply be identified with the finite two-spin labels.