Past exam of the groups and representations course of the University of Oxford / 2020 (A15090W1)

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/8/Physics%20Papers%202019.pdf?cookie_passthrough=1

The linked annual bundle is named “Physics Papers 2019”, but this paper's cover is dated 17 January 2020. The year here follows that cover. Solutions cover every part of all four questions, although only three questions were required.

A group representation on a vector space $V$ over a field $K$ is a group homomorphism $R:G\to \mathrm{GL}(V)$. In particular, $R(gh)=R(g)R(h)$, $R(e)=I$ and $R(g^{-1})=R(g{)}^{-1}$. Choosing a basis identifies it with a matrix representation.

Two group representations $R$ on $V$ and $S$ on $W$ are equivalent representations when an invertible linear map $T:V\to W$ satisfies $TR(g)=S(g)T$ for every $g$. Thus $S(g)=TR(g)T^{-1}$; they differ only by a choice of basis.

A nonzero group representation is a reducible representation if there is a vector subspace $U$ with $0\ne U\ne V$ and $R(g)U\subseteq U$ for every $g$. It is an irreducible representation if it has no such invariant subspace. Reducibility alone does not assert the existence of an invariant complementary vector subspace; that stronger conclusion says that the representation is a semisimple representation.

The Schur lemma states that an intertwiner $T$ between irreducible representations is either zero or an isomorphism. Consequently an intertwiner between inequivalent irreducible representations is zero. For a finite-dimensional irreducible representation over the complex numbers, every commuting endomorphism is scalar:For the first assertion, the kernel and image of $T$ are invariant subspaces, so a nonzero $T$ is injective and surjective. For the scalar conclusion, choose an eigenvalue $\lambda$ of $T$. The intertwiner $T-\lambda I$ has a nonzero kernel, so it must vanish. The finite-dimensional complex hypothesis matters: the scalar conclusion need not hold over the real numbers, as a real planar rotation through $\pi /2$ is an irreducible representation of the cyclic group of order four, while the rotation itself is a commuting nonscalar endomorphism.

Write $P=(P_{ij}{)}_{i,j=1}^3$, where $P_{ij}:{\mathbb{C}}^{m_j}\to {\mathbb{C}}^{m_i}$. Block multiplication makes the commuting condition equivalent toThus every block is an intertwiner. For $i\ne j$, the irreducible representations are inequivalent, so the Schur lemma gives $P_{ij}=0$. On each diagonal block, the same Schur lemma gives $P_{ii}={\lambda }_i{I}_{m_i}$. ThereforeConversely, every displayed matrix commutes with $R(g)$, proving both necessity and sufficiency. This commutant of a semisimple representation has complex dimension three. The parameters may vanish: $P$ was not required to be invertible.

Now $m_1=m_2=n$ and the first two matrix representations agree in the chosen bases. Each of $P_{11},P_{12},P_{21},P_{22}$ is therefore a commuting endomorphism of that common irreducible representation, so each is scalar by the Schur lemma. All blocks connecting either of these copies with the third vanish, since the third irreducible representation is inequivalent. HenceEvery such matrix commutes with $R(g)$. In particular the off-diagonal scalars $b,c$ can mix the two identical copies. The commutant of a semisimple representation is $M_2(\mathbb{C})\oplus \mathbb{C}$, of complex dimension five. If the first two representations were merely equivalent, rather than literally equal, the mixed blocks would instead be scalar multiples of a fixed invertible intertwiner and its inverse.

All three vector spaces now have the same dimension $n$. Each of the nine blocks of $P$ commutes with the common irreducible representation, so the Schur lemma givesConversely $R(g)=I_3\otimes R_1(g)$, so the Kronecker product identity immediately verifies commutation for every $A$. The commutant of a semisimple representation acts arbitrarily on the three-dimensional multiplicity vector space while acting as a scalar on each irreducible factor. Its complex dimension is nine.

Here the star means entrywise complex conjugation, not Hermitian conjugation. Put $s=g_3$ and $r=g_1{g}_3$. Since ${\alpha }^3=1$ and $1+\alpha +{\alpha }^2=0$,The generators are $g_1=rs$, $g_2=r^{2}s$ and $g_3=s$. Moving every occurrence of $s$ to the right using $sr=r^{-1}s$ reduces every word to $r^k$ or $r^{k}s$ with $k=0,1,2$. The three diagonal matrices are distinct, as are the three off-diagonal matrices, soFor an explicit group isomorphism, send $r$ to $(123)$ and $s$ to $(12)$ in the symmetric group $S_3$, composing permutations from right to left. These permutations satisfy the same relations and generate $S_3$. The normal forms above then give an isomorphism; in particular $g_1\mapsto (13)$, $g_2\mapsto (23)$ and $g_3\mapsto (12)$.

Conjugation by $s$ exchanges $r$ with $r^2$, and conjugation by $r$ cycles the three reflections. The conjugacy classes are thereforeTheir sizes are $1,2,3$, respectively. They correspond to the identity, three-cycles and transpositions in $S_3$.

There are three complex irreducible representations, because there are three conjugacy classes. Two are the trivial representation $1$ and the sign representation $\epsilon$: both send $r$ to $1$, while they send $s$ to $1$ and $-1$, respectively. These exhaust the one-dimensional representations, since their scalar images must satisfy $r=r^{-1}$ and $r^3=1$, forcing $r=1$, and then $s^2=1$.

The defining two-dimensional matrix representation $D$ has characterIts character inner product with itself isso it is an irreducible representation. Directly, the two coordinate lines are the distinct eigenspaces of $r$, and $s$ exchanges them, leaving no common invariant line. The sum of squares of irreducible representation dimensions is $1^2+1^2+2^2=6$, proving completeness. In the conjugacy class order of solution, the character table isRepresentatives $r,s$ in this table stand for their full conjugacy classes. This lists all irreducible representations up to equivalence of group representations.

The specified linear maps form a permutation representation, sinceThe trace of a permutation matrix counts fixed coordinate basis vectors. On the identity, three-cycles and transpositions, respectively, this gives ${\chi }_R=(3,0,1)$. Comparing with solution,For an explicit direct sum,Both summands are invariant subspaces under coordinate permutations. The first is fixed pointwise and hence gives the trivial representation; the second has character $(2,-1,0)$ and hence gives $D$. The sign representation has multiplicity zero.

The variables commute, so $x_i{x}_j=x_j{x}_i$. Thus $V$ has dimension six, with basis $x_1^2,x_2^2,x_3^2,x_1{x}_2,x_1{x}_3,x_2{x}_3$, even though nine coefficients occur in the redundant expression in the source. Its representation is the symmetric square of the three-dimensional permutation representation.

The identity fixes all six monomials. A three-cycle fixes none. For a transposition, say $(12)$, exactly $x_3^2$ and $x_1{x}_2$ are fixed; the remaining monomials occur in exchanged pairs. Therefore ${\chi }_{\tilde{R}}=(6,0,2)$ in the conjugacy class order of solution. The character inner products give multiplicitiesConsequentlyThere is also a concrete decomposition: the span of the three squares is one copy of the permutation representation. The span of the three mixed monomials is another, since $x_2{x}_3,x_1{x}_3,x_1{x}_2$ are permuted exactly like their omitted indices $1,2,3$. Within each, the coefficient-sum-zero vector subspace gives a copy of $D$, and the sum of its three basis monomials gives a trivial representation.

A fixed quadratic polynomial must have equal coefficients on each of the two orbits of monomials. Hence its fixed vector subspace isThe two displayed polynomial invariants are linearly independent and account for both singlets. Equivalently, one may use $x_1^2+x_2^2+x_3^2$ and $(x_1+x_2+x_3{)}^2$ as a basis. No constant term belongs to this homogeneous degree-two vector space.

The componentwise product is closed in $G_1\times G_2$. Associativity follows componentwise from associativity in each group:The identity is $(e_1,e_2)$, and $(g_1,g_2{)}^{-1}=(g_1^{-1},g_2^{-1})$. These verify all the group axioms, so this is the direct product of groups.

Write the paper's Kronecker product $\times$ as $\otimes$. Its mixed-product identity and the representation laws of $R_1,R_2$ giveAlso $R(e_1,e_2)=I_{n_1}\otimes I_{n_2}=I_{n_1{n}_2}$, andThus the images are invertible and this is a group representation of dimension $n_1{n}_2$, called the external tensor product of group representations. No irreducibility assumption is needed for this construction.

Use upper indices for the defining representation of SU(3) and lower indices for its dual representation. A Young diagram with row lengths $({\ell }_1,{\ell }_2)$ has highest weight Dynkin labels $(p,q)=({\ell }_1-{\ell }_2,{\ell }_2)$; a full column of height three is a determinant representation and can be removed for SU(3). The Weyl dimension formula reduces toThe six irreducible representations of smallest dimensions are $1,3,\overline{3},6,\overline{6},8$. The following box arrays draw their Young diagrams; filling boxes by tensor-slot labels gives representative Young tableaux:Their associated tensors, in the same order, can be taken as follows:

These low-dimensional SU3 tensor representations have dimensions $1,3,3,6,6,8$ by direct component counting. To check that none were missed, if $p+q\ge 3$ then $(p+1)(q+1)\ge p+q+1$, so the displayed dimension is at least $(p+q+1)(p+q+2)/2\ge 10$. The pairs with $p+q\le 2$ are precisely the six listed pairs.

Under the stated SU(2) embedding, the first two components of the defining SU(3) representation form a doublet, while the third is fixed. Hence $3{\downarrow }_{SU(2)}=2\oplus 1$. Its dual representation restricts in the same way because the SU(2) doublet is self-dual.

For the Adjoint representation, restrict $8=3\otimes \overline{3}-1$. The SU(2) tensor product $2\otimes 2=3\oplus 1$ givesHere subtraction denotes removal of the trivial representation generated by the identity matrix; it is not a negative-dimensional summand. The first SU(3) factor is unchanged, so the requested branching of a representation rules areThe dimensions on the two sides are respectively $9=6+3$, $3=3$, $3=2+1$ and $8=3+2+2+1$.

An element commuting with the full first SU(3) factor has its first component in that factor's center of a group, which is discrete. In the second factor, commutation with every $\mathrm{diag}(U,1)$ forces the doublet block to be scalar and the mixed blocks to vanish, by the Schur lemma. Unitarity and unit determinant then give the connected commuting subgroupThis is the centralizer of the standard SU2 subgroup in SU3 in the second factor; the full centralizer in $G$ also includes the first factor's discrete center of a group ${\mathbb{Z}}_3$. The Lie algebra generator is $iX$ in the anti-Hermitian convention. The opposite sign or a common nonzero rescaling changes the charge convention, not the commuting direction.

In the defining representation, the U(1) charges are $1,1,-2$. In the dual representation they have the opposite signs. In the Adjoint representation a component $T^i{}_j$ has U(1) charge $X_i-X_j$. Thus its upper-right doublet has charge $3$, its lower-left doublet has charge $-3$, and its diagonal blocks have charge zero. Adding the charges to solution givesEach subscript denotes the eigenvalue of $X$, so a state of charge $q$ acquires the phase $e^{iq\theta }$. The connected subgroup generated by this circle and the second SU(2) is $(SU(2)\times U(1{)}_X)/{\mathbb{Z}}_2$: $\theta =\pi$ gives $\mathrm{diag}(-I_2,1)$, also the central element of the embedded SU(2). This finite intersection does not change the displayed Lie algebra branching of a representation rules.

The commuting direction in solution provides the correct candidate Lie algebra for a gauge group, $\mathfrak{su}(3{)}_c\oplus \mathfrak{su}(2{)}_L\oplus \mathfrak{u}(1{)}_Y$. This is a necessary structural condition for the Standard Model, but it does not by itself establish a viable grand unified theory.

The representations in solution do not give a Standard Model fermion generation with the correct weak hypercharges. Use $Q_{\mathrm{electric}}=T_3+Y$. Identifying $(3,2{)}_1$ with the left-handed quark doublet, whose weak hypercharge is $1/6$, fixes $Y=X/6$. But then the doublet in $(1,3)$ has $Y=1/6$, whereas the left-handed lepton doublet requires $Y=-1/2$. Its singlet has $Y=-1/3$, whereas a right-handed electron requires $Y=-1$. The $(\overline{3},1{)}_0$ is a further concrete mismatch: right-handed quarks written as left-handed antiquarks require $Y=-2/3$ or $1/3$, and a rescaling cannot turn zero into either value. These comparisons use the electroweak representation and hypercharge table.

The charge-$-3$ doublet in $(1,8)$ does have the required lepton weak hypercharge under this normalization, but the same Adjoint representation also brings a doublet of the opposite charge and an extra triplet. Therefore merely adding that representation does not supply the missing chiral generation without additional states and a mechanism for their masses.

Furthermore $SU(3)\times SU(3)$ is a product of two Simple Lie groups, permitting two independent gauge couplings; it does not enforce a common strong and electroweak gauge coupling. Thus the embedding exists, but the displayed matter content does not furnish a viable minimal unification of the Standard Model. A more elaborate model would require additional representations, consistent chiral matter, anomaly cancellation, and a specified spontaneous symmetry breaking mechanism. The calculations here do not prove that every possible extension using this product group is impossible.

Differentiate a curve $U(t)\in SU(4)$ through $U(0)=I$, writing $X=U^{\prime }(0)$. Differentiating $U(t{)}^{†}U(t)=I$ gives $X^{†}+X=0$; differentiating $\det U(t)=1$ gives $\mathrm{tr}X=0$. Conversely, for every anti-Hermitian matrix $X$ with zero trace, the matrix exponential $e^{tX}$ is unitary and has determinant $e^{t\mathrm{tr}X}=1$. Thus the real Lie algebra isThe commutator again has zero trace and is anti-Hermitian, so this set is closed under the Lie bracket. Let $E_{jk}$ be the matrix units. A convenient real basis isThere are six independent $A_{jk}$, six independent $B_{jk}$, and three independent $H_a$. Every permitted off-diagonal complex entry is supplied by one real combination of $A_{jk},B_{jk}$; the imaginary diagonal entries have one trace constraint and are spanned by the $H_a$. ConsequentlyFor the rank of a semisimple Lie algebra, the diagonal Cartan subalgebra consists of $i\mathrm{diag}(t_1,t_2,t_3,t_4)$ with real $t_i$ and ${\sum }_i{t}_i=0$, of dimension three. Commuting anti-Hermitian matrices are simultaneously unitarily diagonalizable, so no larger commuting toral Lie subalgebra exists. In the common physics convention one uses the Hermitian generators $-i{A}_{jk},-i{B}_{jk},-i{H}_a$ and writes group elements as $e^{itT}$; these are the same generators with the factor of $i$ moved into the exponential.

The complexification of a Lie algebra gives $\mathfrak{su}(4{)}_{\mathbb{C}}={\mathfrak{sl}}_4(\mathbb{C})$, of type A3 root system. Choose simple roots ${\alpha }_1=e_1-e_2$, ${\alpha }_2=e_2-e_3$, ${\alpha }_3=e_3-e_4$, with squared lengths two. The Cartan matrix and Dynkin diagram areEach line represents a single bond. In the Dynkin label convention $w_a=⟨w,{\alpha }_a^{\vee }⟩$, the simple roots have coordinates equal to the rows of $A$.

In the defining representation, apply the simple coroots $E_{aa}-E_{a+1,a+1}$ to the four coordinate basis vectors. Their weights areEquivalently these are obtained by subtracting ${\alpha }_1,{\alpha }_2,{\alpha }_3$ successively from the highest weight. The dual representation has the negative weights, henceAll listed weight multiplicities are one, so both dimensions are four.

The highest weight $(1,0,1)$ is ${\alpha }_1+{\alpha }_2+{\alpha }_3$, the highest root. Its irreducible representation is therefore the Adjoint representation. Under the action on traceless matrices by conjugation, the off-diagonal matrix unit $E_{ij}$ has weight $v_i-v_j$, while the diagonal Cartan subalgebra has zero weight. ThusThe superscript $[3]$ indicates weight multiplicity three. The twelve nonzero weights each have weight multiplicity one, giving $12+3=15$ in total. This also follows from the tensor product representation $4\otimes \overline{4}=15\oplus 1$: the sixteen differences $v_i-v_j$ contain four zero weights, and removing the identity matrix removes one of them.

The projection deletes the third Dynkin label, so the four weights $v_i$ of solution restrict to $(1,0),(-1,1),(0,-1),(0,0)$. The first three are the defining SU(3) weight system, and the fourth is a singlet. Negating them gives the dual representation. HenceFor the Adjoint representation, the projected nonzero weights consist of the six A2 root system roots, the three defining weights and their negatives. There are also three zero weights. Two zero weights belong to the SU(3) Adjoint representation, leaving one singlet. Therefore the SU4 adjoint branching to SU3 isOne can see all four invariant subspaces directly by writing a traceless matrix asThe traceless part of $A$ is $8$, the column $b$ is $3$, the row $c$ is $\overline{3}$, and the remaining diagonal matrix is proportional to $\mathrm{diag}(1,1,1,-3)$ and is fixed by SU(3). These have dimensions $8+3+3+1=15$. Equivalently restrict $4\otimes \overline{4}-1$ and use $3\otimes \overline{3}=8\oplus 1$.

Take the quarks $q=(u,d,s,c{)}^T$ in the defining $4$ of flavour SU(4) and the antiquarks in $\overline{4}$. The flavour part of a meson is a quark-antiquark pair $M^i{}_j=q^i{\bar{q}}_j$. It transforms as $M\mapsto UM{U}^{-1}$, so the complete four-flavour meson multiplet isIts traceless part $M-\frac{1}{4}(\mathrm{tr}M)I_4$ is the Adjoint representation $15$, the direct four-flavour analogue of the three-flavour meson octet. The trace supplies an additional trivial representation. Thus “the mesons” can refer to the traceless multiplet or to the complete quark model tensor product; keeping both explicit avoids dropping a state.

Using solution, the SU(3) representation contents areDirectly, the upper-left block $q^i{\bar{q}}_j$ with $i,j\le 3$ is the old $3\otimes \overline{3}=8\oplus 1$; the three entries $q^i\bar{c}$ form $3$, the three entries $c{\bar{q}}_j$ form $\overline{3}$, and $c\bar{c}$ is another singlet. The dimensions are $9+3+3+1=16$. This is a classification by flavour symmetry, rather than colour or spin. The much larger charm quark mass breaks the flavour SU(4) approximation strongly; it does not change this representation counting.

Compared with the complete three-flavour meson tensor product, the additional flavour states are exactlyThus there are seven new flavour states. For the pseudoscalar mesons, the six open-charm states are ${\bar{D}}^0,D^{-},D_s^{-}$ and $D^0,D^{+},D_s^{+}$, respectively, using the D meson and Ds meson conventions. A pure $c\bar{c}$ pseudoscalar meson is the idealized flavour state associated with the eta c meson. The quark model flavour calculation alone does not specify spin or spatial excitations, so the same flavour labels also organize other meson species.

If comparing the traceless SU(4) $15$ with the traceless SU(3) meson octet, the extra seven states are instead the same two triplets and the normalized SU(3) singletIts coefficient matrix is traceless and has squared norm one. In particular, $c\bar{c}$ by itself is not a vector in the traceless $15$; its coefficient matrix has trace one.

To relate these two descriptions, let $S_4=(u\bar{u}+d\bar{d}+s\bar{s}+c\bar{c})/2$ be the normalized SU(4) singlet and $S_3=(u\bar{u}+d\bar{d}+s\bar{s})/\sqrt{3}$ the old SU(3) singlet. ThenThis orthonormal change of basis shows explicitly why the complete tensor product contains two light-flavour singlets but adds only one state to the old light-flavour singlet sector.