Past exam of the groups and representations course of the University of Oxford / 2022 (A15090W1)

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Source bundle pages 104–106. The cover dates this paper to 2022 despite the attachment filename.

The matrices obey $Q^5=P^2=I$ and $PQP=Q^{-1}$. Moving every $P$ to the right therefore reduces every product to $Q^k$ or $Q^{k}P$, with $0\le k<5$. The five diagonal matrices $Q^k$ are distinct because $\alpha$ is a primitive fifth root of unity; the five $Q^{k}P$ are distinct and off-diagonal. ThusThis is the dihedral group of the regular pentagon, with $Q$ a rotation and $P$ a reflection.

Conjugation by $P$ interchanges $Q^k$ and $Q^{-k}$, while conjugation by a power of $Q$ fixes rotations. AlsoSince multiplication by $2$ permutes the residues modulo $5$, all five reflections form one conjugacy class. The full list of conjugacy classes isTo prove the representation is irreducible over $\mathbb{C}$, any proper nonzero invariant subspace would be a line invariant under $Q$. Its two distinct eigenvalues force such a line to be one of its coordinate eigenspaces. But $P$ exchanges those two lines, so neither is $G$-invariant. Hence the given complex representation is irreducible.

In a one-dimensional representation, the scalar images of $Q$ and $P$ commute. The relation $PQP=Q^{-1}$ gives $R(Q{)}^2=1$, while $R(Q{)}^5=1$, so $R(Q)=1$ and $R(P)=\pm 1$. These give exactly $R_0$ and $R_1$.

For $k=1,2$, define the two-dimensional representationsThe same invariant subspace argument as in solution proves that each is irreducible. Set $R_2=D_1$ and $R_3=D_2$. Their characters differ at $Q$, so they are inequivalent. The sum of squares of irreducible representation dimensions is already $1^2+1^2+2^2+2^2=10$, proving completeness.

Put $s=2\cos (2\pi /5)=(\sqrt{5}-1)/2$ and $t=2\cos (4\pi /5)=-(\sqrt{5}+1)/2$. Taking traces yields the character tableThe identities $s+t=-1$, $st=-1$, and $s^2+t^2=3$ also verify character orthogonality directly.

A tensor product of group representations has character equal to the product of its factor characters. Multiplying the rows of the character table in solution, using $s^2=2+t$, $t^2=2+s$ and $st=-1$, gives all the Clebsch-Gordan decomposition of group representations:For example, the character of $R_2\otimes R_2$ is $(4,s^2,t^2,0)=(4,2+t,2+s,0)$, the sum of the characters of $R_0$, $R_1$ and $R_3$. Likewise $R_2\otimes R_3$ has character $(4,-1,-1,0)={\chi }_2+{\chi }_3$. The trivial representation and $R_1\otimes R_1=R_0$ account for the remaining entries.

Use the basis of solution for the two doublets, treating the barred left-handed fields as the components to which the printed representation assignment applies. Write $x=({\bar{\psi }}_L^1,{\bar{\psi }}_L^2{)}^T$ and $y=({\psi }_R^1,{\psi }_R^2{)}^T$. The doublet contribution to the Yukawa interaction is $H{x}^{T}Ly$, so invariant tensors obeyThe first equation forces $L_{11}=L_{22}=0$, since their factors are ${\alpha }^{\pm 2}\ne 1$. The second sets $L_{12}=L_{21}$. Thus the doublet invariant tensor is $x_1{y}_2+x_2{y}_1$. The product of either doublet with $R_1$ has no trivial representation, whereas $R_1\otimes R_1=R_0$. ThereforeIf instead the unbarred left-handed doublet is chosen to transform by $D_1$, its barred components transform by $D_1^{\ast }$; in that equivalent basis the upper block is $y{I}_2$. Exchanging the two left components relates these conventions. The singular value decomposition gives the same physical masses in either basis.

Now the doublet invariant tensor would satisfy $D_1(Q{)}^{T}L{D}_2(Q)=L$. Its four entries acquire factors ${\alpha }^3$, ${\alpha }^{-1}$, $\alpha$, and ${\alpha }^{-3}$, respectively. None equals $1$, so $L=0$. Equivalently, $R_2\otimes R_3=R_2\oplus R_3$ has no trivial representation. The mixed doublet and third-family terms are again forbidden because their tensor product of group representations contain no trivial representation. Only the product of the two $R_1$ fields survives:

Let $v_H=⟨H⟩$ be the vacuum expectation value in the convention $M=v_H\lambda$. For canonically normalized fermion kinetic terms, physical charged lepton masses are the nonnegative values in the singular value decomposition of $M$, equivalently the square roots of the eigenvalues of $M^{†}M$.

In case (i),The opposite algebraic eigenvalues $y$ and $-y$ of the doublet block do not represent different physical masses. The unavoidable twofold degeneracy conflicts with the three distinct nonzero measured charged lepton masses, for any assignment of the three families.

In case (ii),One can fit the tau lepton mass by choosing $|v_H{y}_3|$, but the electron and muon remain massless. Thus neither exact dihedral group assignment with this single singlet complex scalar field reproduces the observed spectrum. Additional symmetry-breaking Yukawa interactions or other sources of mass are required; a basis change cannot remove the degeneracy or change the matrix rank.

Normalize the simple roots to $({\alpha }_i,{\alpha }_i)=2$. The A2 root system has $({\alpha }_1,{\alpha }_2)=-1$, so its Dynkin diagram is two nodes joined by one bond:The fundamental weights satisfy $({\omega }_i,{\alpha }_j)={\delta }_{ij}$. Hence the Gram matrix in the fundamental weight basis, the metric for Dynkin label coordinates, isFor weights of a representation with Dynkin labels $a$ and $b$, their inner product is $a^{T}Gb$. In the simple root basis the metric is instead $A$ itself; stating the basis removes this apparent convention ambiguity.

Write $a=a_1$, $b=a_2$ with $a,b\ge 0$, and let $\lambda =a{\omega }_1+b{\omega }_2$ be the highest weight. The positive roots are ${\alpha }_1$, ${\alpha }_2$ and ${\alpha }_1+{\alpha }_2$, and the Weyl vector is $\rho ={\omega }_1+{\omega }_2$. The Weyl dimension formula givesFor the quadratic Casimir operator, use the root-length normalization $C(\lambda )=(\lambda ,\lambda +2\rho )$. Substituting the Gram matrix in solution yieldsTaking the trace of the defining quadratic Casimir operator identity gives $d(a,b)C(a,b)=\dim (\mathfrak{su}(3))c(a,b)=8c(a,b)$, where $c$ is the Dynkin index in the matching normalization. ThereforeIn particular $c(3)=1$ and $c(8)=6$. The common particle-physics convention uses $C_2=C/2$ and $T=c/2$, giving $C_2(3)=4/3$ and $T(3)=1/2$. The subsequent discrete gauge anomaly congruences use the integer Dynkin index $c$, while the sign test for the supplied beta function is independent of a common rescaling of all indices.

The Weyl dimension formula is increasing in each nonnegative Dynkin label. If $a\ge 3$ or $b\ge 3$, the dimension is at least $d(3,0)=10$. Among $0\le a,b\le 2$, the cases with both labels nonzero, other than $(1,1)$, have dimension at least $d(2,1)=15$. Thus the following is the complete list of irreducible representations of dimension strictly less than $10$:Here $6={\mathrm{Sym}}^{2}3$ is the two-index symmetric representation and $8$ is the Adjoint representation. Interchanging $a$ and $b$ gives the conjugate representation, explaining equal dimensions, Casimir eigenvalues and Dynkin indices in each conjugate pair. In the alternative convention of solution, divide the $C$ and $c$ columns by two.

The Dynkin index is additive over a direct sum of group representations, so $n$ copies of the fundamental representation have $c(r_S)=n$. With $c(\mathrm{ad})=6$, the supplied beta function becomesFor positive gauge coupling $g$, asymptotic freedom at this order requires $\beta (g)<0$, hence $n<132$. For nonnegative integer $n$,At $n=132$ the one-loop coefficient vanishes, so this formula alone does not establish asymptotic freedom. This count follows the scalar multiplicity convention of the printed beta function; changing how a real or complex scalar field is counted requires changing its coefficient too.

The two-index symmetric representation of SU(3) is $6$, with Dynkin index $5$. Thus $n$ copies have $c(r_S)=5n$, and the supplied beta function isThe strict sign condition for asymptotic freedom is $n<132/5=26.4$, givingIndeed $132-5(26)=2>0$, while $132-5(27)=-3<0$.

Use the integer normalization $c(3)=1$ from solution. For two fundamental representations, the specified discrete gauge anomaly coefficient is $\mathcal{A}=q_1+q_2$. Consequently all solutions in the finite cyclic group of charges areThere are $n$ ordered charge pairs. The trivial pair is included: the problem does not require a faithful action. If a faithful action on these Weyl fermions were additionally required, one would retain precisely the $q$ coprime to $n$.

This cancels the discrete gauge anomaly defined in the paper. Two left-handed fundamental representations by themselves have a nonzero pure cubic SU(3) gauge anomaly; a complete gauge theory would need additional fermions to cancel it. That separate condition does not alter the requested discrete congruence.

For the fundamental representation and the two-index symmetric representation, the Dynkin indices are $1$ and $5$. Therefore the specified discrete gauge anomaly condition for ${\mathbb{Z}}_4$ isAll ordered solutions areOnly $(1,3)$ and $(3,1)$ give a faithful ${\mathbb{Z}}_4$ action on this pair of Weyl fermions; faithfulness was not imposed. As in solution, cancellation here concerns the stipulated discrete gauge anomaly, not cancellation of the pure cubic SU(3) gauge anomaly. The integer normalization matters: one must not replace $c$ by a half-integer trace index inside an unchanged modular congruence.

The identity commutes with every $h\in H$, so the centralizer $C_G(H)$ is nonempty. If $x,y\in C_G(H)$, then for every $h\in H$,Thus it is closed under multiplication. From $xh=hx$, multiplication by $x^{-1}$ on both sides gives $h{x}^{-1}=x^{-1}h$, so $x^{-1}\in C_G(H)$. These are exactly the subgroup conditions; henceThe proof works for the centralizer of a subset as well; it does not require that $H$ itself be a subgroup.

Write $X=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$. Commuting with all $\mathrm{diag}(U_1,U_2)$ requiresFor $n_1,n_2\ge 2$, the defining complex representations of each factor are irreducible. Schur lemma gives $A=z{I}_{n_1}$ and $D=w{I}_{n_2}$. The two block vector spaces are inequivalent representations of the product group, even when $n_1=n_2$: the first factor acts nontrivially on just the first block. Thus their off-diagonal intertwiners vanish, $B=C=0$. Unitarity and the unit determinant give

The assertion that this full group is always a circle needs correction. Put $d=\gcd (n_1,n_2)$, $a=n_1/d$, $b=n_2/d$. The continuous group homomorphismmaps this centralizer onto the $d$th roots of unity. Its kernel is the identity component of a Lie group, parametrized injectively by $e^{i\theta }\mapsto (e^{ib\theta },e^{-ia\theta })$. Coprimality of $a,b$ proves injectivity; a Bezout identity proves that this parametrization exhausts the kernel. Extending the primitive integer row $(a,b)$ to a determinant-one integer matrix gives coordinates in which the constraint is simply $t^d=1$. ConsequentlyFor a concrete counterexample, take $n_1=n_2=2$. The constraint is $(zw{)}^2=1$, giving two separate circles $zw=1$ and $zw=-1$. The element $\mathrm{diag}(I_2,-I_2)$ lies in the second component, so this centralizer cannot be isomorphic as a Lie group to the connected circle group.

For the intended specialization of this centralizer of two special unitary blocks, $d=\gcd (2,3)=1$ and therefore the full centralizer really is a circle group:We use this primitive integer U(1) charge normalization below. If exactly one $n_i$ equals $1$ and the other is at least $2$, the same block-scalar formula still holds. If $n_1=n_2=1$, both factors are trivial and the centralizer is all of SU(2), another exception to the printed claim.

Write the defining space as $V=A\oplus B$, where $A\cong {\mathbb{C}}^2$ and $B\cong {\mathbb{C}}^3$. Use ordered pairs for the SU(2) and SU(3) factors, in that order. The generator $X_c$ in solution gives U(1) charges $3$ on $A$ and $-2$ on $B$, soTaking the conjugate representation reverses the U(1) charges. The SU(2) doublet is equivalent to its dual representation, while the SU(3) triplet becomes the antitriplet:

For the two-index antisymmetric representation, split the antisymmetric tensor $T^{IJ}=-T^{JI}$ into $T^{ab}$, $T^{a\mu }$ and $T^{\mu \nu }$, with $a,b=1,2$ and $\mu ,\nu =3,4,5$. The exterior square of a direct sum givesThe first term is an SU(2) singlet, the middle is $(2,3)$, and ${\Lambda }^{2}3\cong \overline{3}$ by contraction with the invariant alternating tensor. Adding the U(1) charges of the two indices givesThe dimensions $1+6+3=10$ check the branching of a representation. In the convention $Q_{\mathrm{em}}=T_3+Y$, weak hypercharge is $Y=X_c/6$.

The complexification of a Lie algebra in the Adjoint representation of SU(5) consists of traceless endomorphisms of $V=A\oplus B$. Splitting an endomorphism into its four blocks gives traceless endomorphisms of $A$ and $B$, one relative block-scalar direction, and the two off-diagonal spaces $\mathrm{Hom}(B,A)$ and $\mathrm{Hom}(A,B)$.

The first three pieces transform as $(3,1{)}_0$, $(1,8{)}_0$ and $(1,1{)}_0$. The off-diagonal pieces have U(1) charges equal to output minus input charge, namely $3-(-2)=5$ and $-2-3=-5$. ThusThe dimensions are $3+8+1+6+6=24$. The first three summands are precisely the gauge fields of the local gauge group $H\times U_c(1)$. The extra gauge fields of the SU(5) gauge theory transform asHermiticity relates the two off-diagonal blocks, so these describe twelve real gauge fields, or six complex fields and their conjugates. Globally, the subgroup generated by $H$ and $U_c(1)$ inside SU(5) is $(SU(2)\times SU(3)\times U(1))/{\mathbb{Z}}_6$, since the parametrization $(U_2,U_3,e^{i\theta })\mapsto \mathrm{diag}(e^{3i\theta }{U}_2,e^{-2i\theta }{U}_3)$ has a six-element kernel. This finite quotient does not alter the stated branching of a representation.

The symmetric square of a direct sum givesNow ${\mathrm{Sym}}^{2}2=3$ and ${\mathrm{Sym}}^{2}3=6$, rather than the singlet and antitriplet arising from exterior squares. ThusThe dimensions are $3+6+6=15$.

By contrast, solution gives the following left-handed Weyl fermion content of $\overline{5}\oplus 10$:These are one Standard Model family, with weak hypercharge $Y=q_c/6$. The $15$ contains $Q$ but lacks $L$, $d^c$, $u^c$ and $e^c$; instead it contains an SU(2) triplet and an SU(3) sextet. Equality of total dimensions therefore does not give the same representation content, and $15$ cannot supply the family.

Use the A5 root system with simple roots ${\alpha }_i=e_i-e_{i+1}$ for $1\le i\le 5$, in the hyperplane ${\sum }_{i=1}^6{x}_i=0$. The Dynkin diagram is the five-node chainIts Cartan matrix isDeleting the final node leaves the A4 root system, with simple roots ${\alpha }_1,\dots ,{\alpha }_4$. Concretely, the Lie algebra homomorphism ${\mathfrak{sl}}_5(\mathbb{C})\hookrightarrow {\mathfrak{sl}}_6(\mathbb{C})$ sends $X$ to $\mathrm{diag}(X,0)$ and preserves the Lie bracket. This realizes the required Lie subalgebra and fixes the embedding used below.

For the special unitary group $SU(N)$, denote a one-column Young tableau with $k$ empty boxes by ${\mathcal{C}}_k$. Explicitly the columns needed here areA column ${\mathcal{C}}_k$ is the exterior power ${\Lambda }^{k}V$, with highest weight ${\omega }_k$ and dimension $\left(\genfrac{}{}{0ex}{}{N}{k}\right)$. Contraction with the invariant volume tensor identifies $V^{\ast }\cong {\Lambda }^{N-1}V$ and ${\Lambda }^2{V}^{\ast }\cong {\Lambda }^{N-2}V$. This gives all the requested Dynkin labels, Young tableaux, dimensions and tensors:Here $1\le I,J\le 6$ and $1\le i,j\le 5$. Lower indices denote the dual representation, equivalent to the conjugate representation for these compact groups. The SU(6) $15$ in this question is an exterior square; the SU(5) $15$ in solution is a symmetric square. The ambient group is essential to distinguish them.

For the special unitary group $SU(N)$ define ${\epsilon }_i=e_i-\frac{1}{N}{\sum }_{j=1}^N{e}_j$. These are the weights of a representation in the defining space, with ${\sum }_i{\epsilon }_i=0$. Their Dynkin labels areThe dual representation negates each weight of a representation, and the tensor product of group representations adds them. Consequently the complete weight systems areEvery listed weight of a representation has multiplicity one: the corresponding weight basis vectors are $v_i$, $v^i$, $v_i\wedge v_j$ and $v^i\wedge v^j$. Distinct pairs give distinct projected sums for $N=5,6$. For an explicit enumeration in Dynkin labels, the two defining weight systems areThe weight systems for $\overline{6}$ and $\overline{5}$ are the negatives of the entries in their respective columns, without reversing coordinate order.

For the exterior squares, the full enumerations areThe negatives of the $A_5$ column are exactly the fifteen weights of a representation of $\overline{15}$; the negatives of the $A_4$ column are exactly the ten of $\overline{10}$. Thus the tables specify all eight weight systems, including multiplicities, rather than only their highest weights.

For the embedding in solution, restriction retains the first four Dynkin label coordinates, since the simple coroots of $A_4$ are the first four of $A_5$. In a coordinate-free version of the same projection, let ${\eta }_i$ be the five defining weights of a representation of $A_4$. Restriction gives ${\epsilon }_i\mapsto {\eta }_i$ for $1\le i\le 5$, while ${\epsilon }_6\mapsto 0$.

For the fundamental representation, its first five weights of a representation therefore form a $5$ and the sixth is the zero weight of a representation of a trivial representation. Negating the weights of a representation gives the conjugate representation:For the two-index antisymmetric representation, split the fifteen weights of a representation into the ten with $i<j\le 5$ and the five with $j=6$. Their restricted values are ${\eta }_i+{\eta }_j$ and ${\eta }_i$, respectively. These are precisely the weight systems of $10$ and $5$. Negating again givesFor instance the highest $A_5$ weight of a representation $(0,1,0,0,0)$ restricts to $(0,1,0,0)$ for the $10$, while $(1,0,0,0,-1)$ restricts to $(1,0,0,0)$ for the $5$. The restricted weight systems are disjoint unions of multiplicity-one sets in these cases, and their dimensions check as $6=5+1$ and $15=10+5$.

One left-handed Standard Model family requires the SU(5) representation content $10\oplus \overline{5}$. No single irreducible representation in solution contains both. At the level of containing the required fields, the smallest choice isThe additional SU(5) multiplets are a $5$ and a trivial representation. Another minimal choice using only two of the listed irreducible representations isIts additional multiplets form the conjugate representation of a family. More generally, for multiplicities $(m_6,m_{\bar{6}},m_{15},m_{\overline{15}})$, branching of a representation givesThus containing a family requires $m_{15}\ge 1$ and $m_{\bar{6}}+m_{\overline{15}}\ge 1$.

For a consistent chiral grand unified theory, containment alone is insufficient. Normalize the cubic gauge anomaly coefficient by $A(6)=1$. For the special unitary group $SU(N)$, $A({\Lambda }^2\mathbf{N})=N-4$, so for SU(6) it is $2$. To verify this directly, take a traceless diagonal generator with entries $x_i$: its exterior square has entries $x_i+x_j$, andThe conjugate representation reverses this cubic gauge anomaly. The combination $15\oplus \overline{6}$ has coefficient $2-1=1$, so cannot alone be a quantum-consistent chiral SU(6) gauge theory. The minimal anomaly-free choice with one net chiral gauge spectrum of a family is insteadThe additional nonsinglet fields form a vectorlike gauge spectrum that can acquire an SU(5)-invariant mass, and the two gauge singlets may serve as sterile neutrinos. More generally, precisely one net chiral gauge spectrum of a family with vectorlike extra nonsinglets requires $m_{15}-m_{\overline{15}}=1$ and $m_{\bar{6}}-m_6=2$; these conditions also cancel the cubic SU(6) gauge anomaly. The $15\oplus \overline{15}$ choice above is anomaly-free but has zero net chirality.