Past exam of the groups and representations course of the University of Oxford / 2024 (A15090W1)

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A group representation on a vector space $V$ is a group homomorphism $R:G\to \mathrm{GL}(V)$: $R(gh)=R(g)R(h)$ and $R(e)=I$. A reducible representation has a nonzero proper invariant subspace $W\subset V$, meaning $R(g)W\subseteq W$ for every $g$. Over the complex numbers, a finite-dimensional representation of a finite group is a direct sum of irreducible representations, by Maschke's theorem.

The character of a representation is ${\chi }_R(g)=\mathrm{tr}R(g)$. The cyclic property of the trace makes it a class function:Thus a character is specified by one number per conjugacy class, much less information than all the representation matrices. Moreover,Here character orthogonality says that $m_r$ is the multiplicity of the irreducible representation $r$. Consequently characters determine finite-dimensional complex representations up to equivalence of group representations and compute tensor product of group representations decompositions without constructing a new basis explicitly.

Compose permutations from right to left. The three transpositions are ${\tau }_1=(23)$, ${\tau }_2=(13)$ and ${\tau }_3=(12)$. In cycle notation, direct composition givesThe complete symmetric group is thereforeConjugation relabels the entries of a cycle, so the conjugacy classes are preciselyTheir sizes are $1,3,2$, respectively. The differing orders of group elements also show that these three sets cannot merge under conjugation.

There is one complex irreducible representation for each conjugacy class, hence three. The trivial representation $1$ and sign representation $\epsilon$ are distinct and one-dimensional. The sum of squares of irreducible representation dimensions leavesCall the remaining irreducible representation $2$. The permutation representation on ${\mathbb{C}}^3$ has character equal to the number of fixed objects, namely $(3,1,0)$ on the ordered conjugacy classes $(e,\text{transpositions},\text{three-cycles})$. Removing its constant-vector trivial representation gives $(2,0,-1)$, whose character inner product with itself is one. This constructs $2$ and gives the character tableTo obtain every Clebsch-Gordan decomposition of group representations, multiply the characters pointwise. In particular, ${\chi }_{\epsilon }^2={\chi }_1$, ${\chi }_{\epsilon }{\chi }_2={\chi }_2$ andThus all ordered pairs of irreducible representations are covered by

All three displayed transposition matrices have trace zero. Multiplying two of them givessince ${\alpha }^3=1$. Their traces are $\alpha +{\alpha }^{-1}=-1$, whereas $\mathrm{tr}R(e)=2$. ThereforeBy character orthogonality, $R$ is the irreducible representation $2$ from c.

There is also a direct invariant subspace argument. The two distinct eigenvalues of $R({\tau }_1{\tau }_2)$ force any invariant line to be one of the two coordinate lines. But $R({\tau }_3)$ exchanges those lines. Neither is invariant under the whole group, so this two-dimensional representation is irreducible.

The scalar fields commute, so a degree-$d$ expression in one doublet belongs to a symmetric power, rather than an arbitrary tensor product. Put $x={\varphi }_1$, $y={\varphi }_2$. The generators of a group $r={\tau }_1{\tau }_2$ and $s={\tau }_3$ act byA monomial $x^a{y}^b$ is fixed by $r$ precisely when $a-b$ is divisible by three. Imposing $s$ then pairs the exponents. Through degree four the resulting holomorphic polynomial invariants areIn particular, two explicit invariant terms are ${\varphi }_1{\varphi }_2$ and ${\varphi }_1^3+{\varphi }_2^3$. Their invariance follows directly from ${\alpha }^3=1$ and the interchange of the two components. A quartic example is $({\varphi }_1{\varphi }_2{)}^2$.

For a physical real scalar potential, complex conjugate fields are allowed as well. The Hermitian inner product ${\varphi }^{†}\varphi$ and its square are also invariant, while a holomorphic term can enter with its complex conjugate, for example $b({\varphi }_1^3+{\varphi }_2^3)+\bar{b}({\bar{\varphi }}_1^3+{\bar{\varphi }}_2^3)$. There is still no invariant linear term, because neither the doublet nor its conjugate representation contains a trivial representation. Thus the possible nonconstant total degrees areThe constant term is unrestricted.

For the fundamental representation of the special unitary group $SU(2)$, its central element $-I$ sends $\varphi \mapsto -\varphi$ and $\bar{\varphi }\mapsto -\bar{\varphi }$. Every odd-degree polynomial changes sign, excluding degrees one and three. Further, $SU(2)$ acts transitively on the unit sphere in ${\mathbb{C}}^2$: a unit vector $(u,v{)}^T$ is the first column ofAn invariant polynomial must therefore depend only on ${\varphi }^{†}\varphi$. Through quartic order the general real scalar potential isHence the possible nonconstant degrees are $2,4$; the cubic term permitted by $S_3$ is excluded by $SU(2)$.

If the notation $V(\varphi )$ is instead read as restricting to holomorphic polynomials alone, the distinction matters: a single commuting $SU(2)$ doublet has no nonconstant holomorphic polynomial invariant. The candidate contraction ${ϵ}_{ab}{\varphi }_a{\varphi }_b={\varphi }_1{\varphi }_2-{\varphi }_2{\varphi }_1$ vanishes identically. The quadratic and quartic $SU(2)$ terms above require complex conjugate fields. This also explains why counting a trivial representation in $2\otimes 2$ alone would give an incorrect holomorphic answer.

Use the mathematical Lie algebra convention $U(t)=I+tX+O(t^2)$. Differentiating $U^{†}U=I$ and $\det U=1$ at $t=0$ gives $X^{†}=-X$ and $\mathrm{tr}X=0$. Conversely, for such an $X$, the matrix exponential is unitary and has determinant $\det e^{tX}=e^{t\mathrm{tr}X}=1$. ThusThe four diagonal entries are purely imaginary, and the six entries above the diagonal are arbitrary complex numbers determining the entries below it. This gives $4+2\cdot 6=16$ real parameters before imposing the one real trace constraint, so ${\dim }_{\mathbb{R}}\mathfrak{su}(4)=15$.

A Cartan subalgebra of this compact Lie algebra is a maximal commuting Lie subalgebra; chooseA simple basis is $i{H}_1,i{H}_2,i{H}_3$, whereAny matrix commuting with every diagonal element is itself diagonal, which proves maximality. Commuting skew-Hermitian matrices can also be simultaneously diagonalized, so no larger commuting Lie subalgebra is possible. ConsequentlyThe Hermitian quantum generator convention instead uses the Hermitian $H_a$ and $U=\exp (i{\theta }^a{H}_a)$. The two conventions describe the same Lie algebra.

Use the Hermitian $H_a$ from a to record real weight coordinates: $H_a{e}_j=(w_j{)}_a{e}_j$. Reading off the diagonal entries givesThese are weights in the simple-coroot coordinate system, not coordinates in an orthonormal basis. In the skew-Hermitian convention they mean $(i{H}_a)e_j=i(w_j{)}_a{e}_j$. Intrinsically, the weight on $i\mathrm{diag}(h_1,h_2,h_3,h_4)$ is the functional $h\mapsto h_j$, with ${\sum }_j{h}_j=0$.

The conjugate representation acts by $U^{\ast }$, so a diagonal transformation with entries $e^{i{h}_j}$ becomes one with entries $e^{-i{h}_j}$. Thus its four weights areAll have multiplicity one. In both representations the sum of the weights is zero, consistent with the vanishing trace of every Lie algebra generator.

Write $[r_1,r_2,\dots ]$ for a Young diagram with row lengths $r_i$, and depict an empty box by $\square$. Appending the box of the fundamental representation to another single box gives a row or a column:The corresponding symmetric tensors and antisymmetric tensors have dimensions $4\cdot 5/2=10$ and $4\cdot 3/2=6$, soFor the conjugate representation, the invariant volume tensor identifies $\bar{4}\cong {\Lambda }^{3}4$, represented by a column of three boxes. Adding one box givesA column of four boxes is the determinant representation, which is trivial on $SU(4)$. The remaining Young diagram has Dynkin labels $(1,0,1)$ and dimension fifteen. For example the hook-content formula givesEquivalently, a mixed tensor $M^i{}_j$ transforms by $M\mapsto UM{U}^{†}$ and splits into its trace and traceless matrix parts. Thereforewhere $15$ is the Adjoint representation.

The embedded $SU(3)$ acts on the first three coordinates and fixes the fourth. This immediately gives the branching rulesDistribute the tensor product of group representations over these direct sums and split $3\otimes \bar{3}=8\oplus 1$ by taking the trace:One can locate every summand using the block matrixThe traceless part of $A$ is $8$, $b$ is $3$, $c$ is $\bar{3}$, and $\mathrm{tr}A,d$ are independent trivial representations. Removing the overall trace instead gives $15\downarrow SU(3)=8\oplus 3\oplus \bar{3}\oplus 1$. The dimension check is $16=8+3+3+1+1$.

A meson has quark model flavour content $q_i{\bar{q}}^j$, so four quark flavours give sixteen flavour states inThis counts a flavour set for a specified spin and orbital assignment; it does not count every excited meson. Three quark flavours instead give the familiar nine states $3\otimes \bar{3}=8\oplus 1$.

Taking the fourth quark to be the charm quark, the first nine combinations $q_{\alpha }{\bar{q}}^{\beta }$, with $\alpha ,\beta \in \{u,d,s\}$, are the light sector. The extra seven are $q_{\alpha }\bar{c}$ in $3$, $c{\bar{q}}^{\alpha }$ in $\bar{3}$, and $c\bar{c}$ in $1$ under light-flavour $SU(3)$. For the lowest pseudoscalar mesons, the open-charm states are the three D meson/Ds meson states and their antiparticles, and the hidden-charm state is the eta c meson. The two neutral trivial representation states can mix when the flavour symmetry is broken.

The intended limitation is the accuracy of flavour symmetry: the charm quark is much heavier than the three light quarks, so a quark mass term is far from $SU(4)$ invariant. Exact flavour $SU(4)$ would relate light and charmed states with the same spin, contrary to their large observed mass splittings. However, saying that a four-flavour quark model is never considered would be too strong: flavour $SU(4)$ is used as classification bookkeeping, with much stronger symmetry breaking than light-flavour $SU(3)$.^[ref] This is a flavour special unitary group, distinct from the three-colour gauge group of Quantum chromodynamics.

For a real linear change of coordinates $x^{\prime }=Ax$, one has $d{x}^{\prime }=Adx$. The Euclidean metric therefore transforms toIt agrees with $d{x}^{T}dx$ for every displacement precisely when $A^{T}A=I$. For example, equality of the associated quadratic forms determines their symmetric coefficient matrices by polarization identity. The orthogonal group is defined by this condition, soTaking determinants gives $(\det A{)}^2=1$. Thus these matrices are invertible automatically; both signs of the determinant are allowed, since the metric alone does not specify an orientation.

Write $x=(x_1,x_2,x_3{)}^T$, $v=(x_4,x_5,x_6{)}^T$ and $z=x+iv$. Here products of differentials in a metric tensor are symmetric products, unlike the wedge products used later. ThenIndeed $d{z}_{k}d{\bar{z}}_k=d{x}_k^2+d{x}_{k+3}^2$. A transformation $z^{\prime }=Uz$ preserves this metric whenever $U^{†}U=I$, becauseThus the entire unitary group $U(3)$ preserves it, including its special unitary group $SU(3)$.

To obtain a real matrix embedding, write $U=A+iB$ with $A,B$ real. ThenThe unitarity equations imply $A^{T}A+B^{T}B=I$ and $A^{T}B=B^{T}A$, hence $\rho (U{)}^T\rho (U)=I$. Multiplication of the block matrices gives $\rho (UV)=\rho (U)\rho (V)$, and the blocks recover $U$, proving injectivity. FinallyFor this last identity, over the complex numbers the change of basis from $(x,v)$ to $(z,\bar{z})$ conjugates $\rho (U)$ to $\mathrm{diag}(U,\bar{U})$. This proves that the image lies in the special orthogonal group.

The change of variables in b diagonalizes the complexified action into $z\mapsto Uz$ and $\bar{z}\mapsto \bar{U}\bar{z}$. Consequently the conventional complex branching rule isPrecisely, this is a statement about the complexification of a real vector space carrying the real defining representation. Over the real numbers, the six-dimensional space is the realification of a complex representation $3$; the conjugate summands above are not separate three-dimensional real invariant subspaces.

The block embedding $SO(6)\hookrightarrow SO(7)$ fixes the seventh coordinate. It adds a trivial representation, givingThe dimensions are complex dimensions after complexification of a real vector space, or equivalently $6+1$ as real dimensions before it.

Let ${\rho }_3(P)=P\otimes P\otimes P$ act on the alternating coefficient tensor $\phi$, and setThe coefficient convention here is the one in the source's hint; using the usual covariant pullback gives the same stabilizer subgroup, since an invertible transformation preserves a differential form if and only if its inverse does. The identity matrix fixes $\phi$. If $P,Q$ fix it, then ${\rho }_3(PQ)\phi ={\rho }_3(P){\rho }_3(Q)\phi =\phi$. Multiplying ${\rho }_3(P)\phi =\phi$ by ${\rho }_3(P^{-1})$ shows that $P^{-1}$ fixes it as well. Thus $H$ is a subgroup of $SO(7)$. Its defining equations are polynomial, so it is closed and hence a Matrix Lie group.

For $P(t)=I+tT+O(t^2)\in H$, differentiating $P^{T}P=I$ gives $T^T=-T$. Differentiating the three factors in ${\rho }_3(P(t))\phi =\phi$ givesThe skew-symmetric matrix condition records the ambient special orthogonal group; it must be included when differentiating that group's definition.

These linear conditions are sufficient as well as necessary. On the tensor product, the infinitesimal action isIts three terms commute, so ${\rho }_3(e^{tT})=e^{tD(T)}$. If $D(T)\phi =0$, every positive power of $D(T)$ annihilates $\phi$, giving $e^{tD(T)}\phi =\phi$. Also $T\in \mathfrak{so}(7)$ implies $e^{tT}\in SO(7)$. Therefore $e^{tT}\in H$, proving the asserted Lie algebra description. For the given positive three-form this is the compact G2 Lie algebra ${\mathfrak{g}}_2$.

For each complex coordinate, exchanging the factors of a wedge product of differential forms reverses its sign, givingConsequently the Kähler form is $\omega =d{x}_{14}+d{x}_{25}+d{x}_{36}$, and $\omega \wedge d{x}_7=d{x}_{147}+d{x}_{257}+d{x}_{367}$. Expanding the holomorphic volume form givesThe sign of $d{x}_{426}$ changes on exchanging $4,2$, while reordering $4,5,3$ to $3,4,5$ requires two exchanges. This verifies every sign inUnder $z\mapsto Uz$, unitarity preserves $\omega$:The determinant is the factor by which a transformation acts on a top exterior power, so ${\Omega }^{\prime }=(\det U)\Omega =\Omega$ for $U\in SU(3)$. Extend the real embedding by fixing $y$. It preserves $\omega$, $dy$ and $\Omega$, and therefore $\varphi$. ThusThe induced seven-dimensional representation is exactly the restriction already computed in c. In the conventional complex notation its branching rule isOver the real numbers, the six-dimensional summand is the underlying real space of the complex fundamental representation, and the last real line is the fixed $y$ axis.

The image is in the unitary group, since $(zU{)}^{†}(zU)=\bar{z}z{U}^{†}U=I$ for $z\in U(1)$ and $U\in SU(n)$. The direct product of groups has multiplication $(z,U)(w,V)=(zw,UV)$. Scalar matrices commute with every matrix, soThus $f$ is a group homomorphism; in particular $f(1,I)=I$.

First prove surjectivity. For any $A\in U(n)$, its determinant has modulus one. Choose $z\in U(1)$ with $z^n=\det A$ and define $U=z^{-1}A$. Then $U^{†}U=I$ and $\det U=z^{-n}\det A=1$, so $A=f(z,U)$.

Next, $f(z,U)=I$ is equivalent to $U=z^{-1}I$. This lies in $SU(n)$ precisely when $z^n=1$. Therefore the kernel of a group homomorphism is the central cyclic groupIts generator is $(\zeta ,{\zeta }^{-1}{I}_n)$, which has order $n$. The inverse phases in the two factors are essential: this is an explicitly embedded subgroup, not a quotient by scalar matrices in just one factor. The first isomorphism theorem now givesThis also holds as a Lie group isomorphism: the induced map is smooth and a continuous bijection from a compact quotient space to the Hausdorff unitary group, and its differential is the Lie algebra isomorphism $(it,X)\mapsto itI+X$. For $n=1$ the kernel of a group homomorphism is trivial, consistently with $SU(1)=\{1\}$.

Use the standard chain ordering of the five simple roots of $SU(6)$. A Young diagram of row lengths $r_1\ge \cdots \ge r_6=0$ has highest weight Dynkin labels $a_i=r_i-r_{i+1}$. A full column of six boxes can be removed because it is the trivial determinant representation. The required data areIn the same order, the Young tableaux areEach upper index transforms by $U$, and each lower index by $U^{-1}$. In particular $M\mapsto UM{U}^{-1}$ is the complex Adjoint representation. The column of five boxes can be represented either by $w_i$ or by the totally antisymmetric tensor $B^{i_1\dots i_5}$, using the invariant six-index alternating tensor:Thus it really is the conjugate representation, rather than a second independent six-dimensional choice. Similarly converting the lower index of $M^i{}_j$ into five alternating upper indices gives the displayed hook Young diagram; the removed trace would be its full six-box column component.

The dimension counts areThe first two count independent entries of symmetric tensors and antisymmetric tensors; the last removes one trace from a mixed tensor. The hook-content formula independently gives these dimensions from the Young diagrams.

Let index $0$ denote the first coordinate, which the embedded $SU(5)$ fixes, and let $a,b=1,\dots ,5$ denote the remaining coordinates. The fundamental representation and its conjugate representation immediately giveFor a symmetric tensor, $S^{00}$ is a trivial representation, $S^{0a}=S^{a0}$ is a $5$, and $S^{ab}$ is the fifteen-dimensional symmetric square of a group representation of $5$. For an antisymmetric tensor, $A^{00}=0$, $A^{0a}=-A^{a0}$ is a $5$, and $A^{ab}$ is the ten-dimensional exterior power ${\Lambda }^{2}5$. HenceHere the left-hand $15$ is the antisymmetric tensor of $SU(6)$, while the right-hand $15$ in the first equation is the symmetric tensor of $SU(5)$; equal dimensions alone do not identify representations of different groups.

Finally decompose the traceless block matrixThe traceless part of $D$ is the Adjoint representation $24$, the column $c$ is $5$, the row $b$ is $\bar{5}$, and the one remaining diagonal trivial representation is generated by $\mathrm{diag}(-5,1,1,1,1,1)$. ThusAll dimensions match: $6=1+5$, $21=15+5+1$, $15=10+5$ and $35=24+5+5+1$.

Among the representations in c, the $10$ of $SU(5)$ occurs in the $15$ of $SU(6)$, while the simplest source of $\bar{5}$ is $\bar{6}$. Thus the economical direct sum containing the prescribed family isThe extra $SU(5)$ multiplets are one $5$ and one trivial representation. Calling $R$ a representation does not require it to be an irreducible representation; none of the five individual choices in c contains both required $SU(5)$ summands.

There is a further physical qualification if this is to be the entire chiral matter content of a consistent grand unified theory. Merely containing one Standard Model family does not guarantee cancellation of the cubic $SU(6)$ gauge anomaly. In the convention $A(6)=1$, one has $A(\bar{6})=-1$ and $A(15)=2$, so the displayed sum has $A(R)=1$. With these types of representations, the minimal anomaly-free completion is insteadBeyond one family this adds a vectorlike gauge spectrum $5\oplus \bar{5}$ and two trivial representations. The conjugate pair can receive a mass after the appropriate symmetry breaking, without removing the remaining family.^[ref] The first boxed answer addresses representation containment; the second includes the extra consistency condition for a chiral gauge theory.