Past exam of the groups and representations course of the University of Oxford / 2024 / 1

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.