Past exam of the groups and representations course of the University of Oxford / 2021 / 2

The symmetric group $S_3$ has three conjugacy classes, determined by cycle type: the identity, three transpositions, and two three-cycles. Hence it has three complex irreducible representations. Two are the trivial representation $1$ and the sign representation $\epsilon$. Their dimensions are one, so the sum of squares of irreducible representation dimensions gives $1+1+d^2=6$, whence $d=2$ for the remaining representation $D$.

Write ${\chi }_D=(2,a,b)$ on these classes. Orthogonality to $1$ and $\epsilon$ gives $2+3a+2b=0$ and $2-3a+2b=0$. Thus $a=0,b=-1$, and the character table isThe character of a tensor product representation is the product of the factor characters. Thus ${\epsilon }^2=1$, $\epsilon \otimes D=D$, and ${\chi }_{D\otimes D}=(4,0,1)={\chi }_1+{\chi }_{\epsilon }+{\chi }_D$. These determine all the Clebsch-Gordan decomposition of group representations:

For permutations $\sigma ,\tau$, the prescribed linear maps satisfySince this holds on a basis, $R(\sigma )R(\tau )=R(\sigma \tau )$. Also $R(e)=I$ and $R(\sigma {)}^{-1}=R({\sigma }^{-1})$. Hence $R$ is a representation.

The trace of a permutation matrix counts its fixed coordinate basis vectors. There are respectively $3,1,0$ fixed vectors on the three conjugacy classes, so ${\chi }_R=(3,1,0)={\chi }_1+{\chi }_D$. ConsequentlyIn particular the sign representation is absent.

The two invariant subspaces areCoordinate permutations preserve both equal coordinates and the coordinate sum. Their dimensions add to three and their intersection is zero, so ${\mathbb{C}}^3=U_0\oplus U_2$. On $U_0$, the basis $u_0=(1,1,1{)}^T$ gives $R_{U_0}(\tau )=(1)$ for every transposition.

On $U_2$ choose the basis $u=e_1-e_3$, $v=e_2-e_3$. Directly permuting their entries givesThus, taking images as columns, the matrices of the matrix representation areEach has trace zero and squares to $I$. Their three-cycle products have trace $-1$, so this is exactly the two-dimensional irreducible representation of solution. These are all nonzero invariant subspaces affording individual irreducible representations, since each occurs once in solution.

First choose the same orthonormal basis for the left and right copies of the two-dimensional unitary representation $D$. Then the family representation is $F=D\oplus \epsilon$. Because the barred fermion transforms with $F(g{)}^{†}$ and the scalar field $H$ is a trivial representation, invariance of the Yukawa interaction requiresWrite $\lambda$ in a $2+1$ block form. The off-diagonal blocks would be intertwiners between inequivalent irreducible representations $D$ and $\epsilon$, so the Schur lemma makes them zero. The doublet block must be a scalar matrix, while the singlet block is arbitrary. ThereforeEquivalently, the allowed Yukawa interaction is $H[a({\bar{\psi }}_L^1{\psi }_R^1+{\bar{\psi }}_L^2{\psi }_R^2)+b{\bar{\psi }}_L^3{\psi }_R^3]$, together with its Hermitian conjugate when constructing a real Lagrangian. The tensor product rules explain the two free parameters: both $D^{\ast }\otimes D$ and ${\epsilon }^{\ast }\otimes \epsilon$ contain one trivial representation, while the mixed products contain none.

There is a useful basis qualification. The basis used in solution is not orthonormal: its Gram matrix is $K=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$, and $D(g{)}^{†}KD(g)=K$. If the fields are expressed in exactly those coordinates, with the barred components defined by ordinary conjugate transpose, the invariant doublet coefficient is $aK$, rather than $a{I}_2$. Canonically normalizing the kinetic term returns the boxed form. Independently chosen left and right orthonormal bases replace $a{I}_2$ by $aS$, where $S$ is their fixed unitary intertwiner; the physical degeneracy is unchanged.

After the scalar field obtains a vacuum expectation value $h$, the fermion mass matrix in a canonically normalized basis is $M=h\lambda$. Physical masses are its singular values, not arbitrary entries or signed eigenvalues. ThusThe independent Yukawa coupling $b$ can fit the tau lepton mass relative to the doublet scale, but the electron and muon are predicted to have equal masses. The values supplied by the paper instead give $m_{\mu }/m_e\simeq 105.7/0.511\simeq 207$. Consequently exact unbroken $S_3$ with this representation assignment cannot reproduce all three masses. Complex phases of $a,b$, or changes of basis, cannot split the two equal singular values. Additional breaking of the symmetry or a different field content would be needed; the discrepancy cannot be removed merely by choosing different invariant Yukawa couplings.