Past exam of the groups and representations course of the University of Oxford / 2022 / 1

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.