Past exam of the groups and representations course of the University of Oxford / 2018 / 2

Write $s=g_1$ and $r=g_2$. Direct matrix multiplication givesEvery product reduces to $r^k{s}^{\ell }$, with $0\le k<4$ and $0\le \ell <2$. These eight matrices are distinct and are precisely the listed elements:This is the dihedral group of group order eight. Its element $r^2$ commutes with both generators. Conjugation by $s$ interchanges $r$ and $r^{-1}$, while conjugation by $r$ sends $s$ to $r^{2}s$. It also interchanges $rs$ and $r^{-1}s$. The resulting conjugacy classes areThese are the stated $C_0,\dots ,C_4$ in order. To see that the last two reflection classes do not merge, conjugation changes the exponent in $r^{k}s$ by an even integer or changes its sign, preserving its parity. The five sets exhaust all eight elements.

Both generators are real orthogonal matrices with determinant $+1$. Hence every element obeys $g^{†}g=I_3$ and $\det g=1$. The normal-form calculation establishes closure and inverses, soIn particular the embedding in the special unitary group uses the given real matrices viewed as complex ones.

In a one-dimensional representation, the relation $srs=r^{-1}$ implies $\rho (r)=\rho (r{)}^{-1}$, so $\rho (r)=\delta \in \{\pm 1\}$. Independently $\rho (s)=ϵ\in \{\pm 1\}$. These give four irreducible representationsA two-dimensional representation $E$ is obtained by restricting the given matrices to the $yz$ plane:The eigenvalues of $E(r)$ are $i$ and $-i$. Its two eigenlines are interchanged by $E(s)$, so neither is invariant under the whole group. Thus $E$ is a complex irreducible representation.

Taking traces gives the character table, with class sizes included for the character inner product:For example $⟨{\chi }_E,{\chi }_E⟩=(4+4)/8=1$, and its inner product with each one-dimensional row is $(2-2)/8=0$. The four one-dimensional rows are mutually orthonormal by direct weighted summation. Their squared dimensions sum to $4\cdot 1^2+2^2=8=|G|$, and there are five rows for five conjugacy classes. This proves completeness of the list.

The $x$ axis is an invariant subspace on which $r$ acts as $+1$ and $s$ as $-1$. The $yz$ plane carries the irreducible representation $E$ from b. ThusThe character of a tensor product is the product of the characters, so ${\chi }_{R\otimes R}=(9,1,1,1,1)$. Weighted character inner products giveFor the trivial character, the multiplicity is $(9+1+2+2+2)/8=2$. For any other one-dimensional character, the weighted sum over the nonidentity classes is $-1$, giving $(9-1)/8=1$. For $E$, it is $(9\cdot 2-2)/8=2$.

Equivalently, $E\otimes E$ is the sum of all four one-dimensional representations, and tensoring $E$ by ${\chi }_{-,+}$ gives the same two-dimensional irreducible representation. The dimensions in the boxed decomposition sum to nine, as required.

The defining SU(3) representation restricts to the given matrix representation. All its restricted matrices are real, so the conjugate representation restricts identically. Hence the first two branchings areFor the adjoint representation of SU(3), use $3\otimes \overline{3}=1\oplus 8$. This follows by splitting endomorphisms into scalar and traceless matrices. Subtracting one trivial constituent from c givesAs an independent check, its restricted character is $|{\chi }_R{|}^2-1=(8,0,0,0,0)$, the regular character of $G$. Thus this is precisely the regular representation, whose representation multiplicities equal the corresponding dimensions.