Past exam of the groups and representations course of the University of Oxford / 2019 / 2

Put $r=g_1=\mathrm{diag}(\alpha ,{\alpha }^2)$ and $s=g_2$, using ${\alpha }^{\ast }={\alpha }^{-1}={\alpha }^2$. ThenEvery word can therefore be put in the form $r^k$ or $r^{k}s$, with $k=0,1,2$. These give six distinct matrices:The group isomorphism $r\mapsto (123)$, $s\mapsto (12)$ identifies this group with the symmetric group $S_3$, with permutations composed from right to left.

Conjugation by $s$ exchanges $r,r^2$. Conjugation by $r$ cycles $s,rs,r^{2}s$, because $r(r^{k}s)r^{-1}=r^{k+2}s$. Their conjugacy classes arewith sizes $1,2,3$. There is no mixing between them: their elements have orders one, three and two, respectively.

The matrix $r$ has distinct eigenvalues $\alpha ,{\alpha }^2$. Any one-dimensional invariant subspace for the defining representation would therefore have to be one of its two coordinate eigenspaces. But $s$ exchanges those lines, so neither is invariant under the whole group. Hence $R_2$ is an irreducible representation over the complex numbers.

There are three complex irreducible representations up to equivalence of group representations, because there are three conjugacy classes. The trivial representation and the sign representation already give two inequivalent one-dimensional representations. Together with $R_2$, their dimensions satisfy the sum of squares of irreducible representation dimensions:Thus there are no further irreducible representations. Alternatively the character of $R_2$ is $(2,-1,0)$, and its character inner product with itself is $(4+2)/6=1$, independently confirming irreducibility.

The irreducible representations can all be specified on the generators:The first is the trivial representation and the second is the sign representation. To see directly that these exhaust the one-dimensional representations, their scalar images commute, so the relation $srs=r^{-1}$ forces $r=r^{-1}$. Combining $r^2=1$ with $r^3=1$ gives $r=1$, and then $s$ may be either sign.

Taking traces gives the character table, in the conjugacy class order of solution:Here $\alpha +{\alpha }^2=-1$ gives the middle entry in the last row. The weighted character inner products of distinct rows vanish, and the norm of each row is one. Together with solution, this proves completeness of the table.

Let $S=s=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. Entrywise complex conjugation interchanges the diagonal entries of $r$ and fixes $s$, soSince $r,s$ generate the group, these identities hold for every element. Thus $S$ is an invertible intertwiner, provingThe character of a tensor product representation is the product of the factor characters. Hence ${\chi }_{R_2\otimes R_2}=(4,1,0)$. Its character inner products with the three rows of solution give multiplicities $(1,1,1)$, soFor explicit Clebsch-Gordan decomposition of group representations bases, let $e_{+},e_{-}$ be the coordinate basis. The symmetric tensor $e_{+}\otimes e_{-}+e_{-}\otimes e_{+}$ spans $1$; the antisymmetric tensor $e_{+}\otimes e_{-}-e_{-}\otimes e_{+}$ spans $\epsilon$; and the ordered pair $e_{-}\otimes e_{-},e_{+}\otimes e_{+}$ carries $R_2$, since $r$ acts by $\mathrm{diag}(\alpha ,{\alpha }^2)$ and $s$ exchanges them. In particular $S^2{R}_2=1\oplus R_2$ and ${\Lambda }^2{R}_2=\epsilon$.

Multiplying by the sign representation preserves the character of $R_2$, because its character is zero on all three reflections. Thus $\epsilon \otimes R_2\cong R_2$, and associativity givesThereforeThe dimensions give $8=1+1+3\cdot 2$. Its two one-dimensional summands can also be seen directly: $e_{+}^{\otimes 3}+e_{-}^{\otimes 3}$ and $e_{+}^{\otimes 3}-e_{-}^{\otimes 3}$ transform as $1$ and $\epsilon$, respectively.