Past exam of the groups and representations course of the University of Oxford / 2020 / 2

Here the star means entrywise complex conjugation, not Hermitian conjugation. Put $s=g_3$ and $r=g_1{g}_3$. Since ${\alpha }^3=1$ and $1+\alpha +{\alpha }^2=0$,The generators are $g_1=rs$, $g_2=r^{2}s$ and $g_3=s$. Moving every occurrence of $s$ to the right using $sr=r^{-1}s$ reduces every word to $r^k$ or $r^{k}s$ with $k=0,1,2$. The three diagonal matrices are distinct, as are the three off-diagonal matrices, soFor an explicit group isomorphism, send $r$ to $(123)$ and $s$ to $(12)$ in the symmetric group $S_3$, composing permutations from right to left. These permutations satisfy the same relations and generate $S_3$. The normal forms above then give an isomorphism; in particular $g_1\mapsto (13)$, $g_2\mapsto (23)$ and $g_3\mapsto (12)$.

Conjugation by $s$ exchanges $r$ with $r^2$, and conjugation by $r$ cycles the three reflections. The conjugacy classes are thereforeTheir sizes are $1,2,3$, respectively. They correspond to the identity, three-cycles and transpositions in $S_3$.

There are three complex irreducible representations, because there are three conjugacy classes. Two are the trivial representation $1$ and the sign representation $\epsilon$: both send $r$ to $1$, while they send $s$ to $1$ and $-1$, respectively. These exhaust the one-dimensional representations, since their scalar images must satisfy $r=r^{-1}$ and $r^3=1$, forcing $r=1$, and then $s^2=1$.

The defining two-dimensional matrix representation $D$ has characterIts character inner product with itself isso it is an irreducible representation. Directly, the two coordinate lines are the distinct eigenspaces of $r$, and $s$ exchanges them, leaving no common invariant line. The sum of squares of irreducible representation dimensions is $1^2+1^2+2^2=6$, proving completeness. In the conjugacy class order of solution, the character table isRepresentatives $r,s$ in this table stand for their full conjugacy classes. This lists all irreducible representations up to equivalence of group representations.

The specified linear maps form a permutation representation, sinceThe trace of a permutation matrix counts fixed coordinate basis vectors. On the identity, three-cycles and transpositions, respectively, this gives ${\chi }_R=(3,0,1)$. Comparing with solution,For an explicit direct sum,Both summands are invariant subspaces under coordinate permutations. The first is fixed pointwise and hence gives the trivial representation; the second has character $(2,-1,0)$ and hence gives $D$. The sign representation has multiplicity zero.

The variables commute, so $x_i{x}_j=x_j{x}_i$. Thus $V$ has dimension six, with basis $x_1^2,x_2^2,x_3^2,x_1{x}_2,x_1{x}_3,x_2{x}_3$, even though nine coefficients occur in the redundant expression in the source. Its representation is the symmetric square of the three-dimensional permutation representation.

The identity fixes all six monomials. A three-cycle fixes none. For a transposition, say $(12)$, exactly $x_3^2$ and $x_1{x}_2$ are fixed; the remaining monomials occur in exchanged pairs. Therefore ${\chi }_{\tilde{R}}=(6,0,2)$ in the conjugacy class order of solution. The character inner products give multiplicitiesConsequentlyThere is also a concrete decomposition: the span of the three squares is one copy of the permutation representation. The span of the three mixed monomials is another, since $x_2{x}_3,x_1{x}_3,x_1{x}_2$ are permuted exactly like their omitted indices $1,2,3$. Within each, the coefficient-sum-zero vector subspace gives a copy of $D$, and the sum of its three basis monomials gives a trivial representation.

A fixed quadratic polynomial must have equal coefficients on each of the two orbits of monomials. Hence its fixed vector subspace isThe two displayed polynomial invariants are linearly independent and account for both singlets. Equivalently, one may use $x_1^2+x_2^2+x_3^2$ and $(x_1+x_2+x_3{)}^2$ as a basis. No constant term belongs to this homogeneous degree-two vector space.