Past exam of the introduction to representation theory course of the University of Oxford / 2023 / 2

The character formula for an induced representation isEquivalently, extend the character ${\chi }_W$ by zero outside the subgroup $H$ and sum over all $x\in G$. The factor $1/|H|$ accounts for passing from elements of $G$ to representatives of left cosets of $H$.

For each $h\in g^G\cap H$, count the solutions of $x^{-1}gx=h$. Once a solution $x_0$ is fixed, every solution is uniquely $c{x}_0$ with $c\in C_G(g)$, the centralizer of $g$. Thus each $h$ contributes $|C_G(g)|{\chi }_W(h)$ to the sum in part (a).

The orbit-stabilizer theorem for the conjugation action gives $|C_G(g)|=|G|/|g^G|$. The intersection $g^G\cap H$ is the disjoint union of the conjugacy classes $h_1^H,\dots ,h_{\ell }^H$, and the character ${\chi }_W$ is constant on each such conjugacy class. HenceIf the intersection is empty, the empty sum gives zero. This is the conjugacy-class formula for an induced character.

Order the conjugacy classes of the symmetric group $S_4$ by representatives $1,(12),(12)(34),(123),(1234)$; their sizes are $1,6,3,8,6$. For a three-dimensional rotation through an angle $\theta$, the eigenvalues are $1,e^{i\theta },e^{-i\theta }$, so the trace of a three-dimensional rotation is $1+2\cos \theta$.

Under the identification with the rotational symmetry group of a cube, elements of order two give half-turns, elements of order three give turns through $\pm 2\pi /3$, and elements of order four give turns through $\pm \pi /2$. A half-turn can have an axis through opposite edges or opposite faces, but both kinds have trace $-1$. ThusIts character inner product with itself isBy character orthogonality, a nonzero complex representation has character inner product one with itself exactly when it is an irreducible representation. Therefore $C$ is an irreducible representation of dimension three.

For comparison useful in part (e), the standard representation of the symmetric group has character equal to the number of fixed letters minus one, namely $(3,1,-1,0,-1)$ in this order. Multiplication by the sign representation changes this character to $(3,-1,-1,0,1)$. Thus $C$ is the tensor product of the standard representation of the symmetric group with the sign representation.

The given permutations satisfy $r^4=s^2=1$ and $srs=r^{-1}$. The eight elements $r^j,r^{j}s$ for $0\le j<4$ are distinct: $s\notin ⟨r⟩$, whose only element of order two is $r^2=(13)(24)$. Thus $H$ is the order-eight dihedral group, and we may use the matricesThey obey the same relations and describe a quarter-turn and a reflection of a square.

The eigenvalues of $R$ over the complex numbers are $i$ and $-i$, with respective eigenlines spanned by $(1,-i)$ and $(1,i)$. Since $S$ interchanges these eigenlines, neither is an invariant subspace for $H$. Any proper nonzero subrepresentation of this two-dimensional complex representation would be a line invariant under both $R$ and $S$, and an $R$-invariant line must be one of those eigenlines. None qualifies, so $W$ is irreducible.

For the calculation in part (e), its character isIndeed $R^2=-I$, while a quarter-turn or a reflection has trace zero. The character inner product $(2^2+(-2{)}^2)/8=1$ provides a second irreducibility check.

In the order of conjugacy classes used in part (c), the elements of $H$ have the following types:By part (d), only $1$ and $r^2$ have nonzero character values. Formula (b), or equivalently the element-sum version proved there, therefore giveswith zero on the other three conjugacy classes. Here the centralizer order is $24/3=8$. Thus ${\chi }_V=(6,0,-2,0,0)$.

Let $\epsilon$ denote the sign representation and take the explicit two-dimensional $G$-moduleIts character is ${\chi }_U=(2,0,2,2,0)$. The character of a tensor product of group representations is the pointwise product of their characters, because $\mathrm{tr}(M\otimes N)=\mathrm{tr}(M)\mathrm{tr}(N)$. ConsequentlyBy Maschke's theorem both complex representations are semisimple representations, and characters distinguish finite-dimensional complex semisimple representations. ThereforeThis is the induction of the planar square representation to the symmetric group on four letters.

The dimension of the fixed-point subspace of a group action is the multiplicity of the trivial representation. Its character inner product here isHence $V^G=0$. Equivalently, Frobenius reciprocity identifies this multiplicity with $⟨{\chi }_W,1_H{⟩}_H=(2-2)/8=0$.