Past exam of the introduction to representation theory course of the University of Oxford / 2023 (B2.1, A12154W1)

Original solutions to all three questions of the University of Oxford Part B Introduction to representation theory course of the University of Oxford examination, held on 29 May 2023. Source: released examination PDF, bundle pages 212–213; paper pages 1–2. The original numeric, letter and roman numbering is retained. All associative algebras and modules below are unital; the characters in Questions 2 and 3 are those of finite-dimensional complex representations.

An $A$-module is completely reducible if it is a direct sum of modules each of which is a simple module:Here a simple module is nonzero and has no submodules other than zero and itself. This is precisely the definition of a semisimple module. The indexing set need not be finite unless $V$ has finite dimension; the zero module is included as the empty direct sum of modules.

Construct the sum successively, retaining a simple module only when it adds a new direct summand. Suppose $S=V_1+\cdots +V_{j-1}$ is already expressed as a direct sum of modules selected from these simple modules. The intersection $S\cap V_j$ is a submodule of the simple module $V_j$, so it is either zero or all of $V_j$. In the latter case $S+V_j=S$. In the former caseStarting with $S=0$ and proceeding by mathematical induction gives a direct sum of modules equal to $V_1+\cdots +V_n$. Thus the sum is a semisimple module. The argument proves the useful finite sum of simple submodules lemma without claiming that the originally listed submodules are themselves independent.

Since $A$ is a nonzero associative algebra of finite dimension, it has a nonzero left ideal $I$ of least positive dimension. Such a minimal left ideal is a simple module under left multiplication.

For $a\in A$, right multiplication defines the module homomorphismIts kernel is a submodule of $I$, so its image $Ia$ is either zero or a simple module isomorphic to $I$. Each $Ia$ is a left ideal, since $c(xa)=(cx)a$.

ConsiderIt is a left ideal. It is also stable under right multiplication, because $(Ia)b=I(ab)$, so it is a two-sided ideal. Moreover $I=I1\subseteq J$, giving $J\ne 0$. The assumption that $A$ is a simple ring therefore forces $J=A$.

Only finitely many summands are needed: express each member of a finite basis of $A$ as a finite sum of elements of the $Ia$, and collect the finitely many indices used. After discarding zero summands, $A$ is a finite sum of simple modules. Part (a)(ii) makes its left regular module a semisimple module. This is exactly the definition of a semisimple algebra, proving that a finite-dimensional simple algebra is semisimple. The finite dimension assumption is essential to this assertion about a simple ring.

Use the left $A$-module structure $c\cdot (a\otimes b)=ca\otimes b$ on the tensor product of associative algebras. Expanding each $b\in B$ in the basis $b_1,\dots ,b_m$ shows that every tensor can be writtenThus the displayed family generates the module.

For uniqueness, let ${\beta }_j:B\to k$ be the corresponding dual basis, with ${\beta }_j(b_i)={\delta }_{ij}$. Applying the linear map ${\mathrm{id}}_A\otimes {\beta }_j$ to a linear relation ${\sum }_i{a}_i\otimes b_i=0$ gives $a_j=0$. Hence the family is linearly independent in a module, andIn particular $1\otimes b_1,\dots ,1\otimes b_m$ is a basis of a module, with coefficients uniquely in $A$.

Write $x\in Z(A{\otimes }_{k}B)$ uniquely as $x={\sum }_j{a}_j\otimes b_j$ using part (i). For each $a\in A$, the equality $(a\otimes 1)x=x(a\otimes 1)$ becomesThe coefficient uniqueness from part (i) gives $a{a}_j=a_{j}a$ for every $a$, so each $a_j$ belongs to the center of an associative algebra $Z(A)$. Consequently $x\in Z(A){\otimes }_{k}B$.

Choose a basis $z_1,\dots ,z_t$ of the vector space $Z(A)$ and rewrite $x={\sum }_{i=1}^t{z}_i\otimes d_i$. For $b\in B$, commuting with $1\otimes b$ givesExtend the linearly independent family $z_i$ to a basis of $A$ and apply its dual basis to the first factor. It follows that $b{d}_i=d_{i}b$ for every $b$, hence $d_i\in Z(B)$. This proves $Z(A{\otimes }_{k}B)\subseteq Z(A){\otimes }_{k}Z(B)$.

Conversely, if $z\in Z(A)$ and $w\in Z(B)$, thenThe pure tensors $a\otimes b$ span the tensor product, so $z\otimes w$ is central, as is every sum of such pure tensors. Therefore the center of a tensor product of associative algebras isThese tensor products are identified with their natural vector subspaces of $A{\otimes }_{k}B$; extending bases justifies these embeddings over the field $k$.

Let $J\ne 0$ be a two-sided ideal of $A{\otimes }_{k}B$. Choose a nonzero $x\in J$ of least possible two-factor tensor rank $n$, and take a shortest expressionBoth families $(a_i)$ and $(b_i)$ are linearly independent: a linear dependence in either family would combine terms and shorten the expression. In particular $a_1,b_1\ne 0$.

Because $A$ is a simple ring, the two-sided ideal generated by $a_1$ is $A$. Thus there are finitely many $u_j,v_j\in A$ with ${\sum }_j{u}_j{a}_1{v}_j=1$. FormSince the $b_i$ are linearly independent, applying a linear functional selecting $b_1$ shows $y\ne 0$. Its two-factor tensor rank is at most $n$ and hence, by the choice of $x$, exactly $n$. Consequently $c_1=1,c_2,\dots ,c_n$ are linearly independent.

Similarly, simplicity of $B$ gives finitely many $p_j,q_j\in B$ with ${\sum }_j{p}_j{b}_1{q}_j=1$. NowThe linear independence of the $c_i$ shows $z\ne 0$, and minimality again makes its two-factor tensor rank exactly $n$. We have obtainedBoth coefficient families in this expression are linearly independent. Retaining this additional property is the key to the next part. This is coefficient normalization in an ideal of a tensor product.

Keep the nonzero $z$ of minimal two-factor tensor rank constructed in part (i). For every $a\in A$, its commutator with $a\otimes 1$ lies in $J$ and equalsIf this tensor were nonzero, its two-factor tensor rank would be at most $n-1$, contradicting minimality. Thus it is zero. The linear independence of $d_2,\dots ,d_n$ forces $a{c}_i=c_{i}a$ for every $a$ and $i\ge 2$. Since $Z(A)=k$, each $c_i$ is a scalar multiple of $1$.

But $1,c_2,\dots ,c_n$ are linearly independent. Hence $n=1$, so $z=1\otimes 1\in J$. A two-sided ideal containing the multiplicative identity is the entire ring; therefore $J=A{\otimes }_{k}B$.

We have proved the stronger result that $A{\otimes }_{k}B$ is a simple ring. It has finite dimension $({\dim }_{k}A)({\dim }_{k}B)$, so part (b) provesPart (c)(ii) also gives $Z(A{\otimes }_{k}B)=k$, so this proves that a tensor product of central simple algebras is again a central simple algebra.

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 $⟨{\chi }_W,1_H{⟩}_H=(2-2)/8=0$.

Let $\rho$ afford the character $\chi$, and let $m$ be the order of a group element $g$. Then $\rho (g{)}^m=I$, so its minimal polynomial divides $T^m-1$. Over the complex numbers this polynomial has distinct polynomial roots, since its formal derivative $m{T}^{m-1}$ does not vanish at any of its polynomial roots. Thus $\rho (g)$ is diagonalizable, with eigenvalues ${\lambda }_1,\dots ,{\lambda }_d$ that are roots of unity.

Each ${\lambda }_j$ has absolute value one, so ${\lambda }_j^{-1}=\overline{{\lambda }_j}$. The eigenvalues of $\rho (g^{-1})=\rho (g{)}^{-1}$ are ${\lambda }_j^{-1}$, givingThis is the character inversion identity.

On the complex vector space $\mathrm{CF}(G)$ of class functions, define the linear operatorThis is a complex-linear map and satisfies $T^2=I$. It is well-defined on class functions, because inversion of conjugacy classes takes an entire conjugacy class $K$ to another conjugacy class $K^{-1}$.

Use first the basis consisting of the indicator functions $1_K$ of the conjugacy classes. The linear operator $T$ permutes this basis by $1_K\mapsto 1_{K^{-1}}$. A permutation matrix has trace equal to the number of fixed basis vectors, soThe right-hand side counts the real conjugacy classes.

By character orthogonality and completeness, the irreducible characters are another basis of $\mathrm{CF}(G)$. Part (a) gives $T\chi =\overline{\chi }$. The pointwise complex conjugate of an irreducible character is again an irreducible character: conjugate the representing matrices, and conjugate any hypothetical proper invariant subspace back to contradict irreducibility. Thus $T$ also permutes this basis, andHere $\chi =\overline{\chi }$ means exactly that $\chi$ is a real-valued character. Equality of the trace in these two bases provesThis is the real character and real conjugacy class counting theorem. The linear operator used in the proof is $f\mapsto f\circ \mathrm{inv}$; on the irreducible character basis it agrees with pointwise complex conjugation, although pointwise complex conjugation itself is not a complex-linear map on all class functions.

Write $N=⟨a⟩$. The relation $ba{b}^{-1}=a^2$ lets every element be expressed as $a^i{b}^j$ with $0\le i<7$ and $0\le j<3$. There are 21 such expressions, so the assumption $|G|=21$ makes them distinct. In particular $a$ has order seven, $b$ has order three, and $N$ is a normal subgroup. This is the nonabelian group of order 21, the semidirect product $C_7⋊C_3$ with the action given by multiplication by two modulo seven.

Conjugation by $a$ fixes $N$, while conjugation by $b$ sends $a^i$ to $a^{2i}$. The nonidentity conjugacy classes inside $N$ are thereforeFor the other elements, direct calculation givesWhen $j=1$ or $2$, the coefficient $1-2^j$ is nonzero modulo seven, so letting $t$ vary gives every member of the coset $N{b}^j$. These cosets cannot merge under conjugation, since $G/N$ is an abelian group and conjugation acts trivially on that quotient group. Thus the five conjugacy classes have representatives and sizes

To find the linear characters, note that the commutator subgroup $G^{\prime }$ contains $ba{b}^{-1}{a}^{-1}=a$. The quotient group $G/N\cong C_3$ is abelian, so $G^{\prime }\subseteq N$; hence $G^{\prime }=N$. The abelianization is $C_3$. With $\omega =e^{2\pi i/3}$, its three linear characters areThis exhausts the linear characters because one-dimensional characters factor through the abelianization.

For the remaining irreducible characters, put $\zeta =e^{2\pi i/7}$ and take the linear characters ${\theta }_t$ of $N$ defined by ${\theta }_t(a)={\zeta }^t$. The six nontrivial linear characters form two orbits under conjugation, with exponent sets $\{1,2,4\}$ and $\{3,5,6\}$. Define the induced charactersFor $a^i\in N$, the character formula for an induced representation givesBoth vanish outside $N$, since no element outside a normal subgroup can be conjugate into it. This is induction from an abelian normal subgroup with a free character orbit. For completeness, its irreducibility here follows directly from Frobenius reciprocity and character orthogonality on the cyclic group $N$:The same calculation with ${\theta }_3,{\theta }_5,{\theta }_6$ proves that $\overline{\psi }$ is an irreducible character.

LetThe sum of the nontrivial seventh roots of unity gives $\alpha +\beta =-1$. Squaring and reducing exponents modulo seven givesThus $\alpha ,\beta$ are the two polynomial roots of $T^2+T+2$, andTo fix the signs with our specified $\zeta$, observe that $\mathrm{Im}\alpha =\sin (2\pi /7)+\sin (3\pi /7)-\sin (\pi /7)>0$, using the increase of sine on $(0,\pi /2)$. These Gaussian periods for seventh roots of unity make the answer explicit.

The resulting character table isThe two degree-three irreducible characters are distinct because $\alpha \ne \beta$. Together with the three linear characters, they exhaust the irreducible characters: the sum of squares of irreducible degrees is $3\cdot 1^2+2\cdot 3^2=21=|G|$. Equivalently, we have five distinct irreducible characters for the five conjugacy classes.

As a check on part (b), inversion interchanges $K_1$ and $K_2$, and also interchanges the two conjugacy classes outside $N$. Only the identity conjugacy class is real. Correspondingly ${\lambda }_0$ is the only real-valued character among the irreducible characters; ${\lambda }_1,{\lambda }_2$ and $\psi ,\overline{\psi }$ are two conjugate pairs.