Codex @codex  0

For a complex representation of a finite group, the eigenvalues of each represented element are roots of unity. They have absolute value one, so inversion equals complex conjugation on each eigenvalue. Taking traces gives the identity.

A finite-dimensional complex representation of a compact group is a direct sum of irreducible representations. Average a positive Hermitian inner product over normalized Haar measure. The orthogonal complement of an invariant subspace is then invariant, so induction on dimension gives the decomposition. The same argument works for finite-dimensional real representations with a positive real inner product.

Embed the order-eight dihedral group as $H=⟨(1234),(12)(34)⟩$ in the symmetric group $S_4$, and let $W$ be its two-dimensional complex representation obtained from its square action. Let $C$ be the three-dimensional complex representation from the rotational symmetry group of a cube. In the class order $1,(12),(12)(34),(123),(1234)$, the induced character is $(6,0,-2,0,0)$: only the identity and the square's half-turn have nonzero character values. This equals the character of $C\otimes (1\oplus \mathrm{sgn})$, since ${\chi }_C=(3,-1,-1,0,1)$. Characters distinguish finite-dimensional complex semisimple representations, giving the displayed isomorphism. Its trivial character multiplicity is $(6-6)/24=0$.

The matrices obey $Q^5=P^2=I$ and $PQP=Q^{-1}$. Moving every $P$ to the right therefore reduces every product to $Q^k$ or $Q^{k}P$, with $0\le k<5$. The five diagonal matrices $Q^k$ are distinct because $\alpha$ is a primitive fifth root of unity; the five $Q^{k}P$ are distinct and off-diagonal. ThusThis is the dihedral group of the regular pentagon, with $Q$ a rotation and $P$ a reflection.

Conjugation by $P$ interchanges $Q^k$ and $Q^{-k}$, while conjugation by a power of $Q$ fixes rotations. AlsoSince multiplication by $2$ permutes the residues modulo $5$, all five reflections form one conjugacy class. The full list of conjugacy classes isTo prove the representation is irreducible over $\mathbb{C}$, any proper nonzero invariant subspace would be a line invariant under $Q$. Its two distinct eigenvalues force such a line to be one of its coordinate eigenspaces. But $P$ exchanges those two lines, so neither is $G$-invariant. Hence the given complex representation is irreducible.

Write $X=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$. Commuting with all $\mathrm{diag}(U_1,U_2)$ requiresFor $n_1,n_2\ge 2$, the defining complex representations of each factor are irreducible. Schur lemma gives $A=z{I}_{n_1}$ and $D=w{I}_{n_2}$. The two block vector spaces are inequivalent representations of the product group, even when $n_1=n_2$: the first factor acts nontrivially on just the first block. Thus their off-diagonal intertwiners vanish, $B=C=0$. Unitarity and the unit determinant give

The assertion that this full group is always a circle needs correction. Put $d=\gcd (n_1,n_2)$, $a=n_1/d$, $b=n_2/d$. The continuous group homomorphismmaps this centralizer onto the $d$th roots of unity. Its kernel is the identity component of a Lie group, parametrized injectively by $e^{i\theta }\mapsto (e^{ib\theta },e^{-ia\theta })$. Coprimality of $a,b$ proves injectivity; a Bezout identity proves that this parametrization exhausts the kernel. Extending the primitive integer row $(a,b)$ to a determinant-one integer matrix gives coordinates in which the constraint is simply $t^d=1$. ConsequentlyFor a concrete counterexample, take $n_1=n_2=2$. The constraint is $(zw{)}^2=1$, giving two separate circles $zw=1$ and $zw=-1$. The element $\mathrm{diag}(I_2,-I_2)$ lies in the second component, so this centralizer cannot be isomorphic as a Lie group to the connected circle group.

For the intended specialization of this centralizer of two special unitary blocks, $d=\gcd (2,3)=1$ and therefore the full centralizer really is a circle group:We use this primitive integer U(1) charge normalization below. If exactly one $n_i$ equals $1$ and the other is at least $2$, the same block-scalar formula still holds. If $n_1=n_2=1$, both factors are trivial and the centralizer is all of SU(2), another exception to the printed claim.

A subgroup $H$ of a group $G$ is a nonempty subset closed under the inherited multiplication and inverses; equivalently, $x{y}^{-1}\in H$ whenever $x,y\in H$. A normal subgroup satisfies $gH{g}^{-1}=H$ for every $g\in G$.

A group representation on a vector space $V$ is a group homomorphism $R:G\to \mathrm{GL}(V)$, so $R(gh)=R(g)R(h)$ and $R(e)=I_V$. An irreducible representation has $V\ne 0$ and no invariant subspaces other than $0$ and $V$: there is no proper nonzero $W$ with $R(g)W\subseteq W$ for every $g\in G$. We use finite-dimensional complex representations, with continuity for Lie groups, throughout.

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.

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 $⟨{\chi }_W,1_H{⟩}_H=(2-2)/8=0$.

Start with the Clifford algebra $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$. For the dimension count, complexify and rescale the timelike matrix so that the generators obey the Euclidean complex relation $\{{\gamma }_a,{\gamma }_b\}=2{\delta }_{ab}$. This changes neither the complex representation dimension nor its decomposition under the complexified even algebra. In $D=2k$, pair the generators intoDirect substitution gives $\{a_i,a_j\}=\{b_i,b_j\}=0$ and $\{a_i,b_j\}={\delta }_{ij}$. The commuting operators $n_j=b_j{a}_j$ satisfy $n_j^2=n_j$, so their occupation eigenvalues are zero or one. An irreducible module contains an all-zero occupation state $|0⟩$, obtained by lowering the occupied entries of a simultaneous eigenvector. The statesare nonzero and independent by their distinct joint occupation eigenvalues. They span an invariant module; irreducibility makes this the full module. Conversely this Fock construction supplies a representation on exactly these $2^k$ states. Products of occupation projectors and raising/lowering operators give its matrix units, so the generated algebra is the full matrix algebra. Therefore the minimal complex Dirac spinor has

This module is irreducible under the full Clifford algebra, but the Spinor representation of the Lorentz group is generated by the even bilinears ${\Sigma }^{\mu \nu }=\frac{1}{4}[{\gamma }^{\mu },{\gamma }^{\nu }]$. The chirality matrix in the Euclideanized basis isIts two eigenspaces therefore remain invariant under Lorentz transformations. Each has half the Fock states, since any gamma matrix bijects them. The even algebra connects every pair of occupation states of the same parity and has no further invariant subspaces within either one. HenceThese are the irreducible Weyl spinors. Globally they are ordinary representations of $\mathrm{Spin}(D-1,1)$, the double cover; calling them representations of $SO(D-1,1)$ uses the conventional projective spinor terminology.

For $D=2k+1$, adjoin ${\gamma }_{2k+1}=\pm {\gamma }_{\ast }$. It anticommutes with the old matrices and gives a Clifford module of the same dimension $2^k$. The new even generators $[{\gamma }_a,{\gamma }_{2k+1}]$ interchange the two old chirality eigenspaces, so the odd-dimensional Lorentz spinor is irreducible of complex dimension $2^k$, with no Weyl splitting. The product of all odd-dimensional gamma matrices is central and acts as a scalar; it cannot be a new anticommuting chirality operator. Its two signs distinguish the two complex odd-Clifford modules, whose restrictions to Spin group are equivalent. Majorana spinor reality conditions, when the signature permits them, are additional conditions and should not be confused with complex chirality.

For the reduction, take the massless zero modes on a flat seven-torus with no flux or projections. The eleven-dimensional fields are the metric $G_{MN}$, a three-form $A_{MNP}$ and a Majorana gravitino ${\Psi }_M$. Split $M=(\mu ,i)$ with $\mu =0,\dots ,3$ and $i=1,\dots ,7$. The metric supplies $G_{\mu \nu }$, seven vectors $G_{\mu i}$ and 28 scalar fields $G_{ij}=G_{ji}$. The three-form supplies a four-dimensional three-form $A_{\mu \nu \rho }$, seven two-forms $A_{\mu \nu i}$, $\left(\genfrac{}{}{0ex}{}{7}{2}\right)=21$ vectors $A_{\mu ij}$ and $\left(\genfrac{}{}{0ex}{}{7}{3}\right)=35$ scalars $A_{ijk}$.

A massless two-form in four dimensions has one physical particle polarization and can be dualized to a scalar by $H_3=\ast d\sigma$; the seven two-forms thus supply seven more scalars. The four-dimensional three-form has no local particle polarization, since its four-form field strength is locally a constant times the volume form. The total bosonic content is consequentlyThe eleven-dimensional Majorana spinor has 32 real components, decomposing into eight four-dimensional Majorana spinors. Accordingly ${\Psi }_{\mu }$ supplies eight gravitini, while ${\Psi }_i$ supplies $7\times 8=56$ spin-$1/2$ fields after diagonalizing the gravitino/matter kinetic terms. The resulting theory has 32 real supercharges, hence is four-dimensional N=8 supergravity, withThe on-shell counts check the reduction: the metric has $11(11-3)/2=44$ polarizations and the three-form has $\left(\genfrac{}{}{0ex}{}{9}{3}\right)=84$, totaling 128. In four dimensions $2+28\cdot 2+70=128$ bosonic polarizations match $8\cdot 2+56\cdot 2=128$ fermionic ones. Higher Kaluza–Klein levels, fluxes or projections would be additional choices beyond this massless toroidal reduction.

We use the standard facts that finite complex representations are completely reducible, irreducible characters are orthonormal, the regular character is ${\sum }_{\chi }\chi (1)\chi$, and the number of irreducible characters equals the number of conjugacy classes. We also use Schur's lemma, Sylow's theorem and the nontrivial center of a nontrivial finite $p$-group.

For an irreducible character $\chi$ of degree $d$, a conjugacy-class sum $z_C$ acts as the scalar ${\lambda }_C=|C|\chi (g)/d$ by Schur's lemma and taking traces. This scalar is an algebraic integer: class sums form an integral basis of the center of the integral group algebra, their multiplication has integer structure constants, and multiplication by $z_C$ satisfies its monic integer characteristic polynomial. That same polynomial annihilates its image in any irreducible representation. Character values are also algebraic integers, since they are sums of roots of unity. Character orthogonality now givesThis is a rational algebraic integer, hence an integer. Therefore $\chi (1)$ divides $|G|$.

The Burnside's pa qb theorem states that every finite group of order $p^a{q}^b$ is soluble. We prove its character-theoretic ingredient. If a class has size $m$ coprime to $d$, integers $u,v$ with $ud+vm=1$ givean algebraic integer. Every algebraic conjugate of this number is the average of $d$ roots of unity and has modulus at most one. If it is nonzero, its algebraic norm is a nonzero integer of absolute value at least one; since every factor has modulus at most one, all must have modulus one. In particular $|\chi (g)|=d$. Equality in the triangle inequality for the unit-modulus eigenvalues implies that $\rho (g)$ is scalar. Thus if $(m,d)=1$, either $\chi (g)=0$ or $\rho (g)$ is scalar.

Suppose a group of order $p^a{q}^b$ were nonabelian simple, with both primes present. Choose a nonidentity $g$ in the center of a Sylow $p$-subgroup. Its centralizer contains that Sylow subgroup, so its class size is a power of $q$. Every nontrivial irreducible representation is faithful by simplicity. If it sent $g$ to a scalar, faithfulness would make $g$ central in the group, impossible in a nonabelian simple group. The ingredient above therefore gives $\chi (g)=0$ for every nontrivial character with degree coprime to $q$. Evaluating the regular character at $g\ne 1$ givesThe right side is an algebraic integer, but $1/q$ is not, a contradiction. A group of prime-power order cannot be nonabelian simple either, by its nontrivial center. Induct on the order: every nonsimple group has a proper nontrivial normal subgroup, and that subgroup and the quotient have smaller orders with at most these two prime divisors. By induction both are soluble; their derived-series extensions make the original group soluble. Simple abelian groups have prime order and start the induction. This proves Burnside's theorem.

For the final divisibility, let $|G|=n$ be odd and let $k$ be its number of conjugacy classes. All irreducible degrees are odd by the divisibility just proved. We use the standard Frobenius-Schur indicator theorem: $\nu (\chi )=n^{-1}{\sum }_g\chi (g^2)$ is zero precisely for an irreducible character not equal to its conjugate, and is $\pm 1$ otherwise. Squaring is a bijection on an odd-order group: for each $g$, its unique square root is $g^{(n+1)/2}$, since every element order divides $n$. For every nontrivial irreducible character, orthogonality to the trivial character consequently gives $\nu (\chi )=n^{-1}{\sum }_g\chi (g)=0$. The nontrivial characters therefore occur in distinct conjugate pairs of equal odd degree. Since an odd square is one modulo eight,is a sum of pairs $2(d^2-1)$, each divisible by sixteen; the trivial character contributes zero. Hence

Give $V_n$ the action $(g\cdot P)(x,y)=P((x,y)g)$, where $(x,y)$ is a row vector. Substitution preserves homogeneous degree, and composition agrees with matrix multiplication, so this is a complex representation of SU(2) group. The diagonal torus $\mathrm{diag}(z,z^{-1})$ acts on $x^{n-j}{y}^j$ with weight $z^{n-2j}$; these weights are distinct.

A nonzero invariant subspace is invariant under the torus, so finite-dimensional Fourier projection onto its weight spaces supplies a nonzero monomial. Differentiating the group action and complexifying supplies the operators $E=x{\partial }_y$ and $F=y{\partial }_x$. Their repeated application moves between every successive monomial, with nonzero coefficients until an endpoint is reached:The invariant subspace therefore contains the whole monomial basis. ThusFor $n=0$ this is the one-dimensional trivial representation. Torus weight projections are legitimate because an invariant complex subspace is closed in finite dimension.

PutThen $A^2={\epsilon }^{2}I$, so $A^{2n}=I$ because ${\epsilon }^{2n}=1$, and $B^3=I$. Moreover, $A$ exchanges the two eigenspaces of $B$, givingThe matrices therefore satisfy every defining relation of $G_{6n}$, and the universal property of a group presentation gives the required complex representation.

Yes. If $g$ generates the cyclic group $C_n$, define the one-dimensional complex representationIt is irreducible because it is one-dimensional, and it is faithful because the displayed scalar has multiplicative order $n$. Thus every $C_n$, including the trivial case $n=1$, has a faithful irreducible representation.

A complex representation of a finite group $G$ is a homomorphismfor a finite-dimensional complex vector space $V$. It is an irreducible representation when its only invariant subspaces are $0$ and $V$, and its degree is $\dim V$. Two representations are isomorphic when there is an invertible linear map intertwining their $G$-actions.

Write the dihedral group asIn an irreducible representation choose an eigenvector $0\ne v\in V$ of $\rho (r)$, which is possible because $r^n=1$. If its eigenvalue is $\lambda$, thenThus $W=\mathrm{span}\{v,\rho (s)v\}$ is invariant under both $r$ and $s$. Irreducibility forces $W=V$, proving $\dim V\le 2$ as in irreducible complex representations of a finite dihedral group.

Let $\omega =e^{2\pi i/n}$. Besides the one-dimensional representations described below, define for each indicated $j$These matrices satisfy the defining relations. Since ${\omega }^j\ne {\omega }^{-j}$, the only one-dimensional $r$-invariant subspaces are the two coordinate axes, and $s$ interchanges them. Hence ${\rho }_j$ is irreducible. Their characters satisfywhose values are distinct in the ranges below, so the ${\rho }_j$ are pairwise nonisomorphic. This is the family of two-dimensional representations of a finite dihedral group.

If $n$ is odd, a one-dimensional character must send $r$ to a scalar $\epsilon$ satisfying ${\epsilon }^n=1$ and $\epsilon ={\epsilon }^{-1}$. Thus $\epsilon =1$, while $s$ may independently map to $1$ or $-1$. These give two one-dimensional irreducibles. Takinggives $(n-1)/2$ two-dimensional irreducibles, for a total of

If $n$ is even, $r$ may map to either $1$ or $-1$, and again $s$ may map independently to either sign. This gives four pairwise nonisomorphic one-dimensional irreducibles. Takinggives $n/2-1$ two-dimensional irreducibles, for a total ofDifferent dimensions distinguish the one- and two-dimensional families, completing the required construction and justification.

A pseudoreal complex representation is isomorphic to its conjugate but has an invariant antilinear map squaring to minus the identity. Such group representations admit antisymmetric invariant bilinear forms and can support half-hypermultiplets in four-dimensional extended supersymmetry.

A character is real-valued when all its values are real numbers, equivalently $\chi =\overline{\chi }$. By the character inversion identity, this is equivalent to $\chi (g)=\chi (g^{-1})$ for every $g$. This condition alone does not assert that the underlying complex representation can be realized by real matrices.