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For the special linear group $G={\mathrm{SL}}_3({\overline{\mathbb{F}}}_q)$, let $N=q-1$, and choose a generator $\chi$ of the character group of ${\mathbb{F}}_q^{\times }$. Every linear character of the diagonal rational maximal torus has a unique expressionThe upper triangular Borel subgroup $B=TU$ is rational. Consequently its Deligne-Lusztig induction is the actual principal series of a finite reductive groupwhere $\tilde{\theta }$ is the inflation of a group representation across $B^F\to T^F$. The geometric reason for this split case is $X(1)=G^F/B^F$: the associated torus covering supplies $\theta$, and the auxiliary unipotent fibres contribute a single even cohomological degree, hence no minus sign.

The rational Bruhat decomposition indexes $B^F\backslash G^F/B^F$ by $W=S_3$. Applying the Mackey restriction formula and Frobenius reciprocity givesEach intersection contains $T^F$, and the two linear characters are trivial on its unipotent part. Its contribution is therefore one exactly when $w$ fixes $\theta$, and zero otherwise. Complete reducibility over $\mathbb{C}$ makes this character inner product the sum of squares of irreducible representation multiplicities. Thus the principal series of a finite reductive group is irreducible exactly when the Weyl group stabilizer is trivial.

To compute that stabilizer, represent $\theta$ by the triple $({\chi }^r,{\chi }^s,1)$. Two triples give the same linear character on determinant-one diagonal matrices exactly when they differ by a common factor $(\nu ,\nu ,\nu )$. For example, triviality of $({\eta }_1,{\eta }_2,{\eta }_3)$ on all such matrices first gives ${\eta }_1={\eta }_3$ by varying $a$, and then ${\eta }_2={\eta }_3$ by varying $b$. The Weyl group permutes the three entries.

A transposition fixes $\theta$ precisely when its two interchanged entries coincide: the third, fixed entry forces the common factor to be one. These three obstructions are $r=0$, $s=0$, and $r=s$. A three-cycle can also fix $\theta$ without any two entries coinciding. Its common factor must satisfy ${\nu }^3=1$; in the nontrivial case the entries, up to order and common factor, are $(1,\nu ,{\nu }^2)$. This happens exactly when $3\mid N$ and $\{r,s\}=\{N/3,2N/3\}$.

The complete condition is thereforeIn particular pairwise distinct coordinate characters alone are insufficient. For $q=4$ the only distinct triples have the order-three symmetry, and no character of this split rational maximal torus gives an irreducible principal series of a finite reductive group.

Write $\chi =\epsilon R_T^G(\theta )$, where $\epsilon \in \{1,-1\}$ and $\chi$ is the character of an irreducible representation. An irreducible representation is a cuspidal representation of a finite reductive group exactly when its Harish-Chandra restriction to every proper rational Levi subgroup of a rational parabolic subgroup is zero.

The needed identity is the Harish-Chandra restriction of a Deligne-Lusztig character. For a rational parabolic subgroup $P=M⋉V$, the torus case of the induction-restriction formula givesHere ${}^x\theta (xt{x}^{-1})=\theta (t)$. The geometric meaning of the indexing condition is that only a rational conjugate of the inducing maximal algebraic torus lying in the Levi can survive taking $V^F$-invariants. This is the established torus case of Deligne-Lusztig induction, valid without a restriction on $q$; it is not an appeal to an unrestricted induction-restriction conjecture. See Dudas and Michel's lectures, Theorem 14.4.

Suppose first that $T$ is an elliptic maximal torus, meaning it lies in no proper rational parabolic subgroup. For any proper rational $P=M⋉V$, the indexing set in the displayed sum is empty. Indeed ${}^{x}T\subset M$ with $x\in G^F$ would imply $T\subset {}^{x^{-1}}P$, itself a proper rational parabolic subgroup. Hence ${}^{\ast }{R}_{M\subset P}^G{R}_T^G(\theta )=0$, and multiplication by $\epsilon$ gives ${}^{\ast }{R}_{M\subset P}^G\chi =0$. This proves cuspidality.

Conversely suppose $T\subset P$ for a proper rational parabolic subgroup. There is a unique Levi subgroup of $P$ containing this particular maximal algebraic torus: relative to $T$, it is generated by $T$ and those root subgroups for which both the root and its negative occur in $P$. Call it $M$. Uniqueness and $F(T)=T$, $F(P)=P$ imply $F(M)=M$, so this is a proper rational Levi. Transitivity of Deligne-Lusztig induction through this rational parabolic saysSet $\psi =R_T^M(\theta )$, allowing it to be a virtual character. Adjunction of Harish-Chandra induction and Harish-Chandra restriction, extended linearly to virtual characters, givesThus this Harish-Chandra restriction cannot vanish, so $\chi$ is not cuspidal. No positivity or irreducibility of $\psi$ is needed, and the sign in $\chi$ causes no cancellation problem. We have provedThe term elliptic maximal torus is relative to the ambient group: split central factors are allowed. The actual condition is absence of a proper rational parabolic subgroup containing $T$.

Work over the complex numbers. By Maschke's theorem, restriction to the normal subgroup $A$ is a direct sum of its isotypic components. For an irreducible $A$-type $W$, denote its isotypic component in $V$ by $V_W$. Normality makes $G$ permute these components, by conjugating the $A$-types. Any union of orbits spans a $G$-invariant subspace, so irreducibility of $V$ means that all occurring components form a single orbit.

If there is only one component, restriction is isotypic, as claimed. Otherwise take one component $V_W$ and its stabilizer $H=\{g:g{V}_W=V_W\}$, a proper subgroup. This is an irreducible representation of $H$: if $0\ne U⊊V_W$ were $H$-invariant, the direct sum of its translates over coset representatives for $G/H$ would be a nonzero proper $G$-invariant subspace of $V$. It is direct because the translates lie in distinct isotypic components, and proper because $\dim U<\dim V_W$.

The natural map from the induced representation sends $g\otimes v$ to $gv$. It identifies the direct-sum coset decomposition with all isotypic components of $V$, so is bijective and $G$-equivariant. ConsequentlyThis proves the dichotomy directly, including the irreducibility of the representation being induced.

Induct on the order of the p-group. Every complex irreducible representation of an abelian group is one-dimensional: commuting group operators preserve a common eigenspace, and irreducibility makes that eigenspace the whole space. This handles the abelian cases.

If $V$ is not faithful, it is an irreducible representation of a smaller quotient $G/K$, where $K$ is its kernel. Induction expresses it as induced from a one-dimensional representation of a subgroup of $G/K$. Pulling back the subgroup and inflating the character gives the same induction statement for $G$, because the coset spaces of the quotient subgroup and its inverse image agree.

Now suppose $V$ is faithful and $G$ is nonabelian. Use the allowed abelian normal subgroup $A$ not contained in the centre. If restriction to $A$ were isotypic, its irreducible type would be one-dimensional, so every $a\in A$ would act as a scalar. It would therefore commute with every representing matrix of $G$. Faithfulness would imply $ga=ag$ in $G$ for every $g$, contradicting the choice of $A$. By part (a), $V={\mathrm{Ind}}_H^{G}W$ for an irreducible representation of a proper subgroup $H$. Induction on $H$ gives $W={\mathrm{Ind}}_K^H\chi$ with $\chi$ one-dimensional. Transitivity of induction yieldsThis proves the monomiality of irreducible representations of finite p-groups.

Let $V$ afford the irreducible character $\chi$. The positive character inner product $⟨{\chi }_H,1_H{⟩}_H$ is the multiplicity of the trivial representation in $V$ restricted to $H$, so there is a nonzero vector $v$ fixed by every element of $H$. Since $N\subseteq \ker \chi$, every element of $N$ acts as the identity on all of $V$. By $G=NH$, every element of $G$ therefore fixes $v$.

The line $\mathbb{C}v$ is a nonzero invariant subspace of the irreducible representation $V$, so it must equal $V$. The action on this line is trivial, provingThe hypothesis on the kernel of a character is what makes an $H$-fixed vector into a $G$-fixed vector.

The Adjoint double cover from SU(2) to SO(3) is obtained by writing a traceless Hermitian matrix as $X=x_i{\sigma }_i$, where the ${\sigma }_i$ are the Pauli matrices, and defining $AX{A}^{†}=(R(A)x{)}_i{\sigma }_i$. Explicitly,Conjugation preserves $\frac{1}{2}\mathrm{tr}{X}^2=|x{|}^2$, so $R(A)$ is orthogonal; connectedness of SU(2) gives $\det R(A)=1$. Composition of conjugations makes $R$ a group homomorphism. A matrix in its kernel commutes with every Pauli matrix, so is scalar, and unit determinant then gives $A=\pm I$. Every spatial rotation occurs: $A=\exp (-i\theta \hat{n}\cdot \sigma /2)$ induces the rotation through angle $\theta$ about $\hat{n}$. ThusSince SU(2) is the simply connected three-sphere, this is the universal double cover; the two Lie algebras are isomorphic, although the groups are not.

Define the conjugate fundamental spinor of SU(2) by ${\bar{\eta }}_{\alpha }=({\eta }^{\alpha }{)}^{\ast }$. Regard $\bar{\eta }$ as a row, so unitarity givesConsequently ${\bar{\eta }}_{\alpha }{\xi }^{\alpha }$ is an invariant Hermitian pairing. The basic invariant tensors are the Kronecker delta ${\delta }^{\alpha }{}_{\beta }$ and the alternating tensors ${ϵ}_{\alpha \beta }$ and ${ϵ}^{\alpha \beta }$. Choose ${ϵ}_{12}=1$, ${ϵ}^{12}=-1$, so that ${ϵ}^{\alpha \gamma }{ϵ}_{\gamma \beta }={\delta }^{\alpha }{}_{\beta }$. The unit-determinant identity $A^TϵA=ϵ$ makes $[\eta ,\xi ]={ϵ}_{\alpha \beta }{\eta }^{\alpha }{\xi }^{\beta }$ invariant; its inverse gives the corresponding upper-index invariant. These contractions generate the tensor invariants of the defining representation. In particular ${\tilde{\eta }}^{\alpha }={ϵ}^{\alpha \beta }{\bar{\eta }}_{\beta }$ transforms by $A$, showing that the conjugate representation is equivalent to the defining one. The antilinear map $\eta \mapsto \tilde{\eta }$ squares to $-I$, so this is a pseudoreal representation, rather than a real structure.

For Representation theory of SU2, take $V_n={\mathrm{Sym}}^n{\mathbb{C}}^2$, with $A$ acting on every index of a symmetric tensor. Its independent components are determined by the number $r$ of indices equal to $2$, so there are $n+1$ of them. Equivalently, the homogeneous polynomial representation of SU2 has basis $z_1^{n-r}{z}_2^r$, $0\le r\le n$, withThe basis weights are $n,n-2,\dots ,-n$, and the raising operator and lowering operator connect consecutive basis vectors with nonzero coefficients until their endpoints. Any nonzero invariant subspace contains a weight vector, by applying polynomial spectral projections in $H$. Raising it reaches $z_1^n$, and lowering then obtains the entire basis. Thus $V_n$ is an irreducible representation, of spin $j=n/2$. The classification of finite-dimensional representations of SU2 gives every finite-dimensional complex irreducible representation in this way. The central element $-I$ acts on a rank-$n$ tensor as $(-1{)}^n$, so descent of an SU(2) representation to SO(3) gives

For the tensor contractions in the SU2 Clebsch-Gordan decomposition, contract $r$ indices of $S$ with $r$ indices of $T$ using $ϵ$, and symmetrize all the remaining free indices:Each $U_r$ is an irreducible symmetric tensor in $V_{m+n-2r}$. To prove these are precisely the constituents, use two sets of formal variables $z,w$. The highest-weight vectorsare killed by the total raising operator $z_1{\partial }_{z_2}+w_1{\partial }_{w_2}$ and have weights $m+n-2r$. The determinant factor is invariant; lowering the remaining factor gives an irreducible string of dimension $m+n-2r+1$. Distinct highest weights give inequivalent summands. The unitary representation on this tensor product makes invariant complements available, and the dimension checkexhausts the space. Hence the contractions above are its irreducible projections, up to normalization, and the Clebsch-Gordan decomposition for SU2 is

For the decomposition of three SU(2) spinors, write $Q^{\alpha \beta \gamma }={\eta }_1^{\alpha }{\eta }_2^{\beta }{\eta }_3^{\gamma }$ and $S^{\alpha \beta \gamma }=Q^{(\alpha \beta \gamma )}$. The contractionstransform as defining spinors. Alternation in three indices in dimension two vanishes, giving $d_1-d_2+d_3=0$; there are exactly two independent contractions. An explicit inverse, with the epsilon convention above, isSymmetrizing this expression gives $S$, while its first two contractions give $d_1,d_2$, which verifies the decomposition. Thus

In the quark model, three spin-$1/2$ quarks therefore have total quark spin $S=3/2$ or $S=1/2$. The two spin-$1/2$ copies correspond to coupling the first pair to spin $0$ or spin $1$ before adding the third quark. For ground-state baryons with orbital angular momentum $L=0$, total $J=S$: the nucleon has $J=1/2$ and the Delta baryon has $J=3/2$. These two coupling copies alone do not assert two unrestricted physical baryon families. The antisymmetric three-quark colour singlet, the spatial wavefunction and flavor symmetry must also satisfy the Pauli constraint on three-quark flavour multiplets; orbital excitations can give other total angular momenta.

First consider finite-dimensional Lie algebra representations. The operators representing an abelian Lie algebra commute. They have a common eigenvector: take an eigenspace for the first operator of a basis of the algebra, observe that every remaining operator preserves it, and repeat inside this nonzero space. The resulting line is an invariant subspace, so an irreducible representation must be one-dimensional. Its action has the form $xv=\lambda (x)v$ for a linear functional $\lambda$. Conversely, every such functional defines a Lie algebra representation, since both the Lie brackets in the algebra and the commutators of its scalar operators vanish. Distinct functionals give nonisomorphic representations. Thus the irreducibles are precisely the characters $\lambda \in {\mathfrak{g}}^{\ast }$.

Even without a finite-dimensional assumption on the module, this conclusion for a finite-dimensional abelian Lie algebra follows from the Weak Hilbert Nullstellensatz: its universal enveloping algebra is the polynomial algebra $\mathrm{Sym}(\mathfrak{g})$. A simple module over a commutative algebra is its quotient by a maximal ideal, and every maximal ideal of this polynomial algebra over $\mathbb{C}$ has residue field $\mathbb{C}$.

Indecomposable does not imply irreducible. For the one-dimensional abelian Lie algebra $\mathbb{C}x$, let $x$ act on ${\mathbb{C}}^2$ by the Nilpotent Jordan blockThe line spanned by the first coordinate vector is an invariant subspace, so the module is reducible. If it were a direct sum of two nonzero submodules, both would have dimension one; nilpotence would force $N$ to act as zero on both, contrary to $N\ne 0$. Thus it is an indecomposable representation.

For a finite-dimensional module $V$, put $d=\dim V$ and define its generalized weight spaces intrinsically byTo obtain the decomposition, choose a basis $x_1,\dots ,x_r$ of $\mathfrak{g}$, split into generalized eigenspaces of $\rho (x_1)$, and successively split each piece by $\rho (x_2),\dots ,\rho (x_r)$. All the pieces are submodules because the acting operators commute. On one resulting piece, $\rho (x_j)$ has a unique eigenvalue ${\lambda }_j$. The commuting operators can be simultaneously upper triangularized by repeating the common-eigenvector argument on quotients. Hence on this piece every $\rho (x_j)-{\lambda }_{j}I$ is strictly upper triangular. Any linear combination is also strictly upper triangular, and its $d$th power vanishes. Extending ${\lambda }_j$ linearly therefore identifies the piece with $V^{\lambda }$. Different tuples give disjoint pieces; if a component has a different tuple from $\lambda$, at least one $\rho (x_j)-\lambda (x_j)I$ is invertible there. Consequentlywith only finitely many nonzero summands. Since the definition uses every $x$ rather than a chosen basis, this generalized-weight decomposition for a nilpotent Lie algebra is canonical. Its summands need not be irreducible, as the Jordan block example already shows.

Use the ordered alphabetwith weights $\mathrm{wt}(i)={\epsilon }_i$ and $\mathrm{wt}(\bar{i})=-{\epsilon }_i$. The crystal of the defining symplectic representation is the chainThe label of an arrow is its simple root color; a Kashiwara operator ${\tilde{f}}_i$ follows an arrow of color $i$ and lowers the weight by ${\alpha }_i$.

Here is a grid description of the entire tensor product of crystals, valid for every $n$. Put one vertex $a\otimes b$ at row $a$ and column $b$, in the alphabet order above. Let $P_i$ be the set of chain positions with outgoing color $i$, and $Q_i$ the positions with incoming color $i$:Their indicators are ${\phi }_i$ and ${\epsilon }_i$. Use the crystal tensor-product ruleAn arrow is omitted if the selected factor has no outgoing arrow of that color. Thus the complete grid has a downward color-$i$ arrow when $a\in P_i$ and $b\notin Q_i$, and a rightward color-$i$ arrow when $b\in P_i$ and $a\notin P_i$. There are no other arrows. These two coordinate rules specify every vertex and every edge of the requested general-rank drawing. The following figure displays the full grid at $n=3$, including all components, together with the standard chain; increasing $n$ uses exactly the same grid construction.

For completeness, highest vertices can be read from the raising rule, which selects the first factor when ${\phi }_i(a)\ge {\epsilon }_i(b)$. If $a$ has an incoming edge of color $i$, then ${\phi }_i(a)=0$; either the raising operator acts nontrivially on $a$, or ${\epsilon }_i(b)=1$ and it acts nontrivially on $b$. Thus a highest vertex must have $a=1$. At $a=1$, all ${\phi }_i(a)$ vanish except ${\phi }_1(a)=1$. To be highest, the incoming color of $b$, if any, must consequently be $1$. For $n\ge 2$, the possibilities are exactlyTheir weights are $2{\epsilon }_1=2{\omega }_1$, ${\epsilon }_1+{\epsilon }_2={\omega }_2$, and zero. Each occurs once. Connected highest-weight components of the crystal basis describe the irreducible representations in the tensor product of Lie algebra representations, soThis is the tensor-square decomposition of the defining symplectic representation. The symmetric square is $L(2{\omega }_1)$; the exterior square is the direct sum of its invariant inverse-form line and the primitive exterior square $L({\omega }_2)$. Their dimensions are respectively $n(2n+1)$, $n(2n-1)-1$, and one, summing to $4n^2=\dim (V\otimes V)$. At $n=1$, the chain has two vertices and the highest tensor vertices are only $1\otimes 1$ and $1\otimes \bar{1}$: the decomposition is $L(2{\omega }_1)\oplus L(0)$, of dimensions three and one.

The permutation representation on ${\mathbb{C}}^X$ sends the basis vector $e_x$ to $e_{gx}$. Its character is ${\pi }_X(g)=|X^g|$, because precisely the fixed basis vectors contribute to the trace. Burnside lemma says that the number of orbits is $|G{|}^{-1}{\sum }_g|X^g|$. To prove it, double-count $(g,x)$ with $gx=x$: the count is ${\sum }_g|X^g|={\sum }_x|G_x|$. Each orbit contributes $|Gx||G_x|=|G|$ by the orbit-stabilizer theorem.

A pair is fixed under the diagonal action exactly when both entries are fixed. Since permutation characters are real,by Burnside lemma on the product.

The general linear group ${\mathrm{GL}}_2(q)$ acts on the $q+1$ lines of ${\mathbb{F}}_q^2$. It is doubly transitive: choosing nonzero representatives of any two distinct lines gives a basis, and a linear map can send that basis to representatives of any other ordered pair. Thus there are two orbits on ordered pairs, the diagonal and its complement. The character inner product of this permutation character with itself is two, while the multiplicity of the trivial character is one by transitivity. The sum of squares of irreducible multiplicities is therefore two; the remaining summand is one irreducible representation, of dimension $(q+1)-1$. Its dimension is $\boxed{q}$.

For $S_n$ acting on two-element subsets, an ordered pair of subsets is classified by intersection size $0,1,2$. All three occur for $n\ge 4$, and a permutation carries any pair of a given type to any other. Hence $⟨{\pi }_Z,{\pi }_Z⟩=3$. The natural point permutation representation is the trivial representation plus the standard representation of the symmetric group, of dimension $n-1$: its irreducibility also follows from double transitivity and inner product two.

Embed this point representation into ${\mathbb{C}}^Z$ by $e_i\mapsto {\sum }_{j\ne i}{e}_{\{i,j\}}$. The map is equivariant, and its kernel consists of vectors with $c_i+c_j=0$ for every distinct $i,j$. Three distinct indices force every $c_i=0$, so it is injective for $n\ge 3$. Thus the trivial and standard representations each occur in ${\mathbb{C}}^Z$. The norm three leaves exactly one further irreducible, with multiplicity one. It is the orthogonal complement of that embedded subspace, equivalently the vectors $a_{ij}=a_{ji}$ satisfying ${\sum }_{j\ne i}{a}_{ij}=0$ for every $i$. Its dimension isThe constituent dimensions are $\boxed{1,n-1,n(n-3)/2}$, each once. For $n=3$, complementation identifies two-element subsets with points, so there are only the trivial and 2-dimensional standard constituents.

Equip $E$ with the inner product defining the root system, so the Weyl group acts by orthogonal transformations. As usual, the root system spans its ambient space $E$. Let $U\subseteq E$ be an invariant subspace for the reflection representation of a Weyl group. Its orthogonal complement is also invariant.

For a root $\alpha$, the Weyl reflection satisfiesIf $(u,\alpha )=0$ for every $u\in U$, then $\alpha \in U^{\perp }$. Otherwise choose $u\in U$ with $(u,\alpha )\ne 0$. Both $u$ and $s_{\alpha }(u)$ belong to $U$, and their difference is a nonzero multiple of $\alpha$, so $\alpha \in U$. HenceThe two sets are mutually orthogonal and individually closed under their root reflections. If both were nonempty, this would decompose $R$ into two orthogonal root systems, contradicting that it is an irreducible root system. All roots therefore lie in one of the two subspaces. Because they span $E$, either $U=E$ or $U^{\perp }=E$, the latter giving $U=0$. Thus the reflection representation is an irreducible representation.

The Adjoint representation of a Lie algebra acts on the algebra itself by ${\mathrm{ad}}_X(Y)=[X,Y]$. The Jacobi identity gives $[{\mathrm{ad}}_X,{\mathrm{ad}}_Y]={\mathrm{ad}}_{[X,Y]}$, so this is a Lie algebra representation. In the generator basis, the column labelled $b$ records the coefficients of $[T_a,T_b]$, givingIn the orthonormal compact-algebra convention, the structure constants of a Lie algebra are totally antisymmetric. With the row index written first this is $(t_a^{\mathrm{Ad}}{)}_{bc}=-i{f}_{abc}$.

Applying the representation to the generator commutator and taking a trace against $t_c^R$ givesConsequently, for a nonzero trace normalization,The cyclic property of the trace also shows that this expression is antisymmetric in every pair of indices, justifying the orthonormal convention used for the adjoint matrices.

The summed quadratic Casimir operator has scalar value $C_2(R)$ on the representation under consideration. Taking its trace in two ways yieldsand henceThe constant $C(R)$ is the representation's Dynkin index in this normalization.

Differentiating the tensor product of group representations givesThe generators acting on the two factors commute, so the sum of their squares isFor a semisimple Lie algebra, every element is a sum of commutators. Since the trace of each represented commutator vanishes, all represented generators are traceless matrices. The trace of the cross term is therefore zero. Taking the trace after decomposition into irreducible representations instead gives the tensor-product Casimir trace identitywhere repeated irreducible constituents occur repeatedly in the sum.

This step requires tracelessness, as holds for the special unitary groups used below. For an arbitrary algebra with an abelian factor, the general trace formula has the extra term $2{\sum }_a\mathrm{tr}(t_a^{R_1})\mathrm{tr}(t_a^{R_2})$. For example, one-dimensional representations of an abelian generator with charges one and one have individual Casimirs one, while the product generator has charge two and Casimir four. Thus the identity without the cross term is not a theorem for every Lie algebra as the unrestricted opening wording might suggest.

For the fundamental-antifundamental decomposition for SU(N), write a product tensor as a matrix $M^i{}_j$. It transforms as $M\mapsto UM{U}^{-1}$, and decomposes uniquely asThe first term is a trivial representation, and the second is the invariant traceless subspace, of dimension $N^2-1$. Its infinitesimal transformation is a commutator, so it is the Adjoint representation. This representation is irreducible for $N\ge 2$: an invariant subspace is an ideal of ${\mathfrak{sl}}_N$; commuting a nonzero ideal element with diagonal and elementary matrices produces off-diagonal elementary matrices, whose further commutators generate all off-diagonal matrices and traceless diagonal matrices. A purely diagonal nonzero element also produces an off-diagonal element unless it is scalar, and a traceless scalar in characteristic zero is zero. ThereforeFor the conjugate defining representation the generators are $-t_a^T$, so its trace index and Casimir equal those of the defining representation. With $C(\mathbf{N})=1/2$, the trace formula gives $C_2(\mathbf{N})=(N^2-1)/(2N)$. The product trace identity, with the zero Casimir of the singlet, then readsCancelling the nonzero factor yields

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

If $g$ belongs to the centre of a group, then $\rho (g)$ commutes with $\rho (h)$ for every $h\in G$. The Schur lemma says that an endomorphism of a finite-dimensional complex irreducible representation commuting with the representation is scalar. ThereforeSince $g$ has finite order, ${\lambda }_g$ is a root of unity and has modulus one.

For the order-18 group, the index-two subgroup $P$ is normal and $G=⟨P,t⟩$. Restrict a complex irreducible representation $V$ to the finite abelian group $P$. Simultaneous diagonalization decomposes it into character spaces $V_{\chi }$. The inversion relation makes $t$ exchange $V_{\chi }$ and $V_{{\chi }^{-1}}$. The sum of spaces belonging to any such character orbit is $G$-invariant, so irreducibility leaves only one orbit.

If the character is trivial, all of $P$ is in the representation's kernel. Otherwise $\chi$ has order three and $\chi \ne {\chi }^{-1}$. The only spaces are $V_{\chi }$ and $V_{{\chi }^{-1}}$, and both have the same kernel in $P$. A nontrivial character of $P\cong C_3\times C_3$ maps onto $C_3$, so that kernel has order three. It acts trivially on all of $V$. Therefore the group has no faithful irreducible complex representation. In fact choosing nonzero $v\in V_{\chi }$ makes $\mathrm{span}(v,tv)$ invariant, so these nontrivial-character irreducibles have degree two.

The special unitary group is $\mathrm{SU}(2)=\{g\in M_2(\mathbb{C}):g^{\ast }g=I,\det g=1\}$. The special orthogonal group is $\mathrm{SO}(3)=\{R\in M_3(\mathbb{R}):R^{T}R=I,\det R=1\}$. Conjugation $A\mapsto gA{g}^{-1}$ preserves trace and skew-Hermiticity: for unitary $g$, $(gA{g}^{-1}{)}^{\ast }=g{A}^{\ast }{g}^{-1}=-gA{g}^{-1}$. It therefore defines the representation on the real space $V$ described in the following parts. The bar over the transpose in the PDF is essential; the condition is $A+A^{\ast }=0$, not ordinary antisymmetry over the complex numbers.

An orthonormal basis for the pairing proved below isPut $J_j=\sqrt{2}{E}_j$, so $J_j^2=-I$. For any unit real vector $v$, $J(v)={\sum }_j{v}_j{J}_j$ also squares to $-I$. Thus $g=\cos (\theta /2)I+\sin (\theta /2)J(v)$ belongs to $\mathrm{SU}(2)$. Conjugation fixes $J(v)$ and rotates its perpendicular plane by angle $\theta$ up to the orientation convention of this basis, as follows either by multiplying the displayed matrices or from their anticommutator relation $J(v)J(w)+J(w)J(v)=-2(v\cdot w)I$. Since every three-dimensional proper rotation has an axis and an angle, this proves surjectivity onto $\mathrm{SO}(3)$.

If an element is in the kernel, it commutes with each $E_j$. Commutation with $E_3$ forces it to be diagonal; commutation with $E_1$ forces its two diagonal entries to agree. Thus it is a scalar matrix, and determinant one and unitarity give exactly $\{I,-I\}$. Therefore $\boxed{\mathrm{SO}(3)\cong \mathrm{SU}(2)/\{\pm I\}}$.

The finite-dimensional complex irreducible representations of $\mathrm{SU}(2)$ are ${\mathrm{Sym}}^m({\mathbb{C}}^2)$, $m\ge 0$, of dimension $m+1$. The central element $-I$ acts by $(-1{)}^m$, so precisely the even ones descend. The irreducibles of $\mathrm{SO}(3)$ are $\boxed{{\mathrm{Sym}}^{2\ell }({\mathbb{C}}^2),\ell =0,1,2,\dots }$, viewed as representations of the quotient, with dimensions $2\ell +1$.

Use the full definition of the special unitary group: $A^{†}A=I$ and $\det A=1$. The determinant condition is essential for the alternating tensors; unitarity alone would instead define $U(n)$. Assume $n\ge 2$ when discussing proper invariant subspaces.

Let $V={\mathbb{C}}^n$ be the fundamental representation. A $(j,k)$ tensor belongs to $V^{\otimes j}\otimes (V^{\ast }{)}^{\otimes k}$. Its SU(n) tensor transformation isSince $(A^{-1}{)}^{\delta }{}_{\beta }=\overline{A^{\beta }{}_{\delta }}$, complex conjugation exchanges upper fundamental and lower dual factors. Reordering the two index groups shows that $\overline{T}$ is a $(k,j)$ tensor.

The invariant tensors follow directly. The Kronecker delta transforms to $A^{\alpha }{}_{\gamma }(A^{-1}{)}^{\gamma }{}_{\beta }={\delta }^{\alpha }{}_{\beta }$. For the alternating tensors, the determinant expansion givesThe lower tensor is preserved in the same way using $\det A^{-1}=1$. Thus $\delta$, ${ϵ}^{{\alpha }_1\cdots {\alpha }_n}$ and ${ϵ}_{{\beta }_1\cdots {\beta }_n}$ are invariant.

When $j\ge 2$, the permutation of two upper slots commutes with the tensor product representation. Its $+1$ and $-1$ eigenspaces are nonzero proper invariant subspaces. The same argument uses two lower slots when $k\ge 2$. The remaining rank-two case $(j,k)=(1,1)$ splits into scalar multiples of $\delta$ and traceless second-rank tensors, since tensor contraction is equivariant. This covers every case except the fundamental, dual fundamental and scalar cases specified in the question. Those three are irreducible: the special unitary group acts transitively on unit vectors in its fundamental space, so a nonzero invariant subspace must be the full space, and the dual has the same property.

A symmetric tensor of upper rank $j$ has one component for each multiplicity vector $(r_1,\dots ,r_n)$ with nonnegative entries and $\sum r_i=j$. Equivalently it is a homogeneous degree-$j$ polynomial in $n$ variables. Counting these monomials givesFor a totally antisymmetric tensor, any repeated index gives zero and the remaining components are indexed by increasing $j$-element subsets. HenceThese are the dimensions of the symmetric power and exterior power, respectively.

For SU(4), the two-index antisymmetric representation has complex dimension six. To establish a six-dimensional real representation, one must exhibit a real structure; simply counting six complex components would give twelve real components. Normalize ${ϵ}^{1234}=1$ and define the antilinear mapThe invariant Hermitian form identifies the conjugate indices with lower indices, and the invariant volume tensor then returns an upper pair. Thus $\mathcal{J}$ commutes with SU(4). The identitygives ${\mathcal{J}}^2=1$. Its fixed space hasThree arbitrary complex entries specify all the others, giving exactly six real degrees of freedom. The real structure of the SU4 exterior square therefore provesThe ordinary Hermitian norm restricts to an invariant positive real inner product on this fixed space.