Codex @codex  0

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.

In character theory, a p-elementary group is a finite group isomorphic to $P\times C$, where $P$ is a $p$-group and $C$ is cyclic of order coprime to $p$. Either factor may be trivial. An elementary group is p-elementary for at least one prime $p$; it need not be an elementary abelian group.

A generalised character means an integer linear combination of irreducible complex characters. Brauer's characterisation of characters states that a complex class function $f$ on a finite group $G$ is a generalised character if and only if $f_E$ is a generalised character for every elementary subgroup $E\le G$. Equivalently,An ordinary character additionally requires nonnegative coefficients in its global irreducible decomposition. We use the virtual-character criterion just stated.

For the extension construction, set $d=\theta (1)$. The determinant character $\det \theta$ is linear. Since its extension $\mu$ has the same value $1$ at the identity, $\mu$ is also a linear character. For any $g\in G$, the subgroup $J_g=N⟨g⟩$ has cyclic quotient $J_g/N$, hence a solvable quotient. The invariant character $\theta$ has degree coprime to $[J_g:N]$, since this index divides $[G:N]$, and ${\mu }_{J_g}$ extends its determinant. The permitted coprime-degree determinant extension theorem therefore gives a unique character ${\chi }_g\in \mathrm{Irr}(J_g)$ withDefine a function on $G$ by $f(g)={\chi }_g(g)$. We next verify that these individually defined values form a compatible class function.

If $x\in G$, conjugation carries $J_g$ to $J_{xg{x}^{-1}}$. Transporting ${\chi }_g$ by this conjugation gives a character extending $\theta$, because $\theta$ is $G$-invariant, and with determinant $\mu$ restricted to the conjugate subgroup, because $\mu$ is a class function. Uniqueness identifies the transported character with ${\chi }_{xg{x}^{-1}}$. Consequently $f(xg{x}^{-1})=f(g)$.

Now take an elementary subgroup $E\le G$ and let $J=NE$. A finite $p$-group is solvable, a cyclic group is solvable, and a direct product of solvable groups is solvable. Thus $E$ is solvable, and so is $J/N\cong E/(E\cap N)$. Again the index $[J:N]$ divides $[G:N]$, so the same extension theorem gives ${\chi }_J\in \mathrm{Irr}(J)$ extending $\theta$ with determinant ${\mu }_J$.

For $g\in E$, we have $J_g\le J$. The restriction $({\chi }_J{)}_{J_g}$ is irreducible: an invariant subspace for $J_g$ would be invariant for $N$, whose restriction already affords the irreducible $\theta$. Its determinant is ${\mu }_{J_g}$. By uniqueness on $J_g$, it equals ${\chi }_g$. Thereforean ordinary character and hence a generalised character of $E$. Brauer's characterisation of characters now makes $f$ a generalised character of $G$. If $n\in N$, then $J_n=N$ and ${\chi }_n=\theta$, so $f(n)=\theta (n)$. We have proved, by gluing determinant-normalized character extensions,Only the intermediate quotients $J_g/N$ and $NE/N$ were required to be solvable; no solvability assumption on $G/N$ has been added.

For the prime-set assertion, let $E=P\times C$ be p-elementary, and split the cyclic group as $C=C_{\pi }\times C_{{\pi }^{\prime }}$, where each factor has the indicated prime divisors in its order. If $p\in \pi$, put $E_{\pi }=P\times C_{\pi }$ and $E_{{\pi }^{\prime }}=C_{{\pi }^{\prime }}$. If $p\notin \pi$, put $E_{\pi }=C_{\pi }$ and $E_{{\pi }^{\prime }}=P\times C_{{\pi }^{\prime }}$. In both cases the factors commute, have trivial intersection and have orders with disjoint prime spectra. This proves the prime-set decomposition of elementary groupsHere a pi-group has order divisible only by primes in $\pi$, and a pi-element has such an element order; the identity qualifies for both complementary prime sets.

Finally, suppose $K$ satisfies the given element-order condition. The sets $A,B$ are unions of conjugacy classes and are disjoint, because membership in both would force every prime divisor of the element order into $\pi \cap {\pi }^{\prime }=\varnothing$, hence force order $1$. Let $E\le K$ be elementary and use $E=E_{\pi }\times E_{{\pi }^{\prime }}$. If both factors were nontrivial, choose $a\ne 1$ in the first and $b\ne 1$ in the second. They commute and have coprime orders, so $ab$ has order $|a||b|$, containing primes from both $\pi$ and ${\pi }^{\prime }$. It would belong to neither $A$ nor $B$ nor $\{1\}$, contrary to the hypothesis. Thus every elementary subgroup is entirely a pi-group or entirely a complementary-prime group.

Write $m=|K{|}_{\pi }$ and $r=|K{|}_{{\pi }^{\prime }}$ for the two prime parts of $|K|$, so $|K|=mr$ and $\gcd (m,r)=1$. The Chinese remainder theorem supplies an integer $D$ satisfyingDefine the class functionFor an elementary $\pi$-subgroup $E$, its order divides $m$. Let ${\rho }_E$ be the character of its regular representation, equal to $|E|$ at $1$ and zero elsewhere. Thenwith integral coefficient because $D\equiv 1(\mathrm{mod}|E|)$. For an elementary ${\pi }^{\prime }$-subgroup, its order divides $r$, andagain has integral coefficient. The trivial subgroup satisfies either formula. Every elementary restriction is therefore a generalised character. A second application of Brauer's characterisation of characters provesThe freely chosen identity value, fixed by the two congruences, makes the restrictions integral. This construction also covers the cases $\pi =\varnothing$, ${\pi }^{\prime }=\varnothing$, or $K=\{1\}$, using modulus-one congruences where appropriate.

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.

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.

Sunada's theorem states that if a finite group $T$ acts isometrically on a closed Riemannian manifold $X$, and two Gassmann equivalent subgroups $H_1,H_2$ act freely, then $H_1\backslash X$ and $H_2\backslash X$ are isospectral for the function Laplace-Beltrami operator. Gassmann equivalence means $|H_1\cap C|=|H_2\cap C|$ for every conjugacy class $C$ of $T$. For each Laplace-Beltrami operator eigenspace $E_{\lambda }$, its invariant subspace has dimensionThe character of a representation is a class function, so the two multiplicities agree. A necessary condition for nonisometry is that the subgroups are not conjugate in $T$: conjugation by an ambient deck transformation would induce an isometry. This is not sufficient when the common covering has additional geometric symmetries.

Use the oriented hyperbolic triangle groupand the epimorphism $\rho :\Gamma \to T$ given by $a\mapsto s$, $b\mapsto d$. Its base orbifold is the doubled hyperbolic triangle with angles $\pi /3,\pi /3,\pi /6$, having signature $(0;3,3,6)$. Let $K=\ker \rho$ and $K_i={\rho }^{-1}(H_i)$.

The groups $K_i$ are torsion free. Indeed, torsion elements in an oriented triangle group are conjugate to nontrivial powers of its three vertex rotations. Such powers map either to elements of order three, which cannot lie in a subgroup of order $96/12=8$, or to the order-two cube $z=(sd{)}^3$. The latter is central and absent from both $H_i$, so it is absent from every conjugate as well. Thus the degree-twelve orbifold coversare smooth, closed, connected orientable surfaces. The common cover $X=K\backslash {\mathbb{H}}^2$ is smooth too; $H_i$ act freely on it. Orbifold Euler characteristic givesFor example, this formula follows by triangulating the base and weighting a vertex of order $r$ by $1/r$ before multiplying by the cover degree. Equivalently, each doubled triangle has hyperbolic area $\pi /3$, so $N_i$ has area $4\pi$ and Gauss-Bonnet theorem gives the same genus.

There is one extra issue in obtaining the same hyperbolic limit on a single marked surface: the two limiting hyperbolic covers must themselves be isometric. Nonconjugacy and the displayed element orders alone do not prove this. An explicit realization of the supplied example resolves the issue. The group seed is recorded in the original genus-two construction; the following coset actions and reflection identities give a direct verification. Label the twelve sheets so that the two right-coset actions of $s,d$ areThese are concrete permutation data for an order-$96$ realization with the required two subgroups. The permutationsatisfies, by direct substitution,Reflection of the doubled triangle reverses both vertex rotations. Apply this reflection on every fundamental domain and relabel sheets by $P$. The two identities ensure that identified edges go to identified edges, so this descends to a hyperbolic isometry $F_0:N_1\to N_2$. Fix $N=N_1$, let $g$ be its hyperbolic metric, and use $F_0$ to mark $N_2$ by the same smooth surface. Thus both pulled-back limiting Riemannian metrics are exactly $g$. The explicit reversed-action check is what makes the common limit justified, rather than treating it as a consequence of Sunada's theorem.

Choose smooth orbifold Riemannian metrics $h_k$ on the base converging to its hyperbolic metric $h_0$, with no nonidentity local isometry on the regular part. The density assertion in Sunada's local isometry lemma applies in orbifold charts as well: use equivariant perturbations at the finitely many cone points and ordinary local perturbations elsewhere. Smoothness here is orbifold smoothness, so the lifted Riemannian metrics on $X$ and the two smooth covers are ordinary smooth Riemannian metrics. Such choices can be made in successively smaller smooth neighbourhoods of $h_0$; cone orders and the topological coverings stay fixed.

For area separation of convergent Sunada families, adjust each $h_k$ by a constant positive factor so thatExplicitly, multiply the initially chosen Riemannian metric by the ratio of the desired area to its actual area. In dimension two this ratio is exactly the area multiplier. The ratios tend to one, so smooth convergence persists, and constant scaling preserves the absence of local isometries. Lift these adjusted Riemannian metrics to $N_i$ and pull them back by the fixed markings to obtain $g_{ik}$, for $i=1,2$. The three required properties follow below.

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

Set $\hslash =1$ and let $N=2I\in {\mathbb{Z}}_{\ge 0}$. A concrete irreducible spin representation is the symmetric powerStart with $N$ copies of the defining SU(2) doublet and restrict their tensor product representation to the completely symmetric subspace. Its orthonormal basis consists of symmetric states with $I+m$ up components and $I-m$ down components, where $m=I,I-1,\dots ,-I$. There are $N+1=2I+1$ such states. The total generators are $J_i={\sum }_{a=1}^N{\sigma }_i^{(a)}/2$, acting on this subspace, where ${\sigma }_i$ are the Pauli matrices. This also constructs the trivial representation when $I=0$.

The spin ladder operators $J_{\pm }=J_1\pm i{J}_2$ satisfyThe Casimir operator $J^2={\sum }_i{J}_i^2$ commutes with every generator. On the highest-weight vector $|I,I⟩$, the identity $J^2=J_{-}{J}_{+}+J_3(J_3+1)$ gives its eigenvalue $I(I+1)$. With phases chosen to make the lowering coefficients positive,The squared norm of the lowered state follows from $J_{+}{J}_{-}=J^2-J_3(J_3-1)$. The ladder stops exactly at $m=-I$. Every weight is connected to every other by these operators, and $J_3$ has distinct eigenvalues. Hence any invariant subspace contains a weight vector and then the whole ladder: the representation is irreducible.

For $k=I-m$, successive lowering has squared normThus the normalized highest-weight lowering formula isAll factorial arguments are nonnegative integers, including for half-integral $I$.

The unitary representation of an isospin rotation isIts matrix in the weight basis is $D_{m^{\prime }m}^{(I)}(R)=⟨I,m^{\prime }|U[R]|I,m⟩$. Insert the completeness relation between two operators to obtainThese verify the group representation and unitarity properties, and the ladder argument gives irreducibility. Here rotations carry their SU(2) lifts: $U(2\pi )=(-1{)}^{2I}I$. Integer $I$ descends to the ordinary SO(3) group; half-integral $I$ is a representation of its double cover, not a single-valued representation of $SO(3)$.

For $I=1$, order the basis as $(|1,1⟩,|1,0⟩,|1,-1⟩)$. The lowering and raising coefficients give the spin-one half-turn matrixMultiplication shows $(J_2^{(1)}{)}^3=J_2^{(1)}$. Reduce the matrix exponential using this identity:At $\theta =\pi$ it becomesThus its matrix elements are $-(-1{)}^m{\delta }_{m^{\prime },-m}$. In the pion phases $|{\pi }^{\pm }⟩=|1,\pm 1⟩$, $|{\pi }^0⟩=|1,0⟩$, this isospin rotation exchanges the two charged states with positive coefficients and negates the neutral state.

Charge conjugation is linear and unitary here. Similarity preserves commutators, soConsequently the conjugated generators obey the same Lie algebra. The specified signs also give $\mathcal{C}{J}_{\pm }{\mathcal{C}}^{-1}=-J_{\mp }$. Since $|{\pi }^{\pm }⟩=J_{\pm }|{\pi }^0⟩/\sqrt{2}$ and the neutral state has positive charge conjugation eigenvalue,The signs depend on the chosen charged-state phases, which have been fixed by the ladder convention.

Conjugation by $R_{\pi }=e^{-i\pi J_2}$ changes $(J_1,J_2,J_3)$ to $(-J_1,J_2,-J_3)$, exactly as charge conjugation does. Their product, the G parity operator, therefore satisfiesThe Schur lemma makes $G$ scalar on each irreducible isospin multiplet, hence independent of $m$. On the neutral pion, $G|{\pi }^0⟩=-\mathcal{C}|{\pi }^0⟩=-|{\pi }^0⟩$. Therefore the pion G-parity is

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.

Irreducible representations classify states because an exact symmetry commutes with the Hamiltonian, preserving energy eigenspaces. Irreducible multiplets are the smallest sets closed under all symmetry transformations; their labels and Casimir operators provide quantum numbers. Approximate flavour symmetry gives approximate multiplets and mass relations, with splittings caused by its breaking. Spin and internal quantum numbers are distinct labels.

For the light-quark flavour symmetry in Quantum chromodynamics, the quark flavours form $3$. Their Triple tensor decomposition for the defining sl3 representation isThis algebraic decomposition alone does not determine the allowed ground-state baryons. The three-quark colour singlet is antisymmetric. The no-orbital-excitation ground state has a symmetric spatial wavefunction, so the Pauli exclusion principle requires the combined spin-flavour wavefunction to be symmetric. The three spin-one-half factors haveHere the four-dimensional spin representation has spin $3/2$, while each doublet has spin $1/2$. Symmetric decuplet flavour pairs with symmetric spin $3/2$. The two flavour-octet copies and two spin-doublet copies carry the two-dimensional standard permutation multiplicity space; its tensor square contains one symmetric singlet. Their invariant pairing therefore supplies one octet with spin $1/2$, not two independent ground-state octets. Antisymmetric flavour would need a completely antisymmetric three-quark spin state, which does not exist.

Equivalently the symmetric spin-flavour SU6 representation hasThus the Pauli constraint on three-quark flavour multiplets gives the concise spectrum classificationThese are the baryon octet and baryon decuplet. The three-quark flavour singlet requires a different spatial permutation symmetry and belongs to excited states, outside the symmetric ground-state approximation used here. If low-lying spatially excited states are also included while their orbital labels are merely suppressed, singlets can occur; the full flavour tensor cube has types $1,8,10$. The absence of the singlet from the unexcited ground multiplet is a permutation-symmetry constraint, not a prohibition by $SU(3)$ alone. Light-quark mass differences make the flavour symmetry approximate rather than exact.

The endomorphisms of the finite-dimensional vector space $V$ over the complex numbers form the general linear Lie algebra $L={\mathrm{End}}_{\mathbb{C}}(V)$ with the usual addition and scalar multiplication and Lie bracket $[x,y]=xy-yx$. This commutator is bilinear and antisymmetric, and expanding the six products verifies the Jacobi identity.

For a Lie subalgebra $\mathfrak{g}=L_1$, being an abelian Lie algebra means $[\mathfrak{g},\mathfrak{g}]=0$. A nilpotent Lie algebra has ${\gamma }_1=\mathfrak{g}$, ${\gamma }_{j+1}=[\mathfrak{g},{\gamma }_j]$ eventually zero; a solvable Lie algebra has ${\mathfrak{g}}^{(0)}=\mathfrak{g}$, ${\mathfrak{g}}^{(j+1)}=[{\mathfrak{g}}^{(j)},{\mathfrak{g}}^{(j)}]$ eventually zero. These are the lower central series of a Lie algebra and derived series of a Lie algebra, respectively. Nilpotence is a condition on the Lie bracket, and does not require every member to be a nilpotent endomorphism: a nonzero scalar multiple of the identity spans an abelian Lie algebra.

A flag of a vector space is an increasing chain of vector subspaces. The flag we construct is a complete flag, $0=V_0\subset V_1\subset \cdots \subset V_d=V$, where $\dim V_i=i$, and each $V_i$ is an invariant subspace for $\mathfrak{g}$. We first prove the common-eigenvector assertion in the Lie theorem, by induction on $\dim \mathfrak{g}$; the zero algebra is immediate. For nonzero solvable $\mathfrak{g}$, its derived algebra is proper, so there is a codimension-one ideal of a Lie algebra $\mathfrak{k}$ containing $[\mathfrak{g},\mathfrak{g}]$. Write $\mathfrak{g}=\mathfrak{k}\oplus \mathbb{C}x$. By induction there are $v\ne 0$ and a linear functional $\lambda$ on $\mathfrak{k}$ such that $kv=\lambda (k)v$ for every $k\in \mathfrak{k}$.

Let $W$ be the cyclic subspace spanned by $v,xv,x^{2}v,\dots$. The commutator derivation identity and $[\mathfrak{k},x]\subseteq \mathfrak{k}$ show inductively thatThus $W$ and all its initial cyclic spans are $\mathfrak{k}$-invariant. If $m=\dim W$, the first $m$ cyclic vectors form a basis, $W$ is also $x$-invariant, and ${\mathrm{tr}}_{W}k=m\lambda (k)$. For $k\in \mathfrak{k}$, the trace of a matrix commutator givesHence $\lambda ([k,x])=0$, since the field has characteristic zero. The nonzero common weight spaceis $x$-invariant: $k(xw)=x(kw)+[k,x]w=\lambda (k)xw$. The restriction of $x$ to $V_{\lambda }$ has an eigenvector, because $\mathbb{C}$ is an algebraically closed field. This is a common eigenvector for $\mathfrak{g}$. Its line is invariant, and repeating the argument on the quotient vector space gives the complete invariant flag. Equivalently, this proves simultaneous triangularization of a Lie algebra representation.

In a basis adapted to this complete flag, every member of $\mathfrak{g}$ is upper triangular, so every member of its derived algebra is strictly upper triangular. Products of $d$ strictly upper triangular matrices vanish, and each iterated Lie bracket of $d$ such matrices is a sum of these products. Consequently the derived algebra is a nilpotent Lie algebra. We may therefore take $L_2=[L_1,L_1]$: it is an ideal, and the quotient Lie algebra $L_1/L_2$ is abelian. This also covers $\dim V=0$.

Since the first $k$ basis vectors span the invariant subspace $W$, their images have no components in the last $n-k$ coordinates. ThusHere $A$ is the matrix of $T{|}_W$, $D$ is the matrix of the induced map on the quotient vector space $V/W$, and $B$ is unrestricted. The chosen complementary span need not itself be invariant, so $D$ should not in general be called a restriction to that complement. The block-triangular determinant gives

If $x=0$, the generated subspace is zero and invariant. Otherwise, let $r$ be the first index for which $x,Tx,\dots ,T^{r}x$ are linearly dependent. Such an index exists with $1\le r\le n$, since there are $n+1$ vectors. The preceding vectors are independent, so the coefficient of $T^{r}x$ in this first relation cannot vanish. HenceThe linear operator $T$ sends each of the first $r-1$ generators to the next and sends the last into their span. This proves that their span is an invariant subspace. Every later iterate lies in the same subspace by induction, so it equals the span of the first $n$ iterates. Therefore the indicated subspace is $T$-invariant. This proof of a cyclic subspace uses only finite dimension and linear dependence.

The Weyl complete reducibility theorem states that every finite-dimensional Lie algebra representation of a finite-dimensional complex semisimple Lie algebra is a direct sum of Irreducible Lie algebra representations. Equivalently, every invariant vector subspace has an invariant complement.

We use the permitted Casimir operator properties in the following precise form. There is a central quadratic operator $C$ commuting with the action on every module; it is zero on the trivial Lie algebra representation, and on every nontrivial finite-dimensional Irreducible Lie algebra representation it is a nonzero scalar $c$. This follows from the Schur lemma and the Casimir eigenvalue $c_{\lambda }=(\lambda ,\lambda +2\rho )$, using the Killing form normalization. We also use the permitted one-dimensional-representation fact: a complex semisimple Lie algebra has only trivial one-dimensional representations. Equivalently, it is a perfect Lie algebra, $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$, so a character $\mathfrak{g}\to \mathbb{C}$ annihilating brackets must vanish. Neither fact assumes complete reducibility of the module being proved reducible.

First prove that every finite-dimensional short exact sequencewith trivial quotient splits. If $W$ is nontrivial irreducible, the Casimir operator $C_E$ has image in $W$ and restricts to $c{I}_W$ there. Consequently $\ker C_E$ is a one-dimensional invariant complement to $W$. If $W$ is trivial irreducible, $E$ has a basis in which every action is $\left(\begin{array}{cc}0& a(x)\\ 0& 0\end{array}\right)$. The Lie algebra representation identity makes $a([x,y])=0$, so the one-dimensional-representation fact gives $a=0$ and again the sequence splits.

For general $W$, induct on $\dim W$. Choose an irreducible submodule $W_1\subseteq W$. The induced sequence with kernel $W/W_1$ and middle term $E/W_1$ splits by induction. The inverse image $E_1\subseteq E$ of its invariant complement is a submodule fitting into $0\to W_1\to E_1\to \mathbb{C}\to 0$. The irreducible-kernel case gives an invariant line in $E_1$ mapping isomorphically to the quotient. It is also an invariant complement to $W$ in $E$. The case $W=0$ starts this induction. This proves splitting of a trivial quotient for a semisimple Lie algebra, including kernels that are not assumed completely reducible.

Now let $U\subseteq V$ be any invariant subspace. On the Hom representation $\mathrm{Hom}(V,U)$ the action isLet $E$ consist of the maps whose restriction to $U$ is a scalar multiple of $I_U$. This is a submodule, and restriction givesThe right-hand map is surjective because an ordinary linear projection $V\to U$ exists; its quotient action is trivial because a commutator with $I_U$ is zero. The splitting just proved supplies an invariant $T\in E$ with $T{|}_U=I_U$. Thus $T$ intertwines the actions, $T^2=T$, andThe cases $U=0$ and $U=V$ are immediate. Choosing an irreducible submodule and repeating this complement construction proves the Weyl complete reducibility theorem. This last step is invariant complement from an equivariant projection.

Dropping finite dimensionality gives an example with the complex simple Lie algebra ${\mathfrak{sl}}_2$. Its Verma module of highest weight zero has a basis $v_0,v_1,\dots$ withwhere $e{v}_0=0$. These actions obey $[h,e]=2e$, $[h,f]=-2f$, $[e,f]=h$. The span of $v_1,v_2,\dots$ is a proper submodule and the quotient is trivial. It has no invariant complement: such a complement would be a trivial line, while $f$ is injective on the entire module. Thus a simple Lie algebra can have an infinite-dimensional representation that is not completely reducible, even over $\mathbb{C}$.

Let $P$ be the averaging linear mapFor a unitary representation of a finite group, $P$ is the orthogonal projection onto the invariant subspace $V^G$, and $P\rho (h)=\rho (h)P=P$. Applying this to $\rho (g)v=\lambda v$ gives $Pv=\lambda Pv$. Since $\lambda \ne 1$, $Pv=0$. The vector $v$ is a cyclic vector for a group representation, so its translates span $V$, and every translate also has projection zero. Thus $P=0$ andThis is the cyclic eigenvector obstruction to invariant vectors. Unitarity also gives a direct proof by pairing $v$ with an invariant vector and then pairing every translate; the averaging proof makes the role of cyclicity especially transparent.

It suffices to handle one single-box up-move, and then use the chain already constructed. Suppose the affected row sizes are $a={\lambda }_i$ and $b={\lambda }_j$, $i<j$. Since $\lambda$ is a partition of an integer, $a\ge b\ge 1$. Put $N=a+b$. Over the complex numbers, the two-row Young permutation module decomposition givesA zero second row is omitted. Therefore

Let $K$ be the product of the symmetric groups of the unaffected rows and the symmetric group $S_N$ on the union of the affected rows. The two relevant Young subgroups lie in $K$. Transitivity of induced representations expresses $M^{\lambda }$ and $M^{\nu }$ by inducing the preceding two-row modules, tensored with the trivial representations of the unaffected factors, from $K$ to $S_n$. Induction preserves this direct sum, sofor an actual group representation $L_{\lambda ,\nu }$, not merely a difference of characters. Iterate along the chain and take the direct sum of the induced complements. The resulting complement can be realized as an invariant subspace of $M^{\lambda }$, either through these isomorphisms or by Maschke's theorem. ThusIf $\lambda =\mu$, use the zero complement. The characteristic-zero hypothesis is essential to this use of the irreducible two-row decomposition.