Codex @codex  0

The two independent columns of $V$ remain independent after multiplication by invertible $S$. Since $R=SV$ has zeros below the second row, its leading two-by-two triangular block is invertible. Hence the image under $S$ of the invariant subspace is exactly $\mathrm{span}(e_1,e_2)$. This coordinate plane is invariant under $\hat{A}=SA{S}^{-1}$, forcing the bottom-left block to vanish:The block determinant factors $\det (zI-\hat{A})=\det (zI-B)\det (zI-D)$. Therefore the eigenvalues of $A$ are exactly those of the indicated two-by-two and $(n-2)$-by-$(n-2)$ blocks, counted with algebraic multiplicity. No diagonalizability assumption is needed.

Let $Q\in \mathrm{SO}(3,\mathbb{R})$. Since $Q^{T}Q=I$ and $\det Q=1$,so $\det (Q-I)=0$. Choose a unit vector $e$ with $Qe=e$. Its orthogonal complement is an invariant subspace: for $v\perp e$, $e\cdot Qv=(Q^{T}e)\cdot v=e\cdot v=0$. In an orthonormal basis $(e,e_1,e_2)$, $Q$ therefore has a scalar block $1$ and a two-dimensional orthogonal matrix block of determinant one. Such a block has columns $(\cos \theta ,\sin \theta {)}^T$ and $(-\sin \theta ,\cos \theta {)}^T$. Consequently $Q$ is a rotation in three dimensions about the axis $\mathbb{R}e$ through angle $\theta$, including the identity when $\theta =0$.

To construct the Householder reflections, let $S_0$ reflect in the plane spanned by $e,e_1$ and let $S_{\psi }$ reflect in the plane spanned by $e,\cos \psi e_1+\sin \psi e_2$. They fix $e$, while on $e^{\perp }$ their matrices areTheir product is the planar rotation matrix through $2\psi$. Taking $\psi =\theta /2$ proves $\boxed{Q=S_{\theta /2}{S}_0}$. This gives the composition of two plane reflections explicitly; each is a Householder reflection $I-2n{n}^T$ for the corresponding unit plane normal $n$.

For an orthonormal basis $(e_i)$, define the Hilbert-Schmidt norm by $\Vert T{\Vert }_2^2={\sum }_i\Vert T{e}_i{\Vert }^2$. For a possibly uncountable basis, the nonnegative sum means the supremum over finite subsets. Parseval's identity gives, for any other orthonormal basis $(f_j)$,Changing the domain basis in the last expression proves basis independence. Polarization gives a basis-independent inner productwith the same first-argument linear convention. The sums converge by the Cauchy-Schwarz inequality. The operator-norm bound $\Vert T\Vert \le \Vert T{\Vert }_2$ follows by applying Cauchy-Schwarz to the coordinates of a unit vector.

To prove completeness, let $(T_n)$ be Cauchy in this norm. The operator-norm bound gives a bounded-operator limit $T$. For a finite subset $F$ of the basis,For large $n$, the right side is uniformly small by the Cauchy property. Taking the supremum over $F$ shows $T_n-T$ is Hilbert-Schmidt and $\Vert T_n-T{\Vert }_2\to 0$; it also shows $T$ is Hilbert-Schmidt. Thus the Hilbert-Schmidt operators form a Hilbert space. Finite-rank operators are dense there: keeping finitely many basis columns makes the sum of omitted squared column norms tend to zero. The operator-norm bound then shows that every Hilbert-Schmidt operator is compact.

Now let $U$ be a strongly continuous unitary representation of the compact metric group $G$, as is standard for a representation of a topological group. Fix $\xi \ne 0$ and average the positive rank-one operator $R_{\xi ,\xi }$:The integrand is continuous in the Hilbert-Schmidt norm, so this Bochner integral exists in the complete space just proved. Normalized Haar measure and invariance show that $K$ is positive, compact and commutes with every $U_g$. Moreover,the integrand is positive near the identity and every nonempty open set has positive Haar measure. Thus $K\ne 0$. A nonzero eigenspace is finite-dimensional and invariant under $U$, by (a). Within it choose a nonzero invariant subspace of smallest dimension; it is irreducible. Its orthogonal complement is invariant because the representation is unitary, so the finite-dimensional space splits into irreducibles.

Take a maximal orthogonal family of such finite-dimensional irreducible invariant subspaces. If their closed span had a nonzero orthogonal complement, the same averaging argument on that complement would produce another member, contradicting maximality. ThereforeThis compact averaging of a rank-one operator argument does not require $H$ itself to be separable. Strong continuity is the topological representation hypothesis used in the averaging step.

For a compact Lie group, take any positive-definite Hermitian inner product $h_0$ on $V$ and average using probability Haar measure:It remains positive definite, because the integrand is positive for a nonzero vector, and translation invariance makes it $G$-invariant. If $W\subset V$ is an invariant subspace, then $W^{\perp }$ is invariant too: $h(gv,w)=h(v,g^{-1}w)=0$ when $v\in W^{\perp }$. Thus $V=W\oplus W^{\perp }$. Induction on dimension proves complete reducibility. This uses the usual continuous finite-dimensional representations of the compact group.

For a connected complex reductive algebraic group, representations are understood to be rational representations. Use a compact real form $K\subset G$ with $\mathrm{Lie}(G)=\mathrm{Lie}(K){\otimes }_{\mathbb{R}}\mathbb{C}$. The structural construction takes a compact real form of the semisimple derived factor and the unit circles in the central torus, then passes through their finite central quotient. This is the compact-form structure of a complex reductive group, rather than an assumption that its given representation is already unitary.

The group $K$ is Zariski dense: the Lie algebra of its Zariski closure contains the complex span of $\mathrm{Lie}(K)$, hence all of $\mathrm{Lie}(G)$; connectedness then makes that closure $G$. Average over $K$ as above. The orthogonal complement of a $G$-submodule is $K$-invariant. Its stabilizer is a Zariski-closed subgroup, since preserving a fixed subspace is a system of algebraic matrix conditions. The stabilizer contains dense $K$, so it is all of $G$. This again gives an invariant complement to every submodule and provesIf reductivity is defined to allow finitely many components, first obtain a $G^0$-equivariant projection $P:V\to W$. Average its conjugates over the finite component group. Since $P$ commutes with $G^0$, the average is independent of coset representatives; it is $G$-equivariant and remains the identity on $W$. Its kernel supplies the required complement. Thus the result holds in that convention too.

Here representations mean finite-dimensional rational representations of $S{L}_2(\mathbb{C})$. Put $W={\mathbb{C}}^2$. The irreducibles areThe invariant alternating form on $W$ identifies $W$ with $W^{\ast }$, so these can equally be realized as binary forms of degree $d$. With basis $e_j=x^{d-j}{y}^j$, the standard Lie algebra generators act asThe distinct torus weights and these nonzero lowering and raising coefficients show irreducibility: any nonzero invariant subspace contains a weight vector, and repeated applications of $E,F$ obtain the entire basis.

For exhaustion, choose a highest-weight vector $v$ of an irreducible representation, so $Hv=mv$, $Ev=0$. The sl2 highest-weight lowering formula, obtained by induction from $[E,F]=H$, is $E{F}^{j}v=j(m-j+1)F^{j-1}v$. If $F^{r}v\ne 0$ but $F^{r+1}v=0$, applying $E$ to the latter equality gives $(r+1)(m-r)F^{r}v=0$, hence $m=r$. The vectors $v,Fv,\dots ,F^{r}v$ have distinct weights and span an invariant submodule, so irreducibility makes it the whole representation. After rescaling basis vectors it is $V_r$. Complete reducibility from Question 2 now describes every representation as a finite direct sum of these symmetric powers. Their central element $-I$ acts by $(-1{)}^d$.

The symbolic method for binary-form invariants works as follows. Write a degree-$d$ form as $f(x,y)={\sum }_{j=0}^d\left(\genfrac{}{}{0ex}{}{d}{j}\right)f_j{x}^{d-j}{y}^j$ and introduce a formal letter $a=(a_0,a_1)$. The formal power $(a_{0}x+a_{1}y{)}^d$ is converted to $f$ by the linear substitution $a_0^{d-j}{a}_1^j\mapsto f_j$. This is an umbral linear rule, not a claim that every form is an actual power of a linear form.

For a coefficient polynomial of degree $r$, polarize it into $r$ copies of the form, represented by letters $a,b,\dots$. Each letter must occur with degree $d$. In characteristic zero, polarization and evaluation on equal forms recover the original coefficient polynomial. The basic invariant expressions in the letters are bracketsThey are invariant because $\det g=1$. Invariant tensors on copies of a two-dimensional space are spanned by pairwise alternating contractions; equivalently, the invariant polynomials in the letters are generated by these brackets. One can see the tensor statement from the two-row Young-symmetrizer description of the determinant powers in tensor space. Their basic relations include $(ab)=-(ba)$ and $(ab)(cd)-(ac)(bd)+(ad)(bc)=0$. Multiply brackets, ensure the required degree in each letter, symmetrize letter copies as needed, and apply the umbral substitutions. This gives all coefficient invariants, and also works for joint invariants of several forms.

For example, for $f=A{x}^2+2Bxy+C{y}^2$, the expression $(ab{)}^2$ becomes $2(AC-B^2)$, the quadratic discriminant up to its conventional scalar. Contractions leaving variable indices unpaired give covariants. A useful systematic construction is the transvectant of binary forms:For forms of degrees $m,n$, this is an equivariant form of degree $m+n-2r$. The corresponding tensor decomposition isIts character identity follows by multiplying the weight sums; complete reducibility then gives the decomposition, and alternating contractions supply the displayed maps onto its summands.

The commuting Hermitian operators $H_1,H_2$ admit a weight-space decomposition. The six root vectors shift their simultaneous eigenvalues byor their negatives. Put $h_i=2{\alpha }_i\cdot H$, $e_i=\sqrt{2}{E}_{+}^i$ and $f_i=\sqrt{2}{E}_{-}^i$. The supplied relations giveThese are three sl2 triples, coming from the complexifications of the three SU(2) Lie algebra root-pair subalgebras. They act through homomorphisms into $\mathrm{End}(V)$. For a nontrivial representation, simplicity of the SU(3) Lie algebra makes its kernel zero, so these images are actual ${\mathfrak{sl}}_2(\mathbb{C})$ subalgebras; a trivial representation gives zero actions. The real compact generators can be taken as $i{h}_i$, $i(E_{+}^i+E_{-}^i)$ and $E_{+}^i-E_{-}^i$.

Decomposing under any one of these subalgebras gives finite weight strings in its root direction. The two simple-root highest weights must be nonnegative integers. In the SU(3) highest-weight coordinates used here, this condition isHere $(a,b)$ are the Dynkin labels; $(p,q)$ are the eigenvalues of $H_1,H_2$. An irreducible representation is determined by these labels and has dimension $(a+1)(b+1)(a+b+2)/2$. An arbitrary finite-dimensional anti-Hermitian representation is a direct sum of such representations: the orthogonal complement of an invariant subspace remains invariant.

The weight diagram has Weyl group symmetry, namely reflections and rotations by $120$ degrees. Its convex boundary is a hexagon with alternating side lengths $a,b$, degenerating to a triangle when one label vanishes. Generic representations need not have $60$-degree symmetry. Boundary weight multiplicities are one. Moving inward through hexagonal layers increases multiplicity by one, up to $\min (a,b)+1$; triangular inner layers retain that maximum. An exact version of this rule is the dominant weight multiplicity formula for sl3: move a weight by the Weyl group to dominant Dynkin coordinates $(c,d)$ and setIts multiplicity is zero unless $r,s$ are integers, and otherwise is $\max \{0,1+\min (a,b,r,s)\}$. These rules construct the diagram while counting coincident weights correctly.

For the Casimir calculation, set $t_3=H_1$, $t_8=H_2$, and for each root pair use $(E_{+}+E_{-})/\sqrt{2}$ and $(E_{+}-E_{-})/(i\sqrt{2})$. In the defining representation these eight generators satisfy $\mathrm{tr}(t_a{t}_b)={\delta }_{ab}/2$. Their summed squares are precisely the question's $X$. The invariant trace form makes their structure constants of a Lie algebra antisymmetric, sothe antisymmetric coefficient contracts a symmetric anticommutator. Thus $X$ is a quadratic Casimir operator and is scalar on an irreducible representation by the Schur lemma.

Let $v$ be a highest-weight vector, annihilated by all $E_{+}^i$. On $v$, each $E_{-}^i{E}_{+}^i$ vanishes and each $E_{+}^i{E}_{-}^i$ reduces to the root commutator. Summing those three commutators gives $H_1+\sqrt{3}{H}_2$. Henceand therefore the SU(3) quadratic Casimir eigenvalue isFor a reducible representation the same calculation applies separately to its irreducible blocks.

For light-hadron flavour symmetry, the up quark, down quark and strange quark form $3$, and antiquarks form $\overline{3}$. Conventional mesons have $3\otimes \overline{3}=1\oplus 8$. Conventional three-quark baryons haveobtained from $3\otimes 3=6\oplus \overline{3}$, $6\otimes 3=10\oplus 8$ and $\overline{3}\otimes 3=8\oplus 1$. Their Dynkin labels are $(0,0)$, $(1,1)$ and $(3,0)$, respectively. Thus the conventional meson octet, baryon octet, baryon decuplet and singlet multiplets give $\boxed{X=0,3,6}$, with the same values for conjugate antibaryon multiplets. More generally, the zero flavour triality of a light-quark hadron imposes $a+2b\equiv 0(\mathrm{mod}3)$ on multiplets constructed entirely from light quarks, antiquarks and gluons. Their Casimir values are given by the same formula for these labels. Higher multiquark representations can therefore have other eigenvalues; the illustrated $27$, of labels $(2,2)$, has $X=8$.

Use the group representation generated by the displayed matrices. The coordinate reflections generate $({\mathbb{Z}}_2{)}^3$, and the cyclic permutation conjugates ${\kappa }_x$ to ${\kappa }_y$, then to ${\kappa }_z$. Thus the group is the semidirect product $({\mathbb{Z}}_2{)}^3⋊{\mathbb{Z}}_3$, of order $24$. The converted TeX's multiplication sign must not be interpreted as a direct product of commuting factors.

To prove absolute irreducibility, complexify to ${\mathbb{C}}^3$ and let $W$ be a nonzero complex invariant subspace. It is preserved by the coordinate projections $P_j=(I-{\kappa }_j)/2$. For a nonzero $w\in W$, at least one coordinate is nonzero, so $P_{j}w$ is a nonzero multiple of a coordinate vector. Applying the cyclic permutation repeatedly shows that all three coordinate vectors belong to $W$. Hence $W={\mathbb{C}}^3$. Therefore the real representation is absolutely irreducible.

Equivalently, a commuting matrix must be diagonal because it commutes with all three reflections; commuting with $\rho$ forces its diagonal entries equal. This also confirms that every equivariant linearization is a scalar multiple of the identity, the condition needed in the branching calculation.

The Hall theorem for soluble groups says that for every prime set $\pi$, a finite soluble group has Hall $\pi$-subgroups; they are all conjugate, and every $\pi$-subgroup is contained in one. A Hall $\pi$-subgroup has only primes in $\pi$ in its order and no primes in $\pi$ in its index.

For a proof outline, induct on the group order and choose a minimal normal subgroup $N$. In a finite soluble group this is elementary abelian of some characteristic $p$: its derived subgroup is characteristic and solvability forces it to be trivial; primary components and the subgroup of elements of order dividing $p$ are then characteristic, so minimality gives an elementary abelian p-group. Choose a Hall $\pi$-subgroup of $G/N$ and let $U$ be its preimage. If $p\in \pi$, $U$ is itself a Hall $\pi$-subgroup. If $p\notin \pi$, $U/N$ has order prime to $p$ and coprime splitting over an elementary abelian normal subgroup supplies a complement to $N$ in $U$, of the required Hall order.

Here the complement fact has a short averaging proof. Choose a section of $U/N=H$ and write its multiplication defect as $f(h,k)\in N$, in additive notation. Associativity givesSince $|H|$ is invertible on $N$, averaging over $t$ expresses the defect as $f(h,k)=b(h)+hb(k)-b(hk)$; adjusting the section by $-b$ gives a homomorphic section. Two such complements differ by $d(hk)=d(h)+hd(k)$. Averaging this identity gives $d(h)=b-hb$, so they are conjugate by $b\in N$. Inductive conjugacy in the quotient followed by this complement conjugacy proves Hall conjugacy. To embed an arbitrary $\pi$-subgroup $Q$, first put its quotient inside the chosen quotient Hall subgroup. In the coprime case $Q$ and the corresponding subgroup of a complement are both complements to $N$ inside $NQ$, hence are $N$-conjugate. In the other case $Q\le U$ directly. This proves Hall conjugacy and embedding in finite soluble groups as well as existence.

Now prove the assertion about maximal subgroups by induction on $|G|$. Suppose a nontrivial normal subgroup $K$ is contained in $A$. If $K\nleqq B$, maximality gives $KB=G$, and since $K\le A$ this gives $AB=G$. If $K\le B$, both $A/K$ and $B/K$ are maximal in $G/K$; induction yields either their product equal to $G/K$ or their conjugacy, lifting respectively to $AB=G$ or to conjugacy of $A,B$. The same applies with their roles reversed.

It remains to suppose that no nontrivial normal subgroup of $G$ lies in either maximum. Let $K$ be minimal normal, an elementary abelian p-group. Since $K$ lies in neither maximum, $KA=KB=G$. The intersection $K\cap A$ is normalized by $A$ and by the abelian group $K$, hence by $G$; it is therefore trivial. Likewise $K\cap B=1$. Thus both $A$ and $B$ complement $K$. Their action on $K$ is irreducible, since an invariant subspace would be a proper nontrivial normal subgroup of $G$. It is also faithful: $C_A(K)$ is normalized by $A$ and centralized by $K$, so is a normal subgroup of $G$ contained in $A$ and must be trivial.

If $G/K=1$, these assumptions force $A=B=1$ and $G$ to have prime order. Otherwise choose $L>K$ normal in $G$ with $L/K$ minimal normal in $G/K$. It is an elementary abelian r-group. Under $A\cong G/K$, its copy $R=A\cap L$ is a nontrivial normal subgroup of $A$. The prime $r$ differs from $p$: a p-group acting on the characteristic-p vector space $K$ has nonzero fixed points, by counting its orbits on $K$ modulo $p$. Its fixed subspace would be $A$-invariant, hence all of $K$ by irreducibility, contrary to faithfulness of $R$.

Therefore $A\cap L$ and $B\cap L$ are Sylow r-subgroups of $L=K(A\cap L)=K(B\cap L)$. They are conjugate in $L$. Conjugate $B$ so that these intersections coincide as $R$. Since $R⊲A$, $N_G(R)\ge A$. It cannot equal $G$, since $R$ would then be a nontrivial normal subgroup contained in $A$. Maximality gives $N_G(R)=A$, and the same argument gives $N_G(R)=B$. The conjugated maxima are equal. HenceThis is the factorization-or-conjugacy property of maximal subgroups of a finite soluble group.

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 $I$.