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For a finite-dimensional complex semisimple Lie algebra with a fixed triangular decomposition of a Lie algebra, category $\mathcal{O}$ consists of finitely generated universal enveloping algebra modules that are direct sums of Cartan weight spaces and are locally finite under the positive nilpotent subalgebra. Their weight spaces are finite-dimensional and their weights lie in a finite union of downward positive-root cones. Verma modules, finite-dimensional modules, submodules and quotients belong to this category. Closure under submodules uses the noetherian property of the enveloping algebra, obtained from the Poincaré-Birkhoff-Witt theorem. The category condition does not require the module to be finite-dimensional.

For the defining sl2 Lie algebra module with basis $x,y$, put $A_k=x\otimes x^{3-k}{y}^k$ and $B_k=y\otimes x^{3-k}{y}^k$. The raising operator has $e{A}_k=k{A}_{k-1}$ and $e{B}_k=A_k+k{B}_{k-1}$; the lowering operator has $f{A}_k=B_k+(3-k)A_{k+1}$ and $f{B}_k=(3-k)B_{k+1}$. The highest-weight vectors $A_0$ and $B_0-A_1$ generate irreducible submodules of highest weights four and two. Bases are respectively $(A_0,3A_1+B_0,A_2+B_1,A_3+3B_2,B_3)$ and $(B_0-A_1,B_1-A_2,B_2-A_3)$. Their vectors are independent in every weight space, proving the displayed direct sum without only comparing dimensions.

Over an algebraically closed field of characteristic zero, any two Cartan subalgebras of the symplectic Lie algebra are conjugate by the symplectic group. A Cartan subalgebra $\mathfrak{h}$ is an abelian Lie algebra and equals its centralizer of a Lie subalgebra. The symplectic centralizer decomposition therefore forces all its weight spaces in $V$ to be one-dimensional, nonzero, and paired by opposite weights of a representation. Choose a basis vector in each space, normalizing the perfect pairing with its opposite to one. These vectors form a symplectic basis $\mathcal{B}$. In this basis, $\mathfrak{h}$ is contained in the Lie subalgebra of diagonal matrices with entries $(t_1,\dots ,t_n,-t_1,\dots ,-t_n)$. All those diagonal matrices commute with $\mathfrak{h}$, so equality with its centralizer of a Lie subalgebra forces equality with the entire diagonal Lie subalgebra. Repeating for ${\mathfrak{h}}^{\prime }$ gives a second symplectic basis. The linear map sending one symplectic basis to the other belongs to the symplectic group and conjugates the two diagonal Lie subalgebras, proving the claim.

For a finite-dimensional complex sl2 Lie algebra module $V$, the number of simple summands, counted with multiplicity, is $\dim V_0+\dim V_1$. Indeed, the simple module of highest weight $m\ge 0$ has weights $m,m-2,\dots ,-m$, each once. Exactly one of zero and one appears: zero when $m$ is even, one when $m$ is odd. The classification of finite-dimensional sl2 representations decomposes $V$ into these modules, and dimensions of weight spaces add over the direct sum.

At a primitive cube root of unity, the cubic homogeneous module has basis $x^3,x^{2}y,x{y}^2,y^3$. Its only simple modules occurring as submodules are the one-dimensional subspaces of $⟨x^3,y^3⟩$, on which $K=1$ and $E=F=0$. Its quotient by this two-dimensional submodule is simple, but it has no simple complement: any lift of the highest $K$ weight space contains $x^{2}y$, whose $E$ image is $x^3\ne 0$. Thus the cubic module is not a semisimple module.

For a real inner product space $V$, the displayed map identifies ${\Lambda }^{2}V$ with its Special orthogonal Lie algebra. It sends a wedge of distinct orthonormal basis vectors to the corresponding elementary skew-symmetric matrix, so it is an isomorphism. For an orthogonal transformation $g$, invariance of the inner product gives $\Phi (gu\wedge gv)=g\Phi (u\wedge v)g^{-1}$. Thus the exterior square of the defining representation is the Adjoint representation of the orthogonal group. Complexification identifies their weight spaces, allowing the root system to be read from the exterior-square weights and the zero-weight multiplicity.

For a module with finite-dimensional weight spaces, its formal character is $\mathrm{ch}V={\sum }_{\mu }(\dim V_{\mu })e^{\mu }$ in the completed group algebra of the weight lattice.

Every finite-dimensional complex Lie algebra representation of a nilpotent Lie algebra decomposes into invariant generalized weight spaces. The nilpotence assumption is essential: a general solvable action can have extensions between distinct weights.

Here is a proof. For $h,k\in H$, nilpotence gives $(\mathrm{ad}h{)}^{N}k=0$, hence $(\mathrm{ad}\rho (h){)}^N\rho (k)=0$. By the nilpotent commutator preserves generalized eigenspaces lemma, every generalized eigenspace of $\rho (h)$ is invariant under every $\rho (k)$. Split successively by the acting operators of a basis of $H$. Each resulting block is invariant under $H$, and each basis operator has only one eigenvalue on it. The Lie theorem triangularizes the action on each block; its diagonal entries therefore give one and the same linear functional $\lambda$ on every basis vector of that block. Every $\rho (h)-\lambda (h)I$ is strictly upper triangular there, proving that the block is $V^{\lambda }$. Distinct blocks have distinct weights, so this is a direct sum. On the zero generalized weight space, every acting operator is a nilpotent endomorphism, so the Engel theorem supplies a nonzero common annihilated vector whenever that space is nonzero.

Use the standard split weight-space decomposition in characteristic zero, in particular the usual complex setting: $L={⨁}_{\phi }{L}_{\phi }$, with generalized weights for the nilpotent algebra $H$. The characteristic-zero condition prevents nonzero weight shifts from wrapping around; the unrestricted positive-characteristic version would need additional hypotheses.

For fixed $h\in H$, let $D=\mathrm{ad}h$. On a bracket of vectors from $L_{\phi },L_{\psi }$, the derivation identity givesIterating it gives the binomial expansion of these two commuting actions. At power $2n-1$, each term kills at least one factor, because generalized weight vectors are killed by the $n$th power of the corresponding shifted operator. Therefore the bracket lies in the generalized eigenspace for $(\phi +\psi )(h)$ for every $h$. Stabilization of kernels by dimension $n$ giveswhere $L_{\eta }=0$ when $\eta$ is not a weight.

Nilpotence of $H$ gives $H\subseteq L_0$. If $H$ is a Cartan subalgebra and $L_0/H\ne 0$, the action of $H$ on this quotient consists of nilpotent operators. Engel's theorem supplies a nonzero coset $v+H$ killed by $H$, giving $[H,v]\subseteq H$ with $v\notin H$, a contradiction to self-normalization. Conversely, if $L_0=H$ and $v$ normalizes $H$, then $[h,v]\in H$ for each $h$. Applying further powers of $\mathrm{ad}h$ kills this vector because its restriction to $H$ is nilpotent. Hence $v\in L_0=H$. This proves the equivalence.

The operator $\mathrm{ad}e$ sends $L_{\phi }$ into $L_{\phi +\alpha }$. Repetition follows a consecutive $\alpha$-string; after $r$ steps it must leave the weight set, because $r$ is the largest number of weights in any such string. Thus $\boxed{(\mathrm{ad}e{)}^r=0}$ on every weight space and hence on $L$.

Finally put $E=\mathrm{ad}e$ and $F=\mathrm{ad}f$. Expand $(E+F{)}^n$ as the sum of all words of length $n$, without commuting $E$ and $F$. Acting on $L_{\gamma }$, each word follows weights $\gamma +i\alpha +j\beta$ with $i+j$ increasing by one at each step. Linear independence of $\alpha ,\beta$ in characteristic zero makes the successive weights distinct. A nonzero length-$n$ path would require $n+1$ different weight spaces, but $\dim L=n$. Every word therefore vanishes, giving

We prove Lie's theorem in the required form by induction on $\dim L$. If $L=0$, any nonzero vector works. Otherwise solvability gives $[L,L]\ne L$, so choose a codimension-one ideal $K\supseteq [L,L]$ and $x\in L\setminus K$. By induction there are $v\ne 0$ and a linear functional $\lambda :K\to \mathbb{C}$ with $av=\lambda (a)v$ for all $a\in K$.

Let $W=\mathrm{span}\{v,xv,x^{2}v,\dots \}$. Choose the cyclic basis $v,xv,\dots ,x^{r-1}v$ up to its first linear dependence. The commutator identity $a{x}^j=xa{x}^{j-1}+[a,x]x^{j-1}$, with $[a,x]\in K$, proves inductively thatHence $W$ is invariant under $K$ and $x$, and every $a\in K$ has trace $r\lambda (a)$ on $W$. In particular ${\mathrm{tr}}_W([x,a])=0$ implies $r\lambda ([x,a])=0$. Characteristic zero gives $\lambda ([x,a])=0$.

The simultaneous weight space $V_{\lambda }=\{w:aw=\lambda (a)w\forall a\in K\}$ is nonzero and stable under $x$, becauseOver $\mathbb{C}$, the restriction of $x$ to this nonzero finite-dimensional space has an eigenvector. That vector is an eigenvector of both $x$ and every $a\in K$, hence of every element of $L$.

We use the supplied weight decomposition $V={⨁}_{\lambda \in \Psi (\mathfrak{a})}{V}_{\lambda }$. There is a small distinction to justify: the definition of a toral subalgebra uses the Adjoint representation, while these weight spaces use the defining representation. For the symplectic Lie algebra, adjoint semisimplicity in the symplectic Lie algebra implies that each $a\in \mathfrak{a}$ is diagonalizable on $V$ as well. Here is a direct verification.

Define the symplectic adjoint $a^{\ast }$ by $\omega (av,w)=\omega (v,a^{\ast }w)$. The symplectic Lie algebra condition is $a^{\ast }=-a$. Write the Jordan–Chevalley decomposition in $\mathrm{End}(V)$ as $a=S+N$, with $S$ diagonalizable, $N$ a nilpotent linear map, and $SN=NS$. Uniqueness of the Jordan–Chevalley decomposition, applied to $a^{\ast }=-a$, gives $S^{\ast }=-S$ and $N^{\ast }=-N$. Thus $S,N\in \mathfrak{sp}(V)$. On $\mathrm{End}(V)$, $\mathrm{ad}S$ is diagonalizable by using a matrix unit basis in an eigenbasis of $S$, and $\mathrm{ad}N$ is a nilpotent linear map, sincevanishes for large $m$. These operators commute and preserve $\mathfrak{sp}(V)$, so their restrictions are the Jordan–Chevalley decomposition of $\mathrm{ad}a$. Because $\mathrm{ad}a$ is diagonalizable, its nilpotent part is zero: $\mathrm{ad}N=0$ on $\mathfrak{sp}(V)$.

To conclude that $N=0$ without an assumption on the center of a Lie algebra, observe that each map $T_v(w)=\omega (v,w)v$ lies in $\mathfrak{sp}(V)$. If $v\ne 0$, its image is $kv$, because $\omega$ is a nondegenerate bilinear form. Since $N$ commutes with $T_v$, it preserves $kv$. Hence $Nv\in kv$ for every $v$, forcing $N$ to be a scalar linear map: compare its action on two linearly independent vectors and their sum. A scalar nilpotent linear map is zero. For $V=0$ the assertion is immediate. This proves the needed diagonalizability. Commuting diagonalizable linear maps admit simultaneous diagonalization, giving the displayed weight decomposition.

Now let $v\in V_{\lambda }$ and $w\in V_{\mu }$. For every $a\in \mathfrak{a}$, preservation of the alternating bilinear form givesIf $\lambda +\mu \ne 0$ as a linear functional, choose $a$ on which it is nonzero; then $\omega (v,w)=0$. ThereforeFor $v\ne 0$ in $V_{\lambda }$, the fact that $\omega$ is a nondegenerate bilinear form supplies a $w\in V$ with $\omega (v,w)\ne 0$. Decomposing $w$ into weight spaces, only its $V_{-\lambda }$ component can contribute. Thus $V_{-\lambda }\ne 0$, and $\Psi (\mathfrak{a})=-\Psi (\mathfrak{a})$. The same argument on both sides shows that $\omega :V_{\lambda }\times V_{-\lambda }\to k$ is a perfect pairing, so $\dim V_{\lambda }=\dim V_{-\lambda }$. On $V_0$, the restriction of $\omega$ is a nondegenerate bilinear form that is an alternating bilinear form, so $\dim V_0$ is even.

Choose one representative ${\lambda }_i$ from each pair of nonzero weights of a representation, for $1\le i\le r$. HereThe linear map $\mathfrak{a}\to k^r$ sending $a$ to $({\lambda }_1(a),\dots ,{\lambda }_r(a))$ is injective: an element in its kernel acts by zero on every nonzero weight space, and also on $V_0$, so it is the zero element of $\mathfrak{a}\subseteq \mathrm{End}(V)$. Taking dimensions yieldsThe final inequality follows because each pair contributes at least two to $\dim V=2n$. Equivalently, the nonzero weights of a representation span the dual space ${\mathfrak{a}}^{\ast }$; they need not be linearly independent at this stage.

Choose representatives ${\lambda }_1,\dots ,{\lambda }_r$ as in part (b). We use the symplectic centralizer decompositionTo prove it, a linear map commuting with every $a\in \mathfrak{a}$ must preserve every weight space: if $v\in V_{\lambda }$, then $a(Tv)=T(av)=\lambda (a)Tv$. Conversely, preserving the weight spaces makes a linear map commute with the scalar action of every $a$ on each space.

On $V_0$, the symplectic Lie algebra condition says exactly that the restriction belongs to $\mathfrak{sp}(V_0)$. On $V_{{\lambda }_i}\oplus V_{-{\lambda }_i}$, use the perfect pairing from part (b) to identify $V_{-{\lambda }_i}$ with the dual space $V_{{\lambda }_i}^{\ast }$. The action $A_i$ on $V_{{\lambda }_i}$ can be arbitrary, but its action on $V_{-{\lambda }_i}$ must be $-A_i^{\ast }$, where $A_i^{\ast }$ is the dual map. Indeed,determines $B_i=-A_i^{\ast }$ uniquely. There are no additional conditions between different pairs because $\omega$ vanishes between different pairs. Every choice of these blocks therefore gives an element of the centralizer of a Lie subalgebra. The block correspondence preserves Lie brackets, proving the claimed direct sum of Lie algebras.

The general linear Lie algebra ${\mathfrak{gl}}_m(k)$ is an abelian Lie algebra exactly when $m\le 1$: for $m\ge 2$, the matrix units satisfy $[E_{11},E_{12}]=E_{12}\ne 0$. A nonzero symplectic vector space has a symplectic basis and contains a two-dimensional symplectic subspace. The symplectic Lie algebra acting on that plane, extended by zero on its symplectic orthogonal complement, contains the matrices $H=\mathrm{diag}(1,-1)$ and $E=E_{12}$ with $[H,E]=2E\ne 0$. Thus $\mathfrak{sp}(V_0)$ is an abelian Lie algebra exactly when $V_0=0$.

Since different summands in a direct sum of Lie algebras commute, we obtain the precise criterionEquivalently, there are $2n$ distinct nonzero weights of a representation, each with a one-dimensional weight space. The condition concerns multiplicities in the defining representation; it does not assert that $\dim \mathfrak{a}=n$ for every such $\mathfrak{a}$.

Use the standard structural fact that $\mathfrak{sp}(V)$ is a semisimple Lie algebra over an algebraically closed field of characteristic zero. If $\mathfrak{h}$ is a Cartan subalgebra, part (a)(ii) gives $C_{\mathfrak{sp}(V)}(\mathfrak{h})=\mathfrak{h}$, and $\mathfrak{h}$ is an abelian Lie algebra. Part (c) therefore shows that its weight spaces occur as $n$ pairs of one-dimensional spaces, with no zero weight space.

Choose representatives ${\lambda }_1,\dots ,{\lambda }_n$ of its nonzero weights of a representation. Pick $p_i\ne 0$ in $V_{{\lambda }_i}$ and $q_i\ne 0$ in $V_{-{\lambda }_i}$, scaling $q_i$ so that $\omega (p_i,q_i)=1$. Part (b) shows that all other pairings among different pairs are zero. Each individual weight space is one-dimensional and the bilinear form is alternating, so the ordered basisis a symplectic basis. Every $h\in \mathfrak{h}$ is a diagonal matrix in this basis, with entries ${\lambda }_1(h),\dots ,{\lambda }_n(h),-{\lambda }_1(h),\dots ,-{\lambda }_n(h)$.

Let $D_{\mathcal{B}}$ be the Lie subalgebra consisting of all diagonal matricesEach such matrix lies in the symplectic Lie algebra and commutes with $\mathfrak{h}$. Henceand $\mathfrak{h}=D_{\mathcal{B}}$. In particular, every Cartan subalgebra has dimension $n$; the ${\lambda }_i$ are a basis of ${\mathfrak{h}}^{\ast }$.

Carry out the same construction for ${\mathfrak{h}}^{\prime }$, obtaining a symplectic basis ${\mathcal{B}}^{\prime }=(p_1^{\prime },\dots ,p_n^{\prime },q_1^{\prime },\dots ,q_n^{\prime })$ with ${\mathfrak{h}}^{\prime }=D_{{\mathcal{B}}^{\prime }}$. The unique linear map $g$ sending $p_i$ to $p_i^{\prime }$ and $q_i$ to $q_i^{\prime }$ is invertible and preserves all pairings of basis vectors. By bilinearity, it preserves $\omega$ on all of $V$, so $g$ belongs to the symplectic group $\mathrm{Sp}(V)$. The change of basis givesThus conjugacy of Cartan subalgebras of the symplectic Lie algebra follows by explicitly matching their symplectic bases. If $n=0$, the Lie algebra is zero and the same conclusion holds with $g$ the identity on the zero vector space.

Every $2\times 2$ matrix has the unique decompositionBoth summands define invariant subspaces for the Adjoint representation of a Lie algebra. Scalar matrices commute with everything, so $\mathbb{C}I$ is $L_0$. The traceless matrices form ${\mathfrak{sl}}_2$ with basis $e,h,f$ of weights $2,0,-2$. Their brackets satisfy $[e,f]=h$, $[h,e]=2e$, $[h,f]=-2f$; the raising and lowering operations connect all three weight lines. This is the simple module $L_2$. ThereforeAs a check, its zero-weight space has dimension two and its one-weight space is zero, agreeing with part (ii).

In the simple module $L_m$, weight $0$ occurs once if $m$ is even and never if $m$ is odd. Weight $1$ occurs once if $m$ is odd and never if $m$ is even. Thus each simple summand contributes exactly one to $\dim V_0+\dim V_1$.

Complete reducibility of semisimple Lie algebra representations writes $V={⨁}_m{L}_m^{\oplus a_m}$. Taking the two weight spaces commutes with this finite direct sum, soThe right side counts irreducible summands with multiplicity, independently of a chosen decomposition.

Write ${\mathfrak{g}}_{\mathbb{C}}=L(G_{\mathbb{C}})$ and $\mathfrak{h}=L(H)$. A root is a nonzero weight $\alpha \in {\mathfrak{h}}^{\ast }$ of the adjoint action. Its root space isAt the group level these weights come from nontrivial characters of the Cartan subgroup acting by conjugation. The Cartan root-space decomposition isThe zero-weight space is $\mathfrak{h}$; for the complexification of a compact group the algebra is reductive, with any centre contained in this zero-weight space.

For the product in question,Take the diagonal Cartan subalgebra generated by $H_1=(\mathrm{diag}(1,-1),0)$ and $H_2=(0,\mathrm{diag}(1,-1))$. On $Z=x{H}_1+y{H}_2$, define ${\alpha }_1(Z)=2x$ and ${\alpha }_2(Z)=2y$. The roots are exactlyIf $E=E_{12}$ and $F=E_{21}$, their four root spaces are respectively $\mathbb{C}(E,0)$, $\mathbb{C}(F,0)$, $\mathbb{C}(0,E)$, and $\mathbb{C}(0,F)$. These four lines and the two-dimensional diagonal subalgebra form the full six-dimensional decomposition. There are no mixed roots: the two direct summands commute. This is the root system $A_1\times A_1$.

A maximal torus in a compact connected Lie group is a compact connected abelian Lie subgroup maximal among such subgroups. By the connected abelian classification it is isomorphic to a product of circles. Such subgroups exist: choose one of largest dimension among compact connected abelian subgroups; a proper inclusion of connected Lie subgroups would increase dimension. Write $\mathfrak{t}$ for its Lie algebra.

Here are the main steps of the conjugacy assertion, with the critical arguments supplied. Average an inner product on $\mathfrak{g}$ over the adjoint action to obtain an $\mathrm{Ad}(G)$-invariant positive definite inner product. It defines a bi-invariant Riemannian metric. Compactness makes this metric complete, and the Hopf-Rinow theorem joins the identity to any $g$ by a geodesic. For a bi-invariant metric the Levi-Civita connection on left-invariant fields is ${\nabla }_{X}Y=[X,Y]/2$, so the geodesic starting with $X$ is $\exp (tX)$. Consequently every $g$ has the form $\exp X$.

The Lie-algebra centralizer of $\mathfrak{t}$ is exactly $\mathfrak{t}$. If $X$ commutes with $\mathfrak{t}$, its one-parameter subgroup commutes with $T$. The closure of the subgroup generated by $T$ and $\exp (\mathbb{R}X)$ is compact, connected and abelian, so it is a torus; maximality forces it to equal $T$, giving $X\in \mathfrak{t}$. The torus action on ${\mathfrak{g}}_{\mathbb{C}}$ decomposes into finitely many character weight spaces. The zero-weight space is ${\mathfrak{t}}_{\mathbb{C}}$. Choose $H\in \mathfrak{t}$ outside the finitely many kernels of the nonzero weight differentials. Then the centralizer of $H$ is exactly $\mathfrak{t}$.

For an arbitrary $X\in \mathfrak{g}$, the continuous function $g\mapsto ⟨{\mathrm{Ad}}_{g}X,H⟩$ attains a maximum. Put $Y={\mathrm{Ad}}_{g_0}X$ at a maximizer. Differentiating along $\exp (sZ)g_0$ givesHence $[Y,H]=0$, so $Y\in \mathfrak{t}$. Exponentiating gives $g_0(\exp X)g_0^{-1}=\exp Y\in T$. This adjoint-orbit criterion for maximal-torus conjugacy proves

Restriction of representations is a ring homomorphism $\mathrm{Res}:R(G)\to R(T)$. If a virtual representation restricts to zero, its character vanishes on $T$. Since characters are class functions and every element is conjugate into $T$, it vanishes on all of $G$. The injectivity of the character map proved in Question 3 then givesThe Weyl group is $W=N_G(T)/T$. For $n\in N_G(T)$, the operator $\rho (n)$ intertwines the restriction of $\rho$ with its twist by $t\mapsto nt{n}^{-1}$. Therefore every restricted representation, and every virtual difference of them, is fixed by $W$:

For the special orthogonal group $G=SO(2m+1)$, takewhere $R_{\theta }$ is a real planar rotation. It is a maximal torus: its Lie-algebra centralizer consists of the independent infinitesimal rotations in those planes. Its character lattice is ${\mathbb{Z}}^m$, so, writing $z_i=e^{i{\theta }_i}$,Indeed commuting unitary torus matrices diagonalize simultaneously into one-dimensional characters, and the characters of the circle are exactly the integral powers.

The normalizer permutes the real rotation planes and can reverse each angle independently. Permutations of whole two-dimensional blocks have determinant one. A reflection in one rotation plane reverses its angle and has determinant minus one; combining it with reflection of the remaining fixed line gives an element of $SO(2m+1)$. Thus all independent inversions occur. Conversely a normalizer must permute the distinct nontrivial torus character pairs $\{\pm e_i\}$ and preserve the fixed line, so there are no additional actions. The kernel of this action is exactly $T$, giving

It follows that the restriction identifies $R(SO(2m+1))$ with a subring of the signed-permutation-invariant Laurent polynomials. In this case we can prove equality. Invariance under individual inversions first givesFor one variable, every invariant is an integral combination of $1$ and $z^k+z^{-k}$, and these satisfy the recurrence $s_{k+1}=x{s}_k-s_{k-1}$. Applying this one coordinate at a time proves the assertion over the other Laurent coefficient rings. Permutation invariance and the Fundamental theorem of symmetric polynomials then givewhere $e_k$ are the elementary symmetric polynomials.

Let $V$ be the complexification of the defining $(2m+1)$-dimensional real representation. Its torus weights are $1,z_i,z_i^{-1}$. The characters of its exterior powers obeyFor $k\le m$, choosing $j$ linear terms and $r$ quadratic terms shows that the coefficient $c_k={\chi }_{{\Lambda }^{k}V}$ isIts leading term is $e_k$ with coefficient one; all other terms use $e_j$ with $j<k$. By induction $e_k$ lies in the restriction image, because $c_k$ is the character of a genuine representation and all smaller $e_j$ already lie in the image. Hence every invariant polynomial lies there, proving the representation ring of an odd-dimensional special orthogonal group descriptionThe triangular change of generators is integral, so $[{\Lambda }^{1}V],\dots ,[{\Lambda }^{m}V]$ are algebraically independent polynomial generators. For $m=1$, $c_1=1+z+z^{-1}$ and this reduces to $R(SO(3))=\mathbb{Z}[V_3]$, with the three-dimensional defining representation as generator. These are genuine representations of the orthogonal group, not the half-integral spin weights of its simply connected double cover.

Work over $\mathbb{C}$ with a finite-dimensional semisimple Lie algebra $\mathfrak{g}$. Choose a Cartan subalgebra $\mathfrak{h}$, its root system $R\subset {\mathfrak{h}}^{\ast }$, a positive system $R^{+}$ and its simple roots. Let $W$ be the Weyl group, generated by the corresponding root reflections, and put $\rho =\frac{1}{2}{\sum }_{\alpha \in R^{+}}\alpha$, the Weyl vector. The finite-dimensional irreducible highest-weight representation $L(\lambda )$ is indexed by a dominant integral weight $\lambda$.

If $P$ is the weight lattice, its group ring has formal symbols $e^{\mu }$ with multiplication $e^{\mu }{e}^{\nu }=e^{\mu +\nu }$. The formal character iswhere $L(\lambda {)}_{\mu }=\{v:Hv=\mu (H)v\text{ for all }H\in \mathfrak{h}\}$ is a weight space. For a weight $\nu$, define the Weyl alternant $A_{\nu }={\sum }_{w\in W}\det (w)e^{w\nu }$; the sign $\det (w)=(-1{)}^{\ell (w)}$ is the determinant of $w$ on the real root span, equivalently the parity of its expression in simple reflections.

The Weyl character formula isThe second denominator is the Weyl denominator formula. The quotient can initially be taken in the fraction field of the group ring; the formula says it is the finite character on the left, so its apparent singularities are removable. The shifts by $\rho$ make $\lambda +\rho$ strictly dominant.

Let $V={\mathbb{C}}^n$ and $m\ge 0$. Identify the symmetric power ${\mathrm{Sym}}^{m}V$ with the degree-$m$ part of the symmetric algebra on the basis vectors $e_1,\dots ,e_n$, using formal variables $x_i$ for these vectors. Its monomial basis isThe diagonal maximal torus of the unitary group acts on this monomial by the character $z_1^{a_1}\cdots z_n^{a_n}$. Distinct tuples give distinct characters, so each weight space is one-dimensional.

Suppose $M$ is a nonzero invariant complex subspace. For a nonzero vector $w\in M$, select a monomial with nonzero coefficient and extract it by torus averaging:Fourier orthogonality makes this a nonzero scalar multiple of $x^a$. It belongs to $M$, so $M$ contains a monomial.

Differentiating the unitary group action makes $M$ invariant under $\mathfrak{u}(n)$ and hence under its complexification ${\mathfrak{gl}}_n(\mathbb{C})$. The matrix unit $E_{ij}$ replaces an occurrence of $e_j$ by $e_i$, so its action on the symmetric algebra isWhenever $a_j>0$, this transfers one unit of degree from coordinate $j$ to coordinate $i$ with a nonzero coefficient. Starting with any monomial, such transfers can produce every other tuple of nonnegative integers summing to $m$. Thus $M$ contains every monomial and equals ${\mathrm{Sym}}^{m}V$.

ConsequentlyFor $m=0$ this is the one-dimensional trivial representation, and for $n=1$ every symmetric power is also one-dimensional. This weight-transfer proof of irreducibility of symmetric powers uses neither the Weyl character formula nor the Weyl dimension formula, so no unproved dimension formula enters the argument.

Write the diagonal maximal torus of $U(n)$ as $t=\mathrm{diag}(z_1,\dots ,z_n)$, $|z_i|=1$, and let $\delta =(n-1,n-2,\dots ,0)$. Irreducible complex representations are indexed by dominant integer tuples ${\lambda }_1\ge \cdots \ge {\lambda }_n$, their highest weights. The Weyl character formula in integer-exponent form isThe ratio initially uses distinct $z_i$, and extends to repeated eigenvalues by continuity. Every unitary matrix is conjugate to such a diagonal matrix, so this determines the character on the entire group. The shift $\delta$ differs from the usual Weyl vector $\rho =((n-1)/2,(n-3)/2,\dots ,(1-n)/2)$ by a multiple of $(1,\dots ,1)$; the corresponding determinant factors cancel between numerator and denominator.

We first justify the highest-weight information needed in the proof. Restriction to the torus splits a representation into integer weight spaces. In an irreducible representation choose a lexicographically largest weight $\lambda$ and a nonzero vector $v$ of that weight. A matrix unit $E_{ij}$ with $i<j$ raises the weight by $e_i-e_j$, so it annihilates $v$. The character is invariant under permutations of diagonal coordinates, because permutation matrices conjugate the torus. Hence permutations of $\lambda$ are also weights, and lexical maximality forces ${\lambda }_1\ge \cdots \ge {\lambda }_n$.

Irreducibility implies that $v$ generates the representation under the complexified Lie algebra. Indeed, a Lie-algebra invariant subspace is preserved by exponentials of skew-Hermitian matrices, which generate $U(n)$. Reorder any product of matrix units into lower-triangular units, diagonal units and upper-triangular units, usingInduction on length and on the number of out-of-order pairs gives the required spanning version of the Poincaré-Birkhoff-Witt theorem. Upper-triangular units kill $v$, and diagonal units act by scalars. Every nonempty product of lower-triangular units lowers its weight by a nonzero sum of positive roots. Therefore the top weight space is exactly $\mathbb{C}v$, and its multiplicity is one. This supplies the leading coefficient without assuming a character formula.

For completeness, every dominant integer tuple occurs. Put $d_j={\lambda }_j-{\lambda }_{j+1}\ge 0$ and form the representationIts lexicographically highest weight is $\lambda$, with one-dimensional highest space, represented by the tensor products of $e_1\wedge \cdots \wedge e_j$. A negative determinant exponent is a legitimate one-dimensional unitary-group character. Averaging an inner product over normalized Haar measure gives invariant orthogonal complements, so the tensor representation splits into irreducible summands. One of them has highest weight $\lambda$. Uniqueness will also follow from the formula and Schur orthogonality relations.

Define the alternant $A_{\kappa }(z)=\det (z_i^{{\kappa }_j})$ for a strictly decreasing integer tuple $\kappa$. The denominatoris the Vandermonde determinant, and its modulus is the denominator modulus in the Weyl integration formula. The Laurent polynomial ${\chi }_{\lambda }{A}_{\delta }$ is alternating. Any alternating Laurent polynomial decomposes as a finite summonomials with repeated exponents have zero coefficient, and the remaining permutation orbits are indexed uniquely by strictly decreasing tuples.

The lexicographically highest monomial in the product is $z^{\lambda +\delta }$ with coefficient one. Indeed, the highest monomial of the character is $z^{\lambda }$ with coefficient one, and the unique lexical maximum among permutations of $\delta$ is $\delta$ itself. It follows that $c_{\lambda +\delta }=1$.

Fourier orthogonality on the torus, whose probability measure is ${\prod }_{i}d{\theta }_i/(2\pi )$, gives alternant orthogonality on a unitary torus:Expand both determinants: an integral survives exactly when the two exponent tuples are identical up to permutation. Strict decreasing order then forces $\kappa =\eta$, and its $n!$ matching terms each contribute one.

Finally, Schur orthogonality relations give ${\int }_{U(n)}|{\chi }_{\lambda }{|}^2=1$. Applying the allowed Weyl integration formula to this class function givesSince the coefficient $c_{\lambda +\delta }$ already equals one, every other coefficient vanishes. Thus ${\chi }_{\lambda }{A}_{\delta }=A_{\lambda +\delta }$, proving the formula. If two irreducible representations had the same highest weight, their characters would now coincide; the Schur orthogonality relations imply that they are isomorphic. Apparent singularities at repeated diagonal eigenvalues are removable because the quotient equals the actual finite Laurent-polynomial character. No dimension formula has been assumed.