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The Kostant partition function $P(\nu )$ counts the ways to express a weight $\nu$ as a nonnegative integer combination of positive roots. By the Poincaré-Birkhoff-Witt theorem, $P(\lambda -\mu )$ is the multiplicity of the weight $\mu$ in the Verma module $M(\lambda )$.

Choose a basis $x_1,x_2,x_3$ of the Lie algebra and filter its universal enveloping algebra by word length: $F_{0}R=\mathbb{C}$ and $F_{d}R$ is spanned by products of at most $d$ basis elements. The relations $x_j{x}_i-x_i{x}_j=[x_j,x_i]$ replace a degree-two commutator by degree one, so the degree-one symbols commute.

The Poincaré-Birkhoff-Witt theorem says the ordered monomials $x_1^{a_1}{x}_2^{a_2}{x}_3^{a_3}$ are a basis. Its relevant proof can be seen by reordering an inverted neighboring pair using the displayed relation. Reordering terminates because same-length terms have fewer inversions and bracket terms have smaller word length. Disjoint swaps commute; the only overlapping three-letter reorderings differ by $[x_i,[x_j,x_k]]+[x_j,[x_k,x_i]]+[x_k,[x_i,x_j]]=0$. The Jacobi identity therefore makes the reduced ordered expression independent of the reordering path. Ordered monomials have no remaining relations and form the basis. Their degree-$d$ symbols giveFor generators $m_1,\dots ,m_q$ of the finitely generated module $M$, put $M_d={\sum }_j{F}_{d}R{m}_j$. This is a good filtration, and $\mathrm{gr}M$ is finite over the three-variable polynomial ring. Define the dimension of a filtered module as the Krull dimension of its associated graded module, equivalently the degree of the eventual cumulative polynomial ${\dim }_{\mathbb{C}}{M}_d$, or the Gelfand–Kirillov dimension of a module. Two good filtrations have bounded shifts, so the growth degree is independent of the generators. Sinceits polynomial degree is at most three. Therefore every nonzero finitely generated $R$-module has dimension at most $3$. The zero module may be assigned the usual separate empty-support convention. This bound also follows immediately from its support of a module inside affine three-space.

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.

Let $v_0$ be the highest vector of the Verma module $M(n)$ and set $v_k=f^k{v}_0$. The Poincaré-Birkhoff-Witt theorem implies that $(v_k{)}_{k\ge 0}$ is a basis, so none of these vectors vanishes. The actions areThe raising operator is a locally nilpotent operator: on every vector, repeated application of $e$ eventually removes all its finitely many basis terms. Thus its algebraic exponential of a locally nilpotent operator is defined, and $\exp (e)v_0=v_0$. But the middle factor in the proposed product would have to send this vector towhich has infinitely many nonzero, linearly independent basis coefficients. An element of the algebraic Verma module is a finite linear combination of the $v_k$, so this expression is not in $M(n)$. In particular the lowering operator is not locally nilpotent.

Even for integral $n\ge 0$, $f^{n+1}{v}_0$ is nonzero in the Verma module; it becomes zero only in its finite-dimensional irreducible quotient $L(n)$. Hence $s$ is not defined as the displayed composition of endomorphisms of the algebraic Verma module. Working in a completion would be a different problem and is not part of the module specified here.

Put $U_0=G$ and $U_{j+1}=U_j^p$, where powers mean closed generated subgroups. The finite-quotient arguments of Question 3 pass to a powerful pro-p group: in particular $U_j$ is powerfully embedded, the power map is onto $U_{j+1}$, andFor the root assertion, apply the finite result to every finite quotient and use the finite intersection property on the closed sets of possible roots in the compact group. Each $U_j$ is finitely generated by the appropriate powers of a finite generating tuple, and $U_j/U_{j+1}$ is finite and elementary abelian. Thus the $U_j$ are open. They form an identity-neighbourhood basis: in any finite continuous quotient their images form a terminating lower p-series, so some $U_j$ is contained in the kernel. In particular ${\bigcap }_j{U}_j=1$. Collection also givesOne can see the extra index directly by induction from $[U_a,G]\subseteq U_{a+1}$: collect $[v,w^p]$ for $w\in U_b$. Its leading factor $[v,w{]}^p$ advances one layer, and the remaining iterated commutators, with their binomial coefficients, lie at least as deep. Products and limits preserve the inclusion. This is the usual strongly central property of the lower p-series.

A uniform pro-p group is a finitely generated powerful pro-p group for which all the indices $|U_j:U_{j+1}|$ equal $|G:G^p|$. The preliminary congruence gives surjective linear mapsIf $G$ is uniform, these maps are bijective. If $x\ne y$, write $y=xu$ and choose the unique layer $j$ with $u\in U_j\setminus U_{j+1}$. Bijectivity of ${\theta }_j$ and the power congruence imply $y^p(x^p{)}^{-1}\in U_{j+1}\setminus U_{j+2}$, so $x^p\ne y^p$. Thus the power map is injective. A finite-order element of a pro-p group has $p$-power order; repeated injectivity of the power map excludes every nonidentity such element. Hence a uniform group is a torsion-free group.

Conversely, suppose some ${\theta }_j$ has a nonzero kernel. Choose $x\in U_j\setminus U_{j+1}$ with $x^p\in U_{j+2}$. At step $r\ge j+1$, if the current $x_r$ satisfies $x_r^p\in U_{r+1}$, surjectivity of ${\theta }_r$ supplies $u_r\in U_r$ withSet $x_{r+1}=x_r{u}_r$. The power congruence gives $x_{r+1}^p\in U_{r+2}$. Since the corrections tend to $1$ in the complete profinite group, the $x_r$ converge to $z$, with $z^p=1$ and $z\equiv x(\mathrm{mod}{U}_{j+1})$. In particular $z\ne 1$. A torsion-free group cannot have such an element, so every ${\theta }_j$ is injective and all layer indices are equal. This proves the root correction criterion for uniformity andFor a uniform group the power maps are continuous bijections $U_j\to U_{j+1}$, hence homeomorphisms. Their inverses define the unique roots used below. Comparison at the first differing layer also proves the power congruences in a uniform pro-p group:

The intrinsic Lie algebra of a uniform pro-p group has the same underlying set as $G$, zero equal to the group identity, and operationsHere $x^{\lambda }$ is the limit of integral powers approaching $\lambda$. The first root is taken in $U_n$ and the second in $U_{2n}$; the commutator actually lies in $U_{2n+1}$. The following faithful algebra realization verifies convergence and the Lie algebra axioms, rather than assuming them as part of the definition.

Choose a minimal tuple $a_1,\dots ,a_d$ of topological generators. The layer isomorphisms ${\theta }_j$ show that the images of $a_i^{p^j}$ are bases of $U_j/U_{j+1}$. The ordered-coordinate argument of Question 3 therefore gives a continuous bijectionExistence follows by successive digits and compactness; uniqueness follows by the first digit at which the coordinates differ, because its images are linearly independent in the corresponding power layer. Thus this map is a homeomorphism.

Here are details of the norm needed on the group algebra. In ${\mathbb{Z}}_p[[G]]$ put $b_i=a_i-1$. Its elements have unique ordered expansions ${\sum }_{\alpha }{c}_{\alpha }{b}_1^{{\alpha }_1}\cdots b_d^{{\alpha }_d}$ with $c_{\alpha }\in {\mathbb{Z}}_p$. To verify this description, identify the ordered coordinates of $G$ with ${\mathbb{Z}}_p^d$. Elements of the completed group algebra are integral measures on this space; their values on the binomial polynomials ${\prod }_i\left(\genfrac{}{}{0ex}{}{{\lambda }_i}{{\alpha }_i}\right)$ are precisely the coefficients $c_{\alpha }$. The finite-difference identity pairs the ordered monomial $b^{\beta }$ with this binomial polynomial as ${\delta }_{\alpha \beta }$. The Mahler expansion of a continuous function therefore proves both existence and uniqueness of the measure with the specified coefficients. This also proves that the ordinary rational group algebra embeds in these expansions.

Choose $1/(p-1)<s<1$ and setThis defines the weighted group algebra of a uniform pro-p group. Its multiplication is submultiplicative in the valuation convention, $v(AB)\ge v(A)+v(B)$. Indeed, $a_i^p-1=p{b}_i+\left(\genfrac{}{}{0ex}{}{p}{2}\right)b_i^2+\cdots +b_i^p$ has valuation $1+s$; collection of ordered coordinates gives the same lower bound for $g-1$ when $g\in U_1$. Interchanging two adjacent $b_i$ in a product introduces a correction coming from $[a_i,a_j]-1$, of valuation at least $1+s>2s$. Iterating the interchange rule never lowers the total weight; the corrections have strictly increasing weight, so infinite collection converges. The same argument with $p^j$-th powers showsThe equality follows from the first nonzero coordinate digit: its leading terms are $p^j{\lambda }_i^{\prime }{b}_i$, at least one with unit ${\lambda }_i^{\prime }$, and cannot cancel in the ordered expansion. Completing the rational group algebra in this norm produces a complete normed algebra $A$ in which $G$ embeds faithfully and with its original topology.

Since $v(g-1)\ge s>1/(p-1)$, the seriesconverge on the relevant balls in $A$ and are inverse there. Their convergence follows from $v_p(r!)\le (r-1)/(p-1)$ and $v_p(r)\le (r-1)/(p-1)$; formal one-variable identities prove that they are inverse. Put $X=\log x$ and $Y=\log y$. The Baker--Campbell--Hausdorff formula in $A$ now givesbecause every term of degree at least two tends to zero after this rescaling. Similarly expansion of $\exp (-p^{n}X)\exp (-p^{n}Y)\exp (p^{n}X)\exp (p^{n}Y)$ givesThe leading term is the indicated associative-algebra commutator and all terms of degree at least three tend to zero. These expressions are the logarithms of the unique roots in the intrinsic formulas. The image $\log G$ is compact and hence closed in $A$, so both limits belong to $\log G$, and the inverse logarithm takes the limits back to $G$. Also $\log (x^{\lambda })=\lambda \log x$. Consequently $\log$ identifies the intrinsic operations with addition, scalar multiplication and commutation in $A$. The Jacobi identity follows by expanding the associative products; thus $L_G$ is indeed a Lie algebra over ${\mathbb{Z}}_p$.

The commutator roots in the intrinsic formula belong to $U_1$, so their limit gives $[L_G,L_G]\subseteq \log U_1=p{L}_G$. Iterating the power congruence gives $x^{p^n}{y}^{p^n}\equiv (xy{)}^{p^n}(\mathrm{mod}{U}_{n+1})$. Taking the unique roots shows that intrinsic addition modulo $U_1$ is the original group product modulo $U_1$. Consequently $L_G/p{L}_G$ is the original elementary abelian quotient $G/U_1$, and the images of the $a_i$ form a basis. Successively removing digits expresses every element uniquely as an additive sum ${\sum }_i{\lambda }_i{a}_i$: existence uses completeness, while a nonzero relation would have a first nonzero digit contradicting the basis in $U_j/U_{j+1}$. Hence $L_G$ is a free module of rank $d$ over the p-adic integers and is a powerful Lie lattice.

Conversely a powerful Lie algebra over p-adic integers, for odd $p$, is a finite free ${\mathbb{Z}}_p$-module $L$ with a bilinear alternating Lie bracket satisfying the Jacobi identity and $[L,L]\subseteq pL$. DefineFor completeness, integrality and convergence can be verified without guessing which individual denominators are units. In ${\mathbb{Q}}_p\otimes L$ set $M=p^{-1}L$. Then $[M,M]\subseteq M$, so its integral universal enveloping algebra has an ordered basis by the Poincaré-Birkhoff-Witt theorem. Its coefficientwise $p$-adic norm is submultiplicative, and $L=pM$ consists of elements of valuation at least one. The homogeneous degree-$r$ terms in $\log (\exp X\exp Y)$ have coefficients of valuation at least $-\lfloor (r-1)/(p-1)\rfloor$: expand the logarithm, use $v_p(k!)\le (k-1)/(p-1)$ on each nonconstant exponential factor, and $v_p(\ell )\le (\ell -1)/(p-1)$ on a product of $\ell$ such factors. Therefore evaluating degree $r$ on $X,Y\in pM$ gives valuation at leastwhich tends to infinity and is at least two when $r\ge 2$. Each homogeneous term is a rational Lie polynomial by the formal Baker--Campbell--Hausdorff formula. The ordered enveloping-algebra basis shows that its intersection with ${\mathbb{Q}}_{p}M$ is exactly $M$, so the estimate places the sum in $L$, with the non-linear part in $pL$.

Associativity follows from the identity $\exp (X\ast Y)=\exp X\exp Y$ in the completed universal enveloping algebra. Its identity is $0$, its inverse is $-X$, and $X^{\ast m}=mX$, since the bracket of $X$ with itself is zero. The same coefficient estimate for the group commutator puts it in $pL$. Conjugation in the BCH group is $\exp (\mathrm{ad}X)$; each term $\mathrm{ad}(X{)}^r/r!$ preserves $p^{j}L$, because each bracket adds a factor of $p$ and $v_p(r!)<r$. Thus the submodules $p^{j}L$ are normal subgroups for $\ast$, their intersection is zero, and their finite quotients have $p$-power order. Thus $(L,\ast )$ is a powerful pro-p group, withA module basis topologically generates this group: it generates the quotient $L/pL$, and successive power digits lift that generation through all $p^{j}L$. The BCH construction therefore gives a uniform pro-p group.

Use shifted degreesIts graded Lie bracket is the sign-adjusted Whitehead bracket from Question 2, equivalently the Samelson product on loops. This regrading is essential: the geometric bracket has degree minus one in the unshifted sphere grading. A free graded Lie algebra on a graded vector space $V$ means $\mathbb{L}(V)$ with the universal extension property for linear maps from $V$ to any graded Lie algebra.

Here are the general results used. The Bott–Samelson theorem says that, for a connected based CW complex $X$, the inclusion $X\to \Omega \Sigma X$ induces an isomorphism of graded algebraswith the Pontryagin product. The Milnor–Moore theorem says that for simply connected $Y$, rational loop Hurewicz homomorphism identifies $L_Y$ with the primitives of that loop-homology Hopf algebra and induces $U(L_Y)\cong H_{\ast }(\Omega Y;\mathbb{Q})$. The graded Poincaré-Birkhoff-Witt theorem identifies the associated graded algebra of $U(L)$, for its word-length filtration, with the graded symmetric algebra on $L$; hence its primitive elements of a Hopf algebra are exactly $L$ over $\mathbb{Q}$. Finally the rational Whitehead theorem identifies rational homology equivalences of simply connected CW complexes with rational homotopy equivalences. Rationalization of a topological space induces the corresponding isomorphisms of loop-space homotopy and rational homology.

Suppose first $Y$ has the rational homotopy type of $\Sigma X$. A reduced suspension is simply connected when $X$ is connected; a disconnected $X$ has $H_1(\Sigma X;\mathbb{Q})={\tilde{H}}_0(X;\mathbb{Q})\ne 0$, so it cannot have the rational homology of the simply connected $Y$. Put $V={\tilde{H}}_{\ast }(X;\mathbb{Q})$, in strictly positive degrees. Bott–Samelson theorem gives $H_{\ast }(\Omega Y;\mathbb{Q})\cong T(V)$ as an algebra. To deduce freeness of the primitive Lie algebra, its coproduct must also be considered: the initial Bott–Samelson theorem generators can have reduced-diagonal terms when $X$ has nontrivial cup products.

By Milnor–Moore theorem write this Hopf algebra as $U(L_Y)$. Its algebra indecomposables satisfyThe first identity follows from the enveloping relation: after all positive products are killed, every bracket is killed and no additional linear relation is introduced. Hence each homogeneous basis vector of $V$ has a primitive representative in $H_{\ast }(\Omega Y;\mathbb{Q})$. Choose such representatives $p_v$. Sending $v$ to $p_v$ defines an algebra map from $T(V)$ to itself whose effect on indecomposables is the identity. It is an isomorphism: $p_v=v+$ decomposable terms, and every factor in those terms has smaller positive degree than $v$. Degree induction solves recursively for each original generator and constructs the inverse. This works even with arbitrarily many generators in a degree.

Equip the source tensor algebra with primitive generators. The map is now a Hopf-algebra isomorphism, because the chosen images are primitive. The primitives in this tensor Hopf algebra are exactly $\mathbb{L}(V)$: $U(\mathbb{L}(V))=T(V)$ by the two universal properties, and the graded Graded Poincaré–Birkhoff–Witt theorem gives its primitive subspace. Consequently $L_Y\cong \mathbb{L}(V)$, proving the forward direction without assuming a primitive coproduct for the original generators.

Conversely suppose $L_Y\cong \mathbb{L}(W)$ for a positively graded $W$. Choose a homogeneous basis $w_{\lambda }$ of $W$. Represent its corresponding homotopy class by a based sphere map $S^{|w_{\lambda }|+1}\to Y_{\mathbb{Q}}$, and combine these maps intoThe wedge sum is a reduced suspension, namely $Z=\Sigma ({\bigvee }_{\lambda }{S}^{|w_{\lambda }|})$. The preceding direction, applied to this wedge sum, gives its free rational homotopy Lie algebra on the sphere inclusions. Indeed these particular tensor generators are primitive because the reduced diagonal of each sphere is zero, and loop Hurewicz homomorphism sends the adjoint sphere inclusion to that generator. Naturality of the bracket shows that $F_{\ast }$ sends the free generating basis to the prescribed free generating basis of $L_Y$. It is therefore an isomorphism of free graded Lie algebras, hence an isomorphism in every rational homotopy degree. The rational Whitehead theorem makes $F$ a rational homotopy equivalence. This proves the reverse implication and also shows that a wedge sum of spheres can be chosen as the reduced suspension.

For every chosen connected $X$ with $Y{\simeq }_{\mathbb{Q}}\Sigma X$, the required algebra statement is precisely the Bott–Samelson theorem isomorphism, transported along rational homotopy equivalence:Multiplication is concatenation of tensors and the degree-zero tensor is the unit. All freeness assertions here are graded, with the Koszul signs described above.

Use a finite-dimensional complex semisimple Lie algebra, or the complexification of its compact real form, as required for the root construction. Choose a Cartan subalgebra with basis $H_1,\dots ,H_r$. Simultaneous diagonalization of its adjoint action gives the root-space decompositionThe nonzero weights are the roots of a root system; each root space is one-dimensional in this setting, and roots occur in opposite pairs. This supplies the Cartan and root-vector basis.

The Jacobi identity givesIf the bracket is nonzero and $\alpha +\beta \ne 0$, it is a nonzero vector in the root space of weight $\alpha +\beta$, so that sum is a root. The printed implication needs the opposite-root exception: for $\beta =-\alpha$, $[E_{\alpha },E_{-\alpha }]$ can be a nonzero Cartan element, while zero is not a root. The supplied simple-root commutator itself gives this counterexample. This is the root bracket and the opposite-root exception.

Choose a real linear functional nonzero on every root; positive roots are those on which it is positive, and negative roots are their negatives. The simple roots are positive roots not decomposable into two positive roots. They form a basis ${\alpha }_1,\dots ,{\alpha }_r$ of the root span, and every root is an integer combinationwith all $m_i\ge 0$ for a positive root and all $m_i\le 0$ for a negative root. Positive roots can be decomposed repeatedly until simple roots remain; the finite root set makes this procedure terminate.

For $i\ne j$, ${\alpha }_j-{\alpha }_i$ has simple-root coefficients of both signs and is not a root. It is also nonzero, so the root-space bracket rule givesThe operators $E_i^{\pm },{\hat{H}}_i$ have the SU(2) Lie algebra normalization ${\hat{H}}_i=2J_3$, $E_i^{\pm }=J_{\pm }$. In a finite-dimensional representation, a vector killed by $E_i^{-}$ is a lowest-weight vector in its SU(2) summands. If its ${\hat{H}}_i$ value is $-\lambda$, those summands have spin $\lambda /2$, soOne can also read the termination from successive raising coefficients $\sqrt{(\lambda -k)(k+1)}$ starting at the lowest weight.

Apply this result in the finite-dimensional Adjoint representation, taking the lowest-weight vector to be $E_j^{+}$. The vanishing bracket just proved is its lowering condition. If $[{\hat{H}}_i,E_j^{+}]=a_{ij}{E}_j^{+}$, the lowest-weight result forces $a_{ij}=-n_{ij}$ with $n_{ij}$ a nonnegative integer. It also yieldsThe last identity is exactly the requested nested commutator. This is the SU(2) proof of the Serre relations, and the identities are the Serre relations in these Chevalley basis conventions.

To construct a highest-weight representation, start from its highest-weight vector and apply negative-root operators. Fix an order on the negative roots. The Poincaré-Birkhoff-Witt theorem gives spanning vectorsof weight $w-{\sum }_i{k}_i{\beta }_i$. The spanning assertion follows by commuting all negative-root operators to the left, Cartan operators to the middle and positive-root operators to the right; commutators supply lower-order words, positive operators kill the highest-weight vector and Cartan operators act there by scalars. In the finite-dimensional highest-weight representation, impose the simple-root string relations $(E_i^{-}{)}^{w_i+1}|w⟩=0$ and the commutator/Serre relations. At each weight retain a linearly independent set of the resulting words, discarding zero words and dependencies. This gives a basis, with separate independent vectors retained when a weight has multiplicity greater than one. The construction is finite in the assumed finite-dimensional representation; treating all lowering words as automatically independent would overcount.

For SU(3), put $e_i=E_i^{+}$, $f_i=E_i^{-}$ and $h_i={\hat{H}}_i$. The off-diagonal Cartan entries are minus one, so the Serre relations give $(\mathrm{ad}{e}_1{)}^2{e}_2=(\mathrm{ad}{e}_2{)}^2{e}_1=0$. With $e_3=[e_1,e_2]$, these implyFor $f_3=-[f_1,f_2]$, expand the commutator using Jacobi identity and the distinct-root brackets:since $[h_1,f_2]=f_2$. Similarly,and the same calculation on $f_3$ gives $[h_1,f_3]=-f_3$. ThusThese signs also agree with the explicit model $e_1=E_{12}$, $e_2=E_{23}$, $e_3=E_{13}$, $f_1=E_{21}$, $f_2=E_{32}$, $f_3=E_{31}$.

For the SU(3) Casimir in Chevalley generators, setUsing $[h_1,e_1]=2e_1$, $[h_2,e_1]=-e_1$, and putting Cartan elements on the left, the four terms of $[P,e_1]$ areTheir sum is $3h_1{e}_1$. The remaining terms giveThe last two cancel, so $[C,e_1]=3h_1{e}_1-3h_1{e}_1=0$. Also $h_1$ commutes with every Cartan polynomial and with each $f_n{e}_n$: their raising and lowering weights cancel in the product. HenceOn $|w_1,w_2⟩$, $e_1$ and $e_2$ vanish and therefore $e_3=[e_1,e_2]$ vanishes as well. All three $f_n{e}_n$ terms consequently give zero, leavingThe paper's normalization is $C=3C_2$ relative to the conventional SU(3) quadratic Casimir eigenvalue. For example it gives $4$ in the fundamental representation, $9$ in the octet and $18$ in the decuplet.

Give $U(\mathfrak{g})$ its degree filtration. The Poincaré-Birkhoff-Witt theorem identifies its associated graded ring with ${\mathrm{Sym}}_k\mathfrak{g}$, a polynomial ring in ${\dim }_k\mathfrak{g}$ variables. The Hilbert basis theorem makes this graded algebra Noetherian on both sides.

Here is the required transfer, including why a finite list of symbols gives actual generators. For a right ideal $I$ of a nonnegatively filtered ring $A$, its induced graded ideal $\mathrm{gr}I$ is a homogeneous right ideal of $\mathrm{gr}A$. Choose finite homogeneous generators and lift them to $u_1,\dots ,u_m\in I$. For $u\in I$ of degree $d$, express its leading symbol as a graded right linear combination of these symbols, using homogeneous coefficients of complementary degrees. Lift the coefficients to $A$ and subtract the corresponding ${\sum }_i{u}_i{a}_i$. The remainder has degree strictly below $d$. Induction, terminating because degrees are nonnegative, proves $I={\sum }_i{u}_{i}A$. The identical argument with coefficients on the left treats left ideals.

This ascending filtered-graded transfer of Noetherianity proves that $U(\mathfrak{g})$ is right and left Noetherian. A quotient ring inherits either property: pull an ascending ideal chain back to the original ring, or take images of ideal generators. Part (a) therefore gives $R$ right and left Noetherian. Alternatively, under the broader almost commutative algebra convention, its commutative finitely generated associated graded ring is directly a quotient of a finite-variable polynomial ring, and the same leading-symbol argument proves the conclusion.

An almost commutative algebra over $\mathbb{C}$ has an exhaustive nonnegative increasing multiplicative ascending filtrationfor which the associated graded ringis a commutative finitely generated complex algebra. This includes the customary degree filtration with $F_{0}R=\mathbb{C}$, finitely many degree-one generators, and commutators lowering degree. The finite-generation clause is part of the convention used here and is necessary for the conclusion.

A finitely generated commutative complex algebra is a quotient of a polynomial ring $\mathbb{C}[X_1,\dots ,X_d]$. Repeated application of the central-variable case of the Hilbert basis theorem makes that polynomial ring Noetherian; quotients preserve the ascending chain condition by lifting ideal chains. Thus $\mathrm{gr}R$ is Noetherian.

We prove ascending filtered-graded transfer of Noetherianity directly. Give a left ideal $I\subseteq R$ the induced ascending filtration $I\cap F_{n}R$. Its associated graded module embeds into $\mathrm{gr}R$ as a homogeneous ideal, denoted $\mathrm{gr}I$. Choose finitely many homogeneous generators ${\overline{u}}_1,\dots ,{\overline{u}}_m$ of this ideal, of degrees $d_1,\dots ,d_m$, and lifts $u_j\in I\cap F_{d_j}R$. Homogeneous generators may be chosen by taking homogeneous components of any finite generating set.

For $u\in I\cap F_{n}R$, its degree-$n$ symbol can be expressed aswith terms of negative coefficient degree omitted. Lift each coefficient to $a_j\in F_{n-d_j}R$. ThenInduction on $n$, ending at $F_{-1}R=0$, expresses $u$ in the left ideal generated by the $u_j$. Hence every left ideal is finitely generated. Applying the same argument to right ideals, or to the opposite ring, provesThis degree-lowering proof terminates in finitely many steps and requires no completeness assumption on the filtered ring. Without finite generation of the graded algebra, a counterexample would be the commutative polynomial ring in infinitely many variables, whose ideals $(X_1)⊊(X_1,X_2)⊊\cdots$ do not stabilize.

For the universal enveloping algebra $U(\mathfrak{g})$, use the degree ascending filtration inherited from the tensor algebra of the finite-dimensional Lie algebra $\mathfrak{g}$. For $x,y\in \mathfrak{g}$, the defining relation is $xy-yx=[x,y]\in F_{1}U(\mathfrak{g})$, whereas the product $xy$ has filtration degree at most $2$. Thus the degree-one symbols commute in $\mathrm{gr}U(\mathfrak{g})$. Those symbols generate the graded algebra, so it is a quotient ofThis already proves the required almost-commutative property. The Poincaré-Birkhoff-Witt theorem gives the stronger identification $\mathrm{gr}U(\mathfrak{g})\cong \mathrm{Sym}(\mathfrak{g})$, but injectivity of this map is not needed for the chain-condition argument. Therefore

Use the positive simple-root raising operators $E_{12}=x_1{\partial }_{x_2}$ and $E_{23}=x_2{\partial }_{x_3}$. On the weight space of $2L_1-L_3$, the first maps both $s_{11}\otimes s_{12}$ and $s_{12}\otimes s_{11}$ to $s_{11}\otimes s_{11}$, while the second annihilates both. Consequentlyis a nonzero highest-weight vector; the remaining positive-root operator $E_{13}=[E_{12},E_{23}]$ also annihilates it. Its Dynkin labels are $(2,1)$, and the generated submodule $Z$ is the irreducible highest-weight representation with that highest weight.

Put $F_1=E_{21}=x_2{\partial }_{x_1}$ and $F_2=E_{32}=x_3{\partial }_{x_2}$, acting on both tensor factors. Direct calculation givesDefine the linearly independent tensorsThey both have weight $L_1$, andTherefore $A,B\in Z_{L_1}$. To prove that they exhaust this weight space, its weight differs from the highest weight by $(L_1-L_2)+(L_2-L_3)$. The Poincaré-Birkhoff-Witt theorem expresses all descendants at that difference using $F_2{F}_{1}v$ and $F_1{F}_{2}v$: the third possible operator $E_{31}=[F_2,F_1]$ is their difference. Hence the space has dimension at most two, and the two independent vectors already found give

For a finitely generated algebra over $k$, choose a finite-dimensional generating subspace $V$ containing one. Its Gelfand–Kirillov dimension isChanging generators bounds the two filtrations by constant rescalings of the index, so the value is independent of $V$.

For the noncommutative Krull dimension, set the zero module's dimension to $-1$. A nonzero module has dimension zero exactly when it is an Artinian module. Recursively, dimension at most $d$ means that in every descending chain of submodules the successive factors eventually have dimension less than $d$; use the ordinal version of this definition if necessary. The left Krull dimension of $R$ is that of its left regular module. For commutative Noetherian rings this agrees with the usual prime-chain Krull dimension, but prime-chain counting alone is not its definition for a noncommutative ring.