Alphabeta Math
Pipeline-generated
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

✓ 20 results · all verified · 18 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 2 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Quantized Enveloping Algebras and Quantum Serre Relations

1 · Prerequisites

2 · Summary

This page constructs the Drinfeld–Jimbo quantized enveloping algebra of a symmetrizable Cartan datum and develops its quantum Serre relations, its Hopf structure and its triangular decomposition. The datum with its symmetrizer, lattices and normalization is fixed in Symmetrizable Cartan data for quantum groups, and the qi-integers, factorials and divided powers used throughout are set up in Quantum integers, factorials, Gaussian binomials and divided powers at qi with the Gaussian calculus of The quantum Pascal recurrences, the Gauss product formula and Gaussian integrality and The quantum binomial expansion for q-commuting elements.

Presentation, Serre relations and Hopf structure

The algebra is presented by generators Ei,Fi,Kh and the toral, mixed and quantum Serre relations in The Drinfeld-Jimbo quantized enveloping algebra by generators and relations, with the Hopf foundations Bialgebras, counits and antipodes over a commutative ring and Uniqueness of the antipode. The coproduct of a Serre element is controlled by The coproduct preserves the positive and negative quantum Serre ideals, the bar and Chevalley involutions are checked on all relations in The Chevalley involution, bar involution, and contravariant anti-involution preserve the Drinfeld–Jimbo ideal, and The Drinfeld–Jimbo formulas define a Hopf algebra assembles the coproduct, counit and antipode into a Hopf algebra with the subalgebras of Positive, negative and toral quantum subalgebras and their root gradings.

PBW ranks, duality and the triangular decomposition

The formal shuffle model of The formal quantum shuffle Borel and its Cartan crossed product supplies the coefficientwise Borel, the positive shuffle identity of The quantum Serre sums vanish in the shuffle algebra, and the opposite Serre ideal annihilates the shuffle half and the q-binomial cancellation behind The formal quantum Serre half embeds in the shuffle algebra and is degreewise free, whose inputs include the coideal lemma An augmented coideal ideal of an enveloping algebra is generated by its primitive part and the root-graded bialgebra duality of Lie bialgebras, degreewise duality, and root-graded Manin triples, A root-graded Manin triple gives dual Lie bialgebras and The opposite Borels of a symmetrizable Kac–Moody algebra are root-degreewise dual Lie bialgebras; A nonsingular principal minor of the symmetrized Cartan matrix of size the rank supplies the Cartan coordinates. The classical PBW ranks and the nondegenerate pairing of the halves are established in Generic quantum Serre halves have classical PBW ranks and a nondegenerate Hopf pairing. The generic quantum halves form a Drinfeld–Jimbo crossed double presents the algebra as the crossed double of its two halves and the torus, and Triangular decomposition of a quantized enveloping algebra proves the resulting vector-space decomposition after the local normal-form and opposite-Serre commutator calculations. The total root grading retains all summands with positive degree minus negative degree equal to the prescribed degree, including the additional degree-zero summands. The rank-one string modules used in later pages are computed in Divided-power commutation and the simple Uqi(sl2) string modules.

The companion quantized-enveloping-algebras-and-quantum-serre-relations-examples works through the Uq(sl2) coproduct and antipode, a type-A2 Serre calculation, the double-edge affine A1(1) relation, and a counterexample showing that unsymmetrized q-parameters break the Cartan normalization.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)audited 2026-10-08Open item page →

Bialgebras, counits and antipodes over a commutative ring

Definition

Let R be a commutative ring. A bialgebra over R is a unital associative R-algebra (A,m,η) (Algebras over a commutative ring, central structure maps, and algebra homomorphisms) together with R-algebra homomorphisms Δ:A→A⊗RA (the coproduct) and ε:A→R (the counit) such that Δ is coassociative and the counit axioms hold:

(Δ⊗id⁡A)Δ=(id⁡A⊗Δ)Δ,(ε⊗id⁡A)Δ=id⁡A=(id⁡A⊗ε)Δ.

The counit equations use the canonical identifications R⊗RA≅A≅A⊗RR. The target A⊗RA has the R-algebra structure of The tensor product of R-algebras has multiplication (a⊗b)(a′⊗b′)=aa′⊗bb′.

A Hopf algebra over R is a bialgebra (A,m,η,Δ,ε) together with an R-linear antipode S:A→A satisfying both convolution-inverse equations

m(S⊗id⁡A)Δ=η∘ε=m(id⁡A⊗S)Δ,

where (η∘ε)(a)=ε(a)1A. We do not assume that S is invertible, involutive or multiplicative. Throughout, Δ and ε are algebra homomorphisms, and the displayed order of the antipode factors is our convention.

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-10-08Open item page →

Lie bialgebras, degreewise duality, and root-graded Manin triples

Definition

Let k be a field of characteristic 0. A Lie bialgebra is a Lie algebra a over k together with a linear map δa:a→Λ2a such that

Alt⁡(δa⊗id⁡)δa=0

and δa([x,y])=[x,δa(y)]−[y,δa(x)] for all x,y∈a. The bracket action on Λ2a is [x,u∧v]=[x,u]∧v+u∧[x,v].

Let Q be an additive abelian group. Suppose a=⨁α∈Qaα and c=⨁α∈Qcα are Q-graded vector spaces with finite-dimensional graded pieces, and suppose that, within each space separately, only finitely many pairs of nonzero graded pieces have degrees summing to any fixed degree. A pairing between them is degreewise perfect if aα pairs perfectly with c−α and all other degree pairs are orthogonal. It then identifies c with the restricted graded dual agr′:=⨁α∈Qaα∗. The induced pairing on exterior powers is the determinant pairing: ⟨x1∧⋯∧xm,y1∧⋯∧ym⟩=det⁡(⟨xi,yj⟩)i,j=1m.

For the Lie bialgebra duality below, require both brackets to preserve degree: [aα,aβ]⊆aα+β and [cα,cβ]⊆cα+β. Two such Lie bialgebras are dual when these pairings are degreewise perfect and their cobrackets are transposes of the opposite brackets: ⟨δa(x),y∧z⟩=⟨x,[y,z]⟩ and ⟨x∧x′,δc(y)⟩=⟨[x,x′],y⟩. For x∈aγ, degree preservation and orthogonality make ⟨x,[cα,cβ]⟩ vanish unless α+β=−γ. There are only finitely many such nonzero pairs, and their finite-dimensional perfect pairings identify the transpose with a unique element of (Λ2a)γ under the determinant pairing. The same argument applies with the two algebras exchanged, and linear extension handles arbitrary elements. Thus the transposes lie in the ordinary exterior squares rather than formal infinite sums.

A root-graded Manin triple is a Q-graded Lie algebra d=⨁α∈Qdα with finite-dimensional graded pieces, a symmetric invariant bilinear form B pairing dα perfectly with d−α and satisfying B(dα,dβ)=0 whenever α+β≠0, and graded Lie subalgebras d+ and d− such that d=d+⊕d− as a vector space and B(d+,d+)=B(d−,d−)=0. The cross pairing is degreewise perfect: a vector in (d+)α annihilating (d−)−α also annihilates (d+)−α by isotropy, so it is zero by perfectness on the double. The same argument applies with the two halves exchanged, and finite dimensionality gives perfectness of the cross pairing. All other degree pairs are orthogonal by the condition on B. Require finite degree decompositions within each of d+ and d− separately, as above; a triple satisfying this requirement is called locally finite. No finite-decomposition condition is imposed on the whole double. This permits opposite Borels with infinitely many roots, since each Borel has support in one root cone. The finite-dimensional Manin-triple definition is the special case with finitely many nonzero graded pieces.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)audited 2026-10-08Open item page →

Symmetrizable Cartan data for quantum groups

Definition

A symmetrizable Cartan datum for a quantum group consists of a finite nonempty index set I={1,…,n}, a symmetrizable generalized Cartan matrix A=(aij)i,j∈I (Generalized cartan matrix, Symmetrizable generalized cartan matrix) and a chosen diagonal symmetrizer D=diag⁡(d1,…,dn) with di∈Z>0 and diaij=djaji for all i,j; free abelian groups P∨ and P (Free abelian group on a set) with an integer-valued bilinear pairing ⟨⋅,⋅⟩:P×P∨→Z; simple coroots hi∈P∨ that are linearly independent in P∨⊗ZQ; and simple roots αi∈P that freely generate the root lattice Q=⨁i∈IZαi⊆P (Kac Moody root lattice height and positive cone) and satisfy ⟨αj,hi⟩=aij for all i,j (Realization of a generalized cartan matrix, Minimal realizations exist and are unique up to isomorphism).

The index set is finite and A need not be nonsingular. We do not require the simple coroots to span P∨⊗ZQ, and the pairing is not required to be perfect; choosing P and P∨ is part of the datum, not something determined by the matrix alone.

We fix an indeterminate q over Q, work over Q(q), and set qi=qdi. For h∈P∨, ⟨αi,h⟩∈Z is the exponent in q⟨αi,h⟩. We use the row convention ⟨αj,hi⟩=aij throughout. No fundamental weights with prescribed values on all hi are part of this definition.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-08Open item page →

A nonsingular principal minor of the symmetrized Cartan matrix of size the rank

Statement

Let n≥1 and let B=(bij) be a real symmetric n×n matrix of rank r. For J⊆{1,…,n}, write BJ=(bij)i,j∈J for its principal submatrix. Use the convention that the empty matrix has determinant 1 and is invertible. There is a subset J with ∣J∣=r for which BJ is nonsingular.

If in addition A is a real n×n matrix and D=diag⁡(d1,…,dn) has di>0 with B=DA, then for the same J,

det⁡BJ=(∏i∈Jdi)det⁡AJ.

Thus AJ is also nonsingular and has size r, and ∣{1,…,n}∖J∣=n−r. In particular, if A is symmetrizable and has corank one, then B=DA has corank one and the conclusion gives a nonsingular (n−1)×(n−1) principal submatrix.

Facts & Assumptions

Given: A real symmetric matrix B of rank r; in the second assertion, also A,D with D positive diagonal and B=DA.

[F1]

A symmetrizable generalized Cartan matrix has a positive diagonal symmetrizer D for which B=DA is symmetric (Symmetrizable generalized cartan matrix).

[F2]

For positive size, determinant is defined by the Leibniz formula; we additionally use the local empty-matrix convention stated above (For n≥1, the determinant over a commutative ring by the Leibniz formula, and ∣det⁡A∣ for a real matrix).

[F3]

A real square matrix is invertible exactly when its determinant is nonzero (A finite square real matrix is invertible if and only if its determinant is nonzero); nonsingular means invertible (Invertible square matrices and similarity over a commutative ring).

[F4]

If a symmetric block matrix has an invertible leading block C, its determinant is det⁡(C) times the determinant of the Schur complement (For symmetric M=(ABBTC) with A invertible, a block-unitriangular congruence gives A⊕(C−BTA−1B) and factors det⁡M).

[F6]

A nonzero k-rowed minor of a real matrix forces its rank to be at least k (A matrix has rank at least r exactly when it has a nonzero r-rowed minor).

Proof

technique · Choose a maximal nonsingular principal block and use its Schur complement
1.1givenF2algebra

If r=0, then B=0 and J=∅ works by the stated empty-matrix convention. When B=DA with positive diagonal D, A=0 as well, so the determinant identity is 1=1. Hence assume r>0.

1.2givenF2F3choosealgebra

Since B≠0, either some bii≠0, giving a nonsingular one-by-one principal submatrix, or all diagonal entries vanish and some bij≠0 with i≠j, giving the nonsingular two-by-two principal submatrix with determinant −bij 2. Choose, from the finite family of principal submatrices with nonzero determinant, a set J maximal under inclusion. Then BJ is invertible by [F3]; put k=∣J∣.

2.1step 1.2F4algebra

For any p∉J, maximality makes det⁡BJ∪{p}=0. Write up=(bip)i∈J. The Schur-complement formula [F4] gives 0=det⁡(BJ)(bpp−upTBJ−1up), so bpp−upTBJ−1up=0.

3.1step 2.1F4F8algebra

For distinct p,q∉J, maximality also gives det⁡BJ∪{p,q}=0. Since BJT=BJ, transposing BJBJ−1=I=BJ−1BJ and using [F8] shows that (BJ−1)T is also a two-sided inverse of BJ, hence (BJ−1)T=BJ−1. The two-by-two Schur complement therefore has zero diagonal by step 2.1 and equal off-diagonal entries t=bpq−upTBJ−1uq. Its determinant is −t2, so [F4] and the fact that det⁡BJ≠0 imply t=0. Therefore every entry of the complementary block equals the corresponding entry of WTBJ−1W, where W=BJ,Jc.

4.1step 3.1F5F6F8algebra

Partitioning by J and Jc, the equality in step 3.1 yields B=(IWTBJ−1)BJ(IBJ−1W). By matrix multiplication, every row of B is a linear combination of the k rows of the right factor, so [F5] gives r≤k. Since det⁡BJ≠0, B has a nonzero k-rowed minor, so [F6] gives r≥k. Hence k=r.

5.1F1F5F7step 1.1step 4.1algebra∎

If B=DA, diagonality gives BJ=DJAJ, and [F7] gives det⁡BJ=det⁡DJdet⁡AJ=(∏i∈Jdi)det⁡AJ. The product is nonzero, so the already nonzero det⁡BJ forces det⁡AJ≠0. For a symmetrizable A of corank one, positive diagonal row-scaling preserves rank, hence r=n−1.

Remarks

Symmetry is essential: (0100) has rank 1 but no nonsingular one-by-one principal submatrix. Berkeley's Gabber–Kac example in Ch. 10 §10.4.2.6 assumes the positive-semidefinite corank-one case; the principal-minor argument above needs only symmetry and therefore also applies to indefinite symmetrizable matrices.

DefinitionDefinition: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-08Open item page →

Quantum integers, factorials, Gaussian binomials and divided powers at qi

Definition

Let (I,A,D,P,P∨,q) be a symmetrizable Cartan datum for a quantum group (Symmetrizable Cartan data for quantum groups) and put qi=qdi. For m∈N, define the qi-integer

[m]i:=qim−qi−mqi−qi−1,[0]i:=0,

and the qi-factorial

[m]i!:=∏k=1m[k]i,[0]i!:=1.

For 0≤r≤m, define the Gaussian binomial (mr)i:=[m]i!/([r]i![m−r]i!), and set it to 0 when r<0 or r>m. In the published one-parameter convention, write (mr)t:=(mr,m−r)t for the two-part q-multinomial coefficient of The q-integer, q-factorial and q-multinomial coefficients.

For an element x of a unital Q(q)-algebra and m≥0, define its divided power at qi by x(m):=xm/[m]i!, so x(0)=1 and x(1)=x.

The symmetric convention and the published asymmetric convention are related, for m≥0 and 0≤r≤m, by

[m]i=qi−(m−1)[m]qi2,(mr)i=qi−r(m−r)(mr)qi2.

In particular [m]i is invariant under qi↦qi−1. These quantities depend on the symmetrizer only through qi=qdi.

Facts & Assumptions

Given: A symmetrizable Cartan datum with q indeterminate over Q, and an element x of a unital Q(q)-algebra.

[F1]

The datum has qi=qdi with positive integer di (Symmetrizable Cartan data for quantum groups).

[F2]

The asymmetric q-integer and q-factorial are [m]t=1+t+⋯+tm−1 and [m]t!=∏j=1m[j]t, with [0]t=0 and [0]t!=1; the q-multinomial is the factorial quotient (The q-integer, q-factorial and q-multinomial coefficients).

Verification

technique · Expand the symmetric $q_i$-integer and multiply the resulting finite product
1.1givenF1F2algebra

For m≥1, cancel the nonzero factors qi−qi−1=qi−1(qi2−1) to obtain [m]i=qi−(m−1)(qi2m−1)/(qi2−1)=qi−(m−1)∑j=0m−1qi2j=qi−(m−1)[m]qi2. At m=0 the same identity holds by the zero convention. For m≥1 the quotient is nonzero because its numerator and denominator are nonzero in Q(q) by [F1].

2.1step 1.1F1F2algebra

Multiplying the identity of step 1.1 for j=1,…,m gives [m]i!=qi−m(m−1)/2[m]qi2!; for m=0 this is the equality of empty products. Hence for 0≤r≤m, division by the nonzero factorials is valid and (mr)i=qi−m(m−1)/2+r(r−1)/2+(m−r)(m−r−1)/2(mr)qi2=qi−r(m−r)(mr)qi2, since −m(m−1)+r(r−1)+(m−r)(m−r−1)=−2r(m−r).

3.1step 1.1step 2.1F1algebra∎

Since every [j]i for j≥1 is nonzero, [m]i! is a nonzero scalar and therefore invertible in Q(q); this makes x(m) well-defined. Replacing qi by qi−1 negates numerator and denominator in [m]i, so [m]i is invariant, as are its factorials and Gaussian quotients.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-08Open item page →

An augmented coideal ideal of an enveloping algebra is generated by its primitive part

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let k be a field of characteristic 0 (Field, The characteristic of a ring: the least n≥1 with n⋅1R=0 when one exists, and 0 otherwise) and let l be a Lie algebra over k (Lie algebras over a field). Equip its universal enveloping algebra U(l) with the standard cocommutative Hopf structure (Hopf-algebra structure on U(g)), with coproduct Δ(x)=x⊗1+1⊗x for x∈l and counit ε(x)=0. If J⊴U(l) is a two-sided ideal such that

Δ(J)⊆J⊗U(l)+U(l)⊗J,J⊆ker⁡ε,

then j:=J∩l is a Lie ideal and J=U(l)j=jU(l). Thus J is uniquely determined by j among two-sided ideals satisfying both displayed conditions.

Facts & Assumptions

Given: AC, a characteristic-zero field k, a Lie algebra l, and a two-sided ideal J satisfying the coproduct and augmentation conditions.

[A1]

AC is the assertion that every family of nonempty sets has a choice function (The Axiom of Choice).

[L2]

Under AC, every set, hence any chosen basis, can be well ordered (The well-ordering theorem).

[L3]

For a supplied totally ordered basis of l, ordered monomials form a basis of U(l) and the PBW symbol map identifies gr⁡U(l) with S(l) (Poincaré–Birkhoff–Witt theorem).

[L4]

The standard coproduct and counit are algebra maps, and are primitive on l (Hopf-algebra structure on U(g), Bialgebras, counits and antipodes over a commutative ring).

[L5]

The PBW filtration FnU(l) is exhaustive and is spanned by products of at most n elements of l (PBW filtration on the enveloping algebra).

[L7]

In a characteristic-zero field, every positive integer is nonzero and invertible (Field, The characteristic of a ring: the least n≥1 with n⋅1R=0 when one exists, and 0 otherwise).

Proof

technique · PBW filtration and induction on degree
1.1A1L1L2L3L5construct

By [A1] and [L1], choose a basis of l; by [L2], give it a total well-order. Apply [L3]. In particular l embeds in U(l), and the associated graded algebra of the filtration [L5] is S(l).

1.2L3L4L5algebra

By [L4], Δ is filtered for the total-degree filtration on U(l)⊗U(l): this holds on each degree-one generator because Δ(x)=x⊗1+1⊗x, and hence on products because Δ is an algebra map. Its associated graded coproduct on S(l) is the algebra map making every x∈l primitive, since the two maps agree on the algebra generators.

1.3A1L1L3L4L5construct

Put Jn:=J∩FnU(l) and K:=gr⁡J. The ideal property makes K a graded ideal of S(l). To see its coideal property, use AC and [L1] to choose a complement Wn of Jn−1 in Jn; its image in Fn/Fn−1 is a subspace, so choose a complement there and lift it to Cn⊂Fn. Thus Fn=Fn−1⊕Wn⊕Cn and Jn=Jn−1⊕Wn. The total-degree filtration then has associated graded pieces Wp⊗(Wq⊕Cq) and (Wp⊕Cp)⊗Wq for J⊗U+U⊗J. Taking highest-degree symbols in the assumed containment gives ΔS(l)(K)⊆K⊗S(l)+S(l)⊗K.

1.4L3L4L5givenbasealgebra

Since ε(J)=0 and F0U(l)=k1, we have J0=0. PBW gives F1U(l)=k1⊕l, so K0=0 and K1=j. Also j is a Lie ideal: for x∈l and a∈j, the commutator xa−ax=[x,a] lies in J because J is two-sided, and lies in l by the enveloping relation.

1.5L6givenalgebra

Let Ij be the ideal of S(l) generated by j, and let π:S(l)→S(l/j) be the map induced by the vector-space quotient. The universal properties in [L6] construct inverse algebra maps between S(l)/Ij and S(l/j): maps from either algebra to a commutative unital k-algebra A correspond exactly to linear maps l→A vanishing on j. The maps are inverse because they agree with the identity on the algebra generators. Hence ker⁡π=Ij.

2.1L6L7step 1.3step 1.4step 1.5inductionihalgebra

We prove by induction on n that Kn⊆Ij∩Sn(l). The claim holds in degrees 0 and 1 by step 1.4. For n≥2, take z∈Kn. Its reduced coproduct has, in bidegree (p,q) with p,q>0 and p+q=n, a component in Kp⊗Sq(l)+Sp(l)⊗Kq by step 1.3; both p,q<n, so the induction hypothesis makes its image under π⊗π zero. Thus π(z) has zero reduced coproduct and is primitive in S(l/j). The (n−1,1) component of the coproduct of a homogeneous degree-n element f is the sum obtained by placing each of its n factors in the second tensor slot; multiplying the two slots gives nf. If f is primitive this component is zero, so [L7] forces f=0. Consequently π(z)=0 and z∈Ij.

3.1step 1.3step 1.4step 2.1algebra

The reverse inclusion Ij⊆K follows because K is an ideal and contains K1=j. Hence gr⁡J=K=Ij.

4.1L3L5step 3.1algebra

Since J is two-sided, the left and right ideals U(l)j and jU(l) lie in J, so their associated graded spaces lie in gr⁡J=Ij by step 3.1. Conversely, each degree-n element of Ij is a finite sum of products in Sn−1(l)j, and PBW lifts that sum to an element of U(l)j∩Fn with the same symbol. The right-handed products lift identically. Hence both associated graded spaces equal Ij.

5.1step 1.4step 2.1step 4.1inductiondischarge-induction: step 2.1given∎

For x∈J∩Fn, step 4.1 supplies y∈U(l)j∩Fn with the same degree-n symbol as x. Then x−y∈J∩Fn−1; descending induction, starting with J∩F0=0 from step 1.4, gives x∈U(l)j. The identical argument with the right-generated ideal gives x∈jU(l). Thus J=U(l)j=jU(l), and applying this equality to any other ideal satisfying the same two conditions and the same intersection proves the stated uniqueness.

Remarks

  • The zero Lie algebra is included: then U(0)=k, ker⁡ε=0, and the only admissible ideal is J=0.
  • The augmentation hypothesis is necessary. For nonzero l, the ideal J=U(l) satisfies the coproduct containment, but J∩l=l while U(l)(J∩l)=ker⁡ε≠U(l).
  • The unrestricted “largest ideal with fixed primitive part” claim is false: for l=kx, U(l)=k[x], the ideals 0 and (x2) have the same intersection 0 with l, and 0⊊(x2).
  • AC is used to choose and well-order a basis of arbitrary l, as required by the supplied general PBW theorem, and to split the induced filtration of J in step 1.3. These are the only nonconstructive choices in this proof; the characteristic-zero use is exactly the invertibility of n in step 2.1. No assertion is made in positive characteristic.
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-08Open item page →

Uniqueness of the antipode

Statement

Let (A,m,η,Δ,ε) be a bialgebra over a commutative ring R (Bialgebras, counits and antipodes over a commutative ring) and let S,S′:A→A be antipodes. Then S=S′. Thus a bialgebra admits at most one Hopf algebra structure with its fixed multiplication, unit, coproduct and counit. Every antipode also satisfies S(1A)=1A.

Facts & Assumptions

Given: A bialgebra (A,m,η,Δ,ε) over a commutative ring R and two antipodes S,S′.

[F1]

The multiplication m is associative and the coproduct Δ is coassociative (Bialgebras, counits and antipodes over a commutative ring).

[F2]

The counit identities are (ε⊗id⁡)Δ=id⁡=(id⁡⊗ε)Δ (Bialgebras, counits and antipodes over a commutative ring).

[F3]

The unit map η and the algebra maps Δ,ε are unital, so Δ(1A)=1A⊗1A and η(ε(1A))=1A (Bialgebras, counits and antipodes over a commutative ring).

[F4]

Each antipode T satisfies m(T⊗id⁡)Δ=η∘ε=m(id⁡⊗T)Δ (Bialgebras, counits and antipodes over a commutative ring).

Proof

technique · Use uniqueness of a two-sided inverse in the convolution monoid of endomorphisms
1.1givenF1algebra

For f,g∈Hom⁡R(A,A) define f⋆g:=m∘(f⊗g)∘Δ. Coassociativity of Δ and associativity of m give (f⋆g)⋆h=f⋆(g⋆h) for all f,g,h.

1.2F2algebra

The map e:=η∘ε is a two-sided unit for ⋆: for every f and a, (e⋆f)(a)=∑ε(a(1))f(a(2))=f(a) and (f⋆e)(a)=∑f(a(1))ε(a(2))=f(a), by the two counit identities and R-linearity of f.

1.3givenF3F4algebra

Evaluating either antipode equation for S at 1A, and using Δ(1A)=1A⊗1A and e(1A)=1A, gives S(1A)1A=e(1A)=1A. Hence S(1A)=1A.

2.1step 1.1step 1.2F4algebra∎

By [F4], S⋆id⁡=e=id⁡⋆S′; using [F1] and step 1.2, S=S⋆e=S⋆(id⁡⋆S′)=(S⋆id⁡)⋆S′=e⋆S′=S′. Therefore the antipode is unique, and the stated bialgebra has at most one Hopf structure.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-10-08Open item page →

A root-graded Manin triple gives dual Lie bialgebras

Statement

Let k be a field of characteristic 0, and let (d,d+,d−;B) be a locally finite root-graded Manin triple in the sense of Lie bialgebras, degreewise duality, and root-graded Manin triples. Define δ+:d+→Λ2d+ and δ−:d−→Λ2d− by transposing the opposite brackets under the degreewise perfect pairing induced by B:

⟨δ+(x),y∧z⟩=B(x,[y,z]),⟨u∧v,δ−(y)⟩=B([u,v],y).

Then these maps are well defined and make d+ and d− dual Lie bialgebras.

Facts & Assumptions

Given: A field k of characteristic 0 and a locally finite root-graded Manin triple.

[F1]

The two complementary isotropic subalgebras have finite-dimensional graded pieces, a degreewise perfect cross pairing, and only finitely many degree decompositions within each subalgebra in any fixed degree (Lie bialgebras, degreewise duality, and root-graded Manin triples).

[F2]

The wedge pairing is the determinant pairing on exterior powers, and a Lie bialgebra cobracket is a cocycle satisfying co-Jacobi (The graded exterior algebra ΛV, Lie bialgebras, degreewise duality, and root-graded Manin triples).

Proof

technique · Transpose the bracket and use the Jacobi identity in the double
1.1F1construct

By [F1], for each homogeneous x∈d+ only finitely many opposite-degree pairs of components of d− can bracket to a degree paired with x. The perfect pairings therefore give a unique finite sum δ+(x)∈Λ2d+ satisfying the first transpose identity; the same construction defines δ−, and both maps are linear and preserve total root degree.

1.2F1algebra

Choose dual bases ei and fa in the finitely many homogeneous pieces involved in a fixed calculation, and write [ei,ej]=∑kcijkek and [fa,fb]=∑kdkabfk. Invariance gives B([ei,fa],fk)=B(ei,[fa,fk])=diak and B([ei,fa],ek)=B(fa,[ek,ei])=−cika; since the two subalgebras are isotropic and pair perfectly, these identities determine [ei,fa]=∑k(diakek−cikafk). For each fixed input pair, grading restricts these sums to two fixed finite-dimensional pieces, so they are finite by [F1].

2.1F1F2step 1.1algebra

Pair Alt⁡(δ+⊗id⁡)δ+(x) with y∧z∧w in the opposite subalgebra. By the defining transpose identity, the result is B(x,[y,[z,w]]+[z,[w,y]]+[w,[y,z]])=0 by Jacobi in d−. Degreewise perfectness makes the co-Jacobi expression zero; the identical argument with signs exchanged proves co-Jacobi for δ−.

3.1F1F2step 1.1step 1.2algebra∎

Jacobi in d for ei,ej,fa, paired with fb, gives ∑kcijkdkab=∑k(cikadjkb+cikbdjak−cjkadikb−cjkbdiak). By the transpose definition, this is exactly the coefficient identity for δ+([ei,ej])=[ei,δ+(ej)]−[ej,δ+(ei)]. Interchanging + and − and using Jacobi for ei,fa,fb proves the cocycle identity for δ−. Since the homogeneous bases were arbitrary and the pairings are perfect, both cocycle identities hold for all elements. Together with step 2.1, this proves that the two transposed maps are dual Lie bialgebra structures.

DefinitionDefinition: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-08Open item page →

The Drinfeld-Jimbo quantized enveloping algebra by generators and relations

Definition

Let (I,A,D,P,P∨,q) be a symmetrizable Cartan datum for a quantum group (Symmetrizable Cartan data for quantum groups) and put qi=qdi. Write Ki:=Kdihi and Ki−1:=K−dihi. Let V be the Q(q)-vector space with basis the symbols {Ei,Fi}i∈I∪{Kh}h∈P∨, and let T:=T(V) be its free unital associative Q(q)-algebra (Tensor algebra of a vector space, Universal property of the tensor algebra). Let IDJ be the two-sided ideal of T generated by (The ideal generated by a subset and principal ideals) the following relations:

K0−1,KhKh′−Kh+h′(h,h′∈P∨);

KhEi−q⟨αi,h⟩EiKh,KhFi−q−⟨αi,h⟩FiKh(h∈P∨, i∈I);

EiFj−FjEi−δijKi−Ki−1qi−qi−1(i,j∈I);

Serreij+:=∑r=01−aij(−1)r(1−aijr)iEi1−aij−rEjEir,Serreij−:=∑r=01−aij(−1)r(1−aijr)iFi1−aij−rFjFir(i≠j),

where the Gaussian binomials are those of Quantum integers, factorials, Gaussian binomials and divided powers at qi. The Drinfeld-Jimbo quantized enveloping algebra is

Uq(g):=T/IDJ

(The quotient ring R/I with (r+I)(s+I)=rs+I), with Ei,Fi,Kh denoting the images of the generators. In this quotient, Ki−1 is the two-sided inverse of Ki.

It is graded by the root lattice Q=⨁i∈IZαi (Kac Moody root lattice height and positive cone), with deg⁡Ei=αi, deg⁡Fi=−αi and deg⁡Kh=0; thus Uq(g)=⨁β∈QUq(g)[β].

For every unital associative Q(q)-algebra B, any assignment of elements Ei′,Fi′,Kh′∈B satisfying these relations extends uniquely to a Q(q)-algebra homomorphism Uq(g)→B (A ring homomorphism whose kernel contains a two-sided ideal factors uniquely through the quotient ring).

Facts & Assumptions

Given: A symmetrizable Cartan datum and its root lattice Q; the displayed symbols and relations are formed over Q(q).

[F1]

The root lattice is freely generated by the simple roots, and ⟨αi,h⟩∈Z for h∈P∨ (Symmetrizable Cartan data for quantum groups, Kac Moody root lattice height and positive cone).

[F2]

qi=qdi with di>0, so qi−qi−1≠0 in Q(q) (Quantum integers, factorials, Gaussian binomials and divided powers at qi).

[F3]

The tensor algebra is a direct sum of finite words with concatenation product; assigning group degrees to its generator symbols gives the direct-sum grading by total degree. A linear map from the generating vector space extends uniquely to an algebra homomorphism (Tensor algebra of a vector space, Universal property of the tensor algebra).

[F4]

The two-sided ideal generated by the listed relations is an ideal, its quotient ring is defined by cosets, and maps whose kernel contains that ideal factor uniquely through the quotient (The ideal generated by a subset and principal ideals, The quotient ring R/I with (r+I)(s+I)=rs+I, A ring homomorphism whose kernel contains a two-sided ideal factors uniquely through the quotient ring).

Verification

technique · Check the toral inverse, homogeneity of the relations, and the universal factorization
1.1F2F4algebra

The relations KhK−h−K0 and K0−1 give KhK−h=1=K−hKh in the quotient. In particular Ki−1=K−dihi is a well-defined two-sided inverse, and the mixed commutator relation has a defined coefficient by [F2].

1.2F1algebra

Give each generator its stated degree in Q. The toral relations have degree 0; the KhEi and KhFi relations have degrees αi and −αi; the EiFj relation is homogeneous of degree 0 if i=j and degree αi−αj if i≠j; and the two Serre sums have degrees (1−aij)αi+αj and its negative. Thus every generator of IDJ is homogeneous.

2.1step 1.2F3F4algebra

Every element of the two-sided ideal is a finite sum of products urv with r a homogeneous defining relation. Decomposing u and v into their finite homogeneous components shows that each homogeneous component of every element of IDJ again belongs to IDJ. Hence the ideal is homogeneous and the quotient has the direct-sum root-lattice grading stated above.

3.1F3F4algebra∎

Given an assignment into B satisfying the relations, [F3] extends its values to a unique Q(q)-algebra homomorphism T→B. Every generator of IDJ maps to zero, so the ideal lies in the kernel; [F4] then gives the unique ring factor Uq(g)→B. Since the quotient map and the original map preserve scalars, surjectivity of the quotient map makes the factor preserve scalars as well. Conversely, any such factor is determined by the images of the generators because they generate the quotient.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-10-08Open item page →

The opposite Borels of a symmetrizable Kac–Moody algebra are root-degreewise dual Lie bialgebras

Statement

Let A=(aij)i,j∈I be a finite symmetrizable generalized Cartan matrix over C, with positive symmetrizer D=diag⁡(di) and rank r. Let g=g(A) be defined from a minimal realization, with triangular decomposition g=n−⊕h⊕n+ and Borels b±=h⊕n±. Use the invariant form B=(⋅∣⋅) normalized by (hi∣h)=αi(h)/di and (ei∣fj)=δij/di.

(i) Root-degreewise enveloping-algebra duality. The form pairs gα and g−α perfectly. It induces a canonical degreewise perfect vector-space pairing on U(n+) and U(n−) by PBW symmetrization. Each fixed root-degree component is finite-dimensional, and the pairing identifies U(n−)≅(U(n+))gr′, the restricted graded dual; the reverse identification holds as well. This is a vector-space pairing, not a Hopf pairing for the standard primitive coproducts.

(ii) Dual Borel Lie bialgebras. Let J⊆I be such that the principal block AJ is nonsingular, with ∣J∣=r; such a set is supplied by A nonsingular principal minor of the symmetrized Cartan matrix of size the rank. Choose complementary Cartan coordinates Dj for j∈I∖J with αi(Dj)=δij. In the quadratic Lie algebra d:=g⊕h, with form Bd((x,a),(y,b))=12((x∣y)−(a∣b)), the maps ι+(h+x+)=(h+x+,h) and ι−(h+x−)=(h+x−,−h) embed the Borels as complementary isotropic subalgebras. Thus they form a root-graded Manin triple. The cross pairing is (h∣h′)+12(x+∣x−) on h+x+ and h′+x−; its transpose brackets define dual Lie bialgebra structures on b+ and b−. Both cobrackets vanish on h, and the positive cobracket satisfies δ+(ei)=diei∧hi. The enveloping-algebra vector-space pairing in (i) retains the original invariant-form normalization; it is independent of this rescaled Manin pairing.

Facts & Assumptions

Given: A finite symmetrizable generalized Cartan matrix, a minimal realization, and the associated Kac–Moody algebra over C.

[F1]

The simple roots and coroots are independent, and a minimal realization has dim⁡h=2∣I∣−r (Realization of a generalized cartan matrix).

[F2]

The algebra has the triangular decomposition and separate finite-simple generator Serre presentations of n± (Kac moody algebra associated to a gcm, Contragredient algebra has a triangular decomposition, Serre presentation of a kac moody algebra).

[F3]

Root spaces are finite-dimensional, and gα pairs perfectly with g−α under the invariant form; the Cartan restriction is nondegenerate (Kac moody root spaces are finite dimensional, Invariant bilinear form for a symmetrizable kac moody algebra).

[F4]

A nonsingular principal block of size r exists for A (A nonsingular principal minor of the symmetrized Cartan matrix of size the rank).

[F5]

A countably spanned Kac–Moody half with a supplied countable ordered basis has the PBW ordered-monomial basis, and in characteristic zero PBW symmetrization is a filtered vector-space isomorphism (PBW for countably presented Kac Moody Lie algebras, PBW symmetrization in characteristic zero).

[F6]

A locally finite root-graded Manin triple gives dual Lie bialgebras by transposing the opposite brackets (A root-graded Manin triple gives dual Lie bialgebras).

Proof

technique · PBW symmetrization and the root-graded Manin double
1.1F1F4construct

Put H=span⁡{hi:i∈I} and define ρ:h→CI by ρ(x)=(αi(x))i∈I. By [F1], ρ is onto and ρ(H)=im⁡(AT) of dimension r. For the set J in [F4], projection of this image to CJ is an isomorphism: it is surjective because its restriction to the J-coordinate subspace has matrix AJT, and both spaces have dimension r. Hence im⁡(AT)∩CI∖J=0. For each j∉J, choose Dj with ρ(Dj) the jth coordinate vector. Their span intersects H trivially, and its dimension ∣I∖J∣=∣I∣−r makes it a complement to H; the form normalization gives (hi∣Dj)=δij/di. If r=∣I∣ this is the empty complementary family and h=H.

1.2F2F3given

By [F3], the invariant form has perfect opposite-root pairings and is nondegenerate on the Cartan subalgebra. Also n± are positively and negatively root-graded, respectively.

1.3F2givenalgebra

For fixed β=∑imiαi∈Q+, the degree-β words in the finite simple-generator tensor algebra are finite in number, so the separate Serre presentations [F2] make U(n+)[β] finite-dimensional; the same argument applies to U(n−)[−β]. The height-zero component is C1 on both sides.

1.4F2F5construct

Each half is countably spanned by its finite bracket words. Enumerating those words by length and lexicographic order and retaining the first vectors outside the preceding span gives a countable ordered basis; applying [F5] provides the PBW basis and the symmetrization isomorphism for both halves. This construction uses no choice principle.

1.5F3givenconstruct

For v1,…,vm∈n+ and w1,…,wℓ∈n−, define a pairing on the symmetric algebras to be zero when m≠ℓ, and when m=ℓ put ⟨v1⋯vm,w1⋯wm⟩S=1m!∑σ∈Sm∏t=1m(vt∣wσ(t)), with the empty products paired as 1. It is well defined under permutations of each list.

1.6F3givenalgebra

In a fixed root degree β and symmetric length m, only finitely many tuples of positive roots sum to β. The tensor-product pairings on each such tuple are perfect by [F3]; averaging over Sm identifies coinvariants with invariants because m!≠0 in C, so the induced pairing on Sm(n+)[β]×Sm(n−)[−β] is perfect. Summing over the finitely many lengths 0≤m≤ht⁡(β) gives a perfect pairing on the full symmetric-algebra root components.

1.7F2F3givenalgebra

Let h0 be a second copy of h and define Bd((x,a),(y,b))=12((x∣y)−(a∣b)) on d=g⊕h0 with the componentwise bracket. Jacobi holds componentwise, and [F3] makes this form invariant, symmetric and nondegenerate. The maps ι+ and ι− in (ii) are Lie homomorphisms because h is abelian and the bracket of two Borel elements has zero Cartan component. Their images are complementary: the positive and negative root components split by the triangular decomposition, and the two Cartan copies split into diagonal and antidiagonal subspaces. Each image is isotropic, since the invariant form is orthogonal between the Cartan and nonzero root spaces and vanishes on pairs of positive roots or on pairs of negative roots. The cross pairing is (h∣h′)+12(x+∣x−); it is degreewise perfect by [F3]. Within either Borel, a degree has only finitely many decompositions into degrees in its root cone, and all pieces are finite-dimensional. Thus these images satisfy the same-side finiteness required of the root-graded Manin triple; the whole double need not have finite decompositions.

2.1F5step 1.3step 1.4step 1.5step 1.6algebra

Transport the invariant-form symmetric-power pairing of steps 1.5–1.6 through the PBW symmetrization isomorphisms of step 1.4. They preserve root degree, so they give a perfect pairing of U(n+)[β] with U(n−)[−β] for every β∈Q+. Taking the direct sum of these finite-dimensional dualities gives the restricted graded-dual isomorphism in (i), in both directions; at β=0 it is the pairing ⟨1,1⟩=1.

3.1F3F6step 1.7algebra∎

By [F6], the transposed brackets give dual Lie bialgebra structures on the two Borel copies. For h∈h, every bracket of two elements of b− has either negative root degree or is zero in degree zero, so its cross pairing with ι+(h) vanishes; hence δ+(h)=0. The same argument gives δ−(h)=0. For a Cartan vector h, the determinant pairing gives ⟨ei∧hi,fi∧h⟩=αi(h)/(2di2), whereas ⟨ei,[fi,h]⟩=αi(h)/(2di). There are no other possible degree decompositions of the simple root except its simple-root and Cartan parts. Hence transposition gives δ+(ei)=diei∧hi, proving (ii) with the normalization used by the formal shuffle coproduct.

Remarks

The pairing on enveloping algebras in (i) is transported from symmetric algebras by PBW symmetrization. It is not asserted to satisfy Hopf-pairing adjunction for the standard primitive coproducts. Indeed, in type A2 let e12=[e1,e2] and f12=[f2,f1]. The primitive coproduct gives Δ(f12)=f12⊗1+1⊗f12, so any Hopf pairing with ⟨ei,1⟩=⟨1,ei⟩=0 and ⟨xy,z⟩=⟨x⊗y,Δz⟩ would force both ⟨e1e2,f12⟩ and ⟨e2e1,f12⟩ to vanish. It would then give ⟨e12,f12⟩=0, contrary to the perfect opposite-root pairing in [F3].

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-08Open item page →

The quantum Pascal recurrences, the Gauss product formula and Gaussian integrality

Statement

Fix a symmetrizable Cartan datum and i∈I, and use the conventions of Quantum integers, factorials, Gaussian binomials and divided powers at qi. Write Cm,r:=(mr)i, with Cm,r=0 when r<0 or r>m.

(i) Pascal recurrences. For m≥1 and every integer r,

Cm,r=qi−rCm−1,r+qim−rCm−1,r−1=qirCm−1,r+qir−mCm−1,r−1.

(ii) Symmetry. For 0≤r≤m, Cm,r=Cm,m−r.

(iii) Gauss product formula. For every N≥0, in the polynomial ring Q(q)[z],

∏j=0N−1(1+qi2jz)=∑r=0Nqir(N−1)CN,rzr.

Consequently, for N≥1,

∑r=0N(−1)rqir(N−1)CN,r=0.

(iv) Integrality. For 0≤r≤m,

Cm,r∈qi−r(m−r)Z[qi2]⊆Z[qi±1].

Thus every Gaussian quotient is a Laurent polynomial in qi with integer coefficients, and the Pascal recurrences hold in that Laurent polynomial ring.

Facts & Assumptions

Given: A symmetrizable Cartan datum, a fixed i∈I, and the symmetric qi-integer, factorial and Gaussian quotient from Quantum integers, factorials, Gaussian binomials and divided powers at qi.

[F1]

qi=qdi for an indeterminate q and positive integer di; all symmetric qi-factorials in the quotient are nonzero (Quantum integers, factorials, Gaussian binomials and divided powers at qi).

[F2]

R[z] is the polynomial ring over a commutative ring R, and Z[t±1] consists of finite Laurent sums with integer coefficients (The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution, The Laurent polynomial ring as the principal localisation of Z[t] at t).

Proof

technique · Derive the recurrences from a symmetric $q_i$-integer identity, then use induction
1.1givenF1algebra

For 0≤r≤m, the numerator identity qi−r(qim−r−qi−(m−r))+qim−r(qir−qi−r)=qim−qi−m gives [m]i=qi−r[m−r]i+qim−r[r]i.

2.1step 1.1F1algebra

For 1≤r≤m−1, multiply the identity of step 1.1 by [m−1]i!/([r]i![m−r]i!) and use the factorial quotient to obtain Cm,r=qi−rCm−1,r+qim−rCm−1,r−1. The same formula holds at r=0,m by Cm,0=Cm,m=1 and the out-of-range zero convention; for r<0 or r>m every term is zero. The factorial definition gives Cm,r=Cm,m−r for 0≤r≤m; applying the first recurrence at m−r and using this symmetry gives the second recurrence.

3.1step 2.1F1F2algebra

Put Gm,r:=qir(m−r)Cm,r. The first recurrence in step 2.1 gives, for 1≤r≤m−1, Gm,r=Gm−1,r+qi2(m−r)Gm−1,r−1. Since Gm,0=Gm,m=1, induction on m shows Gm,r is a polynomial in qi2 with integer coefficients; this proves Cm,r∈qi−r(m−r)Z[qi2]. Because di>0 and q is indeterminate, distinct powers of qi are linearly independent over Z, so this evaluation embeds the Laurent polynomial ring and gives the stated inclusion and Laurent-polynomial recurrences.

3.2step 2.1F2algebra

Let PN(z):=∏j=0N−1(1+qi2jz). For N=0 both sides of the Gauss formula are 1. If it holds for N−1, the coefficient of zr in PN is qir(N−2)CN−1,r+qi(r−1)(N−2)+2(N−1)CN−1,r−1; by the first recurrence in step 2.1 this equals qir(N−1)CN,r, since (r−1)(N−2)+2(N−1)=r(N−1)+N−r. Thus induction proves the product formula in Q(q)[z].

4.1step 3.2algebra∎

For N≥1, evaluate the formula of step 3.2 at z=−1. The factor with j=0 makes the product zero, so its right side is the alternating Gaussian sum in the statement and is zero. This proves the final assertion and completes all parts.

DefinitionDefinition: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-08Open item page →

The formal quantum shuffle Borel and its Cartan crossed product

Definition

Let (I,A,D,P,P∨,q) be a symmetrizable Cartan datum (Symmetrizable Cartan data for quantum groups), n=∣I∣, and r=rank⁡A. Choose J⊆I with ∣J∣=r and det⁡(AJ)≠0, as supplied by A nonsingular principal minor of the symmetrized Cartan matrix of size the rank. Work over R=C⟦ℏ⟧ with q=eℏ and qi=ediℏ.

(a) Word space and shuffle product. Let V=⨁i∈ICvi with deg⁡vi=ϵi∈NI. Set Sh(V)=⨁k≥0V⊗k⟦ℏ⟧ and denote z1⊗⋯⊗zk by [z1∣⋯∣zk]. For homogeneous letters, let Σk,l consist of permutations σ mapping original positions to output positions, with σ(1)<⋯<σ(k) and σ(k+1)<⋯<σ(k+l). Define

[z1∣⋯∣zk] [zk+1∣⋯∣zk+l]=∑σ∈Σk,lq−∑i<j, σ(i)>σ(j)⟨deg⁡zi,deg⁡zj⟩[zσ−1(1)∣⋯∣zσ−1(k+l)],⟨ϵi,ϵj⟩=diaij.

This product is associative and makes Sh(V) an NI-graded algebra; every color-degree component is a finite free R-module. Write ⟨V⟩ for the subalgebra generated by the one-letter words [vi].

(b) Cartan derivations and crossed products. Define R-linear derivations on letters by hiV(vj)=aijvj and DjV(vi)=δijvi for j∈I∖J, and extend them to words by summing their action over positions. They extend as commuting derivations of Sh(V) and preserve ⟨V⟩. Let C consist of n commuting symbols Hi corresponding to hiV and n−r commuting symbols Dj corresponding to DjV. Put AC:=lim←⁡NR[C]/ℏNR[C]≅C[C]⟦ℏ⟧, so each coefficient of ℏm is a finite Cartan polynomial. The Cartan crossed-product modules are

V:=⟨V⟩⊗RAC,S:=Sh(V)⊗RAC.

The Cartan symbols are independent polynomial generators; the operators hiV,DjV describe their commutators and are not their images under a Cartan inclusion. The Cartan coefficient algebra AC is ℏ-adically complete; no algebraic-freeness claim is made for the crossed products. Their products extend the products of the two factors and obey Xsx−xXs=δs(x) for the corresponding derivation δs. Equivalently, for multi-indices α,β,

(xXα)(yXβ)=∑0≤t≤α(αt)x δt(y)Xα+β−t,

where (αt)=∏s(αsts) and δt=∏sδsts. The r coroot directions indexed by J are linearly independent because AJ is nonsingular, and the remaining n−r directions supplement the minimal realization of dimension 2n−r (Realization of a generalized cartan matrix); the number of Cartan symbols is n+(n−r)=2n−r.

(c) ℏ-adic Hopf structure and conditional Serre map. In the commutative formal Cartan factor, define exp⁡(ℏZ)=∑m≥0ℏmZm/m! for each Cartan polynomial Z; every coefficient is a finite polynomial, so this is an element of AC. Write ⊗^R for the color-degreewise ℏ-adically completed tensor product: for each total color degree γ, complete ⨁α+β=γS[α]⊗RS[β] in the ℏ-adic topology, then take the direct sum over γ; use the same convention for V and for iterated tensor products. There are finitely many splittings of each γ∈NI. The algebras are complete in each color degree, and multiplication extends on these degreewise completed tensors. Their direct sums are not asserted to be complete for limits with unbounded color support. These formulas make S a topological Hopf R-algebra: Δ:S→S⊗^RS and ε:S→R are continuous algebra maps, Δ is coassociative with both counit identities, and the continuous antipode satisfies both convolution-inverse equations.

ΔS(Xs)=Xs⊗1+1⊗Xs,ΔS([vi1∣⋯∣vim])=∑k=0m[vi1∣⋯∣vik]⊗exp⁡ ⁣(ℏ∑t≤kditHit)[vik+1∣⋯∣vim].

The counit sends every nonempty word and every Xs to 0 and sends 1 to 1. The antipode is the common convolution inverse constructed recursively in word length and Cartan-polynomial degree. The same structure maps make V a topological Hopf subalgebra.

Let ER:=⨁i∈IRei and TR(ER):=⨁k≥0ER⊗Rk be its tensor algebra with concatenation; every assignment of the ei to a unital R-algebra extends uniquely by wordwise multiplication. Let Uℏn+ be the quotient of TR(ER) by the two-sided ideal generated by

∑s=01−aij(−1)s(1−aijs)qiei1−aij−sejeis=0(i≠j),

where the symmetric Gaussian coefficients are evaluated at qi=ediℏ. Let Uℏb+ be the crossed product of this quotient with the commuting Cartan symbols Hi,Dj, using the multiplication rule of part (b) for the induced derivations and acting by [Hi,ej]=aijej and [Dj,ei]=δijei. If the quantum Serre sums vanish in ⟨V⟩, then Hi↦Hi, Dj↦Dj, and ei↦[vi], with the first two images the independent target Cartan symbols in AC define an algebra homomorphism pℏ:Uℏb+→V. This definition makes no assertion that the Serre-vanishing condition holds or that pℏ is injective. All formal exponentials are interpreted ℏ-adically, and tensor completions are taken color-degreewise as specified above; no limit with unbounded color support is placed in either direct-sum algebra.

Facts & Assumptions

Given: A finite symmetrizable Cartan datum, its minimal realization, and the formal parameter ℏ.

[F1]

diaij=djaji, so ⟨ϵi,ϵj⟩=diaij is a symmetric bilinear form (Symmetrizable Cartan data for quantum groups).

[F2]

There is a set J with ∣J∣=r and det⁡(AJ)≠0 (A nonsingular principal minor of the symmetrized Cartan matrix of size the rank).

[F3]

A minimal realization has dimension 2n−r and its simple coroots are linearly independent (Realization of a generalized cartan matrix).

[F4]

Formal power series use coefficientwise addition and finite Cauchy products; C⟦ℏ⟧ is complete and separated in its ℏ-adic topology (Formal power series over a commutative ring and the coefficient-extraction functional [xn]).

[F5]

For a commutative Q-algebra and u divisible by the formal variable, exp⁡(u)=∑m≥0um/m! (Formal exponential, logarithm, and binomial powers over a commutative Q-algebra).

[F6]

The symmetric Gaussian coefficients specialize to Laurent-polynomial coefficients in qi, so their substitution qi=ediℏ lies in R (Quantum integers, factorials, Gaussian binomials and divided powers at qi, The quantum Pascal recurrences, the Gauss product formula and Gaussian integrality).

[F7]

Tensor powers form the direct-sum word space and concatenation is associative (Tensor algebra of a vector space).

[F8]

The bialgebra and Hopf axioms are algebra-map coproduct and counit, coassociativity, both counit identities, and both antipode convolution-inverse equations; here their tensor products are ℏ-adically completed as specified in (c) (Bialgebras, counits and antipodes over a commutative ring).

[F10]

Tensor products of algebras have multiplication (a⊗b)(a′⊗b′)=aa′⊗bb′ (The tensor product of R-algebras has multiplication (a⊗b)(a′⊗b′)=aa′⊗bb′).

[F11]

No axiom of choice is used: J is a single subset of the finite set I supplied by [F2]; all sums are finite and the recursion terminates on finite word length and Cartan degree.

[F12]

A polynomial ring on the finite family of Cartan symbols consists of finite polynomials in those commuting symbols (The polynomial ring R[xi:i∈I] as finitely supported coefficient families on monomials).

[F13]

A formal series over C is a unit exactly when its constant coefficient is nonzero; a submodule of a finite free module over a PID is finite free (A formal power series is a unit exactly when its constant coefficient is a unit, Principal ideal domain, A submodule of a free module of finite rank over a PID is free of no larger rank).

Proof

technique · Verify the shuffle and Ore products by finite combinatorics, check the Hopf maps on word cuts and crossed relations, then construct the antipode by a terminating recursion on word length and Cartan-polynomial degree
1.1F4F7F13algebra

For three word blocks, a shuffle of all letters is uniquely a shuffle of the first two blocks followed by a shuffle with the third, and also uniquely a shuffle of the last two followed by one with the first. In either description every inverted pair of letters occurs exactly once and contributes q−⟨deg⁡zi,deg⁡zj⟩; scalar weights therefore agree, proving associativity. The color degree is additive, and a fixed color multidegree has only finitely many words, each an R-basis vector in the tensor-word space. Thus every graded component of Sh(V) is finite free. Also R is a PID: a nonzero ideal has a least order m, an element of that order is ℏm times a unit by [F13], and every other element is divisible by ℏm, so the ideal is (ℏm). Each ⟨V⟩[α] is a submodule of the finite free word module and hence finite free by [F13]. Tensoring these finite bases with AC shows that each color-degree component of S and V is a finite direct sum of copies of the complete coefficient algebra AC. Consequently each is ℏ-adically complete, and continuous multiplication on a fixed degree split extends to its completion; the finitely many splits of a fixed total degree give the stated completed multiplication.

1.2algebra

On a word, hiV acts by the sum of the aij weights of its letters and DjV by the number of color-j letters. Every shuffle term preserves these color counts, so each operator satisfies Leibniz for the shuffle product; their diagonal actions commute. Since each sends vi to a scalar multiple of itself, it preserves the subalgebra ⟨V⟩.

1.3F4F7F12algebra

Repeatedly commute each Xs past a coefficient using Xsx=xXs+δs(x); the commuting derivations and Leibniz rule give the finite multi-binomial formula in (b). Associativity follows from the two possible reductions of XsXtx, which agree because δsδt=δtδs, and the two reductions of Xsxy, which agree because δs(xy)=δs(x)y+xδs(y). Products of formal ℏ-series have finite coefficient sums and each multi-index sum is finite, so multiplication is well-defined coefficientwise.

1.4F2F3given

The principal minor gives ∣I∖J∣=n−r; the minimal realization has n independent simple coroots and dimension 2n−r, so the n coroot symbols and n−r supplementary symbols give the required count. This uses no claim that J alone spans the Cartan.

1.5F4F5F12algebra

The exponent Z=∑t≤kditHit is a polynomial in commuting Cartan variables. Apply [F5] in the commutative coefficient algebra with formal variable ℏ: at order m the coefficient is the finite polynomial Zm/m!. The coefficient of ℏm in exp⁡(ℏZ)exp⁡(−ℏZ) is Zm∑j=0m(−1)m−j(mj)/m!, equal to 0 for m>0 and 1 for m=0, so Gk is invertible with inverse exp⁡(−ℏZ). Since each Cartan variable is primitive, coefficientwise binomial expansion gives Δ(Gk)=exp⁡(ℏ(Z⊗1+1⊗Z))=Gk⊗Gk.

2.1F1F7F10step 1.5algebra

Fix cuts u=u+u− and v=v+v−, and write Gα=exp⁡(ℏ∑iαidiHi) for a prefix of color degree α. If wβ has color degree β, then Hiwβ=wβ(Hi+∑jβjaij), so Gαwβ=q∑i,jαidiβjaijwβGα=q⟨α,β⟩wβGα. A global shuffle with this cut consists of a shuffle of the prefixes and a shuffle of the suffixes; its cross-cut inversions contribute q−⟨deg⁡u−,deg⁡v+⟩=q−⟨deg⁡v+,deg⁡u−⟩ by [F1]. On the right, multiplying the second tensor factors gives (Gdeg⁡u+u−)(Gdeg⁡v+v−); moving u− past Gdeg⁡v+ contributes the same cross-cut factor. Rewriting the resulting combined prefix exponential in word-times-Cartan normal form contributes q⟨deg⁡u++deg⁡v+,deg⁡u−+deg⁡v−⟩. Rewriting the corresponding Cartan-left factor on the left-hand coproduct gives this identical normal-form factor. The prefix and suffix shuffle weights also agree, so every cut and pair of shuffles has the same coefficient on both sides. Thus Δ preserves the shuffle product.

2.2step 1.2F10algebra

The operators hiV,DjV are additive over a word cut and annihilate the Cartan symbols, so they fix each Cartan exponential. Hence Δ(δs(w))=(δs⊗id+id⊗δs)Δ(w), which is exactly the commutator of Δ(Xs)=Xs⊗1+1⊗Xs with Δ(w). Thus the coproduct preserves every crossed relation and extends as an algebra map to S.

2.3step 1.5F4F8algebra

Both iterated coproducts of a word sum over two cuts; the factors agree because each prefix exponential is group-like, so Δ is coassociative. Define ε as projection to the empty-word, constant-Cartan coefficient. It is an algebra map: positive color degree maps to zero, and every crossed commutator δs(x) with a positive-degree word also maps to zero. The counit identities hold because only the empty-prefix term survives after applying ε to the first factor, only the all-letter prefix term survives after applying it to the second, and every Cartan exponential has counit 1.

3.1step 1.5step 2.3F4F8algebra

Let w be a word and P a Cartan monomial, and induct lexicographically on (word length, Cartan degree). On Cartan polynomials set S(P(X))=P(−X); these are the antipode equations for the primitive commuting Cartan generators. For positive word length, in Δ(wP) the unique term with the full word and full Cartan degree in the first tensor factor is wP⊗Gw. Every other term has a shorter first word or smaller first Cartan degree. Since Gw is invertible, the left convolution equation recursively determines S(wP). The unique term with full word and full Cartan degree in the second tensor factor is 1⊗wP; all other terms have a shorter second word or smaller second Cartan degree, so the right convolution equation recursively determines a right inverse as well. The recursions terminate at every pair of finite word length and Cartan degree; each coefficient of every Gw±1 is a finite Cartan polynomial, so they extend coefficientwise to the formal ℏ-series in the stated modules. The left and right convolution inverses coincide by associativity. For a letter vi, Δ(vi)=1⊗vi+vi⊗Gi, so the equations give S(vi)=−viGi−1=−[vi]exp⁡(−ℏdiHi).

4.1step 2.2step 3.1F8algebra

In Hom⁡R(S⊗^RS,S) with convolution from the completed tensor coalgebra, S∘m is a two-sided convolution inverse of m: since Δ and ε are algebra maps, applying the two antipode equations to ab gives both inverse equations. The map T=m(S⊗S)∘flip is also a two-sided inverse, since (T∗m)(a⊗b)=∑S(b(1))S(a(1))a(2)b(2)=ε(a)ε(b)1 and (m∗T)(a⊗b)=∑a(1)b(1)S(b(2))S(a(2))=ε(a)ε(b)1. These equalities follow by first applying the left and right antipode equations to a and b, respectively. Uniqueness of convolution inverses gives S∘m=T, so S is anti-multiplicative. Hence S(Xs)=−Xs and S([vi])=−[vi]exp⁡(−ℏdiHi) show S(V)⊆V. The coproduct and counit also restrict, so V is a topological Hopf subalgebra.

5.1F6F9givenalgebra∎

The Cartan derivations preserve the two-sided ideal generated by the quantum Serre sums because each sum is homogeneous in the color grading. Under the stated vanishing condition, the free-algebra assignment ei↦[vi] kills every defining Serre generator; its Cartan commutators match the crossed-product derivations. By [F9] it therefore factors uniquely through Uℏb+ to the stated algebra map.

DefinitionDefinition: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-08Open item page →

Positive, negative and toral quantum subalgebras and their root gradings

Statement

Let Uq(g) be the Drinfeld–Jimbo algebra of The Drinfeld-Jimbo quantized enveloping algebra by generators and relations with its root-lattice grading Uq(g)=⨁β∈QUq(g)[β]. Define

Uq+:=⟨Ei : i∈I⟩,Uq−:=⟨Fi : i∈I⟩,Uq0:=⟨Kh : h∈P∨⟩,

the Q(q)-subalgebras generated by the indicated images, and let Q+=⨁iZ≥0αi. Then

Uq±=⨁β∈Q+Uq±[±β],Uq+[β]:=Uq(g)[β]∩Uq+,Uq−[−β]:=Uq(g)[−β]∩Uq−.

The degree-zero component Uq(g)[0] contains Uq0, and for every x∈Uq(g)[β] and h∈P∨,

KhxKh−1=qβ(h)x.

Thus each Kh∈Uq0 acts by conjugation on the degree-β component as the scalar qβ(h). The elements Kh span Uq0 and satisfy KhKh′=Kh+h′ with K0=1; hence Uq0 is a quotient of the group algebra Q(q)[P∨]. The subalgebras Uq± are spanned by finite monomials in the Ei and Fi, respectively. Define the nonnegative and nonpositive product spans by

Uq≥0:=span⁡Q(q)(Uq0Uq+),Uq≤0:=span⁡Q(q)(Uq−Uq0).

This definition asserts no linear independence of these spanning families and no injectivity of the natural maps; those claims belong to Triangular decomposition of a quantized enveloping algebra.

Facts & Assumptions

Given: The Drinfeld–Jimbo presentation is homogeneous in the root lattice, with the degrees of Ei,Fi,Kh specified in The Drinfeld-Jimbo quantized enveloping algebra by generators and relations. The datum provides the pairing of roots and coroots in Symmetrizable Cartan data for quantum groups.

[F1]

Uq(g)=⨁β∈QUq(g)[β] with deg⁡Ei=αi, deg⁡Fi=−αi, and deg⁡Kh=0 (The Drinfeld-Jimbo quantized enveloping algebra by generators and relations).

[F2]

KhEiKh−1=q⟨αi,h⟩Ei and KhFiKh−1=q−⟨αi,h⟩Fi (The Drinfeld-Jimbo quantized enveloping algebra by generators and relations).

[F3]

K0=1 and KhKh′=Kh+h′ (The Drinfeld-Jimbo quantized enveloping algebra by generators and relations).

[F4]

For a ring homomorphism f:R→S, the first isomorphism theorem gives R/ker⁡f≅im⁡f (First isomorphism theorem for rings: R/ker⁡f≅im⁡f).

[F5]

The root lattice is Q=⨁iZαi, and its pairing with P∨ is integer-valued and additive (Symmetrizable Cartan data for quantum groups).

Proof

technique · Use the root grading on finite generator words, and map the group algebra of the torus lattice onto the generated toral subalgebra
1.1F1givenalgebra

Every word in the Ei is homogeneous of degree in Q+; the empty word has degree 0. By generation, every element of Uq+ is a finite sum of such words. Grouping those terms by degree and using the direct grading of Uq(g) gives Uq+=⨁β∈Q+(Uq(g)[β]∩Uq+).

1.2F1givenalgebra

Every word in the Fi is homogeneous of degree in −Q+, including the degree-zero empty word. Grouping finite sums by their degrees in the direct grading gives Uq−=⨁β∈Q+(Uq(g)[−β]∩Uq−).

1.3F3F4construct

Define Q(q)[P∨] to be the vector space of finite sums ∑hchXh with multiplication XhXh′=Xh+h′. By [F3], ϕ(Xh)=Kh extends linearly and multiplicatively to a unital algebra homomorphism into Uq0. Its image is all of Uq0, since the image contains every generator Kh and consists of finite linear combinations of them. Thus [F4] identifies Uq0 with Q(q)[P∨]/ker⁡ϕ, and the Kh span it.

1.4F1F2F3F5algebra

The relations in [F3] give Kh−1=K−h. By [F2], conjugation by Kh acts on a word of degree β by multiplying each Ei factor by qαi(h), each Fi factor by q−αi(h), and each Kh′ factor by 1. Additivity of the pairing in [F5] makes the product scalar qβ(h). Every element is a finite sum of generator-word images; grouping those words by degree and using the direct grading [F1] shows that an element of Uq(g)[β] is a finite linear combination of degree-β word images. Thus the formula holds for every such x. Since the Kh have degree zero, their generated subalgebra Uq0 lies in Uq(g)[0].

2.1givenalgebra∎

By definition, finite monomials in the Ei and Fi span Uq+ and Uq−. Finite sums of products uv with u∈Uq0,v∈Uq+ and with u∈Uq−,v∈Uq0 span the two stated product subspaces. No basis or independence assertion is involved.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-08Open item page →

The quantum binomial expansion for q-commuting elements

Statement

Let t be an indeterminate over Q, and let A be a unital associative Q(t)-algebra. For N≥0 and 0≤r≤N, set

BN,r(t):=[N]t![r]t![N−r]t!,[m]t:=1+t+⋯+tm−1,[m]t!:=∏k=1m[k]t,[0]t:=0,[0]t!:=1,

and set BN,r(t)=0 when r<0 or r>N. If x,y∈A satisfy yx=txy, then for every N≥0,

(x+y)N=∑r=0NBN,r(t)xryN−r.

For a symmetrizable Cartan datum and i∈I, any unital Q(q)-algebra can be viewed as a Q(t)-algebra via t↦qi2. In that algebra, if yx=qi2xy, then

(x+y)N=∑r=0Nqir(N−r)(Nr)ixryN−r,

using the symmetric Gaussian binomials of Quantum integers, factorials, Gaussian binomials and divided powers at qi. If instead yx=t−1xy, the same expansion has coefficients BN,r(t−1)=t−r(N−r)BN,r(t).

Facts & Assumptions

Given: The conventions above, the Gaussian quotient definitions, and the relation yx=txy when the generic expansion is used.

[F1]

The asymmetric q-integer and factorial use [m]t=1+t+⋯+tm−1 and the empty product [0]t!=1; for k≥1, [k]t is a nonzero polynomial, so the Gaussian factorial quotient is defined in Q(t) (The q-integer, q-factorial and q-multinomial coefficients).

[F2]

The symmetric Gaussian coefficient satisfies CN,r=qi−rCN−1,r+qiN−rCN−1,r−1 (The quantum Pascal recurrences, the Gauss product formula and Gaussian integrality).

[F3]

The symmetric and asymmetric Gaussian coefficients satisfy (Nr)i=qi−r(N−r)BN,r(qi2), with qi=qdi and di>0 (Quantum integers, factorials, Gaussian binomials and divided powers at qi).

[F4]

Since q is indeterminate and di>0, qi2=q2di is transcendental; substitution t↦qi2 therefore embeds Q(t) into Q(q) (Quantum integers, factorials, Gaussian binomials and divided powers at qi).

Proof

technique · Induct on the power, using the exact q-Pascal coefficient for the chosen commutation order
1.1givenF1algebra

For k≥1, [k]t−1=t−(k−1)[k]t; multiplying for k=1,…,N and dividing the factorials gives BN,r(t−1)=t−N(N−1)/2+r(r−1)/2+(N−r)(N−r−1)/2BN,r(t)=t−r(N−r)BN,r(t). The exponent equality follows by expanding the three quadratic terms; for N=0,r=0 both coefficients equal 1.

1.2givenF2F3F4algebra

Put DN,r:=qir(N−r)CN,r. Multiplying the recurrence [F2] by qir(N−r) and using [F3] gives BN,r(qi2)=BN−1,r(qi2)+(qi2)N−rBN−1,r−1(qi2) for 0≤r≤N; outside this range all terms vanish. The first exponent becomes r(N−r−1), and the second differs from (r−1)(N−r) by 2(N−r). By the injectivity in [F4], this is the generic recurrence BN,r(t)=BN−1,r(t)+tN−rBN−1,r−1(t).

2.1step 1.2F3algebra

For N=0 the formula is 1=1. Suppose it holds for N−1. From yx=txy, induction on a gives yax=taxya: it is true for a=0, and ya+1x=tayxya=ta+1xya+1. Multiplying the N−1 expansion on the right by x+y and reindexing the terms from the final x gives the coefficient BN−1,r(t)+tN−rBN−1,r−1(t) at xryN−r. By step 1.2 this is BN,r(t), proving the generic expansion. Under t=qi2, [F3] turns this coefficient into qir(N−r)CN,r and gives the symmetric formula.

3.1step 1.1step 2.1algebra∎

If yx=t−1xy, apply the generic expansion of step 2.1 with parameter t−1. Step 1.1 rewrites its coefficients as t−r(N−r)BN,r(t), proving the inverse-parameter formula as well.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-08Open item page →

The Chevalley involution, bar involution, and contravariant anti-involution preserve the Drinfeld–Jimbo ideal

Statement

Let T and IDJ be the free algebra and defining ideal of The Drinfeld-Jimbo quantized enveloping algebra by generators and relations, and put mij=1−aij and Serreij± for its displayed Serre sums. Define ω,ψ,τ:T→T on generators by

ω(Ei)=−Fi,ω(Fi)=−Ei,ω(Kh)=K−h,

ψ(Ei)=Ei,ψ(Fi)=Fi,ψ(Kh)=K−h,

τ(Ei)=Fi,τ(Fi)=Ei,τ(Kh)=K−h.

Here ω fixes Q(q), while ψ and τ apply the field involution q↦q−1 to coefficients; ω and ψ extend as algebra maps, and τ extends as an anti-algebra map. Each map preserves IDJ and descends to Uq(g). All three are involutions; ω is a Q(q)-algebra automorphism, ψ a semilinear algebra automorphism, and τ a semilinear algebra anti-involution. On the Serre sums,

ω(Serreij±)=(−1)aijSerreij∓,ψ(Serreij±)=Serreij±,τ(Serreij±)=(−1)1−aijSerreij∓.

Facts & Assumptions

Given: The free algebra T and the four families of generators of IDJ in the Drinfeld–Jimbo presentation.

[F1]

The ideal is generated by the toral, toral-action, mixed EiFj, and positive and negative Serre relations displayed in The Drinfeld-Jimbo quantized enveloping algebra by generators and relations.

[F2]

The symmetric Gaussian coefficients satisfy (mr)i=(mm−r)i and are invariant under qi↦qi−1; [m]i has the same inversion symmetry (Quantum integers, factorials, Gaussian binomials and divided powers at qi).

[F3]

Assignments of the free generators extend uniquely to algebra maps on T (Universal property of the tensor algebra); reversing the order of each word gives the corresponding anti-algebra extension.

Proof

technique · Check the four relation families and then pass to the quotient
1.1F3construct

The substitution σ(q)=q−1 is an involutive field automorphism of Q(q). By [F3], the displayed assignments extend to a Q(q)-algebra endomorphism ω, a σ-semilinear algebra endomorphism ψ, and a σ-semilinear anti-algebra endomorphism τ of T. Each square fixes every generator and coefficient, so all three squares are the identity on T.

1.2F1algebra

The maps send K0−1 to itself. They send KhKh′−Kh+h′ respectively to K−hK−h′−K−(h+h′), the same relation, and K−h′K−h−K−(h+h′), the same family with its two indices reversed. Thus each image lies in IDJ.

1.3F1algebra

Write a=⟨αi,h⟩, Ri,hE=KhEi−qaEiKh, and Ri,hF=KhFi−q−aFiKh. Then ω(Ri,hE)=−Ri,−hF and ω(Ri,hF)=−Ri,−hE; ψ(Ri,hE)=Ri,−hE and ψ(Ri,hF)=Ri,−hF; and τ(Ri,hE)=−q−aRi,−hF and τ(Ri,hF)=−qaRi,−hE. Each is a scalar multiple of a defining toral-action relation.

1.4F1F2algebra

Let Rij=EiFj−FjEi−δij(Ki−Ki−1)/(qi−qi−1). Direct substitution, including the semilinear inversion of the coefficient in ψ and τ, gives ω(Rij)=−Rji, ψ(Rij)=Rij, and τ(Rij)=Rji. The ratio (Ki−Ki−1)/(qi−qi−1) is unchanged when both numerator and denominator are inverted. Hence these images lie in IDJ.

1.5F1F2algebra

Put m=1−aij. Applying ω to a positive Serre sum contributes m+1 minus signs and preserves the order of factors, so ω(Serreij+)=(−1)m+1Serreij−=(−1)aijSerreij−; the negative case is symmetric. By [F2], ψ fixes the coefficients and each Ei,Fi, so it fixes both Serre sums. The anti-map τ reverses each monomial; reindexing r↦m−r and using [F2] gives τ(Serreij±)=(−1)mSerreij∓. Thus all Serre images belong to IDJ.

2.1F1step 1.1step 1.2step 1.3step 1.4step 1.5algebra∎

The preceding steps show that each map sends every generator of IDJ into that ideal. Since IDJ is two-sided, the algebra maps and the anti-algebra map send the whole ideal into itself. They therefore descend to the quotient; their squares remain the identity there, which makes the descended maps automorphisms or an anti-automorphism of the stated types. This proves the assertion.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-08Open item page →

The coproduct preserves the positive and negative quantum Serre ideals

Statement

Let (I,A,D,P,P∨,q) be a symmetrizable Cartan datum, use the notation qi, Kh and Ki=Kdihi of The Drinfeld-Jimbo quantized enveloping algebra by generators and relations, and put mij=1−aij for i≠j. Let T+ and T− be the free Q(q)-algebras on the symbols Ei,Kh and Fi,Kh, respectively. Let R+ and R− be the two-sided ideals generated by K0−1, KhKh′−Kh+h′, and respectively KhEi−q⟨αi,h⟩EiKh or KhFi−q−⟨αi,h⟩FiKh. Define the toral-action algebras B+=T+/R+ and B−=T−/R−, and let J+ and J− be their two-sided ideals generated by the images of the corresponding Drinfeld–Jimbo Serre elements.

The generator assignments

Δ+(Kh)=Kh⊗Kh,Δ+(Ei)=Ei⊗Ki−1+1⊗Ei,Δ−(Kh)=Kh⊗Kh,Δ−(Fi)=Fi⊗1+Ki⊗Fi

define algebra homomorphisms B±→B±⊗Q(q)B±. For every i≠j, the Serre elements are quasiprimitive:

Δ+(Serreij+)=Serreij+⊗Ki−mijKj−1+1⊗Serreij+,Δ−(Serreij−)=Serreij−⊗1+KimijKj⊗Serreij−.

Consequently Δ±(J±)⊆J±⊗B±+B±⊗J±. The canonical maps B±→Uq(g) send the displayed coproducts of the Serre elements to zero in Uq(g)⊗Uq(g); thus these formulas provide the Serre-family part of the check that the Drinfeld–Jimbo coproduct descends.

Facts & Assumptions

Given: A symmetrizable Cartan datum over Q(q) and the defining relations and Serre sums in The Drinfeld-Jimbo quantized enveloping algebra by generators and relations.

[F1]

For a=aij and m=1−a, symmetrizability gives diaij=djaji, so KjEiKj−1=qiaEi; also KiEjKi−1=qiaEj (Symmetrizable Cartan data for quantum groups).

[F2]

The Drinfeld–Jimbo toral-action relations are those displayed in The Drinfeld-Jimbo quantized enveloping algebra by generators and relations.

[F3]

The positive and negative Serre sums are those displayed in The Drinfeld-Jimbo quantized enveloping algebra by generators and relations.

[F4]

The symmetric Gaussian coefficients are factorial quotients CN,r=(Nr)i=[N]i!/([r]i![N−r]i!) and satisfy CN,r=CN,N−r (Quantum integers, factorials, Gaussian binomials and divided powers at qi).

[F5]

If x,y lie in an algebra and yx=qi2xy, then (x+y)N=∑r=0Nqir(N−r)(Nr)ixryN−r (The quantum binomial expansion for q-commuting elements).

[F6]

For every N≥1, ∑r=0N(−1)rqir(N−1)CN,r=0 (The quantum Pascal recurrences, the Gauss product formula and Gaussian integrality).

[F7]

The tensor product has multiplication (a⊗b)(c⊗d)=ac⊗bd (The tensor product of R-algebras has multiplication (a⊗b)(a′⊗b′)=aa′⊗bb′).

[F8]

The tensor algebra on the free generators admits the unique algebra extension of every generator assignment (Tensor algebra of a vector space, Universal property of the tensor algebra).

[F10]

A two-sided ideal generated by a subset is as in The ideal generated by a subset and principal ideals.

Proof

technique · Work first in the toral-action quotients, expand each coproduct by the q-binomial theorem, and cancel every mixed bidegree by the alternating Gaussian identity [F6]
1.1givenF7F8construct

Write a=aij, m=1−a, Cn,r=(nr)i, and let T±,R±,B±,J± be as in the statement. The algebra maps on T± prescribed by the displayed coproduct assignments exist by the tensor-algebra universal property in [F8], with the target tensor products made into algebras by [F7].

1.2F6algebra

For every N≥1, the coefficient identity [F6] cancels the alternating sum with exponent (N−1)r. For N=0 the corresponding one-term sum equals 1. These two cases will distinguish the mixed terms from the two extreme bidegrees below.

2.1step 1.1F2F7F9algebra

The images of K0−1 and KhKh′−Kh+h′ vanish because Δ(K0)=1⊗1 and Δ(KhKh′)=KhKh′⊗KhKh′=Δ(Kh+h′). For b=⟨αi,h⟩, in the positive algebra Δ(Kh)Δ(Ei)−qbΔ(Ei)Δ(Kh)=0: its two summands cancel by KhEi=qbEiKh in the first tensor factor and in the second tensor factor, while the toral elements commute. In the negative algebra, Δ(Kh)Δ(Fi)−q−bΔ(Fi)Δ(Kh)=0 by KhFi=q−bFiKh in the first and second tensor factors and commutation of toral elements. Thus both maps kill R± and descend to B± by [F9].

3.1step 2.1F1F2F5algebra

In B+, put xi=Ei⊗Ki−1 and yi=1⊗Ei. The toral-action relation gives yixi=qi2xiyi, so [F5] yields Δ+(Ei)N=∑r=0Nqir(N−r)CN,rEir⊗Ki−rEiN−r. Also Ki is grouplike and invertible since KhK−h=1 in B±.

4.1step 3.1F1F2F3F4F5algebra

Expand Δ+(Serreij+)=∑s=0m(−1)sCm,sΔ+(Ei)m−sΔ+(Ej)Δ+(Ei)s and first take the Ej⊗Kj−1 term from the middle factor. Fix r,t≥0 with M=m−r−t≥0, and put k=s−t; the allowed indices are 0≤k≤M. Using [F5] on the two powers of Δ(Ei), the corresponding tensor word is EirEjEit⊗Ki−(r+t)Kj−1EiM, and its coefficient is (−1)t+kCm,rCm−r,tCM,kqi(r+a+2t)M+(M−1)k. Here Cm,sCm−s,rCs,t=Cm,rCm−r,tCM,k follows by cancelling the factorial quotients, and the exponent comes from moving EiM−k past Kj−1 and Ki−t.

4.2step 3.1F1F2F3F4F5algebra

Now take the 1⊗Ej term from the middle factor. Fix u,v≥0 with k=m−u−v≥0; after moving toral factors to the left, the tensor word is Eik⊗Ki−kEiuEjEiv. Its coefficient, summed over r+t=k with t=k−r, is (−1)v+k(mu,v,k)iqik(u+1−k)∑r=0k(−1)rCk,rqi(k−1)r, where (mu,v,k)i=[m]i!/([u]i![v]i![k]i!). Indeed the unsummed Gaussian factor is Cm,v+tCu+r,rCv+t,t=(mu,v,k)iCk,r, and the two q-binomial expansions and toral crossings give exponent ru+tv+t(a+2u)=k(u+1−k)+(k−1)r, using a+u+v=1−k. The prefactor is independent of r and is 1 when k=0.

5.1step 1.2step 4.1F4F6algebra

If M>0, summing the coefficients in step 4.1 over k gives a scalar multiple of ∑k=0M(−1)kCM,kqi(M−1)k=0 by step 1.2. If M=0, only k=0 remains; then r+t=m, and summing over these pairs gives Serreij+⊗Ki−mKj−1, using Cm,r=Cm,m−r. Thus the Ej⊗Kj−1 part contributes exactly the first term in the claimed formula.

6.1step 1.2step 4.2step 5.1F4F6algebra

For k>0, the sum in step 4.2 is zero by step 1.2. For k=0, the only term has u+v=m, and its coefficient is (−1)vCm,v; these terms sum to 1⊗Serreij+. Combining this with step 5.1 proves the positive quasiprimitive identity.

7.1step 2.1step 6.1F2F7algebra

Define ω:B+→B− by Kh↦K−h and Ei↦−Fi. The toral and positive action relations map to the toral and negative action relations (the action relation at −h), so this is a well-defined algebra map. Directly on generators, Δ−ω=flip∘(ω⊗ω)∘Δ+: on Kh both sides equal K−h⊗K−h, and on Ei both sides equal −Fi⊗1−Ki⊗Fi. Moreover ω(Serreij+)=(−1)m+1Serreij− by applying the substitution to its m+1 factors. Applying this identity to the positive quasiprimitive formula gives Δ−(Serreij−)=Serreij−⊗1+KimKj⊗Serreij−.

8.1step 2.1step 6.1step 7.1F2F3F7F9F10algebra∎

The subspace J+⊗B++B+⊗J+ is a two-sided ideal of B+⊗B+ by the tensor multiplication in [F7], and likewise for the negative sign. The two quasiprimitive identities therefore imply Δ±(J±)⊆J±⊗B±+B±⊗J±, since each J± is generated as a two-sided ideal by the Serre elements and Δ± is an algebra map. The canonical maps B±→Uq(g) send every Serre generator to zero; applying their tensor squares to the formulas gives zero, so the Serre relations are preserved by the prescribed coproduct in the full Drinfeld–Jimbo quotient. Once the other defining relation families are checked, the quotient universal property in [F9] gives the descended coproduct.

Source note

The Berkeley text defines its toral-action-only positive Borel and gives the positive quasiprimitivity statement in Lemma 13.1.3.9, then says the cited Jantzen proof is a tedious q-binomial computation; it does not print the computation or the explicit grouplike factors. Jeong–Kang–Kashiwara display the coproduct convention used here in (1.6) but assert the Hopf structure without proving relation preservation. The coefficient cancellations in steps 4.1–6.1 are supplied locally; the original scaffold's claim that Serre-only ideals in the completely free algebra are coideals was replaced because its toral commutations are not valid before quotienting by the toral-action relations.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-08Open item page →

The quantum Serre sums vanish in the shuffle algebra, and the opposite Serre ideal annihilates the shuffle half

Statement

Let Sh(V), ⟨V⟩ and Uℏn+ be as in The formal quantum shuffle Borel and its Cartan crossed product.

(i) In ⟨V⟩⊆Sh(V) the Serre sums vanish: for all i≠j, ∑s=01−aij(−1)s(1−aijs)qi[vi]∗(1−aij−s)∗[vj]∗[vi]∗s=0, where ∗ is the shuffle product of The formal quantum shuffle Borel and its Cartan crossed product.

(ii) Let V∗=⨁iCξi be the dual space with ⟨ξi,vj⟩=di−1δij, let T(V∗)=⨁k≥0(V∗)⊗k⟦ℏ⟧ be the tensor algebra on V∗ with concatenation product, graded by deg⁡ξi=−ϵi, and let Δ:T(V∗)→T(V∗)⊗RT(V∗) be the braided coproduct with Δ(ξi)=ξi⊗1+1⊗ξi, the target carrying the braided product (u⊗v)(u′⊗v′)=q−⟨deg⁡u′,deg⁡v⟩uu′⊗vv′. Let J−⊴T(V∗) be the two-sided ideal generated by the negative Serre elements ∑s=01−aij(−1)s(1−aijs)qiξi1−aij−sξjξis, i≠j. Let ⟨⋅,⋅⟩Sh(V)×T(V∗):Sh(V)×T(V∗)→C((ℏ)),⟨[vi1∣⋯∣vik],ξj1⋯ξjl⟩=δkl ℏ−k∏t=1k⟨vit,ξjt⟩, be the wordwise pairing, extended bilinearly from the tensor-word bases. Then J− annihilates the shuffle half ⟨V⟩: ⟨x,y⟩=0for every x∈⟨V⟩, y∈J−, equivalently the wordwise pairing descends to a bilinear pairing ⟨V⟩×(T(V∗)/J−)→C((ℏ)). The twisted product on T(V∗)⊗RT(V∗) is used only in the definition of the braided coproduct, and no claim is made about the radical of the pairing on all of Sh(V) (see the note after the proof).

Facts & Assumptions

Given: The formal shuffle Borel of a finite symmetrizable Cartan datum and the wordwise pairing stated in part (ii).

[F1]

For i≠j, aij≤0, aii=2, di>0, and diaij=djaji, so ⟨ϵi,ϵj⟩=diaij is a symmetric bilinear form (Symmetrizable Cartan data for quantum groups).

[F2]

qi=qdi, [m]i! is the product of the symmetric qi-integers, and Cm,r=(mr)i=[m]i!/([r]i![m−r]i!) with Cm,0=1 (Quantum integers, factorials, Gaussian binomials and divided powers at qi).

[F3]

Bm,r(t)=[m]t!/([r]t![m−r]t!) are the coefficients in the q-binomial expansion when yx=txy, and Cm,r=Cm,m−r (The quantum binomial expansion for q-commuting elements, The quantum Pascal recurrences, the Gauss product formula and Gaussian integrality).

[F4]

The symmetric Gaussian coefficients are Laurent polynomials in qi and satisfy the q-Pascal and Gauss product identities (The quantum Pascal recurrences, the Gauss product formula and Gaussian integrality).

[F5]

The wordwise quantum shuffle product on Sh(V) uses inversion weight q−⟨deg⁡zi,deg⁡zj⟩ with ⟨ϵi,ϵj⟩=diaij; one-letter words are elements of V (The formal quantum shuffle Borel and its Cartan crossed product).

[F6]

The pairing of part (ii) is diagonal in the tensor-word bases: for tensor words u=[vi1∣⋯∣vik] and v=ξj1⋯ξjl it equals δklδi1j1⋯δikjkℏ−k∏tdit−1; in particular it carries no sign and it is invariant under the interchange i↔j of letters. The letter weights w(i,i)=2di and w(i,j)=diaij=w(j,i) are symmetric.

[F7]

The proof uses only finite word shuffles, coefficient identities, explicit word comparisons and the finitely many permutations of N+1 letters; no axiom of choice is used.

Proof

technique · Write $*$ for the shuffle product and expand the positive Serre expression one word at a time. Then compute the pairing of a shuffle product with the negative Serre element by an explicit permutation count, and compare it with the vanishing alternating sum just obtained
1.1F1F2F3F5givenalgebra

Put N=1−aij, x=[vi], y=[vj], t=qi−2, and Bm,r(t)=0 outside 0≤r≤m. Let Wm=[vi∣⋯∣vi] (with m letters), and let wp be the word with p copies of vi, then vj, then N−p copies of vi (omitting a zero-length block). Repeatedly shuffling one x into a block of m equal x-letters gives x∗m=AmWm, where Am=qi−m(m−1)/2[m]i!: the insertion weights sum to 1+qi−2+⋯+qi−2m=qi−m[m+1]i. In a three-block shuffle Wa∗y∗Wb, with a+b=N, fix wp, and let u (respectively v) count the b-block letters before (respectively after) y, so u+v=b. The prefix interleavings contribute Bp,u(t), the suffix interleavings contribute BN−p,v(t), the u letters before y cross y and the N−p−v letters after it, and each of those u letters crosses those N−p−v suffix x-letters; therefore coeff⁡wp(Wa∗y∗Wb)=∑u+v=bBp,u(t)BN−p,v(t)qi−aij(u+N−p−v)−2u(N−p−v).

1.2F5F6givenalgebra

The pairing is diagonal in the tensor-word bases: for tensor words u=[vi1∣⋯∣vik] and v=ξj1⋯ξjl the defining formula gives ⟨u,v⟩=δklδi1j1⋯δikjkℏ−k∏tdit−1, and ∏tdit−1 depends only on u. Consequently, for x=∑wxw[w] and tensor words y,y′ with letter-sequences u,u′, both sides of ⟨x,yy′⟩=∑k≥0⟨x≤k,y⟩⟨x>k,y′⟩ equal xuu′ℏ−(∣u∣+∣u′∣)(∏s∈uds−1)(∏s∈u′ds−1): the left side because only the word uu′ has nonzero pairing with yy′, the right side because only the cut k=∣u∣ contributes, the product in T(V∗) being concatenation and the cut coproduct of the word [u] being Δ([u])=∑k[u≤k]⊗[u>k]. Extending by bilinearity, this adjunction rule holds for all x∈Sh(V) and all y,y′∈T(V∗).

1.3F5givenalgebra

For homogeneous one-letter words [x1],…,[xm]∈V put w(a,b):=⟨deg⁡xa,deg⁡xb⟩. Then [x1]∗⋯∗[xm]=∑σ∈Smq−∑a<b, σ(a)>σ(b)w(a,b)[xσ−1(1)∣⋯∣xσ−1(m)]. Here σ maps each original letter position to its output position, consistently with [F5]. For m=1 this is the identity. If it holds for m−1, write the product as ([x1]∗⋯∗[xm−1])∗[xm]; inserting [xm] into the k-th position of a word [xσ−1(1)∣⋯∣xσ−1(m−1)] multiplies its coefficient by q−∑a<m, σ(a)≥kw(a,m), exactly the sum of the weights of the pairs (a,m) inverted by the resulting permutation, so the block formula reproduces the displayed expansion; the map (σ,k)↦ the resulting permutation is a bijection onto Sm.

2.1F1F2F3F4F5step 1.1algebra

In the coefficient of wp in ∑s=0N(−1)sCN,sx∗(N−s)∗y∗x∗s, set Q=N−p, write u+v=s, and use AN−sAsCN,s=qi−((N−s)(N−s−1)+s(s−1))/2[N]i! and −aij=N−1. The coefficient simplifies to [N]i!qi−N(N−1)/2+(N−1)Q(∑u=0p(−1)uBp,u(t)qiu(2p−1)−u2)(∑v=0Q(−1)vBQ,v(t)qiv−v2). The quotient definition gives Bm,r(t)=Bm−1,r(t)+tm−rBm−1,r−1(t); induction then yields ∑r=0m(−1)rtr(r−1)/2Bm,r(t)zr=∏j=0m−1(1−ztj). Thus the first factor in parentheses is ∏j=0p−1(1−qi2p−2−2j) and vanishes when p>0; the second is ∏j=0Q−1(1−qi−2j) and vanishes when Q>0. Since p+Q=N≥1, at least one factor vanishes for every p=0,…,N. These words exhaust the color degree Nϵi+ϵj, proving part (i).

2.2F5F6step 1.3algebra

Fix i≠j and N=1−aij, and put w(i,i)=2di, w(i,j)=w(j,i)=diaij; these are the values ⟨deg⁡va,deg⁡vb⟩, and w is symmetric by [F1]. Let z:=∑p=0N(−1)pCN,pξipξjξiN−p; since CN,N−p=CN,p, the element z differs from the printed Serre generator ∑s(−1)sCN,sξiN−sξjξis by the global sign (−1)N, so the two generate the same two-sided ideal. For t∈{0,…,N} let ζt:=[vi]∗t∗[vj]∗[vi]∗(N−t), and for p∈{0,…,N} let wp be the tensor word with p letters vi, then vj, then N−p letters vi. By [F6] and steps 1.2 and 1.3, ⟨ζt,z⟩=ℏ−(N+1)di−Ndj−1∑p=0N(−1)pCN,pM(t,p), where M(t,p):=∑σ∈SN+1, σ(t+1)=p+1q−∑a<b, σ(a)>σ(b)W(a,b) and W(a,b) is the weight w of the letters at positions a,b of the sequence (it,j,iN−t). Then M(t,p)=M(p,t): the map σ↦σ−1 is a bijection from the permutations with σ(t+1)=p+1 onto those with τ(p+1)=t+1, and it preserves the exponent, because the inversion pairs of σ correspond to those of τ=σ−1 with the same pair of letters: if c=σ(b)<d=σ(a) for an inverted pair a<b, then τ(c)=b>a=τ(d), and the weight of the pair {a,b} equals the weight of the letter pair at the positions d,c of the transposed sequence (ip,j,iN−p), since σ carries the position t+1 of j onto the position p+1, so the letter at position σ(x) of the transposed sequence is the letter at position x of the original sequence for every x.

2.3F5step 1.2algebra

The cut coproduct is an algebra map from Sh(V) to its tensor square with product (u⊗v)(u′⊗v′)=q−⟨deg⁡u′,deg⁡v⟩(u∗u′)⊗(v∗v′). Indeed a shuffle followed by a cut consists uniquely of shuffles of the two prefixes and two suffixes; its cross-cut inversions contribute q−⟨deg⁡u−,deg⁡v+⟩, exactly the scalar in this tensor product. On a letter the coproduct is primitive, so it sends every product of letters into ⟨V⟩⊗R⟨V⟩. Write this finite sum as Δ(x)=∑axa′⊗xa′′ with xa′,xa′′∈⟨V⟩; individual prefixes of ambient tensor words need not themselves belong to the generated half. If y annihilates ⟨V⟩, step 1.2 gives ⟨x,yu⟩=∑a⟨xa′,y⟩⟨xa′′,u⟩=0 and likewise ⟨x,uy⟩=0. Thus ⟨V⟩⊥ is a two-sided ideal.

3.1F3step 1.3step 2.2algebra

By part (i) the element ζ:=∑s=0N(−1)sCN,sζN−s is zero in ⟨V⟩⊆Sh(V), so all of its tensor-word coefficients vanish. By step 1.3 the coefficient of wp in ζN−s is M(N−s,p), hence ∑s(−1)sCN,sM(N−s,p)=0 for every p. Substituting t=N−s and using CN,N−t=CN,t (so that (−1)N−t is the global factor (−1)N) gives ∑t(−1)tCN,tM(t,p)=0 for every p; by the symmetry of step 2.2 this is ∑p(−1)pCN,pM(t,p)=0 for every t. Hence step 2.2 gives ⟨ζt,z⟩=0 for every t.

4.1step 2.2step 3.1step 2.3algebra∎

A shuffle product [x1]∗⋯∗[xm] of one-letter words is homogeneous of color degree deg⁡x1+⋯+deg⁡xm, and ⟨V⟩ is generated by the one-letter words, so every x∈⟨V⟩ is a finite sum of such products. Unless the letters are exactly N copies of vi and one copy of vj, their color degree differs from that of z, so the diagonal pairing vanishes. In the remaining case the letter weights agree with the two-letter computation of step 2.2, and step 3.1 gives ⟨[x1]∗⋯∗[xm],z⟩=0. By linearity ⟨x,z⟩=0 for every x∈⟨V⟩, that is, z∈⟨V⟩⊥. Consequently every Serre generator lies in the two-sided ideal ⟨V⟩⊥ of step 2.3, and hence so does the ideal J− they generate: ⟨⟨V⟩,J−⟩=0. This is the stated annihilation and the asserted descent of the wordwise pairing.

Remarks

The literal analogue of (ii) with all of Sh(V) in place of ⟨V⟩ is false, so the restriction of the domain is necessary: the tensor word [vi∣⋯∣vi∣vj] with N copies of vi pairs with the p=0 term of the Serre generator and with no other term, giving ℏ−(N+1)di−Ndj−1≠0. The source's literal full-Sh(V) radical clause therefore fails for the diagonal pairing. The restricted generated-half annihilation used here is proved independently above and supplies the final half-to-half duality. Printed formula (28) also has a single ℏ−1 prefactor; the multiplicative adjunction with generator weights ℏ−1di−1 requires the length-k weight ℏ−k∏tdit−1 used in this item and verified in step 1.2.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-08Open item page →

The generic quantum halves form a Drinfeld–Jimbo crossed double

Statement

Fix a finite symmetrizable Cartan datum over k=Q(q), with qi=qdi and Ki=Kdihi. Let A+ and A− be the independently presented k-algebras on ei and fi, respectively, modulo their separate symmetric quantum Serre relations, and let C=k[P∨] have basis Kh, h∈P∨. Then the vector space D=A−⊗kC⊗kA+ carries a unique associative unital algebra structure in which a−⊗Kh⊗a+ equals the product of the three embedded factors, the factors retain their algebra structures, and Khei=qαi(h)eiKh,Khfi=q−αi(h)fiKh,[ei,fj]=δijKi−Ki−1qi−qi−1. This is the Drinfeld–Jimbo crossed double. Its normal-factor multiplication map to the presented algebra Uq(g) of The Drinfeld-Jimbo quantized enveloping algebra by generators and relations is an algebra isomorphism, whose inverse sends Ei,Fi,Kh to the indicated factor generators. Thus all three abstract factor maps are injective; their images are the subalgebras of Positive, negative and toral quantum subalgebras and their root gradings.

In the presentation U~ obtained by omitting both Serre families, the halves T± are free, the toral factor is C, and multiplication identifies U~=T−⊗C⊗T+. Every Serre element satisfies [Fk,Sij+]=[Ek,Sij−]=0. Writing I±⊴T± for the separate Serre ideals, the full Serre ideal of U~ is exactly I−⊗C⊗T++T−⊗C⊗I+. No nonsingularity of the Cartan matrix and no choice principle is required.

Facts & Assumptions

Given: A finite symmetrizable Cartan datum and its Drinfeld–Jimbo presentation.

[F1]

The toral, toral-action and mixed relations give the Serre-free presentation; adding the separate Serre sums gives Uq(g), and assignments satisfying these relations extend uniquely (The Drinfeld-Jimbo quantized enveloping algebra by generators and relations, Universal property of the tensor algebra, A ring homomorphism whose kernel contains a two-sided ideal factors uniquely through the quotient ring).

[F2]

The integral weight characters are additive, qi=qdi, and qjaji=qiaij by symmetrizability (Symmetrizable Cartan data for quantum groups).

[F3]

For t=qi, the symmetric Gaussian coefficients satisfy CN,r[N−r]i=CN,r+1[r+1]i and ∑s=0N(−1)sCN,st±(N−1)s=0 for N≥1. The first identity is factorial cancellation and the second follows from the Gauss formula and its inverse-parameter substitution (Quantum integers, factorials, Gaussian binomials and divided powers at qi, The quantum Pascal recurrences, the Gauss product formula and Gaussian integrality).

[F4]

Tensoring quotient maps is surjective and has kernel the sum of the factor kernels, by repeated right exactness (Tensoring is right exact).

[F5]

The positive, negative and toral subalgebras are the generated images of the corresponding presentation generators (Positive, negative and toral quantum subalgebras and their root gradings).

Proof

1.1F1F2construct

Put ci=(Ki−Ki−1)/(qi−qi−1). In the free Serre-free presentation reduce EiFj to FjEi+δijci, EiKh to q−αi(h)KhEi, KhFj to q−αj(h)FjKh, KhKg to Kh+g, and K0 to 1. For each monomial use the lexicographic triple consisting of the number of E/F letters, the number of inverted type pairs for F<K<E, and the number of toral letters. Every resulting term decreases this triple: the mixed correction decreases the first coordinate, the other mixed or crossing terms decrease inversions, and toral merging or deletion decreases inversions or the final coordinate. Each rule has finitely many outputs, so the finitely branching reduction tree terminates; an infinite tree of arbitrarily deep descendants would give an infinite decreasing path by choosing its first such child at each stage. The irreducible words are precisely Fi1⋯FiaKhEj1⋯Ejb, with K0 interpreted as the empty toral factor.

1.2F1F2F3algebra

Fix i≠j, set t=qi, a=aij, N=1−a, and Cs=(Ns)i, so Sij+=∑s(−1)sCsEiN−sEjEis. Expanding the commutator in each position and moving toral factors to the right gives [Fi,Eim]=−[m]iEim−1(tm−1Ki−t−(m−1)Ki−1)/(t−t−1) for m≥1: the two geometric sums are ∑r=0m−1t±2r=t±(m−1)[m]i. For k≠i,j, the commutator [Fk,Sij+] is zero because every letter commutes with Fk. For k=j, only the Ej contributes, and moving Kj past the last s copies of Ei gives [Fj,Sij+]=−EiNqj−qj−1((∑s(−1)sCstas)Kj−(∑s(−1)sCst−as)Kj−1)=0. The crossing uses qjaji=ta from [F2], and both sums vanish by [F3] because a=1−N.

2.1F1F2step 1.1algebra

The overlapping reductions are KhKgKl, EiKhKg, KhKgFj, and EiKhFj, together with deletion of K0 against an adjacent crossing or toral product. The first three give respectively Kh+g+l, q−αi(h+g)Kh+gEi, and q−αj(h+g)FjKh+g along both routes. In the fourth, the two routes give the common term q−αi(h)−αj(h)FjKhEi and corrections δijq−αi(h)Khci and δijq−αj(h)ciKh; they agree since the toral elements commute and i=j whenever the correction is present. The zero-index cases and a toral sum h+g=0 agree by character additivity and αi(0)=0. There is no overlapping pair of EF rules, since its shared letter would have to be both E and F. Disjoint reductions commute by distributivity. Inducting on the decreasing triple in step 1.1, these joined first reductions therefore have the same final normal form in every word context. The linear normal-form map kills each relation multiplied on the left and right by arbitrary words, while every word minus its normal form lies in the relation ideal. It consequently induces inverse linear maps between U~ and the freely based normal-word space. This proves the full Serre-free tensor decomposition, including injectivity on arbitrary finite sums.

2.2F1F2F3step 1.2algebra

For k=i, expand in the left and right Ei blocks using step 1.2. The left-block term with s=r and right-block term with s=r+1 both have positive word EiN−1−rEjEir. In the left-block term, the exponent after moving Ki to the right is N−r−1+a+2r=r, since N−1+a=0; the inverse toral exponent is −r. The right-block term has those same exponents. Thus [Fi,Sij+]=−1t−t−1∑r=0N−1(−1)r(Cr[N−r]i−Cr+1[r+1]i)EiN−1−rEjEir(trKi−t−rKi−1)=0. The coefficient identity is [F3], including both endpoints. The assignment ω(Ei)=Fi, ω(Fi)=Ei, ω(Kh)=K−h is an involutive algebra map of the Serre-free presentation: toral actions reverse their signs, and both the mixed commutator and ci change sign. Applying it proves [Ek,Sij−]=0. Toral conjugation of either Serre generator is scalar, since its color degree is homogeneous.

3.1F1F2step 2.1step 1.2step 2.2algebra

In the normal-word space put X=I−CT++T−CI+. It contains the Serre generators and is a two-sided ideal. Same-side multiplication and toral multiplication preserve its two summands, using homogeneous toral conjugation. For the remaining crossed multiplications, expanding [Fk,w] for a positive word w yields positive-word/toral terms, and expanding [Ek,w] for a negative word yields negative-word/toral terms, by the mixed relation. If w=uSij+v, the Leibniz rule gives [Fk,w]=[Fk,u]Sij+v+uSij+[Fk,v] because the middle commutator vanishes by steps 1.2 and 2.2. Moving the resulting toral factors past the homogeneous Serre sum preserves its ideal. Hence [Fk,I+]⊆CI+, and similarly [Ek,I−]⊆I−C. When moving a crossed generator through an arbitrary normal product, these inclusions show that every term still belongs to X; multiplication in the other half only multiplies the existing ideal factor. This checks closure under all Ek,Fk,Kh on both sides. Conversely each summand of X lies in the ambient two-sided Serre ideal, by its definition. Therefore X equals that ideal, with the exact tensor-factor description claimed.

4.1F1F4F5step 1.1step 2.1step 3.1algebra∎

By [F4], the tensor quotient (T−/I−)⊗C⊗(T+/I+) is the quotient of T−⊗C⊗T+ by precisely X. Steps 2.1 and 3.1 therefore identify it linearly with U~/X=Uq(g), proving genuine tensor injectivity. Transport the associative quotient multiplication to this tensor space to define D. Its factor products, normal-factor products and cross-relations are as stated, and the factor embeddings are injective because 1 survives both homogeneous positive-height Serre ideals. Conversely any multiplication with those properties is determined by repeatedly using the crossing rules to rewrite a product of two normal tensors; these rules terminate by step 1.1. Thus the algebra structure is unique. Its generator map to Uq(g) and its inverse respect the defining relations by [F1] and the factor Serre relations, so are mutually inverse algebra homomorphisms. Their factor images are exactly [F5]. This proves all assertions.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-08Open item page →

The Drinfeld–Jimbo formulas define a Hopf algebra

Statement

Let T and IDJ be the free Q(q)-algebra and defining ideal of The Drinfeld-Jimbo quantized enveloping algebra by generators and relations, and write Ki=Kdihi. Let Δ:T→T⊗Q(q)T and ε:T→Q(q) be the algebra homomorphisms determined by

Δ(Kh)=Kh⊗Kh,Δ(Ei)=Ei⊗Ki−1+1⊗Ei,Δ(Fi)=Fi⊗1+Ki⊗Fi,ε(Kh)=1,ε(Ei)=ε(Fi)=0.

Let S:T→T be the Q(q)-linear algebra anti-homomorphism determined by

S(Kh)=K−h,S(Ei)=−EiKi,S(Fi)=−Ki−1Fi.

Then

Δ(IDJ)⊆IDJ⊗T+T⊗IDJ,ε(IDJ)=0,S(IDJ)⊆IDJ.

Hence these maps descend to Uq(g)=T/IDJ and make it a Hopf algebra over Q(q) (Bialgebras, counits and antipodes over a commutative ring): Δ is coassociative, ε is a counit, and

m(S⊗id⁡)Δ=η∘ε=m(id⁡⊗S)Δ.

The antipode S is unique for this multiplication, unit, coproduct and counit (Uniqueness of the antipode), and its square satisfies

S2(Ei)=qi−2Ei,S2(Fi)=qi2Fi,S2(Kh)=Kh.

Facts & Assumptions

Given: A symmetrizable Cartan datum over Q(q) and its Drinfeld–Jimbo presentation.

[F1]

The symmetrizer satisfies diaij=djaji, so the two crossing factors qiaij and qjaji are equal (Symmetrizable Cartan data for quantum groups).

[F2]

The defining ideal is generated by the toral, toral-action, mixed EiFj, and positive and negative Serre relation families (The Drinfeld-Jimbo quantized enveloping algebra by generators and relations).

[F3]

The qi-Gaussian coefficients are symmetric in r and m−r, and qi−qi−1≠0 (Quantum integers, factorials, Gaussian binomials and divided powers at qi).

[F4]

In the toral-action algebras, the Serre coproducts are the quasiprimitive formulas stated in The coproduct preserves the positive and negative quantum Serre ideals; their images in Uq(g)⊗Uq(g) are zero.

[F5]

The tensor product of algebras has multiplication (a⊗b)(c⊗d)=ac⊗bd (The tensor product of R-algebras has multiplication (a⊗b)(a′⊗b′)=aa′⊗bb′).

[F6]

Generator assignments extend uniquely to algebra maps on the free tensor algebra (Universal property of the tensor algebra).

[F8]

A two-sided ideal generated by a subset is as in The ideal generated by a subset and principal ideals.

[F9]

Tensoring an exact sequence ending in zero preserves exactness at the two rightmost terms (Tensoring is right exact).

[F10]

Tensor products over Q(q) have the symmetry isomorphism x⊗y↦y⊗x (Symmetry and associativity isomorphisms for tensor products over a commutative ring).

[F11]

A bialgebra has algebra maps Δ,ε satisfying coassociativity and the counit equations; a Hopf algebra has an antipode satisfying both convolution-inverse equations (Bialgebras, counits and antipodes over a commutative ring).

[F12]

An antipode, if it exists, is unique (Uniqueness of the antipode).

Proof

technique · Check the defining relations after mapping into $U_q(\mathfrak g)\otimes U_q(\mathfrak g)$, use tensor right exactness to recover the free-ideal coideal inclusion, and verify the Hopf equations on generators with an explicit word-induction step for convolution
1.1givenF5F6construct

The displayed assignments extend from the free generators to an algebra map Δ:T→T⊗T by [F6], using the tensor product algebra structure in [F5]. The counit assignment extends to an algebra map by [F6], and reversing words extends the assignment for S to a Q(q)-linear anti-homomorphism.

1.2givenF7F9F10algebra

Let π:T→U:=T/IDJ be the quotient map. Write I⊗T and T⊗I for the images of the corresponding tensor-product maps in T⊗T. Right exactness [F9] gives ker⁡(π⊗id⁡T)=im⁡(I⊗T→T⊗T); by applying [F9] again after the flip [F10], ker⁡(id⁡U⊗π)=im⁡(U⊗I→U⊗T), and π⊗id⁡I:T⊗I→U⊗I is surjective. If z∈ker⁡(π⊗π), then (π⊗id⁡T)z lies in this second kernel, so choose y∈T⊗I with the same image; then z−y∈ker⁡(π⊗id⁡T). Thus ker⁡(π⊗π)=I⊗T+T⊗I. This uses only a lift for one tensor at a time, not a choice function.

1.3givenF2F3algebra

The counit sends K0−1 and KhKh′−Kh+h′ to 1−1=0. It sends each toral-action relation to 0−0, each mixed relation to 0−0−δij(1−1)/(qi−qi−1)=0, and every Serre sum to zero because each monomial contains an E or an F. Thus ε(IDJ)=0.

1.4givenF2F7algebra

The anti-homomorphism S sends K0−1 to itself and KhKh′−Kh+h′ to K−h′K−h−K−(h+h′), another toral generator. For b=⟨αi,h⟩, applying S to the Ei toral-action relation gives −EiKiK−h+qbK−hEiKi=0 in U by the relation at −h; applying it to the Fi relation gives −Ki−1FiK−h+q−bK−hKi−1Fi=0 by the same relation and commutation of the toral elements.

2.1step 1.1F2F5algebra

In U⊗U, the toral relations are preserved: Δ(K0)=1⊗1 and Δ(KhKh′)=Kh+h′⊗Kh+h′. For b=⟨αi,h⟩, the two terms in Δ(Kh)Δ(Ei)−qbΔ(Ei)Δ(Kh) cancel by KhEi=qbEiKh in each tensor factor and commutation of toral elements; the two terms in Δ(Kh)Δ(Fi)−q−bΔ(Fi)Δ(Kh) cancel by the negative toral-action relation in each factor. Therefore (π⊗π)Δ kills the toral and toral-action generators of IDJ.

2.2step 1.4F1F2F3F5algebra

Applying S to Rij=EiFj−FjEi−δijCi and moving toral factors with the defining relations gives S(Rij)=qj2qi−aij(FjEi−EiFj)KiKj−1−δijS(Ci) in U. For i≠j the first factor vanishes by EiFj=FjEi; for i=j its scalar is qi2qi−2=1, FiEi−EiFi=−Ci, and S(Ci)=−Ci, so the two terms cancel. Hence S(Rij)=0 in U.

2.3step 1.4F1F2F3algebra

Let m=1−aij. In S(Serreij+), each of the m+1 substituted generators contributes a minus sign; moving the inserted Ki,Kj to the right contributes exponent s(s−1)+(m−s)(m−s−1)+aijm+2s(m−s)=m(m−1+aij)=0. Reindexing s↦m−s and using the symmetric Gaussian coefficients [F3] contributes the remaining sign, so S(Serreij+)=−Serreij+KimKj. For the negative family, Ki−1Fi=qi2FiKi−1 gives (Ki−1Fi)r=qir(r+1)FirKi−r by induction. Moving all toral factors to the right contributes exponent s(s+1)+(m−s)(m−s+1)+2s(m−s)+aijm=m(m+1+aij)=2m; the equality qjaji=qiaij from [F1] is used when Kj−1 crosses Fi, and Kj−1Fj=qj2FjKj−1 supplies the additional factor qj2. Reindexing s↦m−s and using [F3] gives S(Serreij−)=−qi2mqj2Serreij−Ki−mKj−1. These identities are computed modulo the toral and toral-action relations only; their right sides lie in the two-sided Serre ideal. Thus both Serre families map into IDJ.

3.1step 1.1step 2.1F1F2F3F5algebra

Put a=aij and Ci=(Ki−Ki−1)/(qi−qi−1). Expanding the commutator of the coproducts, the cross terms cancel because EiKj=qi−aKjEi and Ki−1Fj=qiaFjKi−1, where the equality qiaij=qjaji is [F1]. Hence [Δ(Ei),Δ(Fj)]=(EiFj−FjEi)⊗Ki−1+Kj⊗(EiFj−FjEi). In U, EiFj−FjEi=δijCi, and direct expansion gives Δ(Ci)=(Ki⊗Ki−Ki−1⊗Ki−1)/(qi−qi−1)=Ci⊗Ki−1+Ki⊗Ci. Thus the image under (π⊗π)Δ of every mixed relation is zero.

3.2step 2.1F4algebra

For each positive or negative Serre generator, [F4] maps its coproduct to zero in U⊗U. Therefore (π⊗π)Δ kills all remaining generators of IDJ.

3.3step 1.4step 2.2step 2.3F2F7F8algebra

Steps 1.4, 2.2 and 2.3 show that S sends every generator of IDJ into IDJ. Since S reverses products and IDJ is a two-sided generated ideal, S(IDJ)⊆IDJ; the quotient property [F7] gives the descended anti-homomorphism on U.

4.1step 1.1step 1.2step 2.1step 3.1step 3.2F2F5F8algebra

By step 1.2, steps 2.1, 3.1 and 3.2 put every defining generator's coproduct in IDJ⊗T+T⊗IDJ. This sum is a two-sided ideal: for pure tensors, left and right multiplication preserve each summand because IDJ is a two-sided ideal; the product formula in [F5] gives these products. Since Δ is an algebra map, it contains the coproduct of every element of the generated ideal IDJ.

4.2step 1.3step 3.3F11algebra

For Kh, both antipode products are S(Kh)Kh=K−hKh=1 and KhS(Kh)=KhK−h=1. For Ei, they are S(Ei)Ki−1+Ei=−EiKiKi−1+Ei=0 and EiS(Ki−1)+S(Ei)=EiKi−EiKi=0. For Fi, they are S(Fi)+S(Ki)Fi=−Ki−1Fi+Ki−1Fi=0 and Fi+KiS(Fi)=Fi−KiKi−1Fi=0. Since S(1)=1 by construction, the equations hold on the unit as well. These are the two convolution equations of [F11] on every generator and on the unit.

5.1step 1.1step 4.1step 1.3F5F7F11algebra

The maps Δ and ε descend by steps 4.1 and 1.3. For Kh, both iterated coproducts equal Kh⊗Kh⊗Kh. For Ei, each equals Ei⊗Ki−1⊗Ki−1+1⊗Ei⊗Ki−1+1⊗1⊗Ei; for Fi, each equals Fi⊗1⊗1+Ki⊗Fi⊗1+Ki⊗Ki⊗Fi. Thus coassociativity holds on generators. On Kh both counit composites equal Kh; on Ei they give 0+Ei=Ei and Ei⋅1+0=Ei; on Fi they give 0+Fi=Fi and Fi⋅1+0=Fi. Each side of these identities is an algebra homomorphism, so equality on the generators and unit proves equality on all of U, giving the bialgebra axioms in [F11].

5.2step 4.2F11algebra

Write Δ(x)=∑x(1)⊗x(2). If the two convolution equations hold for words x and y, then anti-multiplicativity of S and multiplicativity of Δ give ∑S(x(1)y(1))x(2)y(2)=∑S(y(1))(∑S(x(1))x(2))y(2)=ε(x)ε(y)1. Likewise ∑x(1)y(1)S(x(2)y(2))=∑x(1)(∑y(1)S(y(2)))S(x(2))=ε(y)ε(x)1. Since ε is multiplicative and all words are products of generators, induction gives both identities on every word and hence every element of U. Thus S is an antipode as defined in [F11].

6.1step 3.3step 5.2F12algebra∎

Uniqueness follows from [F12]. Since S reverses products, S2 is an algebra homomorphism, and on generators S2(Ei)=Ki−1EiKi=qi−2Ei, S2(Fi)=Ki−1FiKi=qi2Fi, and S2(Kh)=Kh. This proves the statement.

Source note

JKK §1, display (1.6), printed p. 6, gives exactly the coproduct, counit and antipode convention used here but asserts the Hopf structure without proving relation preservation. Berkeley Ch. 13 §13.1.3, printed pp. 308–309, gives the positive Serre quasiprimitivity/Hopf-ideal results and a different coproduct convention; the convention-sensitive relation calculations are local. The scaffold's claim that the convolution antipode equations extend from generators because all maps involved are algebra or anti-algebra homomorphisms is invalid: m(S⊗id⁡)Δ is not generally an algebra homomorphism. Step 5.2 supplies the required word-induction argument. The free-ideal coideal claim uses the explicitly proved tensor-quotient kernel calculation in step 1.2, rather than an unproved freeness or choice argument.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-08Open item page →

The formal quantum Serre half embeds in the shuffle algebra and is degreewise free

Statement

Let Uℏn+, V, ⟨V⟩, and pℏ:Uℏb+→V be as in The formal quantum shuffle Borel and its Cartan crossed product, with R=C⟦ℏ⟧ and q=eℏ, and let Un+ be the positive half of the Kac–Moody algebra. Assume AC; it is used only to apply the two-sided coideal lemma in step 3.1. For an indeterminate q′, let Uq′n+ be the algebra over C(q′) with the same generators and positive quantum Serre relations, with qi′=q′di. Then:

(i) The composite Uℏn+→Uℏb+→pℏV is an isomorphism of R-algebras onto ⟨V⟩. In particular, Uℏn+ is a free R-module and each root-graded component Uℏn+[α] is a finite-rank free R-module.

(ii) The classical limit is the classical half: Uℏn+/ℏUℏn+≅Un+ as NI-graded algebras, by ei↦eˉi.

(iii) For every α∈Q+, rank⁡RUℏn+[α]=dim⁡CUn+[α]=dim⁡C(q′)Uq′n+[α].

Facts & Assumptions

Given: A finite symmetrizable Cartan datum and the formal Borel and quantum half of The formal quantum shuffle Borel and its Cartan crossed product.

[F1]

The formal Borel defines the conditional algebra map pℏ, gives V=⟨V⟩⊗RAC as an R-module, and gives the coproduct on Cartan symbols and words (The formal quantum shuffle Borel and its Cartan crossed product).

[F2]

For each i≠j, the positive quantum Serre sum vanishes in ⟨V⟩; this is part (i) only of The quantum Serre sums vanish in the shuffle algebra, and the opposite Serre ideal annihilates the shuffle half. The supplier's annihilation statement (ii) is not needed for this embedding proof.

[F3]

At qi=1, the positive Serre relations present Un+, and the classical Borel is generated by its Cartan and positive simple generators (Serre presentation of a kac moody algebra).

[F4]

Under AC, an augmented two-sided coideal ideal J in Ul is generated by its primitive intersection J∩l (An augmented coideal ideal of an enveloping algebra is generated by its primitive part).

[F5]

The opposite classical Borels are degreewise perfectly paired Lie bialgebras (The opposite Borels of a symmetrizable Kac–Moody algebra are root-degreewise dual Lie bialgebras).

[F6]

The formal order on R is additive on nonzero products, and R is a domain (Formal order is non-Archimedean under sums and additive under products over a domain).

[F7]

A finitely generated module over a PID is its torsion submodule plus a finite free summand (A finitely generated PID module is its torsion submodule direct-summed with a finite free module).

[F8]

The symmetric Gaussian binomial is defined by the quantum-factorial quotient (Quantum integers, factorials, Gaussian binomials and divided powers at qi).

[F9]

AC is the assertion that every family of nonempty sets has a choice function (The Axiom of Choice).

[F11]

For a finite-dimensional vector space, the dimension of a quotient by a finite-dimensional subspace is the ambient dimension minus the subspace dimension (Rank-nullity: dim⁡FV=nullity⁡T+rank⁡T).

[F13]

A PID is a domain in which every ideal is principal (Principal ideal domain).

[F14]

Every finitely generated module over a PID is a finite direct sum of cyclic modules (Invariant-factor decomposition of a finitely generated module over a PID).

[F15]

The enveloping algebra is the tensor algebra quotient by the Lie-relator ideal, and Lie algebra maps into associative algebras extend uniquely to it (Universal enveloping algebra, Universal property of the enveloping algebra).

[F16]

Tensoring is right exact, so a quotient presentation remains a quotient by the scalar-extended relation subspace (Tensoring is right exact).

[F17]

The standard coproduct on Ub+ makes its Lie generators primitive and is cocommutative (Hopf-algebra structure on U(g)).

[F18]

A generalized Cartan matrix has a finite nonempty index set I={1,…,n} with n≥1 (Generalized cartan matrix).

[F19]

The symmetrizer entries di are positive integers and qi=qdi (Symmetrizable Cartan data for quantum groups).

[F20]

R=C⟦ℏ⟧ is the formal power-series ring with coefficientwise operations (Formal power series over a commutative ring and the coefficient-extraction functional [xn]).

[F21]

eℏ is the formal exponential ∑m≥0ℏm/m! (Formal exponential, logarithm, and binomial powers over a commutative Q-algebra).

[F22]

A formal power series is a unit exactly when its constant coefficient is a unit (A formal power series is a unit exactly when its constant coefficient is a unit).

[F23]

A submodule of a finite-rank free PID module is finite-rank free (A submodule of a free module of finite rank over a PID is free of no larger rank).

[F24]

Every symmetric Gaussian coefficient is a Laurent polynomial in qi (The quantum Pascal recurrences, the Gauss product formula and Gaussian integrality).

Proof

technique · Reduce the conditional shuffle map to the classical Borel, prove that its kernel has no primitive part by graded Lie-bialgebra duality, and then compare finite presentations over $R$ and over the generic field
1.1F1F2F3F8F18algebra

By [F2], the generator assignment of [F1] defines pℏ:Uℏb+→V. Its restriction to the Cartan coefficient algebra AC is the identity, and its image contains every one-letter word [vi]. Every element of V=⟨V⟩⊗RAC is a finite sum of products of a letter-generated element and an element of AC, so pℏ is surjective. In the symmetric formula, [m]i=qi−(m−1)(1+qi2+⋯+qi2m−2) specializes to m at qi=1; therefore [m]i! specializes to m! and each Gaussian coefficient specializes to (mr). By [F3], the reduced source is Ub+ and reduction defines a surjective graded algebra map p:Ub+→V0:=V/ℏV.

1.2F6F13F20F22givenalgebra

The ring R is a PID: if I≠0 is an ideal, the orders of its nonzero elements have a least value m∈N; choose f∈I of order m. It has the form f=ℏmu with u a unit by [F22], so ℏm∈I, while every g∈I has order at least m and is divisible by ℏm. Thus I=(ℏm); the zero ideal is principal as well, and [F6] says R is a domain, so [F13] applies.

2.1F1F3F17step 1.1algebra

Each generator hi,Dj,ei of the classical Borel has primitive coproduct by [F17], and its image under p is primitive in V0: this follows for Cartan symbols from [F1] and for [vi] by reducing its displayed coproduct modulo ℏ. Hence p commutes with coproduct and counit on generators, and therefore on the generated algebra. Thus p is a surjective Hopf map; since Ub+ is cocommutative by [F17], so is V0. This argument establishes Hopf compatibility only after reduction and makes no Hopf claim about pℏ.

3.1F4F9F15F16step 2.1algebra

Put J=ker⁡p and j=J∩b+. As the kernel of a Hopf map, J is a two-sided ideal and has zero counit. For x∈J, (p⊗p)Δ(x)=0; right exactness of tensor products and surjectivity of p give ker⁡(p⊗p)=J⊗Ub++Ub+⊗J, so J is a coideal. Under the stated AC hypothesis, apply [F4] to obtain J=Ub+j=jUb+ with j a Lie ideal. The universal property [F15] identifies the quotient Ub+/J with Ua, where a=b+/j, and p with the quotient map. AC is used here only through [F4]; all other choices below are finite-dimensional or explicitly specified.

3.2F1F5F23F6step 1.2step 2.1algebra

The first-order skew part δ(xˉ)=((Δ(x)−Δop(x))/ℏ) mod ℏ of the formal Borel coproduct is well-defined on V0: its numerator reduces to zero by step 2.1, and changing a lift by ℏy changes the quotient by Δ(y)−Δop(y), which is zero modulo ℏ because V0 is cocommutative. The quotient is unique because R is a domain and the coefficientwise word–Cartan module is ℏ-torsion-free. To see this, each ⟨V⟩[γ] is a submodule of a finite free word module and hence finite free by [F23] and step 1.2; tensoring each such component with the coefficientwise torsion-free AC in [F1] preserves ℏ-torsion-freeness, as do its coefficientwise completed tensor powers. Put D=Δ−Δop and let c cyclically permute three tensor factors. Coassociativity gives (1+c+c2)(D⊗id⁡)D=0: expanding D gives four permutations of (Δ⊗id⁡)Δ=(id⁡⊗Δ)Δ, whose cyclic sums cancel in pairs. Since D is divisible by ℏ, division by ℏ2 and reduction prove co-Jacobi for δ, and flip gives antisymmetry. For primitive a,b∈V0, subtracting the two algebra-map commutator identities for Δ and Δop, dividing by ℏ, and reducing gives δ([a,b])=[a⊗1+1⊗a,δ(b)]−[b⊗1+1⊗b,δ(a)], the required 1-cocycle rule. Directly from [F1], δ(hˉi)=δ(Dˉj)=0 and δ([vi]‾)=di[vi]‾∧hˉi. These values lie in ⋀2a; since a is generated by their images and the cobracket is a 1-cocycle, δ restricts to a Lie-bialgebra cobracket on a. The transposed bracket in [F5] gives the same formulas on b+: its Cartan cobracket is zero and its explicit rescaled-Manin normalization gives δb+(ei)=diei∧hi. Thus p∣b+:b+→a is a Lie-bialgebra map, because both cobrackets obey the 1-cocycle rule and agree on the Cartan and simple generators.

4.1F1F3F5step 3.2algebra

The graded dual p∗:agr′↪(b+)gr′≅b− is injective and is a Lie-algebra map by step 3.2 and [F5]; every root component is finite-dimensional. Its image contains the full Cartan subalgebra because p maps the classical Cartan basis to the independent polynomial Cartan variables, and it contains each fi because p(ei)=[vi]‾≠0 in the one-dimensional simple-root component. Since b− is generated by its Cartan and the fi by the separate Serre presentation [F3], the image is all of b−. Dualizing each finite-dimensional root component shows p∣b+ is an isomorphism, so j=0 and p:Ub+→V0 is an isomorphism.

5.1F1F3step 4.1algebra

The map p preserves the NI grading and sends ei to [vi]‾. The module decomposition in [F1] identifies the image of ⟨V⟩ in V0 with ⟨V⟩/ℏ⟨V⟩. Restricting the isomorphism of step 4.1 to the positive subalgebra proves Uℏn+/ℏUℏn+≅Un+ and identifies that classical half with ⟨V⟩/ℏ⟨V⟩.

6.1F3F7F14F18F23step 1.2step 5.1algebra

Fix α∈Q+. Since I is finite by [F18], the word space Sh(V)[α] has finitely many words and is finite free over R, so Nα:=⟨V⟩[α] is finite-rank free by [F23]. The presented component Mα:=Uℏn+[α] is finitely generated, since it is a quotient of the finite free span of words of degree α. By [F7] and [F14] over the PID of step 1.2, write Mα≅Rp⊕T, with T≅⨁j=1tR/(ℏmj) for positive integers mj: every nonzero nonunit of R is a unit times a power of ℏ. Since Mα/ℏMα≅Un+[α] by step 5.1, if d=dim⁡CUn+[α], then d=p+t.

7.1F7F10F22step 5.1step 6.1algebra

The surjection pℏ restricts to Mα↠Nα, and Nα has rank d because its reduction is the classical component in step 5.1. The target is torsion-free, so this map kills T; reducing the induced surjection Rp↠Nα modulo ℏ gives p≥d. Together with d=p+t, this forces t=0 and p=d. At α=0, both components are R⋅1 and the map sends unit to unit. If d=0, the decomposition gives Mα=Nα=0. For d>0, choose bases: the surjection is a square matrix whose determinant reduces to a nonzero determinant over C, hence is a unit by [F22]; [F10] makes it invertible. Therefore Mα→Nα is an isomorphism for every α, proving (i) and the first equality in (iii).

8.1F6F11F12F16F19F20F21F24step 1.2step 7.1algebra∎

Put K=Frac⁡(R). Evaluation q′↦eℏ embeds C[q′] into R: for a nonzero polynomial, factor out its maximal power of (q′−1); the remaining factor has nonzero value at 1, while eℏ−1≠0, so its evaluation is nonzero in the domain R. By [F19] and [F21], this substitution sends qi′=q′di to ediℏ; [F12] extends the injection to C(q′)↪K. For fixed α, the relations in that degree are the columns of a finite matrix over C[q′±1]; finiteness follows because there are finitely many words of degree α, and [F24] makes every Gaussian entry Laurent polynomial. Substituting q′=eℏ gives the formal presentation matrix for Mα. By [F16], its quotient after extension to K is the quotient by the same specialized columns. A field embedding preserves which matrix minors vanish, so the generic and formal matrices have the same rank; [F11] and step 7.1 give dim⁡C(q′)Uq′n+[α]=dim⁡K(Mα⊗RK)=d. This proves the remaining equality in (iii) and completes all assertions. The theorem is not an iff statement, so there are no reverse implications to prove.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-08Open item page →

Generic quantum Serre halves have classical PBW ranks and a nondegenerate Hopf pairing

Statement

Let R=C⟦ℏ⟧, K=C((ℏ)), q=eℏ, and qi=ediℏ for a finite symmetrizable Cartan datum. Let H+=Uℏn+, ⟨V⟩ and the shuffle product be as in The formal quantum shuffle Borel and its Cartan crossed product, and put H−=T(V∗)/J− using The quantum Serre sums vanish in the shuffle algebra, and the opposite Serre ideal annihilates the shuffle half. Assume AC (The Axiom of Choice), used only through the formal embedding theorem. For an indeterminate z, let Hz± be the algebras over C(z) presented by the separate symmetric quantum Serre relations, with parameters zi=zdi.

(i) Each H±[±α] is finite free over R, and rank⁡RH±[±α]=dim⁡CU(n±)[±α]=dim⁡C(z)Hz±[±α]. Every family of homogeneous lifts of a basis of the classical component is an R-basis. Likewise, homogeneous word expressions with coefficients rational in z, regular at z=1, which reduce to a classical component basis form a basis of the corresponding generic component. In particular, for a supplied ordered homogeneous basis of n±, ordered monomials in any such regular lifts form a generic PBW basis.

(ii) Both formal halves are graded braided Hopf algebras. Their generators are primitive, their counits kill positive height, and their tensor squares use (u⊗v)(u′⊗v′)=q−⟨deg⁡u′,deg⁡v⟩uu′⊗vv′,⟨ϵi,ϵj⟩=diaij, with negative degrees for H−. The word pairing restricts and descends to a nondegenerate K-valued pairing H+×H− with ⟨ei,fj⟩=δij/(ℏdi) and ⟨1,1⟩=1. It is zero on unequal opposite degrees and satisfies the braided Hopf adjunctions ⟨x,yy′⟩=∑⟨x(1),y⟩⟨x(2),y′⟩,⟨xx′,y⟩=∑⟨x,y(1)⟩⟨x′,y(2)⟩. The generic halves have the same braided Hopf structures and a nondegenerate C(z)-valued pairing normalized by ⟨ei,fj⟩z=δij/(zi−zi−1). Thus opposite graded components are dual. All generic assertions also hold for the corresponding Q(z)-presentations and their rational pairing, with the PBW lift clause using bases of the rational classical Serre form.

Facts & Assumptions

Given: The finite symmetrizable datum, its formal and generic Serre presentations and the formal parameter.

[F1]

The positive formal half is isomorphic to ⟨V⟩, its intrinsic reduction is U(n+), and its finite free components have the stated classical and generic ranks; AC enters only in its coideal argument (The formal quantum Serre half embeds in the shuffle algebra and is degreewise free, The Axiom of Choice).

[F2]

The negative Serre ideal annihilates ⟨V⟩ under the diagonal pairing of word bases with value ℏ−k∏tdit−1 on matching length-k words. The cut coproduct is an algebra map to the scalar-braided tensor square and preserves the generated half (The quantum Serre sums vanish in the shuffle algebra, and the opposite Serre ideal annihilates the shuffle half).

[F3]

The formal word space is degreewise finite free, its scalar form is symmetric, and the Serre coefficients are symmetric Gaussian Laurent polynomials (The formal quantum shuffle Borel and its Cartan crossed product, The quantum Pascal recurrences, the Gauss product formula and Gaussian integrality).

[F4]

The classical half admits an ordered homogeneous basis obtained by enumerating its finite bracket words, and ordered monomials in that basis form its enveloping-algebra basis (PBW for countably presented Kac Moody Lie algebras).

[F5]

Tensoring is right exact, so the tensor quotient kernel is the sum of the two factor kernels; over a field every injection remains injective after scalar extension (Tensoring is right exact, Restriction of scalars and extension of scalars S⊗RM along a ring homomorphism R→S).

[F6]

The fraction field of R is K, and a domain embedding into a field extends to its fraction field (The field of fractions Frac⁡(D)=(D∖{0})−1D of an integral domain).

Proof

1.1F1F3algebra

The identification of free generators ei↦fi carries the positive Serre presentation onto the negative one, reversing degrees; the same holds classically and generically. Thus [F1] gives all three rank equalities and finite freeness on both sides. In a fixed finite free component, lifts of a classical basis have a coordinate matrix whose reduction is invertible over C, so its determinant has nonzero constant term and is an R-unit. The adjugate identity makes this matrix invertible over R, proving the formal lift assertion.

1.2F1F2F3F5algebra

Give the free positive tensor algebra the primitive-generator coproduct into its scalar-braided tensor square. Its map to ⟨V⟩ commutes with the cut coproduct: both are algebra maps by [F2] and agree on each letter. By [F1] its kernel is precisely the positive Serre ideal. Hence the composite of the tensor coproduct with the two quotient maps kills that ideal, and [F5] gives its coideal inclusion and descended coproduct. The same presentation identification gives the negative coproduct. Coassociativity and counit follow on primitive generators and hence on the generated algebras. The color-height grading is connected, so the reduced coproduct of a positive-height homogeneous element has both factors of strictly smaller height. The recursion S(x)=−x−∑S(x′)x′′ and its right-handed counterpart provide left and right convolution inverses by height induction; associativity of convolution makes them equal. Thus both quotients are braided Hopf algebras.

2.1F1F3F4F5F6step 1.1algebra

The substitution z↦eℏ embeds C(z) into K: a nonzero polynomial is (z−1)mp(z) with p(1)≠0, and its value is the nonzero product (eℏ−1)mp(eℏ) in the domain R; fractions then embed by [F6]. In each color degree the generic and formal quotients after extension to K have the same finite word presentation, since the Serre coefficients specialize as in [F3]. A regular rational lift of a classical component basis therefore gives the formal basis of step 1.1 and, after field extension, a generic basis. Applying this degree by degree to the ordered monomials of [F4] proves the PBW monomial clause.

2.2F1F2F3step 1.2algebra

By [F2], the diagonal pairing restricts to ⟨V⟩×H−, and [F1] identifies the first factor with H+. The cut adjunction is the wordwise concatenation identity. For the other adjunction, the coefficient of a word w in a shuffle u∗v equals the coefficient of u⊗v in the primitive braided tensor coproduct of w: both sum over the assignments of letters to the two blocks with inversion scalar q−⟨deg⁡u′,deg⁡v⟩. The degree form is symmetric by [F3], and the diagonal generator weights multiply in the same way on both sides. Thus both adjunctions hold on free representatives, and step 1.2 and the descended pairing make them adjunctions on the quotient coproducts themselves. The empty word gives ⟨1,1⟩=1 and all positive-degree counit pairings vanish.

2.3F1F2F5step 1.1algebra

Fix α∈Q+. If an element of H+[α]⊗RK pairs to zero with every negative class, its realization in the shuffle word space pairs to zero with every negative tensor word, since all those words map to quotient classes. The diagonal word pairing has zero left annihilator, so that realization is zero. The realization remains injective after extension to K by [F1] and [F5]. The opposite components have equal finite dimension by step 1.1; an injective map from one into the other's dual is consequently bijective. Thus both annihilators vanish. For nonhomogeneous elements, separate the finitely many homogeneous components, proving the asserted nondegeneracy.

3.1F3F5step 2.1step 1.2step 2.2step 2.3algebra∎

Define the generic word pairing using shuffle coefficients and matching-word weights ∏t(zit−zit−1)−1. These are rational functions, and under z=eℏ their ratio to the formal pairing in a fixed color degree α is ∏iuiαi, where ui=ℏdi/(ediℏ−e−diℏ) is an R-unit with constant term 1/2. This color-multiplicative rescaling preserves the two adjunctions and nondegeneracy. The specialized generic pairing therefore kills the Serre ideals and is nondegenerate by steps 2.2–2.3; the field embedding of step 2.1 implies the same descent and invertible pairing matrices over C(z). Likewise, coassociativity, coideal inclusions and the coproduct formulas descend generically: in each finite degree their errors vanish after the injective field extension, and the antipode recursion applies there as well. All generic presentations and word pairing coefficients lie in Q(z), so the same injective extension to K proves the statements over that field. This establishes the full braided Hopf pairing with the rational normalization needed for the Drinfeld–Jimbo commutator.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-08Open item page →

Triangular decomposition of a quantized enveloping algebra

Statement

Let Uq(g) be the Drinfeld–Jimbo algebra of a finite symmetrizable Cartan datum, over k=Q(q), and let Uq−,Uq0,Uq+ be its generated subalgebras of Positive, negative and toral quantum subalgebras and their root gradings.

(i) Multiplication is a vector-space isomorphism m:Uq−⊗kUq0⊗kUq+⟶Uq(g),x−⊗x0⊗x+⟼x−x0x+. Products b−Khb+, with b± ranging over supplied bases of the two halves and h∈P∨, form a basis of Uq(g). Arbitrary simple-generator words span the halves; their Serre relations preclude claiming their independence.

(ii) The halves are exactly the algebras presented by their separate Serre relations, and Uq0≅k[P∨]. Thus all three abstract factor maps are injective and the Kh are linearly independent.

(iii) For every β∈Q, the total root grading satisfies Uq(g)[β]≅⨁α,γ∈Q+, α−γ=βUq−[−γ]⊗kUq0⊗kUq+[α]. In particular degree zero contains all equal-positive/negative-degree summands, including FiEi; it is not just the toral subalgebra. As a left Uq0-module this component is free, with rank equal to the possibly infinite sum of the products of the two half-component dimensions. Under AC (The Axiom of Choice), the half-component dimensions are their classical PBW ranks, and products of the generic PBW bases of Generic quantum Serre halves have classical PBW ranks and a nondegenerate Hopf pairing with the toral basis give a PBW basis of the whole algebra. AC enters only through that supplier's formal-embedding proof; (i), (ii) and the graded tensor decomposition are choice-free.

Facts & Assumptions

Given: The datum, its Drinfeld–Jimbo algebra and its generated subalgebras.

[F1]

The independently presented halves A± and free toral algebra C=k[P∨] form an associative algebra on A−⊗C⊗A+; its normal-factor multiplication to Uq(g) is an isomorphism and identifies the factor images with the generated subalgebras. This follows from the locally proved Serre-free normal forms, opposite Serre commutators and factor-ideal identity (The generic quantum halves form a Drinfeld–Jimbo crossed double).

[F2]

The generated subalgebras are root-graded with deg⁡Fi=−αi, deg⁡Ei=αi, deg⁡Kh=0, and toral conjugation on a homogeneous element is given by its additive weight (Positive, negative and toral quantum subalgebras and their root gradings).

[F3]

Under AC, both separately presented generic halves have classical PBW ranks and their regular lifts of ordered classical PBW monomials are generic bases, including over Q(q) (Generic quantum Serre halves have classical PBW ranks and a nondegenerate Hopf pairing, The Axiom of Choice).

Proof

1.1F1algebra

By [F1], the algebra Uq(g) is identified with the tensor space of the independently presented factors, and their injective images are Uq−,Uq0,Uq+. Under these identifications the multiplication map m is exactly the normal-factor isomorphism in [F1]. This proves (i) and (ii) without any additional assumption about a quotient presentation's freeness. For supplied factor bases, finite bilinear expansions show that their pure tensors span the tensor product; coordinate functionals on the supplied bases separate the coefficients of every finite sum of those tensors. Thus their pure tensors, and hence their images b−Khb+, are a basis.

2.1F1F2step 1.1algebra

The normal-factor isomorphism preserves total root degree by [F2], so the degree-β tensor subspace is precisely the direct sum over α−γ=β displayed in (iii). Every algebra element uses finitely many such summands. The equal-degree summands at β=0 include Fi⊗1⊗Ei; both factors are nonzero because their simple-root component presentations have no Serre relations. Their independence from 1⊗C⊗1 follows from the tensor decomposition. Hence Uq(g)[0] contains these additional summands and is not just Uq0.

3.1F1F2F3step 1.1step 2.1algebra∎

Homogeneous bases of the halves can be obtained by enumerating their words and retaining the first vectors outside the previous span, without choice. For homogeneous half basis vectors b− of degree −γ and b+ of degree α, the map C→Uq(g) sending c to cb−b+ identifies a free left toral copy: moving Kh past b− multiplies b−Khb+ by the nonzero scalar q−γ(h). Step 1.1 therefore identifies the fixed total-degree component with the direct sum of these free left toral copies. Its rank is the sum of the products of the two component dimensions, which may be infinite. With the stated AC hypothesis, [F3] identifies those finite half dimensions with the classical PBW ranks and supplies the generic ordered-monomial bases; inserting their products into step 1.1 gives the asserted full PBW basis. This completes all claims with the exact total grading.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-08Open item page →

Divided-power commutation and the simple Uqi(sl2) string modules

Statement

Fix a symmetrizable Cartan datum (I,A,D,P,P∨,q), an index i∈I, and put qi=qdi. Let Uq(g) be the Drinfeld–Jimbo algebra and let Ai be the subalgebra generated by Ei,Fi,Ki±1 (The Drinfeld-Jimbo quantized enveloping algebra by generators and relations, Positive, negative and toral quantum subalgebras and their root gradings); write x(m):=xm/[m]i! for the divided powers of Quantum integers, factorials, Gaussian binomials and divided powers at qi.

(i) The assignment e↦Ei, f↦Fi, k↦Ki, k−1↦Ki−1 defines an algebra isomorphism from the standard rank-one algebra A over Q(q), presented by e,f,k±1 with kek−1=qi2e, kfk−1=qi−2f and [e,f]=(k−k−1)/(qi−qi−1) onto Ai; the images of the monomials FiaKimEib (a,b≥0, m∈Z) are linearly independent in Uq(g), being images of the PBW monomials of the triangular decomposition (Triangular decomposition of a quantized enveloping algebra).

(ii) For all a,b≥0, KiEi(a)Ki−1=qi2aEi(a) and KiFi(b)Ki−1=qi−2bFi(b), and for b≥1 the divided powers satisfy the exact commutation formula EiFi(b)=Fi(b)Ei+Fi(b−1)qib−qi−b∑s=0b−1(qi−2sKi−qi2sKi−1); iterating it computes Ei(a)Fi(b) as a finite sum of monomials Fi(b−t)KimEi(a−t) (0≤t≤min⁡(a,b)) with coefficients in Q(q).

(iii) For every N≥0 there is a simple Ai-module Li(N) of dimension N+1: it is the quotient of the cyclic module M~(N):=Ai/(AiEi+Ai(Ki−qiN)) by the submodule generated by Fi(N+1)wN, where wN is the image of 1, with basis the images vt of Fi(t)wN, 0≤t≤N, on which Kivt=qiN−2tvt, Fivt=[t+1]ivt+1 and Eivt=[N−t+1]ivt−1 (with v−1=vN+1=0); Li(N) is the unique simple Ai-module generated by a vector v≠0 with Eiv=0, Kiv=qiNv and FiN+1v=0.

Facts & Assumptions

Given: A symmetrizable Cartan datum, an index i, and the subalgebra Ai generated by Ei,Fi,Ki±1.

[F1]

EiFi−FiEi=(Ki−Ki−1)/(qi−qi−1); KiEiKi−1=qi2Ei and KiFiKi−1=qi−2Fi; Ki−1=K−dihi is the inverse of Ki (The Drinfeld-Jimbo quantized enveloping algebra by generators and relations).

[F2]

Ai is the subalgebra generated by Ei,Fi,Ki±1; the universal property of Uq(g) assigns a homomorphism to every assignment satisfying the defining relations; the same quotient universal property applies to the explicit rank-one presentation (The Drinfeld-Jimbo quantized enveloping algebra by generators and relations, Positive, negative and toral quantum subalgebras and their root gradings, A ring homomorphism whose kernel contains a two-sided ideal factors uniquely through the quotient ring).

[F3]

[m]i=(qim−qi−m)/(qi−qi−1), [m]i!=∏k=1m[k]i, x(m)=xm/[m]i!, and qi−qi−1≠0 (Quantum integers, factorials, Gaussian binomials and divided powers at qi).

[F4]

Triangular decomposition: the multiplication map Uq−⊗Uq0⊗Uq+→Uq(g) is an isomorphism, so the images of the monomials FiaKimEib are linearly independent and span the rank-one subalgebra Ai (Triangular decomposition of a quantized enveloping algebra); only the unconditional tensor-decomposition clauses (i)–(ii) of that supplier are needed here.

[F5]

The independent coroots and di>0 make mdihi distinct for distinct integers m (Symmetrizable Cartan data for quantum groups). The finite geometric-sum identities ∑s=0b−1qi±2s=(qi±2b−1)/(qi±2−1) hold because q is transcendental and di>0; likewise [m]i≠0 for m≥1 by [F3].

Proof

technique · prove the divided-power identities by a two-term induction on the exponent, then build the cyclic module $\widetilde M(N)$ and its quotient and compute the action on the string vectors; the rank-one normal basis identifies the cyclic quotient and proves its string vectors are independent. In the infinite cyclic module, extend the integer notation to $s\in\mathbb Z$ by $[s]_i=(q_i^s-q_i^{-s})/(q_i-q_i^{-1})$; this agrees with [F3] for $s\ge0$ and gives $[-s]_i=-[s]_i$
1.1F1F3algebra

Weight rules. By [F1], KiEiaKi−1=qi2aEia and KiFibKi−1=qi−2bFib for all a,b≥0 by induction on the exponent via multiplicativity of conjugation; dividing by the nonzero scalars [a]i! and [b]i! of [F3] gives KiEi(a)Ki−1=qi2aEi(a) and KiFi(b)Ki−1=qi−2bFi(b).

1.2F1F3algebra

Commutation induction. Put c:=1/(qi−qi−1) and Tb:=c∑s=0b−1(qi−2sKi−qi2sKi−1) for b≥1. We prove EiFib=FibEi+Fib−1Tb by induction on b. For b=1 this is [F1]. For the induction step, EiFib=(EiFib−1)Fi=(Fib−1Ei+Fib−2Tb−1)Fi=Fib−1(FiEi+c(Ki−Ki−1))+Fib−2Tb−1Fi, and [F1] gives Tb−1Fi=Fi c∑t=1b−1(qi−2tKi−qi2tKi−1)=Fi(Tb−c(Ki−Ki−1)) after reindexing s=t−1; the two correction terms cancel and the identity follows.

2.1F3step 1.1step 1.2algebra

Divided-power formula. Dividing the identity of 1.2 by [b]i! gives EiFi(b)=Fi(b)Ei+Fi(b−1)Tb[b]i,Tb[b]i=1qib−qi−b∑s=0b−1(qi−2sKi−qi2sKi−1), because Fib−1/[b]i!=Fi(b−1)/[b]i and [b]i=(qib−qi−b)/(qi−qi−1); this is the displayed formula of (ii). Multiplying the formula for Ei and for Ei(a−1) on the left and using EiFi(b)=Fi(b)Ei+… repeatedly computes Ei(a)Fi(b) as a finite sum of monomials Fi(b−t)KimEi(a−t) with 0≤t≤min⁡(a,b), since each application of the formula moves one Ei to the right at the cost of one Fi-power.

3.1F1F3F4F5step 1.1step 2.1construct

The cyclic module. Let M~(N):=Ai/(AiEi+Ai(Ki−qiN)) and wN:=1+(AiEi+Ai(Ki−qiN)). Every element of Ai is a finite Q(q)-linear combination of monomials FitKimEia with t,a≥0 and m∈Z: words are rearranged with [F1] and 1.1, and each application of the identity of 2.1 moves one Ei to the right of an Fi-power at the cost of one Fi-power and finitely many K-factors, so the rearrangement terminates. By [F4] these normal monomials are a basis. The left ideal AiEi is their span with a>0. Modulo that span, a product from Ai(Ki−qiN) with positive Ei-exponent still has positive exponent after toral crossing, and its terms with exponent zero are exactly Fitp(Ki)(Ki−qiN) for Laurent polynomials p. Thus the remaining toral factor is quotiented by evaluation Ki=qiN, and the cyclic quotient has basis the images of Fit, t≥0; a toral monomial FitKim maps to qiNmFit, not to zero. Writing vt:=Fi(t)wN for t≥0, the relations [F1] and 1.1 give Kivt=qiN−2tvt (using Ki=Kdihi and αi(dihi)=2di), Fivt=FiFi(t)wN=[t+1]iFi(t+1)wN=[t+1]ivt+1 by [F3], and, using 2.1 with a=1, Eivt=Fi(t−1)1qit−qi−t∑s=0t−1(qiN−2s−qi2s−N)wN, for t≥1, and Eiv0=0; the bracket equals [N−t+1]i by the geometric-sum identity [F5].

4.1F3F4step 3.1algebra

The quotient and its action. Let Li(N):=M~(N)/AiFi(N+1)wN. The submodule AiFi(N+1)wN is spanned by the Fi(t)wN with t≥N+1: it is contained in that span by the action formulas of 3.1, and it contains them because Fi(t)wN=Fi(t−N−1)Fi(N+1)wN up to the nonzero scalar [t]i!/[t−N−1]i![N+1]i! by [F3]. Hence in Li(N) the vectors v0,…,vN satisfy the displayed action formulas with v−1=vN+1=0, and they span Li(N); their nonvanishing and independence follow from the cyclic-module basis in step 3.1, since the specified tail submodule is spanned by the disjoint basis indices t≥N+1.

5.1F3step 4.1algebra

Simplicity and uniqueness. Let W⊆Li(N) be a nonzero submodule and pick 0≠w=∑t≤Tctvt with T maximal and cT≠0. By 3.1, EiTw=[N]i[N−1]i⋯[N−T+1]icTv0 (each application of Ei lowers the index by one with the stated nonzero scalar; all displayed scalars are nonzero because [m]i≠0 for m≥1 by [F3] and the nonvanishing of qi2m−1), so v0∈W; then Fitv0=[t]i!vt puts every vt in W, so W=Li(N) and Li(N) is simple and generated by wN with EiwN=0, KiwN=qiNwN and FiN+1wN=0. Conversely, if V is any simple Ai-module generated by v≠0 with these three properties, then the assignment 1↦v defines a surjection M~(N)→V whose kernel contains Fi(N+1)wN by the third property, hence factors through Li(N); as Li(N) is simple and V≠0, this map is an isomorphism, which gives the asserted uniqueness.

6.1F1F2F4F5algebra∎

Part (i). The generators Ei,Fi,Ki±1 satisfy precisely the explicit rank-one relations by [F1], so sending e,f,k to them gives a surjective map A→Ai by the quotient universal property in [F2]. The same finite rearrangement used in step 3.1 shows that fakmeb spans this presented rank-one algebra; its torus here is generated by k alone, rather than by additional lattice roots of k. In the full algebra, the pure color-i component of either separate half has one word and no Serre relation, so its powers are nonzero and independent. The toral powers Kim=Kmdihi are distinct because the datum's coroots are independent and di>0. By the tensor decomposition [F4], the images FiaKimEib are therefore independent. The spanning rank-one monomials have independent images, proving injectivity and the claimed isomorphism.

5 · Examples, counterexamples and false statements

None yet.

Sources