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Quantized Enveloping Algebras and Quantum Serre Relations — Examples

1 · Prerequisites

2 · Summary

These examples compute the Drinfeld–Jimbo formulas of quantized-enveloping-algebras-and-quantum-serre-relations in small rank and record one counterexample about the parameter normalization.

The quasiprimitive Serre element in type $A_{2}$ expands the two type- $A_{2}$ quantum Serre polynomials, lists every mixed bidegree of their coproducts, and checks the classical specialization at $q = 1$ . The double-edge Serre relation for the cyclic affine type $A_{1}^{(1)}$ performs the same four-letter calculation for the double edge of affine type $A_{1}^{(1)}$ , where the coefficients $[3]_{q} = q^{2} + 1 + q^{- 2}$ produce the alternating Gaussian identity. Coproduct, antipode and $q$ -binomial expansion in $U_{q} ({s l}_{2})$ collects the rank-one formulas: the coproduct, counit and antipode on the generators, the $q$ -binomial expansion of $Δ (E^{n})$ , the failure of involutivity of the antipode, and the non-cocommutativity of the coproduct, with the nonvanishing of $E$ and $K - 1$ proved by an explicit oscillator representation.

Unsymmetrized parameters break the coproduct of the Serre ideal shows that using one and the same parameter at every node of $B_{2}$ destroys the coideal property of the Serre ideal, so that the Cartan normalization $q_{i}^{a_{i j}} = q_{j}^{a_{j i}}$ is essential. An explicit four-dimensional representation detects the failure in the full quotient, including its mixed relations.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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

Unsymmetrized parameters break the coproduct of the Serre ideal

Statement refuted

The following assertion is false: for every generalized Cartan matrix and every assignment of node parameters $q_{i}$ , the standard Drinfeld–Jimbo coproduct formulas descend to the full node-toral presentation, including its mixed and Serre relations, with $K_{i} E_{j} K_{i}^{- 1} = q_{i}^{a_{i j}} E_{j}$ .

Take $A = (\begin{array}{cc} 2 & - 1 \\ - 2 & 2 \end{array})$ , whose standard symmetrizer is $diag (2, 1)$ , but assign $q_{1} = q_{2} = q$ , with $q$ indeterminate. Precisely, let $B$ be the $Q (q)$ -algebra on $E_{1}, E_{2}, F_{1}, F_{2}, K_{1}^{\pm 1}, K_{2}^{\pm 1}$ , with commuting invertible toral generators, relations $K_{i} E_{j} K_{i}^{- 1} = q^{a_{i j}} E_{j}, K_{i} F_{j} K_{i}^{- 1} = q^{- a_{i j}} F_{j}, [E_{i}, F_{j}] = δ_{i j} \frac{K_{i} - K_{i}^{- 1}}{q - q^{- 1}},$ and both symmetric quantum Serre families with the common parameter $q$ . This explicitly changed parameter assignment is the test presentation, not the symmetrized algebra of The Drinfeld-Jimbo quantized enveloping algebra by generators and relations. Put $S = E_{1}^{2} E_{2} - [2]_{q} E_{1} E_{2} E_{1} + E_{2} E_{1}^{2} = 0$ in $B$ .

The assignments $Δ (E_{i}) = E_{i} \otimes K_{i}^{- 1} + 1 \otimes E_{i}$ , $Δ (F_{i}) = F_{i} \otimes 1 + K_{i} \otimes F_{i}$ , $Δ (K_{i}) = K_{i} \otimes K_{i}$ fail to define an algebra map $B \to B \otimes B$ . In an explicit representation of $B$ below, the proposed image of $S$ acts on $v_{0} \otimes v_{1}$ by $q^{- 2} (1 - q) (1 + q^{2}) v_{2} \otimes v_{2} \neq 0.$ Thus the previously detected positive-Borel defect survives in the full quotient. The same proposed formulas already fail the off-diagonal mixed relation $[E_{1}, F_{2}] = 0$ .

Facts & Assumptions

Given: The explicitly stated unsymmetrized full presentation $B$ , with $q$ indeterminate.

[F1]

The symmetric Gaussian coefficients are $[2]_{q} = q + q^{- 1}$ and $[3]_{q} = q^{2} + 1 + q^{- 2}$ (Quantum integers, factorials, Gaussian binomials and divided powers at $q_{i}$ ).

[F2]

The displayed inverse- $K$ positive coproduct and its negative counterpart are the normalized Drinfeld–Jimbo convention; with the correct node parameters their Serre mixed terms cancel (The coproduct preserves the positive and negative quantum Serre ideals, The Drinfeld-Jimbo quantized enveloping algebra by generators and relations).

[F3]

For this matrix the correct symmetry is $q_{1} = q^{2}$ , $q_{2} = q$ ; the assignment tested above violates $q_{1}^{a_{12}} = q_{2}^{a_{21}}$ (Symmetrizable Cartan data for quantum groups).

Counterexample

1.1givenconstruct

On a four-dimensional $Q (q)$ -space with basis $v_{0}, v_{1}, v_{2}, v_{3}$ , let $K_{1}$ have eigenvalues $(1, q^{- 1}, q, 1)$ and $K_{2}$ have eigenvalues $(q^{- 1}, q, q^{- 1}, q)$ . Define $E_{1} v_{1} = v_{2}$ , $F_{1} v_{2} = v_{1}$ , $E_{2} v_{0} = v_{1}$ , $E_{2} v_{2} = v_{3}$ , $F_{2} v_{1} = v_{0}$ , $F_{2} v_{3} = v_{2}$ , and make all other $E / F$ actions zero. Both toral operators are commuting and invertible. An $E_{1}$ arrow changes the pair of toral exponents by $(2, - 2)$ and each $E_{2}$ arrow changes it by $(- 1, 2)$ ; the $F$ arrows reverse these changes. Hence all toral-action relations hold.

2.1step 1.1algebra

The diagonal entries of $[E_{1}, F_{1}]$ are $(0, - 1, 1, 0)$ , and those of $[E_{2}, F_{2}]$ are $(- 1, 1, - 1, 1)$ . Each toral exponent is $0$ or $\pm 1$ , so these are precisely the entries of $(K_{i} - K_{i}^{- 1}) / (q - q^{- 1})$ . For each off-diagonal pair, both operator products $E_{1} F_{2}, F_{2} E_{1}$ and $E_{2} F_{1}, F_{1} E_{2}$ are zero: none of their two-arrow sequences is composable on a basis vector. Thus all four mixed relations hold.

2.2F2step 1.1algebra

On its tensor square put $D_{i} = E_{i} \otimes K_{i}^{- 1} + 1 \otimes E_{i}$ . Direct computation gives $D_{1} (v_{0} \otimes v_{1}) = v_{0} \otimes v_{2}$ and $D_{2} (v_{0} \otimes v_{1}) = q^{- 1} v_{1} \otimes v_{1}$ . Also $D_{1} (v_{1} \otimes v_{1}) = q v_{2} \otimes v_{1} + v_{1} \otimes v_{2}$ , so $D_{1}^{2} D_{2} (v_{0} \otimes v_{1}) = (1 + q^{- 2}) v_{2} \otimes v_{2}$ . Applying $D_{2}$ to $v_{0} \otimes v_{2}$ gives $q v_{1} \otimes v_{2} + v_{0} \otimes v_{3}$ , whose $D_{1}$ image is $v_{2} \otimes v_{2}$ ; thus $D_{1} D_{2} D_{1} (v_{0} \otimes v_{1}) = v_{2} \otimes v_{2}$ . Finally $D_{1}^{2} (v_{0} \otimes v_{1}) = 0$ , so the last Serre term contributes zero.

3.1F1step 1.1step 2.1algebra

All four squares $E_{1}^{2}, E_{2}^{2}, F_{1}^{2}, F_{2}^{2}$ are zero. The compositions $E_{1} E_{2} E_{1}$ and $F_{1} F_{2} F_{1}$ are zero as well: after the first color-1 arrow, the color-2 arrow either vanishes or leads to a vector on which the final color-1 arrow vanishes. These identities give both length-three Serre relations $S_{12}^{\pm} = 0$ . Every term of $S_{21}^{\pm}$ contains a square or cube of the color-2 operator, since its four words are $X_{2}^{3} X_{1}$ , $X_{2}^{2} X_{1} X_{2}$ , $X_{2} X_{1} X_{2}^{2}$ , and $X_{1} X_{2}^{3}$ . Hence both length-four Serre relations vanish too. Steps 1.1 and 2.1 and these checks verify every defining relation of $B$ , so the free-generator assignment factors through a representation of the full quotient.

4.1F1F2F3step 1.1step 3.1step 2.2algebra∎

By [F1] and step 2.2, the proposed coproduct image of $S$ acts as $(1 + q^{- 2} - q - q^{- 1}) v_{2} \otimes v_{2} = q^{- 2} (1 - q) (1 + q^{2}) v_{2} \otimes v_{2}$ , which is nonzero over $Q (q)$ . Since $S = 0$ in the full represented algebra $B$ , this contradicts the relation preservation required of a coproduct algebra map $B \to B \otimes B$ . Independently, expanding the off-diagonal mixed commutator gives $[Δ (E_{1}), Δ (F_{2})] = (q - 1) K_{2} E_{1} \otimes F_{2} K_{1}^{- 1}$ : the cross scalar is $q_{1}^{a_{12}} q_{2}^{- a_{21}} - 1 = q - 1$ . Its action on $v_{1} \otimes v_{1}$ is $(q - 1) v_{2} \otimes v_{0} \neq 0$ . Thus both the full Serre and mixed-relation failures are detected without an assumption of triangular decomposition.

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The quasiprimitive Serre element in type $A_{2}$

Statement

In the Drinfeld–Jimbo algebra of type $A_{2}$ (the Cartan datum $I = {1, 2}$ , $A = (\begin{array}{cc} 2 & - 1 \\ - 1 & 2 \end{array})$ , $d_{1} = d_{2} = 1$ , so $q_{1} = q_{2} = q$ ) the Serre elements

${S e r r e}_{12}^{+} = E_{1}^{2} E_{2} - [2]_{q} E_{1} E_{2} E_{1} + E_{2} E_{1}^{2}, {S e r r e}_{12}^{-} = F_{1}^{2} F_{2} - [2]_{q} F_{1} F_{2} F_{1} + F_{2} F_{1}^{2}, [2]_{q} = q + q^{- 1},$

are quasiprimitive with explicit grouplike factors:

$Δ ({S e r r e}_{12}^{+}) = {S e r r e}_{12}^{+} \otimes K_{1}^{- 2} K_{2}^{- 1} + 1 \otimes {S e r r e}_{12}^{+}, Δ ({S e r r e}_{12}^{-}) = {S e r r e}_{12}^{-} \otimes 1 + K_{1}^{2} K_{2} \otimes {S e r r e}_{12}^{-} .$

In particular every mixed bidegree term cancels. The polynomial positive Serre expression at $q = 1$ is the classical Serre bracket $[e_{1}, [e_{1}, e_{2}]] = 0$ of type $A_{2}$ .

Facts & Assumptions

Given: The Cartan matrix has $a_{12} = a_{21} = - 1$ and symmetrizer $d_{1} = d_{2} = 1$ , so $q_{1} = q_{2} = q$ and the toral actions are those in Symmetrizable Cartan data for quantum groups and The Drinfeld-Jimbo quantized enveloping algebra by generators and relations. The Drinfeld–Jimbo definition supplies the two Serre words; the coproduct/Serre lemma supplies the positive and negative toral-action Borel maps, and their images give the formulas in the Drinfeld–Jimbo quotient. Tensor-product multiplication is as stated in The coproduct preserves the positive and negative quantum Serre ideals and The tensor product of $R$ -algebras has multiplication $(a \otimes b) (a^{'} \otimes b^{'}) = a a^{'} \otimes b b^{'}$ .

[F1]

$[2]_{q} = q + q^{- 1}$ (Quantum integers, factorials, Gaussian binomials and divided powers at $q_{i}$ ).

[F2]

If $y x = q_{i}^{2} x y$ , then $(x + y)^{N} = \sum_{r = 0}^{N} q_{i}^{r (N - r)} {(\binom{N}{r})}_{i} x^{r} y^{N - r}$ (The quantum binomial expansion for $q$ -commuting elements).

[F3]

The normalized positive coproduct assignment is an algebra map on the toral-action Borel (The coproduct preserves the positive and negative quantum Serre ideals).

[F4]

For every $i \neq j$ , the normalized negative coproduct formula is $Δ_{-} ({S e r r e}_{i j}^{-}) = {S e r r e}_{i j}^{-} \otimes 1 + K_{i}^{m_{i j}} K_{j} \otimes {S e r r e}_{i j}^{-}$ (The coproduct preserves the positive and negative quantum Serre ideals).

[F5]

In the classical Kac–Moody algebra, the Serre presentation imposes $(ad e_{i})^{1 - a_{i j}} e_{j} = 0$ (Serre presentation of a kac moody algebra).

[F6]

Tensor-product multiplication is $(a \otimes b) (c \otimes d) = a c \otimes b d$ (The tensor product of $R$ -algebras has multiplication $(a \otimes b) (a^{'} \otimes b^{'}) = a a^{'} \otimes b b^{'}$ ).

[F7]

The toral action is $K_{h} E_{i} K_{h}^{- 1} = q^{⟨ α_{i}, h ⟩} E_{i}$ (The Drinfeld-Jimbo quantized enveloping algebra by generators and relations).

[F8]

The quantum Cartan datum fixes $q_{i} = q^{d_{i}}$ , so $d_{1} = d_{2} = 1$ gives $q_{1} = q_{2} = q$ (Symmetrizable Cartan data for quantum groups).

[F9]

The positive and negative Serre words are the sums in the Drinfeld–Jimbo presentation (The Drinfeld-Jimbo quantized enveloping algebra by generators and relations).

Proof

technique · Expand the positive three-letter words by choosing, at each position, a left coproduct term or a right coproduct term, and collect the six mixed tensor words

1.1F3F6F7F8algebra

The toral rules give $K_{1} E_{1} K_{1}^{- 1} = q^{2} E_{1}$ , $K_{1} E_{2} K_{1}^{- 1} = q^{- 1} E_{2}$ , $K_{2} E_{1} K_{2}^{- 1} = q^{- 1} E_{1}$ , and $K_{2} E_{2} K_{2}^{- 1} = q^{2} E_{2}$ . For $x_{i} = E_{i} \otimes K_{i}^{- 1}$ and $y_{i} = 1 \otimes E_{i}$ , these imply $y_{i} x_{i} = q^{2} x_{i} y_{i}$ .

1.2F4given

Here $m_{12} = 2$ , so [F4] gives $Δ ({S e r r e}_{12}^{-}) = {S e r r e}_{12}^{-} \otimes 1 + K_{1}^{2} K_{2} \otimes {S e r r e}_{12}^{-}$ ; thus the negative expansion has no mixed bidegree terms either.

2.1step 1.1F2F3F6algebra

Applying [F2] at $N = 2$ gives $Δ (E_{i})^{2} = E_{i}^{2} \otimes K_{i}^{- 2} + (1 + q^{2}) E_{i} \otimes K_{i}^{- 1} E_{i} + 1 \otimes E_{i}^{2}$ .

3.1F1F3F6F7F9step 2.1algebra

To collect the positive expansion, for each original word choose $L$ for a position sent to $E_{i} \otimes K_{i}^{- 1}$ and $R$ for one sent to $1 \otimes E_{i}$ . The left word preserves the $L$ letters; the right word preserves the $R$ letters, followed by the $K^{- 1}$ factors from $L$ . Moving a $K_{i}^{- 1}$ across a later $E_{j}$ contributes $q^{- a_{i j}}$ . In bidegree $(2, 1)$ the coefficients of $E_{1} E_{2} \otimes E_{1} K_{1}^{- 1} K_{2}^{- 1}$ , $E_{1}^{2} \otimes E_{2} K_{1}^{- 2}$ , and $E_{2} E_{1} \otimes E_{1} K_{1}^{- 1} K_{2}^{- 1}$ are respectively $1 + q^{- 2} - [2]_{q} q^{- 1}$ , $q^{2} - [2]_{q} q + 1$ , and $- [2]_{q} + q + q^{- 1}$ .

3.2F1F3F6F7F9step 1.1step 2.1algebra

In bidegree $(1, 2)$ the coefficients of $E_{1} \otimes E_{1} E_{2} K_{1}^{- 1}$ , $E_{2} \otimes E_{1}^{2} K_{2}^{- 1}$ , and $E_{1} \otimes E_{2} E_{1} K_{1}^{- 1}$ are respectively $q^{- 1} + q - [2]_{q}$ , $1 - [2]_{q} q + q^{2}$ , and $- [2]_{q} q^{- 1} + q^{- 2} + 1$ . These six coefficient groups exhaust the non-extreme splits of the three-letter Serre words.

4.1F1step 3.1step 3.2F6algebra

Substituting $[2]_{q} = q + q^{- 1}$ makes all six coefficients in steps 3.1–3.2 zero. The all-left and all-right choices contribute exactly ${S e r r e}_{12}^{+} \otimes K_{1}^{- 2} K_{2}^{- 1}$ and $1 \otimes {S e r r e}_{12}^{+}$ , proving the positive formula.

5.1F5algebra∎

At $q = 1$ , the positive Serre polynomial becomes $e_{1}^{2} e_{2} - 2 e_{1} e_{2} e_{1} + e_{2} e_{1}^{2} = [e_{1}, [e_{1}, e_{2}]]$ , which vanishes by the classical type- $A_{2}$ Serre relation [F5]. This is the specialization of the polynomial Serre expression only; it does not assert specialization of the whole algebra over $Q (q)$ .

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The double-edge Serre relation for the cyclic affine type $A_{1}^{(1)}$

Statement

For the double edge $a_{12} = a_{21} = - 2$ (the Cartan matrix $(\begin{array}{cc} 2 & - 2 \\ - 2 & 2 \end{array})$ of the cyclic affine type $A_{1}^{(1)}$ , with $d_{1} = d_{2} = 1$ ) the quantum Serre relation has $m = 1 - a_{12} = 3$ :

${S e r r e}_{12}^{+} = E_{1}^{3} E_{2} - [3]_{q} E_{1}^{2} E_{2} E_{1} + [3]_{q} E_{1} E_{2} E_{1}^{2} - E_{2} E_{1}^{3}, [3]_{q} = q^{2} + 1 + q^{- 2},$

and it is quasiprimitive:

$Δ ({S e r r e}_{12}^{+}) = {S e r r e}_{12}^{+} \otimes K_{1}^{- 3} K_{2}^{- 1} + 1 \otimes {S e r r e}_{12}^{+}, Δ ({S e r r e}_{12}^{-}) = {S e r r e}_{12}^{-} \otimes 1 + K_{1}^{3} K_{2} \otimes {S e r r e}_{12}^{-} .$

The Gaussian coefficients satisfy $\sum_{r = 0}^{3} (- 1)^{r} q^{2 r} {(\binom{3}{r})}_{q} = 0$ . The polynomial positive Serre expression at $q = 1$ is the classical cubic relation $ad (e_{1})^{3} e_{2} = 0$ for this Cartan matrix.

Facts & Assumptions

Given: The matrix has $a_{12} = a_{21} = - 2$ , symmetrizer $d_{1} = d_{2} = 1$ , and the Drinfeld–Jimbo positive and negative Serre words use $m = 3$ . The prescribed coproduct on the positive and negative toral-action Borels is as in The coproduct preserves the positive and negative quantum Serre ideals.

[F1]

$q_{i} = q^{d_{i}}$ , so $q_{1} = q_{2} = q$ (Symmetrizable Cartan data for quantum groups).

[F2]

$[3]_{q} = q^{2} + 1 + q^{- 2}$ (Quantum integers, factorials, Gaussian binomials and divided powers at $q_{i}$ ).

[F3]

If $y x = q_{i}^{2} x y$ , then $(x + y)^{N} = \sum_{r = 0}^{N} q_{i}^{r (N - r)} {(\binom{N}{r})}_{i} x^{r} y^{N - r}$ (The quantum binomial expansion for $q$ -commuting elements).

[F4]

$\sum_{r = 0}^{3} (- 1)^{r} q^{2 r} {(\binom{3}{r})}_{q} = 0$ (The quantum Pascal recurrences, the Gauss product formula and Gaussian integrality).

[F5]

The normalized positive coproduct is an algebra map on the toral-action Borel (The coproduct preserves the positive and negative quantum Serre ideals).

[F6]

The normalized negative formula is $Δ_{-} ({S e r r e}_{i j}^{-}) = {S e r r e}_{i j}^{-} \otimes 1 + K_{i}^{m_{i j}} K_{j} \otimes {S e r r e}_{i j}^{-}$ (The coproduct preserves the positive and negative quantum Serre ideals).

[F7]

The toral action is $K_{h} E_{i} K_{h}^{- 1} = q^{⟨ α_{i}, h ⟩} E_{i}$ (The Drinfeld-Jimbo quantized enveloping algebra by generators and relations).

[F8]

The classical Kac–Moody presentation imposes $(ad e_{i})^{1 - a_{i j}} e_{j} = 0$ (Serre presentation of a kac moody algebra).

[F9]

[F10]

The positive and negative Serre words have Gaussian coefficients and powers $1 - a_{i j}$ (The Drinfeld-Jimbo quantized enveloping algebra by generators and relations).

Proof

technique · Use the q-binomial expansion for $\Delta(E_1)^3$, enumerate left/right choices in each four-letter Serre word, and collect the twelve mixed tensor words by bidegree

1.1F1F5F7F9algebra

For $i, j \in {1, 2}$ the toral relations give $K_{i} E_{i} K_{i}^{- 1} = q^{2} E_{i}$ and $K_{i} E_{j} K_{i}^{- 1} = q^{- 2} E_{j}$ when $i \neq j$ . For $x_{1} = E_{1} \otimes K_{1}^{- 1}$ and $y_{1} = 1 \otimes E_{1}$ , this gives $y_{1} x_{1} = q^{2} x_{1} y_{1}$ .

1.2F6given

Since $m_{12} = 3$ , [F6] gives $Δ ({S e r r e}_{12}^{-}) = {S e r r e}_{12}^{-} \otimes 1 + K_{1}^{3} K_{2} \otimes {S e r r e}_{12}^{-}$ , so the negative expansion also has no mixed bidegree terms.

2.1step 1.1F1F3F5F9algebra

Applying [F3] at $N = 3$ gives $Δ (E_{1})^{3} = E_{1}^{3} \otimes K_{1}^{- 3} + (1 + q^{2} + q^{4}) E_{1}^{2} \otimes K_{1}^{- 2} E_{1} + (1 + q^{2} + q^{4}) E_{1} \otimes K_{1}^{- 1} E_{1}^{2} + 1 \otimes E_{1}^{3}$ .

3.1F2F5F7F9F10step 2.1algebra

For each position of a positive Serre word, choose $L$ for $E_{i} \otimes K_{i}^{- 1}$ or $R$ for $1 \otimes E_{i}$ . The left word preserves the $L$ letters and the right word preserves the $R$ letters, followed by the toral factors from $L$ . Moving $K_{i}^{- 1}$ past a later $E_{j}$ contributes $q^{- 2}$ when $i = j$ and $q^{2}$ when $i \neq j$ . In bidegree $(3, 1)$ the coefficients, for $E_{1}^{3} \otimes E_{2} K_{1}^{- 3}$ , $E_{1}^{2} E_{2} \otimes E_{1} K_{1}^{- 2} K_{2}^{- 1}$ , $E_{1} E_{2} E_{1} \otimes E_{1} K_{1}^{- 2} K_{2}^{- 1}$ , and $E_{2} E_{1}^{2} \otimes E_{1} K_{1}^{- 2} K_{2}^{- 1}$ , are respectively $q^{6} - [3]_{q} q^{4} + [3]_{q} q^{2} - 1$ , $1 + q^{- 2} + q^{- 4} - [3]_{q} q^{- 2}$ , $- [3]_{q} (1 + q^{- 2}) + [3]_{q} (1 + q^{- 2})$ , and $[3]_{q} - (q^{2} + 1 + q^{- 2})$ .

3.2F2F5F7F9F10step 2.1algebra

In bidegree $(2, 2)$ the coefficients, for $E_{1}^{2} \otimes E_{1} E_{2} K_{1}^{- 2}$ , $E_{1}^{2} \otimes E_{2} E_{1} K_{1}^{- 2}$ , $E_{1} E_{2} \otimes E_{1}^{2} K_{1}^{- 1} K_{2}^{- 1}$ , and $E_{2} E_{1} \otimes E_{1}^{2} K_{1}^{- 1} K_{2}^{- 1}$ , are respectively $q^{4} + q^{2} + 1 - [3]_{q} (q^{2} + 1) + [3]_{q}$ , $- [3]_{q} + [3]_{q} (1 + q^{- 2}) - (1 + q^{- 2} + q^{- 4})$ , $1 + q^{- 2} + q^{- 4} - [3]_{q} (1 + q^{- 2}) + [3]_{q}$ , and $- [3]_{q} + [3]_{q} (q^{2} + 1) - (q^{4} + q^{2} + 1)$ .

3.3F2F5F7F9F10step 2.1algebra

In bidegree $(1, 3)$ the coefficients, for $E_{1} \otimes E_{1}^{2} E_{2} K_{1}^{- 1}$ , $E_{1} \otimes E_{1} E_{2} E_{1} K_{1}^{- 1}$ , $E_{1} \otimes E_{2} E_{1}^{2} K_{1}^{- 1}$ , and $E_{2} \otimes E_{1}^{3} K_{2}^{- 1}$ , are respectively $[3]_{q} - [3]_{q}$ , $- [3]_{q} (1 + q^{- 2}) + [3]_{q} (1 + q^{- 2})$ , $[3]_{q} q^{- 2} - (1 + q^{- 2} + q^{- 4})$ , and $1 - [3]_{q} q^{2} + [3]_{q} q^{4} - q^{6}$ . These are all remaining mixed bidegrees.

4.1F2F4F10step 3.1step 3.2step 3.3F9algebra

Substituting $[3]_{q} = q^{2} + 1 + q^{- 2}$ makes all twelve mixed coefficients in steps 3.1–3.3 zero; the last coefficient in step 3.3 is the alternating Gaussian identity [F4]. The all-left and all-right choices give exactly ${S e r r e}_{12}^{+} \otimes K_{1}^{- 3} K_{2}^{- 1}$ and $1 \otimes {S e r r e}_{12}^{+}$ . Thus the positive element is quasiprimitive.

5.1F8algebra∎

At $q = 1$ , $[3]_{q} = 3$ and the positive Serre polynomial becomes $e_{1}^{3} e_{2} - 3 e_{1}^{2} e_{2} e_{1} + 3 e_{1} e_{2} e_{1}^{2} - e_{2} e_{1}^{3} = ad (e_{1})^{3} e_{2}$ , which vanishes by [F8]. This specializes the polynomial Serre expression only, not the entire $Q (q)$ -algebra.

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Coproduct, antipode and $q$ -binomial expansion in $U_{q} ({s l}_{2})$

Example

In the rank-one Drinfeld–Jimbo algebra $U_{q} ({s l}_{2})$ (the Cartan datum $I = {1}$ , $A = (2)$ , $d_{1} = 1$ , $P^{\lor} = Z h_{1}$ , $P = Z α_{1}$ , $⟨ α_{1}, h_{1} ⟩ = 2$ , so $q_{1} = q$ and $K = K_{h_{1}}$ ) with generators $E, F, K^{\pm 1}$ and relations $K E K^{- 1} = q^{2} E$ , $K F K^{- 1} = q^{- 2} F$ , $E F - F E = (K - K^{- 1}) / (q - q^{- 1})$ :

(i) the coproduct, counit and antipode of The Drinfeld–Jimbo formulas define a Hopf algebra are $Δ (E) = E \otimes K^{- 1} + 1 \otimes E$ , $Δ (F) = F \otimes 1 + K \otimes F$ , $Δ (K^{\pm 1}) = K^{\pm 1} \otimes K^{\pm 1}$ , $ε (E) = ε (F) = 0$ , $ε (K) = 1$ , $S (E) = - E K$ , $S (F) = - K^{- 1} F$ , $S (K) = K^{- 1}$ ;

(ii) for every $n \geq 0$ the $q$ -binomial expansion holds: $Δ (E^{n}) = \sum_{r = 0}^{n} q^{r (n - r)} {(\binom{n}{r})}_{q} E^{r} \otimes K^{- r} E^{n - r}$ ; and both antipode identities hold on the generators: $m (S \otimes id) Δ (E) = - E + E = 0$ and $m (id \otimes S) Δ (E) = E K - E K = 0$ , with the same two computations for $F$ and the trivial $K^{\pm 1}$ checks;

(iii) $S^{2} (E) = q^{- 2} E$ and $S^{2} (F) = q^{2} F$ ; since $E \neq 0$ and $K \neq 1$ in $U_{q} ({s l}_{2})$ -- proved below by the oscillator model -- the antipode is not an involution;

(iv) the coproduct is not cocommutative: $Δ (E) - τ Δ (E) = E \otimes (K^{- 1} - 1) + (1 - K^{- 1}) \otimes E \neq 0$ , where $τ$ is the tensor flip.

All four computations use no choice principle. The nonvanishing statements in (iii) and (iv) are proved by an explicit representation of $U_{q} ({s l}_{2})$ on the Laurent polynomial ring, independently of triangular decomposition.

Facts & Assumptions

Given: The rank-one Drinfeld–Jimbo algebra, with generators $E, F, K^{\pm 1}$ and relations as displayed.

[F1]

The Drinfeld–Jimbo algebra of a symmetrizable Cartan datum has the displayed coproduct, counit and antipode, its antipode is unique and $S^{2} (E_{i}) = q_{i}^{- 2} E_{i}$ , $S^{2} (F_{i}) = q_{i}^{2} F_{i}$ , and any assignment of generators satisfying the defining relations extends to an algebra homomorphism (The Drinfeld–Jimbo formulas define a Hopf algebra, The Drinfeld-Jimbo quantized enveloping algebra by generators and relations).

[F2]

The rank-one datum with $P^{\lor} = Z h_{1}$ , $P = Z α_{1}$ and $⟨ α_{1}, h_{1} ⟩ = 2$ has exactly the relations displayed above (The Drinfeld-Jimbo quantized enveloping algebra by generators and relations).

[F3]

If $y x = t x y$ in a unital algebra, then $(x + y)^{N} = \sum_{r} B_{N, r} (t) x^{r} y^{N - r}$ with $B_{N, r} (t)$ the asymmetric Gaussian coefficient; for $t = q_{i}^{2}$ this reads $(x + y)^{N} = \sum_{r} q_{i}^{r (N - r)} {(\binom{N}{r})}_{i} x^{r} y^{N - r}$ (The quantum binomial expansion for $q$ -commuting elements).

[F4]

In $Q (q)$ one has $q \neq 0$ , $q \neq 1$ , $q^{2} \neq 1$ , $q^{2 n} \neq 1$ for $n \neq 0$ , and $(K - K^{- 1}) / (q - q^{- 1})$ is defined (Quantum integers, factorials, Gaussian binomials and divided powers at $q_{i}$ , The Drinfeld-Jimbo quantized enveloping algebra by generators and relations).

[F5]

$M : = Q (q) [X^{\pm 1}]$ is explicitly the space of finite sums $\sum_{n \in Z} c_{n} X^{n}$ , $c_{n} \in Q (q)$ , with coefficientwise addition and product $X^{m} X^{n} = X^{m + n}$ . These finite convolution operations are associative and have unit $X^{0}$ by addition of integer exponents; the formal monomials form a basis by the coefficient-function definition. This is the same finite Laurent construction as The Laurent polynomial ring as the principal localisation of Z[t] at t, here with coefficient field $Q (q)$ ; the $Q (q)$ -linear endomorphisms of $M$ form a unital algebra under composition, and linear functionals on an algebra form a vector space (The Laurent polynomial ring as the principal localisation of Z[t] at t, The endomorphism ring ${End}_{R} (M)$ under addition and composition, Linear functionals and the algebraic dual $V^{*} = L (V, F)$ ).

[F6]

No choice principle is used: the model of step 1.3 is defined by an explicit formula, and every sum below is finite.

Verification

technique · Instantiate the Hopf formulas and expand $\Delta(E^n)$ with the $q$-binomial lemma; then build an explicit oscillator representation of $U_q(\mathfrak{sl}_2)$ on the Laurent polynomial ring and use one of its matrix functionals to separate the elements that occur

1.1F1F2algebra

Part (i). By [F2], $U_{q} ({s l}_{2})$ is the Drinfeld–Jimbo algebra of the rank-one datum, so [F1] gives $Δ (E) = E \otimes K^{- 1} + 1 \otimes E$ , $Δ (F) = F \otimes 1 + K \otimes F$ , $Δ (K^{\pm 1}) = K^{\pm 1} \otimes K^{\pm 1}$ , $ε (E) = ε (F) = 0$ , $ε (K) = 1$ . The antipode is the unique convolution inverse of the identity; on $E$ it is $S (E) = - E K$ because $- E K \cdot K^{- 1} + E = 0$ and $E \cdot K - E K = 0$ compute the two convolution equations of [F1] on $E$ (using $S (K^{- 1}) = K$ ), and on $F$ it is $S (F) = - K^{- 1} F$ by the mirrored computation.

1.2F1F3algebra

Part (ii), first assertion. Put $x : = E \otimes K^{- 1}$ and $y : = 1 \otimes E$ in $U_{q} ({s l}_{2}) \otimes U_{q} ({s l}_{2})$ . Then $y x = (1 \otimes E) (E \otimes K^{- 1}) = E \otimes E K^{- 1} = E \otimes q^{2} K^{- 1} E = q^{2} x y$ , using $E K^{- 1} = q^{2} K^{- 1} E$ , which follows from $K E K^{- 1} = q^{2} E$ by multiplying on the left by $K^{- 1}$ and on the right by $K$ . Since $Δ$ is an algebra homomorphism, $Δ (E^{n}) = Δ (E)^{n} = (x + y)^{n}$ , and [F3] with $q_{1} = q$ gives $Δ (E^{n}) = \sum_{r = 0}^{n} q^{r (n - r)} {(\binom{n}{r})}_{q} E^{r} \otimes K^{- r} E^{n - r}$ , because $x^{r} = E^{r} \otimes K^{- r}$ and $y^{n - r} = 1 \otimes E^{n - r}$ .

1.3F1F4F5construct

The oscillator model. Let $M = Q (q) [X^{\pm 1}]$ and define $Q (q)$ -linear endomorphisms by $\hat{E} (f) : = X f$ , $(\hat{K} f) (X) : = f (q^{2} X)$ , and $\hat{F} (X^{n}) : = λ_{n} X^{n - 1}$ with $λ_{n} : = α q^{2 n} + β q^{- 2 n}$ , $α : = 1 / [(q - q^{- 1}) (1 - q^{2})]$ , $β : = - 1 / [(q - q^{- 1}) (1 - q^{- 2})]$ ; both scalars are nonzero and defined by [F4], and $\hat{K}$ is invertible with ${\hat{K}}^{- 1} (f) (X) = f (q^{- 2} X)$ . Then $\hat{E} \hat{F} - \hat{F} \hat{E}$ acts on $X^{n}$ by the scalar $(q^{2 n} - q^{- 2 n}) / (q - q^{- 1})$ for every $n \in Z$ : indeed $(\hat{E} \hat{F} - \hat{F} \hat{E}) (X^{n}) = λ_{n} X^{n} - λ_{n + 1} X^{n} = (λ_{n} - λ_{n + 1}) X^{n}$ and $λ_{n} - λ_{n + 1} = α q^{2 n} (1 - q^{2}) + β q^{- 2 n} (1 - q^{- 2}) = (q^{2 n} - q^{- 2 n}) / (q - q^{- 1})$ , while $(\hat{K} - {\hat{K}}^{- 1}) / (q - q^{- 1})$ acts on $X^{n}$ by the same scalar; moreover $\hat{K} \hat{E} {\hat{K}}^{- 1} = q^{2} \hat{E}$ and $\hat{K} \hat{F} {\hat{K}}^{- 1} = q^{- 2} \hat{F}$ by direct evaluation on the basis. Hence by the universal property in [F1] there is a unital algebra homomorphism $ρ : U_{q} ({s l}_{2}) \to End (M)$ with $ρ (E) = \hat{E}$ , $ρ (F) = \hat{F}$ , $ρ (K) = \hat{K}$ .

2.1step 1.1F1algebra

Part (ii), antipode identities. Using 1.1, $m (S \otimes id) Δ (E) = S (E) K^{- 1} + E = - E K \cdot K^{- 1} + E = 0$ and $m (id \otimes S) Δ (E) = E \cdot S (K^{- 1}) + S (E) = E K - E K = 0$ , since $S (K^{- 1}) = K$ . The same two computations with $K$ replaced by $K^{- 1}$ and $E$ by $F$ give $m (S \otimes id) Δ (F) = S (F) + S (K) F = - K^{- 1} F + K^{- 1} F = 0$ and $m (id \otimes S) Δ (F) = F + K \cdot S (F) = F - F = 0$ ; for $K^{\pm 1}$ both sides are $K^{\pm 1} K^{\mp 1} = 1$ , and for the unit both are $1$ .

2.2step 1.1F1algebra

Part (iii), first assertion. $S^{2} (E) = S (- E K) = - S (K) S (E) = - K^{- 1} (- E K) = K^{- 1} E K = q^{- 2} E$ and $S^{2} (F) = S (- K^{- 1} F) = - S (F) S (K^{- 1}) = - (- K^{- 1} F) K = K^{- 1} F K = q^{2} F$ , using anti-multiplicativity of $S$ from [F1].

3.1F4F5step 2.2step 1.3algebra

Nonvanishing and separation. Since $\hat{E} (X^{0}) = X^{1} \neq 0$ and $\hat{K} \neq id$ (as $\hat{K} (X^{1}) = q^{2} X^{1} \neq X^{1}$ by [F4]), neither $E = 0$ nor $K = 1$ holds in $U_{q} ({s l}_{2})$ ; likewise $K^{- 1} \neq 1$ . Also $E \notin {span}_{Q (q)} {1, K^{- 1}}$ : if $E = a \cdot 1 + b K^{- 1}$ , applying $ρ$ gives $\hat{E} = a id + b {\hat{K}}^{- 1}$ , and evaluating on $X^{0}$ gives $X^{1} = (a + b) X^{0}$ , whose two sides have disjoint monomial supports, a contradiction. In particular $S^{2} (E) = q^{- 2} E \neq E$ , so $S^{2} \neq id$ , which completes (iii).

4.1step 1.1step 1.3step 3.1F5algebra∎

Part (iv). By 1.1, $Δ (E) - τ Δ (E) = E \otimes K^{- 1} + 1 \otimes E - K^{- 1} \otimes E - E \otimes 1 = E \otimes (K^{- 1} - 1) + (1 - K^{- 1}) \otimes E$ . Let $λ : U_{q} ({s l}_{2}) \to Q (q)$ be the linear functional $λ (g) : = [X^{1}] (ρ (g) (X^{0}))$ (coefficient extraction, [F5]). Then $λ (E) = 1$ , $λ (1) = 0$ and $λ (K^{- 1}) = 0$ by 1.3, so applying $id \otimes λ$ to the displayed element gives $E \cdot λ (K^{- 1} - 1) + (1 - K^{- 1}) \cdot λ (E) = 1 - K^{- 1} \neq 0$ by 3.1; hence $Δ (E) \neq τ Δ (E)$ and the coproduct is not cocommutative.

Remarks

Every displayed computation is finite and uses only [F1]–[F5]; the model of step 1.3 is given by explicit formulas and is the only place where an auxiliary construction is made, and it is choice-free by [F6].

Quantized Enveloping Algebras and Quantum Serre Relations — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Unsymmetrized parameters break the coproduct of the Serre ideal

Statement refuted

Facts & Assumptions

Counterexample

The quasiprimitive Serre element in type A2A2​

Statement

Facts & Assumptions

Proof

The double-edge Serre relation for the cyclic affine type A1(1)A1(1)​

Statement

Facts & Assumptions

Proof

Coproduct, antipode and qq-binomial expansion in Uq(sl2)Uq​(sl2​)

Example

Facts & Assumptions

Verification

Remarks

Sources

The quasiprimitive Serre element in type $A_{2}$

The double-edge Serre relation for the cyclic affine type $A_{1}^{(1)}$

Coproduct, antipode and $q$ -binomial expansion in $U_{q} ({s l}_{2})$