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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]

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

[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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Sources

Richard Borcherds, Mark Haiman, Theo Johnson-Freyd, Nicolai Reshetikhin and Vera Serganova, Berkeley Lectures on Lie Groups and Quantum Groups (standard reference, not scraped)
Benjamin Enriquez, PBW and Duality Theorems for Quantum Groups and Quantum Current Algebras, Journal of Lie Theory 13 (2003), 21–64 (standard reference, not scraped)
Kyeonghoon Jeong, Seok-Jin Kang and Masaki Kashiwara, Crystal Bases for Quantum Generalized Kac-Moody Algebras, arXiv:math/0305390 (standard reference, not scraped)
Alexander Kleshchev, Lectures on Infinite Dimensional Lie Algebras (standard reference, not scraped)

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