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Quantum Lucas and generation of specialized divided powers

Statement

Let $z$ be a primitive root of odd order $ℓ > 1$ . Write $N = ℓ N_{1} + N_{0}$ for $N \in Z$ and $r = ℓ r_{1} + r_{0}$ for $r \geq 0$ , with $0 \leq N_{0}, r_{0} < ℓ$ . The symmetric quantum binomial, interpreted by its Laurent-polynomial product for negative upper index, satisfies ${[\begin{array}{c} N \\ r \end{array}]}_{z} = (\binom{N_{1}}{r_{1}}) {[\begin{array}{c} N_{0} \\ r_{0} \end{array}]}_{z} .$ For a specialized family $X^{(n)}$ with the divided-power multiplication of Divided-power multiplication, toral binomials and the $E_{i}^{(a)} F_{i}^{(b)}$ reordering identity, $X^{(a ℓ + r)} = \frac{(X^{(ℓ)})^{a} X^{r}}{a! [r]_{z}!} (a \geq 0, 0 \leq r < ℓ) .$ In particular the family is generated by $X$ and $X^{(ℓ)}$ ; modulo the ideal generated by $X$ , a divided power vanishes unless its index is divisible by $ℓ$ , and $X^{(a ℓ)} = (X^{(ℓ)})^{a} / a!$ .

Facts & Assumptions

Given: The root $z$ , its odd order, and the divided-power family stated above.

[F1]

Symmetric and ordinary Gaussian binomials satisfy ${[\begin{matrix} N \\ r \end{matrix}]}_{v} = v^{- r (N - r)} G (N, r; v^{2})$ and the quantum Pascal recurrence, with Laurent-integral coefficients for nonnegative indices (The quantum Pascal recurrences, the Gauss product formula and Gaussian integrality, Quantum integers, factorials, Gaussian binomials and divided powers at $q_{i}$ ).

[F2]

The multiplication identity is $X^{(a)} X^{(b)} = {[\begin{matrix} a + b \\ a \end{matrix}]}_{z} X^{(a + b)}$ (Divided-power multiplication, toral binomials and the $E_{i}^{(a)} F_{i}^{(b)}$ reordering identity).

Proof

Given: The data above.

1.1F1givenconstructalgebra

For an indeterminate $Q$ put $P_{N} (t) = \prod_{j = 0}^{N - 1} (1 + Q^{j} t)$ for $N \geq 0$ and $P_{N} (t) = \prod_{j = N}^{- 1} (1 + Q^{j} t)^{- 1}$ for $N < 0$ , with inverses expanded as formal power series. Define $G (N, r; Q)$ by $P_{N} (t) = \sum_{r \geq 0} Q^{r (r - 1) / 2} G (N, r; Q) t^{r}$ . Multiplying the last factor gives the Gaussian Pascal recurrence, so for $N \geq 0$ this is the Gaussian of [F1]. For $N < 0$ , the same recurrence downward from $N = 0$ agrees with the generalized Gaussian product; each fixed coefficient uses finitely many terms of finitely many geometric series and is a Laurent polynomial in $Q$ . Thus evaluation at a root of unity is legitimate, including negative $N$ .

2.1F1step 1.1algebra

Set $Q = z^{2}$ , which has exact order $ℓ$ . Since $ℓ$ is odd, $\prod_{j = 0}^{ℓ - 1} (1 + Q^{j} t) = 1 + t^{ℓ}$ . Periodicity of the factors gives $P_{N} (t) = (1 + t^{ℓ})^{N_{1}} P_{N_{0}} (t)$ , also for $N_{1} < 0$ by formal inversion. The coefficient of $t^{r}$ is $(\binom{N_{1}}{r_{1}}) Q^{r_{0} (r_{0} - 1) / 2} G (N_{0}, r_{0}; Q)$ : the second factor has degree less than $ℓ$ , so its only eligible degree is $r_{0}$ . The difference $r (r - 1) / 2 - r_{0} (r_{0} - 1) / 2$ is divisible by $ℓ$ , because $r - r_{0}$ is divisible by the odd integer $ℓ$ . Consequently $G (N, r; Q) = (\binom{N_{1}}{r_{1}}) G (N_{0}, r_{0}; Q)$ .

3.1F1step 2.1algebra

In the symmetric conversion of [F1], the difference $r (N - r) - r_{0} (N_{0} - r_{0})$ is divisible by $ℓ$ . Therefore the factors $z^{- r (N - r)}$ and $z^{- r_{0} (N_{0} - r_{0})}$ agree, proving the displayed quantum Lucas identity. For $0 < r < ℓ$ , $[r]_{z} \neq 0$ , since otherwise $z^{2 r} = 1$ would contradict the exact order of $z^{2}$ . Hence $[r]_{z}! \neq 0$ for $r < ℓ$ .

4.1F2step 3.1algebra∎

Lucas gives ${[\begin{matrix} a ℓ + r \\ r \end{matrix}]}_{z} = 1$ and ${[\begin{matrix} (a + 1) ℓ \\ ℓ \end{matrix}]}_{z} = a + 1$ . Apply [F2] first repeatedly to $X^{(ℓ)}$ , obtaining $(X^{(ℓ)})^{a} = a! X^{(a ℓ)}$ , and then to $X^{(a ℓ)} X^{(r)}$ , obtaining $X^{(a ℓ + r)} = X^{(a ℓ)} X^{(r)}$ . Finally $X^{(r)} = X^{r} / [r]_{z}!$ for $r < ℓ$ . These prove the generation and quotient assertions, including $r = 0$ and $a = 0$ , using only finite computations.

Depends on

Used by

Dependency tree · two levels

11 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

George Lusztig, Introduction to Quantum Groups (standard reference, not scraped)