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Fraction-field coefficients of an integral LKB class are Laurent polynomials

Statement

Let $c_{i, j} \in Q (q, t)$ for $1 \leq i < j \leq n$ be such that $v = \sum_{i < j} c_{i, j} v_{i, j}$ lies in $H_{2} (\tilde{C}; Z)$ . Then $c_{i, j} \in Λ$ for all $i < j$ ; equivalently the closed surfaces $v_{i, j}$ span $H_{2} (\tilde{C}; Z)$ over $Λ$ .

Facts & Assumptions

Given: the closed surfaces $v_{i, j}$ and dual classes $x_{i, j}$ of Closed LKB basis surfaces have the three required topological types and factors, the pairings of The fork-noodle pairing is well defined and equivariant, and the fraction-field rank and saturated inclusion of The absolute LKB cellular boundary and fraction-field rank and The absolute LKB inclusion obtained by deleting the last puncture is saturated.

[F1]

(Bigelow 2002, Lemma 4.6, printed p. 11.) For all $1 \leq i < j \leq n$ the dual class $x_{i, j}$ is a multiple of $(1 - q)^{2}$ in $H_{2} (\tilde{C}, \partial \tilde{C} \cup \tilde{ν})$ . This is the finite-strip decomposition of source Figure7. Move the two vertical edges, relative to the boundary/end homology, into disjoint U-shaped edges enclosing the prefix punctures $p_{1}, \dots, p_{i}$ and suffix punctures $p_{j}, \dots, p_{n}$ . Cut each U along finitely many radial arcs ending in small puncture neighborhoods. The paired shores have opposite orientations and their lifts differ by $q$ (or $q^{- 1}$ on the opposite oriented U), so each radial piece has coefficient a Laurent unit times $1 - q$ (respectively $1 - q^{- 1}$ ). Pieces lying on the outer boundary or inside an end neighborhood are zero in the indicated relative group. The two U regions are disjoint, so their product pieces never collide and introduce no additional mutual winding. Thus the finite product decomposition factors out $(1 - q) (1 - q^{- 1})$ , a unit multiple of $(1 - q)^{2}$ . The number of pieces depends on the two puncture clusters; it is not an asserted universal eight-piece count.

[F2]

(Bigelow 2002, Lemma 4.5, printed p. 11.) The class $x_{n - 1, n}$ is a multiple of $(1 - q) (1 + q t) (1 - t)$ in $H_{2} (\tilde{C}, \partial \tilde{C} \cup \tilde{ν})$ . The source cuts the square $I \times I$ by the four lines ${x = y}$ , ${x + y = 1}$ , ${x = \frac{1}{2}}$ , ${y = \frac{1}{2}}$ into eight triangles (the antidiagonal in the unit-square coordinates); restricted to the eight pieces the lifted representative represents $1, - q, - t, q t, q t, - q^{2} t, - q t^{2}, q^{2} t^{2}$ times the triangle on the edge $α$ , and the sum of these eight coefficients is $1 - q - t + 2 q t - q^{2} t - q t^{2} + q^{2} t^{2} = (1 - q) (1 + q t) (1 - t) .$

[F3]

The last-column pairings hold up to a Laurent unit: $⟨ v, x_{i, n} ⟩ = (1 - q)^{2} c_{i, n} (n \geq 4, i \leq n - 2), ⟨ v, x_{n - 1, n} ⟩ = (1 - q)^{2} (1 + q t) (1 - t) c_{n - 1, n} (n \geq 2) .$ For $n = 3$ , $⟨ v, x_{1, 3} ⟩ = (1 - q)^{2} (1 + q t) c_{1, 3}$ . After subtracting the adjacent terms, $v^{'} = c_{1, 3} v_{1, 3}$ satisfies $⟨ v^{'}, σ_{2} x_{2, 3} ⟩ = (1 - t) (1 - q)^{2} (1 + q t) c_{1, 3} .$ The class $σ_{2} x_{2, 3}$ is a multiple of $(1 - t) (1 - q) (1 + q t)$ . These are precisely the separate higher-rank and residual three-puncture computations of Bigelow's Lemma 4.4 proof, not one formula for all ranks. They use the diagonal and last-column pairings and the closing factors. The nonunit off-diagonal exception of the corrected surface lemma does not enter a last-column pairing, and it is absent at $n = 3$ .

[F4]

The integers have no zero divisors (The integers have no zero divisors; multiplicative cancellation), polynomial extension preserves this property (A polynomial ring over an integral domain is an integral domain), and localization is the fraction construction of Multiplicative subsets and the localisation $S^{- 1} R$ as equivalence classes of fractions. Localizing $Z [t]$ or $Z [t, q]$ at powers of the variables gives the Laurent domains $Z [t^{\pm 1}]$ and $Λ$ : the denominators are nonzero monomials, so clearing them preserves both equality and nonzero products.

Proof

1.1F4algebra

The exact denominator-removal calculation. Put $R = Z [t^{\pm 1}]$ and $Λ = R [q^{\pm 1}]$ , both domains by [F4]. Evaluation $q \mapsto 1$ has kernel $(1 - q)$ : clear negative $q$ powers and write each $q^{r} - 1$ as $(q - 1) (1 + \dots + q^{r - 1})$ . Suppose $(1 - q) c = A \in Λ$ and $F c = B \in Λ$ , where $F (1, t) \neq 0$ . Then $A F = B (1 - q)$ , and evaluation gives $A (1, t) F (1, t) = 0$ , forcing $A (1, t) = 0$ . Therefore $A = (1 - q) C$ with $C \in Λ$ , and cancellation gives $c = C \in Λ$ . This applies to $F = (1 + q t) (1 - t)$ and to $F = 1 + q t$ , whose evaluated values are $(1 + t) (1 - t)$ and $1 + t$ , both nonzero. No UFD assertion or parameter specialization of the representation is required.

2.1F1F2F3givenstep 1.1algebra

The ranks below three. For $n = 1$ there are no coefficients; the declared absolute rank calculation and localization injection give $H_{2} = 0$ . For $n = 2$ only $c_{1, 2}$ occurs. By [F3] its pairing is a unit times $(1 - q)^{2} (1 + q t) (1 - t) c_{1, 2}$ . Divisibility of the dual class by $(1 - q)^{2}$ in [F1], together with sesquilinearity, gives $(1 + q t) (1 - t) c_{1, 2} \in Λ$ , since $1 - q^{- 1}$ is a unit multiple of $1 - q$ . Divisibility in [F2] gives $(1 - q) c_{1, 2} \in Λ$ , since conjugating any of the three factors changes it only by a unit. Step 1.1 therefore gives $c_{1, 2} \in Λ$ .

2.2F1F2F3step 1.1algebra

Reduction to $n = 3$ and to a single coefficient. Assume $n \geq 4$ and the statement known for $n - 1$ punctures. For $i \leq n - 2$ use the pairing with $x_{i, n}$ and [F1]: since $x_{i, n}$ is divisible by $(1 - q)^{2}$ , sesquilinearity gives $⟨ v, x_{i, n} ⟩ \in (1 - q^{- 1})^{2} Λ$ , and by [F3] $(1 - q)^{2} c_{i, n} \in (q - 1)^{2} Λ$ , so $c_{i, n} \in Λ$ . For $i = n - 1$ , [F1] and [F2] give $(1 + q t) (1 - t) c_{n - 1, n} \in Λ$ and $(1 - q) c_{n - 1, n} \in Λ$ , so $c_{n - 1, n} \in Λ$ by step 1.1. Subtracting the finitely many terms $c_{i, n} v_{i, n}$ leaves an integral class supported in the smaller configuration, which by The absolute LKB inclusion obtained by deleting the last puncture is saturated already lies in the image of $H_{2} ({\tilde{C}}_{n - 1}; Z)$ ; induction on $n$ reduces the statement to $n = 3$ .

3.1F1F2F3step 1.1step 2.2algebra

The case $n = 3$ . By [F3] and [F1], $(1 + q t) (1 - t) c_{2, 3} \in Λ$ ; by [F3] and [F2], $(1 - q) c_{2, 3} \in Λ$ ; step 1.1 gives $c_{2, 3} \in Λ$ . The reflected real-puncture picture gives the identical argument at the first puncture for $c_{1, 2}$ : reflection interchanges the two adjacent classes and inverts the deck variables, an automorphism of $Λ$ preserving the denominator-removal calculation. Thus $c_{1, 2} \in Λ$ . Then $v^{'} = v - c_{1, 2} v_{1, 2} - c_{2, 3} v_{2, 3} = c_{1, 3} v_{1, 3}$ is integral, so it remains to show $c_{1, 3} \in Λ$ from $⟨ v^{'}, σ_{2} x_{2, 3} ⟩ = (1 - t) (1 - q)^{2} (1 + q t) c_{1, 3}$ and $⟨ v^{'}, x_{1, 3} ⟩ = (1 - q)^{2} (1 + q t) c_{1, 3}$ ([F3]). Since $σ_{2} x_{2, 3}$ is a multiple of $(1 - t) (1 - q) (1 + q t)$ , the first pairing lies in $(1 - t) (1 - q^{- 1}) (1 + q t) Λ$ , hence $(1 - q) c_{1, 3} \in Λ$ . The second pairing lies in $(1 - q^{- 1})^{2} Λ$ , hence $(1 + q t) c_{1, 3} \in Λ$ . Step 1.1 with $F = 1 + q t$ gives $c_{1, 3} \in Λ$ .

4.1step 2.1step 2.2step 3.1algebra∎

Spanning. The classes $v_{i, j}$ are independent over $K = Q (q, t)$ by the triangular pairing matrix with Laurent-unit diagonal of Closed LKB basis surfaces have the three required topological types and factors, and the field kernel of the cellular differential has dimension $(\binom{n}{2})$ by The absolute LKB cellular boundary and fraction-field rank; hence they form a $K$ -basis of $K \otimes_{Λ} H_{2} (\tilde{C}; Z)$ . Every integral class $v$ is therefore a $K$ -linear combination $\sum c_{i, j} v_{i, j}$ with $c_{i, j} \in K$ , and steps 2.1, 2.2 and 3.1 show that all coefficients lie in $Λ$ . Thus the closed surfaces $v_{i, j}$ span $H_{2} (\tilde{C}; Z)$ over $Λ$ , as claimed.

Depends on

Used by

The integral LKB module is free of rank n choose two Theorem

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Sources

Bigelow, The Lawrence-Krammer representation, arXiv:math/0204057v1 (standard reference, not scraped)
Paoluzzi and Paris, A note on the Lawrence-Krammer-Bigelow representation, Algebr. Geom. Topol. 2 (2002) 499-518 (standard reference, not scraped)