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The exact grading shift survives braid decategorification

Statement

Fix $n \geq 2$ and the type-A Rouquier categorical braid-line model. Let $χ (C) = \sum_{m} (- 1)^{m} [C^{m}]$ be the alternating class of a bounded cochain complex in the split Grothendieck ring $K_{0}^{s p l i t} ({S B i m}_{n})$ , and use the shift convention $v [X] = [X (1)]$ . Let $Φ : K_{0}^{s p l i t} ({S B i m}_{n}) \to H_{S_{n}}$ be the algebra isomorphism with $Φ ([B_{i}]) = H_{i} = v (T_{i} + 1)$ and Hecke parameter $q = v^{- 2}$ . For $β \in B_{n}$ , let $e (β)$ be its exponent sum, the number of positive letters minus the number of negative letters in any signed word for $β$ , and let $T_{β}$ be the image of $β$ under the group homomorphism $B_{n} \to H_{S_{n}}^{\times}$ sending $σ_{i} \mapsto T_{i}$ . Then

Φ ⁣ (χ (F (β))) = v^{e (β)} T_{β} .

Consequently the normalized map $β \mapsto v^{- e (β)} Φ (χ (F (β)))$ is the standard Hecke braid homomorphism, and its normalization is multiplicative under braid stacking. Homotopy-equivalent complexes have the same Euler class, but need not be isomorphic as graded complexes.

Statement references

Categorified braid-line model with an explicit realization boundary

Decategorification of a Rouquier complex is the Hecke braid generator

Euler classes of Rouquier complexes are homotopy invariant and multiplicative

The Rouquier complex is well defined up to canonical homotopy equivalence

The braid group by Artin presentation

Complexes, homotopies and contractibility in an additive category

Facts & Assumptions

Given: $n \geq 2$ , a braid $β \in B_{n}$ , and the Rouquier complexes and grading convention of Categorified braid-line model with an explicit realization boundary.

[F1]

Decategorification of a Rouquier complex is the Hecke braid generator: in the stated shift convention, $Φ (χ (F_{i})) = v T_{i}$ and $Φ (χ (F_{i}^{- 1})) = v^{- 1} T_{i}^{- 1}$ for each simple generator.

[F2]

Euler classes of Rouquier complexes are homotopy invariant and multiplicative: $χ$ is invariant under homotopy equivalence and satisfies $χ (C \otimes_{R} D) = χ (C) χ (D)$ for the signed tensor totalization of bounded complexes with terms in ${S B i m}_{n}$ .

[F3]

The Rouquier complex is well defined up to canonical homotopy equivalence: signed words representing the same braid determine canonically homotopy-equivalent Rouquier complexes.

[F4]

The braid group by Artin presentation: $B_{n}$ is generated by $σ_{1}, \dots, σ_{n - 1}$ with adjacent braid relations and distant commutation relations; inserting or deleting an inverse pair is also a word equality.

[F5]

Complexes, homotopies and contractibility in an additive category: a cochain complex is contractible when there are maps $h^{m} : C^{m} \to C^{m - 1}$ satisfying $1_{C^{m}} = d^{m - 1} h^{m} + h^{m + 1} d^{m}$ for every $m$ .

[F6]

Categorified braid-line model with an explicit realization boundary: a signed word $σ = σ_{i_{1}}^{ϵ_{1}} \dots σ_{i_{r}}^{ϵ_{r}}$ is assigned the signed tensor totalization $F (σ) = F_{i_{1}}^{ϵ_{1}} \otimes_{R} \dots \otimes_{R} F_{i_{r}}^{ϵ_{r}}$ , and its braid has this complex up to canonical homotopy equivalence.

[A1]

Use $T_{β}$ for the standard homomorphism image specified in the Statement; no physical interpretation of $v$ is assumed.

Proof

1.1A1F1F2F6construct

Choose a signed word $σ = σ_{i_{1}}^{ϵ_{1}} \dots σ_{i_{r}}^{ϵ_{r}}$ representing $β$ . By the word definition and multiplicativity in [F2], $χ (F (σ)) = \prod_{k = 1}^{r} χ (F_{i_{k}}^{ϵ_{k}})$ . Applying the algebra homomorphism $Φ$ and the generator values in [F1] gives $Φ (χ (F (σ))) = v^{\sum_{k} ϵ_{k}} T_{i_{1}}^{ϵ_{1}} \dots T_{i_{r}}^{ϵ_{r}}$ .

2.1F4step 1.1algebra

The exponent sum is unchanged by each defining relation: the two sides of an adjacent braid relation each have sum $3$ , the two sides of a distant commutation relation each have sum $2$ , and an inverse pair has sum $0$ . Thus $e (σ)$ descends to $e (β) \in Z$ and satisfies $e (β γ) = e (β) + e (γ)$ . The product of the $T_{i}^{ϵ_{k}}$ is $T_{β}$ by the specified group homomorphism, so step 1.1 gives the displayed formula independent of the chosen word.

3.1F2F3step 2.1algebra

The comparison in [F3] identifies $F (β γ)$ in the homotopy category with the signed tensor totalization $F (β) \otimes_{R} F (γ)$ . Applying homotopy invariance and multiplicativity from [F2], then applying the algebra homomorphism $Φ$ , gives $Φ (χ (F (β γ))) = Φ (χ (F (β))) Φ (χ (F (γ)))$ . Dividing by $v^{e (β γ)} = v^{e (β)} v^{e (γ)}$ proves multiplicativity of the normalized map.

3.2F1step 2.1algebra

For a positive generator the formula is $v T_{i}$ and for its inverse it is $v^{- 1} T_{i}^{- 1}$ by [F1]. Since $v$ is an indeterminate in the Laurent polynomial coefficient ring, these factors cannot be discarded for nonzero exponent sum. The normalized value is exactly $T_{β}$ by step 2.1.

4.1F2F5F6step 1.1∎

Let $D$ be the two-term complex with $D^{0} = D^{1} = R$ , differential $d^{0} = 1_{R}$ , and all other terms zero. The map $h^{1} = 1_{R} : D^{1} \to D^{0}$ (all other $h^{m} = 0$ ) satisfies the contracting-homotopy equation in [F5], so $D$ is homotopy equivalent to the zero complex. But its graded terms are $R$ in degrees $0$ and $1$ , whereas the zero complex has zero terms; since $R = Q [x_{1}, \dots, x_{n}]$ has $1 \neq 0$ , the complexes are not isomorphic as graded complexes. Their Euler classes are both zero by [F2] and $[R] - [R] = 0$ .

The exact grading shift survives braid decategorification

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources