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Rouquier Complexes and Categorical Braid Relations

1 · Prerequisites

2 · Summary

This page builds Rouquier's 2-braid group for type $A$ inside the homotopy category of bounded complexes of graded $(R, R)$ -bimodules, starting from the type- $A$ Soergel bimodules $B_{i} = R \otimes_{R^{s_{i}}} R (1)$ of the Soergel page. The generator complexes are the two-term complexes $F_{i} = [B_{i} \overset{ε_{i}}{\to} R (1)]$ and $F_{i}^{- 1} = [R (- 1) \overset{η_{i}}{\to} B_{i}]$ , where $ε_{i}$ is the multiplication map and $η_{i} (1) = α_{i} \otimes 1 + 1 \otimes α_{i}$ ; every term is finite free as a left and as a right $R$ -module, so tensor product by these complexes is exact on both sides and descends to the derived tensor functor. The shift dictionary of the page is fixed once and for all: external shifts $(r)$ satisfy $M (r)_{d} = M_{d + r}$ while internal shifts ${r}$ satisfy $M {r}_{d} = M_{d - r}$ , the generator $1 \otimes 1$ of $B_{i}$ has degree $- 1$ and $1 \otimes α_{i}$ has degree $1$ .

The first results make the generators behave like the braid generators. $F_{i} \otimes_{R} F_{i}^{- 1}$ and $F_{i}^{- 1} \otimes_{R} F_{i}$ split as the unit complex $R$ plus two contractible two-term summands, which are exhibited with explicit contracting homotopies and cancelled by homological Gaussian elimination; generators with distant indices commute up to a canonical degree-zero isomorphism; and the three-term braid relation $F_{i} F_{i + 1} F_{i} ≃ F_{i + 1} F_{i} F_{i + 1}$ holds with no grading shift, by splitting the rank-one and rank-two Soergel tensor decompositions and cancelling the contractible summands. Iterating the signed tensor totalization over a signed word therefore attaches to every signed word $σ$ a bounded complex of graded bimodules, the Rouquier complex $F (σ)$ of a braid word.

The page then compares the different word models of one braid. For words $t, u$ with the same product, the canonical comparison $c_{t, u} = μ_{u}^{- 1} \circ μ_{t}$ between the word tensors of standard graph bimodules is a transitive system of degree-zero isomorphisms, and it lifts to a normalized homotopy map $γ_{t, u} ⁣ : F (t) \to F (u)$ that is the unique homotopy class with the prescribed derived image; the normalized maps compose transitively, which makes the Rouquier complex of a braid well defined up to canonical homotopy equivalence. Lifting the comparisons along the derived localization and transporting the graph multiplication produces the compositors $m_{v, w}$ and unit $m_{1}$ of a coherent action of $B_{n}$ on $K^{b} (R -grmod)$ in the strict sense of the pentagon and the two unit triangles; the braid-indexed category with morphisms ${Hom}_{K^{b}} (G_{v}, G_{w})$ has strict object product $v ⊠ w = v w$ , with the morphism product transported by $m$ . It is rigid, with dual label $v^{- 1}$ , and its evaluation is fully faithful onto the full carrier subcategory.

Finally the page decategorifies. The alternating class $χ (C) = \sum_{m} (- 1)^{m} [C^{m}]$ is a homotopy invariant, multiplicative for signed tensor totalizations and additive on cones, and on the generators it takes the values $χ (F_{i}) = [B_{i}] - [R (1)]$ and $χ (F_{i}^{- 1}) = [B_{i}] - [R (- 1)]$ ; under the identification $Φ$ of the split Grothendieck ring with the Hecke algebra, the classes of generator complexes become the normalized Hecke generators, so that a signed word for a braid $β$ satisfies $Φ (χ (F (σ))) = v^{e (σ)} T_{β}$ . The statement concerns classes only and therefore does not determine homotopy types: on the companion examples page the zero-differential complex $Z_{i} = [B_{i} \overset{0}{\to} R (1)]$ shares the class of $F_{i}$ while having different cohomology. No step of the page uses a choice principle: the constructions are termwise canonical, and the one choice made (a representative signed word per braid) is absorbed by the canonical comparisons.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

The positive and negative Rouquier generator complexes

Definition

The setting. Keep $R = Q [x_{1}, \dots, x_{n}]$ with $\deg x_{i} = 2$ , the place-permutation action of $S_{n}$ , the balanced roots $α_{i} = ε_{i} (x_{i} - x_{i + 1})$ with $ε_{i} = (- 1)^{i - 1}$ and the graded $(R, R)$ -bimodules $B_{i} = R \otimes_{R^{s_{i}}} R (1)$ of The Soergel bimodule $B_{i}$ of a simple reflection; recall that $1 \otimes 1 \in B_{i}$ has degree $- 1$ and $1 \otimes α_{i}$ has degree $1$ , that $B_{i}$ is generated as an $(R, R)$ -bimodule by $1 \otimes 1$ , and that $R = R^{s_{i}} \oplus α_{i} R^{s_{i}}$ with $s_{i} (α_{i}) = - α_{i}$ (The standard type-A reflection realization and its polynomial ring, The Soergel bimodule $B_{i}$ of a simple reflection).

The positive generator complex. For $1 \leq i \leq n - 1$ let $F_{i}$ be the bounded cochain complex of graded $(R, R)$ -bimodules $F_{i} : = [B_{i} \overset{ε_{i}}{\to} R (1)],$ with $B_{i}$ in cohomological degree $0$ , $R (1)$ in cohomological degree $1$ and all other terms zero, where $ε_{i} : B_{i} \to R (1)$ is the multiplication map $ε_{i} (r \otimes r^{'}) = r r^{'}$ , i.e. the degree-zero bimodule map determined by $ε_{i} (1 \otimes 1) = 1$ and $ε_{i} (1 \otimes α_{i}) = α_{i}$ .

The negative generator complex. Let $F_{i}^{- 1}$ be the bounded cochain complex of graded $(R, R)$ -bimodules $F_{i}^{- 1} : = [R (- 1) \overset{η_{i}}{\to} B_{i}],$ with $R (- 1)$ in cohomological degree $- 1$ , $B_{i}$ in cohomological degree $0$ and all other terms zero, where $η_{i} : R (- 1) \to B_{i}$ is the bimodule map $η_{i} (1) = α_{i} \otimes 1 + 1 \otimes α_{i}$ .

Well-definedness of the differentials. The map $ε_{i}$ descends from the multiplication $R \otimes_{Q} R \to R$ because $ε_{i} (r a \otimes r^{'}) = r a r^{'} = ε_{i} (r \otimes a r^{'})$ for $a \in R^{s_{i}}$ and $R$ is commutative; it is left and right $R$ -linear and homogeneous of degree zero, since $1 \otimes 1 \mapsto 1$ matches the degrees $- 1$ of $B_{i}$ and of the generator of $R (1)$ , and $1 \otimes α_{i} \mapsto α_{i}$ matches the degree $1 = \deg (α_{i}) - 1$ on both sides. For $η_{i}$ , the bimodule map $R \to B_{i}$ sending $1$ to the class of $x_{i} - x_{i + 1}^{'}$ is well defined: the products $(x_{i} - x_{i + 1}^{'}) (x_{i} - x_{i}^{'})$ and $(x_{i} - x_{i + 1}^{'}) (x_{i + 1} - x_{i + 1}^{'})$ vanish in $B_{i}$ by the explicit computation of GKS Lemma 3.8, while for $a \notin {i, i + 1}$ the factor $x_{a} - x_{a}^{'}$ is already a defining relation of $B_{i}$ , so the element annihilates the kernel ideal of the multiplication $R \otimes_{Q} R \to R$ , and the assignment $1 \mapsto x_{i} - x_{i + 1}^{'}$ extends to a well-defined $R$ -bimodule map. The value of that map at $1$ times the unit $2 ε_{i}$ is $η_{i} (1)$ , because $1 \otimes (x_{i} - x_{i + 1}) + (x_{i} - x_{i + 1}) \otimes 1 = 2 (x_{i} - x_{i + 1}^{'})$ in $B_{i}$ (GKS Lemma 3.8, using $x_{i} + x_{i + 1} = x_{i}^{'} + x_{i + 1}^{'}$ ) and $α_{i} = ε_{i} (x_{i} - x_{i + 1})$ ; a unit multiple of a well-defined bimodule map is a well-defined bimodule map, so $η_{i}$ is well defined. Its value $η_{i} (1)$ is homogeneous of degree $1$ , matching the degree of $1 \in R (- 1)$ , so $η_{i}$ is a degree-zero bimodule map. Both complexes are concentrated in two adjacent cohomological degrees, so $d^{2} = 0$ holds trivially and each of $F_{i}, F_{i}^{- 1}$ is a bounded cochain complex of graded $(R, R)$ -bimodules in the sense of Bounded graded bimodule complexes and signed tensor totalization.

Size and consequences. The bimodules $R$ , $R (1)$ and $R (- 1)$ are free of rank one on both sides and $B_{i}$ is free of rank two on both sides (Soergel generators and Bott–Samelson products are finite free on both sides, the bases being ${1 \otimes 1, 1 \otimes α_{i}}$ on the left and ${1 \otimes 1, α_{i} \otimes 1}$ on the right). Every term of $F_{i}$ and of $F_{i}^{- 1}$ is therefore finite free, hence finite graded projective, as a left $R$ -module and as an underlying right $R$ -module. Applying A bounded two-sided projective bimodule complex defines exact derived tensor functors with the commutative ring $Q$ and $A = B = R$ : the signed totalization with $F_{i}$ is an exact triangulated functor on bounded complexes of finite graded projective $R$ -modules, preserves quasi-isomorphisms between bounded complexes, and descends to the derived tensor functor $F_{i} \otimes_{R}^{L} -$ on $D^{b} (R -grmod)$ ; the same statements hold with $F_{i}^{- 1}$ in place of $F_{i}$ . The right-tensor construction $- \otimes_{R} F_{i}$ is defined separately by signed totalization, and its derived version uses the same two-sided freeness. Commutativity of $R$ does not assert a symmetry between tensor products of arbitrary $(R, R)$ -bimodules.

Recorded convention. The complex $F_{i}$ is the library normalization of Rouquier's positive complex $F_{s} = [A \otimes_{A^{s}} A \to A]$ , in which $A$ sits in degree $1$ and the differential is multiplication: in the library external shift $(r)$ with $M (r)_{d} = M_{d + r}$ one has $F_{i} = T_{i} (1)$ and $F_{i}^{- 1} ≅ T_{i}^{- 1} (- 1)$ for the GKS complexes $T_{i} = [B_{i} (- 1) \to R]$ , $T_{i}^{- 1} = [R \to B_{i} (1)]$ of formula (3.3). The GKS negative differential sends $1$ to $x_{i} - x_{i + 1}^{'}$ , whereas ours is $2 ε_{i}$ times that map. The chain isomorphism $F_{i}^{- 1} \to T_{i}^{- 1} (- 1)$ is multiplication by $2 ε_{i}$ in degree $- 1$ and the identity in degree $0$ . The internal shifts cancel in inverse pairs and agree on the two sides of each positive braid relation. The sign $ε_{i}$ in $α_{i} = ε_{i} (x_{i} - x_{i + 1})$ , and with it the sign of $η_{i}$ , is a unit of $Q$ ; the homotopy classes of the complexes do not depend on the choice of balanced root.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

Opposite Rouquier generator complexes are homotopy inverse

Statement

$F_{i} \otimes_{R} F_{i}^{- 1} ≃ R ≃ F_{i}^{- 1} \otimes_{R} F_{i}$ in $K^{b} (R^{e} -grmod)$ , where $R$ denotes the unit complex concentrated in cohomological degree $0$ with zero differential. Explicitly, the signed tensor totalization of $F_{i}$ and $F_{i}^{- 1}$ has terms $B_{i} (- 1) ⟶ B_{i} \otimes_{R} B_{i} \oplus R ⟶ B_{i} (1)$ in cohomological degrees $- 1, 0, 1$ ; the rank-one splitting $B_{i} \otimes_{R} B_{i} ≅ B_{i} (1) \oplus B_{i} (- 1)$ exhibits two successive invertible differential blocks. Gaussian elimination splits off two contractible two-term complexes, leaving $R$ in degree $0$ . The same argument with the factors exchanged gives $F_{i}^{- 1} \otimes_{R} F_{i} ≃ R$ . All scalars occurring in the contractions are units $ε_{i}^{\pm 1}, 2^{\pm 1} \in Q$ , so the homotopy classes do not depend on the signs of the chosen normalization.

Facts & Assumptions

Given: A simple reflection $s_{i}$ , the bimodule $B_{i} = R \otimes_{R^{s_{i}}} R (1)$ with the generators $u = 1 \otimes 1$ of degree $- 1$ and $w_{0} = 1 \otimes (α_{i} / 2)$ of degree $1$ , the generator complexes $F_{i} = [B_{i} \overset{ε_{i}}{\to} R (1)]$ , $F_{i}^{- 1} = [R (- 1) \overset{η_{i}}{\to} B_{i}]$ of The positive and negative Rouquier generator complexes.

[F1]

The rank-one splitting. The invariant decomposition $R = R^{s_{i}} \oplus α_{i} R^{s_{i}}$ in the middle tensor factor gives a degree-zero bimodule isomorphism $B_{i} \otimes_{R} B_{i} ≅ B_{i} (1) \oplus B_{i} (- 1) .$ The first summand is represented by $r \otimes 1 \otimes r^{'}$ and the second by $r \otimes α_{i} \otimes r^{'}$ . This is the middle-invariant and middle- $α_{i}$ decomposition of The rank-one Soergel bimodule square splits. The outer factors form $R \otimes_{R^{s_{i}}} R$ ; in particular the outer actions of $α_{i}$ need not be equal.

[F2]

The totalization. The signed tensor totalization $K = F_{i} \otimes_{R} F_{i}^{- 1}$ has terms $K^{- 1} = B_{i} (- 1)$ , $K^{0} = B_{i} \otimes_{R} B_{i} \oplus R$ , $K^{1} = B_{i} (1)$ and Koszul differential $d (x \otimes y) = d_{F} (x) \otimes y + (- 1)^{p} x \otimes d_{G} (y)$ ; in particular $d^{- 1} (x) = ε_{i} (x) + x \otimes η_{i} (1)$ for $x \in B_{i} (- 1)$ and $d^{0}$ is $ε_{i}$ on the first tensor factor of $B_{i} \otimes_{R} B_{i}$ and $- η_{i}$ on the $R$ -summand (Bounded graded bimodule complexes and signed tensor totalization, The positive and negative Rouquier generator complexes).

[F3]

Gaussian elimination. If a cochain differential has an invertible block $φ : U \to V$ with respect to fixed biproduct decompositions $X^{n} = A \oplus U$ , $X^{n + 1} = B \oplus V$ , then $X ≃ \overset{ˉ}{X}$ for the reduction $\overset{ˉ}{X}$ obtained by deleting $U, V$ and replacing $d^{n}$ by its Schur complement, and $X ≅ \overset{ˉ}{X} \oplus K$ with $K = [U \overset{φ}{\to} V]$ a contractible two-term complex (Gaussian elimination splits a contractible two-term complex, An invertible cochain differential block and its candidate reduction).

[F4]

Contractibility. A two-term complex $[X \overset{φ}{\to} Y]$ with $φ$ invertible is contractible, and homotopy equivalent complexes have the same homotopy class; $≃$ is transitive (Complexes, homotopies and contractibility in an additive category).

Proof

technique · direct

1.1F2

The terms of $K$ are as in [F2]: over degree $- 1$ only $B_{i} \otimes R (- 1) ≅ B_{i} (- 1)$ contributes, over degree $0$ the two summands $B_{i} \otimes B_{i}$ and $R (1) \otimes R (- 1) ≅ R$ , and over degree $1$ only $R (1) \otimes B_{i} ≅ B_{i} (1)$ .

1.2F1F2algebra

Write $E = R \otimes_{R^{s_{i}}} R \otimes_{R^{s_{i}}} R$ , suppressing the common internal shifts, and denote the three copies of $α_{i}$ by $a, b, c$ . Let $S = R \otimes_{R^{s_{i}}} R$ refer to the outer factors. Since $b^{2} = c^{2}$ and every invariant balances, $E = S \oplus S b$ as an outer bimodule. The map $1 \otimes η_{i}$ sends a source element $x$ to $x (b + c)$ . Because $(b - c) (b + c) = 0$ , its value is also $x (a, c) (b + c)$ : this is immediate on the right $R^{s_{i}}$ -basis $1, α_{i}$ of the source and hence for every $x$ . Its projection to the $S b$ summand is therefore the identity $S \to S b$ under the shifted identification [F1]. This is the invertible block $B_{i} (- 1) \to B_{i} (- 1)$ of $d^{- 1}$ . No equality between the outer $a$ and $c$ is used.

2.1F2F3step 1.2

Apply Gaussian elimination [F3] to that block. It removes the degree $- 1$ term and the middle- $α_{i}$ summand, leaving a complex with $B_{i} (1) \oplus R$ in degree $0$ and $B_{i} (1)$ in degree $1$ . Its degree-zero differential is the restriction of the original $d^{0}$ to the surviving summands: the preceding cancellation changes only the coordinates associated with the eliminated block, and $d^{0} d^{- 1} = 0$ ensures that its eliminated column is zero in the new coordinates.

3.1F1F2F3F4step 2.1

On the middle-invariant summand, $ε_{i} \otimes 1$ sends $r \otimes 1 \otimes r^{'}$ to $r \otimes r^{'}$ . Hence the block $B_{i} (1) \to B_{i} (1)$ of the remaining differential is the identity. A second application of [F3] cancels it and leaves only $R$ in degree $0$ . Both canceled two-term complexes are contractible by [F4], so $F_{i} \otimes_{R} F_{i}^{- 1} ≃ R$ .

4.1F1F2F4step 3.1algebra∎

Taking the opposite bimodule interchanges left and right actions and reverses the order of a tensor product. On complexes, the identification $(X \otimes_{R} Y)^{o p} ≅ Y^{o p} \otimes_{R} X^{o p}$ sends a term of cohomological bidegree $(p, q)$ with the sign $(- 1)^{p q}$ ; direct substitution in the signed tensor differential verifies that it is a chain isomorphism. Reversing the two factors of $B_{i}$ preserves multiplication and the symmetric element $α_{i} \otimes 1 + 1 \otimes α_{i}$ , so $F_{i}^{o p} ≅ F_{i}$ and $(F_{i}^{- 1})^{o p} ≅ F_{i}^{- 1}$ . Applying this additive operation to the equivalence just proved gives $F_{i}^{- 1} \otimes_{R} F_{i} ≃ R$ .

Remarks

The proof follows the route of GKS Lemma 3.11: the tensor product is the displayed three-term complex, its four-term middle term splits by the rank-one square, and two successive Gaussian eliminations cancel the two contractible two-term pieces; the two surviving directions are $R$ in degree $0$ . The two pivots use the middle-factor invariant decomposition; the two outer actions of $α_{i}$ are kept distinct. The ground field $Q$ makes the invariant decomposition available and all normalization scalars invertible. Rouquier's alternative proof of Lemma 3.3 uses the adjoint pairs attached to the split sequence $0 \to R^{s_{i}} \to R \to R^{s_{i}} (2) \to 0$ and Proposition 2.1 of the same paper; that route is not used here.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Rouquier complexes satisfy far commutativity

Statement

For $∣ i - j ∣ > 1$ the swaps $B_{i} \otimes_{R} B_{j} ≅ B_{j} \otimes_{R} B_{i}$ and $R (1) \otimes_{R} R (1) ≅ R (2)$ lift to degree-zero isomorphisms of complexes $F_{i} \otimes_{R} F_{j} ≅ F_{j} \otimes_{R} F_{i}, F_{i} \otimes_{R} F_{j}^{- 1} ≅ F_{j}^{- 1} \otimes_{R} F_{i}, F_{i}^{- 1} \otimes_{R} F_{j}^{- 1} ≅ F_{j}^{- 1} \otimes_{R} F_{i}^{- 1},$ and the same with the two factors exchanged; the signs are the Koszul signs of the total differential, and no grading shift is needed. In particular the corresponding objects of $K^{b} (R^{e} -grmod)$ are isomorphic.

Facts & Assumptions

Given: Indices $i, j$ with $∣ i - j ∣ > 1$ and the generator complexes $F_{i}, F_{j}, F_{i}^{- 1}, F_{j}^{- 1}$ of The positive and negative Rouquier generator complexes.

[F1]

Distant commutativity. There exists a degree-zero isomorphism $B_{i} \otimes_{R} B_{j} ≅ B_{j} \otimes_{R} B_{i}$ of graded $(R, R)$ -bimodules. Compatibility with the generator differentials will be proved below. (Distant Soergel generators commute)

[F2]

Totalization in two factors. For bounded complexes $P, Q$ the signed tensor totalization has degree- $n$ term $⨁_{r + s = n} P^{r} \otimes_{R} Q^{s}$ and differential $d (x \otimes y) = d_{P} (x) \otimes y + (- 1)^{r} x \otimes d_{Q} (y)$ for $x \in P^{r}$ ; internal degrees add, so tensoring with $R (a)$ on either side shifts a bimodule by $(a)$ . (Bounded graded bimodule complexes and signed tensor totalization)

[F3]

The generators use $B_{r} = (R \otimes_{R^{s_{r}}} R) (1)$ , multiplication $ε_{r}$ , and $η_{r} (1) = α_{r} \otimes 1 + 1 \otimes α_{r}$ , where $α_{r} = ε_{r}^{r o o t} (x_{r} - x_{r + 1})$ and $ε_{r}^{r o o t} = (- 1)^{r - 1}$ . (The positive and negative Rouquier generator complexes)

Proof

technique · independent coordinate blocks and the signed flip of complexes

1.1F3algebra

Put $A = Q [x_{i}, x_{i + 1}]$ , $B = Q [x_{j}, x_{j + 1}]$ and let $C$ be the polynomial ring in the remaining coordinates. The disjoint transpositions give $R = A \otimes_{Q} B \otimes_{Q} C$ and $R^{s_{i}} = A^{s_{i}} \otimes B \otimes C$ , $R^{s_{j}} = A \otimes B^{s_{j}} \otimes C$ . For $E = B \otimes C$ , the map $(a \otimes e) \otimes (a^{'} \otimes e^{'}) \mapsto (a \otimes a^{'}) \otimes e e^{'}$ identifies $R \otimes_{R^{s_{i}}} R$ with $(A \otimes_{A^{s_{i}}} A) \otimes E$ ; its inverse sends $(a \otimes a^{'}) \otimes e$ to $(a \otimes 1) \otimes (a^{'} \otimes e)$ . Balancing over $A^{s_{i}}$ and $E$ verifies both maps and their inverse identities. Exchanging the blocks gives the analogous identification for $j$ . Under these maps multiplication and root insertion act only in their own block. Thus $F_{i}^{ϵ} = P_{i}^{ϵ} \otimes B \otimes C$ and $F_{j}^{δ} = A \otimes P_{j}^{δ} \otimes C$ as complexes, with the shifts inherited from the generator definitions.

2.1F2F3step 1.1algebra

In each bidegree the map $(p \otimes b \otimes c) \otimes (a \otimes q \otimes d) \mapsto p a \otimes b q \otimes c d$ identifies the balanced product with $P_{i}^{ϵ} \otimes_{Q} P_{j}^{δ} \otimes_{Q} C$ ; the inverse sends $p \otimes q \otimes c$ to $(p \otimes 1 \otimes 1) \otimes (1 \otimes q \otimes c)$ . The $A, B, C$ balancing relations verify these inverse identities and preservation of both outer actions. Each differential is a bimodule map acting in its own block, so these identifications intertwine the signed total differentials for every choice of signs, including negative cohomological degrees.

3.1F2step 2.1algebra

On the external tensor product, define the flip of a term of cohomological bidegree $(p, q)$ by $x \otimes y \otimes c \mapsto (- 1)^{p q} y \otimes x \otimes c$ . It preserves both outer actions because their block labels move with the blocks. For the component $d_{P} x \otimes y$ , its image has sign $(- 1)^{(p + 1) q}$ , which equals $(- 1)^{p q + q}$ on the corresponding component of the target differential. For $(- 1)^{p} x \otimes d_{Q} y$ , its image has sign $(- 1)^{p + p (q + 1)} = (- 1)^{p q}$ , again the target sign. It is therefore a chain isomorphism, and its square is the identity.

4.1F1step 2.1step 3.1∎

Transport this flip through the block identifications of step 2.1. It yields $F_{i}^{ϵ} \otimes_{R} F_{j}^{δ} ≅ F_{j}^{δ} \otimes_{R} F_{i}^{ϵ}$ for all signs, with zero internal degree and no shift. Its component on $B_{i} \otimes_{R} B_{j}$ realizes the distant bimodule isomorphism of F1 with the required differential compatibility. This proves every displayed case in the category of complexes and hence in its homotopy category. The map is the independent-block flip, not an arbitrary flip of balanced bimodule tensors.

Remarks

The signs are exactly the Koszul signs of the total differential: the swap of two factors of bidegrees $(r, s)$ carries $(- 1)^{r s}$ , and this is the sign under which the two off-diagonal components of the total differential correspond. This is the trivial ( $m_{s t} = 2$ ) case of Rouquier Proposition 3.2 and the last line of GKS Theorem 3.10; no input beyond the distant commutativity of the Soergel generators is used.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Rouquier complexes satisfy the three-term braid relation

Statement

For $1 \leq i \leq n - 2$ , with $s = s_{i}$ and $t = s_{i + 1}$ , $F_{i} \otimes_{R} F_{i + 1} \otimes_{R} F_{i} ≃ F_{i + 1} \otimes_{R} F_{i} \otimes_{R} F_{i + 1} in K^{b} (R^{e} -grmod),$ with no grading shift. More precisely, expanding the two total complexes gives $8$ -term complexes; using $B_{i} \otimes_{R} B_{i} ≅ B_{i} (1) \oplus B_{i} (- 1)$ and the rank-two decompositions $B_{i} B_{i + 1} B_{i} ≅ B_{i, i + 1, i} \oplus B_{i}$ and $B_{i + 1} B_{i} B_{i + 1} ≅ B_{i, i + 1, i} \oplus B_{i + 1}$ with no shift on any summand, each complex is a direct sum of a contractible summand whose extra bimodule is $B_{i}$ , respectively $B_{i + 1}$ , and which has an invertible differential block, and a surviving complex built from $B_{i, i + 1, i}$ ; cancelling the contractible summands is Gaussian elimination, and the two survivors have the same terms and shifts, with differentials identified by the degreewise sign isomorphism $(1, 1, - 1, 1)$ written below. Thus the displayed homotopy equivalence holds. The chain maps and contractions are written explicitly.

Facts & Assumptions

Given: Adjacent indices $i, i + 1$ with $1 \leq i \leq n - 2$ , the complexes $F_{i}, F_{i + 1}$ of The positive and negative Rouquier generator complexes and the Bott–Samelson products $B_{i} B_{i + 1} B_{i} = B_{i} \otimes_{R} B_{i + 1} \otimes_{R} B_{i}$ , $B_{i + 1} B_{i} B_{i + 1}$ .

[F1]

Rank one. $B_{i} \otimes_{R} B_{i} ≅ B_{i} (1) \oplus B_{i} (- 1)$ and $B_{i + 1} \otimes_{R} B_{i + 1} ≅ B_{i + 1} (1) \oplus B_{i + 1} (- 1)$ , with the summands the middle-slot idempotent images of the decomposition $R = R^{s} \oplus α_{s} R^{s}$ , respectively $R = R^{t} \oplus α_{t} R^{t}$ . (The rank-one Soergel bimodule square splits).

[F2]

Rank two. $B_{i} B_{i + 1} B_{i} ≅ B_{i, i + 1, i} \oplus B_{i}$ and $B_{i + 1} B_{i} B_{i + 1} ≅ B_{i, i + 1, i} \oplus B_{i + 1}$ with no additional shift, where $B_{i, i + 1, i} = R \otimes_{R^{W_{i, i + 1}}} R (3)$ is the rank-two longest bimodule (Rank-two type-A Soergel bimodule decompositions, The rank-two longest type-A Soergel bimodule).

[F3]

Gaussian elimination. An invertible differential block $φ : U \to V$ in a fixed biproduct decomposition of two adjacent terms of a cochain complex can be cancelled: the complex is homotopy equivalent to the reduction obtained by deleting $U, V$ and replacing the differential by its Schur complement, and the deleted part is the contractible two-term complex $[U \overset{φ}{\to} V]$ (Gaussian elimination splits a contractible two-term complex, An invertible cochain differential block and its candidate reduction).

[F4]

Totalization. The signed tensor totalization of bounded complexes is associative up to the canonical degree-zero reassociation and has Koszul differential $d (x \otimes y) = d (x) \otimes y + (- 1)^{p} x \otimes d (y)$ ; a shift on a factor is a shift on the tensor product with the same totalization differential (Bounded graded bimodule complexes and signed tensor totalization).

Proof

technique · two explicit Gaussian eliminations to a symmetric four-term complex

1.1F4algebra

Write $S = B_{s}$ , $T = B_{t}$ , $L = B_{i, i + 1, i}$ , and let $m_{s}, m_{t}$ be multiplication. The total complex $K = F_{s} F_{t} F_{s}$ has terms $S T S$ , $(T S) (1) \oplus (S S) (1) \oplus (S T) (1)$ , $S (2) \oplus T (2) \oplus S (2)$ and $R (3)$ in degrees $0, 1, 2, 3$ . Call the degree-one terms $U, V, W$ and degree-two terms $A, B, C$ , in this order. The tensor signs give $d^{0} = (m_{s} \otimes 1 \otimes 1, 1 \otimes m_{t} \otimes 1, 1 \otimes 1 \otimes m_{s})$ ; $d^{1}$ has blocks $U \to A = - m_{t} \otimes 1$ , $U \to B = - 1 \otimes m_{s}$ , $V \to A = m_{s} \otimes 1$ , $V \to C = - 1 \otimes m_{s}$ , $W \to B = m_{s} \otimes 1$ , $W \to C = 1 \otimes m_{t}$ , and the other blocks zero; $d^{2} = (m_{s}, - m_{t}, m_{s})$ .

1.2F1givenalgebra

Put $x = x_{i}$ , $y = x_{i + 1}$ , $z = x_{i + 2}$ and use the coordinate roots $β_{s} = x - y$ , $β_{t} = y - z$ , distinct from the balanced roots in the generator definition. Set $D_{s} (f) = (f - s (f)) / (2 β_{s})$ , the coefficient of $β_{s}$ in $R = R^{s} \oplus β_{s} R^{s}$ . Define $J (r \otimes r^{'}) = - r \otimes β_{t} \otimes 1 \otimes r^{'} - r \otimes 1 \otimes β_{t} \otimes r^{'}, p (r \otimes f \otimes g \otimes r^{'}) = r D_{s} (f g) \otimes r^{'} .$ The map $p : S T S \to S$ is balanced because $D_{s}$ is $R^{s}$ -linear and the middle multiplication is $R^{t}$ -balanced. For $J : S \to S T S$ , unit insertion into $S S$ is $R^{s}$ -balanced, and insertion of $β_{t} \otimes 1 + 1 \otimes β_{t}$ into the middle $T$ is a bimodule map: this element commutes with $R$ , by $R = R^{t} \oplus β_{t} R^{t}$ and $β_{t}^{2} \in R^{t}$ . Both $J$ and $p$ have internal degree zero. Since $s (β_{t}) = β_{t} + β_{s}$ , one has $D_{s} (β_{t}) = - 1 / 2$ and hence $p J (r \otimes r^{'}) = - 2 r D_{s} (β_{t}) \otimes r^{'} = r \otimes r^{'}$ .

2.1F1F2step 1.2algebra

Define $j : L \to S T S$ by $j (r \otimes r^{'}) = r \otimes 1 \otimes 1 \otimes r^{'}$ ; invariants in $R^{⟨ s, t ⟩}$ slide across all three dividers, so $j$ is balanced and degree zero, and $p j = 0$ . The bimodule $S T S$ is generated by $g_{0} = 1 \otimes 1 \otimes 1 \otimes 1$ and $g_{x} = 1 \otimes x \otimes 1 \otimes 1$ : first expand the second middle slot in the $R^{t}$ -basis ${1, z}$ and slide its invariant coefficients to the first middle slot, then expand that slot in the $R^{s}$ -basis ${1, x}$ and slide its invariant coefficients left; the remaining $z$ in the second middle slot slides right because $z \in R^{s}$ . Let $e = J p$ . With $u = x + y - z$ , balancing gives $p (g_{0}) = 0$ , $p (g_{x}) = \frac{1}{2} (1 \otimes 1)$ and $(1 - e) g_{0} = g_{0}, (1 - e) g_{x} = \frac{1}{2} (u g_{0} + g_{0} u),$ since the two inserted $β_{t}$ tensors sum to $u g_{0} + g_{0} u - 2 g_{x}$ . Thus $\ker p = im (1 - e) = im j$ . By F2 and $p J = 1$ , $\ker p$ and $L$ have equal dimensions in each graded degree; these dimensions are finite because $R$ is a polynomial ring with positive-degree variables. The graded surjection $j : L \to \ker p$ is therefore an isomorphism. This establishes the specific decomposition $S T S = j (L) \oplus J (S)$ without assuming splitting maps from the abstract decomposition.

3.1F1F3step 1.1step 1.2step 2.1

In $V = (S S) (1)$ use the coordinate middle decomposition $V_{+} \oplus V_{-} = S (2) \oplus S$ ; replacing the balanced root by its unit multiple $β_{s}$ changes neither summand. Projection to $V_{-}$ is the coefficient map $r \otimes f \otimes r^{'} \mapsto r D_{s} (f) \otimes r^{'}$ . Its composite with the $V$ component of $d^{0}$ is exactly $p$ , so the block $J (S) \to V_{-}$ is $p J = 1$ and the block $j (L) \to V_{-}$ is zero. Cancel this identity pivot by F3. The surviving degree-zero term is $L$ , and its components into $U, V_{+}, W$ are the outer-unit inclusions with signs $+, +, +$ , since $d^{0} j (r \otimes r^{'})$ inserts a middle $1$ in each of those terms.

4.1F1F3F4step 1.1step 3.1algebra

The next pivot is $V_{+} \to A$ : multiplication on $r \otimes 1 \otimes r^{'}$ sends it to $r \otimes r^{'}$ , so it is the identity $S (2) \to S (2)$ . Its component into $C$ is minus the identity. Cancel this pivot by F3; the Schur complement replaces the $C$ row by the sum of the old $C$ and $A$ rows. The resulting complex $H$ is $L ⟶ (T S) (1) \oplus (S T) (1) ⟶ S (2) \oplus T (2) ⟶ R (3),$ with $d_{H}^{0} = (j_{t s}, j_{s t})$ , where $j_{t s} (r \otimes r^{'}) = r \otimes 1 \otimes r^{'}$ and likewise for $j_{s t}$ , and $d_{H}^{1} = (\begin{array}{cc} - m_{t} \otimes 1 & 1 \otimes m_{t} \\ - 1 \otimes m_{s} & m_{s} \otimes 1 \end{array}), d_{H}^{2} = (m_{s}, - m_{t}) .$ The preceding differential retains its $U, W$ components under the elimination, and the following differential retains its $C, B$ components; these are the displayed formulas. Each map has internal degree zero with the written shifts.

5.1step 1.2step 2.1step 3.1step 4.1algebra

For $K^{'} = F_{t} F_{s} F_{t}$ , repeat steps 1.2–3.1 with $s, t$ exchanged; $D_{t} (β_{s}) = - 1 / 2$ gives the same identity pivot. The complement calculation follows by interchanging $x, z$ , which negates both coordinate roots and therefore leaves $J p$ unchanged. Put its survivors into the same order $L$ , $(T S) (1) \oplus (S T) (1)$ , $S (2) \oplus T (2)$ , $R (3)$ . The resulting $H^{'}$ has $d_{H^{'}}^{0} = d_{H}^{0}$ , $d_{H^{'}}^{1} = - d_{H}^{1}$ and $d_{H^{'}}^{2} = - d_{H}^{2}$ , as follows by exchanging $s, t$ in the matrix of step 4.1 and reordering its two rows and columns. Hence the degreewise maps $(1, 1, - 1, 1)$ form an explicit chain isomorphism $q : H \to H^{'}$ .

6.1F3step 3.1step 4.1step 5.1∎

For either identity pivot write the differential block as $(\begin{array}{cc} a & b \\ c & 1 \end{array})$ . The Gaussian chain isomorphism $T$ to the reduced complex plus the identity pair has components $T^{n} = (\begin{array}{cc} 1 & 0 \\ c & 1 \end{array})$ and $T^{n + 1} = (\begin{array}{cc} 1 & - b \\ 0 & 1 \end{array})$ , and identity elsewhere. Its retraction is $π = pr T$ , inclusion $ι = T^{- 1} in$ and homotopy $h = T^{- 1} k T$ , where $k$ is the identity from the pivot target back to its source and zero elsewhere. For the two eliminations set $i = ι_{1} ι_{2}$ , $p = π_{2} π_{1}$ , $h = h_{1} + ι_{1} h_{2} π_{1}$ , and similarly $i^{'}, p^{'}, h^{'}$ for $K^{'}$ . F3 gives $p i = 1_{H}$ , $1_{K} - i p = d h + h d$ and the primed identities. Thus $Φ = i^{'} q p$ and $Ψ = i q^{- 1} p^{'}$ are explicit homotopy-inverse chain maps. All entries are the displayed neighboring blocks and identity pivots, with internal degree zero, so no grading shift is introduced.

Remarks

The local splitting calculation uses coordinate roots, as in Libedinsky §§4.3–4.4. These roots differ by unit signs from the balanced roots of the generator definition; the positive differentials are multiplication and do not change. The abstract rank-two decomposition is used only for the graded dimension comparison in step 2.1. The specific splitting maps and both identity pivots are verified here.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

The Rouquier complex of a braid word

Definition

The construction. For a signed word $σ = σ_{i_{1}}^{ϵ_{1}} \dots σ_{i_{r}}^{ϵ_{r}}, ϵ_{k} \in {\pm 1}, 1 \leq i_{k} \leq n - 1,$ put $F (σ) : = F_{i_{1}}^{ϵ_{1}} \otimes_{R} \dots \otimes_{R} F_{i_{r}}^{ϵ_{r}},$ the iterated signed tensor totalization of Bounded graded bimodule complexes and signed tensor totalization of the Rouquier generator complexes of The positive and negative Rouquier generator complexes, bracketed left to right; the empty word ( $r = 0$ ) is sent to the unit complex $R$ concentrated in cohomological degree $0$ . This is the Rouquier complex of the word. When $ϵ_{k} = + 1$ the letter contributes the positive complex $F_{i_{k}}$ and when $ϵ_{k} = - 1$ it contributes $F_{i_{k}}^{- 1}$ .

Structure. $F (σ)$ is a bounded cochain complex of graded $(R, R)$ -bimodules with degree-zero differentials, concentrated in cohomological degrees $- r_{-}, \dots, r_{+}$ , where $r_{-}$ and $r_{+}$ count the negative and positive letters: the term $F (σ)^{m}$ is the direct sum of the $(\binom{r}{m + r_{-}})$ tensor products of one term of each factor, since the minimal choice has degree $- r_{-}$ and each upper-term choice adds one. It carries a cohomological degree and an internal grading, and the differential is the Koszul totalization differential. A positive letter contributes its unit term $R (1)$ in cohomological degree $1$ and the internal shift $(1)$ to that term; a negative letter contributes its unit term $R (- 1)$ in cohomological degree $- 1$ and the internal shift $(- 1)$ . Every term is a finite direct sum of finite tensor products of copies of $R (\pm 1)$ and $B_{i}$ , hence is finite free on both sides (Soergel generators and Bott–Samelson products are finite free on both sides), so each term is finite graded projective as a left module and as a right module and $F (σ)$ defines a derived tensor functor by A bounded two-sided projective bimodule complex defines exact derived tensor functors.

Status of the notation. The notation records the chosen word $(σ)$ together with the chosen bracket; it asserts nothing about independence of the word, of the sign normalization or of the bracketing. That the homotopy class of $F (σ)$ depends only on the braid represented by $σ$ is the content of The Rouquier complex is well defined up to canonical homotopy equivalence, whose proof uses the braid relations of the generator complexes and could not be stated before those relations were proved.

Convention for comparison. The library's positive generator $F_{i}$ is Rouquier's $F_{s_{i}} = [A \otimes_{A^{s_{i}}} A \to A]$ with the library external shift $(1)$ on both terms, after giving the polynomial variables degree $2$ : $F_{i} = [(R \otimes_{R^{s_{i}}} R) (1) ⟶ R (1)] .$ Rouquier's negative complex of §3.2.4 becomes $F_{i}^{- 1}$ after doubling internal degrees and shifting the whole complex by $(- 1)$ , up to the unit scalar coming from the root normalization. As recorded in The positive and negative Rouquier generator complexes, the complexes of Rouquier, Gorsky–Kivinen–Simental, Khovanov and Elias–Krasner differ from $F_{i}$ and $F_{i}^{- 1}$ by homological or internal shifts, so any comparison with those sources is read through the dictionary recorded there. The empty word corresponds to the trivial braid and to the identity functor.

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Coherent action of a group on a category

Definition

Setting. Let $G$ be a group and let $C$ be a category (Category, object, morphism, domain, codomain, identity, composition, and hom-collection); write $End (C)$ for the strict monoidal category of endofunctors of $C$ and natural transformations between them, with composition of functors as its product (Covariant functor, identity functor, composite functor, and contravariant functor, Natural isomorphism). Composition of endofunctors is strictly associative and its unit is ${i d}_{C}$ , so no associator constraint of $C$ enters anything below.

Data. A coherent action of $G$ on $C$ consists of:

a functor $F_{g} ⁣ : C \to C$ for every $g \in G$ , with $F_{1} = {i d}_{C}$ ;
for all $f, g \in G$ a chosen natural isomorphism $μ_{f, g} ⁣ : F_{f} F_{g} \Rightarrow F_{f g}$ ;
a chosen natural isomorphism $u ⁣ : F_{1} \Rightarrow {i d}_{C}$ , i.e. (since $F_{1} = {i d}_{C}$ ) a natural automorphism of the identity functor.

Pentagon. For all $f, g, h \in G$ the two composites $F_{f} F_{g} F_{h} \Rightarrow F_{f g h}$ agree: $μ_{f g, h} \circ (μ_{f, g} F_{h}) = μ_{f, g h} \circ (F_{f} μ_{g, h}),$ where $μ_{f, g} F_{h}$ is the whiskered natural transformation with components $μ_{f, g, F_{h} X}$ and $F_{f} μ_{g, h}$ the one with components $F_{f} (μ_{g, h, X})$ .

Unit triangles. For all $f, g \in G$ , $μ_{f, 1} = F_{f} u, μ_{1, g} = u F_{g} .$ Here $F_{f} u$ has components $F_{f} (u_{X}) ⁣ : F_{f} F_{1} X \to F_{f} X$ , while $u F_{g}$ has components $u_{F_{g} X} ⁣ : F_{1} F_{g} X \to F_{g} X$ . Since $F_{1}$ is the identity functor, these are natural transformations $F_{f} \Rightarrow F_{f}$ and $F_{g} \Rightarrow F_{g}$ , respectively. They need not agree: for a general natural automorphism $u$ of ${i d}_{C}$ , naturality does not imply $u F_{g} = F_{g} u$ , whose components are $u_{F_{g} X}$ and $F_{g} (u_{X})$ . The two equalities are the left and right unit axioms for a monoidal functor $G \to End (C)$ with unit constraint $u^{- 1} ⁣ : {i d}_{C} \Rightarrow F_{1}$ and the discrete monoidal structure on $G$ .

Relation to a weak action. The underlying functors form a weak action of $G$ on $C$ in the sense of Weak action of a group on a category: the chosen $μ_{f, g}$ in particular exhibit isomorphisms $F_{f g} ≅ F_{f} F_{g}$ for all $f, g$ . The converse does not hold: a weak action records no chosen compositors and imposes no pentagon, and a coherent action is precisely a weak action equipped with chosen compositors and a chosen unit satisfying the pentagon and the two unit triangles above. The pentagon is part of the data; it is not a formal consequence of the existence of isomorphisms $F_{f g} ≅ F_{f} F_{g}$ .

Rouquier's instance. In the application of this page, $C = K^{b} (R -grmod)$ . For $g \neq 1$ , the functor $F_{g}$ is left tensoring by the invertible object $G_{g}$ of Rouquier's rigidification; set $F_{1} = {i d}_{C}$ and use the canonical unit identification $R \otimes_{R} - ≅ {i d}_{C}$ for the empty-word object $G_{1} = R$ . The isomorphisms $μ_{f, g}$ are induced by the unique maps $m_{f, g} : G_{f} \otimes_{R} G_{g} \to G_{f g}$ compatible with the canonical comparisons $c_{t, u}$ of Canonical comparisons between standard graph tensor products in the derived category, with the canonical tensor unit identifications when an index or product is $1$ . The map $m_{1} : G_{1} \to R$ corresponds, under the unit identification, to $u = {i d}_{C}$ . Rouquier's construction produces the pentagon and unit triangles by lifting associative graph multiplication together with the additive internal shifts of signed words: in this normalization the derived models $R_{π (v)} (- e (v))$ are the pullback of the strict $W \times Z$ action along $v \mapsto (π (v), e (v))$ , where $π$ is the permutation projection and $e$ the signed word exponent; the lifts are fixed by localization isomorphisms and normalized uniqueness. The definitions make sense for an arbitrary group and category, and no Hecke algebra enters.

Degenerate cases. If $G$ is the trivial group then $F_{1} = i d$ and the data reduce to natural automorphisms $μ_{1, 1}$ and $u$ of ${i d}_{C}$ . The unit axioms give $μ_{1, 1} = u$ ; with this value the pentagon is automatic. Thus a coherent action of the trivial group may still have any invertible natural automorphism $u$ as its unit; taking $u = i d$ gives the identity example. If $C$ is the one-object category attached to a monoid then the definition reduces to the usual coherence data on that monoid. For a $k$ -linear category, an invertible scalar multiple of the identity is a natural automorphism, so $u$ need not be the identity. Once the compositors are fixed, however, the unit triangle at $f = g = 1$ forces $u = μ_{1, 1}$ .

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Canonical comparisons between standard graph tensor products

Definition

Setting. Keep $R = Q [x_{1}, \dots, x_{n}]$ with $\deg x_{i} = 2$ and the standard graph bimodules $R_{w}$ of Standard graph bimodules, support filtrations and characters: for $w \in S_{n}$ the graded $(R, R)$ -bimodule $R_{w}$ equals $R$ as a graded $Q$ -vector space with $f \cdot x = f x$ and $x \cdot g = w (g) x$ , where $w (g) (λ) = g (w^{- 1} λ)$ , generated by the element $1$ of degree $0$ . Recall the degree-zero isomorphism of graded $(R, R)$ -bimodules $μ_{w, v} : R_{w} \otimes_{R} R_{v} ⟶ R_{w v}, a \otimes b ⟼ a w (b),$ for $w, v \in S_{n}$ , with inverse $1 \mapsto 1 \otimes 1$ (the composition remark of that item, where the assignment is checked to be balanced, $R$ -linear on both sides and surjective between free rank-one left $R$ -modules). Here the internal shifts are those of Associative graded algebras, bimodules, and internal shifts: $μ$ is homogeneous of degree zero.

Word comparisons. Let $t = (x_{1}, \dots, x_{r})$ and $u = (y_{1}, \dots, y_{s})$ be finite words in $S_{n}$ (the empty word allowed) with the same product $x_{1} \dots x_{r} = y_{1} \dots y_{s} = w .$ Fix the left-to-right iteration of $μ$ and write $μ_{t} : R_{x_{1}} \otimes_{R} \dots \otimes_{R} R_{x_{r}} ⟶ R_{w}, a_{1} \otimes \dots \otimes a_{r} ⟼ a_{1} x_{1} (a_{2}) (x_{1} x_{2}) (a_{3}) \dots (x_{1} \dots x_{r - 1}) (a_{r}),$ the composite of the binary multiplications $μ_{x_{1}, x_{2}}, μ_{x_{1} x_{2}, x_{3}}, \dots$ ; for $r = 0$ the source is the unit bimodule $R = R_{e}$ and $μ_{t} = {i d}_{R}$ , and for $r = 1$ it is ${i d}_{R_{x_{1}}}$ . Each $μ_{t}$ is a degree-zero isomorphism of graded $(R, R)$ -bimodules with inverse $1 \mapsto 1 \otimes 1 \otimes \dots \otimes 1$ . The canonical comparison between the two word tensors is $c_{t, u} : = μ_{u}^{- 1} \circ μ_{t} : R_{x_{1}} \otimes_{R} \dots \otimes_{R} R_{x_{r}} ⟶ R_{y_{1}} \otimes_{R} \dots \otimes_{R} R_{y_{s}} .$

Basic properties. The comparisons form a transitive system: $c_{t, t} = i d$ for every word $t$ , and for three words $t, u, v$ with the same product $c_{u, v} \circ c_{t, u} = c_{t, v}$ , because $μ_{v}^{- 1} μ_{u} μ_{u}^{- 1} μ_{t} = μ_{v}^{- 1} μ_{t}$ . In particular $c_{t, u}$ is inverse to $c_{u, t}$ . When one of the words has length one the comparison is the corresponding display: for $r = 1$ , $t = (x_{1})$ , one has $c_{t, u} = μ_{u}^{- 1}$ ; for $s = 1$ , $c_{t, u} = μ_{t}$ . The identifications are associative in the sense that for a threefold product the two iterated comparisons built from the binary $μ$ 's coincide, this being the equality of the explicit formula above under any rebracketing.

Homogeneity of the Hom spaces. For $w, w^{'} \in S_{n}$ , ${\underset{}{Hom}}_{R^{e}} (R_{w}, R_{w^{'}}) = {\begin{cases} ≅ R, & w = w^{'}, \\ 0, & w \neq w^{'}, \end{cases}$ where the underlined Hom is the direct sum over all homogeneous internal degrees. For $w = w^{'}$ the identification sends a homogeneous map $φ$ to $φ (1) \in R$ and the internal degree of $φ$ is the degree of $φ (1)$ ; hence the internal degree-zero part is one-dimensional over $Q$ , spanned by ${i d}_{R_{w}}$ . The vanishing for $w \neq w^{'}$ is the statement that the supports $G r (w) \neq G r (w^{'})$ are distinct graphs, and the description for $w = w^{'}$ is the map remark of Standard graph bimodules, support filtrations and characters.

Small cases. If $r = s = 0$ then $w = e$ and $c_{t, u} = {i d}_{R}$ ; if $t$ and $u$ are both words for $w$ , the comparison $c_{t, u}$ is the isomorphism used to compare the two word models of the same standard bimodule $R_{w}$ . No choice is made: $μ_{t}$ is the unique degree-zero bimodule isomorphism with $μ_{t} (1 \otimes \dots \otimes 1) = 1$ , and $c_{t, u}$ is determined by $t$ and $u$ alone.

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Rouquier generator complexes have canonical derived graph models

Statement

For every $1 \leq i \leq n - 1$ the generator complexes of The positive and negative Rouquier generator complexes are canonically isomorphic in the bounded derived category to shifts of standard graph bimodules: $F_{i} ≅ R_{s_{i}} (- 1), F_{i}^{- 1} ≅ R_{s_{i}} (1) in D^{b} (R^{e} -grmod),$ the isomorphisms being induced by the quasi-isomorphisms $f_{i}$ below. More generally, for a signed word $σ = σ_{i_{1}}^{ϵ_{1}} \dots σ_{i_{r}}^{ϵ_{r}}$ with product $w = s_{i_{1}}^{ϵ_{1}} \dots s_{i_{r}}^{ϵ_{r}}$ and exponent sum $e (σ) = \sum_{k} ϵ_{k}$ , $F (σ) ≅ R_{w} (- e (σ)) in D^{b} (R^{e} -grmod),$ where $F (σ)$ is the iterated signed tensor totalization $F_{i_{1}}^{ϵ_{1}} \otimes_{R} \dots \otimes_{R} F_{i_{r}}^{ϵ_{r}}$ and $R_{w}$ is the standard graph bimodule. All isomorphisms are obtained from the comparison system $c_{t, u}$ of Canonical comparisons between standard graph tensor products and do not depend on the chosen words beyond their permutation product and exponent sum, with the displayed graph models and fixed generator maps understood.

Facts & Assumptions

Given: A simple reflection $s_{i}$ , the bimodule $B_{i}$ with generators $u = 1 \otimes 1$ (degree $- 1$ ) and $w_{0} = 1 \otimes δ_{i}$ , $δ_{i} = α_{i} / 2$ , the element $ρ_{-} = δ_{i} u - w_{0}$ , and the complexes $F_{i}, F_{i}^{- 1}$ of The positive and negative Rouquier generator complexes.

[F1]

The two exact sequences. The multiplication $ε_{i} : B_{i} \to R (1)$ is a surjective degree-zero bimodule map with $ε_{i} (u) = 1$ and $ε_{i} (w_{0}) = δ_{i}$ , and $\ker ε_{i} = R ρ_{-}$ with $R ρ_{-} ≅ R_{s_{i}} (- 1)$ a graded sub-bimodule generated in degree $1$ ; the map $η_{i} : R (- 1) \to B_{i}$ is injective and the quotient $B_{i} / η_{i} (R (- 1)) = B_{i} / R ρ_{+}$ is generated by the image of $u$ with right action $u \cdot g \equiv s_{i} (g) u$ , so that it is isomorphic to $R_{s_{i}} (1)$ ; here $ρ_{+} = δ_{i} u + w_{0}$ satisfies $ρ_{+} \cdot g = g ρ_{+}$ and $ρ_{-} \cdot g = s_{i} (g) ρ_{-}$ (Standard graph bimodules, support filtrations and characters, The positive and negative Rouquier generator complexes).

[F2]

Two canonical chain maps. The assignment $1 \mapsto 1 \otimes α_{i} - α_{i} \otimes 1 = - 2 ρ_{-}$ defines a degree-zero bimodule map $f_{i} : R_{s_{i}} (- 1) \to F_{i}$ , concentrated in cohomological degree $0$ ; it is a chain map because $ε_{i} (ρ_{-}) = 0$ . The assignment $1 \otimes 1 \mapsto 1$ defines the twisted multiplication $ψ : B_{i} \to R_{s_{i}} (1)$ , $ψ (r \otimes r^{'}) = r s_{i} (r^{'})$ , a degree-zero bimodule map with $ψ (η_{i} (1)) = 0$ ; on the quotient $B_{i} / η_{i} (R (- 1))$ it induces the identification with $R_{s_{i}} (1)$ and defines a chain map $g_{i} : F_{i}^{- 1} \to R_{s_{i}} (1)$ , concentrated in cohomological degree $0$ . (Standard graph bimodules, support filtrations and characters, Canonical comparisons between standard graph tensor products)

[F3]

Cohomology of the generator complexes. $H^{0} (F_{i}) = \ker ε_{i} = R ρ_{-} ≅ R_{s_{i}} (- 1)$ , $H^{1} (F_{i}) = coker ε_{i} = 0$ , and $H^{- 1} (F_{i}^{- 1}) = \ker η_{i} = 0$ , $H^{0} (F_{i}^{- 1}) = B_{i} / η_{i} (R (- 1)) ≅ R_{s_{i}} (1)$ ; both complexes are otherwise concentrated in the displayed degrees. [F1]

[F4]

Localization. The localization functor $K^{b} (R^{e} -grmod) \to D^{b} (R^{e} -grmod)$ sends quasi-isomorphisms to isomorphisms, and tensor totalization with a bounded complex of bimodules flat on the tensoring side preserves quasi-isomorphisms. Here all generator complexes and graph models are flat on both sides: the former have finite-free terms and the latter are twisted rank-one regular modules. Consequently the tensor comparisons used below are compatible with localization (The localization functor sends quasi isomorphisms to isomorphisms, Derived category of an abelian category, Bounded above flat tensor complexes preserve quasi isomorphisms).

[F5]

Multiplication of graph bimodules. The balanced assignment $R_{x} \otimes_{R} R_{y} \to R_{x y}$ , $a \otimes b \mapsto a x (b)$ , is a degree-zero isomorphism of graded bimodules with inverse $1 \mapsto 1 \otimes 1$ ; shifts satisfy $M (a) \otimes_{R} N (b) ≅ (M \otimes_{R} N) (a + b)$ (Canonical comparisons between standard graph tensor products, Standard graph bimodules, support filtrations and characters).

Proof

technique · direct

1.1F1F2

The map $f_{i}$ of [F2] is a chain map between the complexes $R_{s_{i}} (- 1)$ concentrated in degree $0$ and $F_{i}$ in degrees $0, 1$ : the only condition is that the composite of $f_{i}$ with the differential $ε_{i}$ vanishes, which holds since $ε_{i} (ρ_{-}) = 0$ by [F1]. It is a bimodule map: for $g \in R$ one has $f_{i} (1 \cdot g) = f_{i} (s_{i} (g)) = - 2 s_{i} (g) ρ_{-}$ and $f_{i} (1) \cdot g = - 2 ρ_{-} \cdot g = - 2 s_{i} (g) ρ_{-}$ by [F1], and $R$ -linearity on the left is clear.

2.1F1F2F3step 1.1

Comparing with [F3], $f_{i}$ induces the identity identification $H^{0} (R_{s_{i}} (- 1)) = R_{s_{i}} (- 1) \to R ρ_{-}$ (up to the unit $- 2$ ) and there are no other cohomology groups on either side; hence $f_{i}$ is a quasi-isomorphism, and so is $g_{i}$ between $F_{i}^{- 1}$ and $R_{s_{i}} (1)$ .

3.1F4step 2.1

By [F4] the quasi-isomorphisms $f_{i}, g_{i}$ become isomorphisms in $D^{b}$ , giving $F_{i} ≅ R_{s_{i}} (- 1)$ and $F_{i}^{- 1} ≅ R_{s_{i}} (1)$ .

4.1F4F5step 3.1

For a signed word, tensoring the isomorphisms of step 3.1 over $R$ and using that the totalization of bimodule complexes is compatible with localization in each variable [F4], together with the shift computation $M (a) \otimes_{R} N (b) ≅ (M \otimes_{R} N) (a + b)$ and the multiplication isomorphism of [F5] iterated over the word, gives $F (σ) ≅ R_{s_{i_{1}}^{ϵ_{1}}} \otimes_{R} \dots \otimes_{R} R_{s_{i_{r}}^{ϵ_{r}}} (- \sum_{k} ϵ_{k}) ≅ R_{w} (- e (σ))$ in $D^{b}$ .

5.1F2F4F5step 4.1∎

With the generator maps $f_{i}, g_{i}$ fixed, tensor their derived isomorphisms (using $f_{i}^{- 1}$ for a positive letter and $g_{i}$ for a negative letter), then compose with the graph multiplication $μ_{t}$ of [F5]. This specifies the word map to $R_{w} (- e (σ))$ with its full shift retained; no replacement of $e (σ)$ by the length of a reduced permutation word is made. Associativity of graph multiplication makes this construction compatible with the canonical rebracketings. Comparisons between different word models are obtained by composing their specified isomorphisms through this common target when their permutation and exponent agree.

Remarks

This is Rouquier's §3.2.1 and §3.2.4 identification of the generators with the standard graph bimodules, in the library normalization: the unit scalar $- 2$ and the shifts $(\mp 1)$ are the exact dictionary entries, so no unit factors are dropped. The word statement is proved here because the later uniqueness and decategorification arguments use the identification $F (σ) ≅ R_{w} (- e (σ))$ in $D^{b}$ as a consequence of the comparison system. The statement is a derived-category statement; it does not assert that $f_{i}$ is a homotopy equivalence, and no homotopy-category identification is made here.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Derived comparisons give unique normalized homotopy maps

Statement

Let $t$ and $u$ be signed words with the same product $w \in B_{n}$ , and let $F (t), F (u)$ be the corresponding word complexes of The Rouquier complex of a braid word. Then:

${Hom}_{K^{b}} (F (t), F (u)) = Q \cdot [γ_{t, u}]$ is one-dimensional over $Q$ , its generator being represented by a morphism of internal degree zero;
the canonical localization map ${Hom}_{K^{b}} (F (t), F (u)) ⟶ {Hom}_{D^{b}} (F (t), F (u))$ is an isomorphism of one-dimensional $Q$ -vector spaces;
the comparison element of Canonical comparisons between standard graph tensor products read through the derived graph models (Rouquier generator complexes have canonical derived graph models) is a nonzero element of the one-dimensional ${Hom}_{D^{b}} (F (t), F (u))$ , and there is a unique element $γ_{t, u} \in {Hom}_{K^{b}} (F (t), F (u))$ mapping to it. In particular $γ_{t, u}$ is a homotopy equivalence with $γ_{u, t} γ_{t, u} = i d$ and $γ_{t, u} γ_{u, t} = i d$ , and its class is the unique normalized comparison between the two words.

Facts & Assumptions

Given: Signed words $t, u$ with product $w$ , the word complexes $F (t), F (u)$ of The Rouquier complex of a braid word, and the generator relations of Opposite Rouquier generator complexes are homotopy inverse, Rouquier complexes satisfy far commutativity, Rouquier complexes satisfy the three-term braid relation.

[F1]

Invertibility of word complexes. Every word complex $F (t)$ is invertible in $K^{b} (R^{e} -grmod)$ : $F (t) \otimes_{R} F (t^{- 1}) ≃ R ≃ F (t^{- 1}) \otimes_{R} F (t)$ , where $t^{- 1}$ is the reversed word with inverted signs; this follows from the generator relations by induction on the length of the word, tensoring the identities $F_{i} \otimes_{R} F_{i}^{- 1} ≃ R$ , $F_{i} \otimes_{R} F_{j} ≅ F_{j} \otimes_{R} F_{i}$ for distant $i, j$ and the three-term relation. By the Artin presentation The braid group by Artin presentation, equal braid words differ by finitely many relation replacements and inverse-pair insertions or deletions: their quotient in the free group is a finite product of conjugates of relators. Tensoring the generator equivalences in those word contexts therefore compares any two words for the same braid. The same relations hold after localization, with tensoring by these two-sided finite-free complexes computed by ordinary totalization. (Opposite Rouquier generator complexes are homotopy inverse, Rouquier complexes satisfy far commutativity, Rouquier complexes satisfy the three-term braid relation, The Rouquier complex of a braid word)

[F2]

The unit and its endomorphisms. A degree-zero bimodule map $R \to R$ is multiplication by its value at $1$ , which must lie in $R_{0} = Q$ . The unit complexes have no possible nonzero chain homotopies, so ${Hom}_{K^{b}} (R, R) = Q \cdot i d$ . Since both objects are modules in degree zero, their degree-zero derived-category Hom is the ordinary module Hom, giving ${Hom}_{D^{b}} (R, R) = Q \cdot i d$ as well: apply the boundary Hom formula of The canonical pair is a t structure with $a = b = 0$ to graded $R^{e}$ -modules, so both cohomological and internal degrees are zero (The homotopy category of chain complexes, Derived category of an abelian category, Standard graph bimodules, support filtrations and characters). For $Z ≃ R$ , an isomorphism in the homotopy category transports ${Hom}_{K^{b}} (R, Z)$ to ${Hom}_{K^{b}} (R, R)$ ; it does not assert a generic identification with all of $H^{0} (Z)_{0}$ .

[F3]

Tensoring with an invertible object. Let $C$ be a monoidal category and let $X \in C$ admit a two-sided inverse $X^{- 1}$ : isomorphisms $X \otimes X^{- 1} \to 1$ and $X^{- 1} \otimes X \to 1$ . Using the associativity and unit isomorphisms of $C$ , these exhibit natural isomorphisms $(- \otimes X^{- 1}) (- \otimes X) ≅ {i d}_{C}$ and $(- \otimes X) (- \otimes X^{- 1}) ≅ {i d}_{C}$ ; hence $- \otimes X$ is an equivalence of categories with quasi-inverse $- \otimes X^{- 1}$ in the sense of Equivalence, quasi-inverse, and adjoint equivalence of categories. By Every equivalence of categories can be equipped as an adjoint equivalence the pair can be equipped as an adjoint equivalence, and the adjunction then gives, by An adjoint equivalence is an adjunction whose unit and counit are natural isomorphisms together with The unit-counit, hom-set, unit-universal, and counit-universal encodings of an adjunction are equivalent, a natural bijection $Hom (U \otimes X, W) ≅ Hom (U, W \otimes X^{- 1}) .$ In $K^{b} (R^{e} -grmod)$ and $D^{b} (R^{e} -grmod)$ the associativity and unit isomorphisms are those of Bounded bimodule tensor is associative, unital, and compatible with cones and their images under localization, so the bijection is available in both categories.

[F4]

Derived graph models. $F (t) ≅ R_{π (w)} (- e (t))$ and $F (u) ≅ R_{π (w)} (- e (u))$ in $D^{b} (R^{e} -grmod)$ , and the comparison $c_{t, u}$ of the two words induces an isomorphism of these models whose class in ${Hom}_{D^{b}}$ is nonzero; since the two Artin relations have equal exponent sums on both sides and inverse pairs have exponent zero, the exponent is invariant on braid words and $e (t) = e (u)$ and the two graph models coincide. (Rouquier generator complexes have canonical derived graph models, Canonical comparisons between standard graph tensor products)

Proof

technique · direct

1.1F1F3

By [F1] the object $F (t)$ is invertible with inverse $F (t^{- 1})$ , so by [F3] the functor $- \otimes_{R} F (t)$ is an equivalence with quasi-inverse $- \otimes_{R} F (t^{- 1})$ and the adjunction gives a natural bijection; applied with $U = R$ and $W = F (u)$ it identifies ${Hom}_{K^{b}} (F (t), F (u))$ with ${Hom}_{K^{b}} (R, F (u) \otimes_{R} F (t^{- 1}))$ . The same argument applies in $D^{b}$ with the derived tensor product.

1.2F1

The complex $Z : = F (u) \otimes_{R} F (t^{- 1})$ is a word complex for the word $u t^{- 1}$ , which represents the trivial braid because $t$ and $u$ represent the same element; by the relations of [F1] it is homotopy equivalent to the unit complex $R$ , and likewise isomorphic to $R$ in $D^{b}$ .

2.1F1F2step 1.1step 1.2

By step 1.2 choose a homotopy equivalence $e : Z \to R$ and its homotopy inverse. Composition with $e$ gives a vector-space isomorphism ${Hom}_{K^{b}} (R, Z) \to {Hom}_{K^{b}} (R, R) = Q$ by [F2]. Combining with step 1.1 proves that ${Hom}_{K^{b}} (F (t), F (u))$ is one-dimensional in internal degree zero.

3.1F2F3step 1.1step 1.2step 2.1

Localizing the equivalence $e$ and its inverse gives the same Hom transport in $D^{b}$ . Together with the localized tensor equivalences of step 1.1, this identifies the target Hom with ${Hom}_{D^{b}} (R, R) = Q$ . The localization square commutes with these transports, and its map on $Hom (R, R)$ sends the identity to the identity. It is therefore an isomorphism; hence so is the localization map on $Hom (F (t), F (u))$ .

4.1F4step 3.1∎

By [F4] the comparison element of the graph models is a nonzero element of the one-dimensional ${Hom}_{D^{b}} (F (t), F (u))$ computed in step 3.1, so it has a unique preimage $γ_{t, u}$ under the localization isomorphism. Applying step 3.1 also to the pairs $(u, t)$ and $(t, t)$ shows that the localization maps ${Hom}_{K^{b}} (F (u), F (t)) \to {Hom}_{D^{b}} (F (u), F (t))$ and ${Hom}_{K^{b}} (F (t), F (t)) \to {Hom}_{D^{b}} (F (t), F (t))$ are isomorphisms; since the comparisons satisfy $c_{u, t} c_{t, u} = c_{t, t} = i d$ by [F4], the unique preimages satisfy $γ_{u, t} γ_{t, u} = i d$ , and symmetrically $γ_{t, u} γ_{u, t} = i d$ . Hence $γ_{t, u}$ is a homotopy equivalence and its class is the unique normalized comparison.

Remarks

The argument is Rouquier's §3.3.1: the invertibility of the word complexes makes $- \otimes_{R} F (t)$ an equivalence and hence $Hom (F (t), F (u)) ≅ Hom (R, F (u) \otimes_{R} F (t^{- 1}))$ one-dimensional, and the localization isomorphism transfers the canonical comparison from $D^{b}$ to a unique homotopy class. The degree-zero requirement is essential: the graded endomorphism object of the unit is the polynomial ring $R$ , not $Q$ , and only its internal-degree-zero part is used. The map $γ_{t, u}$ is normalized by the derived condition of matching $c_{t, u}$ through the graph models, and this normalization pins it down uniquely by step 4.1. No choice principle is needed: the invertibility data are fixed by the generator relations, the equivalence-to-adjunction conversion is constructive, and $γ_{t, u}$ is the unique preimage of $c_{t, u}$ .

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Normalized comparison isomorphisms are transitive

Statement

Let $t, u, w$ be signed words representing the same braid. Then $γ_{u, w} \circ γ_{t, u} = γ_{t, w} in {Hom}_{K^{b}} (F (t), F (w)) = Q \cdot [γ_{t, w}] .$ In particular $γ_{t, t} = i d$ and $γ_{u, t} γ_{t, u} = i d$ , so the maps $γ$ form a transitive system of homotopy equivalences between the word complexes of a fixed braid; consequently the multiplication comparisons of the next theorem are well defined on chosen representatives.

Facts & Assumptions

Given: Signed words $t, u, w$ with the same product, the word complexes $F (t), F (u), F (w)$ of The Rouquier complex of a braid word, and the normalized maps $γ_{t, u}, γ_{u, w}, γ_{t, w}$ of Derived comparisons give unique normalized homotopy maps.

[F1]

Uniqueness. For words $a, b$ with the same product, ${Hom}_{K^{b}} (F (a), F (b))$ is one-dimensional in internal degree $0$ , the localization map to ${Hom}_{D^{b}}$ is an isomorphism, and $γ_{a, b}$ is the unique homotopy class whose derived image is the comparison $c_{a, b}$ of the graph models. (Derived comparisons give unique normalized homotopy maps)

[F2]

Transitivity of the comparisons. The derived comparisons satisfy $c_{u, w} c_{t, u} = c_{t, w}$ and $c_{t, t} = i d$ ; they are the multiplication isomorphisms of the words through the standard graph bimodules. (Canonical comparisons between standard graph tensor products)

Proof

technique · direct

1.1F1

The composite $γ_{u, w} \circ γ_{t, u}$ is an element of ${Hom}_{K^{b}} (F (t), F (w))$ , which by [F1] is one-dimensional in internal degree $0$ ; its derived image is $c_{u, w} c_{t, u}$ because localization is a functor on the homotopy categories in which the $γ$ become isomorphisms.

2.1F1F2step 1.1

By [F2] $c_{u, w} c_{t, u} = c_{t, w}$ , which is the derived image of $γ_{t, w}$ by [F1]; two elements of the one-dimensional space with the same nonzero derived image coincide, so $γ_{u, w} γ_{t, u} = γ_{t, w}$ .

3.1F1F2step 2.1∎

Taking $u = t = w$ and using $c_{t, t} = i d$ gives $γ_{t, t} = i d$ by the same uniqueness argument; then $γ_{t, u}$ and $γ_{u, t}$ are mutually inverse homotopy equivalences because both composites equal the corresponding identity maps, and the identity is the unique degree-zero endomorphism class whose derived image is the normalized identity comparison.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Rouquier complexes form a coherent braid group action

Statement

For every braid $v \in B_{n}$ choose a signed word $t (v)$ representing it, taking $t (1)$ to be the empty word, and put $G_{v} : = F (t (v))$ , with $G_{1} : = R$ . For the action on $K^{b} (R -grmod)$ , set $F_{1} = Id$ and, for $v \neq 1$ , set $F_{v} : = G_{v} \otimes_{R} -$ . For $v, w \in B_{n}$ let $m_{v, w} ⁣ : G_{v} \otimes_{R} G_{w} ⟶ G_{v w}$ be the unique homotopy class whose derived image is the graph-multiplication comparison, transported through the associativity and unit isomorphisms of Bounded bimodule tensor is associative, unital, and compatible with cones; equivalently $m_{v, w}$ is $γ_{t (v) t (w), t (v w)}$ in the normalization of Derived comparisons give unique normalized homotopy maps. Let $m_{1} ⁣ : G_{1} \to R$ be the identity of the unit complex. For $v, w \neq 1$ , the compositor $μ_{v, w} : F_{v} F_{w} \Rightarrow F_{v w}$ is induced by associating $G_{v} \otimes_{R} (G_{w} \otimes_{R} -)$ to $(G_{v} \otimes_{R} G_{w}) \otimes_{R} -$ and then applying $m_{v, w} \otimes_{R} -$ , followed by the left-unit identification $R \otimes_{R} - ≅ Id$ when $v w = 1$ . If either index is $1$ , use the canonical tensor unit identifications, so the compositor is the identity after those identifications; set $u : F_{1} \Rightarrow Id$ to the identity. Then $(F_{v}, μ_{v, w}, u)$ is a coherent action of $B_{n}$ on $K^{b} (R -grmod)$ in the sense of Coherent action of a group on a category: the functors are the exact left tensor functors $G_{v} \otimes_{R} -$ for $v \neq 1$ , with the identity functor at $1$ , and both composites of every pentagon have the same derived image, namely the same associative graph multiplication, so the pentagon commutes by uniqueness in degree $0$ ; both unit triangles hold by the canonical tensor unit identifications. Equivalently, $v \mapsto (F_{v}, μ_{v, w}, u)$ is a monoidal functor from the discrete strict monoidal category $(B_{n}, \cdot)$ to the strict monoidal category of endofunctors of $K^{b} (R -grmod)$ . The construction uses one chosen word per braid and no other choice; no choice principle is needed.

Facts & Assumptions

Given: A choice $v \mapsto t (v)$ of signed word for each braid $v \in B_{n}$ , the word complexes $G_{v} = F (t (v))$ , and the maps $γ$ of Derived comparisons give unique normalized homotopy maps.

[F1]

Well-definedness of the $m_{v, w}$ . For any two choices of words for the same braid the normalized maps agree up to the transitive system, and for the concatenated words $t (v) t (w)$ and $t (v w)$ the class $γ_{t (v) t (w), t (v w)}$ is the unique homotopy class with the prescribed derived image; hence $m_{v, w}$ is independent of the auxiliary choices of the words used to define the concatenation, by transitivity. (Normalized comparison isomorphisms are transitive, Derived comparisons give unique normalized homotopy maps)

[F2]

The relations. The generator complexes satisfy $F_{i} \otimes_{R} F_{i}^{- 1} ≃ R$ , the three-term relation and far commutativity, so the word complex attached to any two words for the same braid is independent of the word up to the canonical comparisons. (Opposite Rouquier generator complexes are homotopy inverse, Rouquier complexes satisfy far commutativity, Rouquier complexes satisfy the three-term braid relation)

[F3]

Associativity and units of the tensor. The balanced tensor totalization is associative and unital up to canonical chain isomorphisms satisfying the pentagon and unit triangles, and compatible with cones. (Bounded bimodule tensor is associative, unital, and compatible with cones)

[F4]

The model of a coherent action. A coherent action consists of functors $F_{g}$ with $F_{1} = i d$ , chosen compositors $μ_{f, g} : F_{f} F_{g} \Rightarrow F_{f g}$ and a unit $u$ satisfying the pentagon and the two unit triangles. (Coherent action of a group on a category)

Proof

technique · direct

1.1F1F2F3

For $v, w \in B_{n}$ the composite $G_{v} \otimes_{R} G_{w} = F (t (v)) \otimes_{R} F (t (w))$ is a word complex for the concatenated word $t (v) t (w)$ , which represents $v w$ ; by [F2] it is canonically compared to $G_{v w} = F (t (v w))$ , so the class $m_{v, w}$ of the statement exists as the normalized comparison and is a homotopy equivalence. If $v, w \neq 1$ , associativity identifies $F_{v} F_{w}$ with $(G_{v} \otimes_{R} G_{w}) \otimes_{R} -$ ; tensoring $m_{v, w}$ with the input complex gives the compositor, followed by the left-unit identification when $v w = 1$ . If an index is $1$ , the canonical unit identification gives the identity compositor.

2.1F1F2step 1.1

The maps $m_{v, w}$ are compatible with replacing the representatives: if $a_{v} : G_{v} \to {\tilde{G}}_{v}$ are their normalized comparisons, then $a_{v w} m_{v, w} = {\tilde{m}}_{v, w} (a_{v} \otimes a_{w})$ , after canonical reassociation. Both sides have the same derived graph multiplication, so normalized uniqueness proves this equality in the correctly typed Hom space.

3.1F1F3step 2.1

Pentagon: after the associativity and unit identifications in [F3], the two composites from $F_{v} F_{w} F_{u}$ to $F_{v w u}$ are induced by the two composites of normalized maps $m$ from $G_{v} \otimes_{R} G_{w} \otimes_{R} G_{u}$ to $G_{v w u}$ . Their derived images are both the triple graph multiplication, so uniqueness in internal degree $0$ makes them agree. This also covers unit indices, where the compositor is the canonical unit identification.

3.2F3F4step 2.1

Unit triangles: since $t (1)$ is empty, the normalized comparisons $m_{v, 1}$ and $m_{1, v}$ are the identity under the right and left tensor unit isomorphisms, respectively. With $F_{1} = Id$ and $u = i d$ , these identifications give separately $μ_{v, 1} = F_{v} u$ and $μ_{1, v} = u F_{v}$ , including $v = 1$ . Thus both unit axioms of [F4] hold.

4.1F2F4step 1.1step 2.1step 3.1step 3.2∎

By steps 1.1, 2.1, 3.1 and 3.2 the functors $F_{v}$ and compositors $μ_{v, w}$ satisfy the pentagon and both unit triangles of [F4]. Since each $G_{v}$ has finite free terms on both sides, $G_{v} \otimes_{R} -$ is exact on bounded complexes of finite graded projectives and descends to the derived category; the identity functor at $1$ is exact as well. Thus these data define the asserted coherent braid group action. The representative system can be specified without a choice axiom: order the finite signed alphabet and take the shortest, then lexicographically least word in each nonempty braid class. The construction uses this system or any given representative system.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

The Rouquier complex is well defined up to canonical homotopy equivalence

Statement

Let $t, u$ be signed words for the same braid $v$ . Then the normalized map $γ_{t, u} ⁣ : F (t) ⟶ F (u)$ of Derived comparisons give unique normalized homotopy maps is a canonical homotopy equivalence, natural with respect to the multiplication maps: for every other braid $v^{'}$ with chosen words $t^{'}, u^{'}$ , the diagrams comparing $F (t) \otimes_{R} F (t^{'})$ with $F (u) \otimes_{R} F (u^{'})$ commute up to the canonical associativity isomorphisms, and $γ_{t, u}$ is the unique homotopy class with the prescribed derived image $c_{t, u}$ . Consequently the object $G_{v}$ of Rouquier complexes form a coherent braid group action is independent of the chosen representative up to canonical homotopy equivalence, and the notation $F (β)$ for the Rouquier complex of a braid element is well defined up to canonical isomorphism in $K^{b} (R^{e} -grmod)$ ; this is an isomorphism statement in the homotopy category, not an equality of complexes.

Facts & Assumptions

Given: Signed words $t, u$ for the same braid $v$ , words $t^{'}, u^{'}$ for a braid $v^{'}$ , and the normalized maps of Derived comparisons give unique normalized homotopy maps.

[F1]

Uniqueness and inverses. ${Hom}_{K^{b}} (F (t), F (u))$ is one-dimensional in degree $0$ , $γ_{t, u}$ is its unique element with derived image $c_{t, u}$ , $γ_{t, t} = i d$ and $γ_{u, t} γ_{t, u} = i d$ . (Derived comparisons give unique normalized homotopy maps, and the transitivity of the same system)

[F2]

The coherent action. The choices $G_{v} = F (t (v))$ and the normalized compositors $m_{v, w}$ assemble into a coherent action; in particular the composite of $m$ 's is associative and unital up to the canonical maps. (Rouquier complexes form a coherent braid group action)

Proof

technique · direct

1.1F1

By [F1] $γ_{t, u}$ is the unique degree-zero class with derived image $c_{t, u}$ and $γ_{u, t}$ is its two-sided homotopy inverse, so $γ_{t, u}$ is a canonical homotopy equivalence.

2.1F1F2step 1.1

Naturality: the composite $F (t) \otimes_{R} F (t^{'}) \to F (u) \otimes_{R} F (u^{'})$ given by $γ_{t, u} \otimes γ_{t^{'}, u^{'}}$ and the composite given by the compositors and the associator both lie in ${Hom}_{K^{b}}$ of one-dimensional degree-zero spaces, and their derived images are the same graph multiplication; by uniqueness they agree up to the canonical associativity isomorphism of [F2].

3.1F2step 2.1∎

The independence of the representative: replacing the word $t (v)$ by another word changes $G_{v}$ by $γ_{t, u}$ , a homotopy equivalence, and these comparisons are compatible with the compositors by step 2.1, so the object is well defined up to canonical isomorphism in $K^{b}$ .

Remarks

The statement is an isomorphism statement: $F (t)$ and $F (u)$ are generally not equal complexes, and the canonical comparison depends on the two words. No choice principle is used: the words are chosen, and the comparisons are then unique in internal degree $0$ . This is Rouquier's well-definedness statement underlying the construction of $G_{v}$ ; the coherent-action theorem is the finite form of the same statement.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

The two-braid category is strict rigid monoidal

Statement

Let $B_{n}$ be the braid-indexed category with an object $v$ for each $v \in B_{n}$ and ${Hom}_{B_{n}} (v, w) : = {Hom}_{K^{b} (R^{e} -grmod)} (G_{v}, G_{w}),$ where $G_{v}$ is the chosen complex of Rouquier complexes form a coherent braid group action. Composition is the carrier composition. Its evaluation $v \mapsto G_{v}$ is fully faithful and has image the full subcategory on those complexes. Keep the braid labels even if two carriers coincide. Define $v ⊠ w = v w$ and, for $f : v \to v^{'}$ , $g : w \to w^{'}$ , define $f ⊠ g = m_{v^{'}, w^{'}} \circ (f \otimes_{R} g) \circ m_{v, w}^{- 1} .$ Then:

$B_{n}$ is strict monoidal, with unit label $1$ and identity associativity and unit constraints (Strict monoidal category);
$v^{- 1}$ is both a left and a right dual of $v$ . Evaluation and coevaluation are the identity of the unit label, with the tensor constraints $m$ furnishing their usual carrier realizations;
the isomorphism classes form a group under $⊠$ , with $[v] [w] = [v w]$ and $[v^{- 1}] = [v]^{- 1}$ . The same identities hold for their classes in the split Grothendieck ring of the additive envelope $B_{n}^{\oplus}$ : its objects are finite formal sums of labels, its morphisms are matrices of the above Hom spaces, and its tensor extends $⊠$ distributively (Split Grothendieck group of an additive category).

No Hecke-algebra identification, no faithfulness and no complete invariant of braids are asserted here; the decategorification of the complexes themselves is the subject of the decategorification proposition of this page.

Facts & Assumptions

Given: the carrier complexes $G_{v}$ and coherent comparisons $m_{v, w}$ of Rouquier complexes form a coherent braid group action, with $G_{1} = R$ .

[F1]

The comparisons are invertible and satisfy the tensor associativity pentagon and canonical unit constraints; they compare $G_{v} \otimes_{R} G_{w}$ to $G_{v w}$ (Rouquier complexes form a coherent braid group action).

[F2]

A strict monoidal category has associative and unital tensor on both objects and morphisms with identity constraints (Strict monoidal category).

[F3]

Left and right duals are evaluation and coevaluation pairs satisfying the triangle identities; an object with both is rigid (Left dual and right dual object, Rigid object and rigid monoidal category).

[F4]

The split Grothendieck group of an essentially small additive category is generated by object isomorphism classes modulo $[X \oplus Y] = [X] + [Y]$ . (Split Grothendieck group of an additive category).

Proof

technique · transport tensor along the coherent carrier comparisons, retaining formal braid labels

1.1F1F2algebra

The category structure is well defined since every Hom and composition is taken from the carrier homotopy category, and the evaluation is fully faithful by its Hom definition. The displayed morphism tensor preserves identity maps and composition: in the composite of two tensor maps the middle factors $m^{- 1} m$ cancel, and tensor composition is componentwise. The pentagon for $m$ makes the two transported products of three morphisms equal; on objects both are the label $v w u$ . The unit constraints for $m$ give $f ⊠ 1_{1} = f = 1_{1} ⊠ f$ . Thus the transported tensor is strictly associative and unital on morphisms as well as on labels, so [F2] applies.

2.1F1F3step 1.1

Set the dual label to $v^{- 1}$ . The product labels $v^{- 1} ⊠ v$ and $v ⊠ v^{- 1}$ both equal $1$ . Take the evaluation and coevaluation maps to be $1_{1}$ in both orders. By the morphism tensor just proved, their triangle composites are $1_{v}$ and $1_{v^{- 1}}$ ; in the carrier category the corresponding evaluation is $m_{v^{- 1}, v} : G_{v^{- 1}} \otimes_{R} G_{v} \to R$ and the corresponding coevaluation is $m_{v, v^{- 1}}^{- 1} : R \to G_{v} \otimes_{R} G_{v^{- 1}}$ , and [F1] supplies the same triangles under evaluation. Hence these are both duals as in [F3].

3.1F1F4step 1.1step 2.1∎

Since every label has this inverse, the isomorphism-class monoid is a group with the stated product and inverse identities. Finite formal sums and matrices form the additive envelope described in the Statement, and the tensor extends bilinearly. In its split Grothendieck group [F4], multiplication respects direct-sum relations, has unit $[1]$ , and satisfies $[v] [w] = [v w]$ and $[v] [v^{- 1}] = [1]$ . This gives the stated class identities without treating the invertible-object category itself as additive.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

Euler classes of Rouquier complexes are homotopy invariant and multiplicative

Statement

Let $C, D$ be bounded cochain complexes of graded $(R, R)$ -bimodules whose terms lie in the type-A Soergel category ${S B i m}_{n}$ (for instance Rouquier word complexes), and define the alternating class $χ (C) : = \sum_{m \in Z} (- 1)^{m} [C^{m}] \in K_{0}^{s p l i t} ({S B i m}_{n}),$ the cochain indexing being translated to the chain indexing of The mapping cone of a chain map by $C_{m} : = C^{- m}$ . Then:

$χ$ is a homotopy invariant: if $C ≃ D$ in $K^{b}$ , then $χ (C) = χ (D)$ , so $χ$ is well defined on isomorphism classes of the homotopy category;
$χ$ is multiplicative: $χ (C \otimes_{R} D) = χ (C) χ (D)$ for the signed tensor totalization and more generally for every finite tensor product;
for every chain map $f ⁣ : C \to D$ one has $χ (Cone (f)) = χ (D) - χ (C)$ , and $χ$ is additive on distinguished triangles of the homotopy category, so it descends to a homomorphism $K_{0}^{t r i} \to K_{0}^{s p l i t} ({S B i m}_{n})$ on the triangulated Grothendieck group of the full triangulated subcategory of complexes with terms in ${S B i m}_{n}$ ;
for every signed word $σ = σ_{i_{1}}^{ϵ_{1}} \dots σ_{i_{r}}^{ϵ_{r}}$ , the iterated signed tensor totalization $F_{i_{1}}^{ϵ_{1}} \otimes_{R} \dots \otimes_{R} F_{i_{r}}^{ϵ_{r}}$ (the word complex $F (σ)$ once that notation is introduced) satisfies $χ (F (σ)) = \prod_{k = 1}^{r} χ (F_{i_{k}}^{ϵ_{k}})$ .

In particular the alternating class of a Rouquier complex depends only on its homotopy class and is compatible with tensor products.

Facts & Assumptions

Given: The ring $R = Q [x_{1}, \dots, x_{n}]$ , the type-A Soergel category ${S B i m}_{n}$ of graded $(R, R)$ -bimodules, bounded cochain complexes $C, D$ with all terms in ${S B i m}_{n}$ , and the alternating class $χ$ of the statement.

[F1]

The category and its idempotents. ${S B i m}_{n}$ is the idempotent completion $Kar ({B S B i m}_{n})$ of the additive category of finite sums of shifted Bott–Samelson bimodules; its objects are the pairs $(M, e)$ with $e \in End (M)$ a degree-zero idempotent, with composition inherited from the ambient category and identity on $(M, e)$ given by $e$ , and it is closed under finite direct sums, internal shifts, tensor products over $R$ , and direct summands (The type-A Soergel category ${S B i m}_{n}$ , The idempotent completion of a preadditive category).

[F2]

Split Grothendieck group. For the additive category ${S B i m}_{n}$ the split Grothendieck group is the free abelian group on isomorphism classes of objects modulo $[X \oplus Y] = [X] + [Y]$ ; in particular $[0] = 0$ , isomorphic objects have equal classes, and the classes of a direct-sum decomposition add up (Split Grothendieck group of an additive category). Equipped with the product $[X] [Y] : = [X \otimes Y]$ this is the split Grothendieck ring of Split Grothendieck rings of the type-A Soergel categories.

[F3]

Cones, shifts and contractibility. For a chain map $f ⁣ : C \to D$ the mapping cone has $Cone (f)_{n} = D_{n} \oplus C [1]_{n} = D_{n} \oplus C_{n - 1}$ , and the shift has $C [k]_{n} = C_{n - k}$ with $d_{n}^{C [k]} = (- 1)^{k} d_{n - k}^{C}$ (The mapping cone of a chain map, The shift of a chain complex). Under the reindexing $C^{m} : = C_{- m}$ of Complexes, homotopies and contractibility in an additive category this is a cochain complex with $Cone (f)^{m} = D^{m} \oplus C^{m + 1}$ and $d (y, x) = (d_{D} y + f^{m + 1} x, - d_{C}^{m + 1} x)$ ; a cochain map is a homotopy equivalence exactly when its (cochain) cone is contractible, i.e. admits a family $h^{n} ⁣ : C^{n} \to C^{n - 1}$ with $1_{C^{n}} = d^{n - 1} h^{n} + h^{n + 1} d^{n}$ for all $n$ (A chain map is a homotopy equivalence exactly when its cone is contractible).

[F4]

Homotopy category and triangles. Morphisms of the homotopy category $K^{b} (R^{e} -grmod)$ of bounded complexes are homotopy classes, so an isomorphism there is a homotopy equivalence (The homotopy category of chain complexes). The ambient category is triangulated by its shifts and distinguished cone triangles (The homotopy category of an abelian category is triangulated), a triangle being distinguished when it is isomorphic to a standard cone triangle (Standard cone triangle in the homotopy category, Distinguished cone triangle in the homotopy category); two distinguished completions of the same map have isomorphic third objects (The cone object of a map is unique up to nonunique isomorphism).

[F5]

Totalization, multiplicativity and the ring. The signed tensor totalization of bounded complexes $C, D$ of graded bimodules has degree- $n$ term $Tot (C \otimes_{R} D)^{n} = ⨁_{p + q = n} C^{p} \otimes_{R} D^{q}$ with Koszul differential $d (c \otimes e) = d_{C} c \otimes e + (- 1)^{p} c \otimes d_{C} e$ (Bounded graded bimodule complexes and signed tensor totalization), its terms are objects of ${S B i m}_{n}$ whenever the terms of $C$ and $D$ are [F1], and $[C^{p} \otimes_{R} D^{q}] = [C^{p}] [D^{q}]$ in the split Grothendieck ring [F2].

[F6]

Triangulated Grothendieck group. $K_{0}^{t r i} (T)$ is the free abelian group on the isomorphism classes of an essentially small triangulated category $T$ modulo $[Y] - [X] - [Z]$ for each distinguished triangle $X \to Y \to Z \to X [1]$ (Grothendieck group of an essentially small triangulated category).

Proof

technique · direct

1.1F1F2

The class $χ (C)$ is well defined: $C$ is bounded, so only finitely many terms are nonzero and the sum is finite; each term lies in ${S B i m}_{n}$ by hypothesis, so each $[C^{m}]$ is a class in the split Grothendieck group; additivity of classes gives $χ (C \oplus D) = χ (C) + χ (D)$ , $χ (C [m]) = (- 1)^{m} χ (C)$ and $χ (0) = 0$ .

1.2F3F2

For a chain map $f ⁣ : C \to D$ between such complexes, the cone term formula gives $χ (Cone (f)) = \sum_{m} (- 1)^{m} ([D^{m}] + [C^{m + 1}]) = χ (D) - χ (C)$ .

1.3F1F2F3

Suppose $(C, d)$ is contractible with contracting homotopy $h$ , so that $1_{C^{m}} = d^{m - 1} h^{m} + h^{m + 1} d^{m}$ for all $m$ . Put $q_{m} : = h^{m + 1} d^{m}$ and $p_{m} : = 1 - q_{m} = d^{m - 1} h^{m}$ ; the second description shows $p_{m} + q_{m} = 1$ , and $q_{m}^{2} = h^{m + 1} d^{m} h^{m + 1} d^{m} = h^{m + 1} (1 - h^{m + 2} d^{m + 1}) d^{m} = h^{m + 1} d^{m} = q_{m}$ using the homotopy identity at degree $m + 1$ and $d^{m + 1} d^{m} = 0$ ; hence $q_{m}, p_{m}$ are idempotents with $q_{m} p_{m} = 0$ and $p_{m} q_{m} = 0$ . Both are degree-zero endomorphisms of the object $C^{m}$ of ${S B i m}_{n}$ , so the Karoubi objects $(C^{m}, q_{m})$ and $(C^{m}, p_{m})$ exist and the maps $u = (q_{m}, p_{m}) ⁣ : C^{m} \to (C^{m}, q_{m}) \oplus (C^{m}, p_{m})$ and $v = (q_{m}, p_{m})^{T} ⁣ : (x, y) \mapsto x + y$ in the reverse direction are inverse isomorphisms, since $v u = q_{m} + p_{m} = 1$ and $u v (x, y) = (q_{m} x + q_{m} y, p_{m} x + p_{m} y) = (x, y)$ for $x = q_{m} x$ , $y = p_{m} y$ . Hence in the split Grothendieck group $[C^{m}] = [(C^{m}, q_{m})] + [(C^{m}, p_{m})]$ .

2.1F1F2step 1.3

Put $α_{m} : = p_{m + 1} d^{m} q_{m} \in Hom ((C^{m}, q_{m}), (C^{m + 1}, p_{m + 1}))$ and $β_{m} : = q_{m} h^{m + 1} p_{m + 1}$ in the reverse direction. From $p_{m + 1} = d^{m} h^{m + 1}$ and $q_{m} = h^{m + 1} d^{m}$ one gets $d^{m} q_{m} = p_{m + 1} d^{m}$ and $q_{m} h^{m + 1} = h^{m + 1} p_{m + 1}$ , hence $α_{m} β_{m} = p_{m + 1} d^{m} h^{m + 1} p_{m + 1} = p_{m + 1} p_{m + 1} p_{m + 1} = p_{m + 1}$ and $β_{m} α_{m} = q_{m} q_{m} q_{m} = q_{m}$ ; as the identities of the Karoubi objects $(C^{m + 1}, p_{m + 1})$ and $(C^{m}, q_{m})$ are $p_{m + 1}$ and $q_{m}$ , the maps $α_{m}, β_{m}$ are mutually inverse isomorphisms, so $[(C^{m + 1}, p_{m + 1})] = [(C^{m}, q_{m})]$ .

2.2F2F5step 1.1

Let $C, D$ be bounded complexes with terms in ${S B i m}_{n}$ . Every term $C^{p} \otimes_{R} D^{q}$ of the signed tensor totalization lies in ${S B i m}_{n}$ , and $[C^{p} \otimes_{R} D^{q}] = [C^{p}] [D^{q}]$ in the split Grothendieck ring, so $χ (C \otimes_{R} D) = \sum_{p, q} (- 1)^{p + q} [C^{p} \otimes_{R} D^{q}] = \sum_{p, q} (- 1)^{p + q} [C^{p}] [D^{q}] = (\sum_{p} (- 1)^{p} [C^{p}]) (\sum_{q} (- 1)^{q} [D^{q}]) = χ (C) χ (D)$ ; both sums are finite because $C$ and $D$ are bounded.

3.1step 1.3step 2.1F2

Combining steps 1.3 and 2.1, $[C^{m}] = [(C^{m}, q_{m})] + [(C^{m}, p_{m})] = [(C^{m}, q_{m})] + [(C^{m - 1}, q_{m - 1})]$ for every $m$ , so $χ (C) = \sum_{m} (- 1)^{m} [(C^{m}, q_{m})] + \sum_{m} (- 1)^{m} [(C^{m - 1}, q_{m - 1})] = \sum_{m} (- 1)^{m} [(C^{m}, q_{m})] - \sum_{m} (- 1)^{m} [(C^{m}, q_{m})] = 0$ , the sums being finite because $C$ is bounded. Thus every contractible complex with terms in ${S B i m}_{n}$ has vanishing Euler class.

4.1step 1.2step 3.1F3F4

If $C ≃ D$ in $K^{b}$ , choose a homotopy equivalence $f ⁣ : C \to D$ ; by [F4] this is an isomorphism in the homotopy category. By [F3] the cone of $f$ is contractible, so step 3.1 gives $χ (Cone (f)) = 0$ , and step 1.2 gives $0 = χ (Cone (f)) = χ (D) - χ (C)$ . Hence $χ (C) = χ (D)$ , and $χ$ is constant on isomorphism classes of the homotopy category; in particular $χ$ of a contractible complex is $0$ .

5.1step 1.2step 4.1F1F4F6

Let $T$ be the replete full subcategory of $K^{b} (R^{e} -grmod)$ generated by bounded complexes with terms in ${S B i m}_{n}$ , extending $χ$ to it by the homotopy invariance of step 4.1. It is essentially small: use the fixed set of Soergel representatives and bounded lists of differential matrices. It is closed under shifts and under cones (the terms of $Cone (f)^{m} = D^{m} \oplus C^{m + 1}$ are sums of objects of ${S B i m}_{n}$ ), so with the inherited triangles it is a triangulated category: the completions and rotations required by the axioms exist in the ambient triangulated category and their objects remain in $T$ by this closure. For a distinguished triangle $X \overset{a}{\to} Y \to Z \to X [1]$ of $T$ , the standard cone triangle of the chain map $a$ is another distinguished completion of the same map, so its third object $Cone (a)$ is isomorphic to $Z$ ; by step 4.1 and step 1.2, $χ (Z) = χ (Cone (a)) = χ (Y) - χ (X)$ . Therefore $χ$ vanishes on every generator $[Y] - [X] - [Z]$ of the kernel of the presentation of $K_{0}^{t r i} (T)$ and, being a function on isomorphism classes by step 4.1, extends to a well-defined group homomorphism $K_{0}^{t r i} (T) \to K_{0}^{s p l i t} ({S B i m}_{n})$ .

6.1step 2.2F2F5∎

Induction on $r$ using step 2.2 gives $χ (F_{i_{1}}^{ϵ_{1}} \otimes_{R} \dots \otimes_{R} F_{i_{r}}^{ϵ_{r}}) = \prod_{k = 1}^{r} χ (F_{i_{k}}^{ϵ_{k}})$ for every finite sequence of signed generators: the case $r = 0$ is $χ (R) = [R] = 1$ , the unit of the split Grothendieck ring, and the induction step applies step 2.2 to the bounded complex $F_{i_{1}}^{ϵ_{1}} \otimes_{R} \dots \otimes_{R} F_{i_{r - 1}}^{ϵ_{r - 1}}$ and the two-term complex $F_{i_{r}}^{ϵ_{r}}$ , whose terms are objects of ${S B i m}_{n}$ ; this is the stated formula for the word complex.

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Decategorification of a Rouquier complex is the Hecke braid generator

Statement

Let $Φ ⁣ : K_{0}^{s p l i t} ({S B i m}_{n}) \to H_{S_{n}}$ be the unique algebra isomorphism of The split Grothendieck group of the Soergel category is the type-A Hecke algebra with $Φ ([B_{i}]) = H_{i} = v (T_{i} + 1)$ and $Φ (v X) = v Φ (X)$ , where $H_{S_{n}}$ is the Hecke algebra over $A = Z [v, v^{- 1}]$ with $q = v^{- 2}$ and standard generators $T_{i}$ , and let $χ$ be the alternating class of Euler classes of Rouquier complexes are homotopy invariant and multiplicative.

(a) $Φ (χ (F_{i})) = H_{i} - v = v T_{i}$ and $Φ (χ (F_{i}^{- 1})) = H_{i} - v^{- 1} = v^{- 1} T_{i}^{- 1}$ ; equivalently $χ (F_{i} (- 1))$ maps to $T_{i}$ and $χ (F_{i}^{- 1} (1))$ maps to $T_{i}^{- 1}$ . In particular $χ (F_{i}) χ (F_{i}^{- 1}) = 1$ , and the powers of $v$ are the exact normalization of the library's shift convention, not an ambiguity.

(b) For every braid $β \in B_{n}$ and every signed word $σ$ for $β$ , $Φ (χ (F (σ))) = v^{e (σ)} T_{β}$ , where $T_{β} : = T_{i_{1}}^{ϵ_{1}} \dots T_{i_{r}}^{ϵ_{r}}$ is the image of $β$ under the standard group homomorphism $B_{n} \to H_{S_{n}}^{\times}$ , $σ_{i} \mapsto T_{i}$ , and $e (σ) = # {ϵ_{k} = 1} - # {ϵ_{k} = - 1}$ depends only on $β$ . Hence $β \mapsto v^{- e (β)} χ (F (β))$ , composed with $Φ$ , is that standard homomorphism: up to the grading normalization, the Rouquier complex of a braid decategorifies to the image of the braid in the Hecke algebra.

Facts & Assumptions

Given: The isomorphism $Φ$ with $Φ ([B_{i}]) = H_{i} = v (T_{i} + 1)$ and $Φ (v X) = v Φ (X)$ , the Euler class $χ$ of Euler classes of Rouquier complexes are homotopy invariant and multiplicative, and the generator complexes of The positive and negative Rouquier generator complexes.

[F1]

Classes of the generators. $χ (F_{i}) = [B_{i}] - [R (1)]$ and $χ (F_{i}^{- 1}) = [B_{i}] - [R (- 1)]$ in $K_{0}^{s p l i t} ({S B i m}_{n})$ ; the unit class is $[R] = 1$ , and $[R (r)] = v^{r}$ under the rule $v [X] = [X (1)]$ . (Euler classes of Rouquier complexes are homotopy invariant and multiplicative, Split Grothendieck rings of the type-A Soergel categories)

[F2]

The Hecke normalization. $Φ$ is an algebra isomorphism with $Φ ([B_{i}]) = H_{i} = v (T_{i} + 1)$ , $Φ (v X) = v Φ (X)$ , $q = v^{- 2}$ ; the quadratic relation of The type-A Hecke algebra in Soergel normalization gives $T_{i} (v^{2} T_{i} - 1 + v^{2}) = 1 = (v^{2} T_{i} - 1 + v^{2}) T_{i}$ , hence $T_{i}^{- 1} = v^{2} T_{i} - 1 + v^{2}$ (The split Grothendieck group of the Soergel category is the type-A Hecke algebra)

[F3]

Multiplicativity and invariance. $χ (C \otimes_{R} D) = χ (C) χ (D)$ for signed totalizations, $χ$ is a homotopy invariant and is multiplicative over finite tensor products; the word complex of a signed word is the corresponding iterated tensor product. (Euler classes of Rouquier complexes are homotopy invariant and multiplicative, The Rouquier complex of a braid word)

[F4]

Word independence. Homotopy equivalent word complexes for the same braid have equal Euler classes, and the sign $e (σ)$ depends only on the braid; the assignment $σ_{i} \mapsto T_{i}$ extends to the group homomorphism $B_{n} \to H_{S_{n}}^{\times}$ because the Hecke algebra is presented by the braid relations and by the invertibility of $T_{i}$ . (The Rouquier complex is well defined up to canonical homotopy equivalence, The braid group by Artin presentation, The type-A Hecke algebra in Soergel normalization)

Proof

technique · direct

1.1F1F2

By [F1] and [F2] we compute $Φ (χ (F_{i})) = Φ ([B_{i}]) - Φ ([R (1)]) = H_{i} - v = v (T_{i} + 1) - v = v T_{i}$ , and likewise $Φ (χ (F_{i}^{- 1})) = H_{i} - v^{- 1}$ ; using $T_{i}^{- 1} = v^{2} T_{i} - 1 + v^{2}$ gives $v^{- 1} T_{i}^{- 1} = v T_{i} - v^{- 1} + v = H_{i} - v^{- 1}$ , so $Φ (χ (F_{i}^{- 1})) = v^{- 1} T_{i}^{- 1}$ .

2.1F1F2step 1.1

Multiplying: $Φ (χ (F_{i})) Φ (χ (F_{i}^{- 1})) = v T_{i} \cdot v^{- 1} T_{i}^{- 1} = 1$ ; equivalently $χ (F_{i} (- 1)) \mapsto T_{i}$ and $χ (F_{i}^{- 1} (1)) \mapsto T_{i}^{- 1}$ since the shift acts by $v^{\pm 1}$ . This proves (a).

2.2F3step 1.1

For a signed word $σ$ , [F3] gives $χ (F (σ)) = \prod_{k} χ (F_{i_{k}}^{ϵ_{k}})$ ; by step 1.1 each positive letter contributes $v T_{i}$ and each negative letter $v^{- 1} T_{i}^{- 1}$ , so $Φ (χ (F (σ))) = v^{# {+ 1}} v^{- # {- 1}} \prod_{k} T_{i_{k}}^{ϵ_{k}} = v^{e (σ)} T_{β}$ where $T_{β}$ is the image of $β$ under the standard homomorphism.

3.1F4step 2.2∎

Word independence: if $σ^{'}$ is another signed word for $β$ then $F (σ) ≃ F (σ^{'})$ by [F4], so $χ (F (σ)) = χ (F (σ^{'}))$ ; and $e (σ) = e (σ^{'})$ because inverse pairs have exponent zero and the two Artin relations have the same exponent on both sides. Hence the assignment $β \mapsto v^{- e (β)} χ (F (β))$ is well defined, and composing with $Φ$ gives the standard homomorphism $σ_{i} \mapsto T_{i}$ .

Remarks

The unit factors $v^{\pm 1}$ are exact and cannot be dropped: the design's shorthand "match $T_{i}^{\pm 1}$ " suppresses them, and the equalities displayed in (a) are the exact normalization of the library's shift convention. The proposition is a decategorification statement at the level of classes; it does not assert that the class map determines the homotopy type, and indeed the counterexample of this pair shows that it does not.

5 · Examples, counterexamples and false statements

None yet.

Rouquier Complexes and Categorical Braid Relations

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

The positive and negative Rouquier generator complexes

Definition

Opposite Rouquier generator complexes are homotopy inverse

Statement

Facts & Assumptions

Proof

Remarks

Rouquier complexes satisfy far commutativity

Statement

Facts & Assumptions

Proof

Remarks

Rouquier complexes satisfy the three-term braid relation

Statement

Facts & Assumptions

Proof

Remarks

The Rouquier complex of a braid word

Definition

Coherent action of a group on a category

Definition

Canonical comparisons between standard graph tensor products

Definition

Rouquier generator complexes have canonical derived graph models

Statement

Facts & Assumptions

Proof

Remarks

Derived comparisons give unique normalized homotopy maps

Statement

Facts & Assumptions

Proof

Remarks

Normalized comparison isomorphisms are transitive

Statement

Facts & Assumptions

Proof

Rouquier complexes form a coherent braid group action

Statement

Facts & Assumptions

Proof

The Rouquier complex is well defined up to canonical homotopy equivalence

Statement

Facts & Assumptions

Proof

Remarks

The two-braid category is strict rigid monoidal

Statement

Facts & Assumptions

Proof

Euler classes of Rouquier complexes are homotopy invariant and multiplicative

Statement

Facts & Assumptions

Proof

Decategorification of a Rouquier complex is the Hecke braid generator

Statement

Facts & Assumptions

Proof

Remarks

5 · Examples, counterexamples and false statements

Sources