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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.

Depends on

Used by

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Sources

Raphaël Rouquier, Categorification of the braid groups, arXiv:math/0409593v1 (30 September 2004), §3 "The 2-braid group" (standard reference, not scraped)
Eugene Gorsky, Oscar Kivinen, José Simental, Algebra and geometry of link homology: Lecture Notes from the IHES 2021 Summer School, Bull. London Math. Soc. 55 (2023) 537-591, §3.1 (standard reference, not scraped)
Nicolas Libedinsky, Gentle introduction to Soergel bimodules I: the basics, São Paulo J. Math. Sci. 13 (2019), arXiv:1702.00039v2, §4 (standard reference, not scraped)