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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.

Depends on

Used by

Equal Euler classes do not by themselves prove homotopy-equivalent complexes Counterexample
Decategorification of a Rouquier complex is the Hecke braid generator Proposition

Dependency tree · two levels

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

Sources

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)
Raphaël Rouquier, Categorification of the braid groups, arXiv:math/0409593v1 (30 September 2004), §3 "The 2-braid group" (standard reference, not scraped)