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The complex of an admissible bigraded curve

Definition

Fix the normalized bigradings ${\tilde{d}}_{i}, {\tilde{b}}_{i}$ of Bigraded string types and their contributions to I^{bigr}. Let $\tilde{c}$ be an admissible bigraded curve in normal form with respect to the fixed basic set and vertical curves (Basic arcs, admissible curves and the standard normal form), with crossing set $cr (\tilde{c})$ and local index $(x_{1}, x_{2}) : = μ^{b i g r} ({\tilde{d}}_{x_{0}}, \tilde{c}; x) \in Z^{2}$ at each crossing $x$ ; here $x_{0}$ denotes the index of the vertical curve $d_{x_{0}}$ containing $x$ , so that $x \in d_{x_{0}}$ , and for a crossing of the underlying curve $c$ the local index of the bigrading is interpreted through the identification of the crossings of $\tilde{c}$ with those of $c$ (Local indices and bigraded intersection numbers). Put $P (x) : = P_{x_{0}} [- x_{1}] {x_{2}} \in C_{m}, L (\tilde{c}) : = ⨁_{x \in cr (\tilde{c})} P (x),$ the direct sum of shifted vertex projectives indexed by the crossings, where $P_{x_{0}} = A_{m} e_{x_{0}}$ is the vertex projective of Finite graded A_m-modules, internal shifts and the vertex projectives and the two shifts are the homological shift $[- x_{1}]$ and the internal shift ${x_{2}}$ .

The differential. For crossings $x, y$ which are the two endpoints of an essential segment of $c$ and satisfy $y_{1} = x_{1} + 1$ , define the component $\partial_{y x} ⁣ : P (x) ⟶ P (y)$ by the following rules, right multiplication meaning the left $A_{m}$ -linear map $A_{m} e_{i} \to A_{m} e_{j}$ , $u \mapsto u a$ , for $a \in e_{i} A_{m} e_{j}$ :

if $x_{0} = y_{0}$ (in which case $x_{2} = y_{2} + 1$ ), then $\partial_{y x}$ is the right multiplication by the return $(x_{0} ∣ x_{0} - 1 ∣ x_{0})$ when $x_{0} > 0$ , and is the zero map when $x_{0} = 0$ (the return $(0 ∣ 1 ∣ 0)$ vanishes);
if $x_{0} = y_{0} \pm 1$ , then $\partial_{y x}$ is the right multiplication by the arrow $(x_{0} ∣ y_{0})$ , which is $(x_{0} ∣ x_{0} - 1)$ or $(x_{0} ∣ x_{0} + 1)$ ;
otherwise $\partial_{y x} : = 0$ .

Put $\partial : = \sum_{x, y} \partial_{y x}$ . For a bigraded $k$ -string $\tilde{g}$ of $\tilde{c}$ the same formulas applied to the crossings and essential segments of $g$ define an object $L (\tilde{g})$ with its differential, and the underlying graded module of $L (\tilde{g})$ is an abelian subgroup of $L (\tilde{c})$ .

Claims. $(L (\tilde{c}), \partial)$ is a bounded complex of finitely generated graded projective left $A_{m}$ -modules with a differential that is degree zero for the internal grading, hence an object of $C_{m} = K^{b} ({proj}^{g r} A_{m})$ ; the object is bounded because there are finitely many crossings; and the deck action translates into the shift rule $L (χ (r_{1}, r_{2}) \tilde{c}) ≅ L (\tilde{c}) [- r_{1}] {r_{2}}$ by the degreewise sign identification described in the proof. These claims are proved below.

Facts & Assumptions

Given: An admissible bigraded curve $\tilde{c}$ in normal form with finitely many crossings $x = (x_{0}; x_{1}, x_{2})$ , its essential segments classified by the six types of Figure 12 and the endpoint types of Figures 13-14, and the vertex projectives $P_{i} = A_{m} e_{i}$ with these typed right multiplication maps.

[L1]

The crossing set is finite, each essential segment has two crossings as endpoints, and for every essential segment with endpoints $x, y$ one has $y_{1} = x_{1} \pm 1$ ; the segment types $1, 1^{'}, 2, 2^{'}$ are the essential ones and the tables of Figures 12-14, in the normalization of Bigraded string types and their contributions to I^{bigr}, record the internal indices at their endpoints: for a segment whose endpoints satisfy $y_{1} = x_{1} + 1$ one has either $x_{0} = y_{0}$ and $x_{2} = y_{2} + 1$ , or $x_{0} = y_{0} \pm 1$ (Basic arcs, admissible curves and the standard normal form, Local indices and bigraded intersection numbers).

[L2]

For $x_{0} > 0$ , $(x_{0} ∣ x_{0} - 1 ∣ x_{0})$ is the return at $x_{0}$ , of internal degree $1$ , and $(x_{0} ∣ y_{0})$ is either the ascending arrow $(x_{0} ∣ x_{0} + 1)$ of internal degree $0$ or the descending arrow $(x_{0} ∣ x_{0} - 1)$ of internal degree $1$ ; the product of two arrow classes is zero whenever it is defined as a path of length two other than a return, the return at vertex $0$ is zero, and every path of length at least three vanishes in $A_{m}$ (The 4m+1 path basis).

[L3]

$P_{i} = A_{m} e_{i}$ is a finitely generated graded projective left $A_{m}$ -module, right multiplication by a homogeneous element $a \in A_{m}$ is a degree-zero map $P_{i} {r} \to P_{j} {r - \deg a}$ when $a$ lies in $e_{i} A_{m} e_{j}$ , the homological shift and the internal shift act as displayed, and $C_{m}$ is the homotopy category of bounded complexes of such modules (Finite graded A_m-modules, internal shifts and the vertex projectives).

[L4]

The deck action adds $(r_{1}, r_{2})$ to the local indices of all crossings: $χ (r_{1}, r_{2})$ replaces $(x_{1}, x_{2})$ by $(x_{1} + r_{1}, x_{2} + r_{2})$ for every crossing $x$ (Local indices and bigraded intersection numbers, The Z^2 cover of the projectivized tangent bundle and bigraded curves).

Proof

technique · direct

1.1L1L2

The composites of two differential components vanish. Let $x, y, z$ be crossings with $\partial_{y x} \neq 0 \neq \partial_{z y}$ , so that $y_{1} = x_{1} + 1$ , $z_{1} = y_{1} + 1$ and the pairs $(x, y)$ , $(y, z)$ are endpoints of essential segments; the composite is right multiplication by the concatenation of the two path labels. If either label is a return, its length is at least three, so it vanishes by [L2]. If both labels are arrows and $x_{0} \neq z_{0}$ , they form a monotone length-two path, which also vanishes. The remaining case would have $x_{0} = z_{0} = y_{0} \pm 1$ . Both segments would then lie in the same region between these adjacent dividing curves and approach $y$ from the same side of $d_{y_{0}}$ . This contradicts transversality at the crossing $y$ , where the two branches of the embedded curve lie on opposite sides of $d_{y_{0}}$ . Thus no such consecutive arrow return occurs, and every composite is zero. Summing the components gives $\partial^{2} = 0$ .

1.2L1L3L4

The shift rule. Under $χ (r_{1}, r_{2})$ the local index of each crossing $x$ becomes $(x_{1} + r_{1}, x_{2} + r_{2})$ by [L4], and the crossing data (which crossings are joined by essential segments, and the segment types) are unchanged because the deck action changes only the bigrading, not the underlying curve or its normal form; the source retains the same path entries, while the homological shift $[- r_{1}]$ multiplies the target differential by $(- 1)^{r_{1}}$ . Map the summand indexed by $x$ by $(- 1)^{r_{1} x_{1}}$ times the identity. For a nonzero entry $x \to y$ one has $y_{1} = x_{1} + 1$ , so the target differential followed by the source sign agrees with the target sign followed by the source differential. These invertible sign maps give the claimed chain isomorphism. An identity on every summand would fail for odd $r_{1}$ .

2.1step 1.1L1L3

The differential is degree zero and the terms are finite graded projective. For a component given by right multiplication by a path $a \in e_{x_{0}} A_{m} e_{y_{0}}$ of internal degree $δ$ , an element of underlying degree $d$ has source shifted degree $d + x_{2}$ and target shifted degree $d + δ + y_{2}$ . The segment tables in [L1] give $δ = x_{2} - y_{2}$ : a return has $δ = 1$ , an ascending arrow has $δ = 0$ , and a descending arrow has $δ = 1$ . Thus both degrees agree. The map is left $A_{m}$ -linear because $(b u) a = b (u a)$ , and its image lies in $A_{m} e_{y_{0}}$ because $a = e_{x_{0}} a e_{y_{0}}$ ; no right-module structure on $A_{m} e_{x_{0}}$ is assumed. Homological degree rises from $x_{1}$ to $y_{1} = x_{1} + 1$ . Each term is a shifted finite graded vertex projective, and there are finitely many crossings; together with step 1.1 this gives a bounded complex of the required bidegree.

3.1step 1.1step 2.1step 1.2∎

Conclusion. $(L (\tilde{c}), \partial)$ is a bounded complex of finite graded projectives with a degree-zero differential, so it defines an object of $C_{m}$ , and the deck action acts by the shift $[- r_{1}] {r_{2}}$ . No choice principle is used; the verifications are finite checks over the segment types.

Depends on

Used by

Dependency tree · two levels

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Sources

Mikhail Khovanov and Paul Seidel, Quivers, Floer Cohomology, and Braid Group Actions, J. Amer. Math. Soc. 15 (2002) 203-271, Section 4a (standard reference, not scraped)