Grid homology for spatial graphs and a skein

exact sequence

Zipei Zhuang

September 29, 2021

Abstract

We defined a grid homology theory for spatial graphs, which is slightlydifferent from the one in [3] . We showed that the skein exact sequenceof singular knots in [10] can be extended to our grid homology for spatialgraphs.

1 Introduction

Knot Floer homology is an invariant for knots in S3, defined using Heegaarddiagrams and holomorphic disks, see [9] [11]. In [8] , knot Floer homology wasgeneralized to singular knots in S3. An skein exact sequence was constructedin [10] , relating the Floer homology of a knot and two resolutions at somecrossing. Iterating the exact sequence, they arrived at a description of the knotFloer homology groups of an arbitrary knot in terms of the knot Floer homologygroups of fully singular knots, which can be explicitly calculated. This givesrise to a cube of resolution for knot Floer homology, which provides conjecturalrelations with Khovanov-Rozansky homology, see [4] [2].

In this paper, we want to generalize the skein exact sequence to the case ofspatial graphs. Heegaard Floer homology of spatial graphs was defined in [1]and in [3] combinatorially using grid diagrams. We adopted [3] ’s construction,with one difference: we represent the vertices of a spatial graph using X’s (seeSection 2 for explicit definition) , while the authors of [3] used O’s . This isin accordance with the definition in [8] for singular knots. This does not needmuch additional work: most arguments in [3] can be used in our definition, withlittle modification. There are also some differences: e.g. compare Prop.1 andProposition 4.21 in [3].

In Section 1, we reviewed the grid diagram representation of a spatial graph;In Section 2, we constructed the grid homology of a spatial graph and provedsome basic properties; In Section 3, we developed a skein exact sequence for aspatial graph at a vertice.

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Figure 1: The graph Reidemeister moves

2 Grid diagrams of a spatial graph

A spatial grgaph is an embedding of an oriented graph in S3. As in the caseof knots and links, we can represent a spatial graph by its diagram on the plane.Two spatial graphs are equivalent if they are connected by a finite number ofgraph Reidemeister moves (see Figure ) and planar isotopies.

The first three graph Reidemeister moves are just the usual Reidemeistermoves for knots and links. Note that the RV moves can be operated only foredges with the same orientation(both incoming or outgoing).

In [3] , a spatial graph (under this equivalence relation) is called a transversespatial graph, indicating that at each vertex v, the incoming and outgoing edgesare separated by a small disk, which intersect the spatial graph at v. Theambient isotopies between two transverse spatial graphs must preserve the disks.For brevity, we omit the terminology ”transverse”: any spatial graph in thispaper is viewed as a transverse spatial graph.

A (planar) grid diagram is an n×n grid. Each square is empty, decoratedwith an O or X such that there is exactly 1 X in each row or column.

If we identify the top boundary segment with the bottom one, and the leftboundary segment with the right one, we get a toroidal grid diagram. Bydefinition, two planar grid diagrmas give rise to the same toroidal grid diagramif and only if they are related by cyclic permutations, i.e. cyclicly permute theorder of their rows or columns.

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Figure 2: A spatial graph and a grid diagram for it

A grid diagram G specifies a spatial graph: Draw oriented segments connect-ing the X-marked squares to the O-marked squares in each column; then draworiented segments connecting the O-marked squares to the X-marked squaresin each row, with the convention that the vertical segments always cross abovethe horizontal ones. Note that under this rule, a vertex with valance > 2 of thespatial graph must correspond to an X in the grid diagram. We will call theseX’s the vertex X’s, and the other standard X.

The following properties has been proved in [3] (Note that we useX-markingsto represent vertices of the spatial graph, while in [3] they use O-markings. Ofcourse this difference does not affect the validity of the proof ):

Theorem 1. (1) Any spatial graph can be represented by a grid diagram.(2)If g and g′ are two grid diagrams representing the same spatial graph,

then g and g′ are related by a finite sequence of graph grid moves.

For an vertex X, let the set of O’s that appear in a row or column with thisX be its flock. A flock is in L-formation, if the O’s are all to the right andbelow of the X(See Figure ). A prefered grid diagram is one that all vertexX’s have their flocks in L-formation.

We need a further lemma in Section 4 , which is proved in [3]:

Lemma 1. Any spatial graph can be represented by a preferred grid diagram.

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3 Grid homology for spatial graphs

3.1 The grid chain complex

From now on we always assume our graph has no vertices with only incomingedges or only outgoing edges. Equvalently, the grid diagrams representing thesegraphs have at least 1 O in each row and column.

Let G be a n×n grid diagram representing a spatial graph G. A grid statefor G is an n-tuple of points x = {x1, ...xn} in the torus, with the property thateach horizontal circle and each vertical circle contains exactly 1 of the elementsof x. The set of grid states for G is denoted S(G). For x,y ∈ S(G), let Rect(x,y)denote the set of rectangles from x to y. A rectangle r ∈ Rect(x,y) is called anempty rectangle if x ∩ Int(r) = y ∩ Int(r) = ∅. The set of empty rectanglesfrom x to y is denoted Rect◦(x,y).

Suppose that G has m O markings, denoted O1, O2, ...Om, andDenote the O markings of G by O1, O2, ...Om, assign a variable Vi to each

Oi. Write R = F[V1, ..., Vm], where F = Z/2Z. Define

C−(G) = the free F[U1, ..., Um]−module generated by S(G) (1)

and ∂− : C−(G) −→ C−(G) is the R-module homomorphism defined by

∂−(x) =∑

y∈S(G)

∑r∈Rect◦(x,y)Int(r)∩X=∅

VO1(r)1 · · ·V On(r)

n · y (2)

The relative Maslov grading M : S(G) −→ Z is defined by

M(x)−M(y) = 1− 2#(r ∩O) + 2#(x ∩ Int(r)) (3)

for any r ∈ Rect(x,y). The relative Alexander grading A : S(G) −→ Z isdefined by

A(x)−A(y) = #(r ∩ X)−#(r ∩O) (4)

for any r ∈ Rect(x,y). The gradings can be extended to C−(G) by definingM(Vi)=−2, A(Vi)=−1 for any i.

As in the case of knots, the homomorphism ∂− is a differential of degree(−1, 0), i.e. it decreases the Maslov grading by 1 and preserves the Alexandergrading.

Proposition 1. Let G be a grid diagram for the spatial graph G.(1)If Oi, Oj lie on the interior of the same edge, the multiplication by Vi is

chain homotopic to multiplication by Vj. More generally, this is true when Oiand Oj can be connected by an arc on the spatial graph which does not intersectany vertex with valance > 2.

(2) Let v be a vertex of G, ei1 , ..., eis the incoming edges at v, and ej1 , ..., ejtthe outgoing edges at v. Choose an Ok on each ek. Then multiplication byUi1 · · ·Uis is chain homotopic to multiplication by Uj1 · · ·Ujt .

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Proof. Choose an X1 ∈ X on the grid diagram. Let Oi1 , ..., Ois be the O’s thatlie on the same line of X1, and Oj1 , ...Ojt the O’s that lie on the same columnof X1.

DefineH1(x) =

∑y∈S(G)

∑r∈Rect◦(x,y)Int(r)∩X=X1

VO1(r)1 · · ·V Om(r)

m · y (5)

For grid states x and z, ψ ∈ π(x, z), let N(ψ) be the number of ways wecan decompose ψ as a composite of two empty rectangles r1 ∗ r2. Then

∂− ◦ H1 +H1 ◦ ∂− =∑

z∈S(G)

∑ψ∈π(x,z)ψ∩X=X1

VO1(r)1 · · ·V Om(r)

m · z (6)

As in the case for knots, when x\x∩z consists of 4 or 3 elements, N(ψ) = 2,hence they contribute nothing to the right of Equation 6 since we wwork inmod2 coefficient.

When x = z, ψ = r1 ∗ r2, in this case r1 and r2intersect along 2 edges andtherefore ψ is an annulus. Recall that our grid diagram contains an X in eachrow and column, therefore ψ is an annulus with height or width equal to 1,which contains X1.

If X1 is the row, then its contribution is multiplication by Vi1 · · ·Vis ; If X1

is the column, then its contribution is multiplication by Vj1 · · ·Vjt . So we have

∂− ◦ H1 +H1 ◦ ∂− = Vi1 · · ·Vis − Vj1 · · ·Vjt (7)

If X1 lies on the interior of an edge, or representing a vertex with exactlyone incoming edge and one outgoing edge, then there is exactly 1 Oi in the samerow as X1 and 1 Oj in the column with X1. The argument above shows thatmultiplication by Oi is chain homotopic to multiplication by Oj . Iterating thisprocedure proves (1). (2) follows immediately from (2) and the above argument.

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Suppose that G has n edges, and G has m O’s. Order the O’s so that thefirst n O’s lie on different edges. By Proposition 1 the multiplication by anyVi is chain homotopic to some Vj , j ≤ m. In particular, they induce the sameaction on the homology group.

Definition 1. The homology of GC−(G) is called the (unblocked) grid homologyof G, viewed as an F[U1, ...Um]-module.

Define the simply blocked grid chain complex to be

GC(G) = GC−(G)/V1 = · · · = Vn = 0 (8)

and the fully blocked grid complex

GC(G)/V1 = · · · = Vm = 0 (9)

The simply blocked grid homology and fully blocked grid homology are respec-tively the homology of these complexes. They are F-vector spaces.

Now we show that GC(G) is independent of the choice of the O1, ..., On.Denote by W the 2-dimensional F-space, with one generator in bigrading

(0, 0), and another generator in bigrading (−1,−1).

Lemma 2. Let G be a spatial diagram with m edges, and G a grid diagram ofG with m O’s. Then there is an isomorphism

GC(G) ∼= GH(G)⊗W⊗(m−n) (10)

Proof. We prove that

H

(GC−(G)

V1 = · · ·Vn+k = 0

)= H

(GC−(G)

V1 = · · ·Vn = 0

)⊗W k (11)

inductively for k ≥ 0.For any j > n, Oj lies on some edge ei, then by Prop.1 multiplication by Vj

is chain homotopic to multiplication by Vi, so

Vj :GC−(G)

V1 = · · ·Vj−1 = 0−→ GC−(G)

V1 = · · ·Vj−1 = 0(12)

is chain homotopic to 0 map. Hence the long exact sequence induced by

0 −→ GC−(G)

V1 = · · ·Vj−1 = 0

Vj−→ GC−(G)

V1 = · · ·Vj−1 = 0−→ GC−(G)

V1 = · · ·Vj = 0−→ 0

(13)becomes into a short exact sequence

0 −→ H

(GC−(G)

V1 = · · ·Vj−1 = 0

)−→ H

(GC−(G)

V1 = · · ·Vj = 0

)−→ H

(GC−(G)

V1 = · · ·Vj−1 = 0

)−→ 0

(14)

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The second arrow preserves the bigrading, and the third is homogeneous ofbidegree (1, 1). Therefore we have

H

(GC−(G)

V1 = · · ·Vj = 0

)= H

(GC−(G)

V1 = · · ·Vj−1 = 0

)⊗W (15)

Iterating this process proves the proposition.

Corollary 1. GC−(G) is a finite dimensional F-space. It is independent of thechoice of the Vi, i = 1, ..., n in the definition.

To show that GH(G), GH−(G) are invariants of the spatial graph G, wehave to prove that they are invariant under grid moves. The proof, however,has no essential difference from that in [3]. The only drawback of our homologygroups is that the usual combinatorial definition for Maslov grading cannot begeneralized to this case, so we confined ourselves with relative Maslov grading,which is easily shown to be invariant under grid moves.

Finally, the homology groups are invariant under orientation-reversing:

Proposition 2. The relatively graded F vector space GH(G) and the relativelygraded F[V1, ...Vn]-module GH−(G) are invariant if we change the orientationof each edge of G.

Proof. Denote by G′ the spatial graph obtained by reversing all the orientationson edges of G. Let G be a grid diagram of G. Reflecting G along some diagonal,we get a grid diagram G′ for G′. The reflection gives rise to a bijection fromS(G) to S(G′), and a bijection of rectangles that appear in the differential. Therelative Maslov grading and Alexander grading depends only on the position ofthe O and X’s, so is also preserved by the reflection.

4 The skein exact sequence

Let G be a spatial graph. By Lemma 1 ,there exists a preferred m×m griddiagram G for G. Suppose that v is a vertex of G, and denote by I(J) the set ofincoming(outgoing) edges at v. Require that |I| ≥ 2, |J | ≥ 2. We separate I, Jinto two disjoint nonempty sets: I = A ∪B, J = C ∪D. Write A = {a1, ..., ai},B = {b1, ...bj}, C = {c1, ...ck}, D = {d1, ..., dl}.

We define G+, G−,RG by modify G locally at v, see Figure 3 . For con-venience, we temporarily call G+(G−) and RG the two resolutions of G at v.Note that this differs from the usual meaning of ”resolution” at a crossing.

Suppose that the vertex v corresponds to Xv ∈ X of G, and let Oa1 , ..., Oai bethe O’s that is adjacent to Xv and lies on the edge a1, ...ai, respectively. WriteOa = {Oa1 , ..., Oai}, Va = V1 · · ·Vi. Similarly define Ob, Oc, Od, and Vb, Vc, Vd.

The main result of this section is the following:

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Figure 3: The resolutions at a vertex

Theorem 2. There are long exact sequences:

· · · −→ GH−(G+) −→ H∗

(GC−(G)

Va + Vb − Vc − Vd

)−→ GH−(RG) −→ GH−(G+) −→ · · ·

(16)

· · · −→ GH(G+) −→ H∗

(GC(G)

Va + Vb − Vc − Vd

)−→ GH(RG) −→ GH(G+) −→ · · ·

(17)

· · · −→ GH−(G−) −→ GH−(RG) −→ H∗

(GC−(G)

Va + Vb − Vc − Vd

)−→ GH−(G−) −→ · · ·

(18)

· · · −→ GH(G−) −→ GH(RG) −→ H∗

(GC(G)

Ua + Ub − Uc − Ud

)−→ GH(G−) −→ · · ·

(19)

Remark. When G+(G−) is a knot/link, the above proposition is just The-orem 4.1 of [10].

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Figure 4: Split the column(row) of Xv into two

Since G is a preferred grid diagram, the Oa, Ob are below A and B, andOc, Od are to the right of A,B. Define RXv

(CXv) to be the row(column) which X

lies on. Ignore X first, split RX into two rows, so that Oa and Ob are separated.Similarly, split CX into two columns, so that Oc and Od are separated(See Fig.4)

Let X0 be the m − 1 X’s in the complement. The original square of Xv

becomes a 2× 2 grid which lie at the intersection of the two new rows and thetwo new columns. Mark the upper -left and lower-right squares by A, and markthe upper-right and lower-leftsquares by B. We call these pairs of squares Aand B respectively. If we add an X on each square of A, we get a grid diagramGA for the spatial graph G+; If we add an X on each square of B, we get a griddiagram GB for the spatial graph R.

Denote by p the intersection point of the two A’s and the two B’s. Considerthe submodule Z of GC−(GB) which is generated by all states containing thecorner p. It is a subcomplex of GC−(GB): For any x ∈ Z,y ∈ GC−(GB),y /∈ Z,an r ∈ Rect(x,y) must contain some B ∈ B, which is not allowed in GC−(GB),since the two B’s have been decorated by X.

Lemma 3. Z is chain homotopic to GC−(G).

Proof. Construct P : GC−(G) −→ Z as follows.For x ∈ S(G), let x ∪ {p} be the state of S(GB) obtained by adding p to

x. This is a bijection from S(G) to the subset of S(GB) containing p. Letx,y ∈ S(G), r ∈ Rect◦(x,y), define P (r) ∈ Rect◦(x∪{p},y∪{p}) as follows: Ifr does not intersect with the interior of CXv and RXv , then r naturally inducesa rectangle P (r) from x∪{p} to y∪{p}; If r intersects with the interior of RXv

,the intersection is a 1× k rectangle, which splits into a 2× k rectangle in GB .Define P (r) to be the rectangle containing this 2× k rectangle. The definitionis similar when r intersects with the interior of CXv

(See Figure 5 ). It is easyto see r −→ P (r) gives a bijection from Rect◦(x,y) to Rect◦(x ∪ {p},y ∪ {p}),

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Figure 5: The correspondence of the rectangles in GC−(G) and Z

and

∂− ◦ P (x) = ∂−(x ∪ {p})

=∑

y∈S(GB)

∑r∈Rect◦(x∪{p},y∪{p})

r∩X0=∅

VO1(r)1 · · ·V Om+1(r)

m+1 y ∪ {p}

= P ◦ ∂−(x)

(20)

This shows that P is a chain complex isomorphism, which obviously pre-serves the ralative Maslov and Alexander grading.

Let Y be the quotient complex GC−(GB)/Z. As a module, Y is generated bystates not containing p, and the differential ∂Y is defined by counting rectanglesthat do not contain elements in X ∪A ∪ B. This shows that Y is a subcomplexof GC−(GA). We claim that the quotient complex

GC−(GA/Y ) ∼= Z (21)

As a module, GC−(GA)/Y is generated by all states that contain p, andthe differential is defined by counting rectangles that do not contain elemnts inX∪A∪B(Rectangles containing elements in B can be omited since their targetsare elements in Y ). This gives the isomorphism.

Define a map

ΦA : Z −→ Y

x −→∑

y∈S(GB)

∑r∈Rect◦(x,y)

r contains exactly 1 Br∩(X∪A)=∅

VO1(r)1 · · ·V Om+1(r)

m+1 y (22)

ΦA is a chain map: As in Prop.1, for ψ ∈ π(x,y), let N(ψ) be the number ofways we can decompose ψ as the composition of 2 empty rectangles, we have

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(∂Y ◦ ΦA + ΦA ◦ ∂Z) (x) =∑

y∈S(GB)

∑ψ∈Rect◦(x,y)

ψ contains exactly 1 Bψ∩(X∪A)=∅

VO1(ψ)1 · · ·V Om+1(ψ)

m+1 y

(23)As in the proof of Prop.1 , when x\x∩y consists of 4 or 3 elements, N(ψ) = 2,

so they contribute nothing to the right of Equation (23). However, this timethere is no longer the x = z case: an annulus containing B must contains an A,which is not allowed. Therefore the right-hand side of is 0 and ΦA commuteswith the differential(Since we work in mod2 coefficient).

Similarly, we can define

ΦB : Y −→ Z

y −→∑

z∈S(GB)

∑r∈Rect◦(y,z)

r contains exactly 1 Ar∩(X∪B)=∅

VO1(r)1 · · ·V Om+1(r)

m+1 z (24)

and prove that ΦB is a chain map as above.

Lemma 4. The composition ΦA ◦ ΦB,ΦB ◦ ΦA are equal to multiplication byVa + Vb − Vc − Vd.

Proof. ΦA ◦ ΦB counts only the four annuli that contain p on their boundary.The vertical annuli contribute Va and Vb, while the vertical ones contribute −Vcand −Vd. The same is true for ΦB ◦ ΦA.

Remark. Since we are working in F-coefficients, the signs appearing aboveare irrevalent. They are suggested by the sign conventions in the Z-coefficientversion.

Lemma 5. The mapping cone C(ΦA) is chain isomorphic to GC−(GB), andthe mapping cone C(ΦB) of ΦB is chain isomorphic to GC−(GB).

Proof. By definition, the underlying module of C(ΦB) is Y ⊕X, and the differ-ential

Y ⊕X −→ Y ⊕X(y, x) −→ (∂Y (y),ΦA(y) + ∂X(x))

(25)

On the other hand, the differential ∂ of GC−(GB) is

Y ⊕X −→ Y ⊕X(y, x) −→ ∂GB

(y) + ∂GB(x)

(26)

where ∂GB(x) = ∂X(x), ∂GB

(y) = ∂Y (y) + ΦA(y), therefore

∂GB(y) + ∂GB

(x) = ∂X(x) + ∂Y (y) + ΦA(y) = (∂Y (y),ΦA(y) + ∂X(x)) (27)

The proof of the latter half statement is similar.

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Figure 6: Resolutions of G and G′

Recall the following properties on mapping cone:

Proposition 3. (1) If f : C −→ C ′ is an injective chain map, then there is a

quasi-isomorphism φ : Cone(f) −→ C′

f(C) ;

(2) Suppose that f : C −→ C ′, g : C ′ −→ C” are two chain maps, then thereis a long exxact sequence:

· · · −→ H(Cone(g)) −→ H(Cone(f)) −→ H(Cone(g ◦ f)) −→ H(Cone(g)) · · ·(28)

Proof of Theorem 2 : By Prop.3(1), the mapping cone of multiplicationby Va + Vb − Vc − Vd : X −→ X is a quasi-isomorphic to X

Va+Vb−Vc−Vd. Then

Equation 16 follows from Lemma 3,4 5 and Prop.3(2). Choose a set {Ve}, onefor each edge e, Equation (2) follows by letting these Ve = 0.

It remains to prove the part for G−. Indeed, this follows directly from (1)(2)by the following observation: If we operate an RIV move on G, we get a spatialgraph G′(equivalent to G). The two resolutions G′+, RG′ of G′ are respectivelyRG and G−(See Fig.6).

References

[1] Yuanyuan Bao. Floer homology and embedded bipartite graphs, 2018. 1

12

[2] Nathan Dowlin. Knot floer homology and khovanov–rozansky homologyfor singular links. Algebraic and Geometric Topology, 18(7):3839–3885, Dec2018. 1

[3] Shelly Harvey and Danielle O’Donnol. Heegaard floer homology of spa-tial graphs. Algebraic & Geometric Topology, 17(3):1445–1525, Jul 2017.(document), 1, 2, 2, 2, 3.1

[4] Ciprian Manolescu. An untwisted cube of resolutions for knot Floer ho-mology. Quantum Topol., 5(2):185–223, 2014. 1

[5] Ciprian Manolescu, Peter S. Ozsvath, and Sucharit Sarkar. A combinatorialdescription of knot Floer homology. Ann. Math. (2), 169(2):633–660, 2009.

[6] Ciprian Manolescu, Peter Ozsvath, Zoltan Szabo, and Dylan P Thurston.On combinatorial link floer homology. Geometry and Topology,11(4):2339–2412, Dec 2007.

[7] Peter S. Ozsvath, Andras I. Stipsicz, and Zoltan Szabo. Grid homologyfor knots and links, volume 208 of Mathematical Surveys and Monographs.American Mathematical Society, Providence, RI, 2015.

[8] Peter Ozsvath, Andras Stipsicz, and Zoltan Szabo. Floer homology andsingular knots. Journal of Topology, 2007. 1

[9] Peter Ozsvath and Zoltan Szabo. Holomorphic disks and knot invariants.Advances in Mathematics, 186(1):58–116, 2004. 1

[10] Peter Ozsvath and Zoltan Szabo. A cube of resolutions for knot floerhomology. Journal of Topology, 2(4):865–910, 2009. (document), 1, 4

[11] Jacob Andrew Rasmussen. Floer homology and knot complements. Pro-Quest LLC, Ann Arbor, MI, 2003. Thesis (Ph.D.)–Harvard University.1

[12] zz. chishi. fgmh.

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