Combinatorial Proofs of Properties of Double-Point Enhanced Grid Homology

We provide a purely combinatorial proof of a skein exact sequence obeyed by double-point enhanced grid homology. We also extend the theory to coefficients over $\mathbb{Z},$ and discuss alternatives to the Ozsv\'ath-Szab\'o $\tau$ invariant.

Theorem 5.1.For each grid diagram of a knot, there exists a homology group GHL S (G; Z), which, as a bigraded Z[U, v]-module, is also a knot invariant.Furthermore, GHL S (G; Z) is the homology of a chain complex, which when we take the homology of the mod 2 version gives us the double-point enhanced grid homology GHL(G).
The subscript S in this notation refers to a sign assignment, a function on certain rectangles in G .We will initially define GHL S (G; Z) in terms of one such function S and then prove it is invariant under change in S.
The second theorem we prove shows that this homology GHL S (G; Z), which we call integral double-point enhanced grid homology, obeys a skein exact sequence.We extend GHL to a link invariant cGHL m (L, a).Omitting the subscript S and Z from the notation, we may state the theorem: Theorem 6.2.Let (L + , L − , L 0 ) be an oriented skein triple, with ℓ and ℓ 0 the number of components of L + and L 0 respectively.If ℓ 0 = ℓ + 1, then there is a long exact sequence where the maps below fit together to be homomorphisms of Z[U, v]-modules: Let J be the 4-dimensional bigraded abelian group J ∼ = Z 4 with one generator in bigrading (0, 1), one generator in bigrading (−2, −1), and two generators in bigrading (−1, 0).
If ℓ 0 = ℓ − 1, then there is a long exact sequence where the maps below fit together to be homomorphisms of Z[U, v]-modules: Finally, for the last two sections we return to F coefficients for simplicity.Section 7 presents some more concrete invariants that can be extracted out of double-point enhanced grid homology, and their potential use in proving that the two invariants do not encode different information.Section 8 computes the double-point enhanced grid homology of alternating knots and torus knots over the field of 2 elements using a spectral sequence, and shows that in these cases the conjecture that GH − (K)[v] ∼ = GHL(K) holds with one caveat.The spectral sequence loses information about the v action, hence these two below theorems are stated under a weaker-than-ideal condition.
Remark 0.1.The family of quasi-alternating knots is a family that contains the alternating knots but is strictly larger (see [OSS Chapter 10].) Theorem 8.2.If K is a torus knot, then GHL(K) ∼ = GH − (K) [v] as bigraded F[U ]modules.
1 Acknowledgements I would like to thank Peter Ozsváth for his advice and support throughout the project, as well as the idea for the project itself.Additionally, I would like to thank Zoltán Szabó, Isabella Khan, Matthew Kendall, Luya Wang, and Joshua Wang -who in particular noticed that Theorem 8.1 holds for quasi-alternating knots rather than only for alternating knots -for helpful correspondences.

Background on Grid Homology
We begin with a brief summary of grid homology, a knot invariant defined in [OSS].For this section, fix a knot K.A grid diagram is an n-by-n grid of squares such that there is one X and one O in each row and each column.We may retrieve a link from a grid diagram as follows.For each row and each column of squares, draw a line segment connecting the X to the O within that row or column respectively.Specify further that whenever two such segments intersect, the vertical segment crosses over the horizontal segments.Then, it is clear that the union of all these segments is a planar link diagram.[OSS Theorem 3.1.3]guarantees that there exists a grid diagram representing any link, in particular K. Let G be such a diagram.
Denote by X the set of all the center points of the X-marked squares, and likewise denote by O the set of all the center points of the O-marked squares.
We consider G to be a fundamental domain of a torus T constructed by gluing opposite sides of G .It is clear that different fundamental domains of this torus represent isotopic links.Fix a coordinate system on a fundamental domain corresponding to the cardinal directions North, South, East, and West.
Call the horizontal circles formed by the edges of the grid of squares as α 1 , . . ., α n , moving further north, and the vertical circles formed by the edges of the grid of squares as β 1 , . . ., β n , moving further east.
Definition 2.1.A grid state x of G is a set of n points on T such that | x ∩α i | = 1 for all i ∈ {1, . . ., n} and | x ∩β i | = 1 for all i ∈ {1, . . ., n}.(In other words, x is a set of n intersection points of the α and β curves, such that each curve is represented once.) We denote the set of all grid states of G by S(G).
Definition 2.2.A rectangle in G . is an embedding of the closed unit disk D 2 into G such that ∂D 2 gets mapped into the union of the α and β curves.Let Then A is a 1-manifold with boundary consisting of two line segments, and has an orientation induced from r by moving clockwise around the boundary ∂r.
. We say the rectangle r connects two grid states x, y ∈ S(G) (or goes from x to y) if ∂A = y − x, and such that the points in y ∩∂A inherit a positive orientation from A.
The set of all rectangles from x to y is denoted by Rect(x, y).
We wish to create two functions M and A from S(G) to Z, which we define as follows.
Definition 2.3.Let P and Q be sets of finitely many points in a fundamental domain for G, which we may embed in R 2 with standard Cartesian coordinates as the rectangle [0, n)×[0, n) such that each square in G is a unit square with integral coordinates for its corners.Then, we define I(P, Q) to be the number of pairs of points (p 1 , p 2 )×(q 1 , q 2 ) ∈ P × Q satisfying p 1 < q 1 and p 2 < q 2 .Now, let Then, we let M (x) := J (x, x) − 2J (x, O) + J (O, O) + 1, M X (x) := J (x, x) − 2J (x, X) + J (X, X) + 1, and We call M (x) the Maslov grading and A(x) the Alexander grading, and we call the pair (M, A) the bigrading of x .
The below proposition is greatly helpful to our future ventures: Proposition 2.1.
• Both M and A are integral-valued functions (note it is only clear from their definitions that they are half-integral valued.)Suppose x and y are two grid states with some rectangle r ∈ Rect(x, y).Then, their Maslov and Alexander gradings are related by the following formulas: The proof is found in [OSS Section 4.3], and is elementary but rather long.Remark 2.1.For the remainder of this paper, let F represent the field of 2 elements Z/2Z.Definition 2.4.Let a domain ψ in G be any formal Z-linear combination of squares in G (which may be defined as the closures of the connected components of Again, the boundary of a domain inherits a clockwise orientation.We say a domain connects two grid states x and z in S(G) if with the z-portions inheriting a positive orientation and the x-portions a negative orientation.Let the set of all domains from x to z be denoted as π(x, z).
Let the multiplicity of the domain ψ at the point p ∈ G be the coefficient of the square containing p in the expression for ψ as a combination of squares (we will only be considering multiplicities at points p not on the boundary of any square.)Let ψ(p) be the multiplicity of ψ at the point p.
We say a domain is decomposed as a juxtaposition of two rectangles r 1 ∈ Rect(x, y) and r 2 ∈ Rect(y, z), and write ψ = r 1 * r 2 , if ψ ∈ π(x, z) and the multiplicities satisfy the following formula: for all p ∈ G.
We are now ready to define grid homology.
Definition 2.5.Let G be a grid diagram.We define the chain complex GC − (G) to be the free [ OSS Chapter 4] demonstrates that ∂ 0 is indeed a differential, that is it is a homogeneous map of bidegree (−1, 0) and ∂ 2 0 = 0, hence the homology of the chain complex (GC − (G), ∂ 0 ) is well-defined.We denote this homology as GH − (G).[OSS Chapter 5] proves that the action of each V i is identical on the level homology, and calling this action by U, that GH − (G) is in fact an invariant of the knot K as a bigraded F[U ]-module; we may write it as GH − (K).
[ OSS Chapter 5] proves that this homology is also a knot invariant GH − (G) is in fact an invariant of the knot K as a bigraded F-vector space.

Grid Moves
We would like a way to relate any two grid representations of the same (oriented) link.We define three types of grid moves: commutations, switches, and stabilizations.
Definition 2.7.Consider two adjacent rows (resp.columns) of a grid diagram in a fundamental domain, and draw the (closed) line segments joining the X-and Omarkings in each row (resp.column).Project these two line segments onto the horizontal (resp.vertical) axis.If either (1) the two projected line segments have disjoint supports or (2) one of the projected line segments completely contains the other, then swapping the two adjacent rows (resp.columns) is called a commutation.
If the two projected line segments share a vertex, then then swapping the two adjacent rows (resp.columns) is called a switch.
Here, we picture a commutation: Definition 2.8.Consider a square marked with an X (resp.an O).Choose one of the following four directions, N E, N W, SE, SW.Subdivide the row and column containing the X (resp.O) so that G is now an n+1-by-n+1 grid diagram, and the square formerly containing the X (resp.O) is now a 2-by-2 grid.Replace the X (resp.O) with two X's in the diagonal of this 2-by-2 grid that does not contain the chosen direction, and an O (resp.X) in a third square of this 2-by-2 grid such that the unmarked square is the one corresponding to the chosen direction.This operation is known as a stabilization of type X : direction (resp.O : direction).Its inverse is known as a destabilization of the corresponding type.
Here, we picture a stabilization of type X : SW : The following theorem will prove extremely useful in showing invariance of doublepoint enhanced grid homology: Theorem 2.1 (generalized from Cromwell, see [OSS Corollary 3.2.3]).Any two grid diagrams of the same (oriented) knot are related by a finite sequence of commutations, switches, and stabilizations and destabilizations of the form X : SW.
We define a differential where ∂ n is defined on grid states by: and extends by linearity.
We shall prove a more general version of this proposition later, see 5.1.For now, we take this for granted, and let GHL(G) denote the homology of the chain complex (GCL(G), ∂).

Ollie Thakar
September 24, 2022 For any toroidal grid diagram on a torus T, we may consider the universal cover of the torus, which we identify with R 2 and its standard (x, y) Cartesian coordinates.Here, lifts of the α i and β j curves, which we may call αi , βj respectively, are the straight lines y = n and x = m as n, m range over Z. Definition 3.2.Consider a rectangle R of width one in R 2 whose sides lie along the αi , βj lines, and such that the projection of R onto T, which we call r, has multiplicity 2 in at least one point, and multiplicity 1 in at least one point.Then, we call r a long rectangle.
Note that r is a domain in G, and it connects grid states analogously to ordinary rectangles.We denote by Rect * (x, y) the set of rectangles and long rectangles from grid state x ∈ S(G) to y ∈ S(G).Definition 3.3.We define the function T : Rect * (x, y) → Z ≥0 as follows.Let r ∈ Rect * (x, y).If r is long, then T (r) = 1, and if r is not long, then Remark 3.1.Note that we can now rewrite the differential ∂ more compactly as Definition 3.4.Let x, z ∈ S(G), and suppose ψ ∈ π(x, z) is a domain, with decomposition ψ = r 1 * r 2 for r 1 ∈ Rect * (x, y) and r 2 ∈ Rect * (y, z).The degree of the decomposition, which we will notate as deg(r 1 , r 2 ), is defined as the sum: Definition 3.5.For a rectangle or long rectangle r ∈ Rect * (x, y), the incoming corners are precisely the members of ∂r∩x, and the outgoing corners are precisely the members of ∂r ∩ y .

For Commutation/Switch Invariance
To prove invariance of grid homology under commutation and switch moves, and also to prove the skein exact sequence, we will require superimposing two grid diagrams G and G ′ differing by a commutation or switch as pictured in the below picture of the relevant portion of this superimposed diagram: We call β i the curved circle belonging to G and γ i the curved circle belonging to G ′ .Let a and b be the two intersections of β i and γ i , with a at the southern end of the bigon containing β i as its western boundary.
Remark 3.2.Very importantly, we may always assume that each bigon contains at least one X-marking in it.We will be counting regions that are forbidden from intersecting X-markings, hence this assumption will markedly simplify our below analysis.
Definition 3.6.(Modified from [OSS] Definition 5.1.1).A pentagon from x ∈ S(G) to y ′ ∈ S(G ′ ) is an embedded disk p in the torus whose boundary is the union of five arcs, each of which lies on an α j , β j , or γ i curve, such that: (1) four of the corners of p are in x ∪ y ′ , (2) at each corner x of p, exactly one of the four quadrants of a small disk surrounding x has multiplicity 1 and the other 3 have multiplicity 0, and (3) Let P be an embedded disk in the universal cover of the torus satisfying conditions (1), (2), and (3), satisfying two extra conditions: (4) that P has width one, and (5), that the projection of P onto the torus, which we call p, has multiplicity 2 in at least one point, and multiplicity 1 in at least one point.Then, we call p a long pentagon from x to y ′ .
Let Pent(x, y ′ ) denote the set of pentagons from x to y ′ , and Pent * (x, y ′ ) denote the set of pentagons and long pentagons from x to y ′ .Definition 3.7.(Modified from [OSS] Definition 5.1.5).A hexagon from x ∈ S(G) to y ∈ S(G) is an embedded disk h in the torus whose boundary is the union of six arcs, each of which lies on an α j , β j , or γ i curve, such that: (1) four of the corners of h are in x ∪ y ′ , and the other two corners are at a and b, (2) at each corner x of h, exactly one of the four quadrants of a small disk surrounding x has multiplicity 1 and the other 3 have multiplicity 0, and (3) Let H be an embedded disk in the universal cover of the torus satisfying conditions (1), (2), and (3), satisfying two extra conditions: (4) that P has width one, and (5), that the projection of H onto the torus, which we call h, has multiplicity 2 in at least one point, and multiplicity 1 in at least one point.Then, we call h a long pentagon from x to y .

Rectangle Decomposition Lemmas
This section contains many useful combinatorial lemmas that will expedite the proofs of the later theorems tremendously.
We will set some consistent notation throughout this section.Fix a grid diagram G .Let x, z ∈ S(G) and let ψ ∈ π(x, z) be a fixed domain.
Observe that if ψ admits at least one decomposition ψ = r 1 * r 2 where r 1 and r 2 are either rectangles or long rectangles, then we must have | x − x ∩ z | = 0, 3, or 4, simply because the initial and final grid states of each rectangle differ by exactly 2 points.We codify this useful fact in the below lemma: Lemma 4.1.Suppose that there exists y ∈ S(G) such that ψ admits at least one decomposition ψ = r 1 * r 2 where r 1 ∈ Rect * (x, y) and r 2 ∈ Rect * (y, z).Then, Suppose that there exists y ∈ S(G) such that ψ admits at least one decomposition ψ = r 1 * r 2 where r 1 ∈ Rect * (x, y) and r 2 ∈ Rect * (y, z).Suppose that r 1 , r 2 are not both long.Then, ψ admits precisely two decompositions ψ = r 1 * r 2 = r ′ 1 * r ′ 2 , such that there exists y ′ ∈ S(G) with r 1 ∈ Rect * (x, y ′ ) and r ′ 2 ∈ Rect * (y ′ , z), and r ′ 1 , r ′ 2 are not both long.Moreover, these two decompositions have the same degree.
Proof.Lift the decomposition ψ = r 1 * r 2 into the universal cover so that r 1 and r 2 are represented by connected polygons.Because the grid states contain precisely one point in each horizontal and vertical circle, we must have that the circles containing the edges of r 1 and r 2 are all different.Hence, it is clear that the only possible corners of x that can be the outgoing corners of any decomposition ψ = r ′ 1 * r ′ 2 are the outgoing corners of r 1 are the outgoing corners of r 2 .Hence, there are clearly precisely 2 decompositions of ψ as a composite of two polygons ψ = r 1 * r 2 = r ′ 1 * r ′ 2 , with r 1 ∈ Rect * (x, y) and r ′ * 1 ∈ Rect(x, y ′ ).Furthermore, r 1 and r ′ 2 share the same support, as do r ′ 1 and r 2 .
Hence, the T terms of the degrees of both decompositions agree.Furthermore, r ′ 1 and r ′ 2 are clearly not both long.Let k = deg(r 1 , r 2 ).We wish to show deg(r ′ 1 , r ′ 2 ) = k.For i = 1, 2, let C(r i ) be the number of corners of r 3−i discounting points of β i ∩ γ i intersecting Int(r i ); suppose without loss of generality that C(r 1 ) ≥ C(r 2 ).Then, there are four cases here: the first is that C(r 1 ) = 0.In this case, clearly x, y and x, y ′ are only different in 2 places.)The remaining cases have C(r 1 ) > 0, so r 1 is not thin, therefore not long.
The second case is that C(r 1 ) = 1; in this case where ψ ∩ (x ∩ z) is counted with multiplicity.The third case is that Int(r 1 ) contains exactly 2 corners of r 2 ; again in this case where ψ ∩ (x ∩ z) is counted with multiplicity.Finally, we could have that Int(r 1 ) contains all 4 corners of r 2 ; in this case That suffices for the proof.
Suppose that in the support of ψ, an entire row or column may not have multiplicity ≥ 2, and that at most one entire row or at most one entire column may have multiplicity 1. (These conditions are achieved, for instance, when only 1 of the two rectangles in the decomposition is allowed to be long.)Suppose that there exists y ∈ S(G) such that ψ admits at least one decomposition ψ = r 1 * r 2 where r 1 ∈ Rect * (x, y) and r 2 ∈ Rect * (y, z).Suppose that r 1 , r 2 are not both long.Then, ψ admits precisely two decompositions ψ = r 1 * r 2 = r ′ 1 * r ′ 2 , such that there exists y ′ ∈ S(G) with r 1 ∈ Rect * (x, y ′ ) and r ′ 2 ∈ Rect * (y ′ , z), and r ′ 1 , r ′ 2 are not both long.Moreover, these two decompositions have the same degree.
Proof.Consider a lift of ψ to the universal cover of the torus such that ψ is represented by a connected L-shaped polygon Q.Then, ψ = r 1 * r 2 , where r 1 and r 2 are represented by rectangles in the universal cover with disjoint interiors (which may not be disjoint when we project back down to the torus.) Since | x −(x ∩ z)| = 3, the two rectangles r 1 and r 2 must share a corner c.Since this corner must be incoming for r 1 and outgoing for r 2 , then the two rectangles must create a 180-degree angle at this corner, and hence their intersection r 1 ∩r 2 is an edge e.The boundary ∂e is thus two points, c and another point, which we shall call d.Clearly, there exists a 270-degree angle at d.In any decomposition of ψ, there cannot be a 270degree angle.Since there are precisely two ways to cut Q at this angle, and each one uniquely specifies a decomposition, then we get ψ has precisely two decompositions , and r ′ 1 , r ′ 2 are not both long by the conditions on the support of ψ.
We must show these two decompositions have the same degree.First, note that any point of x, y, or y ′ inside of Int(Q) must not lie on ∂(r 1 ) ∪ ∂(r 2 ), since this would contradict the fact that grid states contain only 1 point on each horizontal or vertical circle.If Q can embed into a fundamental domain of the torus, then T = 0 for all rectangles in all decompositions, and the local multiplicities are ≤ 1.Thus, so the degrees are the same.
Suppose Q cannot embed into a fundamental domain.Then, by the multiplicity constraints, we must have that one of the two decompositions involves a long rectangle t ∈ Long(x, y) or Long(y, z) for some grid state y, and some other rectangle r ∈ Rect(x, y) or Rect(y, z) such that r ∪ t embeds in a fundamental domain if we delete the annulus contained by t.Suppose the intermediate stage in the other decomposition, 2 is long, then clearly no corner of one can lie in the interior of the other, hence the degrees are clearly the same. Otherwise, We see geometrically that Int(r ′ 1 ) ∩ x must contain precisely one point c of x (a corner of r ′ 2 ) that is not contained in Int(r) or Int(t), since it lies on the edges of both such rectangles.Since t, and hence lie in Int(r) and Int(r 1 ).The degree contribution of the point c in the decomposition r ′ 1 * r ′ 2 is exactly canceled out by the contribution of T (t) in the decomposition r * t.That suffices for the proof.
) Suppose that in the support of ψ, an entire row or column may not have multiplicity ≥ 2, and that at most one entire row or at most one entire column may have multiplicity 1.
Suppose that there exists y ∈ S(G) such that ψ admits at least one decomposition ψ = r 1 * r 2 where r 1 ∈ Rect * (x, y) and r 2 ∈ Rect * (y, x).Then, this decomposition is unique and ψ is an annulus (either horizontal or vertical) of width 1 (such that the multiplicity of ψ in each square is ≤ 1.) Proof.Lift ψ to a connected polygon Q in the universal cover, such that r 1 and r 2 are represented by rectangles R 1 , R 2 in the universal cover projecting onto r 1 , r 2 in the torus.Then, the condition that each grid state must only contain one point in each row or column forces R 1 and R 2 to share an edge.Hence, Q is a rectangle; since all the corners of Q must be members of x, the condition that each grid state must only contain one point in each row or column forces Q to be an annulus.It must have multiplicity 1 and have width one by the conditions on the support and multiplicities of ψ.
Furthermore, such annulus has a unique decomposition since the first rectangle r 1 must have outgoing corners precisely x ∩clo(Q), which is two points; since r 1 is not long, this determines r 1 uniquely.This in turn determines r 2 uniquely.

Pentagon and Hexagon Decomposition Lemmas
We now suppose that G and G ′ are two grid diagrams which have been superimposed as in the previous section.Let x ∈ S(G) and z ∈ S(G ′ ) or S(G), and ψ ∈ π(x, z).
Remark 4.1.A very important warning is that for this entire subsection, we assume that our regions have empty intersection with X .By Remark 3.2, this means that our regions may never contain an entire bigon in their support.This is not strictly necessary for most of the proofs below, but speeds up the arguments nicely.
Definition 4.1.The closest point map I : S(G) → S(G ′ ) is defined by letting I(x) be the grid state in G ′ which matches x in all but one point: the point α j ∩ β i ∈ x is replaced by the point α j ∩ γ i ∈ I(x).
We record here the following useful lemma: Lemma 4.5.For a grid state x ∈ S(G), we have that M (x)− M (I(x)) = −1 + 2|t∩O|, where t is the unique triangular subset of one of the bigons bounded by β i and γ i with two corners in common with Proof.This is demonstrated in the proof of [OSS] Lemma 5.1.3.Definition 4.2.Consider a (possibly long) pentagon or hexagon ψ ∈ π(x, y), where x ∈ S(G) and y ∈ S(G) ∪ S(G ′ ).By slight abuse of notation, let Then, the associated associated rectangular domain Ψ ∈ Rect * (x, I(y)) of ψ is the (possibly long) rectangle from x to I(y) whose multiplicities are identical to those of ψ outside the bigons between β i and γ i .
Denote by R(p) the associated rectangular domain of a (possibly long) pentagon or hexagon p.
Note that Ψ is long if and only if ψ is long, and that T (Ψ) = T (ψ).The following lemma is immediate from the above definition: Lemma 4.6.Let ψ ∈ π(x, z) and suppose that ψ ∩X = ∅.A decomposition ψ = p 1 * p 2 , where p 1 , p 2 are either rectangles, pentagons, or hexagons, and not both long, corresponds to a decomposition of the associated rectangular domain Ψ into two rectangles, not both long.Furthermore, the decompositions of ψ and Ψ have the same degree.
Corollary 4.1.Let ψ ∈ π(x, z) and suppose that ψ ∩ X = ∅.Suppose | x − x ∩ z | = 3 or 4. Suppose ψ admits a decomposition ψ = p 1 * p 2 , where p 1 , p 2 are either rectangles, pentagons, or hexagons, and not both long.Then ψ admits two decompositions as such, and both have the same degree.Furthermore, if ψ = p * r or r * p where p is a long pentagon and r is a (not long) rectangle, then the other decomposition of ψ is also as a long pentagon and a not long rectangle.
Proof.If | x − x ∩ z | = 4, then only one of p 1 or p 2 can have an edge on the curves β i or γ i .Furthermore, the associated rectangular domain Ψ also connects two grid states differing by 4 points.Hence, Lemma 4.2 tells us Ψ admits two rectangular decompositions of the same degree.Since at most one of p 1 or p 2 can have an edge on the curves β i or γ i , clearly each of the two decompositions of Ψ corresponds uniquely to a decomposition of ψ.
If | x − x ∩ z | = 3, the associated rectangular domain Ψ also connects two grid states differing by 3 points.Hence, Lemma 4.2 tells us Ψ admits two rectangular decompositions of the same degree.There are two cases: either, a point a or b at which p 1 or p 2 has a corner is on an edge shared by both rectangles in one of the decompositions of Ψ, or it is not.In the latter case, each of the two decompositions of Ψ corresponds uniquely to a decomposition of ψ.In the former case, one of the two decompositions of Ψ does not correspond to a decomposition of ψ, however we achieve precisely one more decomposition of ψ by removing a portion of one of the bigons from the support of one of p 1 or p 2 and appending it to the other, which is clearly possible since the support of a pentagon or hexagon cannot contain an entire bigon.
The last claim in the above corollary follows simply because any long rectangle in a decomposition of ψ would clearly have to intersect X (here is one instance where assuming ψ ∩ X = ∅ drastically simplifies our argument.)

Sign Assignments
The goal of this section is to define double-point grid homology over the integers, and verify that this is indeed a knot invariant.
Remark 5.1.Note that if r 1 and r 2 are two rectangles with r 1 ∈ Rect(x, y) and r 2 ∈ Rect(z, w), it is possible that r 1 and r 2 have the same support but S(r 1 ) = S(r 2 ).It is therefore important to emphasize that S is a function of the domain, initial, and final grid states.Note further that by the definition of our sign assignments, the proofs of lemmas 4.2 and 4.3 extend to prove the following slightly stronger statement: Lemma 5.1.Under the assumptions of lemma 4.2 or 4.3, the two decompositions of ψ resulting from those lemmas have opposite signs.
Given a sign assignment, we may define a new chain complex over the integers.Let GCL(G; Z) be the free Z[V 1 , . . ., V n , v]-module generated by the grid states of G,

Ollie Thakar
September 24, 2022 with mutliplication by each V i homogeneous of bigrading (−2, −1), and multiplication by v homogeneous of bigrading (2, 0).Define the differential as follows on grid states, extending by linearity: Proposition 5.1.∂ 2 S = 0 for any sign assignment S. Proof.By expanding out definitions, we see Hence, it is sufficient to show that domains ψ ∈ π(x, z) with |D(ψ)| > 0 have exactly two decompositions d ∈ D(ψ) with equal degree and opposite signs.
If The last case is x = z .Lemma 4.4 tells us that in this case, ψ is an annulus, hence ψ ∩ X = ∅, so this also contributes 0 to the equation.
Let GHL S (G; Z) be the homology of this chain complex.We will eventually see that the V i are homotopic to each other, hence calling the induced multiplication U, We wish to prove the following theorem, which is an analog of the invariance of ordinary grid homology over the integers: Theorem 5.1.For each grid diagram, there exists a sign assignment; furthermore, all sign assignments produce isomorphic homology GHL S (G; Z).This homology GHL S (G; Z), as a bigraded Z[U, v]-module, is also a knot invariant.
We divide the proof of this theorem into several steps.First, we lift existence and uniqueness of sign assignments for grid homology to our situation, which is rather simple: Lemma 5.2.For each grid diagram, there exists a sign assignment; furthermore, all sign assignments produce isomorphic homology GHL S (G; Z).
Proof.The existence of a sign assignment follows immediately from [OSS] Theorem 15.1.5.Their proof of this theorem also shows that for any two sign assignments S 1 and S 2 , there exists a function g : S(G) → {−1, 1} such that S 2 (r) = g(x)S 1 (r)g(y) for each r ∈ Rect(x, y).Hence, we may define a x .It is immediate that this is a chain map and an isomorphism of Z[V 1 , . . ., V n , v]-modules, and this suffices for the proof.(This is an analog of the map defined in [OSS] Proposition 15.1.10.) Next, we show that the V i are homotopic.For this, we require a lemma about extended sign assignments, and some definitions.For any long rectangle r ∈ Rect * (x, y), deleting an annulus leaves a rectangle r ′ ∈ Rect * (x, y) which we will call the associated short rectangle.
Lemma 5.3.Each sign assignment S : Rect(G) → {−1, 1} may be extended to an extended sign assignment, such that if r ∈ Rect * (x, y) is long and r ′ ∈ Rect * (x, y) is its associated short rectangle then S(r) = S(r ′ ).
Proof.We use the notation of [OSS Chapter 15].Let Sn denote the spin extension of the symmetric group on n letters (see [OSS Section 15.2]), and let T denote the set of lifts of transpositions in S n ; these are indexed by ordered pairs of distinct integers in {1, . . ., n} and denoted by τi,j for i = j.Here, τi,j and τj,i are the two lifts of the permutation (ij) ∈ S n .Also, z = 1 ∈ Sn is the other lift of the identity 1 ∈ S n .
Let τ : Rect * (G) → T be the map sending r ∈ Rect * (G) to τi,j where the southwest corner of the associated short rectangle r ′ is on β i and the northeast corner of r is on β i .We wish to show the following three conditions hold: • If r 1 * r 2 forms a horizontal annulus of multiplicity 1, then: • If r 1 * r 2 forms a vertical annulus of multiplicity 1, then: Assuming for now that these three conditions hold, we will demonstrate the existence of the extension of our sign assignment S. For each grid state x, let σ x ∈ S n be the corresponding permutation which is determined uniquely by α i ∩ β σx(i) ∈ x .
[OSS Section 15.2] proves that every sign assignment, in particular S, is given, for r ∈ Rect(x, y), as S(r) = τ (r) −1 γ(σ −1 x )γ(σ y ), for some section of the spin extension γ : S n → Sn .Furthermore, the proof of [OSS Proposition 15.2.12] tells us that if S is extended to long rectangles by the same formula, S(r) = τ (r) −1 γ(σ −1 x )γ(σ y ), then S will satisfy the conditions of an extended sign assignment if τ obeys the three above conditions.
Thus, it is sufficient to show that τ obeys the three above conditions.The second and third are immediate from [OSS Section 15.2] since they do not apply to long rectangles.The first condition is immediate in the case when both decompositions of ψ involve one long rectangle, since deleting an annulus from ψ and in turn from each of the long rectangles does not change the map τ and this condition now follows from the equivalent condition for not long rectangles.
Finally, we must show this condition is true when one decomposition, r 1 * r 2 , of ψ involves a long rectangle and the other, r ′ 1 * r ′ 2 , does not.Say that r 1 is long and r ′ 1 has width one.This reduces to eight cases, corresponding to whether r 1 horizontal or vertical, and whether the multiplicity 2 portion of ψ is in the Southwest, Southeast, Northwest, or Northeast corner of r ′ 2 .For each, it is a simple computation involving the relations among the members of T .
Definition 5.3.We define the sign-refined homotopy operator as follows.Let X i ∈ X share a row with O i .Then, define the function: Proposition 5.2.Suppose X i shares a column with O j .Then following equation holds: Proof.By expanding out definitions, we see the left-hand side is given as The last case is x = z .Lemma 4.4 tells us that in this case, ψ is an annulus, hence ψ ∩ X = X i has precisely two solutions, a horizontal thin annulus and a vertical thin annulus.(The multiplicities of these annuli must be 1 since X i (ψ) = 1.)The horizontal thin annulus contributes V i and the vertical thin annulus contributes −V j .
Proposition 5.3.For any k, l, we have V k and V l are chain homotopic.
Proof.The above proposition shows that V i and V j are chain homotopic whenever O i and O j are consecutive O-markings as we traverse the knot.Since the knot has one component, each O k and O l are both members of a finite sequence of consecutive O-markings, which suffices for the proof.
The remainder of the proof of invariance of the sign-refined homology GHL S (G; Z) is to show it is a knot invariant.By Theorem 2.1, it suffices to show that this homology is unchanged under commutations and switches, and also under (de)-stabilizations of type X : SW.

Commutation and Switch Invariance
First, we show this for the commutations and switches; we model our arguments off of those in [OSS Section 15.3].As in Section 2.1, we take two grid diagrams G and G ′ related by a commutation or switch and superimpose them; we borrow here the notation from that section.We will do the computations below for a column commutation or switch; a row commutation or switch proceeds identically.Very concretely, the goal of this section is to prove the following theorem: Theorem 5.2.If two grid diagrams G and G ′ as above differ by a column commutation or switch, then GCL S (G; Z) and GCL S (G ′ ; Z) are quasi-isomorphic chain complexes.
Again, we recall that by Remark 3.2, we may guarantee that each bigon bounded by β i and γ i contains at least one X-marking inside.
First, we must define pentagon and hexagon maps; the pentagon maps will provide the quasi-isomorphism and the hexagon map will be the relevant homotopy operator.For this, we need signs for pentagons and hexagons.
Definition 5.4.For a pentagon p ∈ Pent * (x, y ′ ), let P be its associated rectangular domain.We define the sign of p as follows: S(p) := (−1) M(x)+B(p) S(P ), where B(p) is 1 if p lies to the left of β i and 0 if p lies to the right of β i .Similarly, for p ∈ Pent * (x ′ , y), let P be its associated rectangular domain.We define the sign of p as follows: S(p) := (−1) M(y)+B(p) S(P ).
For a hexagon h ∈ Hex * (x, y), let H be its associated rectangular domain.We define the sign of h simply as S(h) := S(H).
Definition 5.5.The pentagon map P S : GCL S (G; Z) → GCL S (G ′ ; Z) is defined as and we define a similar map P ′ S : GCL S (G ′ ; Z) → GCL S (G; Z) by The hexagon map H S : GCL S (G; Z) → GCL S (G; Z) is defined as Proposition 5.4.The pentagon map P S is a bigraded quasi-isomorphism.
Proof.First, we show P S is bigraded.This is an immediate consequence of the relative formulas for Maslov and Alexander gradings as well as Lemma 4.5.Next, we show P S is a chain map; this amounts to showing that P S •∂ S −∂ S •P S = 0.For a domain ψ ∈ π(x, z ′ ), let D pr (ψ) be the set of all decompositions of ψ as a (possibly long) pentagon and a (not long) rectangle in that order, and D rp (ψ) the reverse.Expanding out the equation, we wish to show In this case, we see immediately that | x − x ∩ z ′ | could be 3, 4, or 1.If it is 4, then the proof of Corollary 4.1 tells us that any ψ with |D rp (ψ) ∪ D pr (ψ)| ≥ 1 has exactly two decompositions r * p and p ′ * r ′ of equal degree corresponding to the two rectangular decompositions of Ψ.To show the contribution of ψ is equal on both sides of the equation, it is sufficient to show that the signs of these two decompositions is equal.By the definition of a sign assignment, and the decompositions of Ψ, we see that The proof of Corollary 4.1 gives us two cases.First, that these two decompositions correspond to different decompositions of Ψ.Here, the two associated rectangular decompositions have opposite signs.However, the B-value of the pentagons in each decomposition is obviously equal, and the Maslov initial grid states of the pentagons in each decomposition differ in parity if and only if one decomposition is in D rp (ψ) and the other is in D pr (ψ).Hence, ψ contributes equally to both sides of the above equation in this case.
Otherwise, these two decompositions correspond to the same rectangular decomposition of Ψ, in which case the decompositions are of the form r * p and p ′ * r ′ , and b(p) = −b(p ′ ), so ψ also contributes equally to both sides of the above equation in this case.
The final case is when | x − x ∩ z ′ | = 1.Then, the associated rectangular domain is an annulus, and must therefore be thin to avoid intersection with X; furthermore, the condition that each bigon contains an X-marking prevents this annulus from having multiplicity 2. Hence, the degrees of all decompositions must be 0, and the remainder of the proof proceeds exactly as in [OSS Lemma 15.3.3].
By identical reasoning, P ′ S is also a bigraded chain map.Finally, we show a homotopy formula.Specifically, we will show that ) is chain-homotopic to the identity.By an analogous argument, we may also show that (−P ′ S ) • P S , which would complete the proof that P S is a quasiisomorphism.
For this formula, we consider a domain ψ ∈ π(x, z) contributing at least one nonzero term to the left-hand side.By the proof of Corollary 4.1, if | x − x ∩ z | = 3 or 4, then there are exactly two decompositions of ψ.If the associated rectangular decompositions of Ψ are different, then we have two possibilities.Either both decompositions correspond to hexagons and rectangles only, in which case ψ contributes zero to the left-hand side by the definition of signs for hexagons.Otherwise, one decomposition consists of two pentagons, with equal B-values, and the relevant Maslov gradings have the same parity.Hence, the two decompositions cancel in this sum.
If the associated rectangular decompositions of Ψ are the same, then one of the decompositions consists of two pentagons with different B-values, and the other consists of a hexagon and rectangle.Hence, the two decompositions cancel in this sum.
The final possibility is x = z; as before the condition that ψ ∩ X = ∅ forces Ψ to be a thin annulus of multiplicity 1 and width 1.Furthermore, geometrically we see it is vertical.Hence, the degrees of all the decompositions must be 0, and we are reduced to the case proven in [OSS Lemma 15.3.4],which is that ψ is unique and contributes the identity to the sum.That proves the theorem, and thus the commutation and switch invariance of GHL S .

Stabilization Invariance
We now let G and G ′ be grid diagrams with G ′ a stabilization of G of type X : SW (recall that by our generalization of Cromwell's Theorem, it is sufficient to consider this case.)Let c ∈ G ′ be the central point of the stabilization.
We label the 2-by-2 region introduced in the stabilization as follows: , and let O 2 be the O-marking sharing a row with X 2 .
Note that S(G ′ ) = I(G ′ ) ∪ N(G ′ ), a disjoint union, where I(G) are those x ′ ∈ S(G ′ ) with c ∈ x ′ , and N(G ′ ) are the other grid states.
There is a natural bijection S(G) → I(G ′ ) which can be thought of as x → x ∪{c}, and this bijection extends to a bijection of rectangles from x to y in S(G) to rectangles from x ∪{c} to y ∪{c} in I(G ′ ).Fix a sign assignment S ′ on G ′ , and define a sign assignment S on G by pulling back S ′ on I(G ′ ).
For a chain complex A, let A[[m, a]] denote A with the bigradings shifted by (+m, +a).
Let I be submodule of GCL S ′ (G ′ ; Z) generated by I(G ′ ), and likewise for N.Then, I is a quotient complex and N is a subcomplex.The bijection of S(G) and I(G ′ ) clearly induces an isomorphism of complexes e : I → GCL S (G; Z) [[1, 1]], since any rectangle in I contributing to ∂ S ′ must not pass through c, lest it also intersect X, hence the T -values of the rectangles are preserved in this correspondence.
Let C be the mapping cone of the map Per [OSS Chapter 5], the homology of C is isomorphic to that of GCL S (G; Z).Now, the proof of [OSS Proposition 15.3.5],which only relies on the equation that we proved earlier, tells us that the map D S : GCL S ′ (G ′ ; Z) → C given by is a quasi-isomorphism.That shows the following proposition, completing the proof of invariance of GHL over the integers: Proposition 5.5.The chain complexes GCL S ′ (G ′ ; Z) and GCL S (G; Z) have isomorphic homologies.
This theorem justifies us using the integer invariant GHL S (K; Z) in the remainder of the paper.

Skein Exact Sequence
The goal of this section is to prove a skein exact sequence for GHL in analogy to the skein exact sequence satisfied by GH − .We first describe this carefully.

Skein Exact Sequence Basics
The idea is that if three links differ in one crossing as in the below picture (called a skein triple), then we can relate their grid homologies by an exact sequence.
with the same differential.The homology of this complex is known as cGH − S (G; Z).The collapsed double-point enhanced grid complex of G is the complex with the same differential.The homology of this complex is known as cGHL S (G; Z).
The proof of the signed version of [OSS Theorem 8.2.5] adapts without variation to the double-point enhanced case to give the following theorem: Theorem 6.1.For any link, the collapsed double-point enhanced grid complex of a grid diagram representing the link is a link invariant as a bigraded Z[U, v]-module.Now, we are ready to state the theorem we wish to prove.We omit the Z and S from the notation for convenience.Let cGHL m (L, a) be the Z-submodule of cGHL(L) consisting of homogenous elements of bidegree (m, a).Theorem 6.2.Let (L + , L − , L 0 ) be an oriented skein triple, with ℓ and ℓ 0 the number of components of L + and L 0 respectively.If ℓ 0 = ℓ + 1, then there is a long exact sequence where the maps below fit together to be homomorphisms of Z[U, v]-modules: Let J be the 4-dimensional bigraded abelian group J ∼ = Z 4 with one generator in bigrading (0, 1), one generator in bigrading (−2, −1), and two generators in bigrading (−1, 0).
If ℓ 0 = ℓ − 1, then there is a long exact sequence where the maps below fit together to be homomorphisms of Z[U, v]-modules: Below, we reproduce figure 9.3 from [OSS], which depicts all four of these grid diagrams simultaneously, and defines for us two crucial points c and c ′ .

Proof of the Theorem
We partition our four chain complexes, as above, into I and N parts depending on whether the grid states contain the marked point c, and I ′ and N ′ parts depending on whether the grid states contain the marked point c ′ , giving the following descriptions of GCL S as mapping cones of the following maps counting some of the distinguished squares in the above diagram.(We omit the ; Z for notational simplicity.) Chain Complex Quotient Complex Map Subcomplex Map Counts ...
Definition 6.2.We define the map T : I ′ (G ′ 0 ) → I(G + ) by the property that T (x) − (T (x) ∩ β i ) = x −(x ∩γ i ).(See [OSS p. 155].)Lemma 6.1.The identification T : Proof.The proof of [OSS Lemma 9.2.3] holds in our situation.Lemma 6.2.The maps As in the proof of [OSS Lemma 9.2.4], we proceed by a now-familiar rectangle counting argument.Consider any juxtaposition of rectangles contributing to the lefthand side of the equation This is a rectangle from x ∈ I(G + ) to y ∈ N(G + ) (where we count rectangles going through Y markings) and then a rectangle from y ∈ N(G + ) to z ∈ I(G + ) (where we count rectangles going through X markings).Thus, y must not contain c, but z does.The only possibility is thus that x = z, and the composite of these two rectangles must be an annulus.The annulus must be width one since otherwise it would intersect X −{X 1 , X 2 , Y 1 , Y 2 }, and similarly must be multiplicity 1.
There are 4 annuli, and since they are all thin, they are empty; the O-markings they pass through are O i for i = 1, 2, 3, 4, giving V 1 + V 2 − V 3 − V 4 since the V 3 and V 4 terms correspond to vertical annuli whereas the others correspond to horizontal annuli.
The other case uses the same decomposition of rectangles.
We get the following commutative square: [OSS Lemma 9.2.5]).Let M 0 , A 0 be the bigradings on G 0 (whose grid states may be naturally identified with those of G + ) and M 0 , ′ A ′ 0 be the bigradings on G ′ 0 (whose grid states may be naturally identified with those of G − .)Endow I ′ and N ′ with bigradings M ′ 0 and A ′ 0 + ℓ0−ℓ−1 2 and endow I and N with bigradings Then, in the above square, the following holds: • Each edge map is homogenous of bidegree (−1, 0).
• The left column is isomorphic as a bigraded chain complex over • The left column is isomorphic as a bigraded chain complex over • The bottom row is isomorphic as a bigraded chain complex over Proof.The proofs of [OSS Lemma 9.2.3] and [OSS Lemma 9.2.5] carry over identically in this situation.
The following corollary is immediate.
Corollary 6.1.The map Note that the mapping cone of this map is precisely the above commutative square.

Defining the New Maps
Define for x ∈ S(G 0 ), the map: And, for i = 1, 2, and x ∈ S(G ′ 0 ), the maps: Further define, for a domain ψ, the quantity Lemma 6.4.The map Proof.This is immediate from the commutation invariance of GCL S as proven in the previous section.Alternatively, the unsigned version of this lemma is proven in [RWW Propositions 9,10,11].(This paper uses a slightly different-looking definition for the map P , allowing pentagons to be long without being thin, however these pentagons always contribute 0 since they contain intersections with X, hence the two maps are in fact identical.)Remark 6.1.h X2,Yi does not vanish on I ′ .This is in stark contrast to the map h X2,Yi defined in [OSS Chapter 9] in the un-enhanced case.Lemma 6.5.
1. h X2 vanishes on N ′ and maps Proof.Both statements follow immediately from the multiplicity conditions on the rectangles in the definitions of the maps.
Lemma 6.6. 1. Suppose r is a rectangle contributing to the sum in h X2 .Then r is not long, and Y 1 (r) = Y 2 (r) = 0. 2. Suppose r ∈ Rect * (x, y) is a rectangle contributing to the sum in h X2,Yi .Then, y ∈ N(G ′ 0 ), so as a consequence, the image of h X2,Yi is within N ′ .
Proof.Both statements follow immediately from the multiplicity conditions on the Y i in the definitions of the maps.Now, we must prove two key lemmas: Lemma 6.7.The following two identities hold: Proof.We start with the first equation.Let ψ = r * t * p be a domain contributing to the left-hand side.Here, since the image of T is in I ′ , a pentagon p contributing to P in the left-hand side has an outgoing corner at c, therefore geometrically we see X 2 (p) = 0 lest Y i (p) ≥ 0 for either i = 1, 2. Now, concatenating p with the triangle t contributing to t therefore gives a rectangle r 2 with an outgoing corner at c ′ , with X 2 (r 2 ) = 1, and Y 1 (r 2 ) = Y 2 (r 2 ) = 0. Furthermore, the image of ∂ I ′ N ′ is some rectangle r = r 1 with an incoming corner at c ′ , and with Y (r 1 ) = 1.Hence, the composite ψ = r 1 * r 2 is a decomposition appearing in the right-hand side.Conversely, for ψ = r 1 * r 2 a decomposition appearing in the right-hand side, we have that the rectangle r 1 = r contributes to ∂ I ′ N ′ , and cutting r 2 along the portion of β i passing through X 2 gives us a decomposition r 2 = t * p such that t contributes to T and p to P in the left-hand side.Furthermore, these two decompositions always have the same degree since the interiors of the domains in each decomposition differ only on the edge of the small triangle t which is part of β i ; however, the only point on β i that could possibly affect the degrees of the decompositions is an outgoing corner of p. Hence, either T (p) = T (r 2 ) = 1 or else T (p) = T (r 2 ) = 0. Furthermore, by the definition of a sign assignment for a pentagon, it is clear that both decompositions have the same sign.Now, we consider the second equation.Let ψ = t * r * p be a domain contributing to the left-hand side.We see geometrically that if X 2 (p) ≥ 1, then it must be the case that either Y 1 (p) or Y 2 (p) is positive.Since we specify in the equation defining P that Y (p) = 0, we must have X 2 (p) = 0. Furthermore, it is not possible for r and p to overlap in a small neighborhood of c ′ , since this would require Y 1 (r) = 1 and then Y (p) > 0. Hence, all local multiplicities are ≤ 1 around c ′ .Suppose c ′ were a 90-degree corner.Then, we must have Y 2 (r) = 1, and p must be a left pentagon.However, geometrically, the only way this is possible is if Y 2 (p) ≥ 1, a contradiction.Also note that X 2 (r) = 0 by the fact that c is an outgoing corner and r cannot be long.In all cases, then, we have X 2 (ψ) = Y (ψ) = 1.
Thus, c ′ is either a 270-degree corner, a 180-degree corner, or in the interior of ψ.In all three cases we may represent ψ by an L-shaped region Q in the universal cover such that a pre-image C ′ of c ′ is either a 270-degree corner, a 180-degree corner, or in the interior of ψ.Indeed, start with t which has a corner at C ′ , and then attach pre-images R and P of r and p such that if r or p has a corner at c ′ then R or P has a corner at C ′ .
We must also show that these three possibilities are also the only three possibilities for a domain contributing to the right-hand side.Suppose ψ = r 1 * r 2 where r 1 is from h + X2 and r 2 is from h Y .Then, r 1 has a corner at c ′ , namely the southwest corner, and no others since Y (r 1 ) = 0.Then, r 2 must contain precisely one of Y 1 or Y 2 , but X 2 (r 2 ) = 0, so either r 2 has a corner at c ′ , leaving ψ a 180-degree corner at c ′ (which clearly forces ψ to be an annulus), or it has an edge passing through c ′ , leaving ψ a 270-degree corner at c ′ .Note there is no 360-degree case; we will return to this at the end of the proof.
We will first show that in each case ψ has two decompositions of equal degree; then, we will appeal to a proof from [OSS] to show that these two decompositions further have the same sign.
In the first 270-degree case, either c ′ is on an edge of r or on p.If c ′ is on an edge of r, which is part of horizontal circle, say, α j , then cutting ψ along α j gives a decomposition ψ = r 1 * r 2 .Since r 1 clearly has c as an outgoing corner and contains X 2 , and we know X 2 (ψ) = X 2 (r) + X 2 (t) + X 2 (p) = 0 + 1 + 0 = 1, that tells us X 2 (r 1 ) = 1.Furthermore, in this case the support of r 1 is a subset of the support of p, hence Y (r 1 ) = 0. Thus, r 1 contributes to h + X2 ; likewise, we must have , then performing this same cut along α j gives a decomposition as in the left-hand side.
The proof of lemma 4.3 ensures that the two decompositions r 1 * r 2 and r ′ 1 * r ′ 2 of ψ as rectangles in GC + (G ′ 0 ) have the same degree.One of these decompositions is r 1 * r 2 , and the other p * r * t differs from r ′ 1 * r ′ 2 by the introduction of the small triangle t and by cutting at β i instead of γ i .No points on β i or γ i can contribute to the degrees of any of these decompositions, for on the left-hand side, p and r border β i and on the right-hand side, r 1 and r 2 border γ i .Thus, both decompositions r 1 * r 2 and p * r * t have the same degree.
Similarly, if c ′ is on an edge of p, then this edge is part of γ i , and we cut ψ along γ i to get a decomposition r 1 * r 2 .In this case, r 1 is the union of t and a portion of r not containing Y 2 ; hence, X 2 (r 1 ) = 1 and Y (r 1 ) = 0, so r 1 contributes to h + X2 .Similarly, r 2 must contribute to h + Y since we must have Y (r 2 ) = 1.Conversely, if ψ = r 1 * r 2 , with Y 2 (r 2 ) = 1, then performing this same cut along γ i gives a decomposition as in the left-hand side.Likewise, these two decompositions differ by the introduction of the small triangle t and by cutting at β i instead of γ i , so they also must have the same degree by the argument in the previous paragraph.
In the 180-degree case on the left-hand side, we must have Y 2 (r) = 1, and since r is not long, the condition that there is a 180-degree corner at c ′ forces p to have an incoming corner at c ′ , and also for x = z; hence, the region ψ must be a horizontal annulus, and it must be thin to avoid intersecting with Y −{Y 1 , Y 2 }.Furthermore, since the local multiplicities of c ′ , which is an incoming corner of p, are all ≤ 1, we must have that the annulus has multiplicity 1.There is clearly a unique decomposition of this annulus into two rectangles r 1 , r 2 such that X 2 (r 1 ) = Y (r 2 ) = 1 and Y (r 1 ) = X 2 (r 2 ) = 0. Conversely, if ψ = r 1 * r 2 has a 180-degree corner at c ′ , either we are in the horizontal case, in which case there is a unique decomposition ψ = t * r * p as in the previous paragraph, or else we are in the vertical case (see the following paragraph); this case is unique for each grid state x.Since the annulus is thin, all constituent polygons must be empty so the degrees of all decompositions are 0.
There is one case remaining for the left-hand side and one case remaining for the right-hand side.For the left-hand side, this case is that c ′ is a 360-degree corner, in which case we must have Y 1 (r) = 1 and therefore that c ′ is an incoming corner at p; again, this forces the initial and final grid states to be equal, so we have a vertical annulus, and it is unique for each grid state x .The remaining case for the right-hand side is a vertical annulus as well.Both annuli are thin, so have degree 0, and contribute 1 power of V 4 to the formula.
What's left to show is that each decomposition in every case has the same sign.Since each of these possibilities is identical to those in the proof of [OSS Theorem 15.5.1], this proof guarantees that each decomposition has the same sign.Indeed, it only uses the defining properties of sign assignments and the relation between the sign of a pentagon and its associated rectangle, all of which carry over to the case where the pentagons and rectangles are potentially long.
This concludes the proof.
Lemma 6.8.The map h X2,Y provides a chain homotopy from e. the following equation holds: If | x −(x ∩ z)| = 3, then Lemma 4.3 shows that |D(ψ)| = 2, and both decompositions have the same degree and opposite signs, since only 1 rectangle in each possible decomposition of ψ could be long.We must show that both decompositions appear in the above formula, and do so exactly once.By analyzing the above formula, we conclude that Y (ψ) = 1 and X 2 (ψ) ≥ 1.Both rectangles in both decompositions clearly have trivial intersections with Y −{Y 1 , Y 2 }.Hence, the only possibilities can be found by considering multiplicities of X 2 and Y i , and are listed below.Note that in the first two cases, r 1 is not forced; it could be from either h X2 or ∂, but this choice is forced when we consider r 2 .
Hence, this case | x −(x ∩ z)| = 3 also contributes 0 to the sum.If | x −(x ∩ z)| = 0, then we must have ψ is an annulus containing X 2 , and it must be thin since Y (ψ) = 1 and ψ ∩ (Y −{Y 1 , Y 2 }) = ∅.Since Y (ψ) = 1, it must have multiplicity 1.Then, clearly, ψ has a unique decomposition into two rectangles, and the possibilities in the table in the previous case still apply to show that this decomposition appears precisely once in the sum on the left-hand side.If the annulus is horizontal, it gives us multiplication by V 2 , and if it is vertical, we get multiplication by −V 4 .
Each such isomorphism comes from composing quasi-isomorphisms Cone(V i ) → GC − /V i induced by projection (c, c ′ ) → π i (c ′ ), and isomorphisms Cone(V i ) → Cone(V j ) given by (c, c ′ ) → (c, H(c) + c ′ ), where H is the homotopy operator from V i to V j .
By the previous subsection, we know that ∂ 1 commutes with all of these maps up to homotopy (when we consider the action of ∂ 1 on the mapping cone as ∂ 1 (c, c ′ ) = (∂ 1 c, ∂ 1 c ′ )), which establishes the result.Now, from [OSS Chapter 7.1] we know that GH(m(K)), where m(K) is the mirror of the knot K, is canonically isomorphic to the dual vector space GH(K) ∨ .Repeating our entire above discussion with sign-refined, we may use the universal coefficient theorem to give ourselves a coefficient field of R rather than F (which we will denote as GH(K; R).
Lemma 7.3.If K is an amphicheiral knot, then there exists at least one nondegenerate bilinear form •, • on GH(K; R) for which ∂ 1 * is self-adjoint and the image of ∂ 1 * is an isotropic subspace.
Proof.We suppress the R-coefficients for convenience here.The invariance of GH on the grid presentation of a knot gives a bigraded isomorphism Ω : GH(K) → GH(m(K)) ∼ = GH(K) ∨ , which in turn induces such bilinear form x, y := Ω(x)(y).
Remark 7.1.We hope that, perhaps, we could prove that ∂ 1 * is zero by computing the signature of one such bilinear form and showing it is positive definite, say; this would require the bilinear form to be symmetric, which is a difficult question.It is another question of interest whether this bilinear form depends on the particular isotopy of K into m(K).
When K is alternating, say, then we know that GH is dimension ≤ 1 in each bigrading, hence the fact that Ω is bigraded forces each such bilinear form to be diagonal with respect to the basis consisting of nonzero homogeneous elements, hence symmetric.Remark 7.2.We still have that ∂ 1 * changes the grading (M aslov, Alexander) by (−3, 0).A python search gives that 18 is the smallest crossing number of a prime knot with GH nonzero in gradings differing by (−3, 0).Proof.This follows immediately from the fact that GCL(G) = ⊕ ∞ n=0 v n GC − (G), which is an expression of GCL as the associated graded object of the filtration F p GC + (G) = ⊕ ∞ n=p v n GC − (G).Clearly, this filtration is respected by ∂.Furthermore, unwinding definitions, it is clear that d 2 ([x]) = v∂ 1 * ([x]).

The connection between GHL(K) and τ
Proof.Let K be quasi-alternating.It is sufficient to show that the spectral sequence with E 2 page GH − (K) [v] converging to GHL(K) collapses at the E 2 page.
By [OSS Chapter 10], we know that the U -torsion part of the grid homology GH − (K) is supported in bigradings (M, A) with M − A = −τ, and the U -nontorsion "tail" is supported in M − 2A = 0, which begins on the line M = A and extends in one direction.
Hence, GH − [v](K), which is what we get when we first take GC − [v](K) and take homology with respect to ∂ 0 , must be supported in copies of these shapes each differing from the previous by an addition of 2 in Maslov grading.In particular, all homogeneous U -torsion elements have the same parity of their Maslov grading.
We now wish to conclude, inductively, that all the higher differentials in the spectral sequence are trivial.
If ξ ∈ GH − [v](K) is U -torsion, then so is d i (ξ) for any i by U -equivariance of the differential.Thus, any torsion elements must be in the kernel of d i for each i since d i reverses the parity of the Maslov grading.Now, note that each differential in the spectral sequence increases the coefficient of v strictly.But, any point on the tail M −2A = 0 is strictly lower in Maslov grading than any point of equal Alexander grading with a higher v coefficient.Thus, the differentials must all be zero there as well.
Proof.Again, we show the spectral sequence collapses at E 2 , and we do so by analyzing the U -torsion and U -non-torsion portions separately.Let K be a positive torus knot K = T p,q .[OSS Theorem 16.2.6]describes the structure of GC(K).In particular, it shows us, by the computation that τ (K) = (p−1)(q−1) 2 , that the infinite tail begins at F (δ −k ,n −k ) , where n −k = − (p−1)(q−1) 2 .Per [OSS Formula 7.6], we can express GH − (K) = F[U ] (−2τ,−τ ) ⊕ i F[U ] (di,si) /U ni .Let ξ i be the nontrivial element in F[U ] (di,si) /U ni that is not a multiple of U.Then, the ξ i generate the U -torsion portion of GH − (K).Call each such F[U ] (di,si) /U ni a "finite tail."Furthermore, any finite tails in GC − (K) contribute precisely two terms to GC(K) per [OSS Formula 7.6], and by this formula, we see that these finite tails must lie in higher Alexander gradings than the tail.Then, clearly any differential d i , for i ≥ 2, which increases the coefficient of v strictly, cannot map any point on the infinite tail to any nontrivial point.Hence it is sufficient to consider the U -torsion parts.
If the map d 2 is nontrivial on the U -torsion part, then some component of it maps some homogeneous U -torsion element ξ = U k ξ i to a point ζ with 1 lower Maslov grading and identical Alexander grading, which is some U, v-linear combination of the ξ j .By grading reasons, none of the ξ j contributing nonzero terms in the expression of ζ is ξ i .The U -equivariance of d 1 tells us that at least one tail containing some ξ j , j = i, must terminate at the same Alexander grading A * at which the tail containing ξ i terminates.But, this requires there to be dimension ≥ 2 of GC(K) at the grading A = A * , contradicting Theorem 16.2.6.Inductively, we may use this same argument to show d k is trivial on the U -torsion part for k > 2. By the construction of the spectral sequence we must have that the sequence collapses on E 2 , as desired.
A similar computation holds for negative torus knots, based on the fact that for a negative torus knot, a U -non-torsion homogeneous element of GH − (K) never lies in the same Alexander grading but greater Maslov grading as a U -torsion homogeneous element of GH − (K).
Remark 8.1.These results can likely be extended without much difficulty to other classes of knots with relatively thin knot Floer homology.

Conclusion
It still remains open whether the homology GHL S (K; Z) encodes different information than GH − S (K; Z) [v].This article shows that these two objects at least obey very similar topological properties.Strategies for perhaps exhibiting an isomorphism between these two objects may relate to using the mirror of a knot, as discussed in the previous section, which may prove effective at least in showing that the non-torsion parts of these objects are isomorphic.Some of the variants of τ for GHL S (K; Z) are possibly sharper topological invariants than τ , so if these homologies are in fact different, we can extract useful data from our ventures.
One pressing open question is to explore if there is an analog of the filtered theory discussed in [OSS Chapters 13 and 14] for double-point enhanced grid homology.The seemingly natural extension of the differential in this case to the double-point enhanced world is markedly not a differential anymore.If in fact the homology GHL S (K; Z) is isomorphic to GH − S (K; Z)[v], then we would expect some filtered theory to exist.
then Lemma 5.1 tells us immediately that any ψ ∈ π(x, z) with |D(ψ)| > 0 has exactly two decompositions d ∈ D(ψ) with equal degree and opposite signs.
then Lemma 5.1 tells us immediately that any ψ ∈ π(x, z) with |D(ψ)| > 0 has exactly two decompositions d ∈ D(ψ) with equal degree and opposite signs.Hence, all such ψ contribute zero to the above sum.
Since the pentagons p and p ′ have initial grid states of opposite parity and clearly B(p) = B(p ′ ), as they have the same support, the signs S(r * p) and S(p ′ * r ′ ) are indeed equal.Next, suppose | x − x ∩ z ′ | = 3.Then, as in the proof of Corollary 4.1, any ψ with |D rp (ψ) ∪ D pr (ψ)| ≥ 1 has exactly two decompositions of equal degree corresponding to the two rectangular decompositions of Ψ.

First
, we must develop a version of grid homology for links.Definition 6.1 (Modified from [OSS Definition 8.2.4]).Let G be a grid diagram representing an ℓ-component link, and suppose O j1 , . . ., O j ℓ are O-markings lying on each component of the link.Then, the collapsed grid complex of G is the complex Per[OSS Chapter 9], we may assume that L + , L − , and L 0 are represented by grid diagrams G + , G − , and G 0 , respectively, which we picture below along with another diagram G ′ 0 also representing L 0 .The below diagram is borrowed from page 153 of[OSS].
z) be a region contributing to the left-hand side of this equation with |D(ψ)| > 0. We have three cases, | x −(x ∩ z)| =0, 3, or 4. If | x −(x ∩ z)| = 4, then Lemma 4.2 shows that |D(ψ)| = 2, and both decompositions have the same degree and opposite signs.Hence, this case clearly contributes 0 to the sum on the left-hand side.