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CABLE KNOTS ARE NOT THIN
SUBHANKAR DEY
Abstract. We prove that the (p, q)-cable of a non-trivial knot is not Floer homologically
thin. Using this and a theorem of Ian Zemke in [33], we find a larger class of satellite
knots, containing non-cable knots as well, which are not Floer homologically thin.
1. Introduction
In his seminal work [31,32], Thurston showed that a knot, based on the geometry on its
complement, is either one of three types: torus, satellite, or hyperbolic. Apart from that
classification, there is a family of knots that are easy to describe diagramatically, namely,
alternating knots, which admit projections onto generic planes, that ‘alternate’ between
under-passes and over-passes. It was proved by Menasco in [17] that an alternating knot
is either a torus or a hyperbolic knot.
Theorem 1 ([17]).If L is a non-split prime alternating link, and if S⊂S3rLis a closed
incompressible surface, then Scontains a circle which is isotopic in S3rLto a meridian
of L.
The above theorem and the fact that the exterior of a satellite knot contains an incom-
pressible torus implies that prime alternating knots are not satellite. Menasco’s proof of
Theorem 1makes direct use of alternating knot diagrams. More recently, Ozsv´ath and
Szab´o defined a larger class of knots, called quasi-alternating knots (see [23, Definition
3.1]).
Definition 2. Let Qdenote the smallest set of links such that
•the unknot is a member of Q.
•if Lis a member of Q, then there exists a projection of Land a crossing cin that
projection such that
(1) both smoothings of Lat c(see Figure 1), L0and L∞are in Q,
(2) det(L) = det(L0) + det(L∞).
The knot Floer homology of knots belonging to this set exhibit the same kind of charac-
teristics as the knot Floer homology of alternating knots, hence the name quasi-alternating.
Specifically, one of the characteristics is that quasi-alternating knots are Floer homolog-
ically thin, which is to say that the knot Floer homology of a quasi-alternating knot is
supported in gradings where the difference between Alexander and Maslov gradings is fixed
[16, Theorem 1.2] and they are completely determined by the signature of the knot and
its Alexander polynomial. Furthermore, Ozsv´ath and Szab´o proved that double branched
1
arXiv:1904.11591v1 [math.GT] 25 Apr 2019
2 SUBHANKAR DEY
LLL
0
8
Figure 1. smoothings of a crossing
covers of quasi-alternating knots are L-spaces, i.e. c
HF(Σ(K)) ∼
=Zdet(K)[23, Proposi-
tion 3.3]. More recently, Gordon and Lidman showed in [3, Theorem 1.2; Theorem 1.3]
that for cable knots, double branched covers are not L-spaces, hence these knots are not
quasi-alternating. This supports the following folklore conjecture:
Conjecture 3. Satellite knots are not quasi-alternating.
One could try to prove Conjecture 3for quasi-alternating knots with fixed determinant.
In this regard, combined results of Greene, Teragaito, Lidman and Sivek show that quasi-
alternating knots with small determinant (≤7) are alternating (see [4], [30], [13]). There-
fore there are no satellite knots with small determinant which are also quasi-alternating by
Theorem 1. However, their methods don’t generalize for knots with higher determinants.
Ideally, one would like to prove an analog of Theorem 1for quasi-alternating knots in the
context of Heegaard Floer homology. But this remains a bit out of reach at the moment.
Meanwhile, we can try to verify the above conjecture for certain classes of satellite knots.
In this regard, it is reasonable to try to show that a satellite knot fails at least one of the
two aforementioned characteristics of quasi-alternating knots: that quasi-alternating knots
are Floer homologically thin, and that their double branched covers are L-spaces.
Gordon and Lidman’s work shows that it’s not known how to characterize double
branched cover of general satellite knots. Therefore in this article, we’ll focus on the
property of Floer homology being thin. Showing that a non-trivial quasi-alternating knot
with trivial Alexander polynomial cannot be satellite is quite straight-forward. Recall that
for a knot Kwith (symmetric) Alexander polynomial ∆K(t) = Pg
i=0 ai(ti+t−i),
ai=χ([
HFK (S3, K, i) = X
α∈\
HFK (S3,K,i)
(−1)mα
where mαis the Maslov grading of α. For Floer homologically thin knots, all elements in
a fixed Alexander grading of its knot Floer homology have same Maslov grading (see [16,
Theorem 1.2]), in other words, [
HFK (S3, K, i) = Z|ai|. This and the fact that knot Floer
homology detects the Seifert genus of a knot (i.e [
HFK (S3, K, g3(K)) 6= 0) imply that K
must be the unknot. Hence satellite knots with trivial Alexander polynomial, for example,
Whitehead doubles, are not quasi-alternating. For a satellite knot with an arbitrary pattern
CABLE KNOTS ARE NOT THIN 3
and having non-trivial Alexander polynomial, this argument cannot be used. But for cable
knots, we can prove the following.
Theorem 4. If the pattern is embedded inside the solid torus in such a way that it lies on
a torus in standard way, in other words if it is Tp,q, then for any non-trivial companion
K, the resulting cable knot Kp,q is not Floer homologically thin.
This theorem verifies Conjecture 3for cable knots, which can alos be inferred from
Gordon and Lidman’s work in [3].
Corollary 5. If Kis non-trivial, Kp,q is not quasi-alternating.
In [33, Theorem 1.1], Zemke proves that if there is a ribbon concordance Cfrom K0
to K1, then there is an injection FC:\
H F K(K0)→\
H F K(K1), which preserves both
Alexander and Maslov gradings. Combined with Zemke’s result, Theorem 4implies the
following corollary.
Corollary 6. If there is a ribbon concordance between a cable knot Kp,q and another knot
K0, then K0is not Floer homologically thin. In particular, K0is not quasi-alternating.
Miyazaki in [18] proved that non-trivial band sums between knots are ribbon concordant
to the trivial band sum between them i.e connected sum of those knots. One can start
with a cable knot Kp,q and place a number of unknots inside a regular neighborhood of K
such that they are unlinked to both the pattern and the companion, and then join them by
some non-trivial bands them and the Tp,q pattern sitting already inside that neighborhood,
such that the bands stays inside the neighborhood. Now if one considers the resulting
knot inside the solid torus as the pattern and take the companion as K, then the resulting
satellite knot P(K) is ribbon concordant to Kp,q by Miyazaki’s result. Hence by Corollary
6,P(K) is not thin and in particular, not quasi-alternating. This gives evidence to the
affirmative answer to the Conjecture 3for a large class of satellite knots containing the
cable knots.
Remark. An alternative way to prove that a knot is not Floer homologically thin is by
comparing τ-invariant, a knot concordance invariant defined by Ozsv´ath and Szabo in [21]
and Rasmussen in [27], and the signature of the knot, since for thin knots τ(K) equals
negative of the half of its signature. In recent years, there has been much work studying
effects of cabling on various knot invariants. To name a few, Shinohara in [28] studied
signature and satellite operation, Hedden in [6], [7], Ina Petkova in [26], Jennifer Hom in
[10] studied τinvariant and cabling operation, Wenzhao Chen in [2] studied Υ invariant
and cabling, Apratim Chakraborty in [1] studied Legendrian knot invariant ˆ
θand cabling.
In particular for τof a cable knot, Hedden provided an inequality in [7], later improved by
Jennifer Hom in [10], in terms of the τof the companion knots. Using that and Shinohara’s
result [28, Theorem 9] about signature of cable knots, one can prove that a certain class
of cable knots are not thin (eg. most of the iterated torus knots). This was pointed out
to the author by Abhishek Mallick. We’ll take a different approach to prove Theorem 4in
full generality.
4 SUBHANKAR DEY
In order to prove Theorem 4, we will use bordered Floer homology package of Lipshitz,
Ozsv´ath and Thurston in [14], which is tailor made to study satellite knots. We’ll discuss
why and how a specific version of the ‘splicing theorem’ in [14] allows us to prove Theorem
4. In Section 2, we will briefly discuss the algebraic structure of bordered Floer homology.
In Section 3, we will use the aforementioned splicing theorem to find two elements in the
knot Floer homology of any cable knot such that the difference between their Alexander
grading and that of their Maslov grading are not same, to deduce that they cannot be
Floer homologically thin.
Acknowledgements. I am grateful to my advisor, Prof. C¸a˘gatay Kutluhan, for his con-
stant encouragement, guidance and inputs throughout the preparation of this paper and
beyond. I would like to thank Prof. Matt Hedden whose initial suggestion to look into
bordered Floer homology led to the result and also for his invaluable feedback on an earlier
draft. Special thanks to Prof. William Menasco and Prof. Xingru Zhang for valuable
discussions time and again and Abhishek Mallick for suggesting me this problem and lots
of helpful discussions.
2. Background on bordered Floer homology
We start by describing the features of bordered Floer setting we will be exploiting to
proof the main theorem. We’ll only be interested in the case when the manifold has torus
boundary. A bordered 3-manifold (with torus boundary) is a compact manifold with (torus)
boundary, along with a diffeomorphism φ:T2→∂Y , upto isotopy fixing a neighborhood
of a point, [8] and [15, Definition 1.4].
There are several versions of pairing theorem [14, Theorem 1.3] that come in handy
to study Heegaard Floer holomogy a closed 3-manifold generated by splicing two mani-
folds with boundaries or a manifold cut along some closed surfaces (see [8],[5]). We’ll be
interested in this specific splicing theorem:
Theorem 7 (Theorem 11.19 in [14]).If K1⊂Y1and after gluing Y1and Y2along their
boundaries ∂Y1=∂Y2=F, produces a null-homologous knot K⊂Y1∪FY2, then there is
a homotopy equivalence of Z-filtered chain complexes
[
CFK (Y, K )'d
CFA(Y1, K1)[
CFD(Y2) (1)
and the following equivalences of F[U]-modules :
gCFK −(Y, K )'CFA−(Y1, K1)[
CFD(Y2) (2)
which respects the gradings, where gCFK −(Y, K )denotes the associated graded object.
To make sense about the [
CFD ,[
CFA,CFA−mentioned above, we start by reviewing the
bordered Floer Homology setting.
For a compact manifold Ywith torus boundary, a bordered Heegaard diagram is a tuple
H: (Σg, αa
1, αa
2, αc
1, αc
2,· · · , αc
g−1, β1,· · · , βg, z) such that
•Σgis a compact, oriented surface of genus g with one boundary component,
CABLE KNOTS ARE NOT THIN 5
•βis a g-tuple of pairwise disjoint circles in the interior of Σg,
•αcis a (g−1) tuple of pairwise disjoint circles in the interior of Σg,
•αais a 2-tuple of pairwise disjoint arcs in Σgwith boundary in ∂Σg,
•zis a base point in ∂Σgrαa,
•αa∩αc=φ,
•both Σgr(αa∪αc) and Σgrβare connected.
To provide an example, given a knot K⊂S3, we can find a bordered Heegaard diagram
of S3rK, by first starting with a specific Heegaard diagram of S3,(Σ,α,β, w) and then
adding an extra basepoint zsuch that (Σ,α,β, z, w) is a doubly pointed Heegaard diagram
of Kin S3i.e w, z are two points on the Heegaard surface such that joining wto zin the
complement of βcurves and joining zto win the complement of αcurves and subsequently
pushing those arcs into the αand βhandle bodies, respectively, produces K.
Given a doubly pointed Heegaard diagram (Σ,α,β, z, w) of Kin S3, first we stabilize
the Heegaard surface by attaching a 2-handle with feet near the base points w, z. Now we
can draw a longitude βgof the knot that goes over that newly attached handle (see [22,
Fig.4])and thus we get a meridian of the knot, living on the attached 2-handle, call it αg.
Hence (Σ0,α∪αg,β∪βg, z, w) is a stabilized doubly pointed knot diagram. Take λ, a closed
curve, parallel to βg, intersecting αgat one point , say p. Let Dbe a small neighborhood
disk around p. Then the complement of int(D) specifies a bordered Heegaard diagram that
represents a bordered Heegaard diagram of S3rK, specifically (Σ0,α∪α0
g∪λ0,β∪βg, z0),
where α0
g=αgr{p}, λ0=λr{p}and z0lies on ∂D, away from the endpoints of the α-arcs.
To get a bordered Heegaard diagram of an n-framed knot complement, instead of con-
sidering λ, we have to take λn, which we can get by winding λaround αg, n times. After
that operation, one of the α-arcs will be λ0
n=λnr{p}, instead of λ0mentioned earlier
(see [12, Section 2.6]).
One can define a doubly pointed bordered Heegaard diagram of Kin a manifold Y, by
finding a bordered Heegaard diagram of Yand then adding an extra basepoint to represent
K⊂Y, just as in a doubly pointed knot diagram.
Given a bordered Heegaard diagram of a manifold H: (Σ,α,β, z), a generator x=
{x1, x2,· · · , xg} ∈ α∩βsuch that exactly one point can lie on each βcircle, exactly one
point can lie on each α-circle and at most one point can lie on each α-arcs. Let G(H) be
the set of all such generators.
The settings in place, we now discuss the algebraic preliminaries regarding bordered
Floer homology that will eventually lead to explaining the components of the splicing
Theorem 7.
Let Zdenotes the boundary of Din a bordered Heegaard diagram. Call (Z,a, M, z )
apointed matched circle, with 4kmarked points a={a1, a0
1, a2, a0
2,· · · , a2k, a0
2k}and M:
ai→a0
i, i = 1,2,· · · ,2k, a pairing of the points and zbeing a base point on Z. For Y, a
compact 3-manifold with torus boundary, we only focus on the case where k= 1 and we
call a={a0, a1, a2, a3}such that M(a0) = a2, M(a1) = a3and the points a0, a1, a2, a3are
labeled on Zin a clock-wise direction.
6 SUBHANKAR DEY
Let αa
1, αa
2denote the arcs from a0to a2and from a1to a3, respectively (See Figure
2). A pointed matched circle represents a compact surface with one boundary component
this way: consider Z × [0,1] and then add bands i.e one dimensional 2-handles with feet
to pairs of matched points on the circle on Z × {0}and then attach a disk to the new
boundary component created after adding the bands. If we cap off the remaining boundary
component with a disk, in general we call that surface F(Z). In particular, when k= 1,
F(Z) is a torus.
For the case in hand i.e when F(Z) represents a torus, A(Z) is an unital algebra over
F2with six ‘Reeb’ elements ρ1, ρ2, ρ3, ρ12, ρ23, ρ123 (in this case, one can think ρ1, ρ2, ρ3to
be the arcs in Zbetween a0, a1, between a1, a2and between a2, a3, respectively) and two
idempotents ι0, ι1such that ι0+ι1= 1 and these generators have these non-zero relations
:
ρ1=ι0ρ1=ρ1ι1ρ2=ι1ρ2=ρ2ι0ρ3=ι0ρ3=ρ3ι1
ι0ρ12 =ρ12ι0=ρ12 ι1ρ23 =ρ23ι1=ρ23 ι0ρ123 =ρ123ι1=ρ123
ρ1ρ2=ρ12 ρ2ρ3=ρ23 ρ12ρ3=ρ1ρ23 =ρ123
Through out the paper, if not mention otherwise, we’ll denote A(Z) by Aonly. For
more in depth discussion on general A(Z), see Chapter 3 of [14] or Chapter 1.4 [15] for a
short exposition.
In [14], Lipshitz, Ozsv´ath and Thurston associates to a bordered 3-manifold (Y, φ :
∂Y →F(Z)), [
CFD(Y) and [
CFA(Y), which are right A∞A(Z) and left dg A(−Z) module,
respectively. When K⊂Y, CFA−(Y, K ) is also a dgA(−Z) module. We’ll concern ourselves
with CFA−(Y, K ) and [
CFD(Y) chain complexes for this paper. We discuss about these
two kinds of modules next.
A vector space Mover F2is said to have a (right) type A structure over Aor is said to
be a right A∞-module over A(where Ais a graded, unital algebra) if Mis equipped with
a
aa
a
z
0
12
3
α
α
1
2
a
b
Figure 2. pointed matched circle
CABLE KNOTS ARE NOT THIN 7
a right action of I(the set of all idempotents in A, in our case I={ι0, ι1}), such that
M=Mι0⊕M ι1, as a vector space and, and multiplication maps
mk+1 :M⊗ A⊗k→M, k ≥0
satisfying the A∞relations
0 =
n
X
i=0
mk−i+1(mi+1(x⊗a1⊗a2⊗ · · · ⊗ ai)⊗ai+1 ⊗ · · · ⊗ ak)+
n−1
X
i=1
mk−1(x⊗a1⊗ · · · ⊗ ai−1⊗aiai+1 ⊗ai+2 ⊗ · · · ⊗ ak)
and the unital conditions
m2(x, 1) = x
mk(x⊗ · · · 1· · · )=0, k > 2.
We say that M is bounded if mk= 0 for sufficiently large k.
Now CFA−(Y, K) is a F[U]-module generated by G(H), where the right action by Iis
defined by
x·ι0=(xif xdoes occupy the arc αa
1
0 otherwise
x·ι1=(xif xdoes occupy the arc αa
2
0 otherwise
The right A∞-module structure on CFA−(Y, K ) is determined by the multiplication maps
mk+1 :CFA−(Y, K )⊗ A⊗k→CFA−(Y, K)
defined by
mk+1(x, ρi1,· · · , ρik) = X
y∈G(H)X
B∈π2(x,y)|ind (B,(ρi1,··· ,ρik))=1
#(MB(x,y,(ρi1,· · · , ρik))Unw(B)y,
m2(x,1) = x,
mk+1(x,· · · ,1,· · · ) = 0 for k >0,
where M(x,y,(ρi1,· · · , ρik)) is described in [15, Chapter 2] as follows: let Sbe a smooth
surface with boundary and 2gpunctures on its boundary and label gof these punc-
tures −, another gpunctures +, and the remaining punctures e. For x,y∈G(H),
M(x,y,(ρi1,· · · , ρik)) consists of maps
u: (S, ∂S )→(Σ r{z})×[0,1] ×R,(α× {1} × R)∪(β× {0} × R))
such that
•at the punctures −,uis asymptotic to x×[0,1] × {−∞},
•at the puncture +, uis asymptotic to y×[0,1] × {∞},
8 SUBHANKAR DEY
•at the punctures labeled e,uis asymptotic to the chords ρi×(1, ti)∈∂Σ×{1} × R,
and t1< t2<· · · < tn,
•uis proper and extends to a proper map from ¯
S→¯
Σ, where both ¯
Sand ¯
Σ are
‘filled in’ by disks,
•the composition of the extended map and projection to [0,1]×Ris a g-fold branched
cover and u|∂S is injective.
If f
M(x,y,(ρ1, ρ2,· · · , ρn)) denotes the moduli space of J-holomorphic maps satisfying
the above properties. Then M(x,y,(ρ1, ρ2,· · · , ρn)) = f
M(x,y,(ρ1, ρ2,· · · , ρn))/R(where
Rdenotes the translation action in the image). Given B∈π2(x,y),MB(x,y,(ρ1,· · · , ρn))
is the set of all such J-holomorphic maps which has the same homology as that of B. See
[14, Chapter 5] for details. The above defined family of multiplication maps {mi}counts
the number of holomorphic representative of such maps.
Now we describe the other module structure mentioned above. A vector space Nover
F2is said to have a (left) type D structure over A, equipped with a left action of Isuch
that N=ι0N⊕ι1N, as a vector space and a map
δ1:N→ A ⊗ N
satisfying the type D condition:
(µ⊗idN)◦(idA⊗δ1)◦δ1= 0
where µ:A ⊗ A → A is the multiplication map in A. Also, we inductively define maps
δk:N→ A⊗k⊗N
such that δ0=idN, δi= (idA⊗(i−1) ⊗δ1)⊗δi−1. We say Nis bounded if δkis zero for
sufficiently large k.
This general definition in place, now [
CFD(Y) is a F-vector space generated by G(H),
where the left action by Iis defined by
ι0·x=(xif xdoes not occupy the arc αa
1
0 otherwise
ι1·x=(xif xdoes not occupy the arc αa
2
0 otherwise
and the left A-module structure on [
CFD(Y) is defined by :
δ1:[
CFD(Y)→ A ⊗ [
CFD(Y)
defined by :
δ1(x) = X
y∈G(H)X
B∈π2(x,y)|ind (B,(ρi1,··· ,ρik))=1
#(MB(x,y,(ρi1,· · · , ρik))ρi1· · · ρiky
CABLE KNOTS ARE NOT THIN 9
Recall that for each x∈G(H), π2(x,x) i.e collection of all J-holomorphic Whitney disks
connecting xto itself, forms a group where the multiplication is given by concatenation
of disks. We can think of Bas living on the Heegaard surface and a linear combination
of the regions in Σ r(α∪β). Elements of π2(x,x) are called periodic domains, which is
naturally isomorphic to H2(Y, ∂ Y ). A non-trivial class B∈π2(x,y) is called positive if
all its local multiplicities are non-negative. The Heegaard diagram His called provincially
admissible if it has no positive periodic domains with multiplicity 0 everywhere along ∂B.
The Heegaard diagram His called admissible if it has no positive periodic domains.
Provincial admissibility of Hensures that the above mentioned maps miand δ1are well-
defined. Admissibility of Hensures that CFA−(Y, K ) and [
CFD(Y) are bounded. Compare
[25, Chapter 4,5] and [14, Chapter 4].
Now the definitions of [
CFD and CFA−in place, we describe the operation between these
two modules, mentioned in Equation (1) and (2). If one of CFA−(Y, K),[
CFD is bounded,
then the box tensor product C F A(Y, K)[
CFD is the F[U]-module CFA−⊗I[
CFD equipped
with the differential :
∂(x⊗y) =
∞
X
k=0
(mk+1 ⊗id|\
CFD )(x⊗δk(y))
The finite-ness of the sum is ensured by bounded-ness of any one of CFA−(Y, K),[
CFD(Y, K).
For the case that we’re interested in, writing δ1map in terms of the elements of A(T2)
helps. Let ρ∅=ι0+ι1= 1 and then rewrite δ1as
δ1=X
i
ρi⊗Di
where iruns over {∅,1,2,3,12,23,123}and Di:[
CFD →[
CFD are called coefficient maps.
In this notation , the differential in CFA−[
CFD can be written like this :
∂(x⊗y) = X
k
mk+1(x, ρi1, ρi2,· · · , ρik)⊗Dik◦ · · · ◦ Di2◦Di1(y)
where kruns over all such sequence i1, i2,· · · , ikof kelements from {ϕ, 1,2,3,12,23,123}
(including the empty sequence when k= 0).
3. Proof of Theorem 4
Through out the proof, whenever we talk about Tp,q ∈D2×S1, we assume that
gcd(p, q) = 1, p > q, |q| 6= 1. If we finish the proof of our main theorem for this case
and for any framing of K, then we can see that if q=mp +i, m > 0,p>i>1, then by
choosing mframing of Kas the companion and Tp,i as the pattern, we’ll be done. For
q=±1 case, one can use the A∞relations described in [10] and [26] and follow the same
strategy that we follow to find the desired elements in the knot Floer homology of the
cable.
10 SUBHANKAR DEY
We start our proof by discussing some materials that we need to borrow from the knot
Floer homology setup. For an integer homology sphere Y, recall that for a knot K⊂Y,
CFK ∞(Y) is a filtered chain complex, gotten from the Heegaard Floer chain complex
of the ambient manifold Yi.e CF ∞(Y) by introducing additional filtration, induced by
Alexander grading, which is dictated by the knot and where CF ∞(Y) is a F[U]-module
over G(H) for a Heegaard diagram Hof Y.
For a fixed Alexander grading j,[
HFK (Y, K, j) is the homology of the chain com-
plex [
CFK (Y, K, j), where [
CFK (Y, K, j) is generated by [x,0, j] where x∈Tα∩Tβ,
((Σ,α,β, w, z) is a doubly pointed knot diagram for K⊂Y) such that
A(x) = hc1(sm(x),[ˆ
F]i
2
where ˆ
Fis a capped-off Seifert surface of Kin Y(since Yis a homology sphere, the
definition is independent of choice of Seifert surface of Kin Y).
We will consider C F K−(K), a chain complex generated by G(H) and the differential in
this bi-filtered chain complex is given by
∂−(x) = X
y∈Tα∩TβX
{φ∈π2(x,y)|µ(φ)=1,nz(φ)=0}
#( ˆ
M(φ)) ·Unw(φ)·y
where ˆ
M(φ) denotes the quotient of the moduli space of J-holomorphic disks representing
the homotopy type of φ,M(φ), divided out by the natural action of Ron this moduli space
and µ(φ) denotes the ‘expected dimension’ of M(φ), see [22] for detailed discussions. Set-
ting U= 0 in the above differential, defines the differential ˆ
∂for [
CFK , and the homology
of that is denoted by \
H F K(S3, K). By g CF K −we denote the associated graded object,
filtered by the Alexander grading i.e j.
Given a null-homologous knot Kin S3, it’s convenient to look at the knot Floer complex
living inside i,j -plane, where jis the additional grading i.e the Alexander grading induced
by the knot and idenotes the power of U. The general rule to draw the knot complex
is this: on i= 0 axis, for a fixed j, we put dim([
HFK (S3, K, j))-no.of points at (0, j)
co-ordinate and then extend that to the whole (i, j) plane by translating through Uand
U−1(eg. multiplying Upushes some [y,0, m] to [y,−1, m −1] and U−1pushes that
to [y,1, m + 1]). That is we assume the complex is reduced which is same as saying
C(i, j) = [
HFK (S3, K, j −i). This is due to the fact that a filtered chain complex is always
filtered chain homotopic to a reduced complex, see [9, Reduction Lemma]. We also draw
the boundary maps ∂∞by arrows emanating from some generator(s), pointing towards the
generator(s) that live(s) in their boundary. A reduced chain complex will ensure that the
arrows will be pointing downwards (i.e when the boundary map will strictly reduce the
Alexander grading), pointing to the left (i.e when the boundary map will strictly reduce
the U-power) or both (i.e when the boundary map points to south-west direction to itself).
CABLE KNOTS ARE NOT THIN 11
One might see CFK −(S3, K) as C(i≤0).The Reduction Lemma implies
dimF[U](CFK −(S3,K)) = dimF([
HFK (S3,K)) = 2n +1,for some n≥0
For a Z⊕Z- filtered complex Cand a given basis, we call an operation on the basis a
filtered change of the basis if that operation replaces some basis element xjby Pm
i=1 aiyi
such that both the filtrations of each aiyiis less than equal to those of xj, i = 1,2,· · · , m.
We call a F[U] basis {ξ0, ξ1, ξ2,· · · , ξ2n+1}of CFK −vertically simplified if
•∂vert (ξ2i−1) = ξ2i(mod U C −) for i= 1,· · · , n.
•A(ξ2i−1)−A(ξ2i) = ki>0.
•ξ0is the generator of the vertical homology.
where ξi∈C(i= 0), i = 0,1,· · · ,2n+ 1 and ∂vert =∂∞|C(i=0) .
Similarly, we can also define a horizontally simplified basis of CFK −{η0, η1, η2,· · · , η2n+1}
where
•∂hor (η2p−1) = Ulp·η2p(mod the associated graded object of C F K−, where j=
A(η2p−1)−1) ,for p= 1,2,· · · , n.
•A(η2p)−A(η2p−1) = lp>0.
•η0is the generator of the horizontal homology.
where ηp∈C(j= 0), p = 0,1,· · · ,2n+ 1 and ∂hor =∂∞|C(j= 0).
Lipshitz, Ozsv´ath, Thurston in [14, Theorem 11.57], Hom in [10, Lemma 2.1] proved
that CFK −always admits vertically and horizontally simplified bases. Also the facts that
C(i= 0) and C(j= 0) are isomorphic and {`1, `2,· · · , `n}={k1, k2,· · · , kn}are same
sets, follow from the symmetry of knot Floer homology under reversing the roles of the
marked points w, z, thus the orientation of the knot, see [22, Proposition 3.8].
Now we recall Lipshitz, Ozsv´ath, Thurston’s algorithm from [14, Theorem 11.26, A.11] to
find the complete set of generators for [
CFD(XK, φn) given CFK −(K), where XK=S3rK
and φndenotes a parametrization of the boundary of the knot exterior, where the knot is
taken to be m-framed.
Theorem 8 (Theorem 11.26, A.11 in [14]).With notation as above, if XKdenotes the
complement of the knot Kwith an integer framing r, then [
CFD(XK, r)has this following
description :
•ι0(\
C F D(XK)) is of dimension 2n+1 and is generated by {ξ0, ξ1,· · · , ξ2n+1}or
{η0, η1,· · · , η2n+1}.
•ι1(\
C F D(XK)) is generated by :
[
i∈{1,2,··· ,n}
{κi
1, κi
2,· · · , κi
`i}[
j∈{1,2,··· ,n}
{λj
1, λj
2· · · , λj
kj}∪{µ1,· · · , µt}, where
•For each vertical arrow of length `i, we have κi
1,· · · , κi
`i(subspace generated by
these is called vertical chain) with following coefficients :
ξ2i−1
D1
−−→ κi
1
D23
←−− · · · D23
←−− κi
`i
D123
←−−− ξ2i
12 SUBHANKAR DEY
•For each horizontal arrow of length kj, we have λj
1,· · · , λj
kj(subspace generated by
these is called horizontal chain) with following coefficients:
η2j−1
D3
−−→ λj
1
D23
−−→ · · · D23
−−→ λj
kj
D2
−−→ η2j
•If t= 2τ(K)−r, then we have another additional set of generators {µ1,· · · , µt}
(subspace generated by these is called unstable chain) with following coefficients:
ξ0
D1
−−→ µ1
D23
←−− µ2
D23
←−− · · · D23
←−− µt
D3
←−− µ0if t > 0
ξ0
D12
−−→ η0if t= 0
ξ0
D123
−−−→ µ1
D23
−−→ µ2
D23
−−→ · · · D23
−−→ µ|t|
D2
−−→ η0if t < 0
where τ(K) = min{p|ip:C(i= 0, j ≤p)→C(i= 0) induces surjection in homology},
which is a concordance-invariant defined by Ozsv´ath and Szab´o in [21]and Rasmussen in
[27], independently.
The gradings are determined as follows:
•The grading set is G/λ−1gr(ρ12)−1gr(ρ23 )−r
•Grading of any element x0in ι0([
CFD(XK, r)), represented by a generator of the
knot Floer homology, is determined by Alexander grading Aand Maslov grading M
of x0in the knot Floer complex : gr(x0) = λM−2A(gr(ρ23))−A
We will discuss about the grading notations later.
Now the settings and the main ingredient in place, we discuss the main strategy to prove
our main result. We’ll produce two non-zero elements in the knot Floer homology of a given
cable knot, using the splicing theorem 7, such that the difference between their Alexander
grading is not equal the difference between their Maslov grading. Thus by [16, Theorem
1.2], they are not Floer homologically thin. To do that we’ll look at CFA−(D2×S1, Tp,q )
and [
CFK (XK) simultaneously to find two elements from both modules such that their box
tensor is well defined. The splicing theorem allows us to find those two elements, living
inside the knot Floer chain complex of the cable knot. Then we have to make sure that
those elements are non-zero in the knot Floer homology of the cable. Then we calculate
their grading to get the desired result.
Observe that to make sure that an element, say γin gC F K−is a non-zero element in
knot Floer homology (i.e the homology of (\
C F K, ˆ
∂)), it is enough to check that all the
elements in ∂−(γ) contains a non-zero Upower and there is no element in gCF K −whose
boundary contains U0·γ.
Petkova in [26] and Hom in [10] used a bordered Heegaard diagram of (p, 1)-pattern knot
in the solid torus and looked at the lifts of the αarcs and the βcurve in the universal
cover of the genus one Heegaard surface i.e Euclidean plane, to count the Whitney disks
between the generators.
To find CFA−(D2×S1, Tp,q ), we follow the same strategy. We will use Ording’s algorithm
from [20, Theorem 3.5] to find a bordered Heegaard diagram of Tp,q in the solid torus, and
CABLE KNOTS ARE NOT THIN 13
then look at the lifts of αarcs and the βcurve in Euclidean plane to find our intended two
elements.
Let Kbe our companion knot in S3. We are going to find the desired generators,
separating our search in two cases : when t= 2τ(K)−ris zero and when tis non-zero.
Case 1. Let t= 2τ(K)−r6= 0. For subcases of this case, we recall the definition of , a
knot concordance invariant, defined by Hom in [11]. To define that, we recall another knot
concordance invariant ν, defined by Ozs´ath, Szab´o, which is ν(K) = min{s|ps:C{max(i=
0, j =s)} → C(i= 0) induces a surjection in homology}, where psis the projection map
onto the ico-ordinate. Recall that ν(K) = τ(K) or τ(K) + 1, see [21]. Then
(K) =
−1 if ν(K) = τ(K)+1
0 if ν(K) = τ(K) and ν(−K2) = τ(−K)
1 if ν(−K) = τ(−K)+1
If (K) = 1, we use [10, Lemma 3.2] to find {ξi}, a vertically simplified F[U] basis of
CFK −, with the following properties, after possible renaming,
•Uk·ξ2is the generator of the homology of C(j= 0), for some k.
•there exists ξ1such that ∂vertξ1=ξ2.
•ξ0is the generator of the homology of C(i= 0).
Here’s a sketch of the proof of [10, Lemma 3.2]: from the definition of , (K) = 1 ⇒
ν(K) = τ(K) but ν(−K) = τ(−K) + 1 i.e ξ0, the generator of the vertical homology, ‘gets
killed’ in the horizontal complex. What we mean by that is : Uτ(K)·ξ0lives in the image
of ∂hor. Also recall the homology of (C(i= 0), ∂vert) is isomorphic to the homology of
(C(j= 0), ∂hor ). That fact and the symmetry of [
HFK with respect to Alexander grading
implies that some ξ2, which lives in [
HFK (S3, K, −τ(K)), will be the generator of the
horizontal complex i.e U−τ·ξ2is the generator of the horizontal homology and also, there
is some ξ1such that ∂vertξ1=ξ2. For a detailed proof, see [10, Section 3].
Now we turn our attention to CFA−(D2×S1, Tp,q ). Petkova and Hom used the following
doubly pointed bordered Heegaard diagram for (p, 1) patterns in D2×S1, see Fig 3.
From the picture, one considers the indicated generator a. Now the observation that
both ξ2and alive in the ι0part of the F-vector spaces [
CFD(XK, r) and CFA−(D2×
S1, Tp,1), respectively, allows to look at a⊗ξ2, which Hom does in [10] to find the τof
the cables Kp,pn+1. Hom shows that a⊗ξ2survives in homology and thus an element of
[
HFK (Kp,pn+1). To do this, both Petkova and Hom find all the generators and the A∞-
relations in CFA−(D2×S1, Tp,1), from the diagram. See [26, Section 4], [10, Section 4.1].
We use Ording’s ‘cat’s cradle’ algorithm from [20] to find a doubly pointed bordered
diagram for (D2×S1, Tp,q) and from that we’ll look for our desired generators.
In [20, Theorem 3.5], given a (1,1) knot, Ording describes an algorithm to find a Hee-
gaard normal form of a genus one knot diagram for the knot. First one starts with finding
a standard form of the knot in torus and then draw the βcurve on the torus, isotopic to
the standard longitude of the torus, such that it misses p(tβ), where tβis the part of the
14 SUBHANKAR DEY
a
b b
b b
1
0
2
3
1
...
p-1 p2p-2
w
z
...
Figure 3. doubly pointed bordered Heegaard diagram of (p,1) torus knot
in D2×S1
knot in the β-handle body (i.e the part one can get by joining wto zwithout crossing β)
and pis the projection map onto the torus. See [20, Figure 3.11] for a step-by-step pictures
obtained after applying the algorithm for T5,3and also for description on standard and
normal forms of (1,1) knots.
One way to view Ording’s algorithm is that one starts with a copy of the standard
longitude and meridian of a torus and starts deforming the longitude in each step, keeping
it isotopic to the longitude but missing the standard form of the (1,1) knot. Now, one can
observe that at each step when the new longitude hits the meridian, two new generators
in the knot Floer homology of the knot, are born. The generator ais the one in knot
Floer homology of Tp,q in S3, which is the generator of the vertical homology i.e the unique
intersection point one starts with at the beginning of the algorithm.
After getting the βcircle for the knot diagram of Tp,q, we can cut out a neighborhood
of the vertices of the fundamental domain of the torus and take the horizontal and vertical
boundary components as the α-arcs (where, one was a longitude and the other was a
meridian of the torus, before cutting out). Now changing the name of win Ording’s
picture to zand placing another basepoint zat the bottom of the picture gives us a genus
one doubly pointed bordered diagram for Tp,q in the solid torus. One might also place the
w’ around the middle of the initial longitude and start the algorithm of finding βwith the
initial points of the standard form of the knot at the middle of the meridian. Observe that
to find the standard (5,3) torus knot, by joining wto zin the complement of βcurve and
CABLE KNOTS ARE NOT THIN 15
then from zto win the complement of the meridian, one has to ‘flip’ Ording’s picture,
which would then look like Figure 6, on the torus and like Figure 7, on the lifted setting.
Below are two examples of doubly pointed bordered Heegaard diagrams of (D2×S1, Tp,q )
on the fundamental domain and then in the lift, where p= 3, q = 2 in the first case (Figure
4,5) and p= 5, q = 3 (Figure 6,7) in the second case.
We describe one more intersection point of the doubly pointed bordered diagram of Tp,q
in the solid torus. Observe that alives in the α0arc. For q > 0 now we describe the
other intersection point we’ll consider from the bordered diagram. It’ll suffice since for
q < 0, the doubly pointed bordered Heegaard diagram would be the reflection with respect
to the meridian in the fundamental domain picture for q > 0 case. If we enumerate the
intersection points of βcurve with α1arc starting from left, we call the first intersection
point b1. See Figure 4,6 for examples when q6= 1 and also Figure 3, when q= 1.
To find the element which survives in the homology, Petkova and Hom found all the A∞
relations from a bordered diagram of (p, 1) pattern in D2×S1. For our case, we will only
be interested in two generators and A∞relations concerning them. For any (p, q), it’s not
always easy to find all the disks and thus all the A∞-relations. Instead we will be looking
into these two specific relations, coming from the bordered Heegaard diagram H(p, q ):
Lemma 9. In CFA−(D2×S1, Tp,q ), these are two specific A∞relations :
m3(a, ρ3, ρ2) = Unw·a
m4(a, ρ3, ρ2, ρ1) = U·b1
0
3
1
2
a
b1
w
z
Figure 4. A genus one bordered Heegaard diagram H(3,2) of T3,2
16 SUBHANKAR DEY
a
b
w
w
w
w
w
w
w
ww w
w
w
w
ww
z
z
z
z
z
z z z
z
z
w
z
w
w
w
1
Figure 5. A part of the lifted bordered Heegaard diagram H(3,2) of T3,2
0
1
23
a
b
1
w
z
Figure 6. A genus one bordered Heegaard diagram H(5,3) of T5,3
CABLE KNOTS ARE NOT THIN 17
w
z
w
w
w
ww
w
w
w
w
w
w
w
w
w
w
w
w
w
zz
z
z
z
z
z
z
z
z
z
z
z
z
z
a
b1
Figure 7. A part of the lifted H(5,3) of T5,3
where nwis the number of w’s in the primitive positive periodic domain of H(p, q), where
by primitive we mean the generator of π2(a, a)∼
=Z.
Proof. For a given genus one doubly pointed bordered Heegaard diagram, we can look at
the fundamental domain of the torus and find a periodic domain joining a, bounded by
α-arcs and βcurve.We start with one of the representative of aon the domain and the
boundary will have ρ3and ρ2crossed, to reach to another representative of aon the other
vertical side. Then by following the whole βcurve, one can reach the arepresentative one
had started with, hence getting the first A∞relation.
Similarly, from the description of b1, one can see a domain bounded by α-arcs to the
right and βcurves to the left, joining aand b1, crossing ρ3, ρ2, ρ1in the process. Also, the
Upower takes care of number of wencountered inside the domain. Combining these, we
get the second A∞relation.
Lemma 10. Multiplicity of win the primitive periodic domain of H(p, q)is vx + 1, where
x, y, u, v are unique positive integers such that p=x+y, q =u+vsuch that vx −uy = 1.
Proof. Instead of looking at the lifts of αarcs, if we look at lifts of the αcircle such that after
pulling tight ˜
β, we get that ˜αand ˜
βintersecting at the lattice points (i.e αand βintersect
18 SUBHANKAR DEY
at only one point), then (Σ, α, β, z , w) becomes a doubly pointed Heegaard diagram of the
knot Tp,q ⊂S3. Hence the intersections of ˜
βand ˜αequals to dimF([
HFK (Tp,q)).
Corollary 2.6 in [29] states that the number of non-zero terms in the Alexander polyno-
mial of Tp,q is 2vx −1, where x, y, u, v are unique positive integers such that p=x+y, q =
u+vsuch that vx −uy = 1. Now since Tp,q are L-space knots (since positive surgery
along torus knots with certain coefficient produces lens space, by [19]), Ozsv´ath and Szab´o
showed in [24] that the L-space knots forms a ‘staircase’ complex and hence each such j
for which [
HFK (K, j)6= 0 is of dimension 1 i.e dimF([
HFK (Tp,q) is equal the number of
non-zero terms in ∆p,q, which is 2vx −1.
Now, in the lifted picture, whenever ˜
βcrosses a ˜α, since it creates two generators in
[
HFK (Tp,q) (and assuming that the complex is reduced i.e there is no disk formed which
doesn’t contain zor w), there has to be a wthat it’ll cross. Thus for 2vx −1 number
of intersections between ˜
βand ˜α, barring the one that’ll generate the d
H F (S3) (i.e the
intersection point denoted a), for every two points, there will be a wcrossed. Hence, the
number of w’s in the primitive periodic domain is equal 2vx−1−1
2+ 2 = vx + 1, where one
of the extra w’s come from the wlying inside the fundamental domain of torus, bounded
by the boundary of the periodic domain. The other wstays inside this periodic domain
because of how the βoccur according the algorithm of Ording in [20]. See Figure 8where
the initial and the end βstrands of the boundary of this periodic domain is drawn.
Next lemma will allow us to consider any element in the knot Floer complex of the cable,
of the form b1⊗ ·, irrespective of (K). The reason is that all elements coming from the
box tensor product of the form b1⊗ · are non-zero in the knot Floer homology of cable.
Lemma 11. In H(p, q), there is no Whitney disk connecting b1that does not contain w’s.
In other words, every A∞relation in C F A−(S1×D2, Tp,q), concerning b1, has a non-zero
Ucoefficients to it, when p>q.
Proof. From the algorithm, observe that to prove the claim of the lemma one has to show
that the second strand of βcurve, i.e the strand that starts from b1,should lie to the left
of w. If that happens, then we can see that since no disks are ‘allowed’ to contain zin it,
the only way any other disk connecting b1can exist if and only if it contains a winside it
and hence the claim.
Now to prove the claim, we observe that since according to the algorithm, p(tβ) misses
the standard form of the torus knot on the fundamental domain and also it is isotopic to
the standard longitude of the torus, from the Figure 8it’s easy to see that βwill cross
wfrom its left right after crossing b1, since each strand of the standard form of the knot
has slope greater than 1 (as p > q) and thus the first strand of the knot will intersect the
horizontal boundary of the fundamental domain to the left of w.
Hom showed in [10] that for a (p, 1) pattern, when (K) = 1, the element a⊗ξ2survives
in the homology and hence becomes a generator of d
H F (S3) and calculated the Alexander
grading of that generator to find the τof (p, pn + 1) cables of K.
CABLE KNOTS ARE NOT THIN 19
01
2
q - 1
.
.
.
...
.
3
q-1
...
p-1
...
...
...
w
z
w
q-1
...
...
...
p-1
z
a
b
1
a
Figure 8. The left most figure shows a normal form of Tp,q on the funda-
mental domain where the i-th strand on the left vertical line, counting from
top, gets identified with p+imod (q−1)-th strand on the right vertical line.
Second picture shows a standard form of Tp,q on the fundamental domain.
The right most picture shows the eventual standard form on a ‘bordered’
fundamental domain, on which one can apply Ording’s algorithm to find
H(p, q) i.e the doubly pointed bordered diagram of Tp.q ⊂D2×S1. The
dotted curve shows the start and the ending of the βcurve
For general (p, q) patterns, we observe that the same is true as well when t6= 0 and
(K) = 1. This will makes sure that when (K) = 1 and t6= 0, then a⊗ξ2will survive in
the homology in the knot Floer complex of the Kp,q , when we take U= 0 in the differential
∂−. In other words, a⊗ξ2is a non-zero element of \
H F K(S3, Kp,q ).
Lemma 12. When (K)=1 and t6= 0,a⊗ξ2is an element that survives in homology in
box tensor.
Proof. First we look at the description of [
CFD(XK, r) generators and we observe the
incoming and outgoing arrows and the corresponding coefficients to and from ξ2in it. The
immediate incoming and outgoing arrows to and from ξ2consist of coefficients D123, D3, D2.
To start with, there is no A∞relation such as mk(a, ρ123,· · · ) = cor mk(d, · · · , ρ123) = a
for some c, d ∈C F D−(D2×S1, Tp,q), since around a,˜
βalways has positive slope in H(p, q)
and since in the algorithm, βalways runs along the standard form of the knot and only
changes its direction around w.
Also, there is no A∞-relation such as mk(c, ρI,· · · ) = awhere c6=ai.e there is no
Whitney disk which starts from aand is bounded by ˜α’s to the left and ˜
β’s to the right
and ends at c(that doesn’t include z), where c6=a.
Also, a A∞relation mk+1(a, ρ3, ρ23 , ρ23,· · · , ρ23) = c, for some cis not possible, since a
Whitney disk in H(p, q), starting from awith α’s to the right and βto the left and having
ρ23 in it, should contain ρ2and ρ1as well.
Combining these observations completes the proof.
20 SUBHANKAR DEY
For this specific subcase, one of the desired generators is a⊗ξ2, while the other one will
be b1⊗κi
`i, for some `. We shall figure out iduring our eventual grading calculation.
At this moment, let us recall the grading scheme in bordered Floer homology from [14,
Chapter 10]. The grading for elements of a bordered Floer complex, gr, takes values in
a non-commutative group G(Z), whose elements are triples of the form (m;i, j) where
m, i, j ∈1
2Z, i +j∈Z, where the half integer mis the Maslov component, the pair (i, j) is
the spinc-component. We’ll also be interested in ˜
G=G(Z)×Z, where the last component
reflects the Ugrading. The group law is defined by :
(m1;i1, j1;n1)·(m2;i2, j2;n2) = (m1+m2+ (i1j2−i2j1); i1+i2, j1+j2;n1+n2)
G(Z) has these grading on Reeb elements :
gr(ρ1)=(−1
2;1
2,−1
2)
gr(ρ2)=(−1
2;1
2,1
2)
gr(ρ3)=(−1
2;−1
2,1
2)
along with this rule that gr(ρIρJ) = gr(ρI)gr(ρJ) and gr(ρI J ) = λgr(ρJ)gr(ρI), (where
IJ ∈ {12,23,123}) where λ= (1; 0,0) ∈G(Z).
Recall that the set of all periodic domain is isomorphic to H2(Y , ∂Y )∼
=Z.
If we call the image of the generator of this group in ˜
Gby g, Then for a multiplication
map mk+1(x, ρi1,· · · , ρik) = Umyin CFA−(Y , K), we have
gr(y) = λk−1gr(x)gr(ρi1)· · · gr(ρik)·u−m∈ hgi\ ˜
G
where gr(u) = (0; 0,0; −1) ∈˜
G(Observe that both λ, u are in the centralizer of ˜
G).
If we call the image of the generator of periodic domains in ˜
G,h, then if DIis a coefficient
map from xto yin [
CFD(Y, r), then we have
gr(y) = λ−1gr(ρI)−1gr(x)∈˜
G/hhi
The box tensor product between [
CFD and CFA−of two manifolds with torus boundary is
then graded by hgi/˜
G\hhi. Every element in this double-coset space is uniquely equivalent
to an element of the form λaub, for some a, b ∈Zi.e the grading of that element takes form
(a; 0,0; −b).
We recall that the z-normalized Maslov grading N, defined by Lipshitz, Ozsv´ath, Thurston
in [14, Section 11.3] can also be realized like this : N=M−2A([Equation 11.13][14] and
N= 0 for the generator of H∗(gC F K−(K)/U = 1) ∼
=Z([14, Equation 11.15]), where
Mdenotes the Maslov grading and Adenotes the Alexander grading of some element in
gC F K−(K). The first co-ordinate afrom the discussion above, is the value of N, upto
an additive constant. The last co-ordinate bfrom above, is the Alexander grading, upto
an additive constant. See [14, 11.9] for an example showing how one can find the exact
CABLE KNOTS ARE NOT THIN 21
Maslov and Alexander grading, using Poincar´e polynomial, using the fact that weighted
Euler characteristics of knot Floer homology is the Poincar´e polynomial of the knot.
Now, if we have two elements in knot Floer homology of a thin knot, from the splicing
formula, whose grading reduce to (a1; 0,0; b1) and (a2;,0,0; b2),then N1=a1+c0=M1−
2A1, N2=a2+c0=M2−2A2and A1=b1+d0, A2=b2+d0(where c0and d0are
some additive constants for Nand A, respectively). Then since M1−M2should be equal
A1−A2, that implies that for thin knots,
a1−a2=b2−b1(3)
We’ll carry out the grading calculations and find two elements in the knot Floer homology,
by the splicing formula, for which Equation (3) fails to hold.
Now we calculate the grading of the two elements a⊗ξ2and b1⊗κi
`i. Since we’re just
interested in showing that the difference in their Maslov grading is not equal the difference
in their Alexander grading, the relative grading will do the job for us.
Let ξ2ibe some element in the vertically simplified basis of CFK −(K) such that :
ξ2i
D123
−−−→ κi
`i
D23
−−→ · · · ξ2i−1and let A(ξ2i) = d, M (ξ2i) = m(by this we mean that
A is the Alexander grading and M is the Maslov grading of the element of the knot
Floer homology of K, that represents ξ2i). Then, using [14, Theorem A.11], we get that
gr(ξ2i) = λm−2d·gr(ρ23 )−d∈G(Z).
Since κi
`i=D123 ·ξ2i, then
gr(κi
`i) = λ−1·gr(ρ123 )−1·λm−2d·gr(ρ23)−d
Lemma 10 and Equation (3) imply that the grading set for CFA−is isomorphic to
gr(a)·˜
G=u−(vx+1) ·gr(ρ23)\˜
G. Theorem 8implies the grading set for [
CFD(XK, r) is
G/λ−1·gr(ρ12 )−1·gr(ρ23)−r, if the framing of the companion knot is r.
Then Lemma 9 implies
gr(b1) = λ·u−1·gr(ρ23 )·gr(ρ1)
∼λ·u−1·uvx+1 ·gr(ρ1)
∼λ·uvx ·gr(ρ1)
From the description of ξ2and using [14, Theroem 2], we get that
gr(ξ2) = λ−2τK+2τK·gr(ρ23 )τK=gr(ρ23)τK,
since the generator of homology of C(j= 0) has Alexander grading −τKand Maslov
grading −2τKand ξ2is a representative of that. Thus, since gr(a) = (0; 0,0; 0),
gr(a⊗ξ2) = gr(ρ23 )τK∼uτK(vx+1) = (0; 0,0; τK(vx + 1)) (4)
22 SUBHANKAR DEY
and
gr(b1⊗κi
`i) = gr(b)·gr(κi
`i) = λ·uvx ·gr(ρ1)·λ−1·gr(ρ123)−1·λm−2d·gr(ρ23)−d
=λm−2d·uvx ·gr(ρ23)−1·λ−1·gr(ρ23 )−d
=λm−2d−1·uvx ·gr(ρ23)−d−1
∼λm−2d−1·u−vdx
= (m−2d−1; 0,0; −vdx)
Now, we can choose m=−2τK, d =−τK, by choosing `= 1 i.e
gr(b1⊗κi
`i) = (−1; 0,0; vxτK) (5)
Getting back to proving that the difference between the relative Maslov grading of those
two elements are not equal the difference between their relative Alexander grading, we see
that if τK6=−1, by comparing with Equation (3) we’re done. Hence, we’re done with the
case when t6= 0, (K) = 1 and τK6=−1.
For t6= 0, (K)=1, τK=−1, we look at the unstable chain of ι1part of [
CFD(XK, r),
mentioned in Theorem 8. If t < 0, then µ1=D123 ·ξ0. If t > 0, then µ1=D1·ξ0. We
calculate the grading of µ1for both cases.
By Theorem 8,gr(ξ0) equals to λ2·ρ23 . Hence if t < 0,
µ1=D123 ·ξ0⇒gr(µ1) = λ−1·gr(ρ123 )−1·gr(ξ0)
=λ−1·gr(ρ123 )−1·λ2·gr(ρ23)
=λ−1·λ−1·gr(ρ1)−1·gr(ρ23 )−1·λ2·gr(ρ23)
∼gr(ρ1)−1
Then
gr(b1⊗µ1) = gr(b1)·gr(µ1)
=λ·uvx ·gr(ρ1)·gr(ρ1)−1
∼λ·uvx
= (1; 0,0; vx)
If t > 0,
µ1=D1·ξ0⇒gr(µ1) = λ−1·gr(ρ1)−1·gr(ξ0)
=λ−1·gr(ρ1)−1·λ2·gr(ρ23 )
=λ·gr(ρ1)−1·gr(ρ23 )
CABLE KNOTS ARE NOT THIN 23
Then
gr(b1⊗µ1) = gr(b1)·gr(µ1)
=λ·uvx ·gr(ρ1)·λ·gr(ρ1)−1·gr(ρ23)
∼λ2·u2vx
= (2; 0,0; 2vx)
Now comparing gr(a⊗ξ2) and gr(b⊗µ1), we get that Equality (3) happens iff vx =−1.
Now dim(\
H F K(S3, Tp,q )) = 2vx −1≥3 (see [29, Corollary 2.6]).That implies that
vx should be greater than or equal to 2. This is because the only torus knot(s) whose
knot Floer homology has rank 3 are ±T2,3. To see this, note that [
HFK (K, g(K)) ∼
=
[
HFK (K, −g(K)) 6= 0 and [
HFK (K, 0) 6= 0 and the only genus 1 fibered knots are ±T2,3
and figure eight knots, as [
HFK (K, g(K)) ∼
=Zimplies that K⊂S3is fibered but the figure
eight knot is not a torus knot.
Hence we have a contradiction to the possible value of vx. So we have a⊗ξ2and b⊗µ1
are non-zero elements (by Lemma 11 and 12) in the knot Floer homology of the cable such
that difference between their Maslov grading cannot be equal the difference between their
Alexander grading.
Now if (K)=0, then we consider an element ξ, which lies in the lowest Alexander
grading in [
HFK (K) i.e A(ξ) = −g. Since (K) = 0 implies τK= 0, there is a ξ0∈[
HFK (K)
such that ∂vertξ0=ξ. Then there is some ξ2sin the vertically simplified basis of CFK −
such that A(ξ2s) = A(ξ) = −gand let M(ξ2s) = M(ξ), (see the Proof of [10, Lemma 2.1]).
We choose our `to be s. Then
gr(b1⊗κs
`s) = (m+ 2g−1; 0,0; vgx)
gr(a⊗ξ2s)=(m+ 2g; 0,0; vgx)
Lemma 13. If (K) = 0, t 6= 0, then a⊗ξ2sis non-trivial in the knot Floer homology of
the cable Kp,q.
Proof. Since (K) = 0 implies τK= 0 and ξ2sis not a generator of either vertical or
horizontal homology, we just have to check that there is no possibility of any A∞relations
in CFA−(D2×S1, Tp,q ) involving a, which has the same coefficients from the vertical and
horizontal chain of [
CFD(XK, r).
To do this, first we check the vertical chain in [
CFD(XK, r) and the coefficient maps from
Theorem (8). We said in the proof of Lemma 12, there is no disk connecting aand starting
with coefficient ρ123, since in the algorithm, the βcurve always lie along the standard form
of the knot, and only changes its direction around w. Hence there is no A∞relation such
as mk(a, ρ123,· · · ) = c, for some c.
Now, we check the vertical chain in [
CFD(XK, r) and the coefficient maps from The-
orem 8. We can see that the only relation involving ρ2, ρ3and ρ23 and involving ais
24 SUBHANKAR DEY
mk(a, ρ3, ρ23,· · · , ρ2) = Ul·a, where lis non-zero, as the said relation indicates a posi-
tive multiple of the primitive periodic domain. Thus the element a⊗ξ2sis non-zero in
[
HFK (Kp,q).
Hence our claim is proved.
We’ll investigate the cases (K) = −1 separately, when we’re done with all the other
cases for t= 0 and t6= 0.
Case 2. Let t= 2τ(K)−r= 0, then we follow the same strategy that we followed for
the last subcase in t6= 0 case. We consider an element ξ0, which is neither a generator
of the vertical homology, nor that of the horizontal homology, and gets ‘killed’ in the
vertical chain complex i.e there is a ξ00 ∈[
HFK (K) such that ξ0∈∂vertξ00 . Then there is
some ξ2sin the vertically simplified basis of CFK −such that A(ξ2s) = A(ξ0) = Aand let
M(ξ2s) = M(ξ0) = M, (see the proof of [10, Lemma 2.1]). We choose our `=s. Then
gr(b1⊗κs
`s)=(M−2A−1; 0,0; −vAx)
gr(a⊗ξ2s)=(m−2A; 0,0; −vAx)
For this case, we can see that Equation (3) will fail to happen and hence the difference
between the Alexander grading and Maslov grading of a⊗ξ2sand b⊗κs
`sare not equal.
We claim this :
Lemma 14. If t= 0, then a⊗ξ2sis non-trivial in the knot Floer homology of the cable.
Proof. Since t6= 0 implies and ξ2sis not a generator of vertical homology, we just have to
check that there is no possibility of any A∞relations in CFA−(D2×S1, Tp,q ) involving a,
which has exact same coefficients from the vertical and horizontal chain of [
CFD(XK, r).
To do that, first we check the vertical chain in [
CFD(XK, r) and the coefficient maps
from Theorem 8. As we’ve said in the proof of Lemma 12, there is no disk connecting a
and starting with coefficient ρ123, since in the algorithm, the βcurve always lie along the
standard form of the knot, and only changes its direction around w. Hence there is no A∞
relation such as mk(a, ρ123,· · · ) = c, for some c.
Now, we check the horizontal chain in [
CFD(XK, r) and the coefficient maps from The-
orem (8). We can see that the only relation involving ρ2, ρ3andρ23 and involving ais
mk(a, ρ3, ρ23,· · · , ρ2) = Un·a, where nis non-zero, as the said relation indicates a positive
multiple of the primitive periodic domain. Thus the element a⊗ξ2sis non-zero in the knot
Floer homology of the cable.
Now, we check the unstable chain in [
CFD(XK, r) and coefficient maps there when t= 0.
There is an incoming D12 map towards η0, the generator of horizontal homology. Even if
ξ2s=η0, one can check that a⊗ξ2swill still be non-trivial in the knot Floer homology of
the cable, since there is A∞relation such as m2(c, ρ12) = a, for any c, in other words, there
is no Whitney disk, connecting aand some other intersection point, which is bounded by
βcurve to the right and α-arcs to the left, starting with a. Hence our claim is proved.
Also, by Lemma 11, b⊗κs
`sis non-zero in the knot Floer homology of the cable.
CABLE KNOTS ARE NOT THIN 25
Now, the remaining case: when (K) = −1, then we can look at −Kinstead. Since
(−K) = −(K) = 1 and also Kp,q = (−K)−p,q i.e we’re already done with the case.
This completes our proof.
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Department of Mathematics, University at Buffalo
E-mail address:subhanka@buffalo.edu
