# Heegaard Floer invariants of Legendrian knots in contact three–manifolds

Paolo Lisca  Dipartimento di Matematica “L. Tonelli”, Università di Pisa
I-56127 Pisa, Italy<EMAIL_ADDRESS>Peter Ozsváth  Department of
Mathematics, Columbia University,
New York 10027, USA<EMAIL_ADDRESS>András I. Stipsicz  Rényi
Institute of Mathematics, Budapest, Hungary and
Department of Mathematics, Columbia University, New York 10027, USA
<EMAIL_ADDRESS>Zoltán Szabó  Department of Mathematics, Princeton
University,
Princeton, New Jersey 08544, USA<EMAIL_ADDRESS>
###### Abstract

We define invariants of null–homologous Legendrian and transverse knots in
contact 3–manifolds. The invariants are determined by elements of the knot
Floer homology of the underlying smooth knot. We compute these invariants, and
show that they do not vanish for certain non–loose knots in overtwisted
3–spheres. Moreover, we apply the invariants to find transversely non–simple
knot types in many overtwisted contact 3–manifolds.

###### keywords:

Legendrian knots, transverse knots, Heegaard Floer homology

57M27 57R58

##  1 Introduction

The reformulation of knot Floer homology (introduced originally in [35, 39])
through grid diagrams [26, 27] provided not only a combinatorial way of
computing the knot Floer homology groups of knots in $S^{3}$, but also showed
a natural way of defining invariants of Legendrian and transverse knots in the
standard contact 3–sphere $(S^{3},\xi_{st})$, see [37]. As it is shown in [29,
44], such invariants can be effectively applied to study transverse simplicity
of knot types. The definition of these invariants relies heavily on the
presentation of the knot through a grid diagram, hence does not generalize
directly to Legendrian and transverse knots in other closed contact
3–manifolds. The aim of the present paper is to define invariants for any
null–homologous Legendrian (and transverse) knot $L\subset(Y,\xi)$ (resp.
$T\subset(Y,\xi)$). As demonstrated by explicit computations, these
constructions give interesting invariants for Legendrian and transverse knots,
even in cases where the ambient contact structure is overtwisted.

Recall [35] that a smooth null–homologous knot $K\subset Y$ in a closed
3–manifold $Y$ gives rise to the _knot Floer homology groups_
$\widehat{\mathrm{HFK}}(Y,K)$ and $\mathrm{HFK}^{-}(Y,K)$ (see also Section 2
for a short review of Heegaard Floer homology). These groups are computed as
homologies of appropriate chain complexes, which in turn result from
doubly–pointed Heegaard diagrams of the pair $(Y,K)$. For an isotopic pair
$K_{1},K_{2}\subset Y$ the corresponding chain complexes are quasi–isomorphic
and hence the homologies are isomorphic. In this sense the knot Floer homology
groups (as abstract groups) are invariants of the isotopy class of the knot
$K$. For the sake of simplicity, throughout this paper we use Floer homology
with coefficients in $\mathbb{F}:=\mathbb{Z}/2\mathbb{Z}$.

In this paper we define the invariants ${\mathfrak{L}}(L)$ and
$\widehat{\mathfrak{L}}(L)$ of an oriented null–homologous Legendrian knot
$L\subset(Y,\xi)$. For an introduction to Legendrian and transverse knots see
[10]. (We will always assume that $Y$ is oriented and $\xi$ is cooriented.)
Recall that the contact invariant $c(Y,\xi)$ of a contact 3–manifold $(Y,\xi)$
— as defined in [34] — is an element (up to sign) of the Floer homology of
$-Y$ (rather than of $Y$). In the same vein, the invariants
${\mathfrak{L}}(L)$ and $\widehat{\mathfrak{L}}(L)$ are determined by elements
in the Floer homology of $(-Y,L)$. If $Y$ admits an orientation reversing
diffeomorphism $\mu$ (like $S^{3}$ does), then the Floer homology of $(-Y,L)$
can be identified with the Floer homology of $(Y,\mu(L))$.

More precisely, we will show that, given an oriented, null–homologous
Legendrian knot $L$ in a closed contact 3–manifold $(Y,\xi)$, one can choose
certain auxiliary data $D$ which, together with $L$, determine a cycle
${\bf{x}}(L,D)$ in a complex defining the $\mathbb{F}[U]$–modules
$\mathrm{HFK}^{-}(-Y,L,\mathbf{t}_{\xi})$ or
$\widehat{\mathrm{HFK}}(-Y,L,\mathbf{t}_{\xi})$ (with trivial $U$–action),
where $\mathbf{t}_{\xi}$ is the ${\mathrm{Spin}}^{c}$ structure on $Y$ induced
by $\xi$ (see Section 2). A different choice $D^{\prime}$ of the auxiliary
data determines $\mathbb{F}[U]$–module automorphisms sending the class
$[{\bf{x}}(L,D)]$ to the class $[{\bf{x}}(L,D^{\prime})]$. Formulating this a
bit more formally, we can consider the set of pairs $(M,m)$, where $M$ is an
$\mathbb{F}[U]$–module and $m\in M$, and introduce an equivalence relation by
declaring two pairs $(M,m)$ and $(N,n)$ equivalent if there is an
$\mathbb{F}[U]$–module isomorphism $f\colon\thinspace M\to N$ such that
$f(m)=n$. Let $[M,m]$ denote the equivalence class of $(M,m)$. Then, our main
result can be stated as follows:

######  Theorem 1.1.

Let $L$ be an oriented, null–homologous Legendrian knot in the closed contact
3–manifold $(Y,\xi)$, and let $\mathbf{t}_{\xi}$ be the ${\mathrm{Spin}}^{c}$
structure on $Y$ induced by $\xi$. Then, after choosing some suitable
auxiliary data, it is possible to associate to $L$ homology classes
$\alpha_{{\mathfrak{L}}}(L)\in\mathrm{HFK}^{-}(-Y,L,\mathbf{t}_{\xi})$ and
$\alpha_{\widehat{\mathfrak{L}}}(L)\in\widehat{\mathrm{HFK}}(-Y,L,\mathbf{t}_{\xi})$
such that

${\mathfrak{L}}(L):=[\mathrm{HFK}^{-}(-Y,L,\mathbf{t}_{\xi}),\alpha_{{\mathfrak{L}}}(L)]$

and

$\widehat{\mathfrak{L}}(L):=[\widehat{\mathrm{HFK}}(-Y,L,\mathbf{t}_{\xi}),\alpha_{\widehat{\mathfrak{L}}}(L)]$

do not depend on the choice of the auxiliary data, and in fact only depend on
the Legendrian isotopy class of $L$.

We can define multiplication by $U$ on the set of equivalence classes $[M,m]$
by setting $U\cdot[M,m]:=[M,U\cdot m]$. We will say that ${\mathfrak{L}}(L)$
is _vanishing_ (respectively _nonvanishing_) and write ${\mathfrak{L}}(L)=0$
(respectively ${\mathfrak{L}}(L)\neq 0$) if
${\mathfrak{L}}(L)=[\mathrm{HFK}^{-}(-Y,L,\mathbf{t}_{\xi}),0]$ (respectively
${\mathfrak{L}}(L)\neq[\mathrm{HFK}^{-}(-Y,L,\mathbf{t}_{\xi}),0]$). Similar
conventions will be used for $\widehat{\mathfrak{L}}(L)$. Let $-L$ denote the
knot $L$ with reversed orientation. It turns out that the pairs
${\mathfrak{L}}(\pm L)$ and $\widehat{\mathfrak{L}}(\pm L)$ admit properties
similar to those of the pair of invariants $\lambda_{\pm}(L)$ of [37].

One useful feature of the invariant ${\mathfrak{L}}(\pm L)$ (shared with
$\lambda_{\pm}$ from [37]) is that it satisfies a nonvanishing property, which
can be formulated in terms of the contact invariant $c(Y,\xi)$ of the ambient
contact 3–manifold $(Y,\xi)$.

######  Theorem 1.2.

If the contact Ozsváth–Szabó invariant
$c(Y,\xi)\in\widehat{\mathrm{HF}}(-Y,\mathbf{t}_{\xi})$ of the contact
3–manifold $(Y,\xi)$ does not vanish, then for any oriented Legendrian knot
$L\subset(Y,\xi)$ we have ${\mathfrak{L}}(L)\neq 0$. If $c(Y,\xi)=0$ then for
$d$ large enough $U^{d}\cdot{\mathfrak{L}}(L)$ vanishes.

As it will be explained later, such strong nonvanishing property does not hold
for $\widehat{\mathfrak{L}}$. Since a strongly symplectically fillable contact
3–manifold has nonzero contact invariant while for an overtwisted structure
$c(Y,\xi)=0$, it follows immediately from Theorem 1.2 that

######  Corollary 1.3.

If $(Y,\xi)$ is strongly symplectically fillable then for any null–homologous
Legendrian knot $L\subset(Y,\xi)$ the invariant ${\mathfrak{L}}(L)$ is
nonvanishing. If $(Y,\xi)$ is overtwisted, then for any Legendrian knot $L$
there is $d\geq 0$ such that $U^{d}\cdot{\mathfrak{L}}(L)$ vanishes. ∎

A stronger vanishing theorem holds for loose knots. Recall that a Legendrian
knot $L\subset(Y,\xi)$ is _loose_ if its complement contains an overtwisted
disk (and hence $(Y,\xi)$ is necessarily overtwisted). A Legendrian knot
$L\subset(Y,\xi)$ is _non–loose_ (or _exceptional_ in the terminology of [8])
if $(Y,\xi)$ is overtwisted, but the complement of $L$ is tight.

######  Theorem 1.4.

If $L\subset(Y,\xi)$ is an oriented, null–homologous and loose Legendrian
knot, then ${\mathfrak{L}}(L)=0$.

Transverse knots admit a preferred orientation, and can be approximated,
uniquely up to negative stabilization, by oriented Legendrian knots [9, 13].
This fact can be used to define invariants of transverse knots:

######  Theorem 1.5.

Suppose that $T$ is a null–homologous transverse knot in the contact
3–manifold $(Y,\xi)$. Let $L$ be a (compatibly oriented) Legendrian
approximation of $T$. Then, $\mathfrak{T}(T):={\mathfrak{L}}(L)$ and
$\widehat{\mathfrak{T}}(T):=\widehat{\mathfrak{L}}(L)$ are invariants of the
transverse knot type of $T$.

The proof of this statement relies on the invariance of the Legendrian
invariant under negative stabilization. After determining the invariants of
stabilized Legendrian unknots in the standard contact 3–sphere and the
behaviour of the invariants under connected sum, in fact, we will be able to
determine the effect of both kinds of stabilization on the invariant
${\mathfrak{L}}$, leading us to

######  Theorem 1.6.

Suppose that $L$ is an oriented Legendrian knot and $L^{+}$, resp. $L^{-}$
denote the oriented positive, resp. negative stabilizations of $L$. Then,
${\mathfrak{L}}(L^{-})={\mathfrak{L}}(L)$ and
${\mathfrak{L}}(L^{+})=U\cdot{\mathfrak{L}}(L)$.

The invariants can be effectively used in the study of non–loose Legendrian
knots. Notice that the invariant of a non–loose Legendrian knot is necessarily
a $U$–torsion element, and can be nonvanishing only if the knot and all its
negative stabilizations are non–loose. In Section 6 a family of non–loose
torus knots in overtwisted contact $S^{3}$’s are constructed, for which we can
determine the invariants ${\mathfrak{L}}$ by direct computation.

Recall that a knot type $K$ in a contact three-manifold is said to be
transversely non-simple if there are two transversely non-isotopic transverse
knots in $Y$ in the topological type of $K$ which have the same self-linking
number. (For more background on transverse non-simplicity, see [10, 13].) In
an overtwisted contact structure however, this definition admits refinements.
(For more on knots in overtwisted contact structures, see [5, 6, 8, 16].)
Namely, it is not hard to find examples of pairs of _loose_ and _non–loose_
(Legendrian or transverse) knots with equal ’classical’ invariants. By
definition their complements admit different contact structures (one is
overtwisted, the other is tight), hence the knots are clearly not
Legendrian/transverse isotopic. Examples for this phenomenon will be discussed
in Section 8. Finding pairs of Legendrian/transverse knots with equal
classical invariants, both loose or both non–loose, is a much more delicate
question. Non–loose pairs of Legendrian knots not Legendrian isotopic were
found in [12]; by applying our invariants we find transverse knots with
similar properties:

######  Theorem 1.7.

The knot type $T_{(2,-7)}\\#T_{(2,-9)}\subset S^{3}$ has two non–loose,
transversely non–isotopic transverse representatives with the same
self–linking number with respect to the overtwisted contact structure
$\xi_{12}$ with Hopf invariant $d_{3}(\xi_{12})=12.$

In fact, by further connected sums we get a more general statement:

######  Corollary 1.8.

Let $(Y,\xi)$ be a contact 3–manifold with $c(Y,\xi)\neq 0$. Let $\zeta$ be an
overtwisted contact structure on $Y$ with
$\mathbf{t}_{\zeta}=\mathbf{t}_{\xi}$. Then, in $(Y,\zeta)$ there are
null–homologous knot types which admit two non–loose, transversely
non–isotopic transverse representatives with the same self–linking number.

######  Remarks 1.9.

(a) Notice that to a Legendrian knot $L$ in the standard contact 3–sphere
$(S^{3},\xi_{st})$ one can now associate several sets of invariants:
$\lambda_{\pm}(L)$ of [37] and ${\mathfrak{L}}(\pm L)$,
$\widehat{\mathfrak{L}}(\pm L)$ of the present work. It would be interesting
to compare these elements of the knot Floer homology groups.

(b) We also note that recent work of Honda-Kazez-Matić [20] provides another
invariant for Legendrian knots $L\subset(Y,\xi)$ through the sutured contact
invariant of an appropriate complement of $L$ in $Y$, using the sutured Floer
homology of Juhász [21]. This invariant seems to have slightly different
features than the invariants defined in this paper; the relationships between
these invariants have yet to be understood.111Added in proof: the relation
between $\widehat{{\mathfrak{L}}}$ and the sutured invariant has been recently
worked out in [42].

(c) In [12] arbitrarily many distinct non–loose Legendrian knots with the same
classical invariants are constructed in overtwisted contact 3–manifolds. In
[14] similar examples were constructed in the standard tight contact $S^{3}$,
using connected sums of torus knots. These constructions, however, do not tell
us anything about transverse simplicity of the knot types.

(d) Note that connected summing preserves transverse simplicity [14] while
torus knots are transversely simple with respect to the standard tight contact
$S^{3}$ [13, 28]. Therefore, in the standard contact 3–sphere there are no
examples such as those of Theorem 1.7.

The paper is organized as follows. In Section 2 we recall facts about open
books and contact structures, Heegaard Floer groups, and the contact
Ozsváth–Szabó invariants. In Section 3 we establish some preliminary results
on Legendrian knots and open books, we define our invariants and we prove
Theorem 1.1. In Section 4 we compute the invariants of a certain (stabilized)
Legendrian unknot in the standard contact $S^{3}$. This calculation is both
instructive and useful: it will be used in the proof of the stabilization
invariance. In Section 5 we prove Theorems 1.2, 1.4 and 1.5. In Section 6 we
determine the invariant $\widehat{\mathfrak{L}}$ for some non–loose Legendrian
torus knots in certain overtwisted contact structures on $S^{3}$. In Section 7
we describe the behaviour of the invariants with respect to Legendrian
connected sum, and then derive Theorem 1.6. In Section 8 we discuss transverse
simplicity in overtwisted contact 3–manifolds, give a refinement of the
Legendrian invariant and prove Theorem 1.7 and Corollary 1.8 modulo some
technical results which are deferred to the Appendix.

Acknowledgements: PSO was supported by NSF grant number DMS-0505811 and
FRG-0244663. AS acknowledges support from the Clay Mathematics Institute. AS
and PL were partially supported by OTKA T49449 and by Marie Curie TOK project
BudAlgGeo. ZSz was supported by NSF grant number DMS-0704053 and FRG-0244663.
We would like to thank Tolga Etgü for many useful comments and corrections.

##  2 Preliminaries

Our definition of the Legendrian knot invariant relies on the few basic facts
listed below.

  * •

There is a one–to–one correspondence between (isotopy classes of) contact
structures and open book decompositions (up to positive stabilization), as it
was shown by Giroux [2, 11, 18].

  * •

An idea of Giroux (cf. [2, 11]) allows one to construct an open book
decomposition of $Y$ compatible with a contact structure $\xi$, which is also
adapted to a given Legendrian knot $L\subset(Y,\xi)$.

  * •

There is a contact invariant $c(Y,\xi)$ of a closed contact 3–manifold
$(Y,\xi)$ originally defined in [34] and recently reformulated by
Honda–Kazez–Matić [19] (cf. also [20]).

We describe these ingredients in more detail in the following subsections. In
Subsection 2.1 we recall how contact 3–manifold admit open book decompositions
adapted to given Legendrian knots. In Subsection 2.2 (which is not logically
required by the rest of this article, but which fits in neatly at this point)
we explain how, conversely, an isotopy class of embedded curves in a page of
an open book decomposition gives rise to a unique isotopy class of Legendrian
knots in the associated contact three–manifold. In Subsection 2.3 we recall
the basics of Heegaard Floer homology, mainly to set up notation, and finally
in Subsection 2.4 we recall the construction of the contact invariant.

###  2.1 Generalities on open books and contact structures

Recall that an open book decomposition of a 3–manifold $Y$ is a pair
$(B,\varphi)$ where $B\subset Y$ is a (fibered) link in $Y$ and $\varphi\colon
Y-B\to S^{1}$ is a locally trivial fibration such that the closure of each
fiber, $S_{t}={\overline{\varphi^{-1}(t)}}$ (a _page_ of the open book), is a
Seifert surface for the _binding_ $B$. The fibration $\varphi$ can also be
determined by its monodromy $h_{\varphi}\colon S_{+1}\to S_{+1}$, which gives
rise to an element of the mapping class group of the page (regarded as a
surface with boundary).

An open book decomposition $(B,\varphi)$ can be modified by a classical
operation called _stabilization_ [41]. The page $S^{\prime}$ of the resulting
open book $(B^{\prime},\varphi^{\prime})$ is obtained from the page $S$ of
$(B,\varphi)$ by adding a 1–handle $H$, while the monodromy of
$(B^{\prime},\varphi^{\prime})$ is obtained by composing the monodromy of
$(B,\varphi)$ (extended trivially to $S^{\prime}$) with a Dehn twist
$D_{\gamma}$ along a simple closed curve $\gamma\subset S^{\prime}$
intersecting the cocore of $H$ transversely in a single point. Depending on
whether the Dehn twist is right– or left–handed, the stabilization is called
_positive_ or _negative_. A positive stabilization is also called a _Giroux
stabilization_.

######  Definition 2.1.

(Giroux) A contact structure $\xi$ and an open book decomposition are
_compatible_ if $\xi$, as a cooriented 2–plane field, is the kernel of a
contact one–form $\alpha$ with the property that $d\alpha$ is a symplectic
form on each page, hence orients both the page and the binding, and with this
orientation the binding $B$ is a link positively transverse to $\xi$. In this
situation we also say that the contact one–form $\alpha$ is compatible with
the open book.

A theorem of Thurston and Winkelnkemper [43] can be used to verify that each
open book admits a compatible contact structure. Moreover, Giroux [2, 11, 18]
proved the following:

  1. 1.

Each contact structure is compatible with some open book decomposition;

  2. 2.

Two contact structures compatible with the same open book are isotopic;

  3. 3.

If a contact structure $\xi$ is compatible with an open book $(B,\varphi)$,
then $\xi$ is isotopic to any contact structure compatible with a positive
stabilization of $(B,\varphi)$.

The construction of an open book compatible with a given contact structure
rests on a _contact cell decomposition_ of the contact 3–manifold $(Y,\xi)$,
see [11, Section 4] or [2, Subsection 3.4]. In short, consider a
$CW$–decomposition of $Y$ such that its 1–skeleton $G$ is a Legendrian graph,
each 2–cell $D$ satisfies the property that the twisting of the contact
structure along its boundary $\partial D$ (with respect to the framing given
by $D$) is $-1$ and the 3–cells are in Darboux balls of $(Y,\xi)$. Then, there
is a compact surface $R$ (a ribbon for $G$) such that $R$ retracts onto $G$,
$T_{p}R=\xi_{p}$ for all $p\in G$ and $T_{p}R\neq\xi_{p}$ for $p\in R-G$. The
3–manifold $Y$ admits an open book decomposition with binding $\partial R$ and
page $R$ which is compatible with $\xi$. We also have the following (see [11,
Theorem 4.28] or [2, Theorem 3.4]):

######  Theorem 2.2.

(Giroux) Two open books compatible with the same contact structure admit
isotopic Giroux stabilizations.∎

The proof of this statement rests on two facts. The first fact is given by

######  Lemma 2.3.

([11, Lemma 4.29] or [2, Proposition 3.7]) Each open book decomposition
compatible with $(Y,\xi)$ admits a sequence of Giroux stabilizations such that
the resulting open book comes from a contact cell decomposition. ∎

The second fact (see the proof of [11, Theorem 4.28]) is that any two contact
cell decompositions can be connected by a sequence of operations of the
following types: (1) a subdivision of a 2–cell by a Legendrian arc
intersecting the dividing set (for its definition see [13]) of the 2–cell
once, (2) an addition of a 1–cell $c^{\prime}$ and a 2–cell $D$ so that
$\partial D=c\cup c^{\prime}$, where $c$ is part of the original 1–skeleton
and the twisting of the contact structure along $\partial D$ (with respect to
$D$) is $-1$, and (3) an addition of a 2–cell $D$ whose boundary is already in
the 2–skeleton and satisfies the above twisting requirement. It is not hard to
see that operations (1) and (2) induce positive stabilizations on the open
book associated to the cell decomposition, while (3) leaves the open book
unchanged.

Thus, an invariant of open book decompositions which is constant under
positive stabilizations is, in fact, an isotopy invariant of the compatible
contact structure. Since in the construction of a contact cell decomposition
for a contact 3–manifold $(Y,\xi)$ one can choose the $CW$–decomposition of
$Y$ in such a way that a Legendrian knot (or link) $L\subset(Y,\xi)$ is
contained in its 1–skeleton $G$, we have:

######  Proposition 2.4.

([11, Corollary 4.23]) Given a Legendrian knot $L$ in a closed, contact
3–manifold $(Y,\xi)$, there is an open book decomposition compatible with
$\xi$, containing $L$ on a page $S$ and such that the contact framing of $L$
is equal to the framing induced on $L$ by $S$. The open book can be chosen in
such a way that $L$ is homologically essential on the page $S$. ∎

Recall (see e.g. [11]) that any open book is obtained via a mapping torus
construction from a pair (sometimes called an _abstract open book_)
$(S,\varphi)$, where $S$ is an oriented surface with boundary and
$\varphi\colon S\to S$ is an orientation–preserving diffeomorphism which
restricts as the identity near $\partial S$. Our previous observations amount
to saying that any triple $(Y,\xi,L)$, where $L\subset Y$ is a Legendrian knot
(or link) is obtained via the standard mapping torus construction from a
triple $(S,L,\varphi)$, where $L\subset S$ is a homologically essential simple
closed curve.

######  Definition 2.5.

Let $(S,\varphi)$ be an abstract open book, and let $L\subset S$ be a
homologically essential simple closed curve. We say that the Giroux
stabilization $(S^{\prime},R_{\gamma}\circ\varphi)$ is _$L$ –elementary_ if
after a suitable isotopy the curve $\gamma$ intersects $L$ transversely in at
most one point.

The construction of the Legendrian invariant rests on the following:

######  Proposition 2.6.

Suppose that $L\subset(Y,\xi)$ is a Legendrian knot in a contact 3–manifold.
If the triple $(Y,\xi,L)$ is associated via the mapping torus construction
with two different triples ${\cal T}_{i}=(S_{i},L_{i},\varphi_{i})$, $i=1,2$,
then $(Y,\xi,L)$ is also associated with a triple $(S,L,\varphi)$, obtained
from each of ${\cal T}_{1}$ and ${\cal T}_{2}$ by a finite sequence of
$L$–elementary Giroux stabilizations.

###### Proof.

The proof of [11, Lemma 4.29] uses $L$–elementary stabilizations only (cf.
[11, Figure 12]) and (1)–(2) after Lemma 2.3 determine $L$–elementary
stabilizations as well, while (3) leaves the open book decomposition
unchanged. ∎

###  2.2 From curves on a page to Legendrian knots: uniqueness

Giroux’s results give a correspondence between Legendrian knots and knots on a
page in an open book decomposition. In fact, with a little extra work, this
correspondence can be suitably inverted. Although this other direction is not
strictly needed for our present applications (and hence, the impatient reader
is free to skip the present subsection), it does fit in naturally in the
discussion at this point. Specifically, in this subsection, we prove the
following:

######  Theorem 2.7.

Let $(B,\varphi)$ be an open book decomposition of the closed, oriented
3–manifold $Y$. Let $\xi=\ker(\alpha)$ be a contact structure, with $\alpha$ a
contact 1–form compatible with $(B,\varphi)$. Suppose that $K\subset
S:=\varphi^{-1}(1)$ is a smooth knot defining a nontrivial homology class in
$H_{1}(S;\mathbb{Z})$. Then the smooth isotopy class of $K$ on the page
uniquely determines a Legendrian knot in $(Y,\xi)$ up to Legendrian isotopy.

The proof of this theorem rests on two technical lemmas.

######  Lemma 2.8.

Let $(B,\varphi)$ be an open book decomposition of the closed, oriented
3–manifold $Y$. Let $K_{t}\subset S:=\varphi^{-1}(1)$ be a smooth family of
knots which are homologically nontrivial on the page and provide an isotopy
from $K_{0}$ to $K_{1}$. Then, there exists a smooth family of contact 1–forms
$\alpha_{K_{t}}\in\Omega^{1}(Y)$ compatible with $(B,\varphi)$ and such that
the restriction of $\alpha_{K_{t}}$ to $K_{t}$ vanishes.

###### Proof.

In view of the argument given in [11, pp. 115–116], it suffices to show that
there exists a smooth family $\lambda_{t}\in\Omega^{1}(S)$ such that, for each
$t$

  1. 1.

$\lambda_{t}=(1+s)d\theta$ near $\partial S$, with coordinates
$(s,\theta)\in[0,1]\times S^{1}$ near each boundary component of $S$;

  2. 2.

$d\lambda_{t}$ is a volume form on $S$;

  3. 3.

$\lambda_{t}$ vanishes on $K_{t}$.

To construct the family $\lambda_{t}$ we proceed as follows. For each $t$,
choose a closed collar $U$ around $K_{t}\subset S$, parametrized by
coordinates $(s,\theta)\in[-1,1]\times S^{1}$, so that $ds\wedge d\theta$ is a
volume form on $U$ with the orientation induced from $S$. Let
$\lambda_{1,t}\in\Omega^{1}(S)$ be of the form $(1+s)d\theta$ near $\partial
S$ and of the form $sd\theta$ on $U$. We have

$\int_{S}d\lambda_{1,t}=\int_{\partial S}\lambda_{1,t}=2\pi|\partial S|.$

Let $\omega$ be a volume form on $S$ such that:

  * •

$\int_{S}\omega=2\pi|\partial S|$;

  * •

$\omega=ds\wedge d\theta$ near $\partial S$ and on $U$.

(The first condition can be fulfilled since $K_{t}\subset S$ is nontrivial in
homology, hence each component of its complement meets $\partial S$.) Let
$U^{\prime}\subset U$ correspond to $[-1/2,1/2]\times S^{1}\subset[-1,1]\times
S^{1}$. We have

$\int_{\overline{S\setminus
U^{\prime}}}\omega-d\lambda_{1,t}=\int_{S}\omega-d\lambda_{1,t}=0.$

Since $\omega-d\lambda_{1,t}=0$ near $\partial(\overline{S\setminus
U^{\prime}})$, by de Rham’s theorem there is a compactly supported 1–form
$\beta_{t}\in\Omega^{1}({\overline{S\setminus U^{\prime}}})$ such that
$d\beta_{t}=\omega-d\lambda_{1,t}$ on ${\overline{S\setminus U^{\prime}}}$.
Let $\tilde{\beta}_{t}$ be the extension of $\beta_{t}$ to $S$ by zero. Then,

$\lambda_{t}:=\lambda_{1,t}+\tilde{\beta}_{t}\in\Omega^{1}(S)$

satisfies (1), (2) and (3) above, and the dependence on $t$ can be clearly
arranged to be smooth. ∎

######  Lemma 2.9.

Let $(B,\varphi)$ be an open book decomposition of the closed, oriented
3–manifold $Y$. Let $\xi=\ker(\alpha)$ be a contact structure, with $\alpha$ a
contact 1–form compatible with $(B,\varphi)$. Let $K\subset(Y,\xi)$ be a
smooth knot contained in the page $S:=\varphi^{-1}(1)$ with $K$ homologically
nontrivial in $S$. Then, the quadruple $(B,\varphi,\alpha,K)$ determines,
uniquely up to Legendrian isotopy, a Legendrian knot

$L=L(B,\varphi,\alpha,K)\subset(Y,\xi)$

smoothly isotopic to $K$.

###### Proof.

Notice first that Giroux’s proof that two contact structures compatible with
the same open book are isotopic shows that the space $C\subset\Omega^{1}(Y)$
of contact 1–forms compatible with $(B,\varphi)$ is connected and simply
connected. In fact, given $\alpha_{0},\alpha_{1}\in C$, one can first deform
each of them inside $C$ to $\alpha_{0,R},\alpha_{1,R}\in C$, where $R\geq 0$
is a constant, so that when $R$ is large enough, the path
$(1-s)\alpha_{0,R}+s\alpha_{1,R}$ from $\alpha_{0,R}$ to $\alpha_{1,R}$ is
inside $C$ (see [11]). This proves that $C$ is connected. A similar argument
shows that $C$ is simply connected. In fact, given a loop
$\mathcal{L}=\\{\alpha_{z}\\}\subset C$, $z\in S^{1}$, one can deform it to
$\mathcal{L}_{R}=\\{\alpha_{z,R}\\}\subset C$. By compactness, when $R$ is
large enough, $\cal L_{R}$ can be shrunk in $C$ onto $\\{\alpha_{1,R}\\}$ by
taking convex linear combinations.

By our assumptions $\alpha\in C$ and by Lemma 2.8 there is a 1–form
$\alpha_{K}\in C$ whose restriction to $K$ vanishes. We can choose a path $P$
in the space $C$ connecting $\alpha_{K}$ to $\alpha$. Then, setting
$\xi_{K}:=\ker(\alpha_{K})$, by Gray’s theorem there is a contactomorphism

$\Phi=\Phi(\alpha,\alpha_{K},P)\colon\thinspace(Y,\xi_{K})\to(Y,\xi).$

We define $L:=\Phi(K)$. Since $\Phi$ is smoothly isotopic to the identity, $L$
is smoothly isotopic to $K$. Hence, to prove the lemma it suffices to show
that changing our choices of $\alpha$, $\alpha_{K}$ or $P$ only changes $L$ by
a Legendrian isotopy. Observe that if $P^{\prime}$ is another path from
$\alpha_{K}$ to $\alpha$, since $C$ is simply connected there exists a family
$P_{t}$ of paths from $\alpha_{K}$ to $\alpha$ connecting $P$ to $P^{\prime}$.
This yields a family of contactomorphisms

$\Phi_{t}\colon\thinspace(Y,\xi_{K})\to(Y,\xi)$

and therefore a family of Legendrian knots $L_{t}:=\Phi_{t}(K)\subset(Y,\xi)$
with $L_{0}=L$. Thus, $L$ only depends, up to Legendrian isotopy, on the
endpoints $\alpha_{K}$ and $\alpha$ of the path $P$. Suppose now that we chose
a different 1–form $\alpha^{\prime}_{K}\in C$ whose restriction to $K$
vanishes. Then, we claim that there is a smooth path $\alpha_{K,s}\subset C$
from $\alpha_{K}$ to $\alpha^{\prime}_{K}$ such that the restriction of
$\alpha_{K,s}$ to $K$ vanishes for every $s$. In fact, such a path can be
found by first deforming each of $\alpha_{K}$ and $\alpha^{\prime}_{K}$ to the
forms $\alpha_{K,R}$ and $\alpha^{\prime}_{K,R}$ obtained by adding multiples
of the standard angular 1–form suitably modified near the binding, and then
taking convex linear combinations (see [11]). Neither of the two operations
alters the vanishing property along $K$, therefore this proves the claim. Now
we can find a smooth family $Q_{s}$ of paths in $C$, such that $Q_{s}$ joins
$\alpha_{K,s}$ to $\alpha$ for every $s$. This produces a family of Legendrian
knots $\phi(\alpha,\alpha_{K,s},Q_{s})(K)\subset(Y,\xi)$, thus proving the
independence of $L$ from $\alpha_{K}$ up to Legendrian isotopy. The
independence from $\alpha$ can be established similarly: if
$\alpha^{\prime}\in C$ and $\xi=\ker(\alpha^{\prime})$ then
$\alpha_{s}:=(1-s)\alpha+s\alpha^{\prime}\in C$ and $\xi=\ker(\alpha_{s})$ for
every $s$. Therefore, we can find a smooth family of paths $R_{s}$ in $C$,
with $R_{s}$ joining $\alpha_{K}$ to $\alpha_{s}$ for every $s$, and proceed
as before. ∎

###### Proof of Theorem 2.7.

It suffices to show that if $K_{t}\subset S$ is a family of smooth knots which
gives a smooth isotopy from $K_{0}$ to $K_{1}$, then, the Legendrian knots
$L_{0}=L(B,\varphi,\alpha,K_{0})$ and $L_{1}=L(B,\varphi,\alpha,K_{1})$
determined via Lemma 2.9 are Legendrian isotopic. By Lemma 2.8 there exists a
smooth family of contact 1–forms $\alpha_{K_{t}}\in\Omega^{1}(Y)$ compatible
with $(B,\varphi)$ and such that the restriction of $\alpha_{K_{t}}$ to
$K_{t}$ vanishes. Applying the construction of Lemma 2.9 for each $t$ we
obtain the required Legendrian isotopy

$L_{t}:=L(B,\varphi,\alpha,K_{t})\subset(Y,\xi).$

∎

###  2.3 Heegaard Floer homologies

The Heegaard Floer homology groups
$\mathrm{HF}^{-}(Y),\widehat{\mathrm{HF}}(Y)$ of a 3–manifold were introduced
in [33] and extended in the case where $Y$ is equipped with a null-homologous
knot $K\subset Y$ to variants
$\mathrm{HFK}^{-}(Y,K),\widehat{\mathrm{HFK}}(Y,K)$ in [35, 40]. For the sake
of completeness here we quickly review the construction of these groups,
emphasizing the aspects most important for our present purposes.

We start with the closed case. An oriented 3–manifold $Y$ can be conveniently
presented by a Heegaard diagram, which is an ordered triple
$(\Sigma,\mathbb{\alpha},\mathbb{\beta})$, where $\Sigma$ is an oriented
genus–$g$ surface, ${\bf{\alpha}}=\\{\alpha_{1},\ldots,\alpha_{g}\\}$ (and
similarly ${\bf{\beta}}=\\{\beta_{1},\ldots,\beta_{g}\\}$) is a $g$–tuple of
disjoint simple closed curves in $\Sigma$, linearly independent in
$H_{1}(\Sigma;{\mathbb{Z}})$. The $\alpha$–curves can be viewed as belt
circles of the 1–handles, while the $\beta$–curves as attaching circles of the
2–handles in an appropriate handle decomposition of $Y$. We can assume that
the $\alpha$– and $\beta$–curves intersect transversely. Consider the tori
${\mathbb{T}}_{\alpha}=\alpha_{1}\times\ldots\times\alpha_{g}$,
${\mathbb{T}}_{\beta}=\beta_{1}\times\ldots\times\beta_{g}$ in the $g^{th}$
symmetric power $\operatorname{Sym}^{g}(\Sigma)$ of $\Sigma$ and define
$\mathrm{CF}^{-}(Y)$ as the free $\mathbb{F}[U]$–module generated by the
elements of the transverse intersection
${\mathbb{T}}_{\alpha}\cap{\mathbb{T}}_{\beta}$. (Recall that in this paper we
assume $\mathbb{F}=\mathbb{Z}/2\mathbb{Z}$. The constructions admit a sign
refinement to $\mathbb{Z}[U]$, but we do not need this for our current
applications.) For appropriate symplectic, and compatible almost complex
structures $(\omega,J)$ on $\operatorname{Sym}^{g}(\Sigma)$,
${\bf{x}},{\bf{y}}\in{\mathbb{T}}_{\alpha}\cap{\mathbb{T}}_{\beta}$, and
relative homology class $\phi\in\pi_{2}({\bf{x}},{\bf{y}})$, we define
${\mathfrak{M}}(\phi)$ as the moduli space of holomorphic maps from the unit
disk $D\subset{\mathbb{C}}$ to $(\operatorname{Sym}^{g}(\Sigma),J)$ with the
appropriate boundary conditions (cf. [33]). Take $\mu(\phi)$ to be the formal
dimension of ${\mathfrak{M}}(\phi)$ and
${\widehat{\mathfrak{M}}}(\phi)={\mathfrak{M}}(\phi)/{\mathbb{R}}$ the
quotient of the moduli space by the translation action of $\mathbb{R}$.

An equivalence class of nowhere zero vector fields (under homotopy away from a
ball) on a closed 3–manifold is called a _Spin c structure_. It is easy to see
that a cooriented contact structure $\xi$ on a closed 3–manifold naturally
induces a ${\mathrm{Spin}}^{c}$ structure: this is the equivalence class of
the oriented unit normal vector field of the 2–plane field $\xi$.

Fix a point $w\in\Sigma-{\bf{\alpha}}-{\bf{\beta}}$, and for
$\phi\in\pi_{2}({\bf{x}},{\bf{y}})$ denote the algebraic intersection number
$\\#(\phi\cap\\{w\\}\times\operatorname{Sym}^{g-1}\Sigma)$ by $n_{w}(\phi)$.
With these definitions in place, the differential
$\partial^{-}\colon\mathrm{CF}^{-}(Y)\to\mathrm{CF}^{-}(Y)$ is defined as

$\partial^{-}{\bf{x}}=\sum_{{\bf{y}}\in{\mathbb{T}}_{\alpha}\cap{\mathbb{T}}_{\beta}}\sum_{\phi\in\pi_{2}({\bf{x}},{\bf{y}}),\mu(\phi)=1}\\#{\widehat{\mathfrak{M}}}(\phi)\cdot
U^{n_{w}(\phi)}\cdot{\bf{y}}.$

With the aid of $w$ the elements of
${\mathbb{T}}_{\alpha}\cap{\mathbb{T}}_{\beta}$ can be partitioned according
to the ${\mathrm{Spin}}^{c}$ structures of $Y$, resulting a decomposition

$\mathrm{CF}^{-}(Y)=\oplus_{\mathbf{t}\in{\mathrm{Spin}}^{c}(Y)}\mathrm{CF}^{-}(Y,\mathbf{t}),$

and the map $\partial^{-}$ respects this decomposition. If the technical
condition of _strong admissibility_ (cf. [33]) is satisfied for the Heegaard
diagram $(\Sigma,\alpha,\beta,w)$ of $(Y,\mathbf{t})$, the chain complex
$(\mathrm{CF}^{-}(Y,\mathbf{t}),\partial^{-})$ results in a group
$\mathrm{HF}^{-}(Y,\mathbf{t})$ which is an invariant of the
${\mathrm{Spin}}^{c}$ 3–manifold $(Y,\mathbf{t})$. Strong admissibility of
$(\Sigma,\alpha,\beta,w)$ for $(Y,\mathbf{t})$ can be achieved as follows:
consider a collection $\\{\gamma_{1},\ldots,\gamma_{n}\\}$ of curves in
$\Sigma$ generating $H_{1}(Y;{\mathbb{Q}})$. It is easy to see that such
$\gamma_{i}$ ($i=1,\ldots,n$) can be found for any Heegaard decomposition.
Then, by applying sufficiently many times a specific isotopy of the
$\beta$–curves along each $\gamma_{i}$ (called ‘spinning’, the exact amount
depending on the value of $c_{1}(\mathbf{t})$ on a basis of
$H_{2}(Y;{\mathbb{Q}})$, cf. [33]) one can arrange the diagram to be strongly
admissible.

By specializing the $\mathbb{F}[U]$–module
$(\mathrm{CF}^{-}(Y,\mathbf{t}),\partial^{-})$ to $U=0$ we get a new chain
complex $(\widehat{\mathrm{CF}}(Y),\widehat{\partial})$, resulting in an
invariant $\widehat{\mathrm{HF}}(Y,\mathbf{t})$ of the ${\mathrm{Spin}}^{c}$
3–manifold $(Y,\mathbf{t})$. In addition, if $c_{1}(\mathbf{t})$ is a torsion
class, then the homology groups $\mathrm{HF}^{-}(Y,\mathbf{t})$ and
$\widehat{\mathrm{HF}}(Y,\mathbf{t})$ come with a ${\mathbb{Q}}$–grading, cf.
[32], and hence split as

$\mathrm{HF}^{-}(Y,\mathbf{t})=\oplus_{d}\mathrm{HF}^{-}_{d}(Y,\mathbf{t}),\ \
\
\widehat{\mathrm{HF}}(Y,\mathbf{t})=\oplus_{d}\widehat{\mathrm{HF}}_{d}(Y,\mathbf{t}).$

Since $\widehat{\mathrm{CF}}(Y)$ is generated over $\mathbb{F}$ by the
elements of ${\mathbb{T}}_{\alpha}\cap{\mathbb{T}}_{\beta}$, the Floer
homology group $\widehat{\mathrm{HF}}(Y)$ and hence also each
$\widehat{\mathrm{HF}}(Y,\mathbf{t})$ are finitely generated
$\mathbb{F}$–modules. There is a long exact sequence

$\ldots\longrightarrow\mathrm{HF}^{-}_{d}(Y,\mathbf{t})\stackrel{{\scriptstyle\cdot
U}}{{\longrightarrow}}\mathrm{HF}^{-}_{d-2}(Y,\mathbf{t})\longrightarrow\widehat{\mathrm{HF}}_{d-2}(Y,\mathbf{t})\longrightarrow\ldots$

which establishes a connection between the two versions of the theory.

Suppose now that we fix two distinct points

$w,z\in\Sigma-\alpha_{1}-...-\alpha_{g}-\beta_{1}-...-\beta_{g},$

where $(\Sigma,\alpha,\beta,w)$ is a Heegaard diagram for $Y$. The ordered
pair of points $(w,z)$ determines an oriented knot $K$ in $Y$ by the following
convention. We consider an embedded oriented arc $\zeta$ in $\Sigma$ from $z$
to $w$ in the complement of the $\alpha$–arcs, and let $\eta$ be an analogous
arc from $w$ to $z$ in the complement of the $\beta$–arcs. Pushing $\zeta$ and
$\eta$ into the $\alpha$– and $\beta$–handlebodies we obtain a pair of
oriented arcs $\zeta^{\prime}$ and $\eta^{\prime}$ which meet $\Sigma$ at $w$
and $z$. Their union now is an oriented knot $K\subset Y$. We call the tuple
$(\Sigma,\alpha,\beta,w,z)$ a Heegaard diagram compatible with the oriented
knot $K\subset Y$. (This is the orientation convention from [35]; it is
opposite to the one from [27].)

We have a corresponding differential, defined by

$\partial^{-}_{K}{\bf{x}}=\sum_{{\bf{y}}\in{\mathbb{T}}_{\alpha}\cap{\mathbb{T}}_{\beta}}\sum_{\phi\in\pi_{2}({\bf{x}},{\bf{y}}),\mu(\phi)=1,n_{z}(\phi)=0}\\#{\widehat{\mathfrak{M}}}(\phi)\cdot
U^{n_{w}(\phi)}\cdot{\bf{y}}.$

Using this map we get a chain complex
$(\mathrm{CFK}^{-}(Y),\partial^{-}_{K})$. This group has an additional
grading, which can be formulated in terms of relative Spinc structures, which
are possible extensions of $\mathbf{t}|_{Y-\nu K}$ to the zero–surgery
$Y_{0}(K)$ along the null–homologous knot $K$. Given $K$, there are infinitely
many relative ${\mathrm{Spin}}^{c}$ structures with a fixed background
${\mathrm{Spin}}^{c}$ structure on the 3–manifold $Y$. By choosing a Seifert
surface $F$ for the null–homologous knot $K$, we can extract a numerical
invariant for relative ${\mathrm{Spin}}^{c}$ structures, gotten by half the
value of $c_{1}(\mathbf{s})$ on $F$ (which is defined as the integral of the
first Chern class of the corresponding ${\mathrm{Spin}}^{c}$ structure of the
0–surgery on the surface ${\hat{F}}$ we get by capping off the surface $F$).
Note that the sign of the result depends on the fixed orientation of $K$. The
induced $\mathbb{Z}$-grading on the knot Floer complex is called its
_Alexander grading_. When $b_{1}(Y)=0$, this integer, together with the
background ${\mathrm{Spin}}^{c}$ structure $\mathbf{t}$, uniquely specifies
the relative ${\mathrm{Spin}}^{c}$ structure; moreover, the choice of the
Seifert surface becomes irrelevant, except for the overall induced orientation
on $K$.

For a null–homologous knot $K\subset Y$ the homology group
$\mathrm{HFK}^{-}(Y,K,\mathbf{s})$ of the above chain complex (with relative
${\mathrm{Spin}}^{c}$ structure $\mathbf{s}$) is an invariant of
$(Y,K,\mathbf{s})$, and is called the _knot Floer homology_ of $K$. The
specialization $U=0$ of the complex defines again a new complex
$(\widehat{\mathrm{CFK}}(Y,K,\mathbf{s}),\widehat{\partial}_{K})$ with
homology denoted by $\widehat{\mathrm{HFK}}(Y,K,\mathbf{s})$. The homology
groups $\widehat{\mathrm{HFK}}(Y,K,\mathbf{s})$ and
$\mathrm{HFK}^{-}(Y,K,\mathbf{s})$ are both finitely generated vector spaces
over $\mathbb{F}$.

An alternative way to view this construction is the following. Using one
basepoint $w$ one can define the chain complex
$(\mathrm{CF}^{-}(Y,\mathbf{t}),\partial^{-})$ as before, and with the aid of
the other basepoint $z$ one can equip this chain complex with a filtration. As
it was shown in [35], the filtered chain homotopy type of the resulting
complex is an invariant of the knot, and the Floer homology groups can be
defined as the homology of the associated graded object.

The restriction of a relative ${\mathrm{Spin}}^{c}$ structure $\mathbf{s}$ to
the complement of $K$ extends to a unique ${\mathrm{Spin}}^{c}$ structure
$\mathbf{t}$ on $Y$. The map induced by multiplication by $U$ changes the
relative ${\mathrm{Spin}}^{c}$ structure, but it preserves the background
${\mathrm{Spin}}^{c}$ structure $\mathbf{t}\in{\mathrm{Spin}}^{c}(Y)$. Thus,
we can view

$\mathrm{CFK}^{-}(Y,K,\mathbf{t}):=\oplus_{\mathbf{s}\
{\mbox{{\small{restricts to}}}}\ \mathbf{t}}\mathrm{CFK}^{-}(Y,K,\mathbf{s}),$
$\mathrm{HFK}^{-}(Y,K,\mathbf{t}):=\oplus_{\mathbf{s}\
{\mbox{{\small{restricts to}}}}\ \mathbf{t}}\mathrm{HFK}^{-}(Y,K,\mathbf{s})$

and

$\widehat{\mathrm{HFK}}(Y,K,\mathbf{t}):=\oplus_{\mathbf{s}\
{\mbox{{\small{restricts to}}}}\
\mathbf{t}}\widehat{\mathrm{HFK}}(Y,K,\mathbf{s})$

as modules over $\mathbb{F}[U]$ (where the $U$–action on
$\widehat{\mathrm{HFK}}(Y,K,\mathbf{t})$ is trivial). If $\mathbf{t}$ is
torsion, then there is an absolute $\mathbb{Q}$–grading on these modules, as
in the case of closed 3–manifolds. As before, the two versions of knot Floer
homologies are connected by the long exact sequence

$\cdots\longrightarrow\mathrm{HFK}^{-}_{d}(Y,K,\mathbf{t})\stackrel{{\scriptstyle\cdot
U}}{{\longrightarrow}}\mathrm{HFK}^{-}_{d-2}(Y,K,\mathbf{t})\longrightarrow\widehat{\mathrm{HFK}}_{d-2}(Y,K,\mathbf{t})\longrightarrow\cdots$

A map
$F\colon\mathrm{HFK}^{-}(Y,K,\mathbf{t})\to\widehat{\mathrm{HF}}(Y,\mathbf{t})$
can be defined, which is induced by the map

$f\colon\mathrm{CFK}^{-}(Y,K,\mathbf{t})\to\widehat{\mathrm{CF}}(Y,\mathbf{t})$

by setting $U=1$ and taking $z$ to play the role of $w$ in the complex
$\widehat{\mathrm{CF}}(Y,\mathbf{t})$. According to the definition, this
specialization simply disregards the role of the basepoint $w$. Indeed, the
fact that a generator of $\mathrm{CFK}^{-}(Y,K)$ belongs to the summand
$\mathrm{CFK}^{-}(Y,K,\mathbf{t})$ is determined only by the point $z$. On the
chain level this map fits into the short exact sequence

$0\to\mathrm{CFK}^{-}(Y,K,\mathbf{t})\stackrel{{\scriptstyle
U-1}}{{\longrightarrow}}\mathrm{CFK}^{-}(Y,K,\mathbf{t})\stackrel{{\scriptstyle
f}}{{\longrightarrow}}\widehat{\mathrm{CF}}(Y,\mathbf{t})\to 0,$ (2.1)

since $f$ is obviously surjective.

######  Lemma 2.10.

Let
$F\colon\mathrm{HFK}^{-}(Y,K,\mathbf{t})\longrightarrow\widehat{\mathrm{HF}}(Y,\mathbf{t})$
be map induced on homology by $f$. The kernel of $F$ consists of elements $x$
of the form $(U-1)y$. Moreover, an element
$x\in\mathrm{HFK}^{-}(Y,K,\mathbf{t})$ is both in the kernel of $F$ and
homogeneous, i.e. contained in the summand determined by a fixed relative
${\mathrm{Spin}}^{c}$ structure, if and only if $x$ satisfies $U^{n}x=0$ for
some $n\geq 0$.

###### Proof.

The long exact sequence associated to Exact Sequence (2.1) identifies $\ker F$
with $(U-1)\cdot\mathrm{HFK}^{-}(Y,K,\mathbf{t})$. The only statement left to
be proved is the characterization of homogeneous elements in $\ker F$. If
$U^{n}x=0$ then

$x=(1-U^{n})x=(1-U)(x+Ux+\ldots+U^{n-1}x),$

hence $x$ is of the form $x=(1-U)y$. Conversely, if $x=(1-U)y$ we may assume
without loss that $y=\sum_{i=0}^{n-1}y_{i}$, where each $y_{i}$ is
homogeneous, $y_{0}=x$ and, for $i=0,\ldots,n-1$, $Uy_{i}$ belongs to the same
relative ${\mathrm{Spin}}^{c}$ structure as $y_{i+1}$. A simple cancellation
argument shows that $y_{1}=Uy_{0},y_{2}=Uy_{1},\ldots$ hence by the finiteness
of the sum we get $U^{n}x=0$. ∎

###  2.4 Contact Ozsváth–Szabó invariants

Next we turn to the description of the contact Ozsváth–Szabó invariant of a
closed contact 3–manifold as it is given in [19]. (See [34] for the original
definition of these invariants.) Suppose that $(B,\varphi)$ is an open book
decomposition of the 3–manifold $Y$ compatible with the given contact
structure $\xi$. Consider a _basis_ $\\{a_{1},\ldots,a_{n}\\}$ of the page
$S_{+1}$, that is, take a collection of disjoint properly embedded arcs
$\\{a_{1},\ldots,a_{n}\\}$ such that $S_{+1}-\cup_{i=1}^{n}a_{i}$ is connected
and simply–connected (therefore it is homeomorphic to a disk). Let $b_{i}$ be
a properly embedded arc obtained by a small isotopy of $a_{i}$ so that the
endpoints of $a_{i}$ are isotoped along $\partial S_{+1}$ in the direction
given by the boundary orientation, and $a_{i}$ intersects $b_{i}$ in a unique
transverse point in int$S_{+1}$, cf. Figure 1 and [19, Figure 2].

2pt $S$ at 94 94 $a_{i}$ at 165 24 $b_{i}$ at 231 24

Figure 1: The arcs $a_{i}$ and $b_{i}$.

Considering ${\overline{a_{i}}}=a_{i}$ and
${\overline{b_{i}}}=h_{\varphi}(b_{i})$ in
$S_{-1}={\overline{\varphi^{-1}(-1)}}$ (where $h_{\varphi}$ denotes a
diffeomorphism representing the monodromy of the given open book), it is shown
in [19] that the triple

$(S_{+1}\cup(-S_{-1}),\\{a_{i}\cup{\overline{a_{i}}}\\}_{i=1}^{n},\\{b_{i}\cup{\overline{b_{i}}}\\}_{i=1}^{n})=(S_{+1}\cup(-S_{-1}),\alpha,\beta)$

is a Heegaard diagram for $Y$. (Notice that we have the freedom of choosing
$h_{\varphi}$ within its isotopy class; this freedom will be used later, cf.
Proposition 2.13.) For technical purposes, however, we consider the triple

$(S_{+1}\cup(-S_{-1}),\\{b_{i}\cup{\overline{b_{i}}}\\}_{i=1}^{n},\\{a_{i}\cup{\overline{a_{i}}}\\}_{i=1}^{n})=(S_{+1}\cup(-S_{-1}),\beta,\alpha),$
(2.2)

which is now a Heegaard diagram for $-Y$. With this choice, and the careful
placement of the basepoint $w$ we can achieve that the proposed chain in the
chain complex for defining the Heegaard Floer group is, in fact, a cycle. More
formally, put the basepoint $w$ in the disk $S_{+1}-\cup_{i=1}^{n}a_{i}$
outside of the small strips between the $a_{i}$’s and the $b_{i}$’s and
consider the element ${\bf{x}}(B,\varphi)=\\{a_{i}\cap b_{i}\\}$ in the chain
complex corresponding to the Heegard diagram (2.2) above (pointed by $w$). It
is not hard to see (cf. [19]) that with these choices the Heegaard diagram is
weakly admissible. In the following we always have to keep in mind the
reversal of the $\alpha$– and $\beta$–curves when working with the contact (or
Legendrian) invariants.

######  Theorem 2.11 ([19], cf. also [20]).

The chain ${\bf{x}}(B,\varphi)$ defined above is closed when regarded as an
element of the Heegaard Floer chain complex $\widehat{\mathrm{CF}}(-Y)$. The
homology class $[{\bf{x}}(B,\varphi)]\in\widehat{\mathrm{HF}}(-Y)$ defined by
${\bf{x}}(B,\varphi)$ is independent (up to sign) of the chosen basis and
compatible open book decomposition. Therefore the homology class
$[{\bf{x}}(B,\varphi)]$ is an invariant of the contact structure $(Y,\xi)$.

The original definition of this homology class is given in [34], which leads
to a different Heegaard diagram. It can be shown that the two invariants are
identified after a sequence of handleslides; though one can work directly with
the above definition, as in [19]. We adopt this point of view, supplying an
alternative proof of the invariance of the contact class (which will assist us
in the definition of the Legendrian invariant).

###### Alternative proof of Theorem 2.11..

In computing $\partial{\bf{x}}(B,\varphi)$ we need to encounter holomorphic
disks which avoid $w$ but start at ${\bf{x}}(B,\varphi)$. By the chosen order
of $a_{i}$ and $b_{i}$ (resulting in a Heegaard diagram of $-Y$ rather than
$Y$) we get that such holomorphic disk does not exist. In fact, there are no
Whitney disks $\phi\in\pi_{2}({\bf{x}},{\bf{y}})$ for any ${\bf{y}}$ with
$n_{w}(\phi)=0$ and whose local multiplicities are all non–negative, cf. [19,
Section 3]. Thus, the intersection point represents a cycle in the chain
complex. The independence of $[{\bf{x}}(B,\varphi)]$ from the basis is given
in [19, Proposition 3.3]. The argument relies on the observation that two
bases of $S_{+1}$ can be connected by a sequence of arc slides [19, Section
3.3], inducing handle slides on the corresponding Heegaard diagrams. (We will
discuss a sharper version of this argument in Proposition 3.2.) As it is
verified in [19, Lemma 3.4], these handle slides map the corresponding
${\bf{x}}$’s into each other.

In order to show independence of the chosen open book decomposition, we only
need to verify that if $(B^{\prime},\varphi^{\prime})$ is the result of a
Giroux stabilization of $(B,\varphi)$, then there are appropriate bases,
giving Heegaard decompositions, for which the map

$\Phi\colon\widehat{\mathrm{CF}}(B,\varphi)\to\widehat{\mathrm{CF}}(B^{\prime},\varphi^{\prime})$

induced by the stabilization satisfies
$\Phi({\bf{x}}(B,\varphi))={\bf{x}}(B^{\prime},\varphi^{\prime})$. Let us
assume that the right–handed Dehn twist of the Giroux stabilization is equal
to $D_{\gamma}$, where $\gamma$ is a simple closed curve in the page
$S^{\prime}$ of $(B^{\prime},\varphi^{\prime})$, and $\gamma_{1}$ is the
portion of it inside the page $S$ of $(B,\varphi)$. Our aim is to find a basis
$\\{a_{1},\ldots,a_{n}\\}$ for $S$ which is disjoint from $\gamma_{1}$. If
$S-\gamma_{1}$ is connected, then choose $a_{1}$ to be (a little push–off of)
$\gamma_{1}$, and extend $\\{a_{1}\\}$ to a basis for $S$. If $S-\gamma_{1}$
is disconnected then the union of any bases of the components (possibly after
isotoping some endpoints along $\gamma_{1}$) will be appropriate.

Now consider the basis for $S^{\prime}$ obtained by extending the above basis
for $S$ with the cocore $a_{n+1}$ of the new 1–handle. Since $\gamma$ is
disjoint from all $a_{i}$ ($i\leq n$), we clearly get that
$\alpha_{n+1}=a_{n+1}\cup{\overline{a_{n+1}}}$ will be intersected only (and
in a single point $y_{n+1}$) by
$\beta_{n+1}=b_{n+1}\cup{\overline{h_{\varphi}(b_{n+1})}}$. Therefore the map
$\Phi$ sending a generator $(y_{1},\ldots,y_{n})$ of
$\widehat{\mathrm{CF}}(B,\varphi,\\{a_{1},\ldots,a_{n}\\})$ to
$(y_{1},\ldots,y_{n},y_{n+1})$ (with the last coordinate being the unique
intersection $\alpha_{n+1}\cap\beta_{n+1}=\\{y_{n+1}\\}$) establishes an
isomorphism

$\Phi\colon\widehat{\mathrm{CF}}(B,\varphi,\\{a_{1},\ldots,a_{n}\\})\to\widehat{\mathrm{CF}}(B^{\prime},\varphi^{\prime},\\{a_{1},\ldots,a_{n},a_{n+1}\\})$

between the underlying Abelian groups. Since $\alpha_{n+1}$ contains a unique
intersection point with all the $\beta$–curves, a holomorphic disk encountered
in the boundary map must be constant at $y_{n+1}$, hence $\Phi$ is a chain
map. Since it maps ${\bf{x}}(B,\varphi)$ to
${\bf{x}}(B^{\prime},\varphi^{\prime})$, the proof is complete. (See also [20,
Section 3].) ∎

######  Remark 2.12.

The basic properties (such as the vanishing for overtwisted and nonvanishing
for Stein fillable structures, and the transformation under contact
$(+1)$–surgery) can be directly verified for the above construction, cf. [19].

In our later arguments we will need that the Heegaard diagram can be chosen to
be strongly admissible, hence we address this issue presently, using an
argument which was first used in [38].

######  Proposition 2.13.

For any ${\mathrm{Spin}}^{c}$ structure $\mathbf{t}$ the monodromy map
$h_{\varphi}$ of the given open book decomposition can be chosen in its
isotopy class in such a way that the Heegaard diagram defined before Theorem
2.11 is strongly admissible for $\mathbf{t}$.

###### Proof.

Recall that strong admissibility of a Heegaard diagram for a given
${\mathrm{Spin}}^{c}$ structure $\mathbf{t}$ can be achieved by isotoping the
$\beta$–curves through spinning them around a set of curves
$\\{\gamma_{1},\ldots,\gamma_{n}\\}$ in the Heegaard surface $\Sigma$
representing a basis of $H_{1}(Y)$. It can be shown that for a Heegaard
diagram coming from an open book decomposition, we can choose the curves
$\gamma_{i}$ all in the same page, hence all $\gamma_{i}$ can be chosen to be
in $S_{-1}$. Since the spinnings are simply isotopies in this page, we can
change a fixed monodromy $h_{\varphi}$ within its isotopy class to realize the
required spinnings. In this way we get a strongly admissible Heegaard diagram
for $(Y,\mathbf{t})$. ∎

##  3 Invariants of Legendrian knots

Suppose now that $L\subset(Y,\xi)$ is a given Legendrian knot, and consider an
open book decomposition $(B,\varphi)$ compatible with $\xi$ containing $L$ on
a page. To define our Legendrian knot invariants we need to analyze the
dependence from the choice of an appropriate basis and from the open book
decomposition as in Section 2, but now in the presence of the Legendrian knot.

### Legendrian knots and bases

Suppose that $S$ is a surface with $\partial S\neq\emptyset$ and
$\\{a_{1},\ldots,a_{n}\\}$ is a basis in $S$. If (after possibly reordering
the $a_{i}$’s) the two arcs $a_{1}$ and $a_{2}$ have adjacent endpoints on
some component of $\partial S$, that is, there is an arc $\tau\subset\partial
S$ with endpoints on $a_{1}$ and $a_{2}$ and otherwise disjoint from all
$a_{i}$’s, then define $a_{1}+a_{2}$ as the isotopy class (rel endpoints) of
the union $a_{1}\cup\tau\cup a_{2}$. The modification

$\\{a_{1},a_{2},\ldots,a_{n}\\}\mapsto\\{a_{1}+a_{2},a_{2},\ldots,a_{n}\\}$

is called an arc slide, cf. [19]. Suppose that $L\subset S$ is a homologically
essential simple closed curve. The basis $\\{a_{1},\ldots,a_{n}\\}$ of $S$ is
_adapted to $L$_ if $L\cap a_{i}=\emptyset$ for $i\geq 2$ and $L$ intersects
$a_{1}$ in a unique transverse point.

######  Lemma 3.1.

For any surface $S$ and homologically essential knot $L\subset S$ there is an
adapted basis.

###### Proof.

The statement follows easily from the fact that $L$ represents a nontrivial
class in $H_{1}(S,\partial S)$. ∎

Suppose now that $\\{a_{1},\ldots,a_{n}\\}$ is an adapted basis for $(S,L)$.
An arc slide $\\{a_{i},a_{j}\\}\mapsto\\{a_{i}+a_{j},a_{j}\\}$ is called
_admissible_ if the arc $a_{i}$ is not slid over the distinguished arc
$a_{1}$. The aim of this subsection is to prove the following

######  Proposition 3.2.

If $\\{a_{1},\ldots,a_{n}\\}$ and $\\{A_{1},\ldots,A_{n}\\}$ are two adapted
bases for $(S,L)$ then there is a sequence of admissible arc slides which
trasform $\\{a_{1},\ldots,a_{n}\\}$ into $\\{A_{1},\ldots,A_{n}\\}$.

###### Proof.

As a first step in proving the statement on arc slides, we want to show that,
up to applying a sequence of admissible arc slides to the $a_{i}$’s, we may
assume $(a_{1}\cup\ldots\cup a_{n})\cap(A_{1}\cup\ldots\cup A_{n})=\emptyset$.
We start with showing that $a_{1}\cap A_{1}=\emptyset$ can be assumed. Suppose
that $a_{1}\cap A_{1}\neq\emptyset$; we will find arc slides reducing
$|a_{1}\cap A_{1}|$. To this end, consider the disk $D^{2}$ obtained by
cutting $S$ along the $a_{i}$’s. Then, $A_{1}\cap D^{2}$ is a collection of
arcs, and (at least) one component intersects $a_{1}$. This component of
$A_{1}$ divides $D^{2}$ into two components $D_{1}$ and $D_{2}$, and one of
them, say $D_{1}$, contains $a_{1}^{-1}$ (or $a_{1}$). Sliding $a_{1}$ over
all the $a_{i}$’s contained in the boundary semicircle of $D_{2}$ (cf. Figure
2(a)) we reduce $|a_{1}\cap A_{1}|$ by one, so ultimately we can assume that
$a_{1}\cap A_{1}=\emptyset$.

2pt $a_{1}$ at 9 406 $A_{1}$ at 77 371 $D_{1}$ at 71 313 $D_{2}$ at 94 406
$a_{1}^{-1}$ [lt] at 134 294

$A_{1}$ at 352 343 $L$ at 359 399 $a_{1}$ at 286 408 $a_{1}^{-1}$ at 431 416
$a_{2}$ at 264 354

$a_{1}$ [lb] at 126 196 $a_{1}^{-1}$ [lt] at 130 72 $A_{1}$ at 40 121 $L$ at
90 176

$A_{2}$ at 319 121 $L$ at 360 179 $a_{2}$ [lt] at 391 59 $a_{2}^{-1}$ [b] at
367 211 $a_{1}$ [rb] at 300 180 $a_{1}^{-1}$ [lb] at 422 185 $(a)$ at 79 254
$(b)$ at 359 254 $(c)$ at 79 35 $(d)$ at 363 35

Figure 2: Arc slides.

Next we apply further arc slides to achieve $a_{i}\cap A_{1}=\emptyset$
($i\geq 2$). For this, let us assume that $a_{2}$ is the first arc
intersecting $A_{1}$ when traversing $A_{1}$ starting from $\partial S$.
(Since $A_{1}$ intersects $L$ exactly once, after possibly starting at the
other end of $A_{1}$ we can assume that it first meets $a_{2}$ and then $L$.)
As before, we can find a segment of $A_{1}$ intersecting $a_{2}$ and dividing
$D^{2}=S-\cup a_{i}$ into two components, one of which contains
$a_{1},a_{1}^{-1}$ (since these are connected by $L$, and the segment we chose
is disjoint from $L$), cf. Figure 2(b). If $a_{2}^{-1}$ is in the same
semicircle as $a_{1}$ (and so $a_{1}^{-1}$) then we can slide $a_{2}$ along
the other semicircle to eliminate one intersection point from $a_{2}\cap
A_{1}$. If $a_{2}^{-1}$ is on the other semicircle, then we cannot proceed in
such a simple way (since $a_{2}$ is not allowed to be slid over neither
$a_{1}$ nor $a_{2}^{-1}$). Now consider the continuation of $A_{1}$, which
comes out from $a_{2}^{-1}$ and stays in the same component. It might go back
to $a_{2}$, and repeat this spiraling some more times, but eventually it will
go to another part of the the boundary of $D^{2}$, producing an arc, which
starts from $a_{2}$ (or $a_{2}^{-1}$) and divides $D^{2}$ in such a way that
$a_{1}$ and $a_{2}^{-1}$ (or $a_{2}$) are on the same side of it. Then,
$a_{2}$ can be slid across the opposite side so as to reduce the intersection
number in question.

After these slides we can assume that
$A_{1}\cap(\cup_{i=1}^{n}a_{i})=\emptyset$. This however allows us to slide
$a_{1}$ until it becomes isotopic to $A_{1}$, cf. Figure 2(c). Consider now
$A_{2}$. If $A_{2}$ intersects some $a_{i}$, then the segment of $A_{2}$
connecting $\partial S$ to the first such intersection (with, say, $a_{2}$)
divides $D^{2}$ into two components, and one of them contains both $a_{1}$ and
$a_{1}^{-1}$ (since $A_{2}$ is disjoint from $L$). If $a_{2}^{-1}$ is on the
same semicircle as $a_{1}$, then sliding over the other semicircle reduces the
number of intersections. The other possibility can be handled exactly as
before.

Finally, we get to the position when $a_{1}$ is isotopic to $A_{1}$ and
$(\cup_{i=1}^{n}a_{i})\cap(\cup_{i=1}^{n}A_{i})=\emptyset$. Now we argue as
follows: consider $A_{2}$ and choose $a_{i}$ such that $a_{i}$ and
$a_{i}^{-1}$ are in different components of $D^{2}-A_{2}$. Such $a_{i}$ exists
because $A_{2}$ is nonseparating. Suppose without loss of generality that
$i=2$. Then on the side of $A_{2}$ not containing $a_{1}$ (and $a_{1}^{-1}$)
we can slide $a_{2}$ until it becomes isotopic to $A_{2}$, see Figure 2(d).
Repeating this procedure for each $A_{i}$ the proof is complete. ∎

### Invariants of Legendrian knots

Consider now a null–homologous Legendrian knot $L\subset(Y,\xi)$ and fix an
open book decomposition $(B,\varphi)$ compatible with $\xi$ and containing $L$
on the page $S_{+1}:={\overline{\varphi^{-1}(1)}}$. Pick a basis
$A=\\{a_{1},\ldots,a_{n}\\}\subset S_{+1}$ such that $a_{1}\cap L$ is a unique
point and $a_{i}\cap L=\emptyset$ ($i\geq 2$). Under the above conditions we
will say that the triple $(B,\varphi,A)$ is _compatible_ with the triple
$(Y,\xi,L)$.

Place the basepoint $w$ as before. Putting the other basepoint $z$ between the
curves $a_{1}$ and $b_{1}$ we recover a knot in $S_{+1}$ smoothly isotopic to
$L$: connect $z$ and $w$ in the complement of $a_{i}$, and then $w$ and $z$ in
the complement of $b_{i}$ within the page $S_{+1}$. This procedure (hence the
ordered pair $(w,z)$) equips $L$ with an orientation. Moreover, if the point
$z$ is moved from one domain between $a_{1}$ and $b_{1}$ to the other, the
orientation induced on $L$ gets reversed. Thus, if $L$ is already oriented
then there is only one compatible choice of position for $z$. Notice that $z$
and $w$ chosen as above determine a knot in $S_{+1}$, unique up to isotopy in
$S_{+1}$. In turn, by Theorem 2.7 the open book decomposition together with
such a knot uniquely determines a Legendrian knot (up to Legendrian isotopy)
in the corresponding contact structure. In short, $(B,\varphi,S_{+1},A,z,w)$
determines the triple $(Y,\xi,L)$.

Recall that when defining the chain complex $\widehat{\mathrm{CF}}$ containing
the contact invariant, we reverse the roles of the $\alpha$– and
$\beta$–curves, resulting in a Heegaard diagram for $-Y$ rather than for $Y$.
For the same reason, we do the switch between the $\alpha$– and $\beta$–curves
here as well. According to our conventions, this change would reverse the
orientation of the knot $L$ as well; to keep the fixed orientation on $L$,
switch the position of the basepoints $w$ and $z$. The two possible locations
of $z$ and $w$ we use in the definition of $\widehat{\mathrm{CF}}(-Y,L)$ are
illustrated in Figure 3; the orientation of $L$ specifies the location of $w$.

2pt $b_{1}$ [b] at 76 69 $a_{1}$ [b] at 100 69 $b_{1}$ [b] at 298 69 $a_{1}$
[b] at 322 69 $L$ at 36 47 $w$ at 87 38 $w$ at 310 56 $L$ at 358 55 $z$ at 147
32 $z$ at 371 32 or at 198 51

Figure 3: There are two regions between $a_{1}$ and $b_{1}$ in $S_{+1}$; the
placement of $w$ is determined by the orientation of $L$.

With ${\bf{x}}={\bf{x}}(B,\varphi)=(a_{i}\cap b_{i})$ as before, it is easy to
see that there are no nonnegative Whitney disks
$\phi\in\pi_{2}({\bf{x}},{\bf{y}})$ with $n_{z}(\phi)=0$ for any ${\bf{y}}$,
hence (as in [19, Section 3]) the intersection point ${\bf{x}}$ can be viewed
as a cycle in both $\mathrm{CFK}^{-}(-Y,L,\mathbf{t})$ and
$\widehat{\mathrm{CFK}}(-Y,L,\mathbf{t})$ for some ${\mathrm{Spin}}^{c}$
structure $\mathbf{t}$ on $Y$. As observed in [35, Subsection 2.3], the
${\mathrm{Spin}}^{c}$ structure $\mathbf{t}$ is determined by the point $z$
only, being equal to $s_{z}({\bf{x}})$ (see [35] for notation). This shows
that the contact invariant $c(Y,\xi)$ lives in the summand of
$\widehat{\mathrm{HF}}(-Y)$ corresponding to $\mathbf{t}$. On the other hand,
it is known [34] that $c(Y,\xi)\in\widehat{\mathrm{HF}}(-Y,\mathbf{t}_{\xi})$,
therefore we conclude $\mathbf{t}=\mathbf{t}_{\xi}$. Since $L$ is
null–homologous, one needs to make sure that the Heegaard diagram is strongly
admissible for $(Y,\mathbf{t}_{\xi})$; this fact follows from Proposition
2.13. Next we will address invariance properties of the knot Floer homology
class represented by ${\bf{x}}(B,\varphi)$.

######  Proposition 3.3.

Let $L\subset(Y,\xi)$ be a Legendrian knot. Let $(B,\varphi,A,w,z)$ and
$(B^{\prime},\varphi^{\prime},A^{\prime},w^{\prime},z^{\prime})$ be two open
books compatible with $(Y,\xi,L)$ and endowed with bases and basepoints
adapted to $L$. Then, there are isomorphisms of $\mathbb{F}[U]$–modules

$\Phi^{-}\colon\mathrm{HFK}^{-}(-Y,L,\mathbf{t}_{\xi})\to\mathrm{HFK}^{-}(-Y,L,\mathbf{t}_{\xi})$

and

$\widehat{\Phi}\colon\widehat{\mathrm{HFK}}(-Y,L,\mathbf{t}_{\xi})\to\widehat{\mathrm{HFK}}(-Y,L,\mathbf{t}_{\xi})$

such that

$\Phi^{-}([{\bf{x}}(B,\varphi,A)])=[{\bf{x}}(B^{\prime},\varphi^{\prime},A^{\prime})]$

and

$\widehat{\Phi}([{\bf{x}}(B,\varphi,A)])=[{\bf{x}}(B^{\prime},\varphi^{\prime},A^{\prime})].$

Recall from Section 1 that, if $M$ and $N$ are $\mathbb{F}[U]$–modules, we
consider $(M,m)$ and $(N,n)$ to be equivalent, and we write $[M,m]=[N,n]$, if
there is an isomorphism of $\mathbb{F}[U]$–modules $f\colon\thinspace M\to N$
such that $f(m)=n$.

In view of Proposition 3.3, we introduce the following

######  Definition 3.4.

Let $L\subset(Y,\xi)$ be an oriented Legendrian knot, $(B,\varphi,A,w,z)$ an
open book decomposition compatible with $(Y,\xi,L)$ with an adapted basis and
basepoints. Then, we define

${\mathfrak{L}}(L):=\left[\mathrm{HFK}^{-}(-Y,L,\mathbf{t}_{\xi}),[{\bf{x}}(B,\varphi,A)]\right].$

Similarly, we define

$\widehat{\mathfrak{L}}(L):=\left[\widehat{\mathrm{HFK}}(-Y,L,\mathbf{t}_{\xi}),[{\bf{x}}(B,\varphi,A)]\right].$

###### Proof of Proposition 3.3.

Suppose for the moment that $(B^{\prime},\varphi^{\prime})=(B,\varphi)$. Then,
Proposition 3.2 provides a sequence of arc slides transforming $A^{\prime}$ to
the chosen $A$. As it is explained in [19], arc slides induce handle slides on
the associated Heegaard diagrams, and the invariance of the Floer homology
element under these handle slides is verified in [19, Lemma 3.4]. Notice that
since we do not slide over the arc intersecting the Legendrian knot $L$ (and
hence over the second basepoint defined by this arc), actually all the handle
slides induce $\mathbb{F}[U]$–module isomorphisms on the knot Floer groups,
cf. [35]. This argument proves the statement in the special case
$(B^{\prime},\varphi^{\prime})=(B,\varphi)$.

We now consider the special case when $(B^{\prime},\varphi^{\prime})$ is an
$L$–elementary stabilization of $(B,\varphi)$. Suppose that the Dehn twist of
the stabilization is along $\gamma\subset S^{\prime}$, with $\gamma_{1}\subset
S$. Since we work with an $L$–elementary stabilization, $|\gamma_{1}\cap
L|\leq 1$. If $\gamma_{1}$ is separating then $\gamma_{1}\cap L=\emptyset$ and
we choose the bases of $S$ and $S^{\prime}$ as in the proof of Theorem 2.11.
If $\gamma_{1}$ is nonseparating and $\gamma_{1}\cap L=\emptyset$ then choose
$a_{2}=\gamma_{1}$ and extend it to an appropriate basis. For $\gamma_{1}\cap
L=\\{{\mbox{pt.}}\\}$ we take $a_{1}=\gamma_{1}$ and extend it further. Denote
by $T$ and $T^{\prime}$ the resulting bases of $S$ and $S^{\prime}$
respectively. The proof of Theorem 2.11 now applies verbatim to show the
existence of automorphisms of $\mathrm{HFK}^{-}(-Y,L,\mathbf{t}_{\xi})$ and
$\widehat{\mathrm{HFK}}(-Y,L,\xi)$ mapping $[{\bf{x}}(B,\varphi,T)]$ to
$[{\bf{x}}(B^{\prime},\varphi^{\prime},T^{\prime})]$. On the other hand, by
the first part of the proof we know that there are other automorphisms sending
$[{\bf{x}}(B,\varphi,A)]$ to $[{\bf{x}}(B,\varphi,T)]$ and
$[{\bf{x}}(B^{\prime},\varphi^{\prime},T^{\prime})]$ to
$[{\bf{x}}(B^{\prime},\varphi^{\prime},A^{\prime})]$. This proves the
statement when $(B,\varphi^{\prime})$ is an $L$–elementary stabilization of
$(B,\varphi)$.

In the general case, since $(B,\varphi)$ and $(B^{\prime},\varphi^{\prime})$
are two open books compatible with $(Y,\xi,L)$, by Proposition 2.6 we know
that there is a sequence of $L$–elementary stabilizations which turns each of
them into the same stabilization $(B^{\prime\prime},\varphi^{\prime\prime})$.
Thus, applying the previous special case the required number of times, the
proof is complete. ∎

######  Remark 3.5.

Let $L\subset(Y,\xi)$ be a Legendrian knot and $(B,\varphi,A,w,z)$ a
compatible open book with adapted basis and basepoints. Then, it follows
immediately from the definitions that the map $f$ from
$\mathrm{HFK}^{-}(-Y,L,\mathbf{s})$ to
$\widehat{\mathrm{HFK}}(-Y,L,\mathbf{s})$ induced by setting $U=0$ sends the
class of ${\bf{x}}(B,\varphi,A)$ in the first group to the class of
${\bf{x}}(B,\varphi,A)$ in the second group. Moreover, the chain map inducing
$f$ can be viewed as the canonical map from the complex
$\mathrm{CFK}^{-}(-Y,L,\mathbf{s})$ onto its quotient complex
$\widehat{\mathrm{CFK}}(-Y,L,\mathbf{s})$. As such, it is natural with respect
to the transformations of the two complexes induced by the isotopies,
stabilizations and arc slides used in the proof of Proposition 3.3. Thus, it
makes sense to write $f({\mathfrak{L}}(L))=\widehat{\mathfrak{L}}(L)$.
Therefore $\widehat{\mathfrak{L}}(L)\neq 0$ readily implies
${\mathfrak{L}}(L)\neq 0$, although the converse does not necessarily hold: a
nonvanishing invariant ${\mathfrak{L}}(L)$ determined by a class which is in
the image of the $U$–map gives rise to vanishing $\widehat{\mathfrak{L}}(L)$.
As we will see, such examples do exist.

######  Corollary 3.6.

Let $L_{1},L_{2}\subset(Y,\xi)$ be oriented Legendrian knots. Suppose that
there exists an isotopy of oriented Legendrian knots from $L_{1}$ to $L_{2}$.
Then, ${\mathfrak{L}}(L_{1})={\mathfrak{L}}(L_{2})$ and
$\widehat{\mathfrak{L}}(L_{1})=\widehat{\mathfrak{L}}(L_{2})$.

###### Proof.

Let $(B,\varphi,A)$ be an open book compatible with $(Y,\xi,L_{1})$ with an
adapted basis, and let $f_{1}$ be the time–1 map of the isotopy. Then, the
triple $(f_{1}(B),\varphi\circ f_{1}^{-1},f_{1}(A))$ is compatible with and
adapted to $(Y,\xi,L_{2})$. The induced map on the chain complexes maps
${\bf{x}}(B,\varphi,A)$ to ${\bf{x}}(f_{1}(B),\varphi\circ
f_{1}^{-1},f_{1}(A))$, verifying the last statement. ∎

######  Remark 3.7.

In fact, we only used the fact that $f_{1}\colon(Y,\xi)\to(Y,\xi)$ is a
contactomorphism mapping $L_{1}$ into $L_{2}$ (respecting their orientation).
In conclusion, Legendrian knots admitting such an identification have the same
Legendrian invariants. The existence of $f_{1}$ with these properties and the
isotopy of the two knots is equivalent in the standard contact 3–sphere, but
the two conditions are different in general.

###### Proof of Theorem 1.1.

The theorem follows immediately from Proposition 3.3 and Corollary 3.6. ∎

##  4 An example

Suppose that $L\subset(S^{3},\xi_{st})$ is the Legendrian unknot with
Thurston–Bennequin invariant $-1$ in the standard tight contact 3–sphere. It
is easy to see that the positive Hopf link defines an open book on $S^{3}$
which is compatible with $\xi_{st}$ and it contains $L$ on a page. A basis in
this case consists of a single arc cutting the annulus. The corresponding
genus–1 Heegaard diagram has now a single intersection point, which gives the
generator of $\mathrm{HFK}^{-}(-S^{3},L)$ (the 3–sphere $S^{3}$ has a unique
${\mathrm{Spin}}^{c}$ structure, so we omit it from the notation). The two
possible choices $w_{1}$ and $w_{2}$ for the position of $w$, corresponding to
the two choices of an orientation for $L$, give the same class defining
${\mathfrak{L}}(L)$, because in this case $w_{1}$ and $w_{2}$ are in the same
domain, cf. Figure 4$(i)$. Let $L^{\prime}$ denote the stabilization of $L$.
The knot $L^{\prime}$ then can be put on the page of the once stabilized open
book, depicted (together with the monodromies) by Figure 4$(ii)$, where the
unknot is represented by the curve with $L^{\prime}$ next to it, and the thin
circles represent curves along which Dehn twists are to be performed to get
the monodromy map.

2pt $L$ [lb] at 12 85 $L^{\prime}$ [lb] at 244 113 $b$ [rt] at 96 106 $a$ [lt]
at 63 106 $w_{1}$ [B] at 79 116 $w_{2}$ [B] at 79 146 $z$ [lb] at 126 113
$b_{1}$ [t] at 326 88 $a_{1}$ [t] at 300 88 $z$ at 358 110 $b_{2}$ [t] at 329
165 $a_{2}$ [t] at 295 164 $w_{1}$ [B] at 311 188 $w_{2}$ [B] at 309 212 $(i)$
at 79 0 $(ii)$ at 311 0

Figure 4: The page of the open book and the basis for the stabilized unknot.

The page on the left represents the open book decomposition for the Hopf band,
while the page on the right is its stabilization. The corresponding Heegaard
diagram (with the use of the adapted basis of Figure 4$(ii)$) is shown in
Figure 5.

2pt $P$ [t] at 98 104 $w_{1}$ [B] at 66 127 $V$ [ll] at 225 139 $T$ [ll] at
261 142 $X_{1}$ [tl] at 334 99 $X_{2}$ [B] at 353 178 $z$ at 262 199 $w_{2}$
at 140 155

Figure 5: Heegaard diagram for the unknot.

We record both possible choices of $w$ by putting a $w_{1}$ and a $w_{2}$ on
the diagram — the two choices correspond to the two possible orientations of
$L$. Incidentally, these two choices also correspond to the two possible
stabilizations with respect to a given orientation, since positive
stabilization for an orientation is exactly the negative stabilization for the
reversed orientation. It still needs to be determined whether $w_{2}$ gives a
positive or negative stabilization.

######  Lemma 4.1.

The oriented knot determined by the pair $z$ and $w_{2}$ represents the
negative stabilization of the oriented unknot (i.e. the stabilization with
$\operatorname{rot}=-1$).

###### Proof.

In order to compute the rotation number of the stabilization given by $w_{2}$,
first we construct a Seifert surface for $L^{\prime}$. To this end, consider
the loop $A_{1}$ given by the upper part of $L^{\prime}$ together with the
dashed line $C_{1}$ of Figure 6.

2pt $L^{\prime}$ at 33 149 $R$ at 30 104 $C_{1}$ at 138 133 $C_{2}$ at 138 75

Figure 6: Computation of the rotation of $L^{\prime}$.

This loop bounds a disk $D_{1}$ in the 3–manifold given by the open book
decomposition, and the tangent vector field along $A_{1}$ obviously extends as
a nonzero section of $\xi$ to $D_{1}$, since $D_{1}$ can be regarded as an
appropriate Seifert surface for the unknot $L$ before stabilization (cf.
Figure 4). Define $A_{2}$ and $D_{2}$ similarly, now using the lower part of
$L^{\prime}$. A Seifert surface for $L^{\prime}$ then can be given by the
union of $D_{1},D_{2}$ and the region $R$ of Figure 6. Extending the tangent
vector field along $L^{\prime}$ to a section of $\xi$ over $C_{1},C_{2}$
first, the above observation shows that the rotation number of $L^{\prime}$ is
the same as the obstruction to extending the above vector field to $R$ as a
section of $\xi$. Notice that along $R$ we have that $\xi=TR$. The region $R$
(with the given vector field on its boundary) can be embedded into the disk
with the tangent vector field along its boundary, hence a simple Euler
characteristic computation shows that the obstruction we need to determine is
equal to $-1$, concluding the proof. ∎

Notice that the monodromy is pictured on the page $S_{+1}$ (which also
contains the knot), but its effect is taken into account on $S_{-1}$, which
comes in the Heegaard surface with its orientation reversed. Therefore when
determining the $\alpha$– and $\beta$–curves in the Heegaard diagrams,
right–handed Dehn twists of the monodromy induce left–handed Dehn twists on
the diagram and vice versa.

The three generators of the chain complex corresponding to the Heegaard
diagram of Figure 5 in the second symmetric product of the genus–2 surface are
the pairs $PX_{1},PX_{2}$ and $TV$. It is easy to see that there are
holomorphic disks connecting $TV$ and $PX_{2}$ (passing through the basepoint
$w_{1}$) and $TV$ and $PX_{1}$ (passing through $w_{2}$); and these are the
only two possible holomorphic disks not containing $z$.

When we use $w_{1}$ as our second basepoint, the two disks out of $TV$ show
that the class represented by $PX_{1}$, and the class of $PX_{2}$ when
muliplied by $U$ are homologous:

$\partial^{-}(TV)=PX_{1}+U\cdot PX_{2}.$

In conclusion, in this case $[PX_{2}]$ generates
$\mathrm{HFK}^{-}(-S^{3},L^{\prime})$ and the invariant $[PX_{1}]$ is
determined by the class of $U$–times the generator.

When using $w_{2}$ as the second basepoint, we see that the class $[PX_{2}]$
will be equal to $U\cdot[PX_{1}]$, hence in this case $[PX_{1}]$ is the
generator. Since $PX_{1}$ represents the Legendrian invariant, we conclude
that in this case the equivalence class modulo automorphisms of the generator
of $\mathrm{HFK}^{-}(-S^{3},L^{\prime})$ (over $\mathbb{F}[U]$) is equal to
the Legendrian invariant of $L^{\prime}$. In summary, we get

######  Corollary 4.2.

Suppose that $L\subset(S^{3},\xi)$ is an oriented Legendrian unknot with
$\operatorname{tb}(L)=-1$. Then ${\mathfrak{L}}(L)$ is represented by the
generator of the $\mathbb{F}[U]$–module $\mathrm{HFK}^{-}(L)=\mathbb{F}[U]$.
If $L^{-}$ is the negative stabilization of $L$ then
${\mathfrak{L}}(L^{-})={\mathfrak{L}}(L)$, while for the positive
stabilization $L^{+}$ we have ${\mathfrak{L}}(L^{+})=U\cdot{\mathfrak{L}}(L)$.
∎

##  5 Basic properties of the invariants

### Nonvanishing and vanishing results

The invariant ${\mathfrak{L}}$ admits a nonvanishing property provided the
contact invariant $c(Y,\xi)$ of the ambient 3–manifold is nonzero (which
holds, for example, when the ambient contact structure is strongly fillable).
When $c(Y,\xi)=0$ (e.g., if $(Y,\xi)$ is overtwisted) then ${\mathfrak{L}}(L)$
is a $U$–torsion element.

###### Proof of Theorem 1.2.

Consider the natural chain map

$f\colon\mathrm{CFK}^{-}(-Y,L)\to\widehat{\mathrm{CF}}(-Y)$

given by setting $U=1$, cf. Lemma 2.10. Let $(B,\varphi,A,w,z)$ be an open
book compatible with $(Y,\xi,L)$ with an adapted basis and basepoints. Since
the map $F$ induced by $f$ on the homologies sends $[{\bf{x}}(B,\varphi,A)]$
to $c(Y,\xi)$, the nonvanishing of ${\mathfrak{L}}(L)$ when $c(Y,\xi)\neq 0$
obviously follows. If the above map sends $[{\bf{x}}(B,\varphi,A)]$ to zero,
then by Lemma 2.10 (and the fact that $[{\bf{x}}(B,\varphi,A)]$ is
homogeneous) we get that $U^{d}\cdot{\mathfrak{L}}(L)=0$ for some $d\geq 0$,
verifying Theorem 1.2. ∎

A vanishing theorem can be proved for a _loose_ knot, that is, for a
Legendrian knot in a contact 3–manifolds with overtwisted complement. Before
this result we need a preparatory lemma from contact topology:

######  Lemma 5.1.

Suppose that $L\subset(Y,\xi)$ is a Legendrian knot such that $(Y,\xi)$
contains an overtwisted disk in the complement of $L$. Then, the complement
$(Y-\nu L,\xi|_{Y-\nu L})$ admits a connected sum decomposition $(Y-\nu
L,\xi_{1})\\#(S^{3},\xi_{2})$ with the property that $\xi_{1}$ coincides with
$\xi|_{Y-\nu L}$ near $\partial(Y-\nu L)$ and $\xi_{2}$ is overtwisted.

###### Proof.

Let us fix an overtwisted disk $D$ disjoint from the knot $L$ and consider a
neighbourhood $V$ (diffeomorphic to $D^{3}$) of $D$, with the property that
$V$ is still disjoint from $L$. By the classification of overtwisted contact
structures on $D^{3}$ with a fixed characteristic foliation on the boundary
[7, Theorem 3.1.1], we can take $\xi_{1}$ on $Y-\nu L$ and $\xi_{2}$ on
$S^{3}$ such that $\xi_{1}\\#\xi_{2}=\xi|_{Y-\nu L}$ and $\xi_{1}$ is equal to
$\xi$ near $\partial(Y-\nu L)$, while $\xi_{2}$ is overtwisted. The statement
of the lemma then follows at once. ∎

###### Proof of Theorem 1.4.

Let us fix a decomposition of $(Y,\xi)$ as before, that is,
$(Y,\xi)=(Y,\xi_{1})\\#(S^{3},\xi_{2})$ with the properties that
$L\subset(Y,\xi_{1})$ and $\xi_{2}$ is overtwisted on $S^{3}$. Consider open
book decompositions compatible with $(Y,\xi_{1})$ and $(S^{3},\xi_{2})$.
Assume furthermore that the first open book has a basis adapted to $L$, while
the second open book has a basis containing an arc which is displaced to the
left by the monodromy. (The existence of such a basis is shown in the proof of
[19, Lemma 3.2].) The Murasugi sum of the two open books and the union of the
two bases provides an open book decomposition for $(Y,\xi)$, adapted to the
knot $L$, together with an arc disjoint from $L$ which is displaced to the
left by the monodromy. Since the basepoint $w$ in the Heegaard diagram is in
the strip determined by the arc intersecting the knot, the holomorphic disk
appearing in the proof of [19, Lemma 3.2] avoids both basepoints and shows the
vanishing of ${\mathfrak{L}}(L)$. ∎

### Transverse knots

Next we turn to the verification of the formula relating the invariants of a
negatively stabilized Legendrian knot to the invariants of the original knot.
We will spell out the details for ${\mathfrak{L}}$ only. Then we will discuss
the implication of the stabilization result regarding invariants of transverse
knots. The effect of more general stabilizations on the invariant will be
addressed later using slightly more complicated techniques.

######  Proposition 5.2.

Suppose that $L$ is an oriented Legendrian knot and $L^{-}$ denotes the
oriented negative stabilization of $L$. Then,
${\mathfrak{L}}(L^{-})={\mathfrak{L}}(L)$.

###### Proof.

The proof relies on the choice of a convenient open book decomposition. To
this end, fix an open book decomposition $(B,\varphi,A)$ compatible with
$(Y,\xi)$ and with a basis adapted to $L$. Place $w$ according to the given
orientation. As shown in [11], after an appropriate Giroux stabilization the
open book also accomodates the stabilization of $L$. The new open book with
adapted basis $(B^{-},\varphi^{-},A^{-})$ together with the new choice of $w$
(denoted by $w^{-}$) is illustrated by Figure 7.

2pt $a$ [t] at 84 22 $b$ [t] at 119 24 $a$ [t] at 328 22 $b$ [t] at 363 24 $w$
[b] at 346 48 $w$ at 101 56 $z$ at 150 47 $z$ at 404 49 $w^{-}$ [b] at 348 113
$a^{\prime}$ [b] at 361 127 $b^{\prime}$ [b] at 329 127 $c$ [tl] at 372 71

Figure 7: Change of the open book after negative stabilization of $L$. The
monodromy changes by a right–handed Dehn twist along the curve $c$.

As in Lemma 4.1, we can easily see that this choice provides the negative
stabilization $L^{-}$. (Recall that the stabilization changed the monodromy of
the open book by multiplying it with the right–handed Dehn twist $D_{c}$.) In
the new page the stabilization of $L$ is determined up to isotopy simply by
changing the basepoint from $w$ to $w^{-}$. (Notice that by placing $w^{-}$ in
the other domain in the strip between $a^{\prime}$ and $b^{\prime}$ the
orientation of the stabilized knot would be incorrect.) Now the corresponding
portion of the Heegaard diagram has the form shown by Figure 8.

2pt $a$ [t] at 174 22 $b$ [t] at 217 24 $w$ at 194 72 $a^{\prime}$ [tl] at 219
197 $w^{-}$ at 194 182 $b^{\prime}$ [tr] at 173 197 $C_{1}$ [l] at 219 105
$C_{2}$ [l] at 202 255 $\beta$ [l] at 251 280 $z$ at 302 116

Figure 8: Stabilization in the Heegaard diagram. The top and bottom boundary
components of the surface and the circles $C_{1}$ and $C_{2}$ should be
thought of as identified via a reflection across the middle (dotted) line of
the picture.

In the picture, the top and bottom boundary components of the surface and the
circles $C_{1}$ and $C_{2}$ should be thought of as identified via a
reflection across the middle (dotted) line of the picture. Moreover, the curve
$\beta$ is only partially represented, due to the action of the monodromy. As
before, in the diagram $w$ gives rise to $L$ while $w^{-}$ to $L^{-}$
(together with the common $z$). It is straightforward from the picture that
$w$ and $w^{-}$ are in the same domain, hence the statement follows. ∎

It follows from Proposition 5.2 that the invariant ${\mathfrak{L}}$ of a
Legendrian approximation provides an invariant for a transverse knot.

###### Proof of Theorem 1.5.

Fix a transverse knot $T$ and consider a Legendrian approximation $L$ of $T$.
By [9, 13], up to negative stabilizations the Legendrian knot $L$ only depends
on the transverse isotopy class of $T$. Therefore by Proposition 5.2 the
equivalence classes $\mathfrak{T}(T)={\mathfrak{L}}(L)$ and
$\widehat{\mathfrak{T}}(T)=\widehat{{\mathfrak{L}}}(L)$ are invariants of the
transverse isotopy class of the knot $T$, and hence the theorem follows. ∎

##  6 Non–loose torus knots in $S^{3}$

In this section we describe some examples where the invariants defined in the
paper are explicitly determined. Some interesting consequences of these
computations will be drawn in the next section. For the sake of simplicity, we
will work with the invariant $\widehat{\mathfrak{L}}$.

### Positive Legendrian torus knots $T_{(2,2n+1)}$ in overtwisted contact
$S^{3}$’s

Let us consider the Legendrian knot $L(n)$ given by the surgery diagram of
Figure 9.

2pt $L(n)$ [l] at 453 55 $+1$ [l] at 458 82 $+1$ [l] at 454 107 $-1$ [l] at
453 131 $-1$ [l] at 514 154 $-1$ [l] at 513 201 $n$ [r] at 55 177

Figure 9: Non–loose Legendrian torus knot $T_{(2,2n+1)}$ in $S^{3}$ ($n\geq
1$).

The meaning of the picture is that we perform contact $(\pm 1)$–surgeries
along the given Legendrian knots, the result being a contact 3–manifold
containing the unframed knot $L(n)$ as a Legendrian knot. (For contact $(\pm
1)$–surgeries and surgery presentations see [3, 30].)

######  Lemma 6.1.

The contact structure $\xi_{n}$ defined by the surgery diagram of Figure 9 is
the overtwisted contact structure on $S^{3}$ with Hopf invariant
$d_{3}(\xi_{n})=1-2n$. The Legendrian knot $L(n)$ is smoothly isotopic to the
torus knot $T_{2,2n+1}$ and is non–loose.

###### Proof.

Figure 10 gives a smooth surgery diagram corresponding to Figure 9.

2pt $-2$ at 28 52 $-1$ at 139 53 $-3$ [l] at 75 55 $-4$ [lt] at 51 80 $-4$
[rb] at 35 96 $0$ [r] at 203 53 $0$ [r] at 232 53 $L(n)$ [l] at 266 56 $(n)$
at 37 81

Figure 10: A smooth version of Figure 9.

The knot type of $L(n)$ and the underlying 3–manifold can be easily
identified. The Kirby calculus moves of Figure 11 show that Figure 10 is
equivalent to the left–hand side of Figure 12.

2pt $-1$ at 97 214 $-2$ at 20 213 $-4$ at 44 233 $-4$ at 22 252 $-3$ at 58 214
$0$ at 139 214 $0$ at 159 214 $(n)$ at 27 234 $L(n)$ at 190 239 $+1$ at 397
178 $L(n)$ at 434 225 $(n)$ at 263 236 $+1$ at 392 243 $+1$ at 379 229 $-1$ at
284 235 $-1$ at 261 255 $+1$ at 236 232 $+1$ at 234 216 $-2$ at 308 231 $-1$
at 367 27 $L(n)$ at 428 83 $+1$ at 235 85 $+1$ at 231 66 $(n)$ at 263 89 $-1$
at 288 91 $-1$ at 267 111 $-2$ at 371 88 $-2$ at 39 71 $-3$ at 82 68 $-2$ at
83 84 $-2$ at 81 150 $n-1$ at 152 114 $L(n)$ at 168 70 $-1$ at 97 14

Figure 11: Kirby moves on Figure 10.

2pt $-1$ [b] at 76 6 $-2$ [rb] at 29 61 $-3$ [r] at 70 61 $-2$ [r] at 70 77
$-2$ [r] at 70 141 $n-1$ [l] at 125 107 $L(n)$ [b] at 139 61 $n$ at 285 55
$L(n)$ [b] at 286 102

Figure 12: $L(n)$ is the torus knot $T_{2,2n+1}$.

Applying a number of “blow–downs” yields the right–hand side of Figure 12,
veryfing that $L(n)$ is isotopic to the positive torus knot $T_{(2,2n+1)}$ in
$S^{3}$.

Figures 9 and 10 can be used to determine the signature $\sigma(X)$ and the
Euler characteristic $\chi(X)$ of the 4–manifold obtained by viewing the
integral surgeries as 4–dimensional 2–handle attachments to
$S^{3}\times[0,1]$. (As it is customary in Heegaard Floer theory, the
4-manifold $X$ denotes the cobordism between $S^{3}$ and the 3–manifold we get
after performing the prescribed surgeries.) In addition, the rotation numbers
define a second cohomology class $c\in H^{2}(X;{\mathbb{Z}})$, and a simple
computation shows

$\sigma(X)=-n-1,\ \ \chi(X)=n+3,\ \ c^{2}=-9n-1.$

Since we apply two $(+1)$–surgeries, the formula

$d_{3}(\xi)=\frac{1}{4}(c^{2}-3\sigma(X)-2\chi(X))+q$

(with $q$ denoting the number of contact $(+1)$–surgeries) computes
$d_{3}(\xi_{n})$ of the contact structure, providing $1-2n<0$ for all
$n\in{\mathbb{N}}$. Since the unique tight contact structure $\xi_{st}$ on
$S^{3}$ has vanishing Hopf invariant $d_{3}(\xi_{st})$, we get that $\xi_{n}$
is overtwisted. Applying contact $(-1)$–surgery along $L(n)$ we get a tight
contact structure, since this $(-1)$–surgery cancels one of the
$(+1)$–surgeries, and a single contact $(+1)$–surgery along the Legendrian
unknot provides the Stein fillable contact structure on $S^{1}\times S^{2}$.
Therefore there is no overtwisted disk in the complement of $L(n)$ (since such
a disk would persist after the surgery), consequently $L(n)$ is non–loose. ∎

As it is explained in [25, Section 6], the Legendrian link underlying the
surgery diagram for $\xi_{n}$ (together with the Legendrian knot $L(n)$) can
be put on a page of an open book decomposition with planar pages, which is
compatible with the standard contact structure $\xi_{st}$ on $S^{3}$. This can
be seen by considering the annular open book decomposition containing the
Legendrian unknot (and its Legendrian push–offs), and then applying the
stabilization method described in [11] for the stabilized knots. The monodromy
of this open book decomposition can be computed from the Dehn twists resulting
from the stabilizations, together with the Dehn twists (right–handed for
$(-1)$ and left–handed for $(+1)$) defined by the surgery curves. Notice that
one of the left–handed Dehn twists is cancelled by the monodromy of the
annular open book decomposition we started our procedure with. This procedure
results in the monodromies given by the curves of Figure 13. We perform
right–handed Dehn twists along solid curves and a left–handed Dehn twist along
the dashed one.

2pt $L(n)$ [l] at 187 159 $(n)$ at 95 243

Figure 13: Monodromy of the open book decomposition compatible with $L(n)$.
Solid curves represent right–handed Dehn twists, while the dashed one
(parallel to $L(n)$) represents a left–handed Dehn twist.

The application of the lantern relation simplifies the monodromy factorization
to the one shown in Figure 14.

2pt $L(n)$ at 149 137 $(n)$ [b] at 30 118

Figure 14: Simplified monodromy.

Notice that in the monodromy factorization given by Figure 13 there are Dehn
twists along intersecting curves, hence these elements of the mapping class
group do not commute. Therefore, strictly speaking, an order of the Dehn
twists should be specified. Observe, however, that although the elements do
not commute, the fact that there is only one such pair of intersecting curves
implies that the two possible products are conjugate, and therefore give the
same open book decomposition, allowing us to suppress the specification of the
order.

Figure 15 helps to visualize the curves on ‘half’ of the Heegaard surface, and
also indicates the chosen basis.

2pt $L(n)$ at 58 77 $a_{4}$ at 75 90 $a_{3}$ at 174 90 $a_{2}$ at 270 90
$a_{1}$ at 370 90 $(n)$ at 408 102

Figure 15: Another view of the monodromy. The diagram also indicates the
chosen basis. (Notice that here $a_{4}$ intersects $L(n)$ in a unique point
and not $a_{1}$.)

The open book decomposition found above equips $S^{3}$ with a Heegaard
decomposition compatible with $L(n)$. The $\alpha$– and $\beta$–curves of this
decomposition are given in Figure 16.

0pt $L(n)$ at 79 39 $\alpha_{4}$ at 227 58 $\alpha_{3}$ at 218 193
$\alpha_{2}$ at 222 236 $\alpha_{1}$ at 223 372 $\beta_{1}$ at 109 406
$\beta_{2}$ at 90 299 $\beta_{3}$ at 93 202 $\beta_{4}$ at 88 100 $n$ at 161
434 $w$ at 128 18 $z$ at 50 211 $L$ at 141 116 $M$ at 155 98 $B_{3}$ at 205
211 $X_{1}$ at 137 200 $C_{2}$ at 157 199 $X_{2}$ at 194 199 $Y_{2}$ at 167
218 $B_{4}$ at 149 218 $Y_{1}$ at 121 217 $B_{1}$ at 86 257 $P$ at 166 303
$B_{2}$ at 218 277 $Q$ at 124 311 $A_{1}$ at 88 381 $A_{2}$ at 133 392 $A_{n}$
at 176 405 $A_{n+1}$ [l] at 218 329 $C_{1}$ at 87 167 $D=D_{1}$ [r] at 91 68

Figure 16: The Heegaard decomposition compatible with $L(n)$.

Recall that we get the $b_{i}$ arcs by the usual perturbation of the $a_{i}$’s
and the action of the monodromy yields a Heegaard decomposition for $S^{3}$
with the distinguished point ${\bf{x}}$ in
${\mathbb{T}}_{\alpha}\cap{\mathbb{T}}_{\beta}$ determining the Legendrian
invariant $\widehat{\mathfrak{L}}(L(n))$. A warning about orientations is in
place. We illustrated the monodromy curves on the page $S_{+1}$ containing the
Legendrian knot $L(n)$, but their action must be taken into account on the
page $S_{-1}$, which comes in the Heegaard surface with its opposite
orientation; hence right–handed Dehn twists take curves to the left on the
Heegaard surface and vice versa. The basepoint $z$ is placed in the ‘large’
region of the page $S_{+1}$, while the point $w$ is in the strip between
$a_{1}$ and $b_{1}$ shown in Figure 16. This choice determines the orientation
on $L(n)$.

Recall that since $L(n)$ is isotopic to the positive torus knot
$T_{(2,2n+1)}$, the Legendrian invariant is given by an element of
$\widehat{\mathrm{HFK}}(-S^{3},T_{(2,2n+1)})$, which is isomorphic to
$\widehat{\mathrm{HFK}}(S^{3},T_{(2,-(2n+1))})$. This latter group is
determined readily from the Alexander polynomial and the signature of the knot
(since it is alternating), cf. [36, 39]. In fact, we have

$\widehat{\mathrm{HFK}}_{n+s}(S^{3},T_{(2,-(2n+1))},s)\cong\left\\{\begin{array}[]{ll}\mathbb{F}&{\text{if
$|s|\leq n$}}\\\ 0&{\text{otherwise.}}\end{array}\right.$ (6.1)

After these preparations we are ready to determine the invariants of the
Legendrian knots discussed above.

######  Proposition 6.2.

The homology class $\widehat{\mathfrak{L}}(L(n))$ is determined by the unique
nontrivial homology class in Alexander grading $1-n$ in
$\widehat{\mathrm{HFK}}(S^{3},T_{(2,-2n-1)})$.

###### Proof.

We claim that in the Heegaard diagram of Figure 16 the point
$A_{1}B_{1}C_{1}D_{1}$ (representing the Legendrian invariant
$\widehat{\mathfrak{L}}(L(n))$) is the only intersection point in Alexander
grading $1-n$.

We sketch the argument establishing this. We can orient every $\alpha$– and
$\beta$–curve in the diagram so that their intersection matrix (whose
$(i,j)^{th}$ entry is the algebraic intersection of $\alpha_{i}$ with
$\beta_{j}$) is

$M=\left(\begin{array}[]{rrrr}n+1&-1&0&0\\\ -1&4&-2&0\\\ 0&-2&2&-1\\\
0&0&-1&1\end{array}\right).$

Note that for this particular diagram, the absolute values of the algebraic
intersection number and the geometric intersection numbers coincide.

A simple calculation shows that there are $16n+19$ intersecton points in
${\mathbb{T}}_{\alpha}\cap{\mathbb{T}}_{\beta}$. To calculate the relative
Alexander gradings of intersection points, it is convenient to organize them
into types: specifically, the intersection points of $\mathbb{T}_{\alpha}$ and
$\mathbb{T}_{\beta}$ correspond to permutations $\sigma$ and then $4$-tuples
in $\prod\alpha_{i}\cap\beta_{\sigma(i)}$, which in terms of the notation of
Figure 16 are given by quadruples of letters; specifically, intersection
points all have one of the following five types

$\begin{array}[]{lllll}ABCD&PQCD&AXYD&ABLM&PQLM.\end{array}$

We begin by calculating relative Alexander gradings of intersection points of
the same type. Consider first the relative gradings between points of the form
$A_{i}\mbox{$\star\star\star$}$ and $A_{i+1}\mbox{$\star\star\star$}$, where
now $\star\star\star$ is a fixed triple of intersection points. Start from an
initial path $\delta$ which travels from $A_{i}$ to $A_{i+1}$ along
$\alpha_{1}$, then back from $A_{i+1}$ to $A_{i}$ along $\beta_{1}$. We add to
this cycle copies of $\alpha_{i}$ and $\beta_{j}$ to obtain a null-homologous
cycle. This can be done since the $\alpha_{i}$’s and $\beta_{j}$’s span
$H_{1}(\Sigma)$, which follows from the fact that the ambient 3–manifold is
$S^{3}$. Concretely, we need to solve the expression

$\sum n_{i}[\alpha_{i}]+\sum m_{j}[\beta_{j}]+[\delta]=0$

for $n_{i}$ and $m_{j}$. In fact, all we are concerned about is the difference
in local multiplicities at $z$ and $w$ in the null-homology of the above
expression. Since $z$ and $w$ are separated by $\alpha_{4}$, and our initial
curve $\delta$ is supported on $\alpha_{1}$ and $\beta_{1}$, this difference
is given by the multiplicity $n_{4}$ of $\alpha_{4}$ in the above expression.
In view of the fact that $\\#\beta_{i}\cap\beta_{j}=0$, $n_{4}$ can be
obtained as the last coefficient of $M^{-1}v$, where $v$ is the vector whose
$i^{th}$ coordinate is $\\#\delta\cap\beta_{j}$ and $M$ is the incidence
matrix. We can find paths of the above type $\delta$ connecting $A_{i}$ and
$A_{i+1}$ whose intersection numbers with the $\beta_{j}$’s are given by the
vector $(1,0,0,0)$, hence giving that the Alexander grading of
$A_{i}\mbox{$\star\star\star$}$ is two smaller than the Alexander grading of
$A_{i+1}\mbox{$\star\star\star$}$. We express this by saying that the relative
gradings of the $A_{i}$ are given by $2i$. Repeating this procedure for the
other curves, we find that relative gradings of $B_{1},...,B_{4}$ (completed
by a fixed triple to quadruples of intersection points) are given by $0$,
$2n$, $4n+2$, and $2n+1$ respectively; for $C_{1}$ and $C_{2}$ the relative
gradings are $0$ and $2n+1$ respectively, for $X_{1}$ and $X_{2}$ they are $0$
and $2n+1$, and they are also $0$ and $2n+1$ for $Y_{1}$ and $Y_{2}$. Putting
these together, we can calculate the relative Alexander gradings of any two
intersection points of the same type.

To calculate the relative Alexander gradings of points of different types, we
use three rectangles. Specifically, intersection points of the form
$PQ\mbox{$\star\star$}$ and $A_{n+1}B_{1}\mbox{$\star\star$}$ (where here
$\star\star$ denotes some fixed pair of intersection points) have equal
Alexander gradings, shown by the rectangle with vertices $PB_{1}QA_{n+1}$,
avoiding both $w$ and $z$. Similar argument relates points of type
$X_{1}Y_{1}\mbox{$\star\star$}$ to the ones of type
$B_{4}C_{1}\mbox{$\star\star$}$. The rectangle $LDMC_{1}$ contains $w$ once,
showing that the Alexander gradings of elements of the form
$LM\mbox{$\star\star$}$ are one less than the Alexander gradings of elements
of the form $C_{1}D\mbox{$\star\star$}$. This allows us to calculate the
relative Alexander gradings of any two intersection points.

Given this information, it is now straightforward to see that there are no
other intersection points in the same Alexander grading as
$A_{1}B_{1}C_{1}D_{1}$, and hence that it represents a homologically
nontrivial cycle in $\widehat{\mathrm{HFK}}(S^{3},T_{(2,-(2n+1))})$. Thus, we
have shown that for all $n\in{\mathbb{N}}$, the class
$\widehat{\mathfrak{L}}(L(n))$ is a nontrivial generator in
$\widehat{\mathrm{HFK}}(S^{3},T_{(2,-(2n+1))})$.

In fact, the absolute Alexander grading of generators is pinned down by the
following symmetry property: the Alexander grading is normalized so that the
parity of the number of points of Alexander grading $i$ coincides with the
parity of the number of points of Alexander grading $-i$. Using this property,
one finds that $A_{1}B_{1}C_{1}D_{1}$ is supported in Alexander grading $1-n$.
∎

######  Remark 6.3.

Recall (cf. [36, 39]) that

$\mathrm{HFK}^{-}(S^{3},T_{(2,-(2n+1))})\cong\mathbb{F}^{n}\oplus\mathbb{F}[U],$
(6.2)

where the top generator of the free $\mathbb{F}[U]$–module is at $(A=n,M=2n)$
while the $n$ generators of the $\mathbb{F}^{n}$ summand are of bi–degrees

$(A=n-1-2i,\quad M=2n-1-2i),\quad i=0,\ldots,n-1.$

It is not hard to show that the above computation implies that
${\mathfrak{L}}(L(n))$ is determined by the unique nonzero $U$–torsion element
of $\mathrm{HFK}^{-}(S^{3},T_{(2,-(2n+1))})$ in Alexander grading $1-n$.

### Negative Legendrian torus knots $T_{(2,-(2n-1))}$ in overtwisted contact
$S^{3}$’s

For $k,l\geq 0$ let us consider the knot $L_{k,l}$ in the contact 3–manifold
$(Y_{k,l},\xi_{k,l})$ given by the surgery presentation of Figure 17. Let
$n=k+l+2$.

2pt $L_{k,l}$ [l] at 453 55 $+1$ [l] at 458 82 $+1$ [l] at 454 107 $-1$ [l] at
453 131 $-1$ [l] at 514 154 $-1$ [l] at 513 201 $l$ at 469 391 $k$ at 46 343

Figure 17: Non–loose Legendrian $T_{(2,-(2n-1))}$ torus knot in $S^{3}$.

######  Proposition 6.4.

The contact 3–manifold specified by the surgery diagram of Figure 17 is
$(S^{3},\xi_{k,l})$ with $d_{3}(\xi_{k,l})=2l+2$, hence $\xi_{k,l}$ is
overtwisted. The Legendrian knot $L_{k,l}$ as a smooth knot is isotopic to the
negative $(2,2n-1)$ torus knot $T_{(2,-(2n-1))}$, and it is non–loose in
$(S^{3},\xi_{k,l})$.

###### Proof.

The simple Kirby calculus argument illustrated in Figure 18

2pt $L_{k,l}$ at 187 255 $-1$ at 106 227 $0$ at 142 226 $0$ at 161 226 $-3$ at
15 200 $-3$ at 38 209 $-n$ at 29 232

$-n$ at 230 258 $-2$ at 265 253 $L_{k,l}$ at 399 249 $-2$ at 295 214 $1$ at
323 196 $1$ at 361 258 $1$ at 343 253

$-n$ at 38 158 $-2$ at 86 167 $-1$ at 115 107 $L_{k,l}$ at 194 157 $-2$ at 133
168

$-n$ at 257 150 $-1$ at 294 108 $1$ at 328 125 $L_{k,l}$ at 381 147 $-1$ at
375 110

$-n$ at 84 68 $0$ at 63 25 $L_{k,l}$ at 182 68 $T_{2,-(2n-1)}\subset S^{3}$
(after two slides) [l] at 233 39

Figure 18: Kirby moves on the diagrams of Figure 17.

shows that the 3–manifold $Y_{k,l}$ is $S^{3}$, while the formula

$d_{3}(\xi_{k,l})=\frac{1}{4}(c^{2}-3\sigma-2b_{2})+q$

of [4] computes the Hopf invariant of $\xi_{k,l}$. As always, $\sigma$ and
$b_{2}$ denote the signature and the second Betti number of the 4–manifold $X$
specified by the underlying smooth surgery diagram, while $c\in
H^{2}(X;{\mathbb{Z}})$ is specified by the rotation numbers of the contact
surgery curves, and $q$ is the number of $(+1)$–surgeries. Simple algebra
verifies that

$d_{3}(\xi_{k,l})=2l+2\geq 2,$

and since the unique tight structure on $S^{3}$ has vanishing Hopf invariant,
we get that $\xi_{k,l}$ is overtwisted. Following the Kirby moves of Figure 18
with the knot $L_{k,l}$ we arrive to the last surgery picture of Figure 18,
and by sliding $L_{k,l}$ twice over the $(-n)$–framed unknot and cancelling
the pair we see that $L_{k,l}$ is, in fact, isotopic to the negative
$(2,2n-1)$ torus knot $T_{(2,-(2n-1))}\subset S^{3}$.

If contact $(-1)$–surgery on $L_{k,l}$ provides a tight contact structure,
then $L_{k,l}$ is obviously non–loose, since any overtwisted disk in its
complement would give an overtwisted disk in the surgered manifold. In our
case, however, contact $(-1)$–surgery simply cancels one of the
$(+1)$–surgeries defining $\xi_{k,l}$, and since a single contact
$(+1)$–surgery on the Legendrian unknot provides the contact boundary of the
Stein 1–handle (cf. [23]), we get that $(-1)$–surgery along $L$ provides a
Stein fillable contact structure. ∎

######  Remark 6.5.

In fact, by stabilizing $L_{k,l}$ on the _left_ and then performing a contact
$(-1)$–surgery we still get a tight contact 3–manifold: it will be Stein
fillable if we perform only one stabilization, and not Stein fillable but
tight for more stabilizations. The tightness of the result of these latter
surgeries were verified in [17] by computing the contact Ozsváth–Szabó
invariants of the resulting contact structures. Notice that this observation
implies that after arbitrary many left stabilizations $L_{k,l}$ remains
non–loose, which, in view of Proposition 5.2, is a necessary condition for
${\mathfrak{L}}(L_{k,l})$ to be nonvanishing. (Finally note that performing
contact $(-1)$–surgery on $L_{k,l}$ after a single _right_ stabilization
provides an overtwisted contact structure (cf. now [17, Section 5])). In
contrast, for the non–loose knots $L(n)$ of the previous subsection (also
having nontrivial ${\mathfrak{L}}$–invariants) the same intuitive argument
does not work, since some negative surgery on the knot $L(n)$ will produce a
contact structure on the 3–manifold $S^{3}_{2n-1}(T_{2,2n+1})$ and since this
3–manifold does not admit any tight contact structure [24], the result of the
surgery will be overtwisted independently on the chosen stabilizations.
Nevertheless, the overtwisted disk in the 3–manifold obtained by only negative
stabilizations cannot be in the complement of the knot $L(n)$, since such
stabilizations are still non–loose (shown by the nonvanishing of the invariant
$\widehat{{\mathfrak{L}}}(L(n))$).

Next we want to determine the classical invariants of $L_{k,l}$. Considering
the problem slightly more generally, and let $L\subset Y$ be a null–homologous
Legendrian knot in a contact 3–manifold $(Y,\xi)$. Assume furthermore that $Y$
is a _rational homology 3–sphere_. The knot $L$ has two “classical
invariants”: the Thurston–Bennequin and the rotation numbers
$\operatorname{tb}(L),\operatorname{rot}(L)$. Recall that the
Thurston–Bennequin invariant of a Legendrian knot is defined as the framing
induced by the contact 2–plane field distribution on the knot, hence, given an
orientation for the knot, it naturally gives rise to a homology class
$\operatorname{TB}\in H_{1}(Y-L;\mathbb{Z})$ (supported in a tubular
neighborhood of $L$). Fixing a Seifert surface $F$ for $L$, the oriented
intersection number of $\operatorname{TB}$ with $F$ provides a numerical
invariant, called the _Thurston–Bennequin number_
$\operatorname{tb}(L)\in\mathbb{Z}$. This intersection number is independent
of the choice of the Seifert surface; moreover, the Thurston–Bennequin number
is independent of the orientation of $L$ since the orientation of
$\operatorname{TB}$ and the orientation of $F$ both depend on this choice.
Analogously, the rotation $\operatorname{rot}$ of $L$ is the relative Euler
class of $\xi$ when restricted to $Y-L$, with the trivialization given by the
tangents of $L$. Again, a Seifert surface for $L$ can be used to turn this
class uniquely into an integer, also denoted by $\operatorname{rot}(L)$. Note
that unlike the Thurston–Bennequin number, the sign of the rotation number
does depend on the orientation of $L$.

Suppose that
${\mathbb{S}}={\mathbb{S}}_{+}\cup{\mathbb{S}}_{-}\subset(S^{3},\xi_{st})$ is
a contact $(\pm 1)$–surgery presentation of the contact 3–manifold $(Y,\xi)$
and $L$ is a Legendrian knot in $(S^{3},\xi_{st})$ disjoint from
${\mathbb{S}}$, null–homologous in $Y$. The Thurston–Bennequin and rotation
numbers of $L$ in $(Y,\xi)$ can be obtained from the Thurston–Bennequin and
rotation numbers of the individual components of ${\mathbb{S}}$ and $L$
through the following data. Let $\operatorname{tb}_{0}$ denote the
Thurston–Bennequin number of $L$ as a knot in the standard contact 3–sphere
(which, in terms of a front projection, is equal to the writhe minus half the
number of cusps). Writing ${\mathbb{S}}=\cup_{i=1}^{n}L_{i}$, let $a_{i}$ be
the integral surgery coefficient on the link component $L_{i}$; i.e.
$a_{i}=\operatorname{tb}(L_{i})\pm 1$ if $L_{i}\in{\mathbb{S}}_{\pm}$. Define
the linking matrix

$M(a_{0},a_{1},...,a_{n})=\left(m_{i,j}\right)_{i,j=0}^{n}$

where

$m_{i,j}:=\left\\{\begin{array}[]{ll}a_{i}&{\text{if $i=j$}}\\\
{\mathrm{lk}}(L_{i},L_{j})&{\text{if $i\neq j$,}}\end{array}\right.$

with the convention that $L=L_{0}$ and ${\mathbb{S}}=\cup_{i=1}^{n}L_{i}$.
Similarly, let $M(a_{1},...,a_{n})$ denote the matrix
$\left(m_{i,j}\right)_{i,j=1}^{n}$. Consider the integral rotation numbers
$r_{0},...,r_{n}$ obtained from our Legendrian knot $L$ and Legendrian
presentation as a link in $S^{3}$.

######  Lemma 6.6.

Suppose that
${\mathbb{S}}={\mathbb{S}}_{+}\cup{\mathbb{S}}_{-}\subset(S^{3},\xi_{st})$ is
a contact $(\pm 1)$–surgery presentation of the contact 3–manifold $(Y,\xi)$
and $L$ is a Legendrian knot in $(S^{3},\xi_{st})$ disjoint from
${\mathbb{S}}$, null–homologous in $Y$. Then the Thurston-Bennequin and
rotation numbers $\operatorname{tb}(L)$ and $\operatorname{rot}(L)$ can be
extracted from the above data by the formulae:

$\operatorname{tb}(L)=\operatorname{tb}_{0}+\frac{\det(M(0,a_{1},...,a_{n}))}{\det(M(a_{1},...,a_{n}))}$
(6.3)

and

$\operatorname{rot}(L)=r_{0}-\langle\left(\begin{array}[]{c}r_{1}\\\ \vdots\\\
r_{n}\end{array}\right),M^{-1}\cdot\left(\begin{array}[]{c}\operatorname{lk}(L_{0},L_{1})\\\
\vdots\\\ \operatorname{lk}(L_{0},L_{n}),\end{array}\right)\rangle,$ (6.4)

where $M=M(a_{1},...,a_{n})$.

###### Proof.

We turn to the verification of Equation (6.4), after a few preliminary
observations. Let $\mu_{i}\subset S^{3}-(L\cup{\mathbb{S}})$ be a meridian for
$L_{i}\subset L\cup{\mathbb{S}}$, and $\lambda_{i}$ be its corresponding
longitude. Recall that $H_{1}(S^{3}-(L\cup{\mathbb{S}}))$ is a free
$\mathbb{Z}$-module, generated by the meridians $\mu_{i}$ (we continue with
the convention that the $i=0$ component of the link $L\cup{\mathbb{S}}$ is
$L$). We can express the homology class of $\lambda_{i}$ in terms of the other
meridians by the expression

$\lambda_{i}=\sum_{j\neq i}\operatorname{lk}(L_{i},L_{j})\cdot\mu_{j}.$

The homology groups of the surgered manifold are gotten from the homology
groups of the link complement by dividing out by the relations
$a_{i}\mu_{i}+\lambda_{i}=0$ for $i=1,...,n$; more precisely,
$H_{1}(S^{3}-(L\cup{\mathbb{S}}))$ is freely generated by
$\mu_{0},...,\mu_{n}$, while $H_{1}(Y-L)\cong\mathbb{Z}$ is obtained from this
free group by dividing out the $n$ relations

$a_{i}\cdot\mu_{i}+\sum_{j\neq
i}\operatorname{lk}(L_{i},L_{j})\cdot\mu_{j}=0,\quad i=1,\ldots,n.$

The rotation numbers $r_{i}$ can be thought of as follows. Let
$e(\xi,L\cup{\mathbb{S}})\in H^{2}(S^{3},L\cup{\mathbb{S}})$ denote the
relative Euler class of $\xi$ relative to the trivialization it inherits along
$L\cup{\mathbb{S}}$. Then, the rotation numbers are the coefficients in the
expansion of the Poincaré dual $\operatorname{PD}[e(\xi,L\cup{\mathbb{S}})]\in
H_{1}(S^{3}-(L\cup{\mathbb{S}}))$ in terms of the basis of meridians
$\operatorname{PD}[e(\xi,L\cup{\mathbb{S}})]=\sum_{i=0}^{n}r_{i}\cdot\mu_{i}$.
Similarly, $H_{1}(Y-L)\cong\mathbb{Z}$ is generated by $\mu_{0}$, the meridian
of $L$, and the rotation number $\operatorname{rot}(L)$ is calculated by
$\operatorname{PD}(e(\xi,L))=\operatorname{rot}(L)\cdot\mu_{0}$. Note also
that $\operatorname{PD}[e(\xi,L)]$ is the image of
$\operatorname{PD}[e(\xi,L\cup{\mathbb{S}})]$ under the inclusion $j\colon
S^{3}-(L\cup{\mathbb{S}})\subset Y-L$. Thus, to find $\operatorname{rot}(L)$,
it suffices to express $j_{*}(\mu_{i})$ in terms of $\mu_{0}$. Write

$\displaystyle\Lambda=\left(\begin{array}[]{c}\operatorname{lk}(L_{0},L_{1})\\\
\vdots\\\ \operatorname{lk}(L_{0},L_{n}),\end{array}\right)$ and
$\displaystyle R=\left(\begin{array}[]{c}r_{1}\\\ \vdots\\\
r_{n}\end{array}\right).$

In view of our presentation for $H_{1}(S^{3}-L)$, we see that for all $i>0$ we
have $\mu_{i}=c_{i}\cdot\mu_{0}$, where $c_{i}$ is the $i^{th}$ entry in the
vector $-M^{-1}\cdot\Lambda$. It follows that

$\sum_{i=0}^{n}r_{i}\cdot\mu_{i}=(r_{0}-\langle
R,M^{-1}\cdot\Lambda\rangle)\cdot\mu_{0},$

estabilishing Equation (6.4).

We turn now to Equation (6.3). Choose $a_{0}$ so that the curve
$\lambda^{\prime}=a_{0}\cdot\mu_{0}+\lambda_{0}$ is null–homologous in $Y-L$.
Letting $\operatorname{TB}$ denote the Thurston–Bennequin framing curve of
$L=L_{0}$, we have that

$\operatorname{TB}=\operatorname{tb}_{0}\cdot\mu_{0}+\lambda_{0}=\operatorname{tb}(L)\cdot\mu_{0}+\lambda^{\prime};$

thus $\operatorname{tb}(L)-\operatorname{tb}_{0}=-a_{0}$. In view of our
presentation for this first homology group, we see that $a_{0}$ is determined
by the condition that

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\left|\begin{array}[]{llll}a_{0}&\operatorname{lk}(L_{0},L_{1})&...&\operatorname{lk}(L_{0},L_{n})\\\
\operatorname{lk}(L_{0},L_{1})&a_{1}&...&\operatorname{lk}(L_{0},L_{1})\\\
\vdots&\vdots&\ddots&\vdots\\\
\operatorname{lk}(L_{n},L_{0})&\operatorname{lk}(L_{n},L1)&\ldots&a_{n}\end{array}\right|$
$\displaystyle=$ $\displaystyle
a_{0}\cdot\det(M(a_{1},...,a_{n}))+\det(M(0,a_{1},...,a_{n})).$

Equation (6.3) follows. ∎

######  Lemma 6.7.

With the orientation given by Figure 17, the Thurston–Bennequin and rotation
numbers of the knot $L_{k,l}$ are given by

$\operatorname{tb}(L_{k,l})=-4(k+l)-6\quad{\mbox{and}}\quad\operatorname{rot}(L_{k,l})=-6l-2k-7.$

###### Proof.

Applying Formulae (6.3) and (6.4) to the surgery diagrams defining $L_{k,l}$,
simple algebra gives the statement of the lemma. ∎

As it is explained in the previous subsection (resting on observations from
[25, Section 6]), the surgery diagram for $\xi_{k,l}$ (together with the
Legendrian knot $L_{k,l}$) can be put on a page of an open book decomposition
with planar pages, which is compatible with the standard contact structure
$\xi_{st}$ on $S^{3}$. The diagram for all choices of $k$ and $l$ is less
apparent, hence we restrict our attention first to $k=0$. In this case we get
the monodromies defined by the curves of Figure 19.

2pt $L_{0,l}$ at 195 158 $l$ at 97 104

Figure 19: Monodromy of the open book decomposition compatible with $L_{0,l}$.

The application of the lantern relation simplifies the monodromy factorization
to the one illustrated in Figure 20.

2pt $L_{0,l}$ at 155 140 $l$ at 389 155

Figure 20: The simplified monodromy of the open book for $L_{0,l}$.

Figure 21 helps in visualizing the curves on ‘half’ of the Heegaard surface.
The diagram also indicates the chosen basis.

2pt $L_{0,l}$ at 47 70 $l+1$ at 349 154

Figure 21: Another view of the open book for $L_{0,l}$.

Notice that in the monodromy illustrated in Figure 19 there are Dehn twists
with intersecting curves, hence these elements of the mapping class group do
not commute. As before, the two products are conjugate, hence there is no need
to record the order.

In a similar spirit the general diagrams could be deduced — since we will not
use them in our computations, we will be content with working on a further
special case when $k=1$ and $l\geq 1$; the result is given by Figure 22.

2pt $L(1,l)$ [lb] at 110 152 $l-1$ at 378 136

Figure 22: The monodromy factorization for $L(1,l)$ with $l\geq 1$.

We just note here that in these cases we need to apply the lantern relation
twice to get commuting Dehn twists as illustrated in Figure 22. For this
reason, these monodromy factorizations contain two left–handed Dehn twists. In
Figure 23 the monodromy for $L(1,l)$ with $l\geq 1$ is represented on ’half’
of the Heegaard surface.

2pt $L(1,l)$ at 39 74 $l-1$ at 185 151

Figure 23: The open book for $L(1,l)$ with $l\geq 1$.

The open book decomposition found above equips $S^{3}$ with a Heegaard
decomposition compatible with $L_{k,l}$; the $\alpha$– and $\beta$–curves of
this decomposition for $k=0$ are given by Figure 24 when $l=0$ and by Figure
25 when $l>0$.

2pt $L_{0,0}$ at 76 41 $w$ at 130 18 $z$ at 44 161

Figure 24: The Heegaard decomposition for $L(0,0)$.

2pt $z$ at 45 306 $w$ at 145 18 $L_{0,l}$ at 81 43 $A$ at 78 661 $B$ at 153
710 $C$ at 153 622 $C^{\prime}$ at 206 622

Figure 25: The Heegaard decomposition for $L(0,l)$, $l>0$.

Recall that we get the $b_{i}$ curves by the usual perturbation of the
$a_{i}$’s and the action of the monodromy yields a Heegaard decomposition for
$S^{3}$ with the distinguished point ${\bf{x}}$ in
${\mathbb{T}}_{\alpha}\cap{\mathbb{T}}_{\beta}$ determining the Legendrian
invariant. As usual, the basepoint $z$ is placed in the ‘large’ region of the
page $S_{+1}$, while the point $w$ giving rise to $\widehat{\mathfrak{L}}$ is
in the strip between $a_{1}$ and $b_{1}$, as indicated in the pictures. It is
easy to see that moving $w$ to the other domain in the strip gives
$\widehat{\mathfrak{L}}=0$, because there is a holomorphic disk running into
${\bf{x}}$.

Recall that $L_{k,l}$ is isotopic to the negative torus knot
$T_{(2,-(2n-1))}$. The group $\widehat{\mathrm{HFK}}(-S^{3},T_{(2,-(2n-1))})$
is isomorphic to the Floer homology
$\widehat{\mathrm{HFK}}(S^{3},T_{(2,2n-1)})$ of the positive torus knot
$T_{(2,2n-1)}$. In this case we have the following (compare Equation (6.1)):

$\widehat{\mathrm{HFK}}_{s-n+1}(S^{3},T_{(2,2n-1)},s)\cong\left\\{\begin{array}[]{ll}\mathbb{F}&{\text{if
$|s|\leq n-1$}}\\\ 0&{\text{otherwise.}}\end{array}\right.$ (6.7)

After these preparations we are ready to determine the invariants of some of
the Legendrian knots discussed above. The computation admits a relatively
simple scheme when $k=0$, so we start with that case.

######  Theorem 6.8.

Let $L_{0,l}$ be given the orientation specified by the diagram of Figure 24
or Figure 25. Then, $\widehat{\mathfrak{L}}(L_{0,l})\neq 0$.

###### Proof.

We show that the intersection point ${\bf{x}}_{0}\in\widehat{\mathrm{CFK}}$,
which determines $\widehat{\mathfrak{L}}(L_{0,l})$, is alone in its Alexander
grading.

Claim: Any other intersection point ${\bf{y}}$ in the Heegaard decomposition
given by Figure 24 or 25 has Alexander grading $A({\bf{y}})$ strictly less
than $A({\bf{x}}_{0})$.

The claim, in turn, will be proved by induction on $l$. For $l=0$ (which means
that we are computing Legendrian invariants of a Legendrian left–handed
trefoil knot in an overtwisted $S^{3}$) the explicit determination of the
Alexander gradings of each of the 19 intersection points verifies the
statement. To do this we only need to find domains connecting any two
intersection points, and their relative Alexander grading is the multiplicity
of the domain containing the basepoint $w$. A straightforward linear algebra
argument similar to the one outlined in the proof of Proposition 6.2 shows
that all top three degress contain unique intersection points (giving rise to
homology elements), while the remaining 16 intersection points are of strictly
smaller Alexander grading.

The argument for the inductive step proceeds as follows: Notice that by
deleting the top–most $\alpha$– and $\beta$–curves (containing the point $A$
of Figure 25) in the Heegaard decomposition ${\mathfrak{H}}_{1}$ (and
destabilizing the Heegaard surface) we get a Heegaard decomposition
${\mathfrak{H}}_{0}$ adapted to $T_{(2,2n-3)}$. Repeating this procedure one
more time we get a Heegaard decomposition ${\mathfrak{H}}_{-1}$, which will be
a decomposition adapted to $T_{(2,2n-5)}$.

Note that the intersection matrix of the $\alpha$– and $\beta$–curves has the
following form

$\left(\begin{array}[]{rrrrrrrr}-2&1&0&0&0&0&0\\\ 1&-2&1&0&0&0&0\\\
0&1&\ddots&\ddots&0&0&0\\\ 0&0&\ddots&-2&1&0&0\\\ 0&0&0&1&-4&2&0\\\
0&0&0&0&2&-2&1\\\ 0&0&0&0&0&1&-1\end{array}\right)$

Now let us group the intersection points of the original diagram into types
$A,B$ and $C$ depending on their coordinate on the top–most $\beta$–curve. It
is easy to see that points in each $A$ and $B$ are in 1-1 correspondence with
intersections in ${\mathfrak{H}}_{0}$, while points in $C$ are in 1-1
correspondence with points in ${\mathfrak{H}}_{-1}$. From the form of this
matrix it is easy to see that the relative gradings within the groups coincide
with the relative gradings in their respective Heegaard diagrams
${\mathfrak{H}}_{0}$ and ${\mathfrak{H}}_{-1}$. As usual, let ${\bf{x}}_{0}$
denote the intersection point representing the Legendrian invariant,
$B_{{\bf{x}}_{0}}$ the point with the same coordinates except on the top-most
$\alpha$–circle, where $A$ is substituted with $B$. Finally,
$C_{{\bf{x}}_{0}}$ will denote the intersection point where on the top–most
$\beta$–circle we choose $C$ (and so on the top–most $\alpha$–circle we should
take $C^{\prime}$) and otherwise we take the same intersection points as in
${\bf{x}}_{0}$. Our inductive assumption is that ${\bf{x}}_{0}$,
$B_{{\bf{x}}_{0}}$ and $C_{{\bf{x}}_{0}}$ are of highest Alexander gradings in
the groups $A,B,C$, respectively. Therefore to conclude the argument we only
need to compare the gradings of these intersection points. This relative
computation can be performed locally near the top–most $\alpha$– and
$\beta$–curves and we get that $A({\bf{x}})-A(B_{{\bf{x}}})=2$ and
$A({\bf{x}})-A(C_{{\bf{x}}})=2$. By the inductive hypothesis this shows that
${\bf{x}}_{0}$ is the unique cycle in the top Alexander grading, concluding
the proof. ∎

Note that the relation found in the proof of Theorem 6.8 amongst the Alexander
gradings of the intersection points of the diagrams ${\mathfrak{H}}_{i}$
($i=\pm 1,0$) is a manifestation of the fact that the symmetrized Alexander
polynomial $\Delta_{n}(t):=\Delta_{T_{(2,2n-1)}}(t)$ satisfies the identity

$\Delta_{n}(t)=(t-t^{-1})\Delta_{n-1}(t)-\Delta_{n-2}(t).$

We consider two more cases, which can be handled by explicit methods:
$L_{1,1}$ and $L_{1,2}$. In these cases the Alexander grading alone is not
sufficient to calculate the homology class; we need a little analysis of
holomorphic disks. We begin with the first of these:

######  Theorem 6.9.

Let $L_{1,1}$ be given the orientation specified by the diagram of Figure 26.
Then, $\widehat{\mathfrak{L}}(L_{1,1})\neq 0$.

Recall that $L_{1,1}$ represents the negative torus knot $T_{(2,-7)}$.

###### Proof.

Consider the adapted Heegaard diagram for $L_{1,1}$ exhibited in Figure 26.

2pt $w$ at 141 18 $z$ at 31 258 $L_{1,1}$ at 77 41 $E$ at 82 68 $R$ at 157 99
$S$ at 139 117 $D_{1}$ at 82 165 $V_{1}$ at 142 199 $D_{2}$ at 172 200 $V_{2}$
at 192 198 $C_{1}$ at 78 257 $C_{2}$ at 148 218 $M_{1}$ at 99 210 $M_{2}$ at
166 218 $C_{3}$ at 184 217 $C_{4}$ at 220 261 $C_{5}$ at 218 273 $C_{6}$ at
215 289 $P_{2}$ at 176 304 $C_{7}$ at 159 293 $P_{1}$ at 136 293 $B_{3}$ at
156 314 $B_{2}$ at 134 313 $Q_{1}$ at 113 312 $X$ at 169 399 $Y$ at 131 409
$B_{1}$ at 85 392 $A_{1}$ at 76 449 $A_{2}$ at 148 497

Figure 26: Adapted Heegaard diagram for $L_{1,1}$.

Analyzing this diagram as in the other cases, we see that there are $149$
elements in $\mathbb{T}_{\alpha}\cap\mathbb{T}_{\beta}$, seven of which with
the same Alexander grading as the point ${\bf{x}}_{0}$ determining
$\widehat{\mathfrak{L}}(L(1,1))$. Indeed, in the notation suggested by that
diagram, the seven intersection points in Alexander grading $A=1$ are

$\begin{array}[]{llll}A_{2}B_{2}C_{7}D_{1}E,&A_{2}D_{1}EP_{2}Q_{1},&A_{2}D_{1}EP_{1}Q_{1},&A_{2}B_{1}C_{6}D_{1}E,\\\
XYC_{6}D_{1}E,&A_{1}B_{2}C_{5}D_{1}E,&A_{1}B_{1}C_{1}D_{1}E={\bf{x}}_{0}.&\end{array}$

Since ${\bf{x}}_{0}$ is a cycle, it can be thought of as determining a
subcomplex of $\widehat{\mathrm{CFK}}(L_{1,1},1)$. The quotient complex then
is generated by the six remaining generators in the given Alexander gradings.
Indeed, it is straightforward to find the positive homotopy classes connecting
these six generators with Maslov index one. There are six, and five of these
are rectangles, and hence admit holomorphic representatives. The sixth
(connecting $XYC_{6}D_{1}E$ to $A_{1}B_{2}C_{5}D_{1}E$) is also a planar
surface, and it is not difficult to verify directly that it, too, always has a
holomorphic representative. We assemble this information of the chain complex
in Figure 27.

$A_{2}D_{1}EP_{2}Q_{1}$$A_{2}B_{1}C_{6}D_{1}E$$A_{2}B_{2}C_{7}D_{1}E$$A_{2}D_{1}EP_{1}Q_{2}$$A_{1}B_{2}C_{5}D_{1}E$$XYC_{6}D_{1}E$
Figure 27: Part of the chain complex for $L_{1,1}$. This is the quotient of
$\widehat{\mathrm{CFK}}(L_{1,1},1)$ by the generator $A_{1}B_{1}C_{1}D_{1}E$.
Arrows indicate differentials.

It is straightforward to verify that this complex is acyclic. It follows at
once that ${\bf{x}}_{0}$ represents the nontrivial homology class in
$\widehat{\mathrm{HFK}}(L_{1,1},1)$. ∎

In a similar vein, we have the following:

######  Theorem 6.10.

Let $L_{1,2}$ be given the orientation specified by the diagram of Figure 28.
Then, $\widehat{\mathfrak{L}}(L_{1,2})\neq 0$.

Recall that $L_{1,2}$ corresponds to $T_{(2,-9)}$.

###### Proof.

The adapted Heegaard diagram is pictured in Figure 28.

2pt $w$ at 141 18 $z$ at 31 258 $L_{1,2}$ at 77 41 $F$ at 84 68 $T$ at 160 98
$S$ at 139 117 $E_{1}$ at 86 165 $R_{1}$ at 142 199 $E_{2}$ at 176 199 $R_{2}$
at 194 198 $D_{5}$ at 148 218 $N_{1}$ at 99 210 $N_{2}$ at 169 217 $D_{4}$ at
187 217 $D_{3}$ at 223 260 $D_{2}$ at 213 294 $V$ at 170 294 $M$ at 155 309
$C_{4}$ at 179 310 $D_{1}$ at 85 257 $C_{1}$ at 123 308 $P_{1}$ at 127 396
$C_{2}$ at 162 388 $P_{2}$ at 180 401 $C_{3}$ at 216 382 $B_{1}$ at 104 491
$Q_{1}$ at 107 396 $B_{2}$ at 138 409 $B_{3}$ at 160 409 $X$ at 176 495 $Y$ at
131 494 $A_{1}$ at 79 544 $A_{2}$ at 153 592 $1$ at 251 452 $2$ at 220 529 $3$
at 146 397 $4$ at 94 381 $5$ at 148 413 $6$ at 172 413 $7$ at 113 410 $8$ at
118 502 $9$ at 100 291 $10$ at 191 295

Figure 28: Adapted Heegaard diagram for $L_{1,2}$. The domains (in the
complement of the $\alpha$– and $\beta$–circles) labeled by $1,...,10$ play a
role in Figure 29.

There are $347$ intersection points, $13$ of which are in the same Alexander
grading as ${\bf{x}}_{0}$. In terms of the numbering conventions from Figure
28, these are

$\begin{array}[]{llll}A_{2}B_{1}C_{3}D_{1}E_{1}F&A_{2}B_{2}C_{2}D_{1}E_{1}F&A_{1}B_{2}C_{4}D_{1}E_{1}F&A_{1}B_{2}C_{1}D_{2}E_{1}F\\\
A_{1}B_{1}C_{2}D_{2}E_{1}F&A_{2}B_{2}C_{3}D_{2}E_{1}F&A_{1}B_{2}MVE_{1}F&A_{1}D_{2}E_{1}P_{1}Q_{1}F\\\
A_{2}D_{1}E_{1}P_{2}E_{1}F&A_{2}D_{1}E_{1}P_{1}Q_{2}F&A_{2}D_{2}E_{1}P_{2}Q_{2}F&XYC_{3}D_{1}E_{1}F\\\
A_{1}B_{1}C_{1}D_{1}E_{1}F={\bf{x}}_{0}.&&&\end{array}$

We can draw a graph whose vertices consist of these generators, and its arrows
denote positive Whitney disks with Maslov index one, as pictured in Figure 29.

$B_{2}C_{3}$$P_{2}Q_{2}$$B_{1}C_{2}$$P_{1}Q_{1}$$B_{2}C_{1}$$A_{1}B_{2}MV$$A_{2}B_{2}C_{4}D_{1}E_{1}$$XYC_{3}$$A_{2}B_{1}C_{3}$$P_{2}Q_{1}$$B_{2}C_{2}$$P_{1}Q_{2}$$B_{1}C_{1}$$6$$1+2$$7$$4$$10$$9$$6+8$$2$$6+3$$5$$3$$1+2+3+4$$5+6$$1+2+4+5$
Figure 29: Complex for $L_{1,2}$. We have displayed here the $13$ generators
in $A=-3$. Arrows represent domains connecting generators. All the thicker
arrows actually represent quadrilaterals; there are only four which are not:
those containing $1$, and those containing $6+8$.

Those domains are written as sums of the $10$ domains labelled in Figure 28.
Note that in that figure, we have abbreviated the names of generators (for
example, dropping $F$, which occurs in each). In that picture, all but four of
the arrows represent rectangles (and hence, immediately, nontrivial
differentials in the chain complex). Inspection of this graph immediately
establishes that there are no other non–negative Whitney disks among the given
generators with Maslov index one (for example, the domain from
$P_{2}Q_{1}=A_{2}D_{1}E_{1}P_{2}Q_{1}F$ to
$B_{2}C_{4}=A_{2}B_{2}C_{4}D_{1}E_{1}$ is gotten as $6+3-2+8+6$, which
contains $2$ with multiplicity $-1$). To show that
$B_{1}C_{1}=A_{1}B_{1}C_{1}D_{1}F$ represents a homologically nontrivial
cycle, it suffices to show that the positive domain $6+8$ supports holomorphic
representatives, which can be done by direct means. (Note that it is an
annulus with an appropriate cut.) ∎

######  Remark 6.11.

Recall that

$\mathrm{HFK}^{-}(S^{3},T_{(2,2n-1)})\cong\mathbb{F}^{n-1}\oplus\mathbb{F}[U],$

where the top generator of the free $\mathbb{F}[U]$–module is at

$(A=-(n-1),\quad M=-2(n-1)),$

while the $n-1$ generators of the $\mathbb{F}^{n-1}$ summand are of bi–degrees

$(A=n-1-2i,\quad M=-2i),\quad i=0,\ldots,n-2.$

(This follows readily from the calculation of $\widehat{\mathrm{HFK}}$, stated
in Equation (6.7).) Once again, the computation above can be adapted to show
that the homogeneous $U$–torsion elements
${\mathfrak{L}}(L_{0,l}),{\mathfrak{L}}(L_{1,1})$ and
${\mathfrak{L}}(L_{1,2})$ in $\mathrm{HFK}^{-}(S^{3},T_{(2,2n-1)})$ are also
nontrivial.

Notice that there is an obvious bijection between the homogeneous $U$–torsion
elements of $\mathrm{HFK}^{-}(S^{3},T_{(2,2n-1)})$ and the Legendrian knots
$L_{k,l}$ with $k+l=n-2$ constructed above. Moreover, this bijection can be
chosen in such a way that if $L_{k,l}$ corresponds to
$x_{k,l}\in\mathrm{HFK}^{-}(S^{3},T_{(2,2n+1)})$ then

$d_{3}(\xi_{k,l})=2A(x_{k,l})-M(x_{k,l}).$

It is reasonable to expect that the Legendrian invariants
$\widehat{\mathfrak{L}}(L_{k,l})$ and ${\mathfrak{L}}(L_{k,l})$ are nonzero
for all $k,l\geq 0$, and hence that ${\mathfrak{L}}(L_{k,l})$ is determined by
$x_{k,l}$. We will not address this general computation in the present paper.

##  7 Connected sums

Suppose that $L_{i}\subset(Y_{i},\xi_{i})$ are oriented Legendrian knots in
the contact 3–manifolds $(Y_{i},\xi_{i})$, $i=1,2$. We want to relate the
invariants of $L_{1}$ and $L_{2}$ with the invariants of the connected sum
$L_{1}\\#L_{2}\subset(Y_{1}\\#Y_{2},\xi_{1}\\#\xi_{2})$ (see [14] for the
definition of the connected sum operation in the contact setting).

Fix open book decompositions $(B_{i},\varphi_{i})$ adapted to
$L_{i}\subset(Y_{i},\xi_{i})$ (with pages $S_{i}$) for $i=1,2$. We may assume
that suitable portions of the open books and the adapted bases appear as in
the left–hand picture of Figure 30, where, as usual, the ordered pairs
$(z_{i},w_{i})$ determine the orientations of the knots. We can now perform
the Legendrian connected sum of $(Y_{1},\xi_{1},L_{1})$ and
$(Y_{2},\xi_{2},L_{2})$ in such a way that we glue the open books as well as
the contact structures and the knots. Specifically, we take the Murasugi sum
of the two open books in the domains containing $z_{2}$ and $w_{1}$, and then
we drop these two basepoints. We can make sure that the resulting oriented
knot $L_{1}\\#L_{2}$ is smoothly determined on a page of the resulting open
book by the ordered pair $(z_{1},w_{2})$, as illustrated in the central
picture of Figure 30. The bases of arcs can also be arranged to be the same as
the ones illustrated.

There is a corresponding map of chain complexes

$\Phi\colon\mathrm{CFK}^{-}(-Y_{1},L_{1})\otimes_{\mathbb{F}[U]}\mathrm{CFK}^{-}(-Y_{2},L_{2})\to\mathrm{CFK}^{-}(-(Y_{1}\\#Y_{2}),L_{1}\\#L_{2})$
(7.1)

which (denoting the intersection point determining ${\mathfrak{L}}(L_{i})$ by
${\bf{x}}(L_{i})$) maps the intersection point
${\bf{x}}(L_{1})\otimes{\bf{x}}(L_{2})$ to the generator