# Characterizing Regular Lagrangians by Lefschetz Fibration

Sauvik Mukherjee  Presidency University, Kolkata, India.
<EMAIL_ADDRESS>

###### Abstract.

In [2] Eliashberg, Ganatra and Lazarev have introduced the notion of Regular
Lagrangians in Weinstein cobordisms and also predicted that Regular
Lagrangians can be characterized by the existence of Weinstein Lefschetz
Fibrations. In this paper we present a proof of this fact.

###### Key words and phrases:

Regular Lagrangians, Weinstein cobordism, Lefschetz Fibration

##  1\. introduction

In [2] Eliashberg, Ganatra and Lazarev have introduced the notion of Regular
Lagrangians in Weinstein cobordisms and also predicted that Regular
Lagrangians can be characterized by the existence of Weinstein Lefschetz
Fibrations. In this paper we present a proof of this fact.

Let us first define the notion of Regular Lagrangians. But in order to do so
we need to understand the notion of Liouville cobordisms.

A Liouville cobordism between contact manifolds $(\partial_{\pm}W,\xi_{\pm})$
is an even dimensional cobordism (say of dimension $2n$)
$(W,\partial_{-}W,\partial_{+}W)$, where $W$ is equipped with a symplectic
form $\omega$ and an expanding Liouville vector field $X$ which is outward
pointing along $\partial_{+}W$ and inward pointing along $\partial_{-}W$,
moreover the contact structure induced by the Liouville form
$\lambda=\iota(X)\omega$ on $\partial_{\pm}W$ coincides with the given ones.

A Liouville cobordism $(W,\partial_{-}W,\partial_{+}W)$ is called a Weinstein
cobordism if there exists a defining Morse function $\phi:W\to\mathbb{R}$ for
which $X$ is gradient-like and $(\omega,X,\phi)$ is called a Weinstein
cobordism structure on $W$. Equivalently if the Liouville form $\lambda$ is
given then $\omega$ and $X$ can be recovered, so we may say $(\lambda,\phi)$
to be a Weinstein structure. A Weinstein cobordism
$(W,\partial_{-}W,\partial_{+}W)$ with $\partial_{-}W=\Phi(empty)$ is called a
Weinstein domain.

In what follows we shall consider exact Lagrangian subcobordism with
Legendrian boundary
$(L,\partial_{-}L,\partial_{+}L)\subset(W,\partial_{-}W,\partial_{+}W)$ in
Weinstein cobordism with contact boundary.

Now consider
$(L,\partial_{-}L,\partial_{+}L)\subset(W,\partial_{-}W,\partial_{+}W)$ an
exact Lagrangian cobordism in a Liouville cobordism and let $N(L)$ be a small
tubular neighborhood of $L\subset W$. Define $\partial^{int}N(L)=\partial
N(L)\cap Int(W)$, $\partial^{ext}_{\pm}N(L)=\partial N(L)\cap\partial_{\pm}W$
and $W_{L}=\overline{W-N(L)}$. As $L$ is exact we can make
$X\pitchfork\partial N(L)$ and hence $(W_{L},\omega_{\mid W_{L}})$ has a
natural Liouville cobordism structure with sutured boundary
$\partial_{+}W_{L}\cup\partial_{-}W_{L}$, where

$\partial_{+}W_{L}=\partial W-\partial^{ext}N(L),\
\partial_{-}W_{L}=\partial^{int}N(L)$

######  Definition 1.1.

([2]) Let $L\subset(W,\omega^{\prime},X^{\prime},\phi^{\prime})$ be an exact
Lagrangian cobordism in a Weinstein cobordism. $L$ is called regular if the
Weinstein structure is homotopic to a Weinstein cobordism structure
$(W,\omega,X,\phi)$ through a homotopy of Weinstein structures for which $L$
remains Lagrangian and $X$ is tangent to $L$. This is equivalent to the
condition that the Liouville form $\lambda:=\iota(X)\omega$ has the property
that $\lambda_{\mid L}=0$.

Such a Weinstein structure is called tangent Weinstein structure to the
regular Lagrangian $L$. It follows that the critical points of $\phi_{\mid L}$
are global critical points for $W$.

It turns out that for a regular Lagrangian $L$ and any Weinstein structure
tangent to $L$ can further be adjusted so that the following happens.

The Weinstein cobordism structure $(W,\omega,X,\phi)$ tangent to $L$ is called
adjusted to $L$ if there exists a regular value $c\in\mathbb{R}$ of $\phi$
such that

  1. (1)

all critical points of $\phi$ in the sublevel set $\\{\phi\leq c\\}$ lies on
$L$ and the indeces of these critical points for $\phi$ and $\phi_{\mid L}$
coincides;

  2. (2)

there are no critical points of $\phi$ on $L\cap\\{\phi\geq c\\}$.

So the handlebody presentation has the following description. First the
handles corresponding to the critical points of $\phi_{\mid L}$ are attached
and then the remaining ones are attached so that the attaching spheres of
these remaining handle attachments do not intersect $\partial_{+}L$.

It turns out that if $\partial_{-}L$ is empty then the Weinstein structure on
$W(L)=\\{\phi\leq c\\}$ is equivalent(radially) to the disjoint union of the
trivial cobordism over $\partial_{-}W$, i.e, $\partial_{-}W\times[0,1]$ and
the canonical Weinstein structure on $T^{*}L$.

If $\partial_{-}L$ is non-empty then the handlebody presentation of $L$ given
by $\phi_{\mid L}$ and the gradient-like vector field $X_{\mid L}\in TL$
defines a weinstein cobordism structure on the cotangent disk bundle $N$ of
$L$ with $\partial_{-}N$ equal to a tubular neighborhood of $\partial_{-}L$ in
$J^{1}(\partial_{-}L)$ and $\partial_{+}N=\partial N-Int\partial_{-}N$. Then
$W(L)$ is obtained by attaching a generalized handle, i.e, by attaching the
Weinstein cobordism $N$ to the trivial cobordism $\partial_{-}W\times[0,1]$
over $\partial_{-}W$ by identifying $\partial_{-}L$ with a neighborhood of a
Legendrian sphere in $\partial_{-}W\times\\{1\\}$.

So in any case the underlying symplectic structure on $W(L)$ near $L$ is given
by the symplectic structure on $T^{*}L$ but the Weinstein structure differs
depending on whether $\partial_{-}L$ is empty or not.

Next we need to define the notion of Weinstein Lefschetz fibration.

######  Definition 1.2.

([4]) A Weinstein Lefschetz fibration is a tuple

$W^{2n}=((W_{0},\lambda_{0},\phi_{0});L_{1},...,L_{m})$

where $W_{0}^{2n-2}$ is a Weinstein domain and $L_{j}$’s are exact
parametrized Lagrangian spheres in $W_{0}^{2n-2}$, where parametrized means
diffeomorphisms $S^{n-1}\to L$ upto precomposition with elements of $O(n)$.
Its total space $|W|$ is given as follows. Consider the Weinstein manifold

$(W_{0}\times\mathbb{C},\lambda_{0}-J^{*}d(\frac{1}{2}|z|^{2}),\phi_{0}+|z|^{2})$

The corresponding Liouville vector field is given by
$X_{\lambda_{0}}+\frac{1}{2}(x\partial_{x}+y\partial_{y})$. Take Legendrian
lifts $\Lambda_{j}\subset(W_{0}\times S^{1},\lambda_{0}+Nd\theta)$ of
$L_{j}\subset W_{0}$ such that $\Lambda_{j}$ projects to small interval around
$\frac{2\pi j}{m}\in S^{1}$. We take $N$ so large that these intervals remains
disjoint. The embedding $S^{1}\to\mathbb{C}$ as circle of radius $N^{1/2}$
pulls back the Liouville form $-J^{*}d(\frac{1}{2}|z|^{2})$ to the contact
form $Nd\theta$. So $\Lambda_{j}$ could be thought as a Legendrian at the
level set $\\{|z|=N^{1/2}\\}\subset W_{0}\times\mathbb{C}$. Then the
restriction of downward Liouville flow defines a map
$\Lambda_{j}\times\mathbb{R}_{\geq 0}\to W_{0}\times\mathbb{C}$ and which
intersects the set $\\{\phi_{0}+|z|^{2}=0\\}$ along a Legendrian
$\Lambda^{\prime}_{j}$. Again $N$ is taken so large that the projection of
$\\{\phi_{0}+|z|^{2}\leq 0\\}$ to $\mathbb{C}$ is contained inside the disc of
radius $N^{1/2}$. Now the total space $|W|$ is defined as the result of
attaching Weinstein handles ([5]) to the Weinstein domain
$\\{\phi_{0}+|z|^{2}=0\\}$ along the Legendrians $\Lambda^{\prime}_{j}$.

Now we end this section with the main theorem of this paper.

######  Theorem 1.3.

Let $L\subset W^{2n}$ be a regular Lagrangian, then there exists a Weinstein
Lefschetz fibration over $\mathbb{C}$ which projects $L$ to a ray in
$\mathbb{R}\subset\mathbb{C}$.

######  Remark 1.4.

Observe that we have only stated the ’only if’ part, the ’if’ part is obvious.

The method of the proof is the following. In section 4 we have constructed a
sequence of complex line bundles $E^{k}$ over $W^{2n}$ along with sections
$s_{k}$ such that $s_{k}$ has a non-degenerate singularity on $L\subset W$ and
the zero locus of $s_{k}$ for large $k$ is Symplectic. The idea of the
construction is similar to [1] but it is different as we needed some
improvisations.

Then we showed that the stable manifold, denoted $L^{\prime}$, of the
Hamiltonian vector field $X_{Im\log s}$ is $C^{0}$-close to $L$ and is
diffeomorphic to $L$, where $s=s_{k}$ for large $k$. Then using results from
section 3 and Weinstein neighborhood theorem we construct a symplectomorphism
which takes $L$ to $L^{\prime}$ and which is identity outside a tubular
neighborhood of $L$. Then we pull back the Weinstein structure by this
symplectomorphism. Obviously under this composition $L$ is mapped to a ray in
$\mathbb{C}$ because the argument is constant along $L^{\prime}$.

We must take $c$ so small that this tubular neighborhood of $L$ is first
considered and then the rest of $W$, where $c\in\mathbb{R}$ is the regular
value as in the definition of Regular Lagrangian.

##  2\. Preliminaries

First we need some notations. Let $z=(z_{1},...,z_{n})\in\mathbb{C}^{n}$ be a
point in $\mathbb{C}^{n}$. Define

  1. (1)

$\frac{\partial}{\partial z_{j}}\equiv\frac{1}{2}(\frac{\partial}{\partial
x_{j}}-i\frac{\partial}{\partial y_{j}})$, $j=1,...,n.$

  2. (2)

$\frac{\partial}{\partial\bar{z}_{j}}\equiv\frac{1}{2}(\frac{\partial}{\partial
x_{j}}+i\frac{\partial}{\partial y_{j}})$, $j=1,...,n.$

  3. (3)

$dz_{j}\equiv dx_{j}+idy_{j},\ j=1,...,n$

  4. (4)

$d\bar{z}_{j}\equiv dx_{j}-idy_{j},\ j=1,...,n$

For $0\leq p,q\leq n$, the differential form

$\omega=\Sigma_{|\alpha|=p,|\beta|=q}\omega_{\alpha\beta}dz^{\alpha}\wedge
d\bar{z}^{\beta}$

is said to be of type or bidegree $(p,q)$, where $\alpha,\beta$ are multi-
indices. Any differential form is the sum of the terms of the form
$\omega_{\alpha\beta}dz^{\alpha}\wedge d\bar{z}^{\beta}$. For
$\omega=\Sigma_{|\alpha|=p,|\beta|=q}\omega_{\alpha\beta}dz^{\alpha}\wedge
d\bar{z}^{\beta}$ define

  1. (1)

$\partial\omega=\Sigma_{j=1}^{n}\Sigma_{\alpha,\beta}\frac{\partial\omega_{\alpha,\beta}}{\partial
z_{j}}dz_{j}\wedge dz^{\alpha}\wedge d\bar{z}^{\beta}$

  2. (2)

$\bar{\partial}\omega=\Sigma_{j=1}^{n}\Sigma_{\alpha,\beta}\frac{\partial\omega_{\alpha,\beta}}{\partial\bar{z}_{j}}d\bar{z}_{j}\wedge
dz^{\alpha}\wedge d\bar{z}^{\beta}$

Now we shall state a theorem from [3] which we need for the proof of 1.3.

######  Theorem 2.1.

(Weinstein Neighborhood Theorem, [3]) Any isotropic and, in particular,
Lagrangian immersion $L\to W$ extends to an isosymplectic immersion $U\to W$,
where $U$ is a tubular neighborhood of the zero-section in the cotangent
bundle $T^{*}L$ endowed with the canonical symplectic structure.

##  3\. $h$-Principle

This section does not have any new result, we just recall some facts from the
theory of $h$-principle which we shall need in our proof.

Let $X\to M$ be any fiber bundle and let $X^{(r)}$ be the space of $r$-jets of
jerms of sections of $X\to M$ and $j^{r}f:M\to X^{(r)}$ be the $r$-jet
extension map of the section $f:M\to X$. A section $F:M\to X^{(r)}$ is called
holonomic if there exists a section $f:M\to X$ such that $F=j^{r}f$. In the
following we use the notation $Op(A)$ to denote a small open neighborhood of
$A\subset M$ which is unspecified.

Let $\mathcal{R}$ be a subset of $X^{(r)}$. Then $\mathcal{R}$ is called a
differential relation of order $r$. $\mathcal{R}$ is said to satisfy
$h$-principle if any section $F:M\to\mathcal{R}\subset X^{(r)}$ can be
homotopped to a holonomic section $\tilde{F}:M\to\mathcal{R}\subset X^{(r)}$
through sections whose images are contained in $\mathcal{R}$. Put differently,
if the space of sections of $X^{(r)}$ landing into $\mathcal{R}$ is denoted by
$Sec\mathcal{R}$ and the space of holonomic sections of $X^{(r)}$ landing into
$\mathcal{R}$ is denoted by $Hol\mathcal{R}$ then $\mathcal{R}$ satisfies
$h$-principle if the inclusion map $Hol\mathcal{R}\hookrightarrow
Sec\mathcal{R}$ induces a epimorphism at $0$-th homotopy group $\pi_{0}$.
$\mathcal{R}$ satisfies parametric $h$-principle if
$\pi_{k}(Sec\mathcal{R},Hol\mathcal{R})=0$ for all $k\geq 0$.

Let $(V,d\alpha_{V})$ and $(W,d\alpha_{W})$ be two exact symplectic manifolds
with $dim(V)\leq dim(W)$. $\mathcal{R}_{isosymp}$ be the relation of
isosymplectic immersions from $V$ to $W$. Let $A\subset V$ be a polyhedron of
positive codimension.

######  Theorem 3.1.

([3]) All forms of local $h$-principle holds for the inclusion

$Hol_{OpA}\mathcal{R}_{isosymp}\to Sec_{OpA}\mathcal{R}_{isosymp}$

on $OpA$.

###### Proof.

Direct consequence of 16.4.2 of [3]. ∎

##  4\. Donaldson’s Construction with a Preassigned Singularity

In this section we follow the construction in [1] with some changes in order
to construct a sequence of sections $s_{k}:W^{2n}\to E^{k}$ which will have a
non-degenerate singularity on $L\subset W$, where $L,W$ are as in 1.3 and $E$
is a hermitian line bundle on $W$ and $E^{k}=E^{\otimes k}$.

Now we shall recall some parts of the local theory as in [1].

A complex structure on a real vector space $V$ is a decomposition of
$V^{*}\otimes\mathbb{C}$ as a complex conjugate subspaces, the complex linear
and anti-linear functionals. If a reference complex structure is fixed then
$V$ becomes $\mathbb{C}^{n}$ and in this case the conjugate subspaces turns
out to be just one forms of type $(1,0)$ and $(0,1)$, i.e,

$Hom_{\mathbb{R}}(\mathbb{C}^{n},\mathbb{C})=\Lambda^{1,0}\oplus\Lambda^{0,1}$

If $J$ is another complex structure then $\Lambda^{1,0}_{J}$ is the graph of a
complex linear map

$\mu:\Lambda^{1,0}\to\Lambda^{0,1}$

and in this case $\Lambda^{0,1}_{J}$ is the graph of
$\bar{\mu}:\Lambda^{0,1}\to\Lambda^{1,0}$. The condition that
$\Lambda^{1,0}_{J}\pitchfork\Lambda^{0,1}_{J}$ is that $(1-\mu\bar{\mu})$ is
invertible. For an almost-complex structure $J$ the corresponding $\mu$ will
be a bundle map i.e, it depends on the base point smoothly and we shall denote
it by $\mu_{z}$ where $z$ is the base point. In this case

$\bar{\partial}_{J}(f)=\bar{\partial}f-\mu(\partial f),$

where $\partial,\ \bar{\partial}$ are the ordinary operators defined by the
standard complex structure on $\mathbb{C}^{n}$.

For $\rho<1$ define $\delta_{\rho}:\mathbb{C}^{n}\to\mathbb{C}^{n}$ by
$\delta_{\rho}(z)=\rho z$. Consider the almost-complex structure
$\tilde{J}=\delta_{\rho}^{*}J$ defined on $\rho^{-1}\Omega$ where $\Omega$ is
the open neighborhood of zero in $\mathbb{C}^{n}$ where $J$ was defined. The
$\tilde{\mu}$ corresponding to $\tilde{J}$ is given by

$\tilde{\mu}_{z}=\mu_{\rho z}$

In this case we get the following estimate

$|\tilde{\mu}_{z}|\leq C\rho|z|,\ |\triangledown\tilde{\mu}|\leq C\rho,$

where $C$ is a constant related to the Nijenhius tensor and its derivative.
Throughout the paper we use the convention that $C$ represents a positive
constant which changes from line to line.

Now consider the Kähler form

$\omega_{0}=\frac{i}{2}\Sigma_{j=1}^{n}dz_{j}d\bar{z}_{j}$

on $\mathbb{C}^{n}$. Set
$A=\frac{1}{4}(\Sigma_{j=1}^{n}z_{j}d\bar{z}_{j}-\bar{z}_{j}dz_{j})$ then
$\omega_{0}=idA$.

If $\xi\to\mathbb{C}^{n}$ be a line bundle with connection having curvature
$\omega_{0}$ there is a trivialization of $\xi$ with respect to which the
connection matrix turns out to be $A$. $A$ defines a coupled
$\bar{\partial}$-operator $\bar{\partial}_{A}$,

$\bar{\partial}_{A}(f)=\bar{\partial}f+A^{0,1}f,$

where $A^{0,1}$ is the $(0,1)$ component of $A$. Similarly we define
$\partial_{A}$. Now observe that

  1. (1)

$\bar{\partial}_{A}(e^{-(Re\Sigma z_{j}^{2})/4}+i)=\frac{e^{-(Re\Sigma
z_{j}^{2})/4}}{4}\Sigma_{j}(z_{j}-\bar{z}_{j})d\bar{z}_{j}+\frac{i}{4}\Sigma_{j}z_{j}d\bar{z}_{j}$

  2. (2)

$\partial_{A}(e^{-(Re\Sigma z_{j}^{2})/4}+i)=-\frac{e^{-(Re\Sigma
z_{j}^{2})/4}}{4}\Sigma_{j}(z_{j}+\bar{z}_{j})dz_{j}-\frac{i}{4}\Sigma_{j}\bar{z}_{j}dz_{j}$

We replace $\omega_{0}$ by $k\omega_{0}$, for a positive integer $k$. The
replacement of $\xi$ by $\xi^{k}$ is same as the dilation with scalar factor
$k^{-1/2}$ on $\mathbb{C}^{n}$.

Now let $(W,\omega)$ be a Liouville manifold with a compatible complex
structure $J$ and a complex line bundle $E\to W$ with a $U(1)$ connection
having curvature $-i\omega$. Let $g$ be the Riemannian metric determined by
$J$ and $\omega$. Then $g_{k}=kg$ is the one determined by $J$ and $k\omega$.

We consider the cubical coordinate chart which is longer in the $y$-coordinate
directions, i.e, for $R>>r$ set

$I^{2n}_{r,R}=\\{(x,y)\in\mathbb{R}^{n}\times\mathbb{R}^{n}:|x_{j}|<r,|y_{j}|<R\\}$

In fact we take $R=k^{-1/2}$ and $r=k^{-1}$. Consider the Darboux chart
$\chi_{p,k}:I^{2n}_{k^{-1}+\epsilon_{k},k^{-1/2}+\epsilon_{k}}\to W$ around
$p\in W$, where $\epsilon^{\prime}_{k}=\epsilon k^{-1}\ and\
\epsilon_{k}=\epsilon k^{-1/2}$. We may further assume that all derivatives of
$\chi_{p}$ are bounded and derivative of $\chi_{p}$ at zero is complex linear
with respect to $J$ on $T_{p}W$. Therefore $\chi^{*}J$ represents a structure
determined by a bundle map $\mu$ over
$I^{2n}_{k^{-1}+\epsilon^{\prime}_{k},k^{-1/2}+\epsilon^{\prime}_{k}}$ as
explained above. The derivatives of $\mu$ satisfies bounds independent of
$p\in W$. So now given $k$ we consider the new chart

$\tilde{\chi}_{p,k}=\chi_{p,k}\circ\delta_{k^{-1/2}}:k^{1/2}I^{2n}_{k^{-1}+\epsilon^{\prime}_{k},k^{-1/2}+\epsilon^{\prime}_{k}}=I^{2n}_{k^{-1/2}+\epsilon_{k},1+\epsilon_{k}}\to
W$

Let $\tilde{\mu}$ be the bundle map representing the almost-complex structure
on this new chart. So $\tilde{\mu}$ satisfies the following bounds

$\begin{array}[]{lcl}|\tilde{\mu}_{z}|&\leq&Ck^{-1/2}|z|,\\\
|\triangledown\tilde{\mu}_{z}|&\leq&Ck^{-1/2}.\\\ \end{array}$

$C$ does not depend on $p$. Observe that
$\tilde{\chi}^{*}(-ik\omega)=-i\omega_{0}$ where $-ik\omega$ is the curvature
on $E^{k}$. So $\tilde{\chi}$ admits a connection preserving bundle map which
we shall also denote by $\tilde{\chi}$. So $\tilde{\chi}:\xi\to E^{k}$ is the
bundle map. So one can think of $\sigma=e^{-(Re\Sigma z_{j}^{2})/4}+i$ as
local sections of $E^{k}$ around $p\in W$.

Let $\bar{\partial}_{A,\tilde{J}}$ denote the $\bar{\partial}$-operator
defined by $\tilde{J}$ on
$k^{1/2}I^{2n}_{k^{-1/2}+\epsilon_{k},1+\epsilon_{k}}$ with connection matrix
$A$. Then

$\bar{\partial}_{A,\tilde{J}}(f)=(\bar{\partial}f+A^{0,1}f)+\tilde{\mu}(\partial
f+A^{1,0}f).$

If we replace $f$ by $e^{-(Re\Sigma z_{j}^{2})/4}+i$ we get the following
estimate

$|\bar{\partial}_{A,\tilde{J}}(e^{-(Re\Sigma
z_{j}^{2})/4}+i)|\leq\frac{|z|}{4}(2e^{-(Re\Sigma
z_{j}^{2})/4}+1)+Ck^{-1/2}|z|^{2}(2e^{-(Re\Sigma z_{j}^{2})/4}+1)$

Furthermore we want the estimate on the derivative. So observe

$\triangledown=\bar{\partial}_{A,\tilde{J}}+\partial_{A,\tilde{J}}=\bar{\partial}_{A}+\partial_{A}+\tilde{\mu}\partial_{A}+\bar{\tilde{\mu}}\bar{\partial}_{A}$

The only term in $\bar{\partial}_{A,\tilde{J}}$ for which $\tilde{\mu}$ is not
a factor is

$\bar{\partial}_{A}(e^{-(Re\Sigma z^{2}_{j})/4}+i)=\frac{e^{-(Re\Sigma
z^{2}_{j})/4}}{4}\Sigma_{j}(z_{j}-\bar{z}_{j})d\bar{z}_{j}+\frac{i}{4}\Sigma_{j}z_{j}d\bar{z}_{j}$

So let us compute
$(\bar{\partial}_{A}+\partial_{A})(\bar{\partial}_{A}(e^{-(Re\Sigma
z^{2}_{j})/4}+i))$. So

$\begin{array}[]{rcl}(\bar{\partial}_{A}+\partial_{A})(\bar{\partial}_{A}f)&=&(\bar{\partial}_{A}+\partial_{A})(\bar{\partial}f+A^{0,1}f)\\\
&=&\bar{\partial}_{A}(\partial f+A^{0,1}f)+\partial_{A}(\partial
f+A^{0,1}f)\\\
&=&(\bar{\partial}^{2}f+\bar{\partial}A^{0,1}f+A^{0,1}\bar{\partial}f+A^{0,1}A^{0,1}f)\\\
&&+(\partial\bar{\partial}f+\partial
A^{0,1}f+A^{1,0}\bar{\partial}f+A^{1,0}A^{0,1}f)\end{array}$

The first term

$(1)\ \bar{\partial}^{2}f=0$

The second term

$\begin{array}[]{rcl}(2)\
\bar{\partial}A^{0,1}f&=&\bar{\partial}(\frac{1}{4}\Sigma_{j}(e^{-(Re\Sigma
z^{2}_{j})/4}+i)z_{j}d\bar{z}_{j})\\\
\Rightarrow(1/2)(\frac{\partial}{\partial_{x_{l}}}+i\frac{\partial}{\partial_{y_{l}}})A^{0,1}f&=&\frac{1}{8}[e^{-(Re\Sigma
z^{2}_{j})/4}+\Sigma_{j}z_{j}\frac{e^{-(Re\Sigma z^{2}_{j})/4}}{4}(-2x_{l})\\\
&&+\frac{i}{4}+i[ie^{-(Re\Sigma
z^{2}_{j})/4}+\Sigma_{j}z_{j}\frac{e^{-(Re\Sigma
z^{2}_{j})/4}}{4}(2y_{l})+(i^{2}/4)]]d\bar{z}_{l}\wedge d\bar{z}_{j}\\\
\Rightarrow\bar{\partial}A^{0,1}f&=&-\frac{e^{-(Re\Sigma
z^{2}_{j})/4}}{4^{2}}\Sigma_{j,l}\bar{z}_{l}z_{j}d\bar{z}_{l}\wedge
d\bar{z}_{j}\end{array}$

The third term

$\begin{array}[]{rcl}(3)A^{0,1}\bar{\partial}f&=&A^{0,1}[-\Sigma_{j}z_{j}\frac{e^{-(Re\Sigma
z^{2}_{j})/4}}{4}d\bar{z}_{j}]\\\ &=&-\frac{e^{-(Re\Sigma
z^{2}_{j})/4}}{4^{2}}\Sigma_{j,l}z_{l}\bar{z}_{j}d\bar{z}_{l}\wedge
d\bar{z}_{j}\end{array}$

The forth term

$\begin{array}[]{rcl}(4)A^{0,1}A^{0,1}f&=&\frac{e^{-(Re\Sigma
z^{2}_{j})/4}}{4^{2}}\Sigma_{j,l}z_{l}z_{j}d\bar{z}_{l}\wedge d\bar{z}_{j}\\\
&&+\frac{i}{4^{2}}\Sigma_{j,l}z_{l}z_{j}d\bar{z}_{l}\wedge
d\bar{z}_{j}\end{array}$

The fifth term

$\begin{array}[]{rcl}(5)\partial\bar{\partial}f&=&\partial[-\Sigma_{j}z_{j}\frac{e^{-(Re\Sigma
z^{2}_{j})/4}}{4}d\bar{z}_{j}]\\\
\frac{1}{2}(\frac{\partial}{\partial_{x_{l}}}-i\frac{\partial}{\partial_{y_{l}}})(\bar{\partial}f)&=&(1/2)[-\frac{e^{-(Re\Sigma
z^{2}_{j})/4}}{4}-\Sigma_{j}z_{j}\frac{e^{-(Re\Sigma
z^{2}_{j})/4}}{4^{2}}(-2x_{l})\\\ &&+i[i\frac{e^{-(Re\Sigma
z^{2}_{j})/4}}{4}+\Sigma_{j}z_{j}\frac{e^{-(Re\Sigma
z^{2}_{j})/4}}{4^{2}}(2y_{l})]]dz_{l}\wedge d\bar{z}_{j}\\\
\partial\bar{\partial}f&=&[-\frac{e^{-(Re\Sigma
z^{2}_{j})/4}}{4}+\frac{e^{-(Re\Sigma
z^{2}_{j})/4}}{4^{2}}\Sigma_{j,l}z_{l}z_{j}]dz_{l}\wedge
d\bar{z}_{j}\end{array}$

The Sixth term

$\begin{array}[]{rcl}(6)\partial
A^{0,1}f&=&\partial[(1/4)\Sigma_{j}z_{j}(e^{-(Re\Sigma
z^{2}_{j})/4}+i)d\bar{z}_{j}]\\\
(1/2)(\frac{\partial}{\partial_{x_{l}}}-i\frac{\partial}{\partial_{y_{l}}})A^{0,1}f&=&(1/8)[(e^{-(Re\Sigma
z^{2}_{j})/4}+i)+\Sigma_{j}z_{j}\frac{e^{-(Re\Sigma
z^{2}_{j})/4}}{4}(-2x_{l})\\\ &&-i[i(e^{-(Re\Sigma
z^{2}_{j})/4}+i)+\Sigma_{j}z_{j}\frac{e^{-(Re\Sigma
z^{2}_{j})/4}}{4}(2y_{l})]]dz_{l}\wedge d\bar{z}_{j}\\\ \partial
A^{0,1}f&=&[(1/4)(e^{-(Re\Sigma z^{2}_{j})/4}+i)-\frac{e^{-(Re\Sigma
z^{2}_{j})/4}}{4^{2}}\Sigma_{j,l}z_{j}z_{l}]dz_{l}\wedge
d\bar{z}_{j}\end{array}$

The seventh term

$\begin{array}[]{rcl}(7)A^{1,0}\bar{\partial}f&=&A^{1,0}[-\Sigma_{j}z_{j}\frac{e^{-(Re\Sigma
z^{2}_{j})/4}}{4}d\bar{z}_{j}]\\\
&=&\Sigma_{j,l}\bar{z}_{l}z_{j}\frac{e^{-(Re\Sigma
z^{2}_{j})/4}}{4^{2}}dz_{l}\wedge d\bar{z}_{j}\end{array}$

The eighth term

$\begin{array}[]{rcl}(8)A^{1,0}A^{0,1}f&=&-(1/4^{2})\Sigma_{j,l}\bar{z}_{l}z_{j}fdz_{l}\wedge
d\bar{z}_{j}\\\ &=&-\frac{e^{-(Re\Sigma
z^{2}_{j})/4}}{4^{2}}\Sigma_{j,l}\bar{z}_{l}z_{j}dz_{l}\wedge
d\bar{z}_{j}-(i/4^{2})\Sigma_{j,l}\bar{z}_{l}z_{j}dz_{l}\wedge
d\bar{z}_{j}\end{array}$

So we get

$\begin{array}[]{rcl}(\partial_{A}+\bar{\partial}_{A})(\bar{\partial}_{A}f)&=&-\frac{e^{-(Re\Sigma
z^{2}_{j})/4}}{4^{2}}\Sigma_{j,l}\bar{z}_{l}z_{j}d\bar{z}_{l}\wedge
d\bar{z}_{j}\\\ &&-\frac{e^{-(Re\Sigma
z^{2}_{j})/4}}{4^{2}}\Sigma_{j,l}z_{l}\bar{z}_{j}d\bar{z}_{l}\wedge
d\bar{z}_{j}\\\ &&+\frac{e^{-(Re\Sigma
z^{2}_{j})/4}}{4^{2}}\Sigma_{j,l}z_{l}z_{j}d\bar{z}_{l}\wedge d\bar{z}_{j}\\\
&&+\frac{i}{4^{2}}\Sigma_{j,l}z_{l}z_{j}d\bar{z}_{l}\wedge d\bar{z}_{j}\\\
&&+[-\frac{e^{-(Re\Sigma z^{2}_{j})/4}}{4}+\frac{e^{-(Re\Sigma
z^{2}_{j})/4}}{4^{2}}\Sigma_{j,l}z_{l}z_{j}]dz_{l}\wedge d\bar{z}_{j}\\\
&&+[(1/4)(e^{-(Re\Sigma z^{2}_{j})/4}+i)-\frac{e^{-(Re\Sigma
z^{2}_{j})/4}}{4^{2}}\Sigma_{j,l}z_{j}z_{l}]dz_{l}\wedge d\bar{z}_{j}\\\
&&+\Sigma_{j,l}\bar{z}_{l}z_{j}\frac{e^{-(Re\Sigma
z^{2}_{j})/4}}{4^{2}}dz_{l}\wedge d\bar{z}_{j}\\\ &&-\frac{e^{-(Re\Sigma
z^{2}_{j})/4}}{4^{2}}\Sigma_{j,l}\bar{z}_{l}z_{j}dz_{l}\wedge
d\bar{z}_{j}-(i/4^{2})\Sigma_{j,l}\bar{z}_{l}z_{j}dz_{l}\wedge d\bar{z}_{j}\\\
&=&\frac{e^{-(Re\Sigma
z^{2}_{j})/4}}{4^{2}}[\Sigma_{j,l}(z_{l}z_{j}-z_{l}\bar{z}_{j}-\bar{z}_{l}z_{j})]d\bar{z}_{l}\wedge
d\bar{z}_{j}\\\ &&+\frac{i}{4^{2}}[\Sigma_{j,l}z_{l}z_{j}d\bar{z}_{l}\wedge
d\bar{z}_{j}-\Sigma_{j,l}\bar{z}_{l}z_{j}dz_{l}\wedge d\bar{z}_{j}]\\\
&&+\frac{i}{4}dz_{l}\wedge d\bar{z}_{j}\end{array}$

So the estimate we need is (using $|z_{j}|\leq|z|$)

$\begin{array}[]{rcl}|\triangledown\bar{\partial}_{A,\tilde{J}}(e^{-(Re\Sigma
z_{j}^{2})/4}+i)|&\leq&\frac{e^{-(Re\Sigma
z^{2}_{j})/4}}{4}\frac{3}{4}{n\choose 2}|z|^{2}+\frac{1}{8}[{n\choose
2}|z|^{2}+2{n\choose 2}]\\\ &&+Ck^{-1/2}[e^{-(Re\Sigma
z^{2}_{j})/4}P^{3}(|z|)+Q^{3}(|z|)]\\\ \end{array}$

where $P^{3},Q^{3}$ are cubic polynomial with constant term zero.

We now define the cut-off function $\beta_{k}:\mathbb{C}^{n}\to\mathbb{R}$ as
a $C^{\infty}$-function with $\beta_{k}=s_{k}$ on
$I^{2n}_{k^{-1/2}-\epsilon_{k},1-\epsilon_{k}}$ and $\beta_{k}=0$ outside
$I^{2n}_{k^{-1/2}+\epsilon_{k},1+\epsilon_{k}}$, where $s_{k}$ is a positive
real sequence so small such that derivative $\triangledown\beta_{k}$ satisfies
$|\triangledown\beta_{k}|\leq Ck^{-1}|z|^{2}$ and hence
$|\triangledown\triangledown\beta_{k}|\leq Ck^{-1}|z|$. So we have

$|\bar{\partial}_{A,\tilde{J}}(\beta_{k}f)|\leq|\beta_{k}||\bar{\partial}_{A,\tilde{J}}f|+|\triangledown\beta_{k}||f|$

Similarly

$|\triangledown(\bar{\partial}_{A,\tilde{J}}\beta_{k}f)|\leq|\triangledown\triangledown\beta_{k}||f|+|\triangledown\beta_{k}||\triangledown_{A}f|+|\beta_{k}||\triangledown(\bar{\partial}_{A,\tilde{J}}f)|+|\triangledown\beta_{k}||\bar{\partial}_{A,\tilde{J}}f|$

So the cut-off function improves the estimates as follows.

$\begin{array}[]{rcl}|\bar{\partial}_{A,\tilde{J}}\beta_{k}(e^{-(Re\Sigma
z_{j}^{2})/4}+i)|&\leq&|\beta_{k}|[\frac{|z|}{4}(2e^{-(Re\Sigma
z_{j}^{2})/4}+1)+Ck^{-1/2}|z|^{2}(2e^{-(Re\Sigma z_{j}^{2})/4}+1)]\\\
&&+|\triangledown\beta_{k}|(e^{-(Re\Sigma z_{j}^{2})/2}+1)^{1/2}\\\
&\leq&\frac{1}{2}s_{k}|z|e^{-(Re\Sigma
z_{j}^{2})/4}+|\beta_{k}|\frac{|z|}{4}+Ck^{-1/2}s_{k}|z|^{2}e^{-(Re\Sigma
z_{j}^{2})/4}\\\ &&+Ck^{-1/2}|\beta_{k}||z|^{2}+Ck^{-1}|z|^{2}e^{-(Re\Sigma
z_{j}^{2})/2}+|\triangledown\beta_{k}|\end{array}$

$\begin{array}[]{rcl}|\triangledown\bar{\partial}_{A,\tilde{J}}\beta_{k}(e^{-(Re\Sigma
z_{j}^{2})/4}+i)|&\leq&|\triangledown\triangledown\beta_{k}|(e^{-(Re\Sigma
z_{j}^{2})/2}+1)^{1/2}\\\
&&+|\triangledown\beta_{k}|\frac{|z|}{4}(2e^{-(Re\Sigma z_{j}^{2})/4}+1)\\\
&&+|\beta_{k}|[\frac{e^{-(Re\Sigma z^{2}_{j})/4}}{4}\frac{3}{4}{n\choose
2}|z|^{2}+\frac{1}{8}[{n\choose 2}|z|^{2}+2{n\choose 2}]]\\\
&&+|\beta_{k}|[Ck^{-1/2}[e^{-(Re\Sigma z^{2}_{j})/4}P^{3}(|z|)+Q^{3}(|z|)]]\\\
&&+Ck^{-1}|z|^{2}|\bar{\partial}_{A,\tilde{J}}f|\\\
&\leq&|\triangledown\triangledown\beta_{k}|e^{-(Re\Sigma
z_{j}^{2})/2}+|\triangledown\triangledown\beta_{k}|\\\
&&+|\triangledown\beta_{k}|\frac{|z|}{4}e^{-(Re\Sigma
z_{j}^{2})/4}+|\triangledown\beta_{k}|\frac{|z|}{4}\\\
&&+s_{k}[\frac{e^{-(Re\Sigma z^{2}_{j})/4}}{4}\frac{3}{4}{n\choose
2}|z|^{2}+\frac{1}{8}[{n\choose 2}|z|^{2}+2{n\choose 2}]]\\\
&&+s_{k}Ck^{-1/2}e^{-(Re\Sigma
z^{2}_{j})/4}P^{3}(|z|)+Ck^{-1/2}|\beta_{k}|Q^{3}(|z|)\\\
&&+Ck^{-1}|z|^{2}|\bar{\partial}_{A,\tilde{J}}f|\end{array}$

We define the smooth section $\sigma_{p}$ of the line bundle $E^{k}$ over $W$
by

$\sigma_{p}=(\beta_{k}\sigma)\circ\tilde{\chi}^{-1}$

around $p\in\tilde{\chi}(I^{2n}_{k^{-1/2}+\epsilon_{k},1+\epsilon_{k}})$ which
we extend by $0$ on all of $W$.

Now fix a finite cover, independent of $k$, of $W$ by charts
$\phi_{s}:O_{s}\to W$, $s=1,...,S$, where $O_{s}\subset\mathbb{C}^{n}$ is
bounded with

$\frac{1}{2}|x-y|\leq d(\phi(x),\phi(y))\leq 2|x-y|$

We may choose nested sets $O^{\prime\prime}_{s}\subset O^{\prime}_{s}\subset
O_{s}$, so that $W$ is covered by $\phi_{s}(O^{\prime\prime}_{s})$. Let
$\Lambda$ be the lattice $\alpha_{k}\mathbb{Z}^{n}+i\mathbb{Z}^{n}$ in
$\mathbb{C}^{n}$, where $\alpha_{k}$ is chosen in such a way that the
translations of the center of $I^{2n}_{k^{-1/2}+\epsilon_{k},1+\epsilon_{k}}$
to the lattice points of $\Lambda$ cover $\mathbb{C}^{n}$. Let $\Lambda_{s}$
be $\phi_{s}(\Lambda\cap O^{\prime}_{s})$. Define the set
$\cup_{1}^{M}\\{p_{i}\\}=\cup_{1}^{S}\Lambda_{s}$. So the images under
$\phi_{s}$ of the translations of
$I^{2n}_{k^{-1/2}+\epsilon_{k},1+\epsilon_{k}}$ to the lattice points of
$\Lambda_{s}$ gives neighborhoods of $p_{i}$’s which cover $W$. Moreover we
arrange that if a neighborhood of $p_{i}$ as mentioned above intersects the
regular Lagrangian $L\subset W$ then $p_{i}\in L$. Later we shall translate
the origins of these coordinate charts except one (the first one) such that
there is only one singularity on $L$.

Now consider the terms of the right hand sides of the estimates for

$|\bar{\partial}_{A,\tilde{J}}\beta_{k}(e^{-(Re\Sigma z_{j}^{2})/4}+i)|\ and\
|\triangledown\bar{\partial}_{A,\tilde{J}}\beta_{k}(e^{-(Re\Sigma
z_{j}^{2})/4}+i)|$

with the understanding that outside of the neighborhoods of $p_{i}$’s they
vanish as $\beta_{k}$ vanishes outside these neighborhoods. We also note that
we can replace $s_{k}$ by $k^{-1/2}s_{k}$ for large $k$.

Our final section $s$ will be

$s=s_{w}=\Sigma_{1}^{M}w_{i}\sigma_{i},\ \sigma_{i}=\sigma_{p_{i}},\
|w_{i}|\leq 1,\ w=(w_{1},...,w_{M})$

Obviously it depends on $k$. Later we shall make some specific choice for
$w_{i}$’s. So we have the following

######  Theorem 4.1.

For any choice of co-efficient vector $w$ with $|w_{i}|\leq 1$ and a very
small choice of $s_{k}$ for large $k$, the section $s=s_{w}$ satisfies

$\begin{array}[]{rcl}|s|&\leq&C\\\
|\bar{\partial}_{E}s|&\leq&C(k^{-1/2}+k^{-1})\\\
|\triangledown_{W}\bar{\partial}_{E}s|&\leq&C(k^{-1/2}+k^{-1})\\\ \end{array}$

Now we shall recall a result from [1].

######  Theorem 4.2.

([1]) If $a^{\prime},a^{\prime\prime}:\mathbb{C}^{n}\to\mathbb{C}$ are
respectively complex linear and anti-linear maps and if
$|a^{\prime\prime}|<|a^{\prime}|$, then the subspace
$ker(a^{\prime}+a^{\prime\prime})\subset\mathbb{C}^{n}$ is symplectic.

In view of 4.1 and 4.2 if we prove the following result then we shall prove
that the zero locus of $s$ is a symplectic submanifold as in [1].

######  Theorem 4.3.

There is an $\epsilon>0$ such that for all large $k$ we can choose $w$ with
$|w_{i}|\leq 1$, so that $s=s_{w}$ satisfies the transversality condition

$|\partial s|>\epsilon$

on the zero locus of $s$ i.e $\mathcal{Z}(s)$.

######  Remark 4.4.

We shall further need that $w_{i}=c+ic,\ or\ -c+ic$ for $c>0$ and shall also
see that such a choice is possible in this context.

Now we shall prove 4.3 which similar to [1] (not identical though).

######  Theorem 4.5.

([1]) Given any $D>0$ there is a number $N=N(D)$, independent of $k$, such
that for any large $k$ we can choose a collection of centers $p_{i}$
satisfying the conditions of the proof of 4.1 in addition with the property
that there is a partition of the set $\Lambda=\\{1,...,M\\}$ into $N$ disjoint
subsets $\Lambda=\Lambda_{1}\cup...\cup\Lambda_{N}$ such that for each
$\alpha$

$d(p_{i},p_{j})\geq D\ if\ p_{i},p_{j}\in\Lambda_{\alpha}$

So now we assume that $D$ is fixed and $p_{i}$’s accordingly as well as the
partition $\\{\Lambda_{\alpha}\\}$. The choice of $D$ will be made at the end
of this section. We think of this partition $\\{\Lambda_{\alpha}\\}$ as the
colouring of the cubes $I^{2n}_{k^{-1/2}+\epsilon_{k},1+\epsilon_{k}}$.

We shall adjust now the co-efficients $w_{i}$’s belonging to the same
$\Lambda_{\alpha}$ starting with $\Lambda_{1}$ with the co-efficient vector
$w_{0}=(w_{1},0,...,0),\ w_{1}=c+ic,\ c>0$ where $p_{1}\in L$. Here it is
different from [1]. In the next stage we shall change the co-efficients
$w_{i}$’s belonging to $\Lambda_{2}$ leaving the others unchanged and proceed
this way. Thus the total number $N$ of stages in this construction depends on
$D$ but not on $k$.

Now we can focus on the adjustment criterion for $w_{i}$’s. The goal is to
achieve the controlled transversality over the cubes belonging to
$\Lambda_{\alpha}$ at the $\alpha$-th stage. We set

$W_{\alpha}=\cup_{i\in\Lambda_{\beta},\beta\leq\alpha}I^{2n}_{i}$

where $I^{2n}_{i}$ is the $i$-th cube
$I^{2n}_{k^{-1/2}+\epsilon_{k},1+\epsilon_{k}}$. So
$\Phi(empty)=W_{0}\subset...\subset W_{N}=W$. We want that for suitable
$\epsilon$, the section $s_{\alpha}$ satisfies $|\partial
s_{\alpha}|>\epsilon$ on $\mathcal{Z}(s_{\alpha})\cap W_{\alpha}$. So the
criterion can be summarized as

  1. (1)

The change in the co-efficient belonging to $\Lambda_{\alpha}$ must achieve
controlled transversality over the cubes belonging to $\Lambda_{\alpha}$.

  2. (2)

The change in the co-efficients belonging to $\Lambda_{\alpha}$ must not
destroy the controlled transversality that has already been achieved over the
cubes belonging to $\Lambda_{\beta}$ for $\beta<\alpha$.

So the precise form of transversality we need is the following

######  Definition 4.6.

([1]) Let $f:U\to\mathbb{C}$ be a smooth map on an open set
$U\subset\mathbb{C}$ and let $w\in\mathbb{C}$. For $\eta>0$ we say that $f$ is
$\eta$-transverse to $w$ over $U$ if for any $z\in U$ such that
$|f(z)-w|\leq\eta$ the derivative satisfies $|(\partial f)_{z}|\geq\eta$.

4.6 is stable under $C^{1}$-perturbations, moreover if $f$ is
$\eta$-transverse to $w$ and $g:U\to\mathbb{C}$ is such that
$|f-g|_{C^{1}}\leq\delta$ then $g$ is $\eta-\delta$-transverse to $w$.

Now we consider any section $s$ of $E^{k}\to W$. On a fixed chart around say
$p_{i}$, $E^{k}\to W$ has a standard trivialization given by $\sigma_{i}$. We
write $s=f_{i}\sigma_{i}$, for a function $f_{i}$ defined on the cube
$I^{2n}_{k^{-1/2}+\epsilon_{k},1+\epsilon_{k}}$. We say that the section $s$
is $\eta$-transverse on the chart if $f_{i}$ is $\eta$-transverse to $0$ on
the cube $I^{2n}_{k^{-1/2},1}$.

######  Lemma 4.7.

If $s=s_{w}$ is the section of $E^{k}$ as described in 4.1 and $f_{i}$’s the
corresponding function then

  1. (1)

$|f_{i}|_{C^{1}(I^{2n}_{k^{-1/2},1})}\leq C$

  2. (2)

$|\bar{\partial}f_{i}|_{C^{1}(I^{2n}_{k^{-1/2},1})}\leq C(k^{-1/2}+k^{-1})$

  3. (3)

If $|\partial f_{i}|>\epsilon$ on $f_{i}^{-1}(0)\cap I^{2n}_{k^{-1/2},1}$,
then for large $k$, $|\partial_{E}s|>C^{-1}\epsilon$ on the intersection of
$\mathcal{Z}(s)$ and the chart.

In view of 4.1, the proofs of 4.7 is same as the proofs of Lemma-18 from [1].

######  Lemma 4.8.

If $s=s_{w}$ be as in 4.1, for any $\alpha$, let $w^{\prime}$ be another
vector which agrees with $w$ except for the co-efficients belonging to
$\Lambda_{\alpha}$ and suppose that $|w^{\prime}_{j}-w_{j}|\leq\delta$ for
$j\in\Lambda_{\alpha}$. Set $s^{\prime}=s_{w^{\prime}}$ and let
$f_{i},f^{\prime}_{i}$ be the functions representing these sections, then for
large $D$

  1. (1)

For any $i$, $|f^{\prime}_{i}-f_{i}|_{C^{1}(I^{2n}_{k^{-1/2},1})}\leq\delta$

  2. (2)

If $i\in\Lambda_{\alpha}$ and $w^{\prime}_{i}=w_{i}+\theta_{i}$ then
$|f^{\prime}_{i}-f_{i}-\theta_{i}|_{C^{1}(I^{2n}_{k^{-1/2},1})}=0$

###### Proof.

We write $s=\Sigma_{j}w_{j}\sigma_{j}$. So on $i$-th smaller cube
$I^{2n}_{k^{-1/2},1}$ our equation $s=f_{i}\sigma_{i}$ gives us $\Sigma_{j\neq
i}w_{j}\sigma_{j}+w_{i}\sigma_{i}=f_{i}\sigma_{i}$ and similarly
$\Sigma_{j\neq
i}w_{j}^{\prime}\sigma_{j}+w_{i}^{\prime}\sigma_{i}=f^{\prime}_{i}\sigma_{i}$.
So we get

$\begin{array}[]{rcl}f^{\prime}_{i}-f_{i}-\theta_{i}&=&w^{\prime}_{i}-w_{i}-\theta_{i}+\Sigma_{j\neq
i}\frac{\sigma_{j}}{\sigma_{i}}(w^{\prime}_{j}-w_{j})\\\
\Rightarrow|f^{\prime}_{i}-f_{i}-\theta_{i}|&\leq&\Sigma_{j\neq
i}|w^{\prime}_{j}-w_{j}||\frac{\sigma_{j}}{\sigma_{i}}|,\ As\
w_{i}^{\prime}-w_{i}-\theta_{i}=0\\\ \end{array}$

Similarly $|f^{\prime}_{i}-f_{i}|\leq\delta+\Sigma_{j\neq
i}|w^{\prime}_{j}-w_{j}||\frac{\sigma_{j}}{\sigma_{i}}|$. Now considering the
derivative we get

$\begin{array}[]{rcl}\triangledown(f^{\prime}_{i}-f_{i}-\theta_{i})&=&\Sigma_{j\neq
i}(w^{\prime}_{j}-w_{j})\triangledown(\frac{\sigma_{j}}{\sigma_{i}})\\\
&=&\Sigma_{j\neq
i}(w^{\prime}_{j}-w_{j})[\frac{1}{\sigma_{i}}\triangledown\sigma_{j}-\frac{\sigma_{j}}{\sigma_{i}^{2}}\triangledown\sigma_{i}]\\\
\Rightarrow|\triangledown(f^{\prime}_{i}-f_{i}-\theta_{i})|&\leq&\Sigma_{j\neq
i}|w_{j}^{\prime}-w_{j}|[\frac{|\triangledown\sigma_{j}|}{|\sigma_{i}|}+\frac{|\sigma_{j}|}{|\sigma_{i}^{2}|}|\triangledown\sigma_{i}|]\\\
\end{array}$

Now as $w^{\prime}_{j}=w_{j}$ for $j\notin\Lambda_{\alpha}$, the summations
$\Sigma_{j\neq i}$ in the above estimates can be replaced by
$\Sigma_{j\in\Lambda_{\alpha}}$. So when $D$ is large
$|\sigma_{j}|=0=|\triangledown\sigma_{j}|$ on the chart
$\tilde{\chi}_{i}(I^{2n}_{k^{-1/2},1})$. ∎

Now define $Q_{p}(\delta)=\log(\delta^{-1})^{-p},\ \delta>0$. The next result
(similar to Theorem-20 of [1]) gives us the suitable $w$ in order to get the
transversality condition. The result is also known as Quantitative
Transversality theorem. We shall however show that we can make specific choice
for $w_{j}$’s in our context which will be needed in the proof of 1.3.

######  Theorem 4.9.

([1]) For $\sigma>0$, let $\mathcal{H}_{\sigma}$ denote the set of functions
$f$ on $I^{2n}_{k^{-1/2},1}$ such that

  1. (1)

$|f|_{C^{0}(I^{2n}_{k^{-1/2},1})}\leq 1$

  2. (2)

$|\bar{\partial}f|_{C^{0}(I^{2n}_{k^{-1/2},1})}\leq\sigma$

Then there is an integer $p$, depending only on the dimension $n$, such that
for any $\delta$ with $0<\delta<1/2$, if $\sigma\leq Q_{p}(\delta)\delta$,
then for any $f\in\mathcal{H}_{\sigma}$ there is a $w\in\mathbb{C}$ with
$|w|\leq\delta$ such that $f$ is $Q(\delta)\delta$-transverse to $w$ over
$I^{2n}_{k^{-1/2},1}$. Moreover $w$ can be taken to be $-(a+ia),\ a>0$ and
$-(a+ib),\ b=-a>0$. The choice between these two options will be clarified
later.

We shall prove 4.9 at the end of this section. For now let us continue with
our construction.

Let us assume that we have reached the $\alpha-1$-th stage of our construction
and we want to reach the $\alpha$-th stage. So we have chosen the
$w^{\alpha-1}$ so that $s_{\alpha-1}$ is $\eta_{\alpha}$-transverse over
$W_{\alpha-1}\subset W$ for some positive $\eta_{\alpha-1}$ with
$0<\eta_{\alpha-1}<\rho$. We will now choose $w_{i}^{\alpha},\
i\in\Lambda_{\alpha}$. We need
$|w_{i}^{\alpha}-w_{i}^{\alpha-1}|\leq\delta_{\alpha}$. We shall set
$\delta_{\alpha}>0$ shortly. With this choice of $w_{i}^{\alpha}$, 4.8 gives
us that $s^{\alpha}$ is $\eta_{\alpha-1}-\delta_{\alpha}$-transverse over
$W_{\alpha-1}$. So if we set $\delta_{\alpha}=\frac{1}{2}\eta_{\alpha-1}$ we
can conclude that $s^{\alpha}$ is still
$\frac{1}{2}\eta_{\alpha-1}$-transverse on $W_{\alpha-1}$. Now we shall
consider the cubes $I^{2n}_{i},\ i\in\Lambda_{\alpha}$. Here the section
$s^{\alpha-1}$ is represented by the function
$f_{i}=f_{i}^{\alpha-1}=\frac{s^{\alpha-1}}{\sigma_{i}}$. By 4.7 $f_{i}$ is
bounded by a fixed constant $C$ over $I^{2n}_{k^{-1/2},1}$ and
$C^{-1}f_{i}\in\mathcal{H}_{\sigma}$ for $\sigma=C^{-1}(k^{-1/2}+k^{-1})$. Now
we apply 4.9 $C^{-1}f_{i}$ for suitably small $\rho$ and
$\delta=C^{-1}\delta_{\alpha}$ as long as

$k^{-1/2}+k^{-1}\leq C\delta_{\alpha}Q_{p}(\delta_{\alpha})$

And thus we get $v_{i},\ i\in\Lambda_{\alpha}$ with
$|v_{i}|\leq\delta_{\alpha}$ such that $f_{i}$ is
$Q_{p}(\delta_{\alpha})\delta_{\alpha}$-transverse to $v_{i}$ and equivalently
$f_{i}-v_{i}$ is $Q_{p}(\delta_{\alpha})\delta_{\alpha}$-transverse to $0$.
Observe that $f_{i}-v_{i}$ represents $s_{w^{\prime}}$ where
$w^{\prime}_{j}=w_{j}^{\alpha-1},\ if\ j\neq i$ and
$w^{\prime}_{i}=w_{i}^{\alpha-1}-v_{i}$. In view of 4.8 for large $D$ we can
make all the changes to the $w_{i}^{\alpha-1},\ i\in\Lambda_{\alpha}$
simultaneously as explained above. This concludes the proof of 4.3.

Now we shall see the proof of 4.9 which is same as the proof of Theorem-20 of
[1]. Although we shall outline the entire proof for completeness but all we
need to show that we can make the choice for $w_{i}$’s in our context. The
choice is different from [1].

Let $P:\mathbb{R}^{n}\to\mathbb{R}$ be a polynomial function of degree $d$.
Let $S\subset\mathbb{R}^{n}$ be the subset

$S=\\{x\in\mathbb{R}^{n}:|x|\leq 1,\ P(x)\leq 1\\}$

Similarly set $S(\theta)=\\{x:|x|\leq 1,\ P(x)\leq 1+\theta\\}$.

######  Proposition 4.10.

([1]) There are constants $C,\ \nu$, depending only on the dimension, such
that for any polynomial $P$ there are arbitrarily small positive $\theta$ so
that $S$ can be decomposed into pieces

$S=S_{1}\cup...\cup S_{A}$

where $A\leq Cd^{\nu}$, in such a way that any pair of points in the same
piece $S_{r}$ can be joined by a path in $S(\theta)$ of length less than
$Cd^{\nu}$.

The proof decomposes into two parts. First we prove the 4.9 for holomorphic
$f$ and then use some estimates to prove the result for approximately
holomorphic $f$.

######  Lemma 4.11.

([1]) Let $f:I^{2n}_{k^{-1/2},1}\to\mathbb{C}$ be a holomorphic map with
$|f(z)|\leq 1$. Then for any $\epsilon$ satisfying $0<\epsilon<1/2$ there is a
complex polynomial function $g$ on $\mathbb{C}^{n}$ of degree less than
$C\log\epsilon^{-1}$ such that $|f(z)-g(z)|,\ |\partial f-\partial
g|\leq\epsilon$ in the interior region.

Now we shall prove 4.9 for holomorphic $f$. So let us assume that $f$ is as in
4.11 and we use 4.11 to approximate $f$ by a polynomial $g$ of degree $d\leq
C\log\epsilon^{-1}$. Define

$S^{f}=\\{z\in I^{2n}_{k^{-1/2},1}\cup\mathbb{C}^{n}:|\partial
f|\leq\epsilon\\}$ $S^{g}=\\{z\in
I^{2n}_{k^{-1/2},1}\cup\mathbb{C}^{n}:|\partial g|\leq 2\epsilon\\}$

Obviously $S^{f}\subset S^{g}$ and therefore $f(S^{f})\subset f(S^{g})\subset
N_{\epsilon}(g(S^{g}))$ where $N_{\epsilon}(g(S^{g}))$ is the
$\epsilon$-neighborhood of $g(S^{g})$. Observe that $S^{g}$ is a semi-
algebraic set of the kind in 4.10, where we have taken
$P=(2\epsilon)^{-2}|\partial g|^{2}$ whose degree is $2(d-1)$. So 4.10 gives a
decomposition of $S^{g}$ into $A$ pieces. Let $z_{1},z_{2}$ are in the same
piece of $S^{g}$ then integrating the derivative of $g$ over a path of length
less than $Cd^{\nu}$ in $P(\theta)$, for suitable $\theta$, joining $z_{1}$ to
$z_{2}$, we have $|g(z_{1})-g(z_{2})|\leq 4\epsilon Cd^{\nu}$. Therefore
$g(S^{g})$ is contained in the union of $A$ discs in $\mathbb{C}$ each of
radius $4\epsilon Cd^{\nu}$. So $f(S^{g})$ and hence $f(S^{f})$ is contained
in the union of slightly larger discs of radius $\epsilon(2Cd^{\nu}+1)$.

The $\epsilon$-transversality between $f$ and $w$ is that $w$ lies outside the
$\epsilon$-neighborhood of $f(S^{f})$ and the neighborhood is contained in the
union of $A$ discs of radius $\epsilon(4Cd^{\nu}+2)$. The total area in
$\mathbb{C}$ of these discs is at most $A\pi(4Cd^{\nu}+2)^{2}\epsilon^{2}$. If
we choose $\rho$ such that the area of the half disc

$\Omega=\\{w\in\mathbb{C}:|w|\leq\rho,\ Im(w)\leq 0\\}$

is biggere than the total area covered by these discs , there is a
$w\in\Omega$ not contained in the $\epsilon$-neighborhood of $f(S^{f})$. The
condition on $\rho$ is

$\frac{1}{2}\pi\rho^{2}>A\pi(4Cd^{\nu}+2)^{2}\epsilon^{2}$

which is $\rho>(2A)^{1/2}(4Cd^{\nu}+2)\epsilon$. Observe that $A$ are bounded
by powers of degree of $P$, hence of the degree $d$ of $g$ which is bounded by
a power of $\log(\epsilon)^{-1}$.

So over all, for arbitrarily small $\epsilon$ there exists a $w\in\mathbb{C}$
such that $f$ is $\epsilon$-transverse to $w$ and $|w|\leq
C\epsilon\log(\epsilon^{-1})^{p}$ for some $p$. Now one needs to re-arrange
the parameters. For fixed $C,p$, the function $h$ given by
$h(\epsilon)=C\epsilon\log(\epsilon^{-1})^{p}$ is increasing for for small
$\epsilon$ and tends to $0$ as $\epsilon$ approaches $0$. If
$\eta=h(\epsilon)$ then

$\eta\log(\eta^{-1})^{-p}=C\epsilon(\frac{\log\epsilon^{-1}}{\log\epsilon^{-1}-p\log\log\epsilon^{-1}-\log
C})^{p}<2C\epsilon,\ (if\ \epsilon\ small)$

By inverting $h$, we conclude that for small $\eta$ there is a $w$ with
$|w|\leq\eta$ such that $f$ is
$\frac{1}{2C}\eta(\log\eta^{-1})^{p}$-transverse to $w$, and by increasing $p$
and assuming $\eta$ to be small, we can replace the factor $\frac{1}{2C}$ by
$1$.

In the above we considered holomorphic $f$. Now we shall consider
approximately holomorphic $f$. Although we have not shown that we can make the
specific choice for $w$ in our context, but we shall show this at the end of
the proof.

Using Hörmander’s weighted $L^{2}$ space method [6] one gets the following

######  Lemma 4.12.

([1]) For each $r<1$ there is a constant $K=K(r)$ such that if $f$ is any
smooth complex valued function on $I^{2n}_{k^{-1/2},1}$, then there is a
holomorphic function $\tilde{f}$ on $rI^{2n}_{k^{-1/2},1}$ such that

$|f-\tilde{f}|_{C^{1}(rI^{2n}_{k^{-1/2},1})}\leq
K(|\bar{\partial}f|_{C^{0}(I^{2n}_{k^{-1/2},1})}+|\triangledown\bar{\partial}f|_{C^{0}(I^{2n}_{k^{-1/2},1})})$

Now if $f\in\mathcal{H}_{\sigma}$, we approximate $f$ by a holomorphic
function $\tilde{f}$ on a slightly smaller region with
$|f-\tilde{f}|_{C^{1}}\leq C\sigma$. Then use the above arguments for
$\tilde{f}$.

Now we show how we can make the specific choice for $w$ in the half disc

$\\{w\in\mathbb{C}:|w|\leq\rho,\ Im(w)\leq 0\\}$

We must show that $f-v_{i}$ is $\tilde{\epsilon}$-transverse to $0$ for some
small $\tilde{\epsilon}>0$. Any coordinate chart intersect only finitely many
adjacent charts and the number depends on the dimension $n$. We only consider
one such intersection. Let $\phi=(\phi_{1},...,\phi_{n})$ be the transition
function. So we take

$f=\frac{\beta^{\prime}[(ce^{-(Re\Sigma\phi_{j}^{2})/4}-d)+i(de^{-(Re\Sigma\phi_{j}^{2})/4}+c)]}{\beta(e^{-(Re\Sigma
z_{j}^{2})/4}+i)}$

where either $c=d>0$ or $d=-c>0$. Here $\beta$ corresponds to the chart under
consideration and $\beta^{\prime}$ corresponds to the adjacent chart. The sign
of $c,d$ changes as we are adding $-v_{i}\sigma_{i}$ because
$w_{i}^{\prime}=w_{i}^{\alpha}-v_{i}$ and at the beginning we set
$w_{0}=(w_{1},0,..,0)$ where $w_{1}=c+ic,\ c>0$. Simplifying we get

$\begin{array}[]{rcl}f&=&\frac{\beta^{\prime}}{\beta(e^{-(Re\Sigma
z_{j}^{2})/2}+1)}[\\{c(1+e^{-(Re\Sigma
z_{j}^{2})/4}e^{-(Re\Sigma\phi_{j}^{2})/4})+d(e^{-(Re\Sigma\phi_{j}^{2})/4}-e^{-(Re\Sigma
z_{j}^{2})/4})\\}\\\ &&+i\\{c(e^{-(Re\Sigma
z_{j}^{2})/4}-e^{-(Re\Sigma\phi_{j}^{2})/4})+d(1+e^{-(Re\Sigma
z_{j}^{2})/4}e^{-(Re\Sigma\phi_{j}^{2})/4})\\}]\\\ \end{array}$

Now observe that for large $k$, i.e, for range of $|x_{j}|$ being very small
$(Assuming\ z_{j}=x_{j}+iy_{j})$, the imaginary part of $f$ for both $c=d>0$
and $d=-c>0$ is non-negative. To see this consider the numerator of the
imaginary part for $c=d>0$

$c[(e^{-(Re\Sigma
z_{j}^{2})/4}-e^{-(Re\Sigma\phi_{j}^{2})/4})+(1+e^{-(Re\Sigma
z_{j}^{2})/4}e^{-(Re\Sigma\phi_{j}^{2})/4})]$ $=c[1+e^{-(Re\Sigma
z_{j}^{2})/4}(1+e^{-(Re\Sigma\phi_{j}^{2})/4})-e^{-(Re\Sigma\phi^{2})/4}]$

First observe that if $e^{-(Re\Sigma z_{j}^{2})/4}>1$ then the imaginary part
is non-negative.

If $e^{-(Re\Sigma z_{j}^{2})/4}<1$ we observe that as $k$ becomes large
$e^{-(Re\Sigma z_{j}^{2})/4}$ becomes very close to $1$. So for large $k$ the
imaginary part is non-negative.

We get the same result for the case $d=-c>0$ by making $k$ large enough.

So the imaginary part of $f-v_{i}$ is bigger than $a$ if we take
$v_{i}=-(a+ia),\ a>0$ and is bigger than $b$ if we take $v_{i}=-(a+ib),\
b=-a>0$ which concludes the argument. For more intersection we may need to
take larger $k$.

##  5\. Main Theorem

In this section we prove 1.3. As in the previous section we consider $s$ for
large values of $k$. So $\mathcal{Z}(s)$ is a symplectic submanifold and hence
$s^{-1}(z)$ for $z$ close to $0$ is also symplectic. This forms a tubular
neighborhood of $\mathcal{Z}(s)$.

Now we consider the $p_{1}\in L$ the center of the first chart in the
construction of $s$ as in the previous section. Obviously it is a singularity
for $s$ and now we’ll figure out the stable manifold corresponding to the
vector field $X_{Im\log s}$, the Hamiltonian vector field corresponding to
$Im\log s=Arg(s)$. But before we do so let us show that it is a hyperbolic
zero for $X_{Im\log s}$.

######  Proposition 5.1.

$p_{1}\in L\subset W$ is a hyperbolic zero for $X_{Im\log s}=X_{\theta},\
where\ \theta=Arg(s)$ and the tangent space to the stable manifold is given by
$\\{x_{j}=y_{j}:j=1,...,n\\}$ at $p_{1}$. Moreover on the set
$\beta_{1}=constant\neq 0$ the stable manifold is given by
$\\{x_{j}=y_{j}:j=1,...,n\\}$.

###### Proof.

On the set $\beta_{1}=constant\neq 0$, $s$ is given by
$s=\beta_{1}(a+ib)(e^{-(Re\Sigma z_{j}^{2})/4}+i)$. So we get

$\begin{array}[]{rcl}\tan\theta&=&\frac{be^{-(Re\Sigma
z_{j}^{2})/4}+a}{ae^{-(Re\Sigma z_{j}^{2})/4}-b}=\frac{A}{B}\\\
\Rightarrow\sec^{2}\theta d\theta&=&(1/B)[(1/2)be^{-(Re\Sigma
z_{j}^{2})/4}\Sigma_{j}(y_{j}dy_{j}-x_{j}dx_{j})]\\\
&&-(A/B^{2})[(1/2)ae^{-(Re\Sigma
z_{j}^{2})/4}\Sigma_{j}(y_{j}dy_{j}-x_{j}dx_{j})]\\\ \Rightarrow
d\theta&=&\frac{B}{2(A^{2}+B^{2})}[be^{-(Re\Sigma
z_{j}^{2})/4}\Sigma_{j}(y_{j}dy_{j}-x_{j}dx_{j})]\\\
&&-\frac{A}{2(A^{2}+B^{2})}[ae^{-(Re\Sigma
z_{j}^{2})/4}\Sigma_{j}(y_{j}dy_{j}-x_{j}dx_{j})]\end{array}$

The second line is achieved from the first by differentiating and the third
line is achieved by multiplying by $\cos^{2}\theta$. Note that because we are
differentiating both sides of an equation we get $d_{A}=d$.

Now let $X_{\theta}=\Sigma_{j}(a_{j}\partial_{x_{j}}+b_{j}\partial_{y_{j}})$
and observe that the symplectic structure on the tubular neighborhood of
$L\subset W$ is represented by $\omega_{st}=\Sigma_{j}dx_{j}\wedge dy_{j}$ on
coordinate charts. Hence Hamiltonian equation
$d\theta=\iota_{X_{\theta}}\omega_{st}$ gives us
$d\theta=\Sigma_{j}(a_{j}dy_{j}-b_{j}dx_{j})$. Solving we get

$a_{j}=\frac{y_{j}e^{-(Re\Sigma z_{j}^{2})/4}}{2(A^{2}+B^{2})}[Bb-Aa]$

Similarly

$b_{j}=\frac{x_{j}e^{-(Re\Sigma z_{j}^{2})/4}}{2(A^{2}+B^{2})}[Bb-Aa]$

Now we compute the partial derivatives of $a_{j}\ and\ b_{j}$ at zero. Observe
that zero in this coordinates corresponds to $p_{1}$. So we get

$(\partial_{x_{j}}a_{j})_{\mid 0}=0,\ (\partial_{y_{j}}a_{j})_{\mid
0}=\frac{Bb-Aa}{2(A^{2}+B^{2})}_{\mid 0}=-\frac{1}{4}$

Similarly

$(\partial_{y_{j}}b_{j})_{\mid 0}=0,\ (\partial_{x_{j}}b_{j})_{\mid
0}=\frac{Bb-Aa}{2(A^{2}+B^{2})}_{\mid 0}=-\frac{1}{4}$

Define $F(z)=(a_{1},b_{1},...,a_{n},b_{n})$, then $DF_{\mid 0}$ has all
eigenvalues equal to $\pm\frac{1}{4}$. The tangent of the stable manifold at
zero corresponds to $-\frac{1}{4}$ which is given by
$\\{x_{j}=y_{j}:j=1,...,n\\}$.

Now let $(x(t),y(t))=(x_{1}(t),y_{1}(t),...,x_{n}(t),y_{n}(t))$ be the flow of
$X_{\theta}$. So

$\partial_{t}x_{j}(t)=a_{j}(x(t),y(t))\ and\
\partial_{t}y_{j}(t)=b_{j}(x(t),y(t))$

Multiplying the first equation $\partial_{t}x_{j}(t)=a_{j}(x(t),y(t))$ by
$x_{j}(t)$ and the second equation $\partial_{t}y_{j}(t)=b_{j}(x(t),y(t))$ by
$y_{j}(t)$ and integrating we get (suppressing the $t$ all along below for
convenient notations)

$\frac{1}{2}(x_{j}^{2}-y_{j}^{2})=0$

which proves the remaining statement. ∎

Now we shall figure out the stable manifold globally. But for this we need to
figure out the flow of the vector field $X_{\theta}$. Let us consider $s$ on a
chart adjacent to the one containing $p_{1}$ and intersecting $L$. Recall that
on these charts $L$ is given by $\\{y_{j}=0,\ j=1,...,n\\}$. Now on this chart
$\tan\theta$ is given by

$\tan\theta=\frac{\beta_{1}be^{-(Re\Sigma
z_{j}^{2})/4}+\beta_{1}a+\beta_{2}de^{-(Re\Sigma\phi_{j}^{2})/4}+\beta_{2}c}{\beta_{1}ae^{-(Re\Sigma
z_{j}^{2})/4}-\beta_{1}b+\beta_{2}ce^{-(Re\Sigma\phi_{j}^{2})/4}-\beta_{2}d}$

where $\beta_{2}$ corresponds to the chart containing $p_{1}$ and $\beta_{1}$
corresponds to the chart adjacent to it and $\phi=(\phi_{1},...,\phi_{n})$ is
the transition function. Obviously $c=d>0$ but the choice of $a,b$ will be
made clear shortly between the two options given by 4.9. Now we proceed as in
the proof of 5.1.

$\begin{array}[]{rcl}\tan\theta&=&\frac{\beta_{1}be^{-(Re\Sigma
z_{j}^{2})/4}+\beta_{1}a+\beta_{2}de^{-(Re\Sigma\phi_{j}^{2})/4}+\beta_{2}c}{\beta_{1}ae^{-(Re\Sigma
z_{j}^{2})/4}-\beta_{1}b+\beta_{2}ce^{-(Re\Sigma\phi_{j}^{2})/4}-\beta_{2}d}=\frac{A}{B}\\\
\Rightarrow\sec^{2}\theta d\theta&=&(1/B)[(1/2)b\beta_{1}e^{-(Re\Sigma
z_{j}^{2})/4}\Sigma_{j}(y_{j}dy_{j}-x_{j}dx_{j})\\\ &&+be^{-(Re\Sigma
z_{j}^{2})/4}\Sigma_{j}(\partial_{x_{j}}\beta_{1}dx_{j}+\partial_{y_{j}}\beta_{1}dy_{j})\\\
&&+\Sigma_{j}\partial_{x_{j}}(\beta_{2}de^{-(Re\Sigma\phi_{j}^{2})/4})dx_{j}+\Sigma_{j}\partial_{y_{j}}(\beta_{2}de^{-(Re\Sigma\phi_{j}^{2})/4})dy_{j}\\\
&&+\Sigma_{j}\partial_{x_{j}}(\beta_{2}c+\beta_{1}a)dx_{j}+\Sigma_{j}\partial_{y_{j}}(\beta_{2}c+\beta_{1}a)dy_{j}]\\\
&&-(A/B^{2})[(1/2)a\beta_{1}e^{-(Re\Sigma
z_{j}^{2})/4}\Sigma_{j}(y_{j}dy_{j}-x_{j}dx_{j})\\\ &&+ae^{-(Re\Sigma
z_{j}^{2})/4}\Sigma_{j}(\partial_{x_{j}}\beta_{1}dx_{j}+\partial_{y_{j}}\beta_{1}dy_{j})\\\
&&+\Sigma_{j}\partial_{x_{j}}(\beta_{2}ce^{-(Re\Sigma\phi_{j}^{2})/4})dx_{j}+\Sigma_{j}\partial_{y_{j}}(\beta_{2}ce^{-(Re\Sigma\phi_{j}^{2})/4})dy_{j}\\\
&&-\Sigma_{j}\partial_{x_{j}}(\beta_{1}b+\beta_{2}d)dx_{j}-\Sigma_{j}\partial_{y_{j}}(\beta_{1}b+\beta_{2}d)dy_{j}]\\\
\Rightarrow d\theta&=&\frac{B}{A^{2}+B^{2}}[(1/2)b\beta_{1}e^{-(Re\Sigma
z_{j}^{2})/4}\Sigma_{j}(y_{j}dy_{j}-x_{j}dx_{j})\\\ &&+be^{-(Re\Sigma
z_{j}^{2})/4}\Sigma_{j}(\partial_{x_{j}}\beta_{1}dx_{j}+\partial_{y_{j}}\beta_{1}dy_{j})\\\
&&+\Sigma_{j}\partial_{x_{j}}(\beta_{2}de^{-(Re\Sigma\phi_{j}^{2})/4})dx_{j}+\Sigma_{j}\partial_{y_{j}}(\beta_{2}de^{-(Re\Sigma\phi_{j}^{2})/4})dy_{j}\\\
&&+\Sigma_{j}\partial_{x_{j}}(\beta_{2}c+\beta_{1}a)dx_{j}+\Sigma_{j}\partial_{y_{j}}(\beta_{2}c+\beta_{1}a)dy_{j}]\\\
&&-\frac{A}{A^{2}+B^{2}}[(1/2)a\beta_{1}e^{-(Re\Sigma
z_{j}^{2})/4}\Sigma_{j}(y_{j}dy_{j}-x_{j}dx_{j})\\\ &&+ae^{-(Re\Sigma
z_{j}^{2})/4}\Sigma_{j}(\partial_{x_{j}}\beta_{1}dx_{j}+\partial_{y_{j}}\beta_{1}dy_{j})\\\
&&+\Sigma_{j}\partial_{x_{j}}(\beta_{2}ce^{-(Re\Sigma\phi_{j}^{2})/4})dx_{j}+\Sigma_{j}\partial_{y_{j}}(\beta_{2}ce^{-(Re\Sigma\phi_{j}^{2})/4})dy_{j}\\\
&&-\Sigma_{j}\partial_{x_{j}}(\beta_{1}b+\beta_{2}d)dx_{j}-\Sigma_{j}\partial_{y_{j}}(\beta_{1}b+\beta_{2}d)dy_{j}]\\\
\end{array}$

Again as in 5.1 let
$X_{\theta}=\Sigma_{j}(a_{j}\partial_{x_{j}}+b_{j}\partial_{y_{j}})$. Then
solving the Hamiltonian equation we get

$\begin{array}[]{rcl}a_{j}&=&\frac{B}{A^{2}+B^{2}}[(1/2)by_{j}\beta_{1}e^{-(Re\Sigma
z_{j}^{2})/4}+be^{-(Re\Sigma z_{j}^{2})/4}\partial_{y_{j}}\beta_{1}\\\
&&+d\partial_{y_{j}}(\beta_{2}e^{-(Re\Sigma\phi_{j}^{2})/4})+\partial_{y_{j}}(\beta_{2}c+\beta_{1}a)]\\\
&&-\frac{A}{A^{2}+B^{2}}[(1/2)ay_{j}\beta_{1}e^{-(Re\Sigma
z_{j}^{2})/4}+ae^{-(Re\Sigma z_{j}^{2})/4}\partial_{y_{j}}\beta_{1}\\\
&&+c\partial_{y_{j}}(\beta_{2}e^{-(Re\Sigma\phi_{j}^{2})/4})-\partial_{y_{j}}(\beta_{2}d+\beta_{1}b)]\\\
&=&\frac{y_{j}\beta_{1}e^{-(Re\Sigma z_{j}^{2})/4}}{2(A^{2}+B^{2})}[Bb-
Aa]+\frac{e^{-(Re\Sigma z_{j}^{2})/4}}{A^{2}+B^{2}}[Bb-
Aa]\partial_{y_{j}}\beta_{1}\\\ &&+\frac{Bd-
Ac}{A^{2}+B^{2}}\partial_{y_{j}}(\beta_{2}e^{-(Re\Sigma\phi_{j}^{2})/4})+\frac{Bc+Ad}{A^{2}+B^{2}}\partial_{y_{j}}\beta_{2}\\\
&&+\frac{aB+bA}{A^{2}+B^{2}}\partial_{y_{j}}\beta_{1}\\\ \end{array}$

Similarly we get $b_{j}$. Now let
$(x(t),y(t))=(x_{1}(t),y_{1}(t),...,x_{n}(t),y_{n}(t))$ be the flow of
$X_{\theta}$. So

$\partial_{t}x_{j}(t)=a_{j}(x(t),y(t))\ and\
\partial_{t}y_{j}(t)=b_{j}(x(t),y(t))$

Multiplying the first equation $\partial_{t}x_{j}(t)=a_{j}(x(t),y(t))$ by
$x_{j}(t)$ and the second equation $\partial_{t}y_{j}(t)=b_{j}(x(t),y(t))$ by
$y_{j}(t)$ and integrating we get (suppressing the $t$ all along below for
convenient notations)

$\begin{array}[]{rcl}\frac{1}{2}(x_{j}^{2}-y_{j}^{2})&=&\int\frac{bB-
aA}{A^{2}+B^{2}}e^{-(Re\Sigma
z_{j}^{2})/4}(x_{j}\partial_{y_{j}}-y_{j}\partial_{x_{j}})(\beta_{1})dt\\\
&&+\int\frac{Bd-
Ac}{A^{2}+B^{2}}(x_{j}\partial_{y_{j}}-y_{j}\partial_{x_{j}})(\beta_{2}e^{-(Re\Sigma\phi_{j}^{2})/4})dt\\\
&&+\int\frac{Bc+Ad}{A^{2}+B^{2}}(x_{j}\partial_{y_{j}}-y_{j}\partial_{x_{j}})(\beta_{2})dt\\\
&&+\int\frac{Ba+Ab}{A^{2}+B^{2}}(x_{j}\partial_{y_{j}}-y_{j}\partial_{x_{j}})(\beta_{1})dt\\\
\end{array}$

First we simplify the integrands. Consider the first term. So

$Bb-Aa=-\beta_{1}(a^{2}+b^{2})+\beta_{2}(bc-
ad)e^{-(Re\Sigma\phi_{j}^{2})/4}-\beta_{2}(bd+ac)$

Now consider the forth term. So

$aB+bA=\beta_{1}(a^{2}+b^{2})e^{-(Re\Sigma
z_{j}^{2})/4}+\beta_{2}(ac+bd)e^{-(Re\Sigma\phi_{j}^{2})/4}+\beta_{2}(bc-ad)$

Hence

$\begin{array}[]{rcl}(Bb-Aa)e^{-(Re\Sigma
z_{j}^{2})/4}+(aB+bA)&=&\beta_{2}(bc-
ad)(e^{-(Re\Sigma\phi_{j}^{2})/4}e^{-(Re\Sigma z_{j}^{2})/4}+1)\\\
&&+\beta_{2}(ac+bd)(e^{-(Re\Sigma\phi_{j}^{2})/4}-e^{-(Re\Sigma
z_{j}^{2})/4})\\\ &=&2bc\beta_{2}(e^{-(Re\Sigma\phi_{j}^{2})/4}e^{-(Re\Sigma
z_{j}^{2})/4}+1)\end{array}$

The last line achieved by putting $c=d>0$ and $b=-a>0$. Similarly considering
the second term we get

$\begin{array}[]{rcl}Bd-Ac&=&\beta_{1}(ad-bc)e^{-(Re\Sigma
z_{j}^{2})/4}-\beta_{1}(bd+ac)-\beta_{2}(c^{2}+d^{2})\\\
&=&-2bc\beta_{1}e^{-(Re\Sigma z_{j}^{2})/4}-2\beta_{2}c^{2}\end{array}$

Again the last line is achieved by putting $c=d>0$ and $b=-a>0$. Now consider
the third term.

$\begin{array}[]{rcl}Bc+Ad&=&\beta_{1}(ac+bd)e^{-(Re\Sigma
z_{j}^{2})/4}-\beta_{1}(bc-
ad)+\beta_{2}(c^{2}+d^{2})e^{-(Re\Sigma\phi_{j}^{2})/4}\\\
&=&-2bc\beta_{1}+2\beta_{2}c^{2}e^{-(Re\Sigma\phi_{j}^{2})/4},\ (putting\
c=d>0,\ b=-a>0)\\\ \end{array}$

Observe that inside the chart, i.e, when $\beta_{2}=0$ and
$\beta_{1}=constant\neq 0$, all the integrands in the equation of the flow
vanishes. So the equation becomes $x_{j}^{2}-y_{j}^{2}=constant$. We shall now
make the constant on the right side in the resulting equation of the flow
negative for large values of $k$ and for suitable translation of the origin
$0$ in this chart.

Stable Manifold$P$$Q$$\cdot\ o^{\prime}$$\cdot\ o^{\prime}$

.
In the above picture first we consider the right side cube, i.e, $P$. The
integrands are non-zero only on the shaded region in the picture. So we try to
determine the signs of the integrands in this region.

Obviously $A^{2}+B^{2}\geq 0$. So consider the sum of first and forth term as
above, i.e, the equation

$\begin{array}[]{rcl}(Bb-Aa)e^{-(Re\Sigma
z_{j}^{2})/4}+(aB+bA)&=&2bc\beta_{2}(e^{-(Re\Sigma\phi_{j}^{2})/4}e^{-(Re\Sigma
z_{j}^{2})/4}+1)\end{array}$

Obviously it is bigger than or equal to zero. So let us determine the sign of

$(x_{j}\partial_{y_{j}}-y_{j}\partial_{x_{j}})(\beta_{1})$

Now observe that for large $k$ the range of $|x_{j}|$ is very small. So the
sign of $-y_{j}\partial_{x_{j}}(\beta_{1})$ determines the sign if we
translate the origin as shown in the picture above. Observe that on the shaded
region $\beta_{1}$ with respect to $x_{j}$ is increasing. Hence
$\partial_{x_{j}}\beta_{1}\geq 0$. So $-y_{j}\partial_{x_{j}}(\beta_{1})<0$ as
on the intersection of the stable manifold and the shaded region, $y_{j}>0$.
So this integrand is negative on the shaded region.

Now consider the second and the third integrands. The second integrand
contains a factor

$(x_{j}\partial_{y_{j}}-y_{j}\partial_{x_{j}})(\beta_{2}e^{-(Re\Sigma\phi_{j}^{2})/4})=e^{-(Re\Sigma\phi_{j}^{2})/4}(x_{j}\partial_{y_{j}}-y_{j}\partial_{x_{j}})(\beta_{2})+\beta_{2}(x_{j}\partial_{y_{j}}-y_{j}\partial_{x_{j}})(e^{-(Re\Sigma\phi_{j}^{2})/4})$

Observe that the third integrand contains a factor
$(x_{j}\partial_{y_{j}}-y_{j}\partial_{x_{j}})(\beta_{2})$. So we compute

$(Bd-
Ac)(e^{-(Re\Sigma\phi_{j}^{2})/4})+(Bc+Ad)=-2bc\beta_{1}(e^{-(Re\Sigma\phi_{j}^{2})/4}e^{-(Re\Sigma
z_{j}^{2})/4}+1)<0$

Now as above the sign is determined by $-y_{j}\partial_{x_{j}}\beta_{2}$. But
this time we have $\beta_{2}$ instead of $\beta_{1}$, which is decreasing on
the shaded region. So $-y_{j}\partial_{x_{j}}\beta_{2}>0$. Hence the sign is
negative.

But we are still left with one term, namely the one containing the factor

$\beta_{2}(x_{j}\partial_{y_{j}}-y_{j}\partial_{x_{j}})(e^{-(Re\Sigma\phi_{j}^{2})/4})$

from the second integrand. Now we know that $Bd-Ac<0$. As
$e^{-(Re\Sigma\phi_{j}^{2})/4}$ is decreasing with respect to $x_{j}$, so if
we take $k$ large and shift the origin in the $y_{j}$ component down enough we
can make this integrand negative and hence we are done with the cube $P$.

We can do similar thing for the cube $Q$ and the global stable manifold is
achieved by observing that in the above argument the role of $(a,b)$ and
$(c,d)$ can be interchanged.

Observe that this stable manifold is diffeomorphic to the regular Lagrangian
$L\subset W$. We call the stable manifold $L^{\prime}$.

Now we shall show that on a tubular neighborhood of the stable manifold and
away from the singularity $p_{1}\in L$, the fibers of $s$ are symplectic sub-
manifolds, i.e $s$ induce a symplectic foliation.

So one needs to figure out $ker(ds)$. On the interior of a cube intersecting
$L$, more precisely on the set $\beta=constant\neq 0$, we compute $ker(ds)$.

$\bar{\partial}[(ae^{-Re(\Sigma z_{j}^{2})/4}-b)+i(be^{-Re(\Sigma
z_{j}^{2})/4}+a)]$ $=\frac{e^{-Re(\Sigma
z_{j}^{2})/4}}{4}\Sigma_{j}[-(ax_{j}+by_{j})dx_{j}+(-bx_{j}+ay_{j})dy_{j}+i(-bx_{j}+ay_{j})dx_{j}+i(ax_{j}+by_{j})dy_{j}]$

$\partial[(ae^{-Re(\Sigma z_{j}^{2})/4}-b)+i(be^{-Re(\Sigma z_{j}^{2})/4}+a)]$
$=\frac{e^{-Re(\Sigma
z_{j}^{2})/4}}{4}\Sigma_{j}[(-ax_{j}+by_{j})dx_{j}+(bx_{j}+ay_{j})dy_{j}-i(bx_{j}+ay_{j})dx_{j}+i(-ax_{j}+by_{j})dy_{j}]$

$A[(ae^{-Re(\Sigma z_{j}^{2})/4}-b)+i(be^{-Re(\Sigma z_{j}^{2})/4}+a)]$
$=2\Sigma_{j}[(x_{j}dy_{j}-y_{j}dx_{j})(be^{-Re(\Sigma
z_{j}^{2})/4}+a)-i(x_{j}dy_{j}-y_{j}dx_{j})(ae^{-Re(\Sigma z_{j}^{2})/4}-b)]$

Combining together we get

$d_{A}s=\Sigma_{j}[\\{-\frac{e^{-Re(\Sigma
z_{j}^{2})/4}}{2}ax_{j}-2y_{j}(a+be^{-Re(\Sigma z_{j}^{2})/4})\\}dx_{j}$
$+\\{\frac{e^{-Re(\Sigma z_{j}^{2})/4}}{2}ay_{j}+2x_{j}(a+be^{-Re(\Sigma
z_{j}^{2})/4})\\}dy_{j}]$ $+i\Sigma_{j}[\\{-\frac{e^{-Re(\Sigma
z_{j}^{2})/4}}{2}bx_{j}+2y_{j}(-b+ae^{-Re(\Sigma z_{j}^{2})/4})\\}dx_{j}$
$+\\{\frac{e^{-Re(\Sigma z_{j}^{2})/4}}{2}by_{j}-2x_{j}(-b+ae^{-Re(\Sigma
z_{j}^{2})/4})\\}dy_{j}]$

$ker(d_{A}s)$ is given by the equation $BZ=0$, where $B$ is the matrix

$\left(\begin{array}[]{ccccc}u_{11}&v_{11}&...&u_{1n}&v_{1n}\\\
u_{21}&v_{21}&...&u_{2n}&v_{2n}\\\ \end{array}\right)$

where $*=2e^{Re(\Sigma z_{j}^{2})/4}(be^{-Re(\Sigma z_{j}^{2})/4}+a)$,
$\tilde{*}=2e^{Re(\Sigma z_{j}^{2})/4}(ae^{-Re(\Sigma z_{j}^{2})/4}-b)$,
$u_{1j}=-ax^{\prime}_{j}-2y^{\prime}_{j}*$,
$v_{1j}=ay^{\prime}_{j}+2x^{\prime}_{j}*$,
$u_{2j}=-bx^{\prime}_{j}+2y^{\prime}_{j}\tilde{*}$,
$v_{2j}=by^{\prime}_{j}-2x^{\prime}_{j}\tilde{*}$ and $Z=(r_{1}\ s_{1}\ ...\
r_{n}\ s_{n})^{T}$, $x^{\prime}_{j}=\frac{e^{-Re(\Sigma
z_{j}^{2})/4}}{2}x_{j}$ and $y^{\prime}_{j}=\frac{e^{-Re(\Sigma
z_{j}^{2})/4}}{2}y_{j}$. By making $r_{2},s_{2},...,r_{n},s_{n}$ arbitrary we
get the equation

$\begin{array}[]{rcl}-r_{1}(ax^{\prime}_{1}+2y^{\prime}_{1}*)+s_{1}(ay^{\prime}_{1}+2x^{\prime}_{1}*)&=&\bigstar_{1}\\\
-r_{1}(bx^{\prime}_{1}-2y^{\prime}_{1}\tilde{*})+s_{1}(by^{\prime}_{1}-2x^{\prime}_{1}\tilde{*})&=&\bigstar_{2}\\\
\end{array}$

Where $\bigstar_{1}=\Sigma_{j=2}^{n}(u_{1j}r_{j}+v_{1,j}s_{j})$ and
$\bigstar_{2}=\Sigma_{j=2}^{n}(u_{2j}r_{j}+v_{2,j}s_{j})$. The coefficient
matrix of this system of equation is given by

$E=\left(\begin{array}[]{cc}-(ax^{\prime}_{1}+2y^{\prime}_{1}*)&(ay^{\prime}_{1}+2x^{\prime}_{1}*)\\\
-(bx^{\prime}_{1}-2y^{\prime}_{1}\tilde{*})&(by^{\prime}_{1}-2x^{\prime}_{1}\tilde{*})\\\
\end{array}\right)$

Whose determinant is given by
$2({x^{\prime}_{1}}^{2}-{y^{\prime}_{1}}^{2})(b*+a\tilde{*})$. So when
$x^{\prime}_{1}\neq\pm y^{\prime}_{1}$ we can solve this system of equations
and the solution space is of dimension $2n-2$, which means outside the cube
containing the singularity $p_{1}$ we can solve this system of equation on the
tubular neighborhood of the stable manifold $L^{\prime}$, provided the tubular
neighborhood is thin enough.

Now we compute $r_{1}$ and $s_{1}$ by assuming $n=4$, i.e, besides
$r_{1},s_{1}$ there are only $r_{2}$ and $s_{2}$, the general case is similar.
We get

$r_{1}=\frac{1}{det(E)}[-abx_{2}y_{1}r_{2}-aby_{1}y_{2}s_{2}+bx_{1}x_{2}r_{2}-bx_{1}y_{2}s_{2}+M]$

and

$s_{1}=\frac{1}{det(E)}[a^{2}y_{1}x_{2}r_{2}+a^{2}y_{1}y_{2}s_{2}-abx_{1}x_{2}r_{2}-a^{2}x_{1}y_{2}s_{2}+N]$

where $M$ and $N$ are terms which contains $*$ and (or, both) $\tilde{*}$ as
factors. Observe that $*$ and $\tilde{*}$ are very small if $a,b$ are small
which is the case.

Observe that

$\begin{array}[]{rcl}\omega((r_{1},s_{1},r_{2},s_{2}),(r^{\prime}_{1},s^{\prime}_{1},r^{\prime}_{2},s^{\prime}_{2}))&=&dx_{1}\wedge
dy_{1}((r_{1},s_{1}),(r^{\prime}_{1},s^{\prime}_{1}))+dx_{2}\wedge
dy_{2}((r_{2},s_{2}),r^{\prime}_{2},s^{\prime}_{2})\\\
&=&(r_{2}s^{\prime}_{2}-r^{\prime}_{2}s_{2})[1+\\{(a^{3}b+2a^{2}b-a^{2}b^{2})x_{1}x_{2}y_{1}y_{2}\\\
&&-(a^{2}b+ab^{2})x_{1}^{2}x_{2}y_{2}\\}+F(*,\tilde{*},a,b,x_{j},y_{j})]\end{array}$

So we see that if $k$ is large then $x$ is very small and $|y_{j}|<2$ (recall
the translations) on the tubular neighborhood of the stable manifold
$L^{\prime}$ and hence $ker(d_{A}s)$ restricted to this tubular neighborhood
is symplectic. As we have observed that on the tubular neighborhood of
$L^{\prime}$, both $r_{1}$ and $s_{1}$ are very small in comparison to $r_{2}$
and $s_{2}$ for high values of $k$, hence $ker(d_{A}s)$ is small perturbation
of the space generated by $r_{2}$ and $s_{2}$ which is the $x_{2}y_{2}$-space.
So its symplectic complement $\perp_{\omega}$ is a small perturbation of the
$x_{1}y_{1}$-plane but the morse function $\phi$ of the Weinstein structure in
this coordinates is given by $\Sigma_{j}y_{j}^{2}$ and hence any vector field
$X$ in the symplectic complement of $ker(ds)$ will have the property

$|X\phi|<\infty$

So parallel transport along a smooth path in the base $\mathbb{C}$ gives rise
to a diffeomorphism between corresponding fibers as in Lemma 6.5 of [4].

Now according to 2.1, the diffeomorphism $L\to L^{\prime}$ can be extended to
an isosymplectic immersion $\chi:N(L)\to N(L^{\prime})$. Note that
$L^{\prime}$ was already contained in a slightly larger tubular neighborhood
of $L$, say $\tilde{N}(L)\supset N(L)$. Now using one-parametric version of
3.1 for $A=L$ and using the fact that $Sp_{2n}$ path-connected, we can
construct a symplectomorphism $\Xi:W\to W$, which is equal to $\chi$ on $N(L)$
and is equal to $id$(identity) outside of $\tilde{N}(L)$ and which sends $L\to
L^{\prime}$.

Define $W_{0}=(s\circ\Xi)^{-1}(0)$ and the Weinstein structure is the pull-
back Weinstein structure by this symplectomorphism $\Xi$. Let us now define
the Lagrangian sphere required in 1.2. Under parallel transport along the
radial paths in $D^{2}\subset\mathbb{C}$ with respect to the symplectic
connection induced by $\omega$, the critical points of $s$ sweeps out a
Lagrangian disc called Lefschetz thimble ([8]) which is the stable manifold of
the Hamiltonian vector field $X_{Im\log s}$. The fiber over $0$ of a Lefschetz
thimble is an exact Lagrangian sphere called the vanishing cycle. Let
$L^{\prime}_{1}$ be the vanishing cycle corresponding to the singularity
$p_{1}$ and let $\Xi^{-1}(L^{\prime}_{1})=L_{1}\subset W_{0}$. Then the
required Weinstein Lefschetz fibration is
$((W_{0},\lambda_{0},\phi_{0});L_{1})$.

## References

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