$\int_{\partial_{j}X(v_{\beta})}\beta\wedge(d\beta)^{\frac{2n-j}{2}}=\int_{\partial_{j-1}^{+}X(v_{\beta})}d\big{(}\beta\wedge(d\beta)^{\frac{2n-j}{2}}\big{)}=\int_{\partial_{j-1}^{+}X(v_{\beta})}(d\beta)^{\frac{2n+2-j}{2}}.$

Conjecturally, the last integral is positive (see Conjecture 5.1 and Examples
5.1-5.2).

Motivated by these observations, we introduce numerical measures of the
stratified concavity/convexity of the Reeb flows $v_{\beta}$ relative to the
boundary $\partial_{1}X$:

$\displaystyle\kappa_{j}^{+}(\beta)$ $\displaystyle=_{\mathsf{def}}\;$
$\displaystyle\int_{\partial_{j}^{+}X(v_{\beta})}(d\beta)^{\frac{2n+1-j}{2}}\;\;\;(=\kappa_{j+1}(\beta)),$
(7.5) $\displaystyle\kappa_{j}(\beta)$ $\displaystyle=\;0$
$\displaystyle\text{\; for an odd }j\in[1,2n-1].$
$\displaystyle\kappa_{j}^{+}(\beta)$ $\displaystyle=_{\mathsf{def}}\;$
$\displaystyle\int_{\partial_{j}^{+}X(v_{\beta})}\beta\wedge(d\beta)^{\frac{2n-j}{2}},$
(7.6) $\displaystyle\kappa_{j}(\beta)$ $\displaystyle=_{\mathsf{def}}\;$
$\displaystyle\int_{\partial_{j}X(v)}\beta\wedge(d\beta)^{\frac{2n-j}{2}}\;\;\;(=\kappa^{+}_{j-1}(\beta))\text{\;
for an even }j\in[2,2n],$

Of course, these quantities depend on the choice of the contact form $\beta$
that represents a given contact structure $\xi$ and whose Reeb vector field
$v_{\beta}$ is boundary generic. They measure how “wrinkled” $\partial_{1}X$
is in relation to the Reeb flow $v_{\beta}$. By the very definition of
$\kappa_{j}^{+}(\beta),\kappa_{j}(\beta)$, they belong to the category of
“boundary data”.

Note that the inequality (7.2) can be written as
$\kappa^{+}_{0}(\beta)\leq\mathsf{diam}_{\mathcal{R}}(\beta)\cdot\kappa^{+}_{1}(\beta)$,
and the inequality(7.3) as
$\kappa^{+}_{0}(\beta)\leq\mathsf{diam}_{\mathcal{R}}(\beta)\cdot\kappa_{2}(\beta)$,
where $\kappa^{+}_{0}(\beta)=_{\mathsf{def}}\int_{X}\beta\wedge(d\beta)^{n}$.

###  7.1. Invariance of the volumes $\\{\kappa_{j}(\beta)\\}$ under special
deformations of $\beta$

######  Definition 7.5.

Let $\mathcal{B}=\\{\beta_{t}\\}_{t\in[0,1]}$ be a family of contact forms on
a compact connected smooth $(2n+1)$-manifold $X$ with boundary.

$\bullet$ We say the family $\mathcal{B}$ is variationally boundary exact if

$\frac{\partial}{\partial t}\beta_{t}\Big{|}_{\partial_{1}X}=d\eta_{t}$

for a family of smooth functions $\eta_{t}:\partial_{1}X\to\mathbb{R}$.

$\bullet$ We say the family $\mathcal{B}$ is variationally boundary
$(2n+1-j)$-exact if, for some smooth family of $(2n-j)$-forms
$\\{\zeta_{t}\\}_{t}$ on $\partial_{1}X$,

(7.7) $\displaystyle\Big{\\{}\frac{\partial}{\partial
t}\Big{[}\beta_{t}\wedge(d\beta_{t})^{\frac{2n-j}{2}}\Big{]}\Big{\\}}\Big{|}_{\partial_{1}X}=d\zeta_{t},\text{
where }j\in[2,2n]\text{ is even, }$ (7.8)
$\displaystyle\Big{\\{}\frac{\partial}{\partial
t}\Big{[}(d\beta_{t})^{\frac{2n+1-j}{2}}\Big{]}\Big{\\}}\Big{|}_{\partial_{1}X}=d\zeta_{t},\text{
where }j\in[1,2n-1]\text{ is odd }.$

Occasionally, we compliment these requirements with the additional one:

(7.9)
$\displaystyle\zeta_{t}\big{|}_{\partial_{j+1}X(v_{\beta_{t}})}=0.\quad\quad\quad\quad\quad\quad\quad\quad\hfill\diamondsuit$

######  Lemma 7.2.

Any variationally boundary exact family $\mathcal{B}$ of contact forms is
variationally boundary $(2n+1-j)$-exact.

If the functions ${\eta_{t}}$ are such that
$\eta_{t}|_{\partial_{j+1}X(v_{t})}=0$, then
$\zeta_{t}|_{\partial_{j+1}X(v_{t})}=0$.

###### Proof.

Note that, for any $t$-family differential forms, their exterior differential
$d=_{\mathsf{def}}d_{t}$ commutes with the partial derivative operator
$\frac{\partial}{\partial t}$ and with the restriction of forms to
$\partial_{1}X$. Consider the derivative

$\frac{\partial}{\partial
t}\Big{[}(d\beta_{t})^{k}\Big{]}=k\;\frac{\partial}{\partial
t}\Big{(}d\beta_{t}\Big{)}\wedge\big{[}(d\beta_{t})^{k-1}\big{]}=k\;d\Big{(}\frac{\partial}{\partial
t}(\beta_{t})\Big{)}\wedge\big{[}(d\beta_{t})^{k-1}\big{]}.$

Let $\eta_{t}:\partial_{1}X\to\mathbb{R}$ be a smooth family of functions from
Definition 7.5. Replacing $\frac{\partial}{\partial
t}(\beta_{t})\big{|}_{\partial_{1}X}$ with $d\eta_{t}$ in these formulas and
using that $d(d\eta_{t})=0$, we see that $\frac{\partial}{\partial
t}\Big{[}(d\beta_{t})^{k}\Big{]}\Big{|}_{\partial_{1}X}=0$. Similarly,

$\frac{\partial}{\partial
t}\Big{[}\beta_{t}\wedge(d\beta_{t})^{k}\Big{]}=\frac{\partial}{\partial
t}\Big{[}\beta_{t}\Big{]}\wedge(d\beta_{t})^{k}+\beta_{t}\wedge\frac{\partial}{\partial
t}\Big{[}(d\beta_{t})^{k}\Big{]}$

Since $\frac{\partial}{\partial
t}\Big{[}(d\beta_{t})^{k}\Big{]}\Big{|}_{\partial_{1}X}=0$, this formula
collapses to $\frac{\partial}{\partial
t}\Big{[}\beta_{t}\Big{]}\wedge(d\beta_{t})^{k}\Big{|}_{\partial_{1}X}$.

Replacing $\frac{\partial}{\partial
t}\Big{[}\beta_{t}\Big{]}\Big{|}_{\partial_{1}X}$ with $d\eta_{t}$ once more,
we get

$\frac{\partial}{\partial
t}\Big{[}\beta_{t}\Big{]}\wedge(d\beta_{t})^{k}\Big{|}_{\partial_{1}X}=d\eta_{t}\wedge(d\beta_{t})^{k}\Big{|}_{\partial_{1}X}=d\Big{[}\eta_{t}\cdot(d\beta_{t})^{k}\Big{]}\Big{|}_{\partial_{1}X}=d\zeta_{t}.$

Thus, (7.7) is valid. If the functions ${\eta_{t}}$ are such that
$\eta_{t}|_{\partial_{j+1}X(v_{t})}=0$, then
$\zeta_{t}|_{\partial_{j+1}X(v_{t})}=0$; so the property (7.9) holds as well.
∎

######  Proposition 7.1.

Let $\mathcal{B}=\\{\beta_{t}\\}_{t\in[0,1]}$ be a variationally boundary
$(2n+1-j)$-exact family of contact forms on a compact connected smooth
$(2n+1)$-manifold $X$. Assume that their Reeb vector fields $v_{\beta_{t}}$
are boundary generic for $t=0,1$, and for almost all $t\in(0,1)$.

$\bullet$ Then the integral $\kappa_{j}(\beta_{t})$ ($j\equiv 0\mod 2$) from
(7.6) and $\kappa_{j-1}^{+}(\beta_{t})$ ($j\equiv 0\mod 2$) from (7) are
$t$-independent. If, in addition, the property (7.9) holds, then
$\kappa^{+}_{j}(\beta_{t})$ ($j\equiv 0\mod 2$) is also invariant within
$\mathcal{B}$.

$\bullet$ If $\mathcal{B}=\\{\beta_{t}\\}_{t\in[0,1]}$ is variationally
boundary exact family, then $\\{\kappa_{j}(\beta_{t})\\}$ are
$\mathcal{B}$-invariant for all even $j\in[2,2n]$. If in addition, (7.9) holds
for the function $\eta_{t}$, then $\kappa^{+}_{j}(\beta_{t})$ ($j\equiv 0\mod
2$) is $\mathcal{B}$-invariant.

###### Proof.

Let $\mathsf{d}$ be the exterior derivative operator on $X\times[0,1]$ and
$d_{t}$ the exterior derivative operator on a typical slice $X\times\\{t\\}$.
Let $v_{t}=_{\mathsf{def}}v_{\beta_{t}}$. Consider the following sets:

$\mathcal{Y}_{j}=_{\mathsf{def}}\bigcup_{t\in[0,1]}\partial_{j}X(v_{t})\times\\{t\\}\subset
X\times[0,1],$
$\mathcal{Y}_{j}^{+}=_{\mathsf{def}}\bigcup_{t\in[0,1]}\partial_{j}^{+}X(v_{t})\times\\{t\\}\subset
X\times[0,1],$

Although $\mathcal{Y}_{j}$ may have singularities, located in some $t$-slices,
by the hypothesis, such special $t$’s have measure zero in $[0,1]$; so one can
integrate a smooth differential form against the rectifiable chain
$\mathcal{Y}_{j}$.

We start with an even $j$. Put $s=\frac{2n-j}{2}$. Then

$\kappa_{j}(\beta_{1})-\kappa_{j}(\beta_{0})=_{\mathsf{def}}\int_{\partial_{j}X(v_{1})}\beta_{1}\wedge(d_{1}\beta_{1})^{s}-\int_{\partial_{j}X(v_{0})}\beta_{0}\wedge(d_{0}\beta_{0})^{s}\stackrel{{\scriptstyle
Stokes}}{{=}}\quad\int_{\mathcal{Y}_{j}}\mathsf{d}\big{(}\beta_{t}\wedge(d_{t}\beta_{t})^{s}\big{)}$
$=\int_{\mathcal{Y}_{j}}(d_{t}\beta_{t})^{s+1}\;+\int_{\mathcal{Y}_{j}}\frac{\partial}{\partial
t}\Big{[}\beta_{t}\wedge(d_{t}\beta_{t})^{s}\Big{]}\wedge dt$

Each of these two integrals vanishes. Indeed, $dt$ is not present in the
$(2s+2)$-form $(d_{t}\beta_{t})^{s+1}$. Thus its restriction to the
$(2s+2)$-chain $\mathcal{Y}_{j}\subset\partial_{1}X\times[0,1]$ vanishes. The
second integral is also zero by the variationally boundary $(2n+1-j)$-exact
hypothesis:

$\int_{\mathcal{Y}_{j}}\frac{\partial}{\partial
t}\big{(}\beta_{t}\wedge(d_{t}\beta_{t})^{s}\big{)}\wedge
dt\stackrel{{\scriptstyle
Fubini}}{{=}}\int_{[0,1]}\Big{(}\int_{\partial_{j}X(v_{\beta_{t}})}\frac{\partial}{\partial
t}\Big{[}\beta_{t}\wedge(d_{t}\beta_{t})^{s}\Big{]}\Big{)}dt$
$=\int_{[0,1]}\Big{(}\int_{\partial_{j}X(v_{\beta_{t}})}d_{t}\zeta_{t}\Big{)}dt=0,$

since integrating the exact form $d_{t}\zeta_{t}$ over the cycle
$\partial_{j}X(v_{\beta_{t}})\subset\partial_{1}X$ results in zero. Therefore,
$\kappa_{j}(\beta_{1})=\kappa_{j}(\beta_{0})$ for this particular even
$j\in[2,2n]$.

By Lemma 7.2, the second bullet’s claim follows for all _even_ $j\in[2,2n]$.

Using Stokes’ theorem identity $\kappa_{j}(\beta)=\kappa_{j-1}^{+}(\beta)$
($j\equiv 0\mod 2$), we conclude that, for _odd_ $j\in[1,2n-1]$, the
quantities $\\{\kappa_{j}^{+}(\beta)\\}_{j}$ are $\mathcal{B}$-deformation
invariants as well.

If we add property (7.9) to property (7.7), then a similar computation shows
that $\kappa^{+}_{j}(\beta_{1})=\kappa^{+}_{j}(\beta_{0})$ for an even $j$.
This time, Stokes’ theorem is applied to $\mathcal{Y}_{j}^{+}$. Its boundary
consists of tree parts $\partial_{j}^{+}X(v_{0}),\partial_{j}^{+}X(v_{1})$,
and their complement $\delta\mathcal{Y}_{j}^{+}=\mathcal{Y}_{j+1}$. Therefore,

$\kappa^{+}_{j}(\beta_{1})-\kappa^{+}_{j}(\beta_{0})=\int_{\mathcal{Y}^{+}_{j}}\mathsf{d}\big{(}\beta_{t}\wedge(d_{t}\beta_{t})^{s}\big{)}\pm\int_{\delta\mathcal{Y}^{+}_{j}}\beta_{t}\wedge(d_{t}\beta_{t})^{s}$
$=\int_{\mathcal{Y}^{+}_{j}}(d_{t}\beta_{t})^{s+1}\;+\int_{\mathcal{Y}^{+}_{j}}\frac{\partial}{\partial
t}\Big{[}\beta_{t}\wedge(d_{t}\beta_{t})^{s}\Big{]}\wedge
dt\;\pm\;\int_{\delta\mathcal{Y}^{+}_{j}}\beta_{t}\wedge(d_{t}\beta_{t})^{s}.$

Since $dt$ is not present in the first and third integrals, they vanish. As a
result,

$\kappa^{+}_{j}(\beta_{1})-\kappa^{+}_{j}(\beta_{0})=\int_{[0,1]}\Big{(}\int_{\partial^{+}_{j}X(v_{\beta_{t}})}\frac{\partial}{\partial
t}\Big{[}\beta_{t}\wedge(d_{t}\beta_{t})^{s}\Big{]}\Big{)}dt$
$=\int_{[0,1]}\Big{(}\int_{\partial^{+}_{j}X(v_{\beta_{t}})}d_{t}\zeta_{t}\Big{)}dt\;\stackrel{{\scriptstyle
Stokes}}{{=}}\;\int_{[0,1]}\Big{(}\int_{\partial_{j+1}X(v_{\beta_{t}})}\zeta_{t}\Big{)}dt=0$

by (7.9). ∎

Consider now the following quantities (see Definition 7.3):

(7.10) $\displaystyle\mathcal{K}_{j}^{+}(\xi)$
$\displaystyle=_{\mathsf{def}}\;\inf_{\\{\beta\leadsto\xi\;|\;\mathsf{av}_{\mathcal{R}}(\beta)=1\\}}\big{\\{}|\kappa_{j}^{+}(\beta)|\big{\\}}$
$\displaystyle\text{\; for an odd }j\in[1,2n-1],$
$\displaystyle\mathcal{K}_{j}^{+}(\xi)$
$\displaystyle=_{\mathsf{def}}\;\inf_{\\{\beta\leadsto\xi\;|\;\mathsf{av}_{\mathcal{R}}(\beta)=1\\}}\big{\\{}|\kappa_{j}^{+}(\beta)|\big{\\}}$
$\displaystyle\text{\; for an even }j\in[2,2n],$ (7.11)
$\displaystyle\mathcal{K}_{j}(\xi)$
$\displaystyle=_{\mathsf{def}}\;\inf_{\\{\beta\leadsto\xi\;|\;\mathsf{av}_{\mathcal{R}}(\beta)=1\\}}\big{\\{}|\kappa_{j}(\beta)|\big{\\}}$
$\displaystyle\text{\; for an even }j\in[2,2n].$

They measure how “wrinkled” the Morse stratification
$\\{\partial_{j}^{\pm}X(v_{\beta})\\}_{j}$ could be with respect to a given
contact structure $\xi=\ker\beta$ on $X$. As a practical matter, the
computation of
$\\{\mathcal{K}_{j}^{+}(\xi)\\}_{j},\\{\mathcal{K}_{j}(\xi)\\}_{j}$ could be
quite challenging…

######  Proposition 7.2.

Let $\beta$ be a contact form on a compact manifold $X$ and let the function
$h:X\to\mathbb{R}$ be as in (4).

$\bullet$ Any contact vector field $u=hv_{\beta}+w$, such that its component
$w$ satisfies equation (5.2), integrates to the $1$-parameter family of
contactomorphisms $\Phi^{t}(u):X\to X$ that preserve all the forms
$\\{(d\beta)^{k}|_{\partial_{2}X(v_{\beta})}\\}_{k}$ and
$\\{\beta\wedge(d\beta)^{k}|_{\partial_{2}X(v_{\beta})}\\}_{k}$, the manifold
$X$, and the Morse stratification $\\{\partial_{j}^{\pm}X(v_{\beta})\\}_{j}$
of its boundary.

$\bullet$ As a result, for an odd $j\in[3,2n-1]$, the form
$(d\beta)^{\frac{2n+1-j}{2}}\big{|}_{\partial_{j}X(v_{\beta})}$ is
$\Phi^{t}(u)$-invariant. Similarly, for an even $j\in[2,2n]$, the form
$\beta\wedge(d\beta)^{\frac{2n-j}{2}}\big{|}_{\partial_{j}X(v_{\beta})}$ is
$\Phi^{t}(u)$-invariant.

###### Proof.

Consider a contact vector field $u=hv_{\beta}+w$, such that its component $w$
satisfies equation (5.2). Then $\mathcal{L}_{u}\beta=dh(v_{\beta})\,\beta$.
Since $v_{\beta}$ is tangent to $\partial_{1}X$ along
$\partial_{2}X(v_{\beta})$, $dh(v_{\beta})|_{\partial_{2}X(v_{\beta})}\equiv
0$. Therefore, $\mathcal{L}_{u}\beta\,||_{\partial_{2}X(v_{\beta})}\equiv 0$.

On the other hand, by Corollary 6.1, the diffeomorphisms $\Phi^{t}(u)$ map
$v_{\beta}$-trajectories to $v_{\beta}$-trajectories. Therefore, $\Phi^{t}(u)$
preserves the Morse stratification $\\{\partial_{j}^{\pm}X(v_{\beta})\\}_{j}$.
In particular, $\partial_{2}X(v_{\beta})$ is invariant under $\Phi^{t}(u)$,
which is equivalent to $u$ being tangent to $\partial_{2}X(v_{\beta})$. Hence,
$\mathcal{L}_{u}\beta\,||_{\partial_{2}X(v_{\beta})}=0$ implies
$\beta|_{\partial_{2}X(v_{\beta})}$ being $\Phi^{t}(u)$-invariant. As a
result, $d\beta|_{\partial_{2}X(v_{\beta})}$ is $\Phi^{t}(u)$-invariant as
well. And so are all the forms
$\\{(d\beta)^{k}|_{\partial_{2}X(v_{\beta})}\\}_{k}$ and
$\\{\beta\wedge(d\beta)^{k}|_{\partial_{2}X(v_{\beta})}\\}_{k}$. Using the
invariance of the Morse stratification, we conclude that, for an odd
$j\in[3,2n-1]$, the form
$(d\beta)^{\frac{2n+1-j}{2}}\big{|}_{\partial_{j}X(v_{\beta})}$ is
$\Phi^{t}(u)$-invariant. Similarly, for an even $j\in[2,2n]$, the form
$\beta\wedge(d\beta)^{\frac{2n-j}{2}}\big{|}_{\partial_{j}X(v_{\beta})}$ is
$\Phi^{t}(u)$-invariant. ∎

Proposition 7.2 leads instantly to the following assertion.

######  Corollary 7.2.

Let $\beta$ be a contact form on a compact manifold $X$ of dimension $2n+1$
and let $h:X\to\mathbb{R}$ be as in (4). Consider a contact vector field
$u=hv_{\beta}+w$, such that its component $w$ satisfies the equation (5.2);
$u$ integrates to a $1$-parameter family of contactomorphisms
$\\{\Phi^{t}(u):X\to X\\}_{t}$.

$\bullet$ Pick an even $j\in[2,2n]$. Let $Z$ be a compact submanifold of
$\partial_{2}X(v_{\beta})$ of dimension $2n+1-j$. Then

$\int_{\Phi^{t}(u)(Z)}\beta\wedge(d\beta)^{\frac{2n-j}{2}}=\int_{Z}\beta\wedge(d\beta)^{\frac{2n-j}{2}}.$

$\bullet$ Pick an odd $j\in[3,2n-1]$. Let $Z$ be a compact submanifold of
$\partial_{2}X(v_{\beta})$ of dimension $2n+1-j$. Then

$\int_{\Phi^{t}(u)(Z)}(d\beta)^{\frac{2n+1-j}{2}}=\int_{Z}(d\beta)^{\frac{2n+1-j}{2}}.$

$\diamondsuit$

Let $Z$ be a space with a measure $\mu$. Recall that a measure-preserving
transformation $\Phi:(Z,\mu)\to(Z,\mu)$ has an infinite return property if,
for any set $U$ whose measure $\mu(U)>0$, there exists a subset $V\subset U$
such that $\mu(V)=\mu(U)$ and for each point $z\in V$, $\Phi^{k}(z)\cap
U\neq\emptyset$ for infinitely many $k$. The Poincaré Recurrence Theorem
claims that if the measure $\mu$ is finite, then this infinite return property
holds.

######  Corollary 7.3.

We adopt the hypotheses of Corollary 7.2. Then, for any
$\Phi^{t}(u)$-invariant compact manifold $Z\subset\partial_{2}X(v_{\beta})$
such that the forms $\beta\wedge(d\beta)^{\frac{2n-j}{2}}\big{|}_{Z}$
($j\equiv 0\mod 2$) or $(d\beta)^{\frac{2n+1-j}{2}}\big{|}_{Z}$ ($j\equiv
1\mod 2$) vanish on a set in $Z$ of the Lebesgue measure zero, the
diffeomorphism $\Phi^{t}(u):Z\to Z$ has the Poincaré infinite return property
with respect to the measure. $\diamondsuit$

Figure 2. By Theorem 7.3,
$\kappa_{1}^{+}(\beta_{X})=\kappa_{2}(\beta_{X})=-\int_{\mathcal{Y}_{1}^{+}}d\beta_{Y}$.
The figure illustrates the geometry of $3$-folds $X$ and $Y$ and of an
embedding $\Psi:X\to Y$ that leads to this formula. Although the relevant for
the formula loci are: the surface $\Psi(\partial_{1}^{+}X)$, the solid
$\mathcal{X}^{+}_{1}\subset Y$, and the surface
$\mathcal{Y}^{+}_{1}\subset\partial_{1}^{+}Y$, they are not shown in the
figure due to artistic limitations; instead, we have chosen to show similar
loci of one dimension lower: the curve $\Psi(\partial_{2}^{+}X)$, the surface
$\mathcal{X}^{+}_{2}$, and the curve $\mathcal{Y}^{+}_{2}$.

######  Theorem 7.3.

(Holography for the $\kappa_{j}^{+}$-volumes under the contact embeddings)

Let $X$ and $Y$ be compact smooth $(2n+1)$-manifolds, equipped with contact
forms $\beta_{X},\beta_{Y}$ and their Reeb vector fields $v_{X}$, $v_{Y}$,
respectively. We assume that $v_{X}$ and $v_{Y}$ are boundary generic and
$v_{Y}$ is traversing. Let $\Psi:X\hookrightarrow\mathsf{int}(Y)$ be a smooth
embedding such that $\beta_{X}=\Psi^{\ast}(\beta_{Y})$. For $j\in[1,2n+1]$,
consider the $(-v_{Y})$-guided projections
$\Pi_{Y}:\Psi\big{(}\partial_{j}^{+}X(v_{X})\big{)}\to\partial_{1}^{+}Y(v_{Y})$
and denote their images by $\mathcal{Y}^{+}_{j}$.555Here, as before, we assume
that $\Psi(X)$ is “semi-transparent” for the $(-v_{Y})$-flow.

Then the integrals $\\{\kappa_{j}^{+}(\beta_{X})\\}_{j}$ from (7) and
$\\{\kappa_{j}(\beta_{X})\\}_{j}$ from (7.6) can be recovered from the
$\Pi_{Y}$-projections $\\{\mathcal{Y}^{+}_{j}\\}_{j}$ and from the knowledge
of the contact form
$\beta_{Y}^{\partial+}=_{\mathsf{def}}\beta_{Y}||_{\partial_{1}^{+}Y(v_{Y})}$.
Specifically, for an odd $j\in[1,2n-1]$,

$\kappa_{j}^{+}(\beta_{X})=-\int_{\mathcal{Y}_{j}^{+}}(d\beta_{Y})^{\frac{2n+1-j}{2}}$

and, for an even $j\in[2,2n]$,

$\kappa_{j}(\beta_{X})=-\int_{\mathcal{Y}_{j-1}^{+}}(d\beta_{Y})^{\frac{2n+2-j}{2}}.$

###### Proof.

We consider the union $\mathcal{X}_{j}^{+}$ of downward $(-v_{Y})$-directed
trajectories through the points of $\Psi(\partial^{+}_{j}(X)(v_{X}))$. Note
that $\dim(\mathcal{X}_{j}^{+})=2n+2-j$. Let

$\mathcal{Y}_{j}^{+}=_{\mathsf{def}}\mathcal{X}_{j}^{+}\cap\partial_{1}^{+}Y(v_{Y})$

—the $\Pi_{Y}$-image of $\partial_{j}^{+}X(v_{X})$—, and let (see Fig. 2)

$\mathcal{Z}_{j}^{+}=_{\mathsf{def}}\partial(\mathcal{X}_{j}^{+})\setminus(\Psi\big{(}\partial_{j}^{+}X(v_{X})\big{)}\cup\mathcal{Y}_{j}^{+})$

.

We claim that, for an odd $j$,
$\int_{\mathcal{Z}_{j}^{+}}(d\beta_{Y})^{\frac{2n+1-j}{2}}=0$ since
$v_{Y}\in\ker(d\beta_{Y})$ and $\mathcal{Z}_{j}^{+}$ of dimension $2n+1-j$
consists of portions of $v_{Y}$-trajectories.

(a) For an odd $j\in[1,2n-1]$, we get

$\kappa_{j}^{+}(\beta_{X})=\int_{\partial^{+}_{j}X(v_{X})}\big{(}d\beta_{X}\big{)}^{\frac{2n+1-j}{2}}=$
$=\int_{\partial^{+}_{j}X(v_{X})}\Psi^{\ast}\big{(}d\beta_{Y}\big{)}^{\frac{2n+1-j}{2}}=\int_{\Psi\big{(}\partial^{+}_{j}(X)(v_{X})\big{)}}\big{(}d\beta_{Y}\big{)}^{\frac{2n+1-j}{2}}.$

By Stokes’ theorem, applied to $\mathcal{X}_{j}^{+}$, and using that
$d\big{(}(d\beta_{Y})^{\frac{2n+1-j}{2}}\big{)}=0$ and that
$\int_{\mathcal{Z}_{j}^{+}}(d\beta_{Y})^{\frac{2n+1-j}{2}}=0$, we get

$\kappa_{j}^{+}(\beta_{X})=\int_{\Psi\big{(}\partial^{+}_{j}(X)(v_{X})\big{)}}(d\beta_{Y})^{\frac{2n+1-j}{2}}=-\int_{\mathcal{Y}_{j}^{+}}(d\beta_{Y})^{\frac{2n+1-j}{2}}.$

This validates the claim of Theorem 7.3 for odd $j$.

(b) For an even $j\in[2,2n]$, by Stokes’ theorem, we get

$\kappa_{j}(\beta_{X})=\int_{\partial_{j}X(v_{X})}\beta_{X}\wedge(d\beta_{X})^{\frac{2n-j}{2}}=\int_{\partial_{j-1}^{+}X(v_{X})}(d\beta_{X})^{\frac{2n+2-j}{2}}$
$=\int_{\partial_{j-1}^{+}X(v_{X})}\Psi^{\ast}\big{(}(d\beta_{Y})^{\frac{2n+2-j}{2}}\big{)}=\int_{\Psi\big{(}\partial^{+}_{j-1}(X)(v_{X})\big{)}}(d\beta_{Y})^{\frac{2n+2-j}{2}}.$

Now we are facing the situation, described by the case (a). Therefore,

$\kappa_{j}(\beta_{X})=-\int_{\mathcal{Y}_{j-1}^{+}}(d\beta_{Y})^{\frac{2n+2-j}{2}},$

and thus is determined by $\mathcal{Y}_{j-1}^{+}$, the $\Pi_{Y}$-projection of
the locus $\Psi\big{(}\partial^{+}_{j-1}(X)(v_{X})\big{)}$. ∎

######  Example 7.3.

_Let $\dim X=\dim Y=5$. We adopt the notations and hypotheses of Theorem 7.3.
Then $\kappa_{1}(\beta_{X})=0=\kappa_{3}(\beta_{X})$ and_

$\kappa_{1}^{+}(\beta_{X})=-\int_{\mathcal{Y}_{1}^{+}}(d\beta_{Y})^{2}=\kappa_{2}(\beta_{X}),\quad\kappa_{3}^{+}(\beta_{X})=-\int_{\mathcal{Y}_{3}^{+}}d\beta_{Y}=\kappa_{4}(\beta_{X}).$

_In contrast, we do not know whether $\kappa^{+}_{2}(\beta_{X})$ and
$\kappa^{+}_{4}(\beta_{X})$ can be recovered in terms of the
$\Pi_{Y}$-generated boundary data residing on $\partial_{1}Y$._ $\diamondsuit$

Combining Proposition 7.1 with Theorem 7.3 leads to the following claim.

######  Corollary 7.4.

Let $X$ and $(Y,\beta_{Y})$ be as in Theorem 7.3. Consider a $t$-family
$\\{\Psi_{t}:X\hookrightarrow\mathsf{int}(Y)\\}_{t\in[0,1]}$ of orientation-
preserving contact embeddings such that, for an even $j\in[2,2n]$, the
$t$-family of forms
$\big{\\{}\Psi_{t}^{\ast}\big{(}\beta_{Y}\wedge(d\beta_{Y})^{\frac{2n-j}{2}}\big{)}\big{\\}}$
is variationally boundary $(2n+1-j)$-exact in the sense of Definition 7.5.

Then
$\int_{\mathcal{Y}_{j-1}^{+}(t)}(d\beta_{Y})^{\frac{2n+2-j}{2}}=-\kappa_{j}(\Psi_{t}^{\ast}\big{(}\beta_{Y}))=-\kappa^{+}_{j-1}(\Psi_{t}^{\ast}\big{(}\beta_{Y}))$
is $t$-independent. $\diamondsuit$

##  8\. Shadows of Legandrian submanifolds of $(X,\beta)$ on the screen
$\partial_{1}^{+}X(v_{\beta})$

Let $\dim X=2n+1$. We consider a cooriented contact structure $(X,\xi)$, its
contact form $\beta$, and its traversing (boundary generic) Reeb vector field
$v_{\beta}$. We denote by $\mathcal{I}(d\beta)$ the isotropic space of
$d\beta$, that is, the maximal vector subspace of $\ker(\beta)$ such that
$d\beta(w_{1},w_{2})=0$ for any
$w_{1},w_{2}\in\mathcal{I}(d\beta)\subset\ker(\beta)$. For a contact form
$\beta$, $\dim(\mathcal{I}(d\beta))=n$. By definition, $L\subset X$ is sub-
Legendrian manifold, if $d\beta|_{L}\equiv 0$; that is, if the tangent space
$TL\subset\mathcal{I}(d\beta)$.

Note that $\beta|_{L}$ is a closed $1$-form. Hence, one gets an element
$[\beta_{L}]\in H^{1}(L;\mathbb{R})$, provided $\partial L=\emptyset$. If the
boundary $\partial L\neq\emptyset$, we always assume that $\partial
L\subset\partial_{1}X$. In such a case, we getting an element $[\beta_{L}]\in
H^{1}(L,\partial L;\mathbb{R})$.

Our next goal is to describe the images $L^{\dagger}$ of sub-Legendrian
compact submanifolds $L\subset X$ ($\dim(L)\leq n$) both in
$\partial_{1}^{+}X(v_{\beta})$ under the $(-v_{\beta})$-flow and in the
trajectory space $\mathcal{T}(v_{\beta})$ under the obvious quotient map
$\Gamma:X\to\mathcal{T}(v_{\beta})$. Note that these shadows of $L$ do not
change when we apply a $t$-family ($t$ being sufficiently small) of local
diffeomorphisms $\phi^{t}$ (as in Definition 4.3), generated by the vector
field $v_{\beta}$. Since $\beta$ and $d\beta$ are invariant under $\phi^{t}$,
we conclude that $\phi^{t}(L)$ is also sub-Legendrian.

By Lemma 4.3, $\mathsf{int}(\partial_{1}^{+}X(v_{\beta}))$ is an open
symplectic manifold, with the symplectic form
$\omega^{\dagger}=_{\mathsf{def}}d\beta|_{\partial_{1}^{+}X(v_{\beta})}$.
Recall that the form $\omega^{\dagger}$ degenerates along the boundary
$\partial_{2}X(v_{\beta})=\partial(\partial_{1}^{+}X(v_{\beta}))$, since there
$v_{\beta}\in\ker(d\beta)\cap T(\partial_{1}X)$.

Let $\mathcal{X}_{2}(v_{\beta})$ denote the union of the downward
$(-v_{\beta})$-trajectories through the locus $\partial_{2}X(v_{\beta})$, and
$L(v_{\beta})$ the union of the downward $(-v_{\beta})$-trajectories through
the points of a sub-Legandrian submanifold $L\subset X$. Put
$L^{\dagger}=_{\mathsf{def}}L(v_{\beta})\cap\partial_{1}^{+}X(v_{\beta})$ (see
Fig. 3).

######  Definition 8.1.

We say that a sub-Legandrian manifold $L\subset X$ is in weakly general
position to the $v_{\beta}$-flow, if
$L\setminus(L\cap\mathcal{X}_{2}(v_{\beta}))$ is dense in $L$. $\diamondsuit$

Figure 3. The $(-v_{\beta})$-shadow
$L^{\dagger}\subset\partial_{1}^{+}X(v_{\beta})$ of a closed Legandrian
manifold $L\subset X$, shown as a bold loop in the interior of the bulk $X$.
In the figure, $L^{\dagger}$ consists of two arcs. The surface $L(v_{\beta})$
consists of $(-v_{\beta})$-trajectories falling down from $L$. Note the locus
$\delta L(v_{\beta})\subset X$ that consists of $(-v_{\beta})$-trajectories
falling down from $L^{\dagger}\cap\partial_{2}^{+}X(v_{\beta})$.

######  Lemma 8.1.

The locus
$\mathsf{int}(L^{\dagger})=L^{\dagger}\setminus(L^{\dagger}\cap\mathcal{X}_{2}(v_{\beta}))$
is an open (possibly disconnected) Lagrangian submanifold of the open
$(2n)$-manifold

$\partial_{1}^{+}X(v_{\beta})\setminus(\mathcal{X}_{2}(v_{\beta})\cap\partial_{1}^{+}X(v_{\beta})).$

That is, for any $x\in\mathsf{int}(L^{\dagger})$, the tangent space
$T_{x}(\mathsf{int}(L^{\dagger}))\subset\mathcal{I}_{x}\big{(}\omega^{\dagger}\big{)}$,
where $\mathcal{I}_{x}\big{(}\omega^{\dagger}\big{)}\subset
T_{x}(\partial_{1}^{+}X(v_{\beta}))$ is the isotropic subspace of the
symplectic form
$\omega^{\dagger}=d\beta|_{\mathsf{int}(\partial^{+}_{1}X(v_{\beta}))}$ at
$x$.

###### Proof.

By Lemma 4.3, the $2$-form
$\omega^{\dagger}=d\beta|_{\partial^{+}_{1}X(v_{\beta})}$ is symplectic on
$\mathsf{int}(\partial_{1}^{+}X(v_{\beta}))$. Since $d\beta$ is
$v_{\beta}$-invariant, under the differential of the $(-v_{\beta})$ projection
on $\partial_{1}^{+}X(v_{\beta})$, the space
$\mathcal{I}_{x}\big{(}\omega^{\dagger}\big{)}$ is the isomorphic image of the
space $\mathcal{I}_{x}\big{(}d\beta\big{)}$. Therefore,
$\mathsf{int}(L^{\dagger})$ is a Lagrangian submanifold of the open manifold
$\partial_{1}^{+}X(v_{\beta})\setminus(\mathcal{X}_{2}(v_{\beta})\cap\partial_{1}^{+}X(v_{\beta}))$.
∎

Remark 8.1. Since $L$ is embedded and $v_{\beta}$ is transversal to $\xi$,
$\mathsf{int}(L^{\dagger})$ is _immersed_ in the open $(2n)$-manifold
$\partial_{1}^{+}X(v_{\beta})\setminus(\mathcal{X}_{2}(v_{\beta})\cap\partial_{1}^{+}X(v_{\beta}))$.
Note that, for $n\geq 2$, at each point of self-intersection, the tangent
spaces of the two local branches of $\mathsf{int}(L^{\dagger})$ must be
tangent at their self-intersections. Indeed, assume that $L$ hits a
$v_{\beta}$-trajectory $\gamma$ at a pair of points $x\neq y$. Let
$z\in\gamma\cap\mathsf{int}(\partial_{1}^{+}X(v_{\beta}))$. Since the
$n$-distribution $\mathcal{I}_{x}\big{(}d\beta\big{)}\subset\ker(\beta)$ is
$v_{\beta}$-invariant, the Reeb flow takes the isotropic subspace
$\mathcal{I}_{y}(d\beta)$ to the isotropic subspace $\mathcal{I}_{x}(d\beta)$
and thus the flow-governed projections of both subspaces on
$T_{z}(\partial_{1}^{+}X(v_{\beta}))$ must coincide. $\diamondsuit$

######  Lemma 8.2.

Let $\gamma$ be a Reeb $v_{\beta}$-trajectory through a point $a_{\star}\in
X$, which hits $\partial_{1}^{+}X(v_{\beta})$ transversally at a point
$b_{\star}$. Let $L^{\dagger}$ be a given Lagrangian submanifold (with respect
to the symplectic form
$\omega^{\dagger}=_{\mathsf{def}}d\beta|_{\mathsf{int}(\partial^{+}_{1}X(v_{\beta}))}$)
of $\partial_{1}^{+}X(v_{\beta})$, which contains $b_{\star}$. Then, in the
vicinity of $a_{\star}$, there exists a unique germ of Legendrian submanilold
$L\subset X$ that contains $a_{\star}$ and whose $(-v_{\beta})$-governed
projection is the germ of $L^{\dagger}$ at $b_{\star}$.

###### Proof.

By the Darboux theorem, in the vicinity of $a_{\star}$, we can pick the
coordinates $(z,x_{1},y_{1},\hfill\break\;\ldots,\;x_{n},y_{n})$ such that
they vanish at $a_{\star}$, $\beta=dz+\sum_{i=1}^{n}x_{i}\,dy_{i}$,
$d\beta=\sum_{i=1}^{n}dx_{i}\wedge dy_{i}$, and $v_{\beta}=\partial_{z}$. In
these coordinates, the germ at $a_{\star}$ of the Legendrian submanilold $L$
in question is given, in its parametric form, by some smooth functions
$z(\vec{t}),\,x_{1}(\vec{t}),\,y_{1}(\vec{t}),\,\ldots\,,x_{n}(\vec{t}),\,y_{n}(\vec{t}),$
where $\vec{t}=(t_{1},\ldots,t_{n})\in\mathbb{R}^{n}$. We assume that the
tangent to $L$ vectors

$\tau_{i}=\Big{(}\frac{\partial z}{\partial t_{i}},\frac{\partial
x_{1}}{\partial t_{i}},\frac{\partial y_{1}}{\partial
t_{i}},\,\ldots,\frac{\partial x_{n}}{\partial t_{i}},\frac{\partial
y_{n}}{\partial t_{i}}\Big{)},$

$1\leq i\leq n$, are linearly independent in the vicinity of $a_{\star}$. Put

$\tilde{\tau}_{i}=\Big{(}\frac{\partial x_{1}}{\partial t_{i}},\frac{\partial
y_{1}}{\partial t_{i}},\,\ldots,\frac{\partial x_{n}}{\partial
t_{i}},\frac{\partial y_{n}}{\partial t_{i}}\Big{)}.$

We first will prove the assertion of the lemma for the case when
$\partial_{1}^{+}X(v_{\beta})$ is equal to the hyperplane $H=\\{z=const\\}$,
where the point $\gamma\cap H$ resides in the interval of $\gamma$, bounded by
$a_{\star}$ and $b_{\star}$.

The property of $L$ being Legendrian can be expressed by the two sets of
equations:

(8.1) $\displaystyle\\{\beta(\tau_{i})=0\\}_{1\leq i\leq
n}=\Big{\\{}-\frac{\partial z}{\partial
t_{i}}=\sum_{q=1}^{n}x_{q}\frac{\partial y_{q}}{\partial
t_{i}}\Big{\\}}_{1\leq i\leq n}$ (8.2)
$\displaystyle\\{d\beta(\tau_{i},\tau_{j})=0\\}_{1\leq i<j\leq n},$

where the first group of equations depends on $z$ and the second does not.

Assume for a moment that the second group (8.2) is satisfied by some functions
$x_{1}(\vec{t}),\,y_{1}(\vec{t}),\hfill\break\,\ldots\,,x_{n}(\vec{t}),\,y_{n}(\vec{t}).$
Then we can solve the equations of the first group (8.1) for $z(\vec{t})$,
provided that the RHS of equation (8.1) is a gradient vector field whose
components are $\Big{\\{}\sum_{q=1}^{n}x_{q}\frac{\partial y_{q}}{\partial
t_{i}}\Big{\\}}_{1\leq i\leq n}$, or that

(8.3) $\displaystyle\Big{\\{}\frac{\partial\tilde{\tau}_{i}}{\partial
t_{j}}=\frac{\partial\tilde{\tau}_{j}}{\partial t_{i}}\Big{\\}}_{1\leq i<j\leq
n}.$

Moreover, the solution $z(\vec{t})$ of (8.1) with the property
$z(\vec{0})=a_{\star}$ will be unique.

By a direct computation, the second group of equations (8.2) coincides with
the group of equations in (8.3). In turn, the equations in (8.3) claim that
$\\{d\beta(\tilde{\tau}_{i},\tilde{\tau}_{j})=0\\}_{1\leq i\neq j\leq n}$. In
other words, they claim that the vectors $\\{\tilde{\tau}_{i}\\}_{1\leq i\leq
n}$ belong to $n$-space $\mathcal{I}so(d\beta|_{H})\subset H$, so that the
$n$-manifold $L_{H}^{\dagger}$, given by the functions
$\\{x_{1}(\vec{t}),\,y_{1}(\vec{t}),\,\ldots\,,x_{n}(\vec{t}),\,y_{n}(\vec{t})\\}$
is Lagrangian in the hypersurfuce $H$. The same argument shows that the germ
of any Lagrangian submanifold $L_{H}^{\dagger}\subset H$, passing through the
point $\vec{0}\in H$, is the shadow, under the $z$-projection, of the germ of
unique Legandrian $L$, containing the origin of $\mathbb{R}^{2n+1}$.

It remains to prove that any germ of a Lagrangian $L_{H}^{\dagger}\subset H$
(with respect to $d\beta|_{H}$) at the point $\gamma\cap H$, with the help of
$(-v_{\beta})$-flow, projects onto a germ of a Lagrangian
$L^{\dagger}\subset\partial_{1}^{+}X(v_{\beta})$ (with respect to
$d\beta|_{\partial_{1}^{+}X(v_{\beta})}$) at the point
$b_{\star}=\gamma\cap\partial_{1}^{+}X(v_{\beta})$. The latter observation
follows from the transversality of $\gamma$ to $\partial_{1}^{+}X(v_{\beta})$
at $b_{\star}$ and the $v_{\beta}$-invariance of the form $d\beta$. ∎

For a given Lagrangian submanifold
$L^{\dagger}\subset\mathsf{int}(\partial_{1}^{+}X(v_{\beta}))$, consider the
(discontinuous) function
$\Psi^{\beta}_{L^{\dagger}}:L^{\dagger}\to\mathbb{R}_{+}$, defined by

$\Psi^{\beta}_{L^{\dagger}}(x)=\int_{[x,\;C_{v_{\beta}}(x)]}\beta,$

where $C_{v_{\beta}}$ is the causality map. Note that, for a connected
component
$L^{\ddagger}\subset\partial_{1}^{+}X(v_{\beta})\setminus(\mathcal{X}_{2}(v_{\beta})\cap\partial_{1}^{+}X(v_{\beta}))$,
the function $\Psi^{\beta}_{L^{\ddagger}}$ is continuous and smooth.

Put $\mathcal{D}_{L^{\ddagger}}^{\beta}=\sup_{x\in
L^{\ddagger}}\big{\\{}\Psi^{\beta}_{L^{\ddagger}}(x)\big{\\}}.$ One may think
of this number as the “Reeb diameter” of the component $L^{\ddagger}$.
Evidently,
$\mathsf{diam}_{\mathcal{R}}(\beta)\geq\mathcal{D}^{\beta}_{L^{\ddagger}}$.

Let us revisit a given Legandrian manifold $L\subset X$ whose
$(-v_{\beta})$-projection is $L^{\dagger}$. For any
$x\in\mathsf{int}(L^{\dagger})$, take the first point $y(x)\in\gamma_{x}$ that
belongs to $L$ and consider the function

$\psi^{\beta}_{L}(x)=_{\mathsf{def}}\int_{[x,y(x)]}\beta,$

where $[x,y(x)]\subset\gamma_{x}$ is the segment of the trajectory
$\gamma_{x}$, bounded by $x$ and $y(x)$.

Evidently, $\psi^{\beta}_{L}(x)\leq\Psi^{\beta}_{L^{\ddagger}}(x)$ for any
$x\in L^{\ddagger}\subset L^{\dagger}$.

######  Corollary 8.1.

Let $\beta$ be a contact form whose Reeb vector field is traversing and
boundary generic. Let $L\subset X$ be a Legandrian submanifold in weakly
general position to “the waterfall” $\mathcal{X}_{2}(v_{\beta})$. Let
$L^{\dagger}$ be its $(-v_{\beta})$-guided projection on
$\partial_{1}^{+}X(v_{\beta})$.

Consider the connected components $\\{L^{\ddagger}_{\kappa}\\}_{\kappa}$ of
the Lagrangian manifold $\mathsf{int}(L^{\dagger})$, immersed in
$\partial_{1}^{+}X(v_{\beta})\setminus(\mathcal{X}_{2}(v_{\beta})\cap\partial_{1}^{+}X(v_{\beta})).$
In each component $L^{\ddagger}_{\kappa}$ we pick a base point $b_{\kappa}$,
distinct from the points of self-intersection of $L^{\dagger}$.

Then the knowledge of the shadow $L^{\dagger}$, together with the set the base
points $\\{b_{\kappa}\\}_{\kappa}$ and the set of values
$\\{\psi^{\beta}_{L}(b_{\kappa})\\}_{\kappa}$, makes it possible to
reconstruct the embedding $L\subset X$.

###### Proof.

The reconstruction of $L$ resembles the process of analytic continuation.

Let $\pi_{v_{\beta}}:L\to L^{\dagger}$ be the $(-v_{\beta})$-guided
projection. Let $L^{\dagger}_{\times}\subset L^{\dagger}$ denotes the subset
consisting of points that have multiple $\pi_{v_{\beta}}$-preimages in $L$,
and let $\mathcal{L}^{\dagger}_{2}=L^{\dagger}\cap\mathcal{X}_{2}(v_{\beta})$.

Knowing $b_{\kappa}\in L^{\ddagger}_{\kappa}$ makes it possible, with the help
of $\psi^{\beta}_{L}(b_{\kappa})$, to locate a unique point $a_{\kappa}\in
L\cap\gamma_{b_{\kappa}}$. There is a unique germ $L_{\kappa}$ of $L$ in the
vicinity of $a_{\kappa}$ that projects onto the germ of
$L^{\ddagger}_{\kappa}$ at $b_{\kappa}$. Let $U^{\dagger}_{\kappa}$ be an open
neighborhood of $b_{\kappa}$ in $L^{\dagger}$ over which the reconstruction of
$L$ is achieved. Take a point $x\in L^{\dagger}$ that belongs to the boundary
of $U^{\dagger}_{\kappa}$ and such that $x\notin
L^{\dagger}_{\times}\cup\mathcal{L}^{\dagger}_{2}$. Then by Lemma 8.2, there
exists a unique lift $L^{\bullet}$ to $X$ of $L^{\dagger}$ over some open
neighborhood $U_{x}\subset L^{\dagger}$ of $x$, such that the point
$y=\gamma_{x}\cap L$ belongs to $L^{\bullet}$. Since, over $U_{x}\cap
U^{\dagger}_{\kappa}$, both $L^{\bullet}$ and $L$ are the lifts of
$L^{\dagger}$ that share the point $y$, again by Lemma 8.2, we conclude that
$L^{\bullet}=L$ over $U_{x}$. As a result, the reconstruction of $L$ over
$U_{x}\cup U^{\dagger}_{\kappa}$ is available. Therefore, the maximal open set
in $L^{\dagger}$, over which the lift of $L^{\dagger}$, subject to the
constraint “$a_{\kappa}$ belongs to the lift”, is possible is a connected
component $L^{\ddagger}$ of the open set
$L^{\dagger}\setminus(L^{\dagger}_{\times}\cup\mathcal{L}^{\dagger}_{2})$.
Since $L$ is in weakly general position with $\mathcal{X}_{2}(v_{\beta})$, the
union of such lifts, taken over all $\kappa$’s, is a dense subset
$L\setminus(L\cap\pi_{v_{\beta}}^{-1}(\mathcal{L}^{\dagger}_{2}\cup
L^{\dagger}_{\times}))$ of $L$. Therefore, taking the closure of this union,
produces the given $L\subset X$. ∎

Remark 8.1. Note that the very existence of Legandrian $L\subset X$ insured
that the assembly instructions in the proof of Corollary 8.1 were consistent.

Even when the Reeb vector field $v_{\beta}$ is convex (so the waterfall
$\mathcal{X}_{2}(v_{\beta})=\partial_{2}^{-}X(v_{\beta})$) and the
$v_{\beta}$-trajectories $\gamma_{x}$ depend continuously on
$x\in\partial_{1}^{+}X(v_{\beta})$, we do not know whether any closed
Lagrangian submanifold $L^{\dagger}\subset\partial_{1}^{+}X(v_{\beta})$ is the
$\pi_{v_{\beta}}$-shadow of a Legandrian manifold $L\subset X$. Corollary 8.1
claims only that, if such $L$ exists, then choosing a point $a_{\star}\in L$
makes it unique. It seems that to tackle the question we need to be able to
estimate from above the function $\Psi^{\beta}_{L^{\dagger}}(x)$ (or at least
the Reeb diameter $\mathcal{D}_{L^{\dagger}}^{\beta}$ of $L^{\dagger}$) in
terms of the form $\beta$ in the vicinity of
$L^{\dagger}\subset\partial_{1}^{+}X(v_{\beta})$. $\diamondsuit$

######  Proposition 8.1.

Let $\beta$ be a contact form on $X$, and $v_{\beta}$ its traversing boundary
generic Reeb vector field. Assume that Property $\mathsf{A}$ from Definition
3.1 is valid. Let $\phi:(L,\partial L)\hookrightarrow(X,\partial_{1}X)$ be a
Legandrian embedding of a compact manifold $L$.

Then the following boundary-confined data:

  * •

the causality map
$C_{v_{\beta}}:\partial_{1}^{+}X(v_{\beta})\to\partial_{1}^{-}X(v_{\beta})$,

  * •

the restriction $(f^{\bullet})^{\partial}:\partial_{1}X\to\mathbb{R}$ of the
Lyapunov function $f^{\bullet}:X\to\mathbb{R}$ such that
$df^{\bullet}(v_{\beta})=1$,

  * •

the restriction $\beta^{\partial}=\beta||_{\partial_{1}X}$ of the contact form
$\beta$,

  * •

the Lagrangian666with respect to the symplectic form
$d\beta|_{\mathsf{int}(\partial_{1}^{+}X(v_{\beta}))}$ immersion
$\mathsf{int}(L^{\dagger})\subset\partial_{1}^{+}X(v_{\beta})\setminus(\mathcal{X}_{2}(v_{\beta})\cap\partial_{1}^{+}X(v_{\beta}))$,

  * •

the set of base points $\\{b_{\kappa}\\}_{\kappa}$ and the values
$\\{\psi^{\beta}_{L}(b_{\kappa})\\}_{\kappa}$ as in Corollary 8.1,

allow to reconstruct the Legandrian embedding $\phi:L\hookrightarrow X$, up to
a contactomorphism of $X$ that is the identity on $\partial_{1}X$.

###### Proof.

By Corollary 4.6, the hypotheses in the first three bullets allow for a
reconstruction of $(X,v_{\beta},f^{\bullet},\beta)$ up to a diffeomorphism
that is the identity on $\partial_{1}X$. With this quadruple in place, the
base points $\\{b_{\kappa}\in\mathsf{int}(L^{\dagger})\\}_{\kappa}$ and the
values $\\{\psi^{\beta}_{L}(b_{\kappa})\\}_{\kappa}$, determine uniquely a
point $y(b_{\kappa})\in L$ on each trajectory $\gamma_{b_{\kappa}}$. By
Corollary 8.1, the assertion follows. ∎

######  Lemma 8.3.

Let $X$ be a compact connected smooth $2n+1$-dimensional manifold, $\beta$ a
contact form, and $v_{\beta}$ its Reeb vector field, which is traversing.

Let $L\subset X$ be a sub-Legandrian $(2k+1)$-dimensional closed submanifold
of $X$, where $2k+1\leq n$. We assume that $L$ is in weakly general position
with respect to $\mathcal{X}_{2}(v_{\beta})$. We denote by $L(v_{\beta})$ the
set of downward trajectories originating in $L$, and by $\delta L(v_{\beta})$
the set of downward trajectories originating in $L$ and hitting the locus
$\partial_{2}^{+}X(v_{\beta})$ (see Fig. 3).

Then, for any $k>0$, we get

$\int_{L(v_{\beta})}(d\beta)^{k+1}=0,\quad\text{ and
}\quad\quad\int_{L^{\dagger}}\beta\wedge(d\beta)^{k}=0.$

In contrast, for $k=0$, we get

$\int_{L(v_{\beta})}d\beta=0,\quad\text{ and
}\quad\int_{L^{\dagger}}\beta=-\int_{\delta L(v_{\beta})}\beta\;\leq\;0.$

###### Proof.

Since $T_{x}L\subset\mathcal{I}_{x}\subset\ker\beta$, we see that
$\beta\wedge(d\beta)^{k}|_{L}=0$. Thus, $\int_{L}\beta\wedge(d\beta)^{k}=0$.

Note that, at a generic point $x\in L(v_{\beta})$, the tangent space
$T_{x}(L(v_{\beta}))$ is spanned by $v_{\beta}(x)$ and the image of a tangent
space $T_{y}L\subset\mathcal{I}_{y}$ under $(-v_{\beta})$-flow that takes
$y\in L$ to $x$. Then, $(d\beta)^{k+1}|_{T_{x}(L(v_{\beta}))}=0$, since
$v_{\beta}\in\ker(d\beta)$ . Therefore, $\int_{L(v_{\beta})}(d\beta)^{k+1}=0$.

Let $\delta L(v_{\beta})$ be the set of downward trajectories originating in
$L$ and intersecting $\partial_{2}^{+}X(v_{\beta})$ (see Fig. 3). Again, at a
generic point $x\in\delta L(v_{\beta})$, the tangent space $T_{x}(\delta
L(v_{\beta}))$ is spanned by $v_{\beta}(x)$ and a $2k$-dimensional subspace
$\mathcal{K}_{x}$ of $\mathcal{I}_{x}$. Thus, for $k>0$,
$\beta\wedge(d\beta)^{k}|_{T_{x}(\delta L(v_{\beta}))}=0$ since
$v_{\beta}\in\ker\beta$ and $d\beta|_{\mathcal{K}_{x}}=0$. Therefore,
$\int_{\delta L(v_{\beta})}\beta\wedge(d\beta)^{k}=0$.

The boundary $\partial(L(v_{\beta}))=\delta L(v_{\beta})\cup
L^{\dagger}\cup-L$. In fact, $L^{\dagger}$, away from the set $\delta
L(v_{\beta})\cap\partial_{1}^{+}X(v_{\beta})$ of measure zero, is an immersed
submanifold of $\partial_{1}^{+}X(v_{\beta})$.

Recall that $d(\beta\wedge(d\beta)^{k})=(d\beta)^{k+1}$. Therefore, for $k>0$,
by the Stokes’ theorem and using that $\int_{L}\beta\wedge(d\beta)^{k}=0$,
$\int_{\delta L(v_{\beta})}\beta\wedge(d\beta)^{k}=0$, and
$\int_{L(v_{\beta})}(d\beta)^{k+1}=0$, we conclude that
$\int_{L^{\dagger}}\beta\wedge(d\beta)^{k}=0$ as well.

In the special case $k=0$, using similar analysis and Stokes’ theorem, we get

$\int_{L^{\dagger}}\beta=-\int_{\delta L(v_{\beta})}\beta\;\leq\;0$

since $\beta(v_{\beta})=1$. ∎

######  Corollary 8.2.

(A holographic property of sub-Legandrian links in relation to a concave
boundary)

Let $\beta$ be a contact form whose Reeb vector field $v_{\beta}$ is
traversing on $X$ and boundary generic. Let a sub-Legandrian link777a
collection of embedded disjoint loops $L\hookrightarrow X$ be in weakly
general position with respect to $\mathcal{X}_{2}(v_{\beta})$. Let
$L^{\dagger}\subset\partial_{1}^{+}X(v_{\beta})$ be the image of $L$ under the
$(-v_{\beta})$-flow guided projection $X\to\partial_{1}^{+}X(v_{\beta})$.

Then $\int_{L^{\dagger}}\beta<0$, if and only if, there exists a down-flow
trajectory $\gamma_{\star}$ that originates at $L$ and hits the concavity
locus $\partial_{2}^{+}X(v_{\beta})$.

On the other hand, if $\int_{L^{\dagger}}\beta\geq 0$, then
$\int_{L^{\dagger}}\beta=0$, and there is no such trajectory $\gamma_{\star}$.
In such a case, the $L^{\dagger}$ is a collection of closed immersed curves.

###### Proof.

Since $\beta(v_{\beta})=1$, $\int_{\delta L(v_{\beta})}\beta>0$ when $\delta
L(v_{\beta})$ has a positive $1$-dimensional measure. This is equivalent to
the existence of downward trajectory that connects $L$ to
$\partial_{2}^{+}X(v_{\beta})$.

By Lemma 8.3, in the case $k=0$, we get that $\int_{\delta
L(v_{\beta})}\beta>0$ if and only if $\int_{L^{\dagger}}\beta<0$. ∎

We conclude Section 8 with a few natural questions in hope that experts may
know the answers…

######  Question 8.1.

A $C^{\infty}$-small perturbation $\beta^{\prime}$ of a contact form $\beta$
is a contact form. Is a $C^{\infty}$-small perturbation $v^{\prime}$ of a Reeb
vector field $v_{\beta}$ a Reeb vector field for some contact form? For some
contact form $C^{\infty}$-close to $\beta$? Equivalently, is the map from the
space of contact forms to the space of their Reeb fields open? $\diamondsuit$

######  Question 8.2.

Assuming that a Reeb vector field $v_{\beta}$ is traversing, how to
differentiate between tight and overtwisted contact structures (see [E], [BEM]
for the definition) in terms of boundary data? $\diamondsuit$

######  Question 8.3.

Let $\beta$ be a contact form on a compact manifold $X$ with boundary.
Assuming the the Reeb vector field $v_{\beta}$ is traversing, does any
Lagrangian888with respect to the symplectic form
$\omega^{\dagger}=_{\mathsf{def}}d\beta|_{\mathsf{int}(\partial^{+}_{1}X(v_{\beta}))}$
submanifold $L^{\dagger}$ of the open manifold
$\partial_{1}^{+}X(v_{\beta})\setminus(\mathcal{X}_{2}(v_{\beta})\cap\partial_{1}^{+}X(v_{\beta}))$
is the image, under the $(-v_{\beta})$-directed projection, of a Legandrian
manifold $L\subset X\setminus\mathcal{X}_{2}(v_{\beta})$? The same question,
assuming that $v_{\beta}$-flow is convex. $\diamondsuit$

Acknowledgments: I am grateful to Yakov Eliashberg for very enlightening
conversations. Many discussions with Gunther Uhlmann helped me to organize
this project; it is my pleasant duty to thank Gunther for the much needed
encouragement.

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