John Erik Forn\ae ss
Nessim Sibony
math0402432.tex





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\begin{document}




\centerline{HARMONIC CURRENTS OF FINITE ENERGY AND LAMINATIONS}

\medskip

\centerline{by}

\medskip

\centerline{John Erik Forn\ae ss\footnote{The first author is supported by an NSF grant.
Keywords: Harmonic Currents, Laminations.
2000 AMS classification. Primary: 58F23;
Secondary 57R30, 30F10} and Nessim Sibony
}



\bigskip



%\vspace{.5cm}

%\centerline{         by }

%\vspace{.5cm}


%\centerline{ John Erik Forn\ae ss$^*$ and Nessim Sibony}

%\vspace{.5cm}



%\today

%\tableofcontents

%\vspace{.5cm}


Abstract: We introduce, on a complex K\"{a}hler manifold $(M,\omega),$
a notion of energy for harmonic currents of bidegree $(1,1).$
This allows us to define $\int T \wedge T \wedge \omega^{k-2},$
for positive harmonic currents. We then show that for a lamination
with singularities of a compact set in $\PP^2$ there is a unique
positive harmonic current which minimizes energy. If $X$ is a compact
laminated set in $\PP^2$ of class ${\mathcal C}^1$ it carries a
unique positive harmonic current 
$T$ of mass $1.$ The current $T$ can be obtained
by an Ahlfors type construction starting with a arbitrary leaf of $X.$



%\maketitle



%\authors{John Erik Forn\ae ss, Nessim Sibony}



%$^*$ The first author was supported by an NSF grant. 


\section{Introduction}

Let $X$ be a compact set in a complex manifold $M.$ If $X$ is laminated
by Riemann surfaces, a result due to L. Garnett implies the existence
of a positive current $T$ of bidimension $(1,1)$ which is harmonic
i.e. such that $i\partial \overline{\partial}T=0.$ Moreover in a flow box
$B,$ the current can be expressed as

$$
T=\int h_\alpha [V_\alpha]d\mu(\alpha),\;\;\;\;\;\;\;\;\;\;(1)
$$

\noindent the functions $h_\alpha$ are positive and harmonic on the 
local leaves $V_\alpha$, and $\mu$ is a measure on the transversal.
On the other hand there is a well known problem. Does there
exist a compact laminated set $X$ in $\PP^2$ which is not a compact
Riemann surface? See [CLS], [Gh]
and [Z] where the problem is dicussed. If such an $X$ does not exist
then the closure of any leaf $L$ of a holomorphic lamination in $\PP^2$
will contain a singularity. 

\bigskip

In this paper we study positive harmonic currents directed by a laminated
set with singularities. More precisely we consider compact sets $X$ laminated
by Riemann surfaces out of an exceptional set $E.$ We will assume
that $E$ is locally pluripolar and that $\overline{X \setminus E}=X.$
We will call such a set $(X,{\mathcal L},E)$ a laminated compact set
with singularities.
We consider on such sets harmonic currents $T$ of order $0$ and bidegree
$(1,1)$ in $\PP^2.$ The current $T$ can be written in the form

$$
T=c\omega+ \partial S + \overline{\partial S}
$$

\noindent where $\omega$ is the standard K\"{a}hler form on $\PP^2, c
\in \RR$, $S$ is a $(0,1)$ current. It turns out that $\overline{\partial}S$
depends only on $T$ and that for a positive closed current one can
define the energy $E(T)$ of $T$ by the following integral

$$
E(T)=\int \overline{\partial} S \wedge \partial \overline{S}
$$

\noindent and that $0 \leq E(T)<\infty.$ It is hence possible to introduce
a Hilbert space of classes of currents of finite energy. With this in hand
the integral

$$
Q(T)=\int T \wedge T
$$

\noindent makes sense for positive harmonic currents (not just the ones
associated to $(X,{\mathcal L},E)$) and has the
usual meaning when $T$ is smooth.
We then prove (Theorem 2.22):

\begin{theo}
Let $(X,{\mathcal L},E)$ be a laminated compact set with singularities.
There is a closed positive laminated current on $X$ or
there is a $\underline{unique}$ positive harmonic laminated current $T$
on $X$ minimizing energy.
\end{theo}

We then study the geometric intersection of laminated currents. We show,
see Proposition 2.16 and Theorem 5.2:

\begin{theo}
If $X$ is a ${\mathcal C}^1$ laminated compact set in $\PP^2$, then
$X$ carries a $\underline{unique}$ laminated positive harmonic current $T$. 
The class of the current
$T$ is extremal in the cone of positive harmonic currents. Moreover
$\int T \wedge T=0$.
\end{theo}

When $X$ is not ${\mathcal C}^1$ we have to assume "finite transverse
energy" to get the result. The main tool is that the quadratic form
$Q$ is negative definite on the hyperplane $\{T; \int T \wedge \omega=0\}.$

\medskip

The current can be obtained using a variant of Ahlfors' technique. Let
$\Phi:\Delta \rightarrow L$ be the universal covering map from the unit disc
to a leaf $L.$ Let $G_r(z)=\frac{1}{2\pi} \log^+ \frac{r}{|z|}.$
Define $T_r:= (\Phi)_*(G_r[\Delta]), r<1.$ If $A(r)$ is the mass of $T_r$
we get that 

$$
T=\lim 
_{r \rightarrow 1} \frac{T_r}{A(r)}.
$$

For that purpose we need to estimate the derivative of $\Phi$,
the estimates are valid for any laminated set.

The main question that is left open in our approach is to estimate

$$
c(\PP^2):= \inf \{ \int T \wedge T; \int T \wedge \omega=1,T \geq 0,
i\partial \overline{\partial}T=0\}.
$$
If $c(\PP^2)>0,$ then there is no ${\mathcal C}^1$ laminated set in $\PP^2.$
  


 
\section{Harmonic currents}

A subset $Y$ of a complex manifold $M$ is laminated by Riemann surfaces 
if it admits
an open covering $\{U_i\}$ and on each $U_i$ there is a homeomorphism
$\phi_i=(h_i,\lambda_i):U_i \rightarrow \Delta \times T_i$ where $\Delta$ is the
unit disc and $T_i$ is a topological space. The $\phi_i^{-1}$ are holomorphic
in $z.$ Moreover, 
$$\phi_{ij}(z,t)=\phi_j \circ \phi_i^{-1}(z,t)=(h_{ij}(z,t),\lambda_{ij}(t))$$
\noindent where the $h_{ij}(z,t)$ are holomorphic with respect to $z.$
When $T_i$ is in a Euclidean space and $\phi_{i}$ extend to ${\mathcal
C}^k$ diffeomorphisms, we say that the lamination is ${\mathcal C}^k.$
We call the $U_i$ flow boxes, $\{\lambda_i=t_0\}$ is a plaque.
A leaf is a minimal connected set such that if $L$ intersects a plaque,
then $L$ contains the plaque. We only consider the case when we have
a compact set $X$ contained in a K\"{a}hler manifold $M$ of dimension $k.$
We assume that $X$ contains a closed set $E$ such that $\overline{X \setminus E}=X$ and
$X \setminus E$ is a lamination ${\mathcal L}$ by Riemann surfaces.
We call such a triple $(X,{\mathcal L},E)$ a lamination with singularities,
$E$ is the singular set.
We will say that $(X,{\mathcal L},E)$
is oriented if there are continuous nonvanishing $(1,0)$ form
$\gamma_j,j=1,\dots,k-1$ defined on $X \setminus E$ such that $\gamma_j
 \wedge [\Delta_\alpha]=0$
for every plaque $\Delta_\alpha,$ $[\Delta_\alpha]$ denotes the current
of integration on the disc $\Delta_\alpha.$ 
We only consider sets $E$ which are pluripolar, more precisely,
for any $p\in E$, there are small balls $B(p,r)$ and $u,$ a plurisubharmonic
function on $B(p,r)$ such that $E \cap B(p,r)
\subset \{u=-\infty\}$ and $u$ is not identically $-\infty$
on $K \cap B(p,r).$ When $E \cap B(p,r)=\{u=-\infty\}$ we say
that $E$ is locally complete pluripolar. 
A positive  current $T$ of bidimension $(1,1)$ with support in $X$
is said to be directed by ${\mathcal L}$ if on any open
set $U$ where ${\mathcal L}$  is defined by nonvanishing
continuous $(1,0)$ forms $\gamma_j,j=1 \dots,k-1 $
i.e. $\gamma_j \wedge [\Delta_\alpha]=0,$
we have
$$ T \wedge \gamma_j=0.$$


To introduce the notion of a minimal set for $X$ we need the following 
assumption on the family of leaves $L.$ 

$$
 {\mbox{There is a neighborhood }}\; {\cal V}\; {\mbox{of E,
so that no leaf is contained}}\\
{\mbox{ in}}\; {\cal V}. \;\;\;\;\;\;\;\;(*)
$$


\begin{definition}
A minimal set for $(X,{\cal L}, E)$ is a compact subset $Y\subset X$
such that $Y$ is not contained in $E$,
moreover $Y\setminus E$ is a union of
leaves $L$ and for every leaf $L \subset Y\setminus E$, $\overline{L}=Y.$ 
\end{definition}

\begin{proposition}
Let $(X,{\cal L},E)$ satisfy assumption $(*).$ Then there are minimal sets
$Y \subset X.$ Two different minimal sets intersect only on $E.$ If M is a 
surface where the Levi problem is solvable and $E$ is locally contained
in a complex hypersurface, then any 
two minimal sets intersect.
\end{proposition}


{\bf Proof:}
Let ${\cal V}$ be an open neighborhood of $E$ such that no leaf is contained
in ${\cal V}.$ Let $(X_\alpha)$ be an ordered decreasing chain saturated for 
${\cal L}.$  Let $X'_\alpha:= X_\alpha \cap (X \setminus \cal V) .$
Then $X'_\alpha \neq \emptyset.$ Hence $\cap X_\alpha \neq \emptyset.$
So Zorn's Lemma applies. This shows that minimal sets exist.\\

It follows that each $M \setminus Y$
is locally pseudoconvex away from $E$, and since $E$ is locally contained
in a complex hypersurface,
$M\setminus Y$ is pseudoconvex [GR]. If the Levi problem is solvable
on $M,$ i.e. if pseudoconvex domains are Stein,
each $M\setminus Y$ is Stein. Since Stein manifolds cannot have two ends,
it follows that any two minimal sets must intersect. \\


\CQFD\\



\begin{remark}
Oka solved the Levi problem in $\PP^k.$ See [E] for a proof
of this and some generalizations. 
The condition on $E$ in the above Proposition, 
can be relaxed to assuming that $E$ is meager
in the sense of [GR].
\end{remark}


\begin{example}
Let ${\cal L}_\alpha$ be the foliation in $\PP^2$ defined by 
$wdz-\alpha zdw=0,$ with $\alpha$ irrational. Then $Y_c=
\overline{\{|z|=c|w|^\alpha\}}$ is minimal for every $c,$
the closure is in $\PP^2.$ The associated positive closed current
$T$ is defined by $\pi^* T=i\partial \overline{\partial} u, u(z,w,t)=
\log (\max \{|z||t|^{\alpha-1}, |w|^\alpha\})$ if $\alpha>1$ and is directed
by ${\mathcal L}_\alpha.$
\end{example}







\begin{definition}
Let $M$ be a complex manifold of dimension $k$. For $0\leq
p,q\leq k,$ let $T$ be a $(p,q)$ current of order $0.$
We say that $T$ is harmonic if $i \partial \overline{\partial} T=0.$
\end{definition}

\medskip

Observe that if $T$ is harmonic then $\overline{T}$, the
conjugate, is also harmonic. A current is real if $T=
\overline{T},$ in which case $p=q.$

\subsection{Decomposition of harmonic currents}
 We want to prove a representation theorem for real harmonic 
currents on compact K\"{a}hler manifolds.





\medskip




There is also a notion of $\Box-$harmonic forms in the $\overline{\partial}$
literature [FK]. These are smooth $(p,q)$ forms $\Omega$ for which
$\Box \Omega=(\overline{\partial} \overline{\partial}^*+
\overline{\partial}^*\overline{\partial}) \Omega=0.$
These forms consist of the common null space
of $\overline{\partial}$ and $\overline{\partial}^*$.
Note that when $p=q$, $\overline{\Box}=\Box$ so the conjugate of 
a $\Box-$harmonic
form is also $\Box-$harmonic. Since $\Box-$ harmonic forms are $\overline{\partial}$
closed they are also harmonic in the sense of currents as defined above.
Recall that on a compact K\"{a}hler manifold, for a closed current $u$ of
bidegree $(p,q),$ the following are equivalent:\\

\noindent i) $u$ is exact,\\
ii) $u$ is $\overline{\partial}$ exact,\\
iii) $u$ is $\partial \overline{\partial}$ exact.\\

See Demailly [De], p 41 for smooth forms. The proof is the same
for currents since cohomology groups for currents and smooth forms
are the same. This follows from the deRham Theorem and the fact that
for K\"{a}hler manifolds the Dolbeault cohomology groups
$\oplus_{p+q=k} H^{p,q}(X,\CC)$ and $H^k_{{\mbox{DR}}}$ are isomorphic
[De] p.42.


\begin{proposition}
Let $T$ be a harmonic $(p,q)$ current on a compact
K\"{a}hler manifold $M$ of dimension $k.$ Then
$$
T=\Omega+\partial S+\overline{\partial} {R}\;\;\;\;\;\;\;(2)
$$
\noindent where $\Omega$ is a unique 
closed smooth $\Box$-harmonic
form of bidegree $(p,q)$, and $S$ is a current
of bidegree 
$(p-1,q),$ $R$ is of bidegree $(p,q-1).$
When $dT$ is of order $0,$ we can choose $S,R$ depending linearly on $T$ and 
$$
L:T \rightarrow (\Omega,S,R)
$$
is continuous in the topology of currents. When $T$ is real we can choose
$R=\overline{S}.$
\end{proposition}

{\bf Proof:}
The current $\overline{\partial}T$ is $\partial-$
closed, hence $d-$ closed and is $\overline{\partial}$ exact. It follows
from Lemma 8.6 in [De] that $\overline{\partial}T$ is $\partial \overline{
\partial}$ exact so there is an $S_0$ of order $0$ and
of bidegree $(p-1,q)$
such that 
$$
\overline{\partial}T = \overline{\partial}\partial S_0.
$$

Let $\Omega$ be a smooth $\Box-$ harmonic representative
of the Dolbeault cohomology class of $T-\partial S_0$.
(See [De] and [V].)

 
Then $T-\partial S_0-\Omega $ is
$\overline{\partial}$ exact. Hence
$$
T=\Omega+\partial S_0+\overline{\partial}R.
$$

The choices of $S_0$ and $R$ can be made linearly since there is an
explicit inverse for $\partial \overline{\partial}$ and hence
$\overline{\partial}$, on currents cohomologous to zero. See
Dinh-Sibony [DS] where an explicit kernel is given.
If $T$ is real we obtain that $T$ can be expressed as claimed.


The current $T$ acts on $H^{n-p,n-q},$ the Dolbeault cohomology group,
because of the $\partial \overline{\partial}$ lemma. Hence $\Omega$
is uniquely determined by $T.$

\CQFD\\


\medskip



\begin{prop}
 If $T$ is as in $(2)$, then $S,R$ are not unique, but any other
$S',R'$ can be obtained as $S'=S+\omega'+\partial v+
\overline{\partial} u,$ similarly for $R'.$
\end{prop}

{\bf Proof:} The cohomology class $\Omega$ is defined
uniquely. If $S',R'$ is another solution 
we get $\partial(S-S')+\overline{\partial} (R-R')
=0$. 
Assume $\partial \sigma+\overline{\partial}\overline\sigma'=0.$
Then $\sigma$ is harmonic. Using the above construction for a harmonic
$(p-1,q)$ form, we get

$$
S-S'=\sigma=\omega'+\partial v+\overline{\partial} u.$$

Hence

$$
S=S'+(\omega'+\partial v+\overline{\partial} u ).$$


\CQFD\\

\begin{coro}
Let $T$ be a harmonic current of degree $(1,p)$ on $(M,\omega).$
Let $T=\Omega+\partial S+\overline{\partial} R$ be any decomposition
as in (1). Then $\overline{\partial}
S$ is uniquely determined by $T.$ If $p=1$ and $T$ is real, then
$T$ is closed if and only if $\overline{\partial} S=0.$
\end{coro}

{\bf Proof:}
If $S',R'$ also satisfy $T=\Omega+\partial S'+\overline{\partial} R'$
then $S'=S+\omega'+\overline{\partial} u$ for bidegree reasons. Consequently
$\overline{\partial}S'=\overline{\partial}S.$
Assume that $T$ is a real $(1,1)$ current. If $T$ is closed, then
$T=\Omega+i\partial \overline{\partial}u$, i.e. $\overline{\partial}S=0.$
Conversely, if $\overline{\partial}S=0$ and $T=\Omega+\partial S+
\overline{\partial S},$

$$
<T,\overline{\partial}\theta>=-<\overline{\partial}T,\theta>
=<-\overline{\partial}\partial S,\theta>=0.$$

Hence $T$ is closed.\\

\CQFD\\


\subsection{Energy of harmonic currents}

In this paragraph we introduce a notion of energy of harmonic currents of
bidegree $(1,1)$ on $\PP^k.$ On $\PP^k$ we consider the standard 
K\"{a}hler form $\omega$ normalized so that $\int \omega^k=1.$
Recall that $H^{p,q}=0$
except when $p=q$ in which case $H^{p,p}$ is generated by $\omega^p.$

\medskip

We showed above that if $T$ is a real harmonic $(1,1)$ current on $\PP^k,$ then
it can be represented as

$$
T=c\omega+\partial S+\overline{\partial}\overline{S}
$$

\noindent with $S$ of bidegree $(0,1)$. We have $c\in \RR$, 
$c=\int T \wedge \omega^{k-1}.$
We define the energy $E(T)=E(T,T)$ of $T$ as
$$
E(T,T)=\int \overline{\partial}S\wedge \partial \overline{S}\wedge \omega^{k-2}
$$
\noindent when $\overline{\partial}S \in L^2$.
 Observe that
$\overline{\partial}S$
is a $(0,2)$ form. Hence $0\leq E(T,T) < \infty.$ We have seen in
Corollary 2.8 that the energy depends on $T$ only. 

\medskip

We define ${\cal H}_e$ to be the space of real harmonic $(1,1)$ currents on
$\PP^k$ of 
finite energy. We consider on ${\cal H}_e$ the (real) inner product and 
semi norm

\bea
<T_1,T_2>_e & = & \left[ \int T_1\wedge \omega^{k-1}\right]
 \left[ \int T_2\wedge \omega^{k-1}\right]+\frac{1}{2}
\int \overline{\partial}S_1\wedge \partial \overline{S}_2 \wedge \omega^{k-2}\\
& + & \frac{1}{2}\int \overline{\partial}S_2\wedge \partial \overline{S}_1
 \wedge \omega^{k-2}\\
\|T\|^2_e & = & \left| \int T \wedge \omega^{k-1}\right|^2+
\int \overline{\partial} S \wedge \partial \overline{S}\wedge \omega^{k-2}\\
\eea

\begin{lem}
Let $T=c\omega+\partial S+\overline{\partial S},
\overline{\partial}S\in L^2,$ a $(1,1)$ real harmonic current
of order $0$ in $\PP^k.$ Then
$\|T\|_e=0$ if and only if  $T=i \partial \overline{\partial}u,$
for $u \in L^1,$ $u$ real.
\end{lem}

{\bf Proof:}
If $\|T\|_e=0$, then $\overline{\partial}S=0$ and $c=0.$ Then
$T$ is closed, 
hence $T$ is exact and therefore $T=i\partial \overline{\partial}u$
[De]. Regularity of the Laplace equation shows that $u\in L^1.$
Conversely, suppose that $T=i \partial 
\overline{\partial} u,$ $u\in L^1,$ $u$ real, so
we can set $S=
\frac{1}{2}i\overline{\partial} u.$ Then $\overline{\partial}S=0,$
hence $\int \overline{\partial}S \wedge
\partial \overline{S}=0.$ Clearly also, $\int T \wedge \omega=0$
so  $\|T\|_e=0.$\\

\CQFD\\

\medskip

\begin{lem}
There is a constant $C=C_k$ so that 
if $T$ is a real harmonic $(1,1)$ current of order $0$ on $\PP^k, k \geq 2,$
with finite energy,
then there is an element $\tilde{T}$ in the equivalence class of 
$T$, i.e. $\|T-\tilde{T}\|_e=0,$ which can be written
as $\tilde{T}=c\omega+\partial S+ \overline{\partial S}$
with $S,\partial S, \overline{\partial}S \in L^2,$
$\|S\|_{L^2}, \|\partial S\|_{L^2}, \|\overline{\partial} S\|_{L^2}
\leq C \|T\|_e.$ Hence $T=\tilde{T}+i\partial \overline{\partial}u$
and $i\partial \overline{\partial }u$ is of order $0.$
\end{lem}

{\bf Proof:} We can write $T=c\omega+ \partial S_1+ \overline{ \partial S}_1$
with $\overline{\partial}S_1\in L^2.$ 
Since $ \overline{\partial} S_1 $ is in $ L^2$, we can find an
$S\in L^2_{01}$ for which 
$\overline{\partial} S_1=\overline{\partial} S.$
Moreover, $\partial S\in L^2$ as well since, by Hodge theory
we gain one derivative
by solving the $\overline{\partial}$ problem 
$\overline{\partial} S=\overline{\partial} S_1.$


Since $H^{(0,1)}(\PP^k)=0,$ there is a distribution $v$ for which 
$\overline{\partial}v= S_1-S.$ 
Therefore we have the decomposition

\bea
T & = &  c\omega+[\partial(S+\overline{\partial}v)]+\overline{
[\partial(S+\overline{\partial}v)]}\\
& = & c\omega+\partial S+\overline{\partial S}+\partial
\overline{\partial}v+\overline{\partial}\partial \overline{v}\\
& = & \tilde{T}+i \partial \overline{\partial} \left( \frac{
v-\overline{v}}{i} \right)\\
\eea

The distribution  $u:= \frac{v-\overline{v}}{i}$ is real.
Since $T,\tilde{T}$ have order $0$, $i \partial \overline{\partial}u$
is also of order $0.$


Since $\|\partial{\overline{\partial}}u\|_e=0,$ it follows that
$\tilde{T}:=  c\omega+\partial S+\overline{\partial S}$ is in the equivalence
class of $T$ as desired.

The $L^2$ estimates are classical [FK].\\


\CQFD\\

Let $H_e$ denote the quotient space of equivalence classes $[T]$
in ${\mathcal H}_e.$


\begin{proposition} The space $H_e$ is a 
real Hilbert space. Every element $[T]$ in $H_e$ can be represented
as
$$
T=c\omega+\partial S+\overline{\partial} \overline{S}
$$

\noindent where $S$ is a $(0,1)$ form in $L^2,$ with $\partial S$ and
$\overline{\partial}S$
in $L^2.$ Convergence in $H_e$ implies weak convergence of currents:
If $[T_n] \rightarrow 0$ in $H_e$ then there are representatives $\tilde{T}_n
\in [T_n]$ such that $\tilde{T}_n \rightarrow 0$ 
in the weak topology of currents. In fact the mass norms $\|\tilde{T}_n\|
\rightarrow 0.$
\end{proposition}

{\bf Proof:}
We show first that $H_e$ is complete. Let $\{[T_n]\}$ be a Cauchy sequence
of equivalence classes, $\lim_{n,m \ra \infty} \|T_n-T_m\|_e =0.$
We can suppose $\|T_{n+1}-T_n\|_e<\frac{1}{2^n}.$ Inductively, we can 
(Lemma 2.10) choose 
representatives $\tilde{T}_n$ so that 

$$
\tilde{T}_{n+1}=\tilde{T}_n+c_n \omega+\partial S_n+ \overline{\partial S_n},
|c_n|, \|\partial S_n\|_{L^2}, \|S_n\|_{L^2}, 
\|\overline{\partial}S_n\|_{L^2}
\leq C \frac{1}{2^n}.
$$

Hence $\{[T_n]\}$ converges in $H_e.$ This shows that $H_e$ is complete.
The last statement is similar.\\

\CQFD\\

We will next introduce a notion of wedge product of real
harmonic currents. Let $T,T'$ be representatives of equivalence classes
as above,

$$
T=c\omega+\partial S+ \overline{\partial S},
T'=c'\omega+\partial S'+ \overline{\partial S'}.
$$

Then a formal calculation gives:

\bea
\int T \wedge T'\wedge \omega^{k-2} & = &
cc'\int \omega^k+
\int \partial S \wedge \overline{\partial S'}\wedge \omega^{k-2}+
\int \overline{\partial S} \wedge {\partial S'}\wedge \omega^{k-2}\\
& = & <T,\omega^{k-1}><T',\omega^{k-1}>-
\int \overline{\partial} S \wedge \partial \overline{S'}\wedge \omega^{k-2}\\
& - & 
\int \partial \overline{S} \wedge \overline{\partial} S'\wedge \omega^{k-2}\\
\eea

Notice that if $T,T'$ have finite energy, the last expression
is well-defined. We define in this case the quadratic form $Q(T,T')$
for currents $T,T'$ of finite energy:

$$
Q(T,T')=
<T,\omega^{k-1}><T',\omega^{k-1}>-
\int \overline{\partial} S \wedge \partial \overline{S'}\wedge \omega^{k-2}-
\int \partial \overline{S} \wedge \overline{\partial} S'\wedge \omega^{k-2}
$$

\noindent and motivated by the formal calculation, we define

$$
\int T \wedge T'\wedge \omega^{k-2}:=Q(T,T')
$$

\noindent when $T,T'$ are harmonic $(1,1)$ current on $\PP^k$
with finite energy.
Recalling the definition of energy,
we get

$$
\int T \wedge T\wedge \omega^{k-2}=Q(T,T)= <T,\omega^{k-1}>^2
-2E(T,T).
$$

Note that $Q(T,T')$ is well defined on equivalence classes in $H_e.$

\begin{theo}
Any positive harmonic current $T$ of bidegree $(1,1)$ on $\PP^k$ is
of finite energy.  Also $\int T \wedge T \wedge \omega^{k-2}\geq 0$
for these currents.
The quadratic form $Q(T,T')$
is continuous on $H_e.$
\end{theo}

{\bf Proof:} Assume first that $T$ is smooth, 
$T=c\omega+\partial S
+\overline{\partial}\overline{S}$ with $c \geq 0$
and $S$ smooth. We get after integration by parts:

\bea
\int T \wedge T \wedge \omega^{k-2} & = & 
c^2\int \omega^k+2\int \partial S \wedge \overline{\partial S}\wedge 
\omega^{k-2}\\
& = & c^2 \int \omega^k-2\int \overline{\partial} S \wedge \partial 
\overline{S} \wedge \omega^{k-2}\\
& \geq & 0\\
\eea

So
$$
2\int \overline{\partial} S \wedge \partial \overline{S}\wedge \omega^{k-2}
\leq |<T,\omega^{k-1}>|^2.
$$

In general, we can still write $T=c\omega+\partial S
+ \overline{\partial} \overline{S}$ by Proposition 2.6.
Let $S_\epsilon$ be a regularization of $S$ and define
$T_\epsilon=c\omega+\partial S_\epsilon+\overline{\partial}
\overline{S}_\epsilon.$ Here we use the classical regularization
for a current $S$, $$S_\epsilon= \int_{U(k+1)}\rho_\epsilon(g)g_*Sd\nu(g),$$
\noindent where $\nu$ is the Haar measure on $U(k+1)$ acting as automorphisms
of $\PP^k$ and $
\rho_\epsilon$ is an approximation of unity in $U(k+1).$

Then $T_\epsilon$ is still positive and harmonic. We get that
$$
2\int \overline{\partial} S_\epsilon \wedge \partial \overline{S}_\epsilon
\wedge \omega^{k-2} \leq |<T,\omega^{k-1}>|^2\;\;\;\;\;\; (3)
$$

Since $S_\epsilon \rightarrow S$ weakly,
 $\overline{\partial}S_\epsilon
\rightarrow \overline{\partial}S$ converges weakly in $L^2$, because
$\{{\overline{\partial}S}_\epsilon\}_\epsilon$ is bounded in $L^2.$
Hence $\overline{\partial}S\in L^2$ and
$$
\int \overline{\partial}S \wedge \partial \overline{S}\wedge \omega^{k-2}
\leq \underline{\lim}\int \overline{\partial}S_\epsilon
\wedge \partial \overline{S}_\epsilon\wedge \omega^{k-2}
\leq 1/2 |<T,\omega^{k-1}>|^2.$$

\noindent Hence $T$ has finite energy and $\int T \wedge T \wedge \omega^{k-2}\ge 0.$
 

If $T=c\omega+\partial S+\overline{\partial} \overline{S}$ and
$T'=c\omega+\partial S'+\overline{\partial} \overline{S'}$ then
$Q(T,T')=cc'\int \omega^k-
\int \overline{\partial}S \wedge \partial \overline{S'}\wedge \omega^{k-2}-
\int \overline{\partial}S' \wedge \partial \overline{S}\wedge \omega^{k-2}.$
It is clear that $|Q(T,T')| \leq 2 \|T\|_e\|T'\|_e.$ Hence $Q$
is continuous on $H_e.$\\


\CQFD\\


\begin{remark}
Since our regularization is a convolution, we also have $\overline{\partial}
S_\epsilon \rightarrow \overline{\partial}S$ in $L^2$ and hence
$E(T_\epsilon,T_\epsilon)\rightarrow E(T,T).$
\end{remark}

\begin{coro}
On $H_e$, the quadratic form
$$
Q(T_1,T_2)=\int T_1 \wedge T_2 \wedge \omega^{k-2} 
$$
is strictly negative definite on the hyperplane
${\cal H}=\{T; \int T \wedge \omega^{k-1}=0\}.$ Consequently, if $T,T'$
are positive harmonic currents, non proportional, then
$\int T \wedge T'\wedge \omega^{k-2}>0.$ If
$\int T \wedge \omega^{k-1}=1$ and $T \geq 0$ then $T$ is 
non closed if and only
if $0 \leq \int T \wedge T \wedge \omega^{k-2}<1.$  
\end{coro}

{\bf Proof:}
If $\int T \wedge \omega^{k-1}=0,$ then
$$
Q(T,T):=
-2\int \overline{\partial}S \wedge \partial \overline{S} \wedge \omega^{k-2}
\leq 0,$$ it is zero only if $T=0$ in $H_e.$ Hence $Q$ is strictly
negative definite on ${\mathcal H}.$ 
Suppose the space generated by $T',T$ is of dimension $2.$ There is an $a>0$,
$\int (T'-aT)\wedge \omega^{k-1}=0. $
Hence 
\bea
0 & > & Q(T'-aT,T'-aT)\\
& = & Q(T',T')+a^2 Q(T,T)-2aQ(T',T)\\
\eea

\noindent Since $Q(T',T'), Q(T,T) \geq 0$ by Theorem 2.12, it
follows that $Q(T',T)>0.$ 

The last part is an immediate consequence of Corollary 2.8.\\



\CQFD\\



\begin{prop}
The function $T \rightarrow Q(T,T)$ is upper semi continuous
in the weak topology on positive harmonic currents and is strictly
concave on $\{T \wedge \omega=1\}.$
\end{prop}

{\bf Proof:}
If $T_n \rightarrow T$ weakly, we
have seen as above in the proof of Theorem 2.12 
that $\int \overline{\partial}S \wedge \partial \overline{
S} \leq \underline{\lim}\int \overline{\partial}S_n
 \wedge \partial \overline{S}_n$, so $Q$ is upper semi continuous.
Concavity is clear because if $\int (T-T')\wedge \omega=0$,
$Q(T-T',T-T')<0,$ so $2Q(T,T')>Q(T,T)+Q(T',T').$ Hence
$Q(\frac{T+T'}{2}, \frac{T+T'}{2})> \frac{1}{2} Q(T,T)+
\frac{1}{2} Q(T',T').$ \\


\CQFD\\
 
\begin{prop}
In $\PP^k, k \geq 2,$
positive harmonic currents $T$ of bidegree $(1,1)$ satisfying
$Q(T,T)=0$ are extremal on $H_e$ in the cone of positive harmonic currents.
\end{prop}

{\bf Proof:}
Assume that $0\leq T'\leq cT$ for some $c>0.$

\bea
\int T' \wedge T \wedge \omega^{k-2} & = &
\lim \int T_\epsilon' \wedge T \wedge \omega^{k-2}\\
& \leq & \lim \int cT_\epsilon \wedge T \wedge \omega^{k-2}\\
& = & c \int T \wedge T \wedge \omega^{k-2}\\
& = & 0\\
\eea

Hence $[T']$ is proportional to $[T]$ by Corollary 2.14.\\

\CQFD\\


\begin{prop}
The map $T \rightarrow \overline{\partial}S$ is not continuous for the
weak topology on positive harmonic currents $T$ of bidegree $(1,1)$
and $L^2$ topology on $\overline{\partial}S.$ 
\end{prop}

{\bf Proof:}
Let $T=\omega+\epsilon(\partial S+\overline{\partial S})$ for a smooth
$(0,1)$ form $S$ supported in the unit bidisc. For $\epsilon>0$ small
enough, $T$ is positive and
$\int T \wedge T= \int \omega \wedge \omega -2\epsilon^2 
\int \overline{\partial}S
\wedge \partial {\overline{S}}<1$.
If the map $T\rightarrow \overline{\partial}S$
with weak topology on $T$ and $L^2$ topology on $\overline{\partial}S$
were continuous then if $T_n \rightarrow T_0$,
$\int T_n \wedge T_n \rightarrow \int T_0 \wedge T_0.$
Let $f$ be an endomorphism of $\PP^2$ of algebraic degree $d.$
Then
$f^*:  {{\cal H}_e \rightarrow \cal H}_e$ is a linear map of norm $d$
if the algebraic degree of $f$ is $d.$ Indeed
\bea
|\int f^*T \wedge \omega|^2 & = & |\int T\wedge f_*\omega|^2\\
& = & d^2|\int T \wedge \omega|^2\\
\eea
because $f_*\omega \sim d\omega.$ We also have

$$
\left|\int f^*(\overline{\partial}S \wedge \partial \overline{S})\right|
=d^2 \left| \int \overline{\partial}S \wedge \partial 
\overline{S}\right|
$$

\noindent This can be obtained by smoothing.


Therefore
$E(f^*T/d)=E(T)$ so
$\int f^*T/d \wedge f^*T/d=\int T \wedge T.$

Let $f[z:w:t]=[z^2:w^2:t^2]$ and $T=\omega+\epsilon(\partial S+
\overline{\partial S})$ as above. Then

\bea
1 & > & \int T \wedge T \\
& = & \int (f^n)^*T/2^n \wedge (f^n)^*T/2^n \\
& = & \int (f^n)^*\omega/2^n \wedge
(f^n)^*\omega/2^n-2\int \overline{\partial} (f^n)^*(S)/2^n
\wedge {\partial} (f^n)^* (\overline{S})/2^n.\\
\eea

\noindent If we choose $S=a(|z|,|w|)d\overline{z}$
it is easy to check that
$(f^n)^*S/2^n
\rightarrow 0$ weakly. Hence $(f^n)^*T/2^n$ converges
weakly to a closed current $A=\lim \frac{(f^n)^* \omega}{2^n}$
whose class in $H^{(1,1)}$ is $\omega.$
If $T \rightarrow \overline{\partial} S$ were continuous
the second integral would converge to zero.
Since $\int \omega \wedge \omega =1,$ we get 
that
the map $T \rightarrow \overline{\partial} S$ is not
continuous. \\

\CQFD\\

This is the justification for introducing the norm
on finite energy currents, which gives a different topology than
weak topology.

\medskip


L. Garnett has shown in [G] the existence of positive currents
$T$, satisfying $i \partial \overline{\partial} T=0$ and directed
by foliations. In [BS] a version of this result is given allowing
leaves to intersect. Here we are interested in constructing laminar
currents for a foliation with singularities that are only holomorphic
motions in flow boxes, the holomorphic case in treated in
[BS].

 
\begin{theo}
Let $(X,{\mathcal L},E)$ be a directed set with singularities
in $\PP^2.$ Then there is a laminated harmonic positive current
$T$ of the form $T=\int_\alpha h_\alpha [\Delta_\alpha]d\mu(\alpha)$
in flow boxes. Here $\mu(\alpha)$ is a measure on transversals,
$h_\alpha$ are strictly positive harmonic functions,
uniformly bounded above and below by strictly positive constants,
$h_\alpha$ are Borel measurable with respect to $\alpha.$
\end{theo}

{\bf Proof:} Let $\gamma$ be a continuous $(0,1)$ form such that
$\gamma \wedge [\Delta]=0$ for every plaque $\Delta.$ In [BS]
Theorem 1.4 a current $T \geq 0$ supported on $X$
satisfying $i \partial \overline{\partial} T=0$ and
$T \wedge \gamma =0$ is constructed. It is shown that the current
is laminar in flow boxes when the foliation is holomorphic.
We consider here the general case.

The Theorem follows from the next Lemma.


 
\begin{lem}
A positive harmonic current, directed by a holomorphic motion in a 
polydisc is a laminar current.
\end{lem}




{\bf Proof of the Lemma:} Let $B$ be a flow box.
In $B$, $T=\|T\|i\gamma \wedge \overline{\gamma}$
where $\|T\|$ is a positive measure. Assume $z=0$ is a transversal 
and $\pi$ is the projection along leaves in $B,$ on that transversal.
Let $(\nu_w)$ be the desintegration of $\|T\|$ along $\pi.$ 
If $\phi$ is a test form of bidegree $(1,1)$ supported on $B$,

$$
<T,\phi>=\int <\nu_w i\gamma \wedge \overline{\gamma},\phi> d\mu(w)
$$

\noindent with $\mu=\pi_*(\|T\|).$
This applies when $\phi=i \partial \overline{\partial} f$ and when
$f$ has support on an arbitrary open set of plaques, we have:
$<T,i \partial \overline{\partial}f>=0.$ Hence

\bea
0 & = & \int <\nu_wi\gamma \wedge \overline{\gamma},i\partial
\overline{\partial} f>d\mu(w)\\
& = & \int <\nu'(w)[D_w],\Delta_w f>
d\mu(w).\\
\eea

\noindent where $\Delta_w f$ denotes the Laplacian along the leaf through
$w.$ This extends to test functions which are continuous and
${\mathcal C}^2$ along leaves.
Hence $<\nu'(w)[D_w],\Delta_w f>=0$
$\mu$ a.e., so $\nu'(w)$ is a harmonic function on the leaf
$[D_w].$  \\

\CQFD\\


\begin{remark}
It follows from a Theorem by Skoda ${\mbox{[Sk]}}$ that no positive
harmonic $(1,1)$ current can have mass on a set of $2-$
dimensional Hausdorff measure $\Lambda_2=0.$
\end{remark}

\begin{theo} Let $(X,{\cal L},E)$ be a laminated set with singularities
in $\PP^2.$ There is a $\underline{\mbox{unique}}$ equivalence class
of harmonic currents 
directed by ${\cal L}$ of mass one and minimal energy.
\end{theo}

{\bf Proof:} Let $C_1=\{T; T \geq 0, \int T \wedge \omega=1,
T\; {\cal L}-\mbox{directed, harmonic}\}.$ Then $C_1$ is compact
in the weak topology of currents. From Theorem 2.18 we know that $C_1$
is nonempty.
 
The energy is a lower semi continuous function on $C_1$ by Proposition 2.15.
Since $C_1$ is compact  it now follows that $E(T,T)$ 
takes on a minimum value $c$ on $C_1$

If $E(T,T)=c$ and $\int T \wedge \omega=1$, then
$Q(T,T)=1-2c.$
If $T,T'$ are two elements of non colinear equivalence classes
of currents where the minimum $c$
is reached, then $Q(\frac{T+T'}{2}, \frac{T+T'}{2})>1-2c$
by strict concavity of $Q$ (Proposition 2.15),
a contradiction. So the minimum is unique.\\

\CQFD\\


We next show that under mild extra hypotheses, minimal equivalence classes
contain only one current.
Recall that a current on a laminated compact $X$ 
which in local flow boxes has the form
$\int h_\alpha [V_\alpha]d\mu(\alpha)$ for a positive measure
$\mu(\alpha)$ on the space of plaques, and $h_\alpha>0,$ harmonic
functions on plaques $V_\alpha$ is said to be a laminated positive harmonic
current. The current is closed and laminated if the $h_\alpha$
are constant.

\begin{theo}
Let $(X,{\mathcal L},E)$ be a directed set with singularities.
Suppose $E$ is locally complete pluripolar with $\Lambda_2(E)=0$
in $\PP^2$. Assume there is no nonzero
positive closed laminated current on $X.$ Consider the convex compact
set $C$ of laminated positive harmonic $(1,1)$ currents of mass $1$. Then 
there is a unique element $T$ in $C$ minimizing energy. The current $T$
is extremal in $C.$
\end{theo}

{\bf Proof:}
We know that $C$ is nonempty. Let $T_1,T_2 \in C$ be two minimizing currents.
We know by remark 2.20 that $T_j$ has no mass on $E.$
By Theorem 2.21, $[T_1]=[T_2]$ so 
$T_1-T_2=i\partial\overline{\partial}u$, hence $T_1-T_2$ is closed.

In a flow box we have $T_j=\int h_j^\alpha[V_\alpha]d\mu_j(\alpha),
j=1,2.$
Let $\nu(\alpha)=\mu_1+\mu_2,$ so $\mu_j=r_j(\alpha)\nu$. Then

$$
T_1-T_2= \int \left(h_1^\alpha r_1(\alpha)-h_2^\alpha r_2(\alpha)\right)
[V_\alpha]d\nu(\alpha).$$

Since $d(T_1-T_2)=0$ it follows that $$h_1^\alpha r_1(\alpha)-
h_2^\alpha r_2(\alpha) \equiv c(\alpha)$$ 

We decompose the measure $c(\alpha)\nu(\alpha)$ 
on the space of plaques, $c(\alpha)
\nu(\alpha)=\lambda_1-\lambda_2$ for positive mutually singular
measures $\lambda_j.$ Then

$$T_1-T_2= \int [V_\alpha]\lambda_1(\alpha)-\int [V_\alpha]\lambda_2(\alpha)
=T^+-T^-$$ for positive closed currents $T^\pm.$ These locally defined
currents fit together to global positive closed currents on $K\setminus E.$
Observe that the mass of $T^\pm$ is bounded by the mass of $T_1+T_2$.

Since $E$ is
locally complete pluripolar the trivial extensions of $T^\pm$ are also closed. 

 Consequently $T^\pm \equiv 0$ and $T_1=T_2.$
The fact that $T$ is extremal follows from the strict concavity of $Q$.\\


\CQFD\\


\begin{coro}
Let $(X,{\cal L},E)$ be a laminated set with singularities in $\PP^2$ with
$\Lambda_2(E)=0.$ One of the
following statements holds:\\
i) There is a closed positive laminated 
current of mass $1$ on $X.$\\
ii) There is a unique positive laminated harmonic current of mass one
on $X.$\\
iii) There is a positive laminated harmonic current $T$ of mass one on $X$
such that $\int T \wedge T >0.$ In particular the current $T_0$ 
minimizing energy satisfies $\int T_0 \wedge T_0>0.$
\end{coro}

{\bf Proof:} Assume i) and ii) fail. Then there are two
positive harmonic non closed currents $T_1,T_2$ of mass one which
are not colinear. We can assume that $\int (T_1-T_2)\wedge \omega =0.$ 
By Corollary 2.14,
$Q(T_1-T_2,T_1-T_2)<0.$ If $Q(T_1,T_1)$ or $Q(T_2,T_2)$ is strictly
positive, we are done since $\int T \wedge T= Q(T,T).$ If not, then
$Q(T_1,T_2)=\int T_1 \wedge T_2 >0.$ But then if $T:=\frac{T_1+T_2}{2},$
$Q(T,T)>0.$\\

\CQFD\\



\section{Intersections of laminar currents}




\subsection{${\cal C}^1$ laminations}

Here we assume that $X\subset \mathbb P^2$ is a laminated compact
covered by finitely many flow boxes $B_i.$
We suppose that $X$ locally extends to a ${\cal C}^1$ lamination of
an open neighborhood.
We can assume that $p:=[0:0:1]\in B_1 \subset X$ and that the leaf $L$ through
$p$ has the form $w={\cal O}(z^2)$. So we will assume that the lamination of $X$ is of the
form $w=w_0+f_{w_0}(z), f_{w_0}(0)=0,$ where the map $\Psi(z,w_0)=(z,
w_0+f_{w_0}(z))$ is a ${\cal C}^1$ diffeomorphism in a neighborhood
of $p.$

\medskip

Let $\Phi_\epsilon([z:w:t])=[z,w+\epsilon z:t]$ denote
a family of automorphisms of $\mathbb P^2.$ Notice that each
of these automorphisms fixes the $w$ axis.

For two graphs $L_1,L_2$ given by $w=g_1(z),w=g_2(z), z\in S
,$ we define the vertical distances over $S$ between the two
as $d_S^{max}(L_1,L_2)=\sup_{z\in S}|f_1(z)-f_2(z)|$
and $d_S^{min}(L_1,L_2)=\inf_{z\in S}|f_1(z)-f_2(z)|.$



\begin{theo}
There exists an integer $N$ independent of $\epsilon $
so that in any of the flow boxes $B_i$
local leaves $L_1$ and $\Phi_\epsilon(L_2)$ can at most
intersect in $N$ points, counted with multiplicity. Moreover there exist
neighborhoods $U_\epsilon$ of ${\mbox{Id}}$ in $U(3)$ so that the
same conclusion holds for $\Psi_1(L_1)$ and $\Psi_2(\Phi_\epsilon(L_2))$,
$\Psi_1, \Psi_2\in U_\epsilon.$  
\end{theo}


{\bf Proof:} Fix a $\delta>0.$ Let $L_{w}$ denote the leaf
through $[0:w:1]$ and let $L^\epsilon_w$ denote
its image under $\Phi_\epsilon.$ 
Say $L_{w_0}$ is given by $w=w_0+f_{w_0}(z)$ 
and $L^\epsilon_{w_0}$ is given by
$w=w_0+f_{w_0}(z)+\epsilon z.$ Note that the vertical
distance $d_S^{\mbox{max}}$ between  $L_{w_0}$ and $L^\epsilon_{w_0}$
is $|\epsilon| \delta$ at the boundary $S$ of the disc $|z|\leq 
\delta.$
Because the lamination is of class ${\mathcal C}^1,$ there
exists a constant $C>1$ so that if $(0,w_0),(0,w_1)\in B_1,$
then 

$$
\sup_{|z|\leq \delta}
 |w_0+f_{w_0}(z)-w_1-f_{w_1}(z)| \leq C
\inf_{|z|\leq \delta}
 |w_0+f_{w_0}(z)-w_1-f_{w_1}(z)|$$

If $L_1$ and $\Phi_\epsilon(L_2)$ intersect in a flow box $B_i$,
then $L_1$ and
$L_2$  must be at most $a|\epsilon|$ apart in $B_i$ for some
fixed $a.$ Let $c>0$ be any small constant. If
$L_1$,  $\Phi_\epsilon(L_2)$ intersect in $N$ points
in $B_i$ and $N$ is sufficiently large then
$L_1$ and  $\Phi_\epsilon(L_2)$ can be at most a distance
$c |\epsilon|$ apart from each other in $B_i.$ Note that the
same conclusion holds if the number of intersections is counted for
$\Psi_1(L_1)$ and $\Psi_2(\Phi_\epsilon(L_2))$ for a small enough
$U_\epsilon.$  
However, there is a
path of at most a fixed length along these leaves ending
in the flow box containing $[0:0:1]$.
It follows that continuing these leaves to this
neighborhood, they will have to stay at most $bc |\epsilon|$ apart,
for a fixed constant $b.$ Choosing $c$ small enough, we get
$bc<\frac{\delta}{2C}$. Let $L_1=\{w=w_1+f_{w_1}(z)\},$ 
$L_2=\{w=w_2+f_{w_2}(z)\}.$
Then $|w_2+f_{w_2}(z)+\epsilon z-w_1-f_{w_1}(z) |
<\frac{|\epsilon|\delta}{2C}$ when $|z| \leq \delta.$ Hence


\bea
d_{|z| \leq \delta}^{\mbox{max}}(L_1,L_2) & \leq &
C d_{|z| \leq \delta}^{\mbox{min}}(L_1,L_2)\\
 & \leq & C d_{z=0}^{\mbox{min}}(L_1,L_2)\\
& = & C d_{z=0}^{\mbox{max}}(L_1,\Phi_\epsilon (L_2))\\
& \leq & C d_{|z| \leq \delta}^{\mbox{max}}(L_1,\Phi_\epsilon (L_2))\\
& \leq & \frac{|\epsilon|\delta}{2}.\\
\eea

Applying this estimate when $|z|=\delta,$ we get 

\bea
\frac{|\epsilon|\delta}{2} & > &  \frac{|\epsilon|\delta}{2C}\\
& > & |w_2+f_{w_2}(z) +\epsilon z-w_1-f_{w_1}(z)|\\
& \geq & |\epsilon| \delta -|w_2+f_{w_2}(z)-w_1-f_{w_1}(z)|\\
& \geq & |\epsilon|\delta -\frac{|\epsilon|\delta}{2}\\
& = & \frac{|\epsilon|\delta}{2},\\
\eea

\noindent a contradiction.\\


\CQFD\\


\subsection{Laminations by holomorphic motions}

Now we consider the case of laminations which are not ${\mathcal C}^1.$
We recall the following result
by Bers-Royden [BR].

\begin{proposition}
We are given a lamination of a neighborhood of the unit polydisc
in $\CC^2.$ Assume that the leaves are of the following
form:

$$
L_t, t\in \CC, |t|<C, w=F_t(z), F_t(0)=t,F_0(z)\equiv 0.
$$

The map $\Phi(z)(t)=F_t(z)$ is a holomorphic motion and we have
the estimate:

$$
\frac{1}{K}|t-s|^{\frac{1+|z|}{1-|z|}} \leq |F_t(z)-F_s(z)|
\leq {K}|t-s|^{\frac{1-|z|}{1+|z|}}.
$$
\end{proposition}




\begin{theo}
Let $X\subset \mathbb P^2$ be a compact subset laminated
by Riemann surfaces. Then there exists a holomorphic family
$\Phi_\epsilon:\mathbb P^2 \rightarrow \mathbb P^2$ for
$\epsilon\in \mathbb C, \Phi_0\equiv \; {\mbox{Id}}$ with the
following properties. There are finitely many flow boxes
$\{B_i\}_{i=1,\dots,\ell}$ covering $X$ and an $\epsilon_0>0$ and a constant
$A$ such that if $L,L'$ are any local leaves in any flow box
$B_i$ then:\\
If $0<|\epsilon|<\epsilon_0$, the number of intersection points counted
with multiplicity of $L,\Phi_\epsilon(L')$ is at 
most $A \log \frac{1}{|\epsilon|}.$ Moreover there exist
neighborhoods $U_\epsilon$ of ${\mbox{Id}}$ in $U(3)$ so that the
same conclusion holds for $\Psi_1(L_1)$ and $\Psi_2(\Phi_\epsilon(L_2))$,
with $\Psi_1,\Psi_2 \in U_\epsilon.$
\end{theo}

{\bf Proof:}
We first choose a finite cover by flow boxes, $B_i$.
We can do this so that for each flow box there is a
linear change of coordinates in $\mathbb P^2$ so
that $[z:w:t]=[0:0:1]\in B_i \cap K$. Moreover, we can arrange
that if $L$ is any local leaf intersecting $\Delta(0,2)$
then $L\cap  \Delta(0,2)$ is contained in a local leaf
$\tilde{L}$ of the form $\{w=f_{\alpha}(z), |z|<3\},
(0,\alpha)\in \tilde{L},$
and $\|f_\alpha\|_\infty<3.$ Moreover we can assume that
each $\|f'_\alpha\|<.1$ and that $f'_0(0)=0.$ Redefining the flow boxes, we
can let
$B_i$ denote the union of those graphs over $|z|<3$
intersecting $\Delta(0,2).$ We can assume that
the smaller flow boxes $B_i'$ consisting of those
graphs over $|z|<1$ for which the graph is in $\Delta^2(0,1)$
already cover $K.$\\


Next we fix the coordinates $z,w,t$ on $\mathbb P^2$ used for
the first flow box $B_1$. Define the family $\Phi_\epsilon$ by
$$
\Phi_\epsilon [z:w:t]= [z:w+\epsilon z:t].
$$

\begin{lem}
There exists a $\; \delta>0$  and $C>0$ so that if
$w=f_\alpha (z), w=f_\beta (z)$ are two local leaves in $B_1$, then
$$
\frac{|\alpha-\beta|^2}{C}
 \leq |f_\alpha (z)-f_\beta (z)| \leq C|\alpha-\beta|^{1/2}, 
\; \forall \; z, |z| \leq \delta.
$$
\end{lem}

{\bf Proof:} This is a special case of the Bers-Royden result.\\

\CQFD\\

\begin{lem}
Let  $\epsilon_0>0$ be small enough. Then if
$L_1,L_2$ are leaves in the first flow box then
$d^{max}_{\{|z|\leq \delta\}}(\Phi_\epsilon (L_1),L_2)\geq |\epsilon|^3$
for all $|\epsilon|\leq \epsilon_0.$
\end{lem}

{\bf Proof:}
Let $L_i$ be given by $w=f_i(z), f_i(0)=w_i.$ Then
 $\Phi_\epsilon(L_1)$ is the graph $w=f_1(z)+\epsilon z.$
Suppose that $|f_1(z)+\epsilon z-f_2(z)| \leq |\epsilon|^3$
for all $|z|\leq \delta.$
Then, we get that $|w_1-w_2| \leq |\epsilon|^3.$ Hence, by the
previous Lemma, we have that $|f_1(z)-f_2(z)| \leq C|\epsilon|^{\frac{3}{2}}$
for all $|z| \leq \delta.$ Hence if $|z|=\delta,$

\bea
|\epsilon|^3 & \geq &  |f_2(z)+\epsilon z-f_1(z)|\\
& \geq & |\epsilon |\delta-   |f_2(z)-f_1(z)|\\
& \geq & |\epsilon| \delta- C|\epsilon|^{\frac{3}{2}}\\
& \geq & |\epsilon|(\delta-C \sqrt{|\epsilon|}),\\
& \Rightarrow & \\
\epsilon_0^2 & \geq & |\epsilon|^2 \geq  \delta-C \sqrt{|\epsilon|}
 \geq  \delta-C \sqrt{\epsilon_0}\\
& \Rightarrow & \\
\epsilon_0^2+C \sqrt{\epsilon_0} & \geq & \delta,\\
\eea

\noindent a contradiction if $\epsilon_0$ is small enough.\\

\CQFD

The following lemma is well known.

\begin{lem}
a) There is a constant $0<c<1$ so that the following holds:
Let $g$ be a holomorphic function on the unit disc
with $|g|<1$ and suppose that $g$ has $N$ zeroes in $\Delta(0,1/2).$
Then $|g|\leq c^N$ on  $\Delta(0,1/2).$\\
b) Let $g$ denote a holomorphic function on the unit disc and suppose
that $|g|<1$ and that $|g|< \eta<1$ on  $\Delta(0,1/4).$ Then
 $|g|< \sqrt{\eta}$ on  $\Delta(0,1/2).$
\end{lem}

{\bf Proof:} To prove a) set $$c= sup_{|\alpha|\leq 1/2, |z|\leq 1/2}
\frac{|z-\alpha|}{|1-z\overline{\alpha}|}<1.$$ To prove b)
observe that $\log |g|< \log \eta$ when $|z| < \frac{1}{4}.$
Hence by subharmonicity
$$
\log |g| \leq \max\{ \log \eta \frac{\log|z|}{\log 1/4}, \log \eta\}.
$$

This implies that if $|z|=1/2$, then $\log |g| \leq \frac{\log \eta}{2}.$

 \CQFD\\

{\bf Continuation of the Proof of Theorem 3.4:} Pick $\rho>0.$
Let $p\in X$. Since every leaf is dense, there is a (nonunique) continuous
curve $\gamma_p(t), 0 \leq t \leq 1$ from $\gamma(0)=p$ to a point $
\gamma_p(1)=(0,w_p)\in B_1'$ which
is contained in the leaf through $p.$ By continuity, for every
$q\in K$ close enough to $p,$ the curve $\gamma_q$  can be chosen
so that $\mbox{dist}(\gamma_q(t),\gamma_p(t))\leq \rho$
for all $0\leq t\leq 1.$

\medskip
A chain of flow boxes is a finite collection $C=\{B_{i(j)}\}_{j=1}^k$.
Let $p \in X$. We say that the leaf through $p$
follows the chain $\{B_{i(j)}\}_{j=1}^k$ if there are local
leaves $L_j \subset B_{i(j)}, \hat{L}_j:=L_j \cap B'_{i(j)},
p\in \hat{L}_1, \hat{L}_j \cap \hat{L}_{j+1}\neq \emptyset\; \forall \; j<k,
i(k)=1.$\\

By compactness there are finitely many chains of flow boxes $C_1,
\dots, C_\ell$ such that for each $p\in X$, there is an open neighborhood
$U(p)$ and a chain $C_r$ so that the leaf through $q$
follows $C_r$ for any $q\in U(p)\cap X.$\\


We will apply Lemma 3.6 repeatedly along a chain. We need to apply Lemma
3.6
at most a fixed number of times $m$ depending on the length
of each chain.
Note that every time we switch flow box there is a change of coordinates
which distorts distances by at most a factor $C>1.$

\begin{lem}
Let $\epsilon$ be sufficiently small and suppose
that $N=N(\epsilon)$ is an integer such that
$C^2c^{\frac{N}{2^m}}\leq |\epsilon|^3.$ Then
no local leaves of the laminations $L_1,\Phi_\epsilon(L_0)$
can intersect more than $N$ times in any flow box.
\end{lem}

{\bf Proof:}
Suppose that  local components of 
$L_1$ and $\Phi_\epsilon(L_0)$ intersect in more than
$N$ points in some local flow box $B'_s$. Then these local graphs
differ by at most $c^N.$ Using Lemma 3.6 they differ
by at most $c^{\frac{N}{2}}$ in a suitable larger flow box.
Changing to the coordinates of another flow box might increase
the difference to $C c^{\frac{N}{2}}.$ Applying Lemma 3.6
the difference increases to at most $C^{\frac{1}{2}}
c^{\frac{N}{4}}$ and after another change of flow box
to $C^{\frac{3}{2}}c^{\frac{N}{4}}.$ Following the leaves along a chain of flow
boxes we see inductively that the distance between continuations
of the leaves grows at most
like $C^2 c^{\frac{N}{2^k}}$ after $k$ steps.
Hence once we are in the first flow box,
the leaves differ
by at most $|\epsilon|^3.$ By the above
lemmas, this is impossible for any pair of leaves.\\

\CQFD\\

There is a constant $A$ so that for all small enough $\epsilon$
local leaves $L_1,\Phi_\epsilon(L_0)$ have at most
$N_\epsilon:=A \log \frac{1}{|\epsilon|}$ intersection points. The contruction
is stable under small perturbations by $\Psi_1,\Psi_2$
close to the identity.\\

\CQFD\\


\section{Construction from discs. Ahlfors type construction.}


In this paragraph we consider a laminated set $(X,{\mathcal L},E)$
in a compact complex manifold $M.$
 
We want to construct harmonic currents using the Ahlfors exhaustion
technique.

\subsection{When leaves are not uniformly Kobayashi hyperbolic}

We consider only the case when $X$ is not a compact
Riemann surface, possibly singular. Consider the universal covering
for each leaf. We can assume that the covering is $\CC$ or the
unit disc $\Delta.$ Let $\phi:\Delta \rightarrow L$ be a covering map.
If $|\phi'(0)|$ is not uniformly bounded, then using the Brody
technique, one can construct an image of $\CC$. The part of the image not
in $E$ is locally
contained in a leaf of $X$.

The Ahlfors exhaustion technique furnishes a positive closed current
of mass $1$ directed by the lamination. So we get the following proposition:

\begin{proposition}
Let $(X,{\mathcal L},E)$ be a laminated set with singularities. If there
is no positive closed current on $X,$ laminated on $X \setminus E$, then
there is a constant $C$ such that $|\phi'(0)| \leq C.$ 
\end{proposition}

When the $|\phi'(0)|$
are uniformly bounded, we say that the leaves are uniformly hyperbolic.



\subsection{The case with no positive closed current
directed by $(X,{\mathcal L},E)$} Let $X$ be a minimal laminated compact 
set in $\PP^2$. Suppose that
$X$ does not contain any non constant holomorphic image of $\CC$. Let $B_i$ 
be a 
covering of $K\setminus E$ by flow boxes which in local coordinates are of
the form $w_i=f_\alpha(z_i), |z_i|<1$ [but the graphs extend uniformly
to $|z_i|<2$]. Let $\Phi:\Delta\rightarrow L$ denote the universal
covering of an arbitrary leaf. We say that $x\in \Delta$ is a center point
if $\Phi(x)=(0,w_i)$ in some $B_i.$ We can normalize $\Phi$
for any center point, say $\Phi_x:\Delta\rightarrow L, \Phi_x(0)=\Phi(x).$
[i.e. we move $x$ to $0$ with an automorphism of the unit disc.]
Let $w_i=f_x(z_i)$ denote the associated graph in the flow box.
 Denote by $U_x:=\Phi_x^{-1}(\{(z_i,f_x(z_i));|z_i|<1\}).$ If $\Phi$
is a multisheeted covering, we let $U_x$ denote the connected
component containing $0.$ Then $U_x\subset
\Delta$ is a relatively compact open subset of $\Delta$ containing
$0.$ Let $0<r_x\leq R_x<1$ denote the largest, respectively smallest
radii such that $\Delta(0,r_x)\subset U_x\subset \Delta(0,R_x).$

\begin{lem}
$R_x\leq 1/2$.
\end{lem}

{\bf Proof:} Since $\Phi^{-1}_x$ maps $\Delta(0,2)$ into $\Delta(0,1)$
and sends $0$ to $0$ this follows from the Schwarz' Lemma.\\

\CQFD\\

\begin{lem}
Fix a finite number of flow boxes $\overline{B}_i \cap E=\emptyset,
i=1,\dots,\ell.$ Then $\inf \{r_x; x\; {\mbox{is a center point}},
\phi(x) \in \cup_{i=1}^\ell B_i\}>0.$
In fact the same holds if we inf over all leaves and all covers of the leaves
by discs.
\end{lem}

{\bf Proof:}
Fix an $x$ and a covering $\Phi:\Delta \rightarrow L$ of the leaf
through $x.$ Note that $\Phi_x^{-1}(\Delta(0,1)) \supset \Delta(0,r_x)$
hence by the Koebe 1/4 Theorem, $[\Phi_x^{-1}]'(0)=\alpha, |\alpha|\leq 4r_x.$
Hence  $\Phi'_x(0)=\frac{1}{\alpha}, \left| \frac{1}{\alpha}\right|
\geq \frac{1}{4r_x}.$ If $r_x \rightarrow 0$ then using the Brody
technique we construct an image of $\CC$ contained in $X.$\\

\CQFD\\

Next we prove a density Theorem for the above minimal laminations
with only Kobayashi hyperbolic leaves:

\begin{theo}
Assume $E =\emptyset.$ Fix a finite cover of $X $
by flow boxes $B_i.$
There are constants $R, N$ so that if $\Phi:\Delta\rightarrow X$
is a covering of any leaf, then $\Phi(\Delta_{kob}(x,R)\cap B_i)$
intersects at most $N$ of the graphs in $\{|z_i|<\frac{3}{2}\}$ and contains
at least one complete graph over $|z_i|<1.$
\end{theo}

{\bf Proof:} 
Let $\eta\in \cup B_j$ and $L_\eta$ the leaf through $\eta.$
There exist finitely many curves $\gamma_i^\eta\subset L_\eta,
\gamma_i^\eta(0)=\eta,
\gamma_i^\eta(1)\in B_i.$ In fact, for every $\zeta\in U(\eta)\cap X$
there are curves $\gamma_i^\zeta \subset L_\zeta$ depending continuously
on $\zeta$ and landing in $B_i$ for a small enough neighborhood $U(\eta).$
There exists a finite number $M_\eta$ so that if $\zeta\in U(\eta)\cap X,
t\in \left[\frac{j-1}{M_\eta},\frac{j}{M_\eta}\right],$ then
$\gamma_i^\zeta(t)\in \Delta_{i,\zeta,j}$ where $\Delta_{i,\zeta,j}$
is one of the unit discs in one of our chosen local flow boxes. 
It follows that if $\Phi:\Delta\rightarrow
L$ is any covering of any leaf and $\Phi(x_0)\in U(\eta),$ then
$\Phi(\Delta_{kob}(x_0,R_\eta))$ contains a disc in each flow box if $R_\eta$
is
large enough. By compactness of $X,$ there is an $R={\mbox{max}}\; R_\eta>0$
so that
if $\Phi:\Delta \rightarrow X$ is a covering of any leaf and $x \in 
\Delta$, then $\Phi(\Delta_{kob}(x,R))$ contains at least one
complete graph in each flow box. There is by Lemma 4.2 an $R'>R$
so that if $\zeta\in \Phi(\Delta_{kob}(x,R)) \cap B_i$ (or a slight extension),
then the whole graph in $B_i$ is contained in $\Phi(\Delta_{kob}(x,R')).$
It follows from Lemma 4.3 that there is an $N$ so that no
$\Phi(\Delta_{kob}(x,R))$ can intersect any flow box in more than
$N$ graphs.\\


\CQFD\\

\begin{coro}
Let $R$ be sufficiently large. Then if $D$ is any maximal disc
of center $p$ contained in an annulus $1-1/R^k<|x|<1-1/R^{k+1}$, 
and assume $\Phi(p) \in \cup B_j$, then the image
$\Phi(D)$ always will contain a full disc in each $B_i$ and will intersect
at most $N$ graphs in any $B_i$.
\end{coro}

{\bf Proof:} This follows since the Poincar\'e radius of these discs
increase to infinity independently of $k$ as $R\rightarrow \infty$.\\

\CQFD\\

\begin{coro} For large $R$ the area of $\Phi(\{
 1-1/R^k<|x|<1-1/R^{k+1}\})$ grows like $R^k$ as $k \rightarrow 
\infty.$
Moreover the length of $\Phi(\{|x|=1-1/R^{k+1}\})$ is on the
order of $R^k,$ and $|\Phi'(x)| \sim \frac{1}{1-|x|}$ near
$\partial \Delta,$ so
$$
 \int_\Delta(1-|x|)|\Phi'(x)|^2d\lambda(x)=\infty
$$
\end{coro}

{\bf Proof:} The map $\Phi$ is an isometry for the Kobayashi distance.
The Kobayashi distance  on leaves is comparable to any given Hermitian
metric. The map $\Phi$ expands a disc of radius $\epsilon$
centered at $1-2\epsilon$ to a disc of radius about $1$ in the
Kobayashi metric. Hence $|\Phi'(x)| \sim \frac{1}{1-|x|}.$\\


\CQFD\\


\begin{theo}
Let $\phi:\Delta \rightarrow M$ be a holomorphic map where $M$ 
is a compact complex Hermitian manifold. If
$$
 \int_\Delta(1-|x|)|\phi'(x)|^2d\lambda(x)=\infty,
$$
\noindent then there is a positive harmonic current $T$, supported on
$\overline{\phi(\Delta)}.$
If $\phi(\Delta)$ is contained in a leaf of a lamination ${\cal L}$, then the current
$T$ is directed by the lamination.
\end{theo}

{\bf Proof:}
Assume that $\phi(0)=p.$ Define 
$$
G_r(x):=\frac{1}{2\pi}\log^+ \frac{r}{|x|}, T_r:=(\phi)_*(G_r[\Delta]
), r<1.
$$ 
If $\theta$ is a $(1,1)$ test form on $M$
$$
<T_r,\theta>=\frac{1}{2\pi}\int_{\Delta}\log^+ \frac{r}{|x|}\phi^*(\theta).
$$
So $T_r$ is positive of bidimension $(1,1)$. The mass of $T_r$
is comparable to \\
$\int_{\Delta}\log \frac{r}{|x|}|\phi'(x)|^2 \sim
\int_{\{|x|<r\}}(r-|x|)|\phi'(x)|^2=:A(r).$ A direct computation gives
$$
i \partial \overline{\partial} T_r=\phi_*(\nu_r)-\delta_p,
$$
\noindent where $\delta_p$ is the Dirac mass at $p$ and $\nu_r$ is the Lebesgue measure on the circle
of radius $r.$ Let $T$ be a cluster point of $T'_r:=\frac{T_r}{A(r)}.$
Since $A(r) \rightarrow \infty$ we have $i \partial \overline{\partial}T=0.$

\medskip

If $\phi(\Delta)$ is a leaf of a lamination ${\cal L}$, there is a $(1,0)$
form $\gamma$ such that $[\phi(\Delta)]\wedge \gamma=0.$ Hence
$T_r \wedge \gamma=0.$ 
Hence $<T,i\partial_b\overline{\partial}_bf>=0$ and we have a decomposition
of $T$ as in Theorem 2.18.\\


\CQFD\\


\begin{remark} If we assume that $\lim_{r \rightarrow 1}
(1-r) \int_{D_r} |\phi'(z)|^2=\infty,$ it follows from a result of Ahlfors
that $\underline{\lim}_{r\rightarrow 1} \frac{\ell(\phi(\partial
\Delta_r))}{{\mbox{Area}}(\phi(\Delta_r))}\rightarrow 0,$ $\ell$ represents
length. Hence one can choose the current $T$ to be closed.\\
In the usual Ahlfors procedure to construct a closed current
starting from an image of $\CC$ one has to first extract good subsequences from
$\frac{\Phi_*[\Delta_R]}{{\mbox{Area}}\; \Phi_* [\Delta_R]}$ when
$R \rightarrow \infty.$ Then cluster points of these give closed
currents. In our case, there is no need to first take a subsequence.
\end{remark}



\begin{theo}
There is no nonvanishing holomorphic 
vector field
along leaves on a laminated
compact with only hyperbolic leaves.
\end{theo}

{\bf Proof:} If we pull back the vector field to a disc covering
a leaf, we get a holomorphic function on the unit disc going uniformly
to infinity at the boundary, since as we have seen $|\phi'(x)| \sim \frac{1}
{1-|x|}$ and the vector field is bounded in flow boxes.\\

\CQFD\\

\begin{proposition}
If a positive harmonic current gives mass to a leaf, then this leaf is
a compact Riemann surface.
\end{proposition}

\begin{lem}
Let $T$ be a laminated harmonic current. Let $\phi$
denote the covering map $\Delta \rightarrow L.$ If $H$ denotes the
analytic continuation of $h \circ \phi$ in a flow box, then we have the
estimate:
$$
c (1-|x|) \leq H  \leq C \frac{1}{(1-|x|)}.
$$
\end{lem}

{\bf Proof:}
The estimate 
follows from Harnack's inequality and the Hopf Lemma.\\

\CQFD\\

{\bf Proof of the Proposition:} We assume first that the leaf is hyperbolic.
Let $\phi, H$ be as in the Lemma.
Suppose at first that $H$ is unbounded. Then we can choose a sequence
$p_n \rightarrow \partial \Delta$  and $H(p_n) \rightarrow \infty$ such
that $H$ is uniformly large on $\Delta(p_n,R)$
by Harnack, where $R$ is as in Theorem 4.4.
Hence $T$ will have infinite mass on a flow box.
If $H$ is bounded and nonconstant, we can choose $\theta_n$ so that
$lim_{r \rightarrow 1}H(re^{i \theta_n})$ are different.
We can again choose $p_n \rightarrow \partial \Delta$  so that
$\phi(\Delta(p_n,R))$ are disjoint and again $T$ will have infinite mass
on a flow box. If $H$ is constant, we get a positive
closed current and the leaf has an analytic closure. The same
argument applies to the case when the leaf is not hyperbolic.\\


\CQFD\\

\section{Vanishing of $\int T \wedge T.$}

\begin{definition}
The harmonic laminated current has finite transverse energy if
in some local flow box
$\int \log |\alpha-\beta| d\mu(\alpha) d\mu(\beta)<\infty.$
\end{definition}

Having finite transverse energy is well defined and independent of the 
choice of flow box.

\medskip

Recall that $\Phi_\epsilon([z:w:t])=[z:w+\epsilon z:t].$ If $T$
is a current, let $T_\epsilon:= (\Phi_\epsilon)_*(T).$ 


\begin{theo}
If a harmonic current for a
laminated compact in $\PP^2$ has finite transverse energy, then
the geometric intersection $T \wedge T_\epsilon \rightarrow 0.$
The same conclusion holds for ${\mathcal C}^1-$laminations without the
hypothesis of finite transverse energy. In both cases we have
$\int T \wedge T=0.$  
\end{theo}

{\bf Proof:}
We calculate the geometric wedge product $T \wedge T^\epsilon$ in a flow box.
Set $T=\int h_\alpha[\Delta_\alpha]d\mu(\alpha),
T^\epsilon =\int h_\beta^\epsilon[\Delta^\epsilon_\beta]d\mu(\beta).$
Let $\phi$ be a test function supported in a flow box.
To avoid confusion, we index with $g$ when wedge products are geometric
during the proof. 
We have

$$
<T \wedge T^\epsilon,\phi>_g=
\int \sum_{p \in J_{\alpha,\beta}}\phi h_\alpha(p)h^\epsilon_\beta(p)
d\mu(\alpha)d\mu(\beta)$$

\noindent where $J_{\alpha,\beta}$
consists of the intersection points of $\Delta_\alpha$ and
$\Delta^\epsilon_\beta.$ Assume at first that $\mu$ has finite
transverse energy. Using the estimate on the size of $J_{\alpha,\beta}$
in Theorem 3.3, we get:

\bea
|(T \wedge T^\epsilon)_g (\phi)| & \leq & C_1
\|\phi\|_\infty   \int_{{\mbox{dist}}(\Delta_\alpha,\Delta_\beta)\leq
C \epsilon} A \log \frac{1}{|\epsilon|}d\mu(\alpha)d\mu (\beta)\\
& \leq & C_2\|\phi\|_\infty
\int_{{\mbox{dist}}(\Delta_\alpha,\Delta_\beta)\leq C \epsilon}
 \log \frac{1}{dist(\Delta_\alpha,\Delta_\beta)}d\mu(\alpha)d\mu(\beta)\\
& \rightarrow & 0 \; as\; \epsilon \rightarrow 0\\
\eea

In the ${\mathcal C}^1$ case the number of intersection points is bounded by 
$N$ independent of $\epsilon$ (Theorem 3.1). Hence

\bea
|(T \wedge T^\epsilon)_g(\phi)| & \leq &
C \|\phi\|_\infty \int_{{\mbox{dist}}(\Delta_\alpha,\Delta_\beta)
\leq C\epsilon}Nd\mu(\alpha)d\mu(\beta)\\
& \rightarrow & 0\\
\eea

\noindent since $\mu$ has no pointmasses by Proposition 4.10.
Next we show that $Q(T,T)=\int T \wedge T=0.$ It suffices to show
by Theorem 2.12 that $Q(T,T_\epsilon) \rightarrow 0$ or even that
for smoothings $T^\delta, T_\epsilon^{\delta'}$ that
$Q(T^\delta,T_\epsilon^{\delta'}) \rightarrow 0$ when
$\delta,\delta'$ are sufficiently small compared to $\epsilon$
and $\delta,\delta',\epsilon \rightarrow 0.$ 

\medskip

Note that the estimate on the geometric wedge product is stable under
small translations of $T,T_\epsilon$. This is what allows us to
smooth.






Let $\phi$ be a test function supported in some local
flow box. 
As above, the value of the geometric wedge product
on $\phi$ is:

$$
<T \wedge T^\epsilon,\phi>_g=
\int \sum_{p \in J_{\alpha,\beta}}\phi h_\alpha(p)h^\epsilon_\beta(p)
d\mu(\alpha)d\mu(\beta)$$

We can write this as

$$
<T \wedge T^\epsilon,\phi>_g=
\int \left(
\int_{\Delta^\epsilon_\beta}[\phi h_\alpha h^\epsilon_\beta](p)
i\partial \overline{\partial} \log|w-f_\alpha(z)|\right) 
d\mu(\alpha)d\mu(\beta)$$

The same applies when we do this for slight translations within small
neighborhoods $U(\epsilon)$ of the identity in $U(3)$
and
their smooth averages $T^\delta,$


$$
<T^\delta \wedge T_\epsilon,\phi>_g=
\int \left(
\int_{\Delta^\epsilon_\beta}[\phi h^\epsilon_\beta](p)
T^\delta \right) d\mu(\beta)=<T_\epsilon,\phi T^\delta>.$$

Averaging also over small translations of $T^\epsilon$ we get

$$
<T^\delta \wedge T^{\delta'}_\epsilon,\phi>_g=
<T^{\delta'}_\epsilon,\phi T^\delta>.$$

We still have that $<T_\epsilon^{\delta'},\phi T^\delta> \rightarrow 0$
when $\delta,\delta' << \epsilon, \epsilon \rightarrow 0.$ If we apply this
to $\phi=1$, we get
$<T^{\delta'}_\epsilon,T^\delta>=Q(T^{\delta'}_\epsilon,T^\delta)
\rightarrow 0.$ Hence
$Q(T,T)=0.$\\

\CQFD\\




\begin{coro}
If a laminated compact set in $\PP^2$
carries a positive closed laminar 
current $T$, then $T$ has infinite transverse energy.
\end{coro}

{\bf Proof:}
If $T \neq 0$ has finite energy, then $0=\int T \wedge T=
|\int T \wedge \omega|^2-E(T,T)$ but $E(T,T)=0$
since $T$ is closed. Hence $T=0$, a contradiction.\\


\CQFD\\

J. Duval has independently obtained this Corollary.
Hurder and Mitsumatsu proved that there is no ${\mathcal C}^1$
lamination in $\PP^2$ which carries a positive closed current
[HM].


\begin{coro}
If $(X,{\mathcal L})$ is a ${\mathcal C}^1$ lamination on $\PP^2$
with only hyperbolic leaves,
then $$T=\lim_{r \nearrow 1}\phi_*\frac{
\left(\log^+ \left(\frac{r}{|z|}\right)[D_r]\right)}{A_r}$$
\noindent uniformly with respect to $\phi.$
\end{coro}

{\bf Proof:}
We know from [HM] that there is no positive closed current directed by 
${\mathcal L}$. It follows from Corollary 2.23 and Theorem 5.2 that there
is a unique harmonique current of mass $1$ on $(X,{\mathcal L})$.
Hence the result follows.\\

\CQFD\\



\section{Examples of harmonic current}


In this section we investigate harmonic currents on $\PP^2$
of the form $T=i \partial u \wedge \overline{\partial u}.$
Our main result is that if $u \in {\cal C}^2_{\RR}(\PP^2)$
and $i \partial \overline{\partial} T=0,$ then $u$ is constant,
hence $T\equiv 0.$ We also compute the energy of some positive
harmonic currents. 

\medskip

Let $M$ be a complex manifold of dimension $m.$
For $1 \leq k \leq m,$
 we define ${\cal P}^{(k)}_-(M)$ as the cone of upper semicontinuous
real functions $v$ on $M$ such that for every $p\in M,$
there is an open neighborhood $U$ of $p$ and $\{v_n\}\subset {\cal C}^2(U)$ such
that $v_n \searrow v$ in $U$ and
$(-1)^k(i\partial \overline{\partial} v_n)^k \leq \epsilon_n \omega^k, \epsilon_n \searrow 0.$ We say
that $U$ is associated to  $v.$ Here 
$\omega$ denotes a strictly positive hermitian form. 
Notice that this condition implies that when $\epsilon_n=0,$
not all eigenvalues of $i\partial \overline{\partial} v_n$ can be strictly negative.
We define ${\cal P}^{(k)}_+:=-{\cal P}^{(k)}_-(M)$ and ${\cal P}^{(k)}
={\cal P}_+^{(k)}\cap {\cal P}_-^{(k)}.$
In particular, $P^{(1)}_-$ consists of the plurisubharmonic functions
and $P^{(1)}$ are the pluriharmonic functions. In dimension $2$, a
smooth function
$v$ belongs to $P^{(2)}_-$ if and only if its Levi form has at most one
eigenvalue of each sign. These functions then, also belong to 
$P^{(2)}_+$ and hence $P^{(2)}.$ Pseudoconvex domains are usually 
characterized by plurisubharmonic functions, i.e. $P^{(1)}_-.$ 
We show here that $P^{(2)}_-$ works as well, and that there are
similar results for $P^{(k)}_-, k>2.$




 
 Let $\Phi:M \rightarrow N$ be a holomorphic map between complex manifolds.
If $v\in {\cal P}^{(k)}_-(N),$ then
$v \circ \Phi \in {\cal P}_-^{(k)}(M).$ In particular, if $k> {\mbox{dim}}\;
N,$ any upper semicontinuous $v$ is in $P_-^{(k)}(N),$ hence
$v \circ \Phi \in P^{(k)}_-(M).$

\medskip

We give some examples of compact complex 
manifolds for which $P^{(2)}(M) \neq \RR:$\\

\smallskip

1. Tori: Let $T$ be a torus. Then $T=\CC^k$ mod a lattice
generated by $\{v_i\}_{i=1}^{2k}.$ Let $\pi:\CC^k \rightarrow \RR$,
$\pi(\sum x_i v_i)=x_i.$ Let $v=\phi(x_1)$ where $\phi$ is a smooth function supported in
$]0,1[.$ Then $(i\partial \overline{\partial} v)^2=0.$\\

\smallskip

2. Hopf manifolds. Let $M=\CC^2/<\phi>$ where $<\phi>$ denotes the
group generated by $\phi(z_1,z_2)=(\alpha_1 z_1,\alpha_2 z_2)$ with
$\alpha_1,\alpha_2$ fixed, $0<|\alpha_1| \leq |\alpha_2| <1.$ Fix
$r$ such that $|\alpha_1|=|\alpha_2|^r.$ Define
$$v(z_1,z_2)=\frac{|z_1|^2}{|z_1|^2+|z_2|^r}$$
The function $v$ is well defined on $M$ and $(i\partial \overline{\partial} v)^2=0.$\\

\smallskip

3. For any surface admitting a projection on a Riemann surface $P^{(2)}(M)
\neq \RR,$ for example ruled surfaces. Actually the Hopf surfaces above admit
such a projection $\CC^2/(\phi) \rightarrow \PP^1,$ 
$(z_1,z_2) \rightarrow [z_1^q:z_2^p]$ if
$\alpha_1^q=\alpha_2^p.$\\

\medskip


For $k>1,$ set $z=(z_1,\dots,z_{k-1}), |z|= \max \{|z_1|,\dots,|z_{k-1}|\}$
and let $H^r_{k-1}$ denote the Hartogs figure:

$$
H^r_{k-1}:=\{(z,w)\in \CC^{k-1} \times \CC= 
\CC^k; |z|\leq 1+r,|w|\leq 1\}\setminus \{r<|w|\leq 1, |z|< 1\}.
$$

Let $\hat{H}^r_{k-1}:= \{(z,w)\in \CC^k; |z|\leq 1+r,|w|\leq 1\}$.

\begin{definition} Let $2 \leq k \leq m.$ We say that an open set 
$N\subset M$ is\\
$(k-1)-$pseudoconvex if whenever  $\Phi:U \rightarrow M$ is a biholomorphic 
map
of a neighborhood $U\supset \hat{H}^r_{k-1}$ onto its image and
$\Phi(H^r_{k-1}) \subset N,$ then $\Phi(\hat{H}^r_{k-1}) \subset N.$
\end{definition}

\begin{remark} In the case $k=2$, the definition is  
equivalent to $N$ being pseudoconvex.
\end{remark}


\begin{prop}
Let $M$ be a complex manifold of dimension $m \geq 2.$ Let $N$ be a
connected open 
set in $M$. Assume $v\in P_-^{k}(M), 2\leq k \leq m$ and $v<0$
on $N,$ $v_{|\partial N}\equiv 0, v \leq 0$ on $M.$
Let ${\cal U}=\{U_\alpha\}$ be a cover of $M$ associated to $v.$ If
$\Phi(\hat{H}^r_{k-1})\subset U_\alpha$ and $\Phi({H}^r_{k-1})\subset N,$
then $\Phi(\hat{H}^r_{k-1})\subset N.$ In particular if $k=2$ then $N$
is pseudoconvex in $M$.
\end{prop}

{\bf Proof:} Assume that $\Phi({\hat{H}}^r_{k-1})\setminus N
\neq \emptyset.$  Then also, $\Phi({\hat{H}}^r_{k-1})\cap
\partial N \neq \emptyset.$ There is a biholomorphic map
$\Phi:U \rightarrow M$ onto its image, ${\hat{H}}^r_{k-1}\subset U,$
$\Phi(H^r_{k-1}) \subset N.$
We can then assume that $M=U,$ $\Phi={\mbox{Id}},$
$H^r_{k-1}\subset N$ and that there is an interior point of ${\hat{H}}^r_{k-1}$
in $\partial N.$ 
Assume that $v_n \searrow v,v_n \in {\cal C}^2(U), (-1)^k(i\partial \overline{\partial} v_n)^k \leq 
\epsilon_n \omega^k.$
We also assume that $\partial N$ contains a point $(z_0,w_0),
|z_0|,|w_0|<1.$ 
Then $v<0$ on ${H^r_{k-1}} $ and $v(z_0,w_0)=0.$ This is where we use
that $v_{|\partial N} \equiv 0.$ We would like
to get a contradiction.

Let $X:=\{|z|\leq 1, r< |w|\leq 1\}.$
Define the function $u$ by

$$
u(z,w):= \eta(1-\sum_{j=1}^{k-1}|z_j|^2)-\frac{\epsilon}{|w|^2}+\delta
$$

\noi where $\eta<<\epsilon<\delta<<1.$ We will choose the constants so that
$v < u$ on $\partial X$ and $u(z_0,w_0)<0.$ First observe that if $\delta$ is small enough,
then automatically $v<u$ on all of $\partial X$ except possibly where
$|w|=1$ and $|z|<1.$ Fix any such $\delta.$ Let $\epsilon<\delta$ be chosen
big enough that $-\frac{\epsilon}{|w_0|^2}+\delta<0.$ Since
$\epsilon<\delta$ and $v\leq 0$ on $|w|=1$ and $|z|<1$,
 $v<u$ on all of $\partial X$ if we choose $\eta=0$.
To finish the choice of constants, $\eta, \epsilon,\delta$ choose
$\eta>0$ small enough that $v<u$ still on $\partial X$ and in, addition 
$u(z_0,w_0)<0.$ Then $i\partial \overline{\partial} (-u) \geq a \omega$ on $X$ for some constant
$a>0.$ 

\medskip

Next choose $n$ large
enough so that $v_n<u$ on $\partial X$. We have $v_n(z_0,w_0)
\geq v(z_0,w_0)=0>u(z_0,w_0).$ Then if we add a strictly
positive constant $c_n$ to $u$ we can assume that

\bea
v_n & < & u+c_n \; {\mbox{on}}\; \partial X\\
v_n & \leq & u+c_n \; {\mbox{on}}\; X\\
v_n(z_1^n,w_1^n) & = & u(z_1^n,w_1^n)+c_n, (z_1^n,w_1^n) \in X\\
\eea

\bigskip

This implies that $i\partial \overline{\partial} v_n(z_1^n,w_1^n)\leq i 
\partial \overline{\partial} u(z_1^n,w_1^n).$ 
Hence $i\partial\overline{\partial}  (-v_n)(z_1^n,w_1^n) \geq a\omega.$ This implies
that

\bea
\epsilon_n \omega^k & \geq & (-1)^k (i\partial \overline{\partial} v_n)^k(z_1^n,w_1^n)\\
& = &  (i \partial \overline{\partial} (-v_n))^k(z_1^n,w_1^n)\\
& \geq & a^k \omega^k,\\
\eea

\noindent a contradiction.\\

\CQFD\\

\begin{coro}
If $M$ is a compact manifold and $v\in P^{(2)}(M),$ then
$K(v): =\{p;v(p)=\max_M v\}$
is pseudoconcave, i.e. $M \setminus K(v)$ is
pseudoconvex. If $v \in P^{(k)}(M) \cap {\cal C}^2,$ then
$M\setminus K(v)$ is $k-1$ pseudoconvex.
\end{coro}

\begin{remark}
Let $v$ be a ${\mathcal C}^2$ function on a compact complex manifold
of dimension $m.$ Stokes' Theorem implies that if $(-1)^k
(i \partial \overline{\partial} v)^k \leq 0,$ then $(dd^c v)^k =0.$ In the case
of compact K\"{a}hler manifolds, Stokes' Theorem applied to
$(-1)^k(i\partial \overline{\partial} v)^k \wedge \omega^{m-k}$ shows that the same
conclusion holds.
\end{remark}


\begin{remark}
The proof above shows that if an upper semicontinuous function $v$
is locally a decreasing limit of ${\cal C}^2$ functions $v_n$ such that
at each point $i\partial \overline{\partial} v_n$ has $m-1$ nonnegative eigenvalues,
then $K(v)$ is pseudoconcave. Namely, we get by the above construction
with a Hartogs figure of dimension two:
\bea
v_n & < & u+c\; {\mbox{on}}\; \partial X,\\
v_n & \leq & u+c  \; {\mbox{on}}\;X\\
v_n(z_1,w_1) & = & u(z_1,w_1)+c \; {\mbox{at some point of }}\; X.\; 
{\mbox{Hence}}\\
i \partial \overline{\partial}(v_n-u)(z_1,w_1) & \leq & 0\\
\eea
\noi which contradicts that one eigenvalue is nonnegative.
\end{remark}


\begin{coro}
Let $v$ be a continuous function on $\PP^m$ such that $v\in {\cal P}_-^{(2)}$,
then $v$ is constant. In particular there are no
nonconstant functions in ${\mathcal C}^2_{\RR}$ such that $T=
i\partial v \wedge \overline{\partial v}$ satisfies
$i \partial \overline{\partial} T=0.$
\end{coro}

{\bf Proof:}
We know that $K(v)$ is pseudoconcave. We show next that also
$K(-v)$ is pseudoconcave. It suffices to show that $-v$ is also
a decreasing limit of ${\cal C}^2$ functions $w_n$, 
$(i\partial \overline{\partial} w_n)^2 \leq \epsilon_n \omega^2.$
For this, let $v_n$ be such a sequence for $v.$ Taking a subsequence if
necessary we can assume that $v \leq v_n \leq v+1/2^n.$ Set $w_n=-v_n+1/n.$
 Since on $\PP^m$ the Levi problem has a positive
solution, this implies that the complements of $K(v)$ and $K(-v)$ are
both Stein. But then the intersection of the two domains is a Stein
manifold of dimension $>1$ with two ends, unless $K(v)$ and $K(-v)$ have
a nonempty intersection. But then $v$ must be constant.\\

\CQFD\\

\begin{coro}
If the Levi problem is solvable on a compact complex manifold $M$, then
$P^{(2)}$ only contains constant functions.
Hence there is no nonconstant holomorphic
map from $M$ to a manifold with nontrivial
$P^{(2)}(M).$
\end{coro}


 Recall [BS] that given a positive closed current 
$S$ on $M,$ an upper semicontinuous 
function $\phi$ defined on ${\mbox{Supp}}(S)$ is $S$-plurisubharmonic if for
every $p \in {\mbox{Supp}}(S)$ there is an open set $U,$ $p\in U$ and a 
sequence  $\phi_n \in {\cal C}^2(U),$ such that
$\phi=\lim_{\searrow}\phi_n$ on $U \cap S$ and
$i\partial \overline{\partial} \phi_n\wedge S \geq 0$.
A function $\phi$ is $S-$ pluriharmonic if both $\phi$ and
$-\phi$ are \\
$S-$plurisubharmonic.

\begin{theo}
Let $S$ be a positive closed current of bidegree $(1,1)$ in $\PP^2.$ Assume
$\phi$ is $S-$plurisubharmonic and let $K(\phi)=
\{p\in {\mbox{Supp}}(S); \phi(p)= \max \phi\}. $ Then $\PP^2 \setminus
K(\phi)$ is pseudoconvex. If $\phi$ is $S-$pluriharmonic, then $\phi$ is constant.
\end{theo}


{\bf Proof:}
Recall that for $S-$plurisubharmonic functions, the local maximum principle
is valid [BS], Prop. 3.1. We claim that $\PP^2
\setminus K(\phi)$ is pseudoconvex. 
We modify the proof of Proposition 6.3. Assume that 
$\PP^2
\setminus K(\phi)$ is not pseudoconvex.
We can assume in local coordinates that $K(\phi)$
contains a point $(z_0,w_0), |z_0|,|w_0|<1$ and that $K(\phi)$
does not intersect the Hartogs figure $H=\{(z,w); |w| \leq r<1, |z|\leq 
1+\delta,\}\cup \{(z,w); 1\leq |z| \leq 1+\delta, |w|\leq 1\}.$
We can also assume that on a fixed neighborhood of $\hat H=
\{(z,w);|z|\leq 1+\delta, |w|\leq 1\}$
there is a sequence of $\mathcal C^2$ functions $\phi_n
\searrow \phi$ on Supp$(S)$, $i\partial \overline{\partial}
 \phi_n \wedge S\geq 0.$
We can assume $\phi=0$ on $K(\phi).$
Then $\phi<0$ on ${H} \cap {\mbox{Supp}}(S)$ 
and $\phi(z_0,w_0)=0.$ 

Let $X:=\{|z|\leq 1, r< |w|\leq 1\}.$
Define the function $u$ by

$$
u(z,w):= \eta(1-|z|^2)-\frac{\epsilon}{|w|^2}+\delta
$$



\noi where $\eta<<\epsilon<\delta<<1.$ We will choose the constants so that
$\phi < u$ on $\partial X \cap {\mbox{Supp}}(S)$.
First observe that if $\delta$ is small enough,
then automatically $\phi<u$ on all of $\partial X \cap {\mbox{Supp}}(S)$ 
except possibly where
$|w|=1$ and $|z|<1.$ Fix any such $\delta.$ Let $\epsilon<\delta$ be chosen
big enough that $-\frac{\epsilon}{|w_0|^2}+\delta<0.$ Since
$\epsilon<\delta$ and $\phi\leq 0$ on $|w|=1$ and $|z|<1$,
$(z,w)\in {\mbox{Supp}}(S),$ $\phi<u$ on all of 
$\partial X \cap {\mbox{Supp}}(S)$ if we choose $\eta=0$.
To finish the choice of constants, $\eta, \epsilon,\delta$ choose
$\eta>0$ small enough that $\phi<u$ still on $\partial X \cap
{\mbox{Supp}}(S)$ and in, addition 
$u(z_0,w_0)<0.$ 

\medskip

Next choose $n$ large
enough so that $\phi_n<u$ on $\partial X \cap {\mbox{Supp}}(S)$. 
Then if we add a strictly
positive constant $c$ to $u$ we can assume that

\bea
\phi_n & < & u+c \; {\mbox{on}}\; \partial X \cap {\mbox{Supp}}(S)\\
\phi_n & \leq & u+c \; {\mbox{on}}\; X \cap {\mbox{Supp}}(S)\\
\phi_n(z_1,w_1) & = & u(z_1,w_1)+c, (z_1,w_1) \in X \cap  {\mbox{Supp}}(S)\\
\eea

\bigskip

Now, $-u$ is plurisubharmonic, so $i\partial \overline{\partial}
(-u) \wedge S\geq 0.$ Hence
$\phi_n-u$ is $S-$plurisubharmonic so this contradicts
the local maximum modulus principle for $S-$ plurisubharmonic
functions.

\medskip

If $\phi$ is $S-$ pluriharmonic, then $K(\phi)$ and $K(-\phi)$ 
intersect, hence $\phi$ is constant.\\

\CQFD\\


\begin{prop}
If $v \in {\cal P}_-^{(k)}(M)$ then $v$ satisfies the local maximum 
principle.
\end{prop}

{\bf Proof:}
Recall that the local maximum principle says that
for every ball $\max_B v \leq \max_{\partial B} v.$
This follows, since the Hartogs figure argument is local. In fact, let
$K$ denote the compact set at which the maximum is reached.
Let $p\in \partial K$ and use a Hartogs figure there.\\

\CQFD\\

There are positive closed currents $T$ on $\PP^2$
of the form $T=i\partial u \wedge \overline{\partial u},$
$u$ continuous except at one point and such that $\int T \wedge
T \neq 0,$ for example: $u=\log^+|z|$ in $\CC^2$ if
$[z:w:t]$ are the homogeneous coordinates in $\PP^2.$ 

\bigskip

\begin{prop} Consider $C=\{T \geq 0,
i\partial \overline{\partial} T=0, \int T \wedge \omega=1\}.$
Then $\inf_ {T\in C} \int T \wedge T \leq 1-\frac{1}{2\pi^2}.$
\end{prop}

{\bf Proof:}
Let $u(|z|^2), v(|w|^2)$ be ${\mathcal C}^\infty$ real valued functions
with support in the unit interval. Define $\psi(z,w):=u(|z|^2)+
iv(|w|^2).$ Let $T:=i \partial \psi \wedge \overline{\partial \psi}$
on ${\mathbb C}^2.$ Then $T \geq 0$ and $T \wedge T=0.$ Moreover, $T$
is pluriharmonic on ${\mathbb C}^2.$ 


We want to decompose $T$ as in Proposition 2.6.

\bea
i\partial \psi \wedge \overline{\partial \psi}
& = & i(u' \overline{z}dz+iv' \overline{w}dw) \wedge
(u'zd \overline{z}-i v' wd\overline{w})\\
& = & i (u')^2 z\overline{z}dz \wedge d\overline{z}
+i (v')^2 w\overline{w}dw \wedge d\overline{w}\\
& + &  u'v' z\overline{w} d\overline{z}\wedge dw
+ u'v' \overline{z}w dz \wedge  d\overline{w}\\
\eea

Let 
\bea
U(z) & := & \frac{i}{\pi} \int \log|z-x|(u'(x \overline{x}))^2x \overline{x}
dx \wedge d \overline{x}\\
V(w) & := & \frac{i}{\pi} \int \log|w-y|(v'(y \overline{y}))^2y \overline{y}
dy \wedge d \overline{y}\\ 
\eea

Then

\bea
i \partial \psi \wedge \overline{\partial \psi}
& = & i \partial \overline{\partial} U(z)+ i \partial
 \overline{\partial} V(w)+
\overline{\partial}(uv'\overline{w}dw)-uv'' w\overline{w}
d\overline{w}\wedge dw\\
& - & uv' d\overline{w}\wedge dw
+{\partial}(uv'wd\overline{w})-uv" w\overline{w}
dw \wedge d\overline{w}-uv' dw \wedge d\overline{w}\\
\eea

Hence:

\begin{lem}
On ${\mathbb C}^2,$ 
\bea
T & = & i \partial \psi \wedge \overline{\partial \psi}\\
& = & i \partial \overline{\partial}U(z)+ i \partial \overline{\partial}V(w)
+\overline{\partial}(uv' \overline{w}dw)+\partial(uv' wd\overline{w}).\\
\eea
\end{lem}

Let $A:= i/\pi \int (u')^2|z|^2 dz \wedge d \overline{z},
B:= i/\pi \int (v')^2|w|^2 dw \wedge d \overline{w}.$
Then $U(z)=A \log |z|, |z|>1, V(w)=B \log |w|, |w|>1.$
We decompose $T$ further:
Let $h:= U(z)+V(w)-\frac{1}{2}(A+B) \log (1+|z|^2+|w|^2).$
Then

\begin{lem}
On ${\mathbb C}^2:$ 
\bea
T & = & i \partial \overline{\partial}h(z,w)+\frac{1}{2}(A+B) \omega
+ \overline{\partial}(uv'\overline{w}dw)
+\partial(uv'wd\overline{w})\\
\omega & := & i \partial \overline{\partial} \log(1+|z|^2+|w|^2).\\
\eea
\end{lem}
We extend $T$ to ${\mathbb P}^2$ as $\tilde{T}$, the trivial
extension. We need to know that $T$ has finite mass near
$[0:1:0]$ and $[1:0:0]$ for $\tilde{T}$ to be well defined.
To extend first across the line at infinity, $\eta=0$ away from
$[0:1:0]$ and $[1:0:0]$, we extend the three parts individually.
First $\omega$ extends as the K\"{a}hler form, also called $\omega.$
The form $uv'wd\overline{w}$ has compact support and extends
as $S$ trivially.
Next we investigate $h$ near $[0:1:0].$ We calculate in local coordinates.
$[z:w:1]=[Z:1:t], $ to get
$h(z,w)=\tilde{h}(Z,t))= U(Z/t)+V(1/t)-\frac{1}{2}(A+B)
\log (1+|Z/t|^2+|1/t|^2).$ When $|Z/t|>1$ we have
$U(Z/t)=A \log |Z/t|, V(1/t)=B \log |1/t|.$ So $\tilde{h}(Z,t)
=A \log|Z|-A \log |t|-B\log |t|
-\frac{1}{2}(A+B) \log (1+|Z|^2+|t|^2)+ \frac{1}{2}(A+B)\log |t|^2,$
$\tilde{h}(Z,t)=A \log |Z|-\frac{1}{2}\log(1+|t|^2+|Z|^2).$
Hence $\tilde{h}(Z,t)$ extends smoothly across $\eta=0$ except possibly
at $[0:1:0]$ and $[1:0:0].$ In particular $i \partial \overline{\partial}
\tilde{h} $ extends trivially at $\eta=0$ except
at $[1:0:0]$ and $[0:1:0].$ Next we calculate in a neighborhood
of $[0:1:0].$ We get


$$\tilde{h}(Z,t)=U(Z/t)+A \log |t|-\frac{A+B}{2}\log (1+|t|^2+|Z|^2).$$

The function $U(Z/t)+A\log|t|=:\phi(Z,t)$ is plurisubharmonic
when $t \neq 0$ and equals $A\log |Z|$ when $|Z/t|>1$ or $t=0, Z \neq 0.$
So $\phi(Z,t)$ is plurisubharmonic away from the origin. Hence $\phi$ has
a well defined plurisubharmonic extension through $(0,0)$ by setting
$\phi(0,0)=-\infty.$
It follows that $\tilde{h}$ is a global quasiplurisubharmonic function on
$\mathbb P^2$ with poles at $[0:1:0], [1:0:0].$ Hence, 

\begin{lem}
The trivial extension $\tilde{T}$ is given by
$\tilde{T}=\frac{A+B}{2}\omega+\partial S+\overline{\partial S}
+i \partial \overline{\partial}\tilde{h}, S=uv'wd\overline{w}.$
\end{lem}

It is easy to check that $\tilde{T}$ is pluriharmonic on $\PP^2.$

\medskip

{\bf End of Proof of Proposition 6.11:} This follows since $\partial \overline{\partial}\tilde{h}$
has no mass on the line at infinity. Hence $\int \overline{\partial}S
\wedge \partial \overline{S}=AB,$$
\int T \wedge T= \left(\frac{A+B}{2}\right)^24\pi^2-2AB$ and
$\int T \wedge \omega=\frac{A+B}{2} 2\pi$ so if we let
$T_1=\frac{T}{\int T \wedge \omega}$ we find
$$\int T_1 \wedge T_1 = 1-\frac{2AB}{\left(\frac{A+B}{2}\right)
4\pi^2}.$$ The minimum is reached for $A=B$ and equals
$1-\frac{1}{2\pi^2}.$\\ 

\CQFD\\

\begin{proposition}
Let $M$ be a complex surface and $\rho\in {\mathcal C}^2_{\RR}(M).$
Assume that $\partial \rho$ is non vanishing on $X=
\{\rho=0\}$ and that
$(i\partial \overline{\partial} \rho)^2=0$ on $X$ and
also that $i\partial \overline{\partial} \rho \wedge \partial \rho
\wedge \overline{\partial} \rho={\mathcal O}(\rho^2)$. Then
$T=i\delta_{\{\rho=0\}} \partial \rho \wedge \overline{\partial}\rho 
$ is a smooth positive harmonic current. Moreover $T \wedge T=0.$ 
\end{proposition}

{\bf Proof:}
Choose a $\chi\in {\mathcal C}^\infty_0(-1,1)$, $\chi \geq 0, \int \chi=1.$
Let $$T_\epsilon =i \frac{1}{\epsilon} \chi\left(\frac{\rho}{\epsilon}\right)
\partial \rho \wedge \overline{\partial} \rho.$$
Then we have 
$$
i \partial \overline{\partial} T_\epsilon
=-\frac{1}{\epsilon^2}\chi'\left(\frac{\rho}{\epsilon}\right)
\partial \rho \wedge \overline{\partial}\partial \rho \wedge
\overline{\partial} \rho-\chi \left(\frac{\rho}{\epsilon}\right)
\overline{\partial} \partial \rho 
\wedge \partial \overline{\partial} \rho.
$$

Clearly then $i \partial \overline{\partial} T_\epsilon \rightarrow 0.$\\

\CQFD\\






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\bigskip


\noindent John Erik Forn\ae ss\\
Mathematics Department\\
The University of Michigan\\
East Hall, Ann Arbor, MI 48109\\
USA\\
fornaess@umich.edu\\

\noindent Nessim Sibony\\
CNRS UMR8628\\
Mathematics Department\\
Universit\'e Paris-Sud\\
Batiment 425\\
Orsay Cedex\\
France\\
nessim.sibony@math.u-psud.fr\\
\end{document}











END PAPER



\section{Further Material}

\subsection{Smoothing of harmonic laminated currents}


We could get formulas $\tilde{T}\wedge \tilde{T}=$ integral over close to the
nal as in [HM].

\medskip

Let $T$ be a positive harmonic current on $\mathbb P^k$, ${\mathcal G}=
{\mbox{Aut}}(\mathbb P^k)$ has a natural invariant measure $\nu.$ Let
$\rho_\epsilon$ be an approximation of identity supported near 
${\mbox{Id}}\in {\mathcal G}.$ We define

$$
T_\epsilon:= \int \rho_\epsilon(g) g_*(T)d\nu(g).
$$

It is classical that $T_\epsilon$ is smooth, clearly $T_\epsilon$ is harmonic
and $T_\epsilon \rightarrow T$ as $\epsilon \rightarrow 0.$ If $T$
is laminated we get if $(\chi_j)$ is a partition of unity associated to flow
boxes.

$$
T_\epsilon=\sum_j \int \chi_j(z)\rho_\epsilon(g)(g_*T)d\nu(g).
$$

But on $B_j$ open set containing strictly the support of $\chi_j$ we
have:

$$
T=\int h_\alpha [\Delta_\alpha]d\mu(\alpha).
$$

Hence

$$
T_\epsilon=\sum \int \rho_\epsilon(g)h_\alpha(g^{-1})g_*[\Delta_\alpha]
\chi_jd\mu^j(\alpha)d\nu(g).
$$

$h_\alpha(g^{-1})$ is well defined since ${\mbox{Supp}}(\chi_j) \subset
\subset B_j,$ proviede $\epsilon$ is small enough.

\medskip

The point here is that the $g_*[\Delta_\alpha]$ when $g$ varies are not
disjoint. Do you know how to calculate that $\int T_\epsilon
\wedge T_\epsilon \rightarrow 0?$
I am not sure about my argument in p.2.



\begin{theo}
Let $K$ be laminated in $\mathbb P^2$, lamination given on $K$ by
a holomorphic vector field with no singularities on $K.$
Then if $T$ is a harmonic laminated current with transverse measure,
$\int T_\epsilon \wedge T_\epsilon \rightarrow 0.$
\end{theo}

{\bf Proof:} Since the lamination is holomorphic, the local intersection
numbers are bounded uniformaly independent of $\epsilon.$ [See section 9]
Since the harmonic functions are uniformly bounded and the transvers
measures have no point masses, the result follows.\\

\CQFD\\ 

\subsection{Global harmonic functions on laminated compacts}

Let $T$ be a laminated compact $K$ with finite transverse energy
in $\PP^2$, $[0:1:0]\notin K.$
For any fixed $z$ define $H(z)=\int_{\alpha,\beta} \log
|f_\alpha-f_\beta|d\mu(\alpha)d\mu(\beta)$.

\begin{lem}
$H$ is a nonzero harmonic function on $\CC$ which goes to
infinity like $\log |z|,$ i.e. impossible.
\end{lem}

We can also define $\tilde{H}=\int H_\alpha d\mu(\alpha)$ for each fixed
$z.$ This must be a constant harmonic function.
Moreover, if the lamination has isolated singularities,
this harmonic function extends across.
In the case of the other harmonic function, it might have a pole
at singularities.

\begin{theo}
Let $K$ be a proper compact laminated subset of $\PP^2.$
Then $K$ cannot support a laminated closed positive current
with finite transverse energy. 
\end{theo}
 
{\bf Proof:}
We can assume that $[0:1:0]\notin K$.
Let $\mu$ denonte local transverse measures. For $z_0\in \CC$
let $\{f_\alpha\}$ denote the leaves near $z_0$ with multiplicity.
Let $$\Phi(z_0):=\int_{\alpha,\beta}\log
|f_\alpha-f_\beta|d\mu(\alpha)d\mu(\beta).$$
We show that $\Phi(z)$ is harmonic, and $\Phi(z) \rightarrow
\infty$ when $z \rightarrow \infty.$ This will be a contradiction.

\bigskip

Note that $\log|f_\alpha-f_\beta|$ is harmonic except at branch points.
Note that $f_\alpha, f_\beta$ branches at different $z_0$ unless they
are far away in the $w$ direction.

\medskip

If we are near a vertical branch point, parametrize the leaves,
writing them as vertical graphs. Then for each $\alpha$ and $z,$
$\{f_\alpha\}$ has finitely many values (the branch number).
For any $\alpha \neq \beta$ $\pi (f_\alpha-f_\beta)(z),$ the product taken
over all local branches is symmetric and nonzero, hence holomorphic.
This proves the theorem since there are no point masses of $\mu$
(this shows harmonicity). Next we show that $\Phi \rightarrow \infty$
as $z\rightarrow \infty.$

\medskip

There are finitely many local leaves tangent to the line at infinity.
They are away from each other, so we only need to deal with having one
or zero. If there are none, then a leaf can be parametrized
as $w=f_\alpha(z)$, $f_\alpha=\alpha z+o(z).$ Then we get that
$|f_\alpha-f_\beta|(z)=\log |(\alpha-\beta)z+o(z)|,$
so $\log|f_\alpha-f_\beta| \rightarrow \log |\alpha-\beta|+\log |z|$
which implies unboundedness. A tangency doesn't change this.\\

\CQFD\\

We can also define $\int H_\alpha d\mu(\alpha)$  where we
integrate over vertical slices or over any line through any
point. This must be a harmonic function, hence constant.
Since for two distinct points we also get the line through those
two points, this constant is independent of which point we use.

\medskip

Let $T=c\omega+\partial S + \overline{\partial S}$. We try to find a
formula for $S$ using global integration over slices.

\bea
\overline{\partial}(\partial T) & = & 0\\
\partial T & = & \overline{\partial} S_{20}\\
S_{20} & = & \partial S_{10}\\
\partial T & = & \partial \overline{\partial}S_{10}\\
T-\overline{\partial}S_{10} & = & \partial R+c\omega\\
\partial T & = & \partial h_\alpha \wedge [\Delta_\alpha]\\
& = & \partial h_\alpha \wedge \partial \overline{\partial} \log |w|\\
& = & \overline{\partial} (\partial h_\alpha \wedge
\partial \log |w|\\
S_{20} & = & \partial h_\alpha \wedge \partial \log |w|\\
S_{10} & = & \log |w| \partial h_\alpha\\
\eea

\noindent where we integrate over global slices.

$S= \int \log|w-f_\alpha| \partial h_\alpha d\mu(\alpha)$
integrating over $w$ with fixed $z.$ This is a global formula.

\subsection{Entropy of Laminations}

\subsubsection{Topological Entropy}

Let $L$ be a leaf and let $D$ be a transversal. Let 
$\phi:\Delta\rightarrow L$ denote a covering.
We define $h(L,D):= \overline{\lim}_R \frac{\log \# (B(p,R)\cap D}{R}.$

\begin{theo}
If all leaves are biholomorphic to $\Delta$, then $h(L,D)=1$.
\end{theo}

{\bf Proof:}
Let $R'$ denote a sufficently large number. Then any Kobayashi disc
of radius $R'$ in $\Delta$ covers a disc across each flow box, and
intersects at most $M$. Moreover, the annuli
$1-1/(R')^{k}\leq |x| <1-1/(R')^{k+1}$ is covered by about $(R')^k$ such 
discs. The estimate follows.\\

\CQFD\\

Let $D_\epsilon(L,R)$ denote those leaves starting at near $p\in D\cap L$
and having distance at most $\epsilon$ from $L$ when the distance in $L$
from $p$ is at most $R$.
Let $N_\epsilon(L,R)$ denote the max number of disjoint ones.
We define $h_{\mbox{top}}=\lim_{\epsilon \rightarrow 0} 
\overline{\lim}_{R \rightarrow \infty} \frac{\log N_\epsilon(L,R)}{R}.$

\begin{theo}
The topological entropy is independent of the slice.
\end{theo}

In the case of singular points of the lamination, avoid a ball around the
singular points.

 
\subsubsection{Metric Entropy}

For this part we need to have an ergodic measure. Let $T \geq 0$ be a
harmonic current. This is a closed convex set and pick an extremal
$T.$ Then there is no proper transverse set of leaves of positive measure
which is invariant, which should mean ergodicity.

\medskip

Let $x$ be in a transversal. Define

$$
h(T):= \frac{\log \mu(D_\epsilon(L,R))}{R}.
$$

Show this is independant of $x$ a.e. Show $h(T)\leq h_{\mbox{top}}$
Show $\exists \! T, h(T)=h_{\top}.$

\begin{lem}
For ${\mathcal C}^1$ laminations $h_{top}<\infty.$
\end{lem}


\subsection{Explicit Calculations}

  


We want to write a pluriharmonic current $T=c\omega+i \partial
\overline{\partial} u+ \partial S^g_{0,1}
+\overline{\partial}R^g_{1,0}.$

\medskip

Plan, we find approximate solutions which capture the main singularity
locally, then patch with cut-off functions and see what is left. Then
repeat the process.

\medskip

We cover $K$ by finitely many open sets $U_j$. We assume that there are local
holomorphic coordinates $(z_j,w_j)$ in $U_j$. We assume that the leaves of
$K\cap U_j$ are of the form $w_j=f^{\alpha_j}(z_j),
\alpha_j \in A_j,|z_j|<10.$ We can use as parameter-space $A_j$,
$\alpha_j=f^{\alpha_j}(0)$, i.e. the intersection of the leaf with the
$w_j$ axis. We can suppose that $U_j \cap (z_j=0)=
\{|w_j|<10\}.$ We also assume that the leaves are almost parallel to the
$z_j$ axis and that $(f^{\alpha_j})'$ and  $(f^{\alpha_j})''$ are small.

\bigskip

There exists a finite measure $\mu_j$ on $A_j$ and a family of pluriharmonic
functions $H^{\alpha_j}(z_j,w_j)=H^{\alpha_j}(z_j)$ defined when $|z_j|<10,$
$H^{\alpha_j}(0)=1.$ We can write $H^{\alpha_j}(z_j)=
\Re F^{\alpha_j}(z_j), F^{\alpha_j}(0)=1.$ Then $\overline{\partial}
 H^{\alpha_j}= \frac{1}{2}\overline{\partial F^{\alpha_j}}$
and ${\partial} H^{\alpha_j}= \frac{1}{2}{\partial F^{\alpha_j}}.$


\bigskip

Let $A_{k \ell}\subset A_k$ denote those leaves in $U_k$ which intersect
leaves in $U_\ell.$ Notice that a leaf in $U_k$,
$w_k=f^{\alpha_k}(z_k)$ can be projected to the $z_\ell$ axis,
$z_k \rightarrow (z_k,f^{\alpha_k}(z_k) \rightarrow
z_\ell= a z_k+\epsilon f^{\alpha_k}(z_k)$ where $a$ is close
to one and $\epsilon$ is small. Then the image of the leaf is convex.
Hence the leaves of $U_k$ and $U_\ell$ have well defined connected or empty
intersections. Also $A_{k \ell}$ is open in $A_k.$ There is a map $\gamma_{k\ell}: A_{k\ell}
\rightarrow A_{\ell k}\subset A_\ell$ sending a leaf to its continuation
in $U_\ell.$ This continuation is unique. 
There is a unique map $\lambda_{k\ell}: A_{k\ell} \rightarrow (0,\infty)$
so that $H^{\gamma_{k\ell}(\alpha_k)}=\lambda_{k\ell} H^{\alpha_k}$
on the overlap of the leaves. We then also obtain that

$$
\gamma_{k\ell}^* (d\mu_\ell) = \lambda_{\ell k} d\mu_k.
$$

For test forms $\phi_{1,1}\in {\cal C}^\infty_0 (U_j)$ we get:

\bea
T^g_{1,1}(\phi_{1,1}) & = & \int_{A_j}( H^{\alpha_j}[\Delta_{\alpha_j}
:=\{w_j=f^{\alpha_j}(z_j)\}])(\phi_{1,1})\\
 & = & \int_{A_j}[\Delta_{\alpha_j}
]( H^{\alpha_j}\phi_{1,1})\\
& = & \int_{A_j} \left[\int_{\Delta_{\alpha_j}}
H^{\alpha_j}(z_j,w_j)\phi_{1,1}\right] 
d\mu_j(\alpha_j)\\
\eea

\bea
[\Delta_{\alpha_j}
]( H^{\alpha_j}\phi_{1,1}) & = &\left( i\partial \overline{\partial}
\log|w_j-f^{\alpha_j}(z_j)|^2\right)( H^{\alpha_j}\phi_{1,1})\\
T^g_{1,1}(\phi_{1,1}) & = &
\int_{A_j} 
[\left( i\partial \overline{\partial}
\log|w_j-f^{\alpha_j}(z_j)|^2\right)( H^{\alpha_j}\phi_{1,1})]
d\mu_j(\alpha_j)\\
\eea


\subsubsection{$T$ as current with measure coefficients}

We will write in $U_j$ 

\bea
T & = & i \lambda_{11} dz \wedge d\overline{z}+
i \lambda_{12} dz \wedge d\overline{w}+
 i\lambda_{21} dw \wedge d\overline{z}+
i \lambda_{22} dw \wedge d\overline{w}\\
\eea

Let $\psi$ be a smooth function with compact support.
Then we define $\lambda_{11}$ by duality:

\bea
\int \psi \lambda_{11} & = &   T(\psi idw\wedge \overline{w})\\
& = & \int_{A_j} \int_{\Delta_\alpha} H_\alpha \psi idw\wedge \overline{w}
d\mu_\alpha.\\
\eea

This represents $T$ as a current with finite measure coefficients.
We can similarly represent $\partial T$ as a current with measure
coefficients, replacing $H_\alpha$ with $\partial H_\alpha.$ These
measure coefficients are also finite measures. The same applies to
$\overline{\partial} T$.

\begin{lem}
For any rotation $\epsilon$, the functions
$\log |w^\epsilon_j-f^{\epsilon,\beta_j}|^2$ are integrable with respect
to any of the above measures. For fixed $\epsilon$ this integrability
is uniform (out of a fixed neighborhood of the identity).
\end{lem}

We can hence define a family of currents:

\bea
\partial \log |w^\epsilon_j-f^{\epsilon,\beta_j}|^2 \wedge T^g_{11}
& := & \partial [\log |w^\epsilon_j-f^{\epsilon,\beta_j}|^2 \wedge T^g_{11}]
- \log |w^\epsilon_j-f^{\epsilon,\beta_j}|^2 \wedge \partial T\\
\overline{\partial} \log |w^\epsilon_j-f^{\epsilon,\beta_j}|^2 \wedge T^g_{11}
& := & \overline{\partial} [\log |w^\epsilon_j-f^{\epsilon,\beta_j}|^2 \wedge T^g_{11}]
- \log |w^\epsilon_j-f^{\epsilon,\beta_j}|^2 \wedge \overline{\partial} T\\
& & H_{\beta_j}\partial \log |w^\epsilon_j-f^{\epsilon,\beta_j}|^2 \wedge T^g_{11}\\
S^g_{01} \wedge T& :=  &\int_{\beta_j}H_{\beta_j}\partial \log |w^\epsilon_j-f^{\epsilon,\beta_j}|^2 \wedge T^g_{11}d\mu(\beta_j)\\
\eea

Next, let $\chi_j$ be a partition of unity subordinate
to the cover $U_j$. In fact we can assume that $\chi_j$ is supported in
$\{|z_j|,|w_j|<1/10\}.$

Then 

\bea
T^g_{1,1}(\phi_{1,1}) & = & T^g_{1,1}((\sum \chi_k)\phi_{1,1})\\
& = & \sum_k T^g_{1,1}( \chi_k\phi_{1,1})\\
& = &
\sum \int_{A_k} 
[\left( i\partial \overline{\partial}
\log|w_k-f^{\alpha_k}(z_k)|^2\right)( H^{\alpha_k}\chi_k\phi_{1,1})]
d\mu_k(\alpha_k)\\
\eea

We then get that 

\bea
T^g_{1,1}(\omega) & = & \\
T^g_{1,1}\wedge \omega  & = & 
\sum \int_{A_k} 
[\left( i\partial \overline{\partial}
\log|w_k-f^{\alpha_k}(z_k)|^2\right)( H^{\alpha_k}\chi_k\omega)]
d\mu_k(\alpha_k)\\
T^g_{1,1} & = & c\omega + \cdots\\
c & = & \sum \int_{A_k} 
[\left( i\partial \overline{\partial}
\log|w_k-f^{\alpha_k}(z_k)|^2\right)( H^{\alpha_k}\chi_k\omega)]
d\mu_k(\alpha_k)\\
& = & \sum \int_{A_k} 
[\left( \log|w_k-f^{\alpha_k}(z_k)|^2\right) i\partial
 \overline{\partial}( H^{\alpha_k}\chi_k)\wedge \omega]
d\mu_k(\alpha_k)\\
\eea

Next we define 

\subsubsection{$u^g$}

\bea
u^g_{00} & := & \sum_k \int_{A_k} \left( \log |w_k-f^{\alpha_k}(z_k)|^2
\right) (H^{\alpha_k}\chi_k)d\mu_k(\alpha_k)\\
(\overline{\partial} u^g)_{01}(\phi_{21}) & = &
\sum_k \int_{A_k} \left( \overline{\partial} \log |w_k-f^{\alpha_k}(z_k)|^2 \right)
(H^{\alpha_k} \chi_k  \phi_{12}) d \mu_k (\alpha_k)\\
& + &  \sum_k \int_{A_k} \left( \log |w_k-f^{\alpha_k}(z_k)|^2 \right)
\overline{\partial} (H^{\alpha_k} \chi_k)\wedge  \phi_{21} d \mu_k (\alpha_k)\\
i\partial \overline{\partial}u^g_{1,1}(\phi_{11} 
& = & i \sum_k \int_{A_k} \left(\partial \overline{\partial} 
\log |w_k-f^{\alpha_k}(z_k)|^2 \right)
(H^{\alpha_k} \chi_k  \phi_{11}) d \mu_k (\alpha_k)\\
& - & i \sum_k \int_{A_k} \left( \overline{\partial} 
\log |w_k-f^{\alpha_k}(z_k)|^2 \right)
\partial(H^{\alpha_k} \chi_k ) \wedge\phi_{11}) d \mu_k (\alpha_k)\\
& + &  i\sum_k \int_{A_k} \left( \partial\log |w_k-f^{\alpha_k}(z_k)|^2 \right)
\overline{\partial} (H^{\alpha_k} \chi_k)\wedge   \phi_{11}
 d \mu_k (\alpha_k)\\
& + &  i\sum_k \int_{A_k} \left( \log |w_k-f^{\alpha_k}(z_k)|^2 \right)
\partial\overline{\partial} (H^{\alpha_k} \chi_k)\wedge  \phi_{11}
 d \mu_k (\alpha_k)\\
\eea


We conclude:

\begin{proposition}
\bea
T^g_{11}(\phi_{11}) & = &
(i\partial \overline{\partial} u^g)_{11}(\phi_{11})\\
& + & i \sum_k \int_{A_k} \left( \overline{\partial} 
\log |w_k-f^{\alpha_k}(z_k)|^2 \right)
\partial(H^{\alpha_k} \chi_k ) \wedge\phi_{11}) d \mu_k (\alpha_k)\\
& - &  i\sum_k \int_{A_k} \left( \partial\log |w_k-f^{\alpha_k}(z_k)|^2 \right)
\overline{\partial} (H^{\alpha_k} \chi_k)\wedge   \phi_{11}
 d \mu_k (\alpha_k)\\
& - &  i\sum_k \int_{A_k} \left( \log |w_k-f^{\alpha_k}(z_k)|^2 \right)
\partial\overline{\partial} (H^{\alpha_k} \chi_k)\wedge  \phi_{11}
 d \mu_k (\alpha_k)\\
\eea
Here the first term on the right side has measure coefficients,
the second and third has $L^p$ coefficients for any $p<2$ and
the fourth has $L^p$ coefficients for any $p<\infty.$
\end{proposition}


\subsubsection{$S^{g,i}_{01}$, $R^{g,i}_{10}$}



\bea
S^{g,1}_{01}(\phi_{21}) & = & 
\sum \int_{A_k} 
[\left( i \overline{\partial}
\log|w_k-f^{\alpha_k}(z_k)|^2\right)( H^{\alpha_k}\chi_k\phi_{21})]
d\mu_k(\alpha_k)\\
 T^g_{11}(\phi_{11})& = &
\partial S^{g,1}_{01}(\phi_{11})\\
& + &
\sum \int_{A_k} 
[\left( i\overline{\partial}
\log|w_k-f^{\alpha_k}(z_k)|^2\right)( H^{\alpha_k}\partial \chi_k\wedge 
 \phi_{1,1})]
d\mu_k(\alpha_k)\\
& + &
\sum \int_{A_k} 
[\left( i\overline{\partial}
\log|w_k-f^{\alpha_k}(z_k)|^2\right)( \chi_k \partial H^{\alpha_k}
\wedge 
 \phi_{1,1})]
d\mu_k(\alpha_k)\\
R^{g,1}_{10}(\phi_{12}) & = &  \sum \int_{A_k} 
[\left( i
\log|w_k-f^{\alpha_k}(z_k)|^2\right)( H^{\alpha_k}\partial \chi_k\wedge 
 \phi_{12})]
d\mu_k(\alpha_k)\\
& + &
\sum \int_{A_k} 
[\left( i
\log|w_k-f^{\alpha_k}(z_k)|^2\right)( \chi_k \partial H^{\alpha_k}
\wedge 
 \phi_{12})]
d\mu_k(\alpha_k)\\
 T^g_{11}(\phi_{11})& = &
\partial S^{g,1}_{01}(\phi_{11})\\
& + & 
\overline{\partial}R^{g,1}_{10}(\phi_{11})\\
& - & \sum \int_{A_k} 
[\left( i
\log|w_k-f^{\alpha_k}(z_k)|^2\right)( \overline{\partial} H^{\alpha_k}
\wedge \partial \chi_k 
\wedge 
 \phi_{11})]
d\mu_k(\alpha_k)\\
& - &\sum \int_{A_k} [\left( i
\log|w_k-f^{\alpha_k}(z_k)|^2\right) 
 H^{\alpha_k}\overline{\partial}\partial \chi_k\wedge 
 \phi_{11}]
d\mu_k(\alpha_k)\\
&- &\sum \int_{A_k} 
[\left( i
\log|w_k-f^{\alpha_k}(z_k)|^2\right)(\overline{\partial}
 \chi_k \wedge \partial H^{\alpha_k}
\wedge 
\phi_{11})]
d\mu_k(\alpha_k)\\
\eea

\bea
S^{g,2}_{0,1}(\phi_{21}) & := & \sum_s  
\frac{1}{2}\sum \int_{A_k} [\left( i \log|w_s-f^{\alpha_s}(z_s)|^2\right)
(\overline{\partial}  \chi_k  [(w_s-f^{\alpha_s})\psi^{\alpha_s,\alpha_k}] 
\wedge 
\Lambda_s \phi_{21})]
d\mu_s(\alpha_s)\\
R^{g,2}_{1,0}(\phi_{12}) & := &
\sum_s
- \frac{1}{2}\sum \int_{A_k} 
[\left( i
\log|w_s-f^{\alpha_s}(z_s)|^2\right) \overline{
[(w_s-f^{\alpha_s})\psi^{\alpha_s,\alpha_k}]}
\wedge \partial \chi_k 
\wedge 
\Lambda_s \phi_{12})]
d\mu_s(\alpha_s)\\
\eea

\begin{definition}
\bea
S^{g,3}_{0,1} : = S^{g,1}_{0,1}+S^{g,2}_{0,1}\\
R^{g,3}_{1,0} : = R^{g,1}_{1,0}+R^{g,2}_{1,0}\\
\eea
\end{definition}

We then show, if $\sigma$ denotes a ${\cal C}^\infty$ form:

\begin{theo}
$$T= \partial S^{g,3}_{0,1}+ \overline{\partial} R^{g,3}_{1,0} +\sigma$$
\end{theo}


We define 

\bea
S^{g,4}_{01}(\phi_{21}) & = & -\sum_k \int_{A_k} \left( \log
 |w_k-f^{\alpha_k}(z_k)|^2 \right)
\overline{\partial} (H^{\alpha_k} \chi_k)\wedge   \phi_{21}
 d \mu_k (\alpha_k)\\
R^{g,4}_{10}(\phi_{12}) & = & -\sum_k \int_{A_k} \left( 
\log |w_k-f^{\alpha_k}(z_k)|^2 \right)
\partial(H^{\alpha_k} \chi_k ) \wedge\phi_{12}) d \mu_k (\alpha_k)\\
(\partial S^{g,4}_{11})(\phi_{11}) & = &
 -\sum_k \int_{A_k} \left( \log
 |w_k-f^{\alpha_k}(z_k)|^2 \right)
\overline{\partial} (H^{\alpha_k} \chi_k)\wedge   \partial\phi_{11}
 d \mu_k (\alpha_k)\\
 & = &
 \sum_k \int_{A_k} \partial[\left( \log
 |w_k-f^{\alpha_k}(z_k)|^2 \right)
\overline{\partial} (H^{\alpha_k} \chi_k)\wedge  \phi_{11}]
 d \mu_k (\alpha_k)\\
 & - &
\sum_k \int_{A_k} \left(\partial \log
 |w_k-f^{\alpha_k}(z_k)|^2 \right)
\overline{\partial} (H^{\alpha_k} \chi_k)\wedge \phi_{11}
 d \mu_k (\alpha_k)\\
 & - &
\sum_k \int_{A_k} \left( \log
 |w_k-f^{\alpha_k}(z_k)|^2 \right)
\partial\overline{\partial} (H^{\alpha_k} \chi_k)\wedge  \phi_{11}
 d \mu_k (\alpha_k)\\
(\partial S^{g,4}_{11})(\phi_{11})
 & = &
\sum_k \int_{A_k} \left(\partial \log
 |w_k-f^{\alpha_k}(z_k)|^2 \right)
\overline{\partial} (H^{\alpha_k} \chi_k)\wedge \phi_{11}
 d \mu_k (\alpha_k)\\
 & - &
\sum_k \int_{A_k} \left( \log
 |w_k-f^{\alpha_k}(z_k)|^2 \right)
\partial\overline{\partial} (H^{\alpha_k} \chi_k)\wedge  \phi_{11}
 d \mu_k (\alpha_k)\\
{\mbox{Hence}} & &
-i\sum_k \int_{A_k} \left(\partial \log
 |w_k-f^{\alpha_k}(z_k)|^2 \right)
\overline{\partial} (H^{\alpha_k} \chi_k)\wedge \phi_{11}
 d \mu_k (\alpha_k)\\
& = & i(\partial S^{g,4}_{11})(\phi_{11})\\
 & + &i
\sum_k \int_{A_k} \left( \log
 |w_k-f^{\alpha_k}(z_k)|^2 \right)
\partial\overline{\partial} (H^{\alpha_k} \chi_k)\wedge  \phi_{11}
 d \mu_k (\alpha_k)\\
(\overline{\partial}R^{g,4}_{11})(\phi_{11}) & = &
 -\sum_k \int_{A_k} \left( \log
 |w_k-f^{\alpha_k}(z_k)|^2 \right)
 (H^{\alpha_k} \chi_k)\wedge  \overline{ \partial}\phi_{11}
 d \mu_k (\alpha_k)\\
 & = &
 \sum_k \int_{A_k} \overline{\partial}[\left( \log
 |w_k-f^{\alpha_k}(z_k)|^2 \right)
{\partial} (H^{\alpha_k} \chi_k)\wedge  \phi_{11}]
 d \mu_k (\alpha_k)\\
 & - &
\sum_k \int_{A_k} \left(\overline{\partial} \log
 |w_k-f^{\alpha_k}(z_k)|^2 \right)
{\partial} (H^{\alpha_k} \chi_k)\wedge \phi_{11}
 d \mu_k (\alpha_k)\\
 & - &
\sum_k \int_{A_k} \left( \log
 |w_k-f^{\alpha_k}(z_k)|^2 \right)
\overline{\partial}{\partial} (H^{\alpha_k} \chi_k)\wedge  \phi_{11}
 d \mu_k (\alpha_k)\\
(\overline{\partial}R^{g,4}_{11})(\phi_{11}) 
 & = &
 \sum_k \int_{A_k} \overline{\partial}[\left( \log
 |w_k-f^{\alpha_k}(z_k)|^2 \right)
{\partial} (H^{\alpha_k} \chi_k)\wedge  \phi_{11}]
 d \mu_k (\alpha_k)\\
 & - &
\sum_k \int_{A_k} \left(\overline{\partial} \log
 |w_k-f^{\alpha_k}(z_k)|^2 \right)
{\partial} (H^{\alpha_k} \chi_k)\wedge \phi_{11}
 d \mu_k (\alpha_k)\\
 & - &
\sum_k \int_{A_k} \left( \log
 |w_k-f^{\alpha_k}(z_k)|^2 \right)
\overline{\partial}{\partial} (H^{\alpha_k} \chi_k)\wedge  \phi_{11}
 d \mu_k (\alpha_k)\\
{\mbox{Hence}} & &
 i \sum_k \int_{A_k} \left( \overline{\partial} 
\log |w_k-f^{\alpha_k}(z_k)|^2 \right)
\partial(H^{\alpha_k} \chi_k ) \wedge\phi_{11}) d \mu_k (\alpha_k)\\
& = & -i(\overline{\partial}R^{g,4}_{11})(\phi_{11})\\
 & - &i
\sum_k \int_{A_k} \left( \log
 |w_k-f^{\alpha_k}(z_k)|^2 \right)
\overline{\partial}{\partial} (H^{\alpha_k} \chi_k)\wedge  \phi_{11}
 d \mu_k (\alpha_k)\\
\eea

\bea
(\overline\partial S^{g,4}_{11})(\phi_{11}) & = &
 -\sum_k \int_{A_k} \left( \log
 |w_k-f^{\alpha_k}(z_k)|^2 \right)
\overline{\partial} (H^{\alpha_k} \chi_k)\wedge   \overline\partial\phi_{11}
 d \mu_k (\alpha_k)\\
 & = &
 \sum_k \int_{A_k} \overline\partial[\left( \log
 |w_k-f^{\alpha_k}(z_k)|^2 \right)
\overline{\partial} (H^{\alpha_k} \chi_k)\wedge  \phi_{11}]
 d \mu_k (\alpha_k)\\
 & - &
\sum_k \int_{A_k} \left(\overline\partial \log
 |w_k-f^{\alpha_k}(z_k)|^2 \right)
\overline{\partial} (H^{\alpha_k} \chi_k)\wedge \phi_{11}
 d \mu_k (\alpha_k)\\
(\overline\partial S^{g,4}_{11})(\phi_{11})
 & = &
\sum_k \int_{A_k} \left(\overline\partial \log
 |w_k-f^{\alpha_k}(z_k)|^2 \right)
\overline{\partial} (H^{\alpha_k} \chi_k)\wedge \phi_{11}
 d \mu_k (\alpha_k)\\
 {\mbox{Hence}} & &
-i\sum_k \int_{A_k} \left(\overline\partial \log
 |w_k-f^{\alpha_k}(z_k)|^2 \right)
\overline{\partial} (H^{\alpha_k} \chi_k)\wedge \phi_{11}
 d \mu_k (\alpha_k)\\
& = & i(\partial S^{g,4}_{11})(\phi_{11})\\
 ({\partial}R^{g,4}_{11})(\phi_{11}) & = &
 -\sum_k \int_{A_k} \left( \log
 |w_k-f^{\alpha_k}(z_k)|^2 \right)
 (H^{\alpha_k} \chi_k)\wedge  { \partial}\phi_{11}
 d \mu_k (\alpha_k)\\
 & = &
 \sum_k \int_{A_k} {\partial}[\left( \log
 |w_k-f^{\alpha_k}(z_k)|^2 \right)
{\partial} (H^{\alpha_k} \chi_k)\wedge  \phi_{11}]
 d \mu_k (\alpha_k)\\
 & - &
\sum_k \int_{A_k} \left({\partial} \log
 |w_k-f^{\alpha_k}(z_k)|^2 \right)
{\partial} (H^{\alpha_k} \chi_k)\wedge \phi_{11}
 d \mu_k (\alpha_k)\\
 ({\partial}R^{g,4}_{11})(\phi_{11}) 
 & = & -
\sum_k \int_{A_k} \left({\partial} \log
 |w_k-f^{\alpha_k}(z_k)|^2 \right)
{\partial} (H^{\alpha_k} \chi_k)\wedge \phi_{11}
 d \mu_k (\alpha_k)\\
 {\mbox{Hence}} & &
 i \sum_k \int_{A_k} \left({\partial} 
\log |w_k-f^{\alpha_k}(z_k)|^2 \right)
\partial(H^{\alpha_k} \chi_k ) \wedge\phi_{11}) d \mu_k (\alpha_k)\\
& = & -i({\partial}R^{g,4}_{11})(\phi_{11})\\
 \eea

Hence

\begin{prop}
\bea
T^g_{11}(\phi_{11}) & = &
(i\partial \overline{\partial} u^g)_{11}(\phi_{11})\\
& + & i(\partial S^{g,4}_{11})(\phi_{11})\\
& - & i(\overline{\partial}R^{g,4}_{11})(\phi_{11})\\
& + & i
\sum_k \int_{A_k} \left( \log
 |w_k-f^{\alpha_k}(z_k)|^2 \right)
\partial\overline{\partial} (H^{\alpha_k} \chi_k)\wedge  \phi_{11}
 d \mu_k (\alpha_k)\\
u^g_{00} & \in & L^p,\; p<\infty\\
\overline{\partial} u^g & \in & L^p\; p<2\\
i\partial \overline{\partial} u^g && {\mbox{has measure coefficients}}\\
S^{g,4}_{01} & \in & L^p\; p<\infty\\
\partial S^{g,4} & \in & L^p\; p<2\\
R^{g,4}_{10} & \in & L^p\; p<\infty\\
\overline{\partial} R^{g,4} & \in & L^p\; p<2\\
T \geq 0 & \Rightarrow & \\
\overline{\partial}S^{g,4} & \in & L^2\\
\partial R^{g,4} & \in & L^2\\
 & \Rightarrow & \\
\exists \tilde{S}^{g,4}_{01} & \in & W^1\\
\overline{\partial}  \tilde{S}^{g,4}_{01} & = & \overline{\partial}S^{g,4}\\
S^{g,4} & = & \tilde{S}^{g,4}_{01} +\lambda_{00}, \lambda_{00}
 \in L^p,p<\infty\\
\lambda_{00} & = & \overline{\partial}\tilde{u}_{00},  \tilde{u}_{00}\in
 W^1.\\
\eea
\end{prop}

We estimate $\overline{\partial}S^{g,4}.$ Suppose $T$ has finite energy.
Then $\overline{\partial} S \in L^2.$
The main local terms are in the form

$F= \int_\alpha \frac{ F^\alpha(z)}{w-f^\alpha(z)}d\mu(\alpha),$
$F^\alpha(z)$ holomorphic.
Since $F(z,w)\in L^2$ it follows that
$$\int_\alpha \frac{ F^\alpha(z)}{\overline{w-f^\alpha(z)}}d\mu(\alpha)
\; and$$

$$
\int_\alpha \frac{ F^\alpha(z)}{|w-f^\alpha(z)|}d\mu(\alpha)
$$ $$
\int_{\alpha,\beta} \int_D F\overline{F}<\infty.
$$
  
\noindent also are in $L^2$.

We have the following formula valid almost everywhere in $(z,w):$

\bea
|F|^2(z,w) & = &  \int \frac{F^\alpha{\overline{F}}^\beta}{(w-f^\alpha)
(\overline{w-f^\beta})}d\mu(\alpha)d\mu(\beta)\\
& \leq &   \int \frac{|F^\alpha||{{F}}^\beta|}{|w-f^\alpha|
|w-f^\beta|}d\mu(\alpha)d\mu(\beta)\\
& < & \infty\\
\eea

This is closely related to the condition that the transverse measure
has finite energy: $\int_{\alpha,\beta}\log|f^\alpha-f^\beta|d\mu(\alpha)
d\mu(\beta)<\infty.$

\begin{lem}
Assume that $\mu$ has finite transverse energy.
Then $u^g$ is integrable with respect to $i\partial \overline{\partial} u^g$.
\end{lem}

{\bf Proof:}
\bea
\int u^g i\partial \overline{\partial} u^g & \geq & 
\int \int_\alpha \int_\beta \log|w-f^\alpha| dd^c \log |w-f^\beta|
d\mu(\alpha) d\mu(\beta)\\
& = & \int_{\alpha,\beta} \log |f^\alpha-f^\beta|d\mu(\alpha)d\mu(\beta)\\
& = & E(\mu)\\
\eea

Here $E(\mu)$ denotes the energy of $\mu.$

\CQFD\\

Hence the current $(dd^c u^g)^2$ is well defined. So it seems that
$T \wedge T$ is well defined. It should also be the limit
of what one gets by rotations. $T^\epsilon \wedge T \rightarrow T \wedge T.$

\begin{prop}
If $T$ has finite energy, then we can define $S^{g,4}_{01} \wedge
i \partial \overline{\partial} g$ and hence
$i\partial S^{g,4}_{01} \wedge
i \partial \overline{\partial} g,$ similarly for
 $R^{g,4}_{10} \wedge
i \partial \overline{\partial} g$ and hence
$i\overline{\partial} R^{g,4}_{01} \wedge
i \partial \overline{\partial} g.$
We can also define $i\partial S \wedge i\overline{\partial} R$
directly. The same goes for $i\partial S \wedge i\partial S $
and  $i\overline{\partial} R \wedge i\overline{\partial} R$. The last
two have integral $0.$
\end{prop}


The condition that $\overline{\partial}S^{g,4}\in L^2$ means
that 

We can do this integral by first integrating away $w$.
This leads to finiteness of the integral:

$$
\int_{\alpha,\beta} \log |f^\alpha-f^\beta|(z) F^\alpha \overline{F}^\beta
d\mu(\alpha) d\mu(\beta).
$$

\begin{coro}
The signed measures $\int F^\alpha d\mu^\alpha$ of any harmonic laminar
current on $\PP^2$ have finite energy.
\end{coro}



Note that we can define the form $\partial S^{g,4}\wedge \overline{\partial}
R^{g,4}$ pointwise. The coefficients are in $L^p,p<1.$
In fact they are in $L^1:$

\begin{theo}
$\partial S^{g,4}\wedge \overline{\partial S^{g,4}}+
\overline{\partial}S^{g,4}\wedge \partial \overline{S^{g,4}}
\in L^1.$
\end{theo}

{\bf Proof} We estimate the contributions from any pair $\alpha,\beta.$
Say one local leaf is $w=0$ the other is $w=f(z).$
By Holders continuity of holomorphic motion we know that
$|f'(0)| \leq C |f(0)|^\epsilon$ for some small fixed $\epsilon.$
the key estimate then becomes:

\bea
\partial \log |w|^2 \wedge \overline{\partial} \log |w-f|^2
& + & \overline{\partial} \log |w|^2\wedge
\partial \log |w-f|^2\\
& = & dw\wedge d \overline{w} \left[
\frac{1}{w(\overline{w}-\overline{f})}-\frac{1}{(w-f)\overline{w}}\right]
-\cal{O}(\frac{f'}{w(w-f)})\\
& \in & L^1.\\
\eea

\CQFD\\

\subsubsection{Currents on holomorphic motion}

\begin{prop}
Let $T$ be a positive current supported by a holomorphic
motion on the unit polydisc $\Delta^2.$ (So we have a collection
of graphs over the $z-$ axis.) Let $\pi$ be the projection in the transverse
direction, (along leaves to the $w$ axis?)
we assume that $\pi$ is ${\cal C}^1.$  \\
(i) If $T$ is closed and directed by the holomorphic motion then
$$
T=\int [V_\alpha]d\mu(\alpha)
$$
(ii) If $T$ is harmonic and directed by the holomorphic motion
then
$$
T=\int h_\alpha [V_\alpha] d\mu(\alpha)
$$
\noindent where $h_\alpha>0$ is harmonic.
\end{prop}

{\bf Proof:}
A current is directed if it is in the closed convex hull of the
local discs of the lamination. If $\pi$ is ${\cal C}^1$, then $d\pi$
is continuous and hence $T \wedge d\pi=0. (1)$
We are going to determine the extreme points of the closed convex set
of positive closed (resp. harmonic) currents satisfying (1).
Let $\chi$ be a function with small support on a transversal.
Assume $\chi\geq 0$ and smooth. The current $T_\chi=\chi\circ \pi T\geq 0,$
is closed when $T$ is closed. If $T$ is harmonic $d^c T_\chi=
T \wedge \chi'\circ \pi d\pi+\chi \circ \pi d^cT=\chi\circ \pi d^cT.$
$dd^c T=\chi\circ \pi dd^cT+\chi' \circ \pi d\pi \wedge d^cT=0$ for
type reasons. As a consequence the extreme points have a support intersecting
a transversal at only one point.\\

\CQFD\\

\begin{question}
If the holomorphic motion is only continuous there is a one form
$\gamma,$ continuous so that the currents supported by the lamination
satisfy $T \wedge \gamma=0.$ What are the extreme points in this
case?
\end{question}


  




 
Next step: Calculate $\overline{\partial}S^{g,4}_{01} ,
\partial R^{g,4}_{10}$ and 
$\overline{\partial}S^{g,4}_{01} \wedge \partial R^{g,4}_{10}.$
Show that since  
$\int \overline{\partial}S^{g,4}_{01} \wedge \partial R^{g,4}_{10}$
is finite assuming $\|T\|_e<\infty$, it follows that the transverse
measures of the harmonic current has finite energy also.
 

\subsection{$ T^g_{11}(\phi_{11})-\partial S^{g,1}_{01}(\phi_{11})
-\overline{\partial}R^{g,1}_{10}(\phi_{11})$ in local coordinates}

Observe that 

\begin{lem}
\bea
 T^g_{11}(\phi_{11}) & - & \partial S^{g,1}_{01}(\phi_{11})
-\overline{\partial}R^{g,1}_{10}(\phi_{11})\\
& = &  T^g_{11}(\phi_{11})-i\partial \overline{\partial}u^g_{00}(\phi_{11})\\
& - & i\partial S^{g,4}_{01}(\phi_{11})
+i\overline{\partial}R^{g,4}_{10}(\phi_{11})\\
& = & 
\sum \int_{A_k} [\left( i
\log|w_k-f^{\alpha_k}(z_k)|^2\right) 
 {\partial}\overline{\partial} ( H^{\alpha_k}\chi_k)\wedge 
 \phi_{11}]
d\mu_k(\alpha_k)\\
& = & 
- \sum \int_{A_k} 
[\left( i
\log|w_k-f^{\alpha_k}(z_k)|^2\right)( \overline{\partial} H^{\alpha_k}
\wedge \partial \chi_k 
\wedge 
 \phi_{11})]
d\mu_k(\alpha_k)\\
& - &\sum \int_{A_k} [\left( i
\log|w_k-f^{\alpha_k}(z_k)|^2\right) 
 H^{\alpha_k}\overline{\partial}\partial \chi_k\wedge 
 \phi_{11}]
d\mu_k(\alpha_k)\\
&- &\sum \int_{A_k} 
[\left( i
\log|w_k-f^{\alpha_k}(z_k)|^2\right)(\overline{\partial}
 \chi_k \wedge \partial H^{\alpha_k}
\wedge 
\phi_{11})]
d\mu_k(\alpha_k)\\ 
T^g_{11}(\phi_{11}) & \sim & c \omega\\
c & = &  
[\sum \int_{A_k} \left( i
\log|w_k-f^{\alpha_k}(z_k)|^2\right) 
 {\partial}\overline{\partial} ( H^{\alpha_k}\chi_k)](\omega)
d\mu_k(\alpha_k)\\
\eea
\end{lem}

Let $U_j$ denote one of the local coordinate neighborhoods.
For any $k$ we can write $$w_k-f^{\alpha_k}(z_k)=
[w_j-f^{\alpha_j}(z_j)] \Phi_{\alpha_j,\alpha_k}(z_j.w_j)$$

\noindent where $\alpha_k=\gamma_{jk}(\alpha_k)$
and $\Phi_{\alpha_j,\alpha_k}(z_j.w_j)$ is an invertible holomorphic
function with uniform estimates.


The functions $\frac{H^{\alpha_k}}
{\lambda_{jk}}$ and $H^{\alpha_j}$ agree on the $\Delta_\alpha$.
Hence they are of the form

$$\frac{H^{\alpha_k}}
{\lambda_{jk}}= H^{\alpha_j}+ \Re[ (w_j-f^{\alpha_j})\psi^{\alpha_j,\alpha_k}]
.$$

\subsubsection{The smooth parts in local coordinates}



\begin{theo}
\bea
T & = & \partial [S^{g,1}_{01}+S^{g.2}_{01}-\sigma^j_2+\sigma^j_6+\sigma^j_7]
+\overline{\partial} [R^{g,1}_{10}+R^{g,2}_{10}-\sigma^j_3+\sigma^j_5
+\sigma^j_8]\\
\eea
\end{theo}

\bea
(\sigma^j_2)_{01}(\phi_{2,1}) & = & 
\sum_s  
\frac{1}{2}\sum \int_{A_k} [\left( i \log|\Phi_{\alpha_j,\alpha_s}|^2\right)
(\overline{\partial}  \chi_k  [(w_s-f^{\alpha_s})\psi^{\alpha_s,\alpha_k}] 
\wedge 
\Lambda_s \phi_{21})]
d\mu_s(\alpha_s)\\
(\sigma^j_6)_{01}(\phi_{12}) & = & 
 -  \frac{1}{2}\sum \int_{A_k} 
[\chi_k \overline{
\psi^{\alpha_j,\alpha_k}}\left( i\overline{\partial
(w_j-f^{\alpha_j}(z_j))}\right)] \wedge 
 \phi_{12}
d\mu_j(\alpha_j)\\
(\sigma^j_7)_{01}(\phi_{21}) & = & -\sum \int_{A_k} ( i \overline{
\left[\frac{\partial \Phi_{\alpha_j,\alpha_k}(z_j,w_j)}{\Phi_{\alpha_j,\alpha_k}}\right]}
 [H^{\alpha_k} \chi_k]
\wedge  \phi_{21}) 
d\mu_k(\alpha_k)\\
(\sigma^j_3)_{10}(\phi_{12}) & = &
\sum_s
- \frac{1}{2}\sum \int_{A_k} 
[\left( i
\log|\Phi_{\alpha_j,\alpha_s}|^2\right) \overline{
[(w_s-f^{\alpha_s})\psi^{\alpha_s,\alpha_k}]}
\wedge \partial \chi_k 
\wedge 
\Lambda_s \phi_{12})]
d\mu_s(\alpha_s)\\
(\sigma^j_5)_{10}(\phi_{21}) & = &
\frac{1}{2}\sum \int_{A_k}[\chi_k \psi^{\alpha_j,\alpha_k}
\left( i \partial 
(w_j-f^{\alpha_j}(z_k))\right)] 
\wedge 
  \phi_{21}
d\mu_j(\alpha_j))\\
(\sigma^j_8)_{10}(\phi_{12}) & = & 
-\sum \int_{A_k} \left( i 
\log|\Phi_{\alpha_j,\alpha_k}(z_j,w_j)|^2\right)
(\partial [H^{\alpha_k} \chi_k]
\wedge  \phi_{12})]
d\mu_k(\alpha_k)\\
\eea

\newpage

\subsubsection{Integrals $\partial S,R \wedge T$}

\bea
\int \partial S^{g,1,\epsilon}_{01} \wedge T^g_{1,1} & = & 
\int S^{g,1,\epsilon}_{01} \wedge \partial T\\
& = & \sum_k \int_{\Delta_{\alpha_k}}
[\sum_\ell \int_{A_\ell} i \overline{\partial}
\log |w^\epsilon_\ell-f^{\beta_\ell,\epsilon}|^2 H^{\beta_\ell,\epsilon}
\chi_{\ell,\epsilon} (-\partial (H_k \chi_k))d\mu(\beta_\ell)d\mu(\alpha_k)\\
& = & \sum_k \int_{\Delta_{\alpha_k}}
[\sum_\ell \int_{A_\ell} i \partial\overline{\partial}
\log |w^\epsilon_\ell-f^{\beta_\ell,\epsilon}|^2 H^{\beta_\ell,\epsilon}
\chi_{\ell,\epsilon} (- (H_k \chi_k))d\mu(\beta_\ell)d\mu(\alpha_k)\\
& + & \sum_k \int_{\Delta_{\alpha_k}}
[\sum_\ell \int_{A_\ell} i \overline{\partial}
\log |w^\epsilon_\ell-f^{\beta_\ell,\epsilon}|^2 \partial(
H^{\beta_\ell,\epsilon}
\chi_{\ell,\epsilon}) ( (H_k \chi_k))d\mu(\beta_\ell)d\mu(\alpha_k)\\
& = & \mbox{Geometric product}\\
& + &  \sum_k \int_{\Delta_{\alpha_k}}
[\sum_\ell \int_{A_\ell} i \overline{\partial}
\log |w^\epsilon_\ell-f^{\beta_\ell,\epsilon}|^2 \partial(
H^{\beta_\ell,\epsilon}
\chi_{\ell,\epsilon}) ( (H_k \chi_k))d\mu(\beta_\ell)d\mu(\alpha_k)\\
\eea
 


\bea
\int \partial S^{g,1,\epsilon}_{01} \wedge T^g_{1,1} & = & 
\int S^{g,1,\epsilon}_{01} \wedge \partial T\\
& = & \sum_k \int_{\Delta_{\alpha_k}}
[\sum_\ell \int_{A_\ell} i \overline{\partial}
\log |w^\epsilon_\ell-f^{\beta_\ell,\epsilon}|^2 H^{\beta_\ell,\epsilon}
\chi_{\ell,\epsilon} (-\partial (H_k \chi_k))d\mu(\beta_\ell)]d\mu(\alpha_k)\\
& = & -\sum_k \int_{\Delta_{\alpha_k}}
[\sum_\ell \int_{A_\ell} i \overline{\partial}
[\log |w^\epsilon_\ell-f^{\beta_\ell,\epsilon}|^2 H^{\beta_\ell,\epsilon}
\chi_{\ell,\epsilon} (\partial (H_k \chi_k))]d\mu(\beta_\ell)]d\mu(\alpha_k)\\
& + & \sum_k \int_{\Delta_{\alpha_k}}
[\sum_\ell \int_{A_\ell} i 
\log |w^\epsilon_\ell-f^{\beta_\ell,\epsilon}|^2  \overline{\partial}
(H^{\beta_\ell,\epsilon}
\chi_{\ell,\epsilon}) (\partial (H_k \chi_k))d\mu(\beta_\ell)]d\mu(\alpha_k)\\
& + & \sum_k \int_{\Delta_{\alpha_k}}
[\sum_\ell \int_{A_\ell} i 
\log |w^\epsilon_\ell-f^{\beta_\ell,\epsilon}|^2 H^{\beta_\ell,\epsilon}
\chi_{\ell,\epsilon}  \overline{\partial}(\partial (H_k \chi_k))d\mu(\beta_\ell)]d\mu(\alpha_k)\\
& = & \sum_k \int_{\Delta_{\alpha_k}}
[\sum_\ell \int_{A_\ell} i 
\log |w^\epsilon_\ell-f^{\beta_\ell,\epsilon}|^2  \overline{\partial}
(H^{\beta_\ell,\epsilon}
\chi_{\ell,\epsilon}) (\partial (H_k \chi_k))d\mu(\beta_\ell)]d\mu(\alpha_k)\\
& + & \sum_k \int_{\Delta_{\alpha_k}}
[\sum_\ell \int_{A_\ell} i 
\log |w^\epsilon_\ell-f^{\beta_\ell,\epsilon}|^2 H^{\beta_\ell,\epsilon}
\chi_{\ell,\epsilon}  \overline{\partial}(\partial (H_k \chi_k))d\mu(\beta_\ell)]d\mu(\alpha_k)\\
\eea






\bea
\int \overline{\partial} R^{g,1\epsilon}_{10} \wedge T^g_{1,1} & = & 
\sum_k \int_{\Delta_k} 
(i\overline{\partial} \log |w^\epsilon_\ell-f^{\beta_\ell,\epsilon}|^2
\partial (H^{\beta_\ell,\epsilon}\chi_{\ell,\epsilon}) (H_k\chi_k)
d\mu(\beta_\ell)d\mu(\alpha_k)\\
& + & \sum_k \int_{\Delta_k} 
(i \log |w^\epsilon_\ell-f^{\beta_\ell,\epsilon}|^2
\overline{\partial}
\partial (H^{\beta_\ell,\epsilon}\chi_{\ell,\epsilon}) (H_k\chi_k)
d\mu(\beta_\ell)d\mu(\alpha_k)\\
\eea





\bea
\int \overline{\partial} R^{g,1\epsilon}_{10} \wedge T^g_{1,1} & = & 
\int R^{g,1\epsilon}_{10} \wedge \overline{\partial}
T^g_{1,1}\\
& = & -\sum_k \int_{\Delta_k} \sum_\ell
i \log |w^\epsilon_\ell-f^{\beta_\ell,\epsilon}|^2
{\partial}
(H^{\beta_\ell,\epsilon}\chi_{\ell,\epsilon}) \overline{\partial}(H_k\chi_k)
d\mu(\beta_\ell)d\mu(\alpha_k)\\
\eea




\bea
\int [\partial S^{g.1} +\overline{\partial} R^{g.1}]\wedge T & = &
 \sum_k \int_{\Delta_{\alpha_k}}
[\sum_\ell \int_{A_\ell} i 
\log |w^\epsilon_\ell-f^{\beta_\ell,\epsilon}|^2  \overline{\partial}
(H^{\beta_\ell,\epsilon}
\chi_{\ell,\epsilon}) (\partial (H_k \chi_k))d\mu(\beta_\ell)]d\mu(\alpha_k)\\
& + & \sum_k \int_{\Delta_{\alpha_k}}
[\sum_\ell \int_{A_\ell} i 
\log |w^\epsilon_\ell-f^{\beta_\ell,\epsilon}|^2 H^{\beta_\ell,\epsilon}
\chi_{\ell,\epsilon}  \overline{\partial}(\partial (H_k \chi_k))d\mu(\beta_\ell)]d\mu(\alpha_k)\\
& - &\sum_k \int_{\Delta_k} \sum_\ell
i \log |w^\epsilon_\ell-f^{\beta_\ell,\epsilon}|^2
{\partial}
(H^{\beta_\ell,\epsilon}\chi_{\ell,\epsilon}) \overline{\partial}(H_k\chi_k)
d\mu(\beta_\ell)d\mu(\alpha_k)\\
\eea

\bea
(\Delta T^g)(\phi_{11}) & := & 
(T^g_{11}-\partial S^{g,1}_{01}- \overline{\partial}R^{g,1}_{10})(\phi_{11})\\
& = & -\sum \int_{A_k} 
[\left( i
\log|w_k-f^{\alpha_k}(z_k)|^2\right)( \overline{\partial} H^{\alpha_k}
\wedge \partial \chi_k 
\wedge 
 \phi_{11})]
d\mu_k(\alpha_k)\\
& - &\sum \int_{A_k} [\left( i
\log|w_k-f^{\alpha_k}(z_k)|^2\right) 
 H^{\alpha_k}\overline{\partial}\partial \chi_k\wedge 
 \phi_{11}]
d\mu_k(\alpha_k)\\
&- &\sum \int_{A_k} 
[\left( i
\log|w_k-f^{\alpha_k}(z_k)|^2\right)(\overline{\partial}
 \chi_k \wedge \partial H^{\alpha_k}
\wedge 
\phi_{11})]
d\mu_k(\alpha_k)\\
\eea

\bea
\int T^{g,\epsilon} \wedge T & = & \int [\partial S^{g.1} +\overline{\partial} R^{g.1}]\wedge T\\
& + & \int \Delta T^g \wedge T\\
& = &
 \sum_k \int_{\Delta_{\alpha_k}}
[\sum_\ell \int_{A_\ell} i 
\log |w^\epsilon_\ell-f^{\beta_\ell,\epsilon}|^2  \overline{\partial}
(H^{\beta_\ell,\epsilon}
\chi_{\ell,\epsilon}) (\partial (H_k \chi_k))d\mu(\beta_\ell)]d\mu(\alpha_k)\\
& + & \sum_k \int_{\Delta_{\alpha_k}}
[\sum_\ell \int_{A_\ell} i 
\log |w^\epsilon_\ell-f^{\beta_\ell,\epsilon}|^2 H^{\beta_\ell,\epsilon}
\chi_{\ell,\epsilon}  \overline{\partial}(\partial (H_k \chi_k))d\mu(\beta_\ell)]d\mu(\alpha_k)\\
& - &\sum_k \int_{\Delta_k} \sum_\ell
i \log |w^\epsilon_\ell-f^{\beta_\ell,\epsilon}|^2
{\partial}
(H^{\beta_\ell,\epsilon}\chi_{\ell,\epsilon}) \overline{\partial}(H_k\chi_k)
d\mu(\beta_\ell)d\mu(\alpha_k)\\
& - &  \sum_k \sum_\ell \int_{A_k} 
[\left( i
\log|w_\ell-f^{\beta_\ell}(z_\ell)|^2\right)( \overline{\partial} \partial
[H^{\beta_\ell} \chi_\ell] (H^{\alpha_k} \chi_k) d \mu(\beta_\ell)
d\mu_k(\alpha_k)\\
& = & 0 \; {\mbox{mod sign}}\\
\eea
$\int T^\epsilon \wedge T=0???$




\subsection{Cross-Laminations}

{\bf Find examples of Cross larminations}
{\bf Take a map from the unit disc to $\PP^2$ coming from perturbing
the universal coverin of a nonhyperbolic compact Riemann surface.
Then one can use the Ahlfors trick to contruct a harmonic current
on the closure. What are the properties of this harmonic current?}

{\bf For JE: It will be interesting to get rid of the hypothesis $Y \setminus
E$ is a lamination and just assume that $T=\sum T_j$ each $T_j=
\int [D_\alpha^j] d\mu^j(\alpha)$, $[D^j_\alpha]$ graphs on a fixed direction
$e_j.$}



Until we have a better word, I'll use the expression Cross-lamination
or X-lamination to describe laminations where leaves are allowed to cross.

\medskip

Let $w=f(z)=\sum_n a_nz^n$ be a graph of a holomorphic function,
$|z|<2.$
We say that $f$ is $\underline{horizontal}$ if $|a_0|\leq 10,
|a_1| \leq 2, |a_n| \leq \frac{1}{3^n},n \geq 2.$ We say that $f$ is
$\underline{local}$ if the graph intersects the closed unit bidisc.

\medskip

We put a topology on the space ${\cal H}$ of horizontal graphs,
$f=\sum_n a_n z^n,g=\sum_n b_n z^n.$


$$
d(f,g)= \sup_n\{|a_n-b_n|\}.
$$

Then ${\cal H}$ is a separable metric space (polynomials in ${\cal H}$
with rational coefficients are dense). Moreover ${\cal H}$ is sequentially
compact, so compact.

\medskip

A local horizontal chart of an $X-$lamination is a compact subset
${\cal H}' \subset {\cal H}.$ Similarly we  define local vertical
charts ${\cal V}'\subset {\cal V}.$ We define ${\cal H} \cap {\cal V}$
as the space of graphs $w=f(z),z=g(w)$ intersecting in the unit bidisc,
such that $f(z)$ extends analytically to $\tilde{f}$ on a larger disc so that
the graph $z=g(w)$ is a subset of the graph of $\tilde{f}$ and vice versa.

\medskip

We will next restrict to $f\in {\cal H}$ for which $f$ extends to a "large"
disc and

$\frac{1}{1+\epsilon}<|f'(z)|<1+\epsilon$
when $(z,w)\in \Delta^2(0,2).$ Then we can identify $f$ with $g$ in ${\cal V}$
where $g$ is the inverse function. This is a homeomorphism.

\begin{definition}
An $X-$lamination is a compact subset of $\PP^2$ which has local ${\cal H}$
and ${\cal V}$ as above. Moreover, the leaves in neighborhing charts
are analytic continuations of each other and the parametrizations vary
continuously.
\end{definition}

\begin{definition}
An $X-$lamination is $\underline{minimal}$ if all leaves are dense.
\end{definition}


An $X-$lamination is locally the image of a compact metric space $M$ times
the unit disc.

\medskip

A positive closed current on $X$ is locally of the form $\int_{M_j}
[\Delta_\alpha] d\mu_j(\alpha)$, here $\mu_j$ is a transverse measure
on $M_j$. Likewise a harmonic
current on $X$ is locally of the form $\int_{M_j} h_\alpha[\Delta_\alpha]
d\mu_j(\alpha).$ Here $h_\alpha>0$ is harmonic. Here the measures
$h_\alpha d\mu_j(\alpha)$ are invariant.

\medskip

\begin{theo}
There exists a nontrivial positive harmonic current on any $X-$lamination.
\end{theo}


 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection{Remarks and questions}

 As a consequence we don't know whether $Q$ has a
minimum on $C,$ and what is the infimum.
The main problem seems to me to show that $Q(T,T)>0$ for every $T$
in $C.$ 

\bigskip

We are back to the question: Is it possible that $T:=i\partial u
\wedge \overline{\partial u},$ $u$ complex valued, satisfies
$i \partial \overline{\partial}T=0$ because obviously $\int T \wedge T=0.$
The only thing we know, now, is that if it exists it's extremal.

\begin{question} How to define $T$ as a dual element of $H^{1,1}$:
Let $S$ be an element of $H^{1,1}$. We can consider that $H^{1,1}=
\{\phi_{1,1}; \overline{\partial}\phi=0\}/\{\overline{\partial}
\psi_{1,0}\}.$ We define $T(S)$ by the formula 
$T(S)=\int T \wedge \phi_{1,1}$ for any representative of $S$. Hence
we need to show that $<T, \overline{\partial}\psi_{1,0}>=0$
for all forms $\psi.$ Hence we have to show that $<\overline{\partial}T,
\psi_{1,0}>=0$ for all $\psi.$ This is not the case though.
\end{question}


\begin{question} Let $\Phi:\Delta \rightarrow L$ be a covering
of a leaf. Are all harmonic limit currents equal?
\end{question}

\begin{question}
Are all transverse measures of harmonic currents absolutely
continuous with respect to each other?
\end{question}


\begin{question}
Can there exist a domain in $\PP^n$ with smooth boundary
such that the boundary has two pieces and the domain has a smooth
defining function with at least two $\underline{strictly\; positive}$ eigenvalues
in the $\underline{complex\; tangent}$  direction? Not in $\PP^3$,
since the domain is strictly pseudoconvex, hence Stein.
\end{question}


\begin{question}
Let $M$ be a complex manifold of dimension $2$. Assume $M$ has a smooth
plurisubharmonic exhaustion function $v$. Suppose $M$ has just one end. Is $M$ Stein
(or a modification of Stein)? In the counterexample by Grauert there
are two ends. One has to study level sets of $v$, first case assume
$(dd^c v)^2\equiv 0.$
\end{question}


\begin{question}
Can there exist a smooth domain in $\PP^4$ with two ends
and a defining function with two strictly positive tangential eigenvalues?
\end{question}

\begin{conjecture} (Hodge)
Let $V$ be projective. Suppose that $\alpha$ is a $(p,p)$ closed form
for which $\int_X \alpha \in \ZZ$ for any submanifold $X.$
Then $\alpha$ is represented by $\sum n_i [X_i]$ for complex
subvarieties of dimesion $p.$
\end{conjecture}

\begin{problem}
Understand positive closed $(2,2)$ currents
in $\BB^3.$ If $T \geq 0$ has bidimension $(1,1)$ in $\BB^3$,
is $T=\lim c_n[V_n]$ for curves $V_n?$
\end{problem}

Methods for making $(2,2)$ currents:\\
$T=T_1 \wedge T_2.$\\
$T$ current, $u$ continuous on $T.$ Take $dd^c (uT).$


\begin{problem}
Solve $(dd^c u)^2=-1 $ on $\BB^2$ with $u, \nabla u$ given on boundary.
Is $u$ unique?
\end{problem}

\begin{problem}
Find a substitute for the Ahlfors Lemma for $\CC^2.$
Needed for dynamics.
\end{problem}

\begin{remark}
Harmonicity of $\partial u \wedge \overline{\partial u}$ means
$\partial \overline{\partial}u \wedge \partial \overline{\partial}\overline{u}=0$, not $(\partial \overline{\partial}u)^2=0.$
\end{remark}

\begin{remark}
i) $u=|w|^2+z+\overline{z}.$ Then $(\partial \overline{\partial} u)^2=0$
but $\partial \overline{\partial} u \wedge \partial u
\wedge {\overline{\partial}}u \neq 0.$
So it does not satisfy the Frobenius condition.\\
ii) $u=z+z\overline{w}.$ Then $\partial \overline{\partial} u \wedge \partial u=0
$ but $\partial \overline{\partial}u \wedge \partial \overline{\partial}\overline{u} \neq 0$
\end{remark}

\begin{remark}
Primary goal: Find $u$ complex so that $T=i\partial u \wedge \overline{\partial u}$ is an integrable distribution, $\partial \overline{\partial u}
\wedge \partial u \wedge \overline{\partial u} \equiv 0$.
Harmonicity of $T, \partial \overline{\partial}u \wedge \partial \overline{\partial}\overline{u}=0$ is a $\underline{different}$ and secondary condition.
\end{remark}

\begin{question} If $\int \omega \wedge S=1$ and $\int \omega \wedge T=1$
for $S,T$ harmonic and smooth, is $S \wedge T=1?$
\end{question}


Answer: No. Let $S_{01}$ be a smooth current, $T:=\omega+\epsilon \partial S
+\epsilon \overline{\partial}\overline{S}$, if $\epsilon$ is small, $T>0.$
If $\overline{\partial}S \neq 0, $ i.e. $T$ is non closed, then
$\int T \wedge T=\int \omega^2-\epsilon^2 \int \overline{\partial}S
\wedge \partial \overline{S}<1.$ Actually, since
$\int (T-\omega)\wedge \omega=0$ we get from Corollary 4.24 that $\int
T \dot{\wedge}T<1$ if $T$ is nono closed. $T \geq 0$
harmonic. If $C=\{T \geq 0, \int T \wedge \omega=1, T \;{\mbox{harmonic}}
\}.$




\begin{problem}
Can there be more than one harmonic current on a minimal laminated compact?
If $T_1(\phi)>T_2(\phi)$ and $T_j(\psi)=1,$ can we move the mass
around and get a contradiction?
One can add $\phi+ i\partial \overline{\partial} g$ to try to
spread  out $\phi$ to be equal to a multiple of $\psi.$ What is the codimension of $L_0$ in $L?$
\end{problem}

\begin{problem}
Find positive harmonic current $T \neq 0$ on $\mathbb P^2$
which is nowhere strictly positive.
\end{problem}

Can we show that harmonic currents are extremal?

\begin{problem}
Show that there exists no laminated compact set in $\PP^2$
which is not ${\cal C}^1$ and carries a closed positive laminated
current.
\end{problem}

\begin{problem}
Show that there exists no laminated compact set in $\PP^2$
which is a holomorphic lamination and carries a positive laminated
harmonic current.
\end{problem}

\begin{problem}
Let $T=\omega+ i \partial S-i\overline{\partial S}\geq 0$ be a laminated
current. Since $\int \overline{\partial}S \wedge \partial \overline{S}<\infty$
does it follow that the energy of the transverse measure is finite?
\end{problem}


\begin{problem} Let $L$ be a complex subspace in $\PP^k$ of
dimension $\ell_0=k-\left[\frac{k+2}{2}\right].$ Then there
exists a smooth foliation by complex subspaces of dimension
$\ell_0$ in some neighborhood of $L.$
\end{problem}


\begin{conjecture} There exists a smooth nonconstant $v$ on $\PP^k,
 (dd^c v)^\ell \equiv 0$ iff $\ell >\left[ \frac{k+2}{2}\right].$
Given by B\'ezout: $K^+(v)$ is $\ell-1$ pseudoconcave. $K^+(-v)$
also. They should intersect if $2(\ell-1)\leq k.$
\end{conjecture}


\begin{question}
Suppose $\Omega_1,\Omega_2\subset \PP^n$ are $\ell-1-$pseudoconcave. For which
$\ell$ must $\Omega_1,\Omega_2$ intersect?
\end{question}

\begin{theo}
If $\Omega$ has an exhaustion function $\rho,$ a Morse function with
Levi form with at least two strictly positive eigenvalues,
then $\Omega$ cannot have two ends.
\end{theo}

\begin{theo}
If $\Omega \subset \PP^k$ has an exhaustion function which is the local
maximum of functions with
$\ell>\left[\frac{k+2}{2}\right]$ positive eigenvalues, then
$\Omega$ cannot have two ends.
\end{theo}


\begin{question}
If $u=u_1+iu_2$ is it possible that $T_u$ is harmonic?
The problem is to define $\{T\}$ for harmonic currents. This is
done in Claim 3.6 so $\{T\}$ is the class of $T$ in
$\{\partial \overline{\partial}\;{\mbox{closed forms}}/
\{\partial S+\overline{\partial S'}\}.$
Proposition $1$ shows that this space is one-dimensional and $\{T\}
=c\{\omega\}.$ The problem is that we don't know
$\int T \wedge S=\int \{T\} \wedge \{S\}$ if $T,S$ are harmonic
we only know that if one of them is harmonic and the other is closed.
According to representation in Proposition $1$ 
\bea
T & = & c \omega+\partial S+\overline{\partial S}\\
\int T\wedge T  & = & c^2 \int \omega^2+2 \int \partial S \wedge
\overline{\partial S}\\
\int \partial S \wedge \overline{\partial S} & = & \int S \wedge \partial
\overline{\partial } \overline{S}\\
\eea

So if  $ S \wedge \partial
\overline{\partial } \overline{S}\geq 0$ we are done.
That is we can prove:
\end{question}

\begin{claim} If $T_u$ is harmonic on $\PP^2$, $T_u=c\omega+
\partial S + \overline{\partial} \overline{S},$
$\partial S \wedge  \overline{\partial S} \geq 0,$ then $c=0.$
\end{claim}

{\bf Proof:} 

\bea
\int T_u \wedge T_u & = & c^2 \int \omega^2+2\int \partial S \wedge
\overline{\partial}\overline{S}\\
& \geq & c^2\\
\int T_u \wedge T_u & = & \int i\partial u \wedge 
\overline{\partial}\overline{u} \wedge i\partial u \wedge 
\overline{\partial}\overline{u}\\
& = & 0\\
\eea

In particular, $c =0.$\\

\CQFD\\


\begin{question}
Find examples in $\PP^2$ of $u \in {\cal C}^2_\CC (\PP^2)
$ such that $\partial u \wedge \overline{\partial}\overline{u}
\wedge \partial \overline{\partial} u=0.$
\end{question}

\begin{problem}
If $X$ is a $K3$ surface, is $P^2(X) \neq \mathbb R?$
\end{problem}


{\bf ERASE THE NEXT EXAMPLE. EXAMPLE 3.2 IS MUCH BETTER}

\begin{example}
Let $S =-i\overline{\partial} \log (1+|z|^2+|w|^2)-i\overline{\partial}
 \log (1+|z|^2).$ Then $\partial S$ is harmonic. Moreover $\partial S
+\overline{\partial S}+\omega \geq 0$ but $[\partial S
+\overline{\partial S}+\omega] \wedge [\partial S
+\overline{\partial S}+\omega] $ is not identially zero.
\end{example}


\begin{example}
The lamination of a hyperbolic H\'enon map is a lamination with one
singular point and dense leaves.
\end{example}

This current can be given by the Ahlfors trick. If a leaf of a laminated set
is a disc, is there a way to construct a harmonic current by using
some version of the Ahlfors trick?

\begin{question}
Can one construct harmonic currents in $\PP^3$ with 2 dimensional
leaves? Can there exist ${\cal C}^\infty_{1,1}$ with $\partial 
\overline{\partial} \phi>0$ along leaves?
\end{question}

Berndtsson-Sibony has found $T$ with $\partial \overline{\partial}T
\wedge \omega=.$


A lamination $K$ is said to be a limit of algebraic curves if there is a
sequence of smooth algebraic curves $\Sigma_n$ converging to $K$
in the Hausdorff metric and locally $\Sigma_n$ are small perturbations
of leaves in the ${\cal C}^1$ norm.


\begin{lem}
A harmonic positive current, laminated can be written
$$
T=\int_\alpha \partial \overline{\partial} (\psi_\alpha)d\sigma_\alpha+
\int_\alpha \partial \overline{\partial} (\phi_\alpha)d\sigma_\alpha
$$
(tubular integrals, global $\psi_\alpha$, like $\log G_\alpha$ and smooth
$\phi_\alpha.$
$$
T=\partial \overline{\partial}({\mbox{global function}})+
\Re \int_\alpha \frac{\partial G^\alpha}{G^\alpha}\overline{\partial}
\tilde{u}_\alpha+\partial \overline{\partial}({\mbox{tame}})
$$
\end{lem}

 Things to do:\\

Question 1: Uniqueness of harmonic currents in the singular case.
[Look at Furstenberg example and suspension construction for nonunique
examples]\\

Question 2: Uniqueness of our minimal sets with singularities.\\

Question 3: Ergodic theory for harmonic currents in spaces with
singularities+ Entropy.\\

Show that if $T$ is harmonic, laminar on a minimal set with
singularities then $T \wedge T$ is dirac masses at the singular
points. ($T_\epsilon \wedge T_\epsilon \rightarrow 0$ elsewhere.)



\begin{theo}
Let $(Y,{\cal L},E)$ be a minimal lamination with singularities.
There is at most one positive
closed current $T$ with no mass on $E$ and of the form in the above Theorem
on $Y \setminus E$, of mass $1$.
\end{theo}

{\bf Proof:}
To prove the uniqueness we show that every such $T$ is extremal. Assume that $S \leq cT,c>0$. Then $S=hT$ where $h \geq 0$ is a
measurable function since the harmonic functions are constant on leaves
as $T$ is closed.
 Recall that in a coordinate chart 
on $Y \setminus E$ we have

$$
T=\int [V_\alpha] d\mu(\alpha)
$$

The function $h$ is constant on leaves.



 But since leaves are dense the function is continuous on a set
of arbitrarily large measure on a transversal, hence $h$ is constant.\\

THIS IS FALSE BECAUSE THE SET WHERE h IS CONTINUOUS IS NOT TOTALLY
INVARIANT.THE DENSITY ARGUMENT DOES NOT APPLY.

{\bf Here is a sketch of a possible argument:\\ SOUNDS MORE PLAUSIBLE TO
ME. THERE ARE HOMEOMORPHISMS $\Phi$ OF THE 2 TORUS MINIMAL ie EVERY POINT HAS
DENSE ORBIT BUT WITH SEVERAL INVARIANT PROBABILITY MEASURE. (Furstenberg)
Take $F\times_\Phi T.$ Take one loop $\gamma$ around torus $T$.
Suspension. Leaves are dense, each invariant probability measure gives
rise to a current.

Step 1: Let $f$ be a Borel measurable function on an open set $U$ in $\CC$
and let $\mu$ be a probablility measure on $U.$
We define $p\in S_\mu$ to be a point of $Strong\; continuity$
if there is for every $\epsilon>0$  a sequence of discs $\Delta(p,r_n)$ 
such that $r_n \rightarrow 0$ and $\frac{\mu(\{q\in \Delta(p,r_n);
|f(q)-f(p)|<\epsilon\})}{\mu(\Delta(p,r_n))}>1-\epsilon.$


\medskip

Step 2: Show that the set of points of $Strong\; continuity$
has full $\mu$ measure.

\medskip

Step 3: Relax the condition of Strong continuity to quasiconformal discs:
A point $p \in U$ is a point of $quasiconformal\; strong \; continuity$
if there is an open subset $V, p\in V\subset U$ and a quasiconformal map
$\Phi:V \rightarrow \Delta$ such that $\Phi(p)$ is a point
of Strong continuity for $f\circ \Phi^{-1}, \Phi_* \mu.$ Note that
using $\Phi=Id$ we get that $Strong\; continuity \Rightarrow
quasiconformal\; strong \; continuity$ so the set of \\
$quasiconformal\; 
strong \; continuity$ also has full $\mu$-measure.

\medskip

Step 4: Observe that  $quasiconformal\; strong \; continuity$ is preserved
along leaves of a lamination. This shows that $h$ is constant(?).}

{\bf I don't see how to take advantage of the quasiconformality.}





   

  
 IS THE FOLLOWING TRUE?I DOUBT IT

\begin{theo}
There is only one closed positive current on a minimal set.
\end{theo}

{\bf Proof:} The quotient of the currents is a function constant
on leaves and the leaves are dense, the function is continuous on
a set of arbitarily large measure...\\

\CQFD\\


 Dear Nessim,

 Here is an idea to show that if T is a closed
1,1 form on a bidisc which is if the convex hull
of local discs, then T is in the hull of global discs.

 Let $D_k={D^k_n}_n$ denote a dyadic decomposition of
the unit square in the w-axis.
 Let $ U^k_n$ denote the saturated set of leaves through
$D^k_n$.
 Let $T^k_n=T|U^k_n/\| T|U^k_n \|.$
 Then these are all closed in $U^k_n$ and are in
the convex hull of local discs in $U^k_n.$

 Consider all limit currents. They will be
closed, living on individual leaves.
 The numbers $\| T|U^k_n\|$ define a transverse
measure.

 Best regards,

 John Erik

Dear John Erik,
I just faxed you few pages on harmonic currents.It seems the main 
quesion is wether Twedge T could be zero. And the question of 
uniqueness for C1 minimal Laminations with singularities.
Best regards
Nessim
How to prove uniqueness of harmonic currents
for ${\cal C}^1$ laminations without
singularities (see Theorem 4.26)
 
Remarks from Nessim April 16, 2003:

Ad definition of $T \wedge \tau:$
$$
\overline{\partial}T(\alpha)=T \wedge \tau \wedge \alpha.$$
$\tau $ only exists on the support of $T.$ [Similarly,
for automorphisms of $\CC^3$ one can define Green functions
living on support of wedge prodict of two currents.]

\begin{question}
Given a holomorphic motion on a bidisc.
Suppose that $T$ is a positive closed form directed by the current.
Then $T$ is in the convex hull of local discs, but is it in
the hull of global discs? [See Theorem 4.1] If we assume instead that
$T$ is locally in the convex hull of discs, then $T$ is globally
(in flow boxes) in the convex hull of discs. There are two types of
closed laminated currents: $i \partial \overline{\partial}T=0$
and $i \partial_b \overline{\partial}_b T=0.$
\end{question}

\bigskip

This is true for ${\cal C}^1$ laminations:
Namely, project by $\pi$ along leaves to their intersection points with
the $w$ axis. Then one can show that
$\chi \circ \pi T$ is closed. See Proposition 4.50.
This implies that the
extreme points are points, so coming from the
global discs.
Same argument applies to harmonic currents.

\bigskip

One can try to prove that a $C^1$ holomorphic lamination
has no current of finite energy [whatever definition of
finite energy].

\bigskip

Write to Nessim why the case of transverse measures with
finite energy leads to $T \wedge T$ well defined pointwise.
This should lead to $T_\epsilon \wedge T \rightarrow 0$ so
$T \wedge T=0.$
Can one also define $T \wedge T$ directly for harmonic measures
using finite energy and a global definition?

\begin{question}
How can we define the harmonic current on a laminated set
in $\PP^2$ when the lamination is only holomorphic motion?
\end{question}

 
Dear John Erik,
I dont see what Lp ,$p<2$ does here Lemma 4.21 in the
paper says everything is in L2 so if we neglect ddcu
the pointwise product is in L1.But may be you don`t
neglect ddcu?
Concerning the approach to uniqueness step one is fine
:FEDERER p159
I don`t see how you do step 4,so if you know to do it
proof is complete if we assume the lamination is C1
out of the singular set.
Indeed the question is :What is a laminated current
for a lamination. In a flow box it should be in the
convex hull of  small discs.If the lamination is C1
one shows it can be represented as an integral that we

always wrote.If we have only a holomorphic motion I
don`t know how to do it.
In the C1 case one shows that the extreme pts are what
we expect, by multiplying with a test function composed
with the projection on a transversal.
So a good exercise.
Best regards
Nessim

\medskip

General question: Is the harmonic current associated to a laminated sets
with singularities in arbitrary complex manifolds unique?


We assume that $K$ is a compact subset of $\PP^2$ which is laminated
by discs. The following is a loose argument why this is impossible.

\bigskip

Let ${\cal L}$ be a leaf, parametrized by $\Phi:\Delta
\rightarrow {\cal L}.$ We can write in homogenous coordinates:
$\Phi(x)=[Z(x):W(x):T(x)].$
Assume that $K$ is in general position whatever that means...
 Assume that the intersections with the axes are transverse and that
the pairwise intersection points of the axes $[1:0:0]$ etc. are not
in $K.$
Fix small tubular neighborhoods of each axis, $U_z,U_w,U_t.$
We consider the regions of $\Delta$ mapped to these tubular neighborhoods.
Roughly speaking the set of points mapped to $U_z,U_w,U_t$ are given
by a sequence of discs $\Delta_n^z,\Delta_n^w,\Delta_n^t\subset \Delta$
respectively. The discs are roughly evenly spaced, so in annular
regions of the form $1-1/2^k<|x|<1-1/2^{k+1}$ there should be about
$2^k$ of each of these three kinds. Moreover the discs should have about
radius $1/2^k.$ 

\medskip

We can write down a simple model map of this situation:

\bea
Z(x) & : = & \Pi_{r \geq 0}(1-2x^{(2^k)})\\
W(x) &  : = & \Pi_{r \geq 0}(1+2x^{(2^k)})\\
T(x) &  : = & \Pi_{r \geq 0}(1+2ix^{(2^k)})\\
\eea

We calculate the intersection points with the $w-$ axis and show that this
cannot be a bounded set.

Let $x_{k,\ell}=\left(\frac{1}{2}\right)^{\frac{1}{2^k}}
e^{\frac{2\pi i \ell}{2^k}}, k=0,\dots, \ell=0,\dots,2^k-1.$
The $w$ coordinates of these points are given by 

\bea
w_{k,\ell} & = & \frac{W(x_{k,\ell})}{T(x_{k,\ell})}\\
& = &  \frac{\Pi_{r \geq 0}(1+2x_{k,\ell}^{2^r})}{
\Pi_{r \geq 0}(1+2ix_{k,\ell}^{2^r})}\\
& = & 
 \frac{\Pi_{r \geq 0}(1+2[\left(\frac{1}{2}\right)^{\frac{1}{2^k}}
e^{\frac{2\pi i \ell}{2^k}}]^{2^r})}{
\Pi_{r \geq 0}(1+2i [\left(\frac{1}{2}\right)^{\frac{1}{2^k}}
e^{\frac{2\pi i \ell}{2^k}}]^{2^r})}\\
& = & \Pi_{r=0}^{k-1} \frac{1+2e^{2\pi i \ell 2^{r-k}}(\frac{1}{2})^{2^{r-k}}}
{1+2ie^{2\pi i \ell 2^{r-k}}(\frac{1}{2})^{2^{r-k}}}
\frac{2}{1+i} \pi_{r>k}\frac{1+2 (\frac{1}{2})^{2^{r-k}}}{1+2i (\frac{1}{2})^{2^{r-k}}}\\
& = & C_0 \Pi_{r=0}^{k-1} \frac{1+2e^{2\pi i \ell 2^{r-k}}(\frac{1}{2})^{2^{r-k}}}
{1+2ie^{2\pi i \ell 2^{r-k}}(\frac{1}{2})^{2^{r-k}}}\\
& = & C_0 \pi_{s=1}^k \frac{1+2e^{2\pi i \ell 2^{-s}}2^{-1/2^{s}}}{1+2ie^{2\pi i \ell 2^{-s}}2^{-1/2^{s}}}\\
\eea

If we now fix $\ell$ and let $k \rightarrow \infty$ we see that the general term in the
infinite product approaches $\frac{1+2}{1+2i}$ hence the product diverges to infinity, a
contradiction since $[0:1:0]$ is assumed not to be in $K.$

\bigskip


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%




\subsection{OLD THEOREM AND PROOF:}

\begin{theo}
Let $K$ be a compact laminated set in $\PP^2$. Then there exists 
a harmonic current $T$ on $K$,  of the form 
$$T=\int_\alpha H^\alpha [\Delta_\alpha] d\mu(\alpha)$$
\noindent in flow boxes, where the $\mu(\alpha)$ are transverse measures
on the leaves and $H^\alpha$ are strictly positive harmonic functions, 
bounded uniformly below and above.
The function $H^\alpha$ is Borel measurable as a function on $K.$
\end{theo} 
[Do this when $K$ has singularities.]

{\bf Proof: (Sketch)}
We can define the space $L$ of continuous $(1,1)$ forms along leaves.
This contains a subspace $L'_0$ of forms which are of the form
$i\partial_b \overline{\partial}_b f$ where $f$ is a continuous
function on $K$ which is ${\cal C}^2$ along leaves.
Set $L_0$ equal to the closure of $L_0'$ in sup-norm in $L_0.$

\bigskip

It is easy to find a form $\lambda$ in $L$ which is strictly positive
everywhere. Hence, by the maximum principle it can not be in $L_0.$
Therefore there exists by Hahn-Banach, a continuous linear functional
$\phi:L\rightarrow \CC$ which is positive and which vanishes
on $L_0.$

\bigskip

Next we describe $\phi$ on local flow boxes. 

\bigskip

Making a continuous local change of coordinates which is holomorphic
on leaves, we can assume that the flowbox is
given by $\Delta*X$ where $X$ is a compact subset of the $w-$ axis.
We can identify $L,L_0',L_0$ with continuous functions on
$\Delta* X$ and $L_0'$ are those which are Laplacians in the
$z$ direction of continuous functions which are ${\cal C}^2$ along
the $z-$direction. By the Riesz representation theorem there
exists a measure $d\mu$ on $X$ in this local flow box such that
$\phi(f)=\int f d\mu$ for all continuous functions (forms) of compact
support say.
We will describe this measure more precisely.

\bigskip

Let $\xi(w)$ be a local cut-off function.
Let $\tilde{L}$ denote the set of ${\cal C}^2$ functions $g(z)$
with compact support on the unit disc $\Delta$.
We consider the restriction of $\phi$ to functions of the form
$\xi(w) h(z),$ $h$ continuous. Call this $\phi_\xi (h).$ Then
$\phi_\xi(\Delta g)=0,$ $g \in \tilde{L}.$ By Riesz there exists a measure
$\mu_\xi$ on the unit disc so that $\phi_\xi(h)=\int h d\mu_\xi.$
Notice that if we smooth $d\mu_\xi$ to $d\mu_\epsilon$, then
still $\int \Delta g d\mu_\epsilon=0.$ This means that
$\mu_\epsilon= \lambda_\epsilon dz\wedge d\overline{z}$ for
a harmonic function $\lambda_\epsilon.$ Hence
$d\mu=\lambda_\xi dz \wedge d\overline{z}$ for a harmonic
function $\lambda_\xi.$ The theorem follows....add details...

\CQFD\\
For harmonic laminated currents $T$ we have $\overline{\partial}
T=\tau \wedge T.$ {\bf What does this mean?}

{\bf Let $T$ be a harmonic currrent. Define $C_T^{1,2}:=
\{\phi_{0,1}\wedge T; \phi\in {\mathcal C}^\infty_{0,1}(\PP^2)\}$, which is 
$some$  space of currents.
We can then define $\tau\in (C_T^{1,2})^*,$ the dual space, by
$\tau(\phi_{0,1}\wedge T):= T(\overline{\partial}\phi).$ Then we adopt
the ad hoc notation $\tau\wedge T$ to be the current which acts
by the rule $(\tau\wedge T)(\phi)=\tau(\phi_{0,1}\wedge T)= T(\overline{\partial}\phi)=\overline{\partial}T(\phi)$ which justifies the notation
$\tau \wedge T= \overline{\partial} T.$
The space $C_T^{1,2}$ is in local coordinates very close to the space
of $\{h_\alpha\phi_{0,1}^\alpha [\Delta_\alpha]\}_\alpha$ (if we don't
worry too much about how the forms depend on $\alpha$).
Then the dual object  $\tau$ acts on this
space by integration of $h_\alpha\overline{\partial}\phi_{0,1}^\alpha$
or equivalently  $\phi_{0,1}^\alpha \wedge \overline{\partial}h_\alpha$ on 
each $[\Delta_\alpha]$ and then integrating the result agains a 
transverse measure $\mu_\alpha.$ Hence we can think of $\tau$ as an average
of some objects $\tau_\alpha$ where $\tau_\alpha$ is a dual element
to the space $\hat{C}^{1,2}_T=\{C_{h_\beta[\Delta_\beta]}^{1,2}\}_\beta=
\{\{\phi^\beta_{0,1}\wedge h_\beta[\Delta_\beta]; \phi\in 
{\mathcal C}^\infty_{0,1}(\PP^2)\}\}_\beta.$ Can such an object be called
a current?}

Example: $T=h_1[\Delta_1]+h_2[\Delta_2]$ for pluriharmonic functions
in a neighborhood. Then 
\bea
\overline{\partial}T & = & 
\frac{\overline{\partial} h_1}{h_1}h_1[\Delta_1]
+ \frac{\overline{\partial} h_2}{h_2}h_2[\Delta_2]\\
& \neq  (?)& [\frac{\overline{\partial} h_1}{h_1}
+ \frac{\overline{\partial} h_2}{h_2}]
\wedge [h_1[\Delta_1]+h_2[\Delta_2]]\\
\eea

YES ARE EQUAL THIS IS CLEAR WHEN $[\Delta_1]$EQUALS $[\Delta_2]$ and
otherwise their intersection is only a discrete set and they are equal
I still don't understand: If $h_1$ is a function on $[\Delta_1]$,
say $h_1= 2x_1,$ then $\frac{{\overline{\partial}}h_1}{h_1}= \frac{d\overline{z}_1}{2x_1}$
should be thought of as a form on $\CC^2$. Hence 
$\frac{d\overline{h}_1}{h_1}[\Delta_2] \neq 0.$
NO THE FUNCTION h LIVES ONLY ON DELTA you CAN WHEN YOU FACTOR DERVATIVES
MULTIPLY EACH ONE BY THE INDICATRIX FUNCTION OF THE DISC WHERE IT
LIVES.OTHERWISE YOU WOULD BE ABLE WITHOUT CHANGING THE CURRENT TO CHANGE
THE"ACTION" ON ANOTHER DISC.







The same holds true for smoothings using the
Riesz representation theorem (Berndtsson-Sibony).

\bea
|\int \overline{\partial} T \wedge \alpha| & = &
|\int  T \wedge \tau \wedge \alpha|\\
& \leq & \left( \int \alpha \wedge \overline{\alpha} \wedge T 
\right)^{\frac{1}{2}} c\\
\eea

Special case:
\bea
T & = & \partial u \wedge \overline{\partial u},\\
\partial \overline{\partial} T & = & 0\\
\overline{\partial} T & = & \tau \wedge \partial u 
\wedge \overline{\partial u}\\
\eea

\bea
\overline{\partial} T & = & \tau \wedge T\\
{\partial} T & = & \overline{\tau} \wedge T\\ 
\partial \overline{\partial} T & = & \partial \tau \wedge T
+\tau \wedge \partial T\\
& = & \partial \tau \wedge T +\tau \wedge \overline{\tau} \wedge T\\
& = & [\partial \tau+ \tau \wedge \overline{\tau}] \wedge T\\
& = & 0.
\eea
 
For harmonic currents

\bea
\tau & = & \frac{\overline{\partial} H}{H}\\
\partial \tau & = & -\frac{\overline{\partial} H}{H^2} \wedge \partial H\\
\eea

\begin{lem}
If $T=i \partial u \wedge \overline{\partial u}$ and $\partial T=
\tau \wedge T$ then the distribution $\alpha \wedge T=0$ is integrable.
\end{lem}

{\bf Proof:}
\bea
d\alpha \wedge T & = & d(\alpha \wedge T)-\alpha \wedge dT\\
& = & 0-\alpha \wedge \tau \wedge T\\
& = & 0\\
\eea

\CQFD\\


\subsection{Smooth harmonic currents of rank $1$}

We investigate smooth positive harmonic currents $T$ with $T \wedge T=0.$

\begin{example}
Let $T=H(z,w) dw \wedge d\overline{w}$. If $H$ is harmonic in $z$
for each fixed $w$, then  $\partial \overline{\partial} T=0.$
\end{example}

For a more general local description, we can write $T=H(z,w)
(dw+f(z,w)dz) \wedge (d\overline{w}+\overline{f(z,w)}d\overline{z}).$
To calculate the condition
of harmonicity at the origin, note at first that we can assume
that $f=f_z=f_w=0$ at the origin.

\newpage

\bea
\overline{\partial} T & = & [H_{\overline{z}}d\overline{z}+
H_{\overline{w}}d\overline{w}]\wedge (dw+f(z,w)dz) \wedge 
(d\overline{w}+\overline{f(z,w)}d\overline{z})\\
& + &  H[f_{\overline{z}}d\overline{z}+
f_{\overline{w}}d\overline{w}]\wedge dz \wedge 
 (d\overline{w}+\overline{f(z,w)}d\overline{z})\\
& - & H (dw+f(z,w)dz) \wedge \overline{f_w(z,w)}d\overline{w}\wedge 
d\overline{z}\\
\partial \overline{\partial} T & = &
H_{\overline{z}z}dz \wedge d\overline{z}\wedge dw\wedge d\overline{w}\\
& - & 
[H_{\overline{z}}d\overline{z}+
H_{\overline{w}}d\overline{w}]\wedge f_w dw \wedge dz \wedge d\overline{w}\\
& + & [H_{\overline{z}}d\overline{z}+
H_{\overline{w}}d\overline{w}]\wedge
dw \wedge \overline{f}_zdz \wedge d\overline{z}\\
& + & 
H_wdw \wedge [f_{\overline{z}}d\overline{z}+
f_{\overline{w}}d\overline{w}]\wedge dz \wedge 
 (d\overline{w}+\overline{f(z,w)}d\overline{z})\\
& + & Hf_{\overline{z}w}dw \wedge d\overline{z}\wedge dz \wedge 
d\overline{w}\\
& + & H[f_{\overline{z}}d\overline{z}+
f_{\overline{w}}d\overline{w}]\wedge dz \wedge 
 (\overline{f}_wdw \wedge d\overline{z})\\
& + & Hdw \wedge \overline{f}_{\overline{w}z}dz \wedge d\overline{w}\wedge
d\overline{z}\\
& = & [H_{z\overline{z}}-H_{\overline{w}}\overline{f}_z
-H_w f_{\overline{z}}-Hf_{\overline{z}w}+H f_{\overline{w}}\overline{f}_w
-H\overline{f}_{z\overline{w}}]
dz \wedge d\overline{z}\wedge dw \wedge d\overline{w}\\
\eea

We have proved:

\begin{lem}
Suppose that $T=H(z,w)(dw+f(z,w)dz) \wedge \overline{(dw+f(z,w)dz)}$
with $f(0,0)=f_z(0,0)=f_w(0,0)=0$. Then $\partial \overline{\partial}T=0$
at the origin if and only if 
$[H_{z\overline{z}}-H_{\overline{w}}\overline{f}_z
-H_w f_{\overline{z}}-Hf_{\overline{z}w}+H f_{\overline{w}}\overline{f}_w
-H\overline{f}_{z\overline{w}}](0,0)=0.$
\end{lem}

\begin{example}
\bea
T & = & H (dw+\overline{z}dz) \wedge (d\overline{w}+zd\overline{z})\\
\overline{\partial} T & = & \overline{\partial}H \wedge 
(dw+\overline{z}dz) \wedge (d\overline{w}+zd\overline{z})
+ H d\overline{z}\wedge dz \wedge (d\overline{w}+zd\overline{z})\\
\partial \overline{\partial} T & = &
\partial   \overline{\partial}H \wedge 
(dw+\overline{z}dz) \wedge (d\overline{w}+zd\overline{z})
+ \overline{\partial}H \wedge (dw+\overline{z}dz) \wedge
dz \wedge d\overline{z}\\
& + & 
 H_w dw\wedge  d\overline{z}\wedge dz \wedge (d\overline{w}+zd\overline{z})\\
& = & 
\partial   \overline{\partial}H \wedge 
(dw+\overline{z}dz) \wedge (d\overline{w}+zd\overline{z})\\
& - & H_{\overline{w}}dz \wedge d\overline{z}\wedge dw \wedge d\overline{w}\\
& - & H_w dz \wedge d\overline{z}\wedge dw \wedge d\overline{w}\\
& = & [H_{z\overline{z}}-zH_{z\overline{w}}-\overline{z}
H_{w\overline{z}}+|z|^2H_{w\overline{w}}-H_w -H_{\overline{w}}]
 dz \wedge d\overline{z}\wedge dw \wedge d\overline{w}\\
\eea
In particular, if $H=H(z)$ is harmonic, then
$T=H(z)(dw+\overline{z}dz) \wedge (d\overline{w}+zd\overline{z})$
is a harmonic current.
\end{example}

\begin{example}
Any current of the form $T=(dw+f(z)dz) \wedge \overline{(dw+f(z)dz)}$
is harmonic.
\end{example}


{\bf Proof: (incomplete)} 
We can write $T=c\omega+dd^c \phi$ for a locally integrable function $\phi$
and a constant $c>0$, where $\omega$ is the standard K\"ahler form
on $\PP^2.$  We can assume that $c=1.$ Let $T=S+dd^c \psi$ be any
decomposition as above. Then we get

\bea
\int S \wedge S & = & \int S \wedge [\omega+dd^c(\phi-\psi)]\\
(SKIP)  & = & \int S \wedge \omega+\int d[S \wedge d^c(\phi-\psi)]\\
& = & \int S \wedge \omega\\
(SKIP) & = & \int [\omega+dd^c(\phi-\psi)]\wedge \omega\\
(SKIP) & = & \int \omega \wedge \omega +\int d[\omega \wedge d^c(\phi-\psi)]\\
& = & \int \omega \wedge \omega\\
& = & 1\\
\eea


\medskip



Let $\phi_\epsilon$ be a sequence of automorphisms
converging to the identity. Then by continuity of the operation of
intersection of cohomology classes

\bea
\{T\}\wedge \{T\} & = & \lim_\epsilon
 \{\phi_\epsilon T\}\wedge \{T\}\\
& = & \lim_{\epsilon \rightarrow 0}  \{\phi_\epsilon T \wedge T\}\\
\eea

Here we have used: Let $(T_j)$ be a sequence of smooth positive closed
currents, $T_j \rightarrow T,$ $\{T_j\}=\{T\}$ and $\pi^* T_j=dd^c v_j,
v_j \searrow .$ Then we have
\bea
\{\phi^*_\epsilon T\}\wedge \{T\} & = & \{\phi^*_\epsilon T_j\}\wedge \{T\}\\
& = & \{\phi_\epsilon T_j \wedge S\}\\
& = & \{\phi_\epsilon^* T_j \wedge T\}\\
& = & \{\phi_\epsilon^* T \wedge T\}\\
\eea

The last relation holds because $\phi^*(v_j)$ decreases to a local
potential of $\phi^* T$ and that that one is integrable with respect to
$T.$

\bigskip

\begin{lem}
We have to show that $\phi^*_\epsilon T$ is wedgeable with respect to $T$
and that $\phi_\epsilon T \wedge T \rightarrow 0.$
Potential of $T$ is $u(z,w)=\int \log |w-a_\alpha(z)| d\mu(\alpha).$
One has to check that if $\mu$ has no atoms the potentials $u_\epsilon$
of $T_\epsilon$ are integrable with respect to $T.$

$$
\{T\}\wedge \{T\}  =  \lim_\epsilon
 \{\phi_\epsilon T\}\wedge \{T\}$$

where $\phi_\epsilon$ is such that $\phi_\epsilon {\cal L}$ is transverse
to ${\cal L}.$ Then the current $\phi_\epsilon^*
T \wedge T$ is well defined.
\end{lem}

 We have to show that the local potential
for $\phi_\epsilon^* T$ is locally integrable with respect to $T,$ and
that the mass is small.

\medskip


$$
\int u \sigma_T = \int \log |c'-c+\epsilon z|d\mu(c) d\mu(c') idz \wedge
d\overline{z}.
$$

We assume that $z$ varies on the unit disc. Define 

$$
H(\zeta) := \int_{|z|<1} \log |\zeta+z|   idz \wedge
d\overline{z}$$

Then

$$
\int u \sigma_T = \log \epsilon +\int H\left(\frac{c-c'}{\epsilon}\right)
d\mu(c) d\mu(c').
$$

Since $H$ is continuous there is no problem of integrability. So we get

$$
\{T\}\wedge \{T\}  =  \lim_\epsilon
 \{\phi_\epsilon T\wedge T\}.$$

Since $\phi_\epsilon T \wedge T$ is defined pointwise we get in a coordinate
chart for ${\cal L}$


$$
\phi_\epsilon T\wedge T=\int (\phi_\epsilon)^*[\Delta_{c'}]\wedge
[\Delta_c] d\mu(c) d\mu(c').
$$



Since $\mu$ has no point mass it is clear that the total mass
for a fixed chart converges to zero when $\epsilon \rightarrow 0.$ It
seems to me that the argument is more or less complete for ${\cal C}^1$
lamination since one can make a change of coordinates. Otherwise
one has to be more careful. 

 
\medskip

Assume next that we have an arbitrary current $T$ directed by
a lamination ${\cal L}.$


{\bf Proof:}
In local coordinates, let $w=f(z),w=g(z)$ be leaves of
$T$ and $\phi_\epsilon T$ respectively. The contribution to the
potential from $w=g(z)$ is $\log |w-g(z)|$. Note that
the two leaves intersect in finitely many points 
and to finite order and that for all nearby leaves
this is the same. So we are done.\\

\CQFD\\

\begin{lem}
Hence we are lead to the problem of finding a $\phi_\epsilon$ for which
${\cal L}$ and $\phi_\epsilon {\cal L}$ have no common leaf.
In $\PP^2$ let ${\cal L}_0$ be the unique minimal lamination
contained in ${\cal L}.$ (Uniqueness follows from the positive solution
of the Levi problem.) Every leaf in ${\cal L}_0$
is dense in ${\cal L}_0.$ Fix $p \in {\cal L}_0$ and choose $\Phi_\epsilon$
which fixes $p$ and don't send the local leaf through $p$ to itself.
Since leaves in ${\cal L}_0$ are dense, this implies there is no common
leaf in ${\cal L}_0$ (and $\phi_\epsilon {\cal L}_0$. There is also no common leaf $L$ for 
$\Phi_\epsilon {\cal L}$ and ${\cal L}$ because, $\overline{L}
\supset {\cal L}_0$ and the laminations will then coincide on
${\cal L}_0$, contradicting the previous construction.
\end{lem}


\bigskip

\begin{theo}
Suppose that $T_1$, $T_2$ are positive harmonic currents on a laminated set.
Then if they are linearly independent then there exists a harmonic current $T$
on the laminated set with $\int T \wedge T>0$.
\end{theo}

{\bf Proof:}
We can assume that $T_1$, $T_2$ have mass one and that $ \int T_j \wedge 
T_j =0 $,j=1,2. Then
$T_1-T_2 =i\partial S-i \overline{\partial S}$ and $
\int T_1-T_2 \wedge  T_1-T_2 <0 $. This implies that $\int T_1 \wedge
T_2<0.$ But then $\int (T_1+T_2 ) \wedge  (T_1+T_2 )>0$ which contradicts
positivity.\\

\CQFD\\


\medskip

Given a positive current $T$ of bidegree $(1,1)$,
define

$$
I(T) =\{\alpha; \alpha\; {\mbox{continuous,}}\; \alpha\wedge T=0\}.
$$

\begin{example}
Let $u$ be a complex valued ${\cal C}^2$ function. Let $T_u:=i 
\partial u \wedge \overline{\partial u}.$ 
$I(T) = \{\partial u,\overline{\partial u}\}$ on  $\{\partial u\neq 0\}$.
The distribution defined by $I(T)$ is integrable on $\{\partial u\neq 0\}$
i.e. there are complex discs through points where $\partial u \neq 0$
if and only if 
$$
\partial  \overline{\partial}u \wedge i \partial u \wedge \overline{\partial}
\overline{u}=0 \equiv \partial  \overline{\partial}\overline{u}
 \wedge i \partial u \wedge \overline{\partial}
\overline{u}=0.
$$
This follows from Frobenius theorems, since we need the ideal
$I(T)$ to be stable under derivation.
\end{example}

\begin{claim}
If the current $T_u$ is closed on $\PP^2$ then $\{T_u\}=0.$
\end{claim}

{\bf Proof:}
If $T$ is closed, $\{T_u\}=c\omega.$ Hence
\bea
c^2 \int \{T_u\}\wedge \{T_u\} & = & c^2 \int \omega\wedge \omega\\
& = & c^2 \int (\omega+dd^c \alpha ) \wedge \omega, \alpha \in 
L^1_{\mbox{loc}}\\
& = & c \int T_u \wedge \omega\\
& = & c \int T_u \wedge (\omega+dd^c \alpha)\\
 & = & \int T_u \wedge T_u\\
& = & \int i \partial u \wedge \overline{\partial u}
\wedge  i \partial u \wedge \overline{\partial u}\\
& = & 0.\\
\eea


\CQFD\\

The argument is valid if $\nabla u \in L^4(\PP^2).$ If not there
is a counterexample.
$T=i\partial \log^+|z| \wedge \overline{\partial} \log^+|z|$ in
$\CC^2(z,w).$ Then $T=\frac{i}{2}\partial \overline{\partial} (\log^+|z|)^2$
in $\CC^2.$ Hence $T$ is closed in $\CC^2$ with a closed extension
to $\PP^2.$ $\int T \wedge T \neq 0$ cohomologically.
Potential in $\PP^2$ $dd^c \log(|z|v|t|)=\pi^*T,$ $T$ is a rank $1$
current. Locally $T=i\partial u \wedge \overline{\partial u}$
Compute $\nabla u$ near $[0:1:0].$





Let $T=c\omega+\partial S+ \overline{\partial S}+ 
\partial \overline{\partial}u.$ We assume that $T$ is
positive. We know $\overline{\partial}S$
is in $L^2$ and is unique, although $S$ itself is not unique.
Next, if $f$ is a holomorphic selfmap of $\mathbb P^2$, then
$f_*T=  f_*\omega+\partial f_*S+  \overline{\partial f_*S}+ 
\partial \overline{\partial}f_*u.$
Since $\omega$ is a closed positive current, the same is true
for $f_*\omega$ so we can write $f_*\omega=a\omega+dd^c v.$ Hence we get
a formula:

$$
f_*T= a \omega+\partial f_*S+  \overline{\partial f_*S}+
\partial \overline{\partial} \tilde u.
$$

Notice that $f_*\overline{\partial} S$ being a $(0,2)$ form has the same
$L^2$ norm as $\overline{\partial} S $ except for multiplicity. 
[Nope: The image current has self crossings, this applies to
Hence $\overline{\partial} f_*S \wedge \partial \overline{f_*S}$ as well.
Locally one can think of $f_*S$ as a finite sum of pieces, pairwise
intersecting...]
$E$ is preserved up to a constant multiple.
 


\begin{coro}
Let $T=\int H_\alpha [\Delta_\alpha] d\mu(\alpha)$
in local coordinates denote a harmonic laminated current.
Then 
\bea
\overline{\partial}S & = &  \int \overline{\partial}H_\alpha
\wedge \frac{d(\overline{w-f_\alpha(z)}}{\overline{w-f_\alpha(z)}}
d\mu(\alpha)+ fd\overline{z} \wedge d\overline{w}\\
& = & 
 \int \overline{\partial}H_\alpha
\wedge \overline{\partial} \log |w-f_\alpha(z)|^2
d\mu(\alpha)+ fd\overline{z} \wedge d\overline{w}\\
\eea
\noindent in local coordinates,
$f$ antiholomorphic. {\bf ... arbitrary?}
\end{coro}

The form $S$ can be chosen as a good solution of the $\overline{\partial}$
problem. We can in general write
$S= - \int  \log |w-f_\alpha(z)|^2\overline{\partial}H_\alpha
d\mu(\alpha)+Fd\overline{z}+ad\overline{z}+bd\overline{w}$

\noindent where $F$ is antiholomorphic with $\frac{\partial F}{\partial 
\overline{w}}=-f$ and $a,b$ are functions with $\frac{\partial b}{
\partial \overline{z}}=\frac{\partial a}{
\partial \overline{w}}$.

[Here we write locally $\Delta_\alpha$ as a graph $w=f_\alpha(z)$
and $H_\alpha(z,w)$ is the pluriharmonic extension of $H_\alpha$
on $\Delta_\alpha$ independent of $w$.]

This should mean that $S$ doesn't carry the main singularity.
The main singularity is in the function $u.$ So maybe the main
singularity is contained in $u=\int H_\alpha \log |w-f_\alpha|^2 d\mu(\alpha)
?$

Local formula

$$
T = \int\left[\partial (\log \overline{\partial}H)+
 \overline{\partial} (\log \partial H)+
\partial \overline{\partial} (\log H)\right].
$$

\begin{theo}
The function $u$ is integrable with respect to 
$\overline{\partial}\partial u$ if and only if
the transverse measure $\mu(\alpha)$ has finite transverse
energy. {\bf ??}
\end{theo}

\begin{theo}
Let $T$ be laminated current with transverse measure $\mu.$
Then $T \wedge T$ can be defined if and only if the transverse
measure has finite transverse energy. On the other hand
$T  \dot{\wedge} T$ is always defined.
\end{theo}





Next we explore the fact that $\overline{\partial}S \in L^2.$
In a local flow box this means that

$\int_{|z|,|w|<1,\alpha, \beta} \frac{H_{\alpha,\overline{z}}
H_{\beta,z}}{(\overline{w-f_\alpha(z)})(w-f_\beta(z))}
dV(z,w) d\alpha d\beta<\infty.$

\begin{lem}
$$\int_{|w|<1}\frac{1}{(\overline{w-f_\alpha(z)})(w-f_\beta(z))}
=\log \frac{1}{|f_\alpha(z)-f_\beta(z)|}+{\cal O}(1).$$
\end{lem}

{\bf Proof:} Clear.\\

\CQFD\\

\bea
T & = & \omega+
\partial S+ \overline{\partial S}+\partial \overline{\partial}u\\
& = &  \omega+
\partial S'+ \overline{\partial S'}+\partial \overline{\partial}u'\\
& \Rightarrow & \\
\partial  \overline{\partial S} & = & \partial  \overline{\partial S'}
\eea

Set $  \overline{\partial S}-\overline{\partial S'}=:V$. Then
in local coordinates $V=a(z,w)d\overline{z} \wedge d\overline{w}$ with
$a(z,w)$ antiholomorphic since $\partial V=0.$ Let us
replace $V$ by its conjugate, so
$V=a(z,w)dz \wedge dw$ and $a$ is a holomorphic function [in local
coordinates]


In a local flow box away from $E,$ we can write $K=\{(z,w)
; |z|<1, |w|<1, (z,w)\in F\}$ after a homeomorphic change of coordinates,
holomorphic along leaves. Forms in $L$ can be written 
$A(z,w)dz \wedge d\overline{z},$ $A$ continuous. We restrict $\Phi$ to 
$\{A(z,w)dz \wedge d\overline{z},
A \in {\mathcal C}(F).$ Then $\Phi$ is represented by a measure
$ d\mu_w(z)d\nu(w)$ where $d\mu_w(z)$ is a probability measure
on $\{(z,w);|z|<1\}.$ Since $\Phi$ annihilates all functions
$A(z,w)=f(w)\Delta_zg(z),$ $f$ continuous, $g$ smooth with compact support,
it follows after a smoothing argument in the $z$ direction that 
$$
d\mu_w(z)=H_w(z)idz \wedge d\overline{z}, H_w \; {\mbox{
harmonic}}.$$

Patching, it follows that we can represent $\Phi$ on $L$ by the desired 
formula

$$
\Phi= \int H^\alpha [\Delta_\alpha]d\mu(\alpha)
$$

\noindent in local coordinates away from $E.$ The linear functional
defines naturally a $(1,1)$ current, which by construction is
harmonic and positive.\\

\subsection{ $T \wedge T$}


\begin{theo} Let $T$ be a laminated harmonic current of a ${\mathcal C}^1$
laminated compact in $\PP^2.$ Then $T \wedge T=0.$
\end{theo}

{\bf Proof:}
We use the approach in [HM].
([HM]) uses a homotopy formula of deRham:
$RT-T=d(AT)+A(dT)$ for smoothing $R.$ If $T$ is closed this becomes:
$RT-T=d(AT).$ This breakes down for harmonic currents. Maybe there is
a direct smoothing.



How to define $T \wedge T$ for harmonic currents.
There is a decomposition $T=c\omega+i\partial S+\overline{i \partial S}+
i \partial \overline{\partial}u,$ $S, \partial S, \overline{\partial}S$
in $L^2$, $u \in L^1$. If $S \perp {\mbox{Im}}\partial$ and $u$ is normalized,
$\int u=0,$ the map
$$
T \rightarrow (c,S,u)
$$
\noindent is continuous for the weak topology of currents, i.e.
from $H_e$ (strong topology) to $\RR \times {\cal D}_{0,1}\times L^1.$

\medskip


When $T,T'\in H_e$, We define 

$$T \wedge T' \wedge \omega^{k-2}
= \int (c\omega + \partial S+\overline{ \partial S})
\wedge (c'\omega + \partial S'+\overline{ \partial S'}),
\partial S, \partial S'\in L^2.$$
The integral is independant of $\partial S, \partial S'\in L^2.$

\section{new}



\begin{theo}
Let $K$ be a laminated set in $\mathbb P^2$. We assume that
$K$ is a holomorphic lamination, say given locally by a holomorphic
vector field. Fix $p\in K$ and let $L$ be the leaf through $p.$
Suppose that $\Phi_\epsilon$ is a holomorphic family of
automorphisms of $\mathbb P^2$ which fix $p$ and
for which the angle between $\Phi_\epsilon(L)$ and $L$ at $p$
is of order $\epsilon$ for all small $\epsilon.$
Then there exists a neighborhood
$U$ of $p$ and an $\epsilon_0>0$ so that if
$L_1,L_2$ are two local leaves intersecting $U$, then
$L_1,\Phi_\epsilon(L_2)$ have at most one intersection point in $U$ 
for $0<|\epsilon|< \epsilon_0$ and
the angle between $L_1,\phi_\epsilon(L_2)$ at such an intersection point
is on the order of $\epsilon.$
\end{theo}

{\bf Proof:}
We fix a small neighborhood of $p$. We can assume that in local
coordinates $(z,w)$ the lamination is given by $w={\mbox{constant}}$
and $p=0$. The automorphisms can locally be written as
$\Phi_\epsilon(z,w)=(z+\epsilon f(\epsilon,z,w), w+\epsilon g(\epsilon,z,w))$
for holomorphic functions $f,g.$
Then $\Phi_\epsilon(z,0)=
(z+\epsilon f(\epsilon,z,0), \epsilon g(\epsilon,z,0)).$
The transversality condition is equivalent to $\frac{\partial g}{\partial z}
(0,0,0) \neq 0.$ By scaling we can assume that
 $\frac{\partial g}{\partial z}
(0,0,0) =1.$
We can make a local coordinate change $G_\epsilon(\epsilon, z,w)=
(\eta,\zeta)=(z+\epsilon f(\epsilon,z,w),w)$ in the domain of
$\Phi_\epsilon.$ This reparametrizes leaves, but leaves
are still horizontal. With $\tilde{\Phi}_\epsilon(\eta,\zeta)
:= \Phi_\epsilon \circ G_\epsilon^{-1}(\eta,\zeta),$
we get $\tilde{\Phi}(\eta,\zeta)= 
(\eta, \zeta+\epsilon \tilde{g}(\epsilon, \eta,\zeta))$ with the 
nondegeneracy condition that $\frac{\partial \tilde{g}}{\partial \eta}
(0,0,0)=1.$ Note that $L_1$ and $\Phi_\epsilon(L_2)$
intersect in $U$ if and only if $L_1$ and $\tilde{\Phi}_\epsilon(L_2)$
intersect. 

\medskip

We are led to the situation where we have a local lamination
in $(z,w)$ coordinates of horizontal leaves and biholomorphic
images $\Phi_\epsilon(z,w)=(z,w+\epsilon g(\epsilon,z,w))$
for which $\frac{\partial g}{\partial z}(0,0,0)=1$ and we
want to count intersections between leaves and images of leaves.

\medskip


The number of intersection points is still
unchanged if we reorder the leaves.
Let $H_\epsilon(z,w)=(z,w+\epsilon g(\epsilon,0,w))=(\tilde{z},\tilde{w}).$
For each $\epsilon$
this maps leaves to leaves. 
We replace $\Phi_\epsilon$ by $\tilde{\Phi}_\epsilon \circ
H_\epsilon^{-1}(\tilde{z},\tilde{w}).$ Then we still need to
count intersections between horizontal lines and images
of horizontal lines. 

\medskip

We are led to the problem of counting intersections of horizontal
lines and their images in the case

$$\Phi_\epsilon(z,w)=(z,w+\epsilon z(1+{\mathcal O}(z,w,\epsilon))).$$

Fix a small $\delta>0.$ Suppose first that
$w_0,\eta_0$ are close to the origin independent of $\epsilon$
and that $|w_0-\eta_0|\leq \frac{3}{2}|\epsilon|\delta.$ Then by comparing
with the solution of the equation $w_0+\epsilon z =\eta_0$, we see there
is exactly one intersection point of $\Phi_\epsilon
(\{w=w_0\})$ and $\{w=\eta_0\}$ in the disc $|z|<2\delta.$
On the other hand if $|w_0-\eta_0|\geq \frac{3}{2}|\epsilon|\delta$
there is no intersection point for $|z|<\delta.$\\

\CQFD\\

\begin{remark}
In fact we have proved that if there is a local intersection
of $L_1$ and $\Phi_\epsilon(L_2)$ then the leaves
are at a distance of order at least $|\epsilon| \delta$
on the circle around the intersection point of radius
$\delta/2$.
\end{remark}

\subsection{More}

Remark that in the holomorphic case, Camacho- Lins Neto-Sad ([CLS])
proved the non existence of a holomorphic lamination in $\PP^2$
supporting 
a directed positive closed current.
Locally the foliation is given by $\omega_i=0, d\omega_i=0.$ 
We recall the argument.

We can write $\omega_j= g_{ij}\omega_i$. Hence $dg_{ij}$ are zero along 
leaves so the functions are constant along leaves. Use a partition of unity
to define $\beta_j= \sum_i \rho_i \frac{dg_{ij}}{g_{ij}}.$
Then $\Omega=d\beta_j$ is globally defined and $\Omega=\Omega^{1,1}+
\Omega^{2,0}$.
Hence ([GH]) 
$\Omega-c\omega=d\gamma.$ Up to scaling we can assume that
$c=0$ or $c=1.$ If $c=0,$ $d\beta_j-d\gamma=0$ $\Omega_{| \cal L}
=d\gamma_{| \cal L}=0.$
The distribution is given by a global vector field.
So assume $c=1.$  Now if we assume that 
$T$ is a closed positive laminated current
$0 < <T,\omega>=<T,\Omega> $ but $<T,\Omega>=
\int [\Delta_\alpha]\mu_\alpha d\rho_i \wedge \frac{dg_{ij}}{g_{ij}}=0.$
Contradiction. 

\begin{theo}
Let $K$ be a laminated set of a bidisc. Let $T$
be a closed positive current which is in the convex
hull of local discs in leaves. Then $T$
is of the form $T=\int [\Delta_\alpha]\mu(\alpha)$
for global discs $\Delta_\alpha.$
\end{theo}

{\bf Proof:} We can assume that the lamination covers the
whole bidisc.\\

Suppose that $T_k=\sum_j c^k_j [\Delta^k_j], c^k_j>0$
and that $T_k \rightarrow T$ weakly. Let $\pi:K \rightarrow
\Delta(w)$ denote projection along leaves.

For any compact set $L\subset \Delta(w)$, we can let
$\mu(L):=<T, \chi_L\circ \pi dz \wedge d\overline{z}>.$
This defines a transverse measure. Let

$$
\tilde{T}:= \int \mu(\alpha)[\Delta_\alpha].
$$

We only need to show that $\tilde{T}=T.$



\begin{theo}
Let $K$ be a compact set in a $2-$ dimensional complex 
manifold $X.$ Suppose $(K,{\mathcal L},E)$
is an oriented lamination with singularities. 
Then there exists a positive harmonic current $T$ on $K$ which on 
$K \setminus E$ is of the form
$$
T=\int H^\alpha [\Delta_\alpha]d\mu(\alpha)
$$
\noindent in flow boxes, where $\mu(\alpha)$ is a measure on leaves
and the $H^\alpha$ are strictly positive harmonic functions on leaves,
bounded
uniformly below and above. The function $H^\alpha$ is Borel measurable
as a function on $K$. If $\Lambda_2(E)=0$
($2$-dimensional Hausdorff dimension), then $T$ has no mass on $E.$
\end{theo}

{\bf Proof:}
We can assume that $\gamma$ is continuous on $K$, vanishing
only on $E$ and is strictly positive out of $E.$ 

\medskip

Let $F$ be the space of bounded continuous $(1,1)$ forms along leaves
on $K$. This contains
the space of forms $F'_0$ which are of the form $i \partial_b
\overline{\partial}_bf$ where $f$ is continuous with compact support on
$K \setminus E$ and ${\cal C}^2$ along leaves.
Set $F_0$ equal to the closure of $F'_0$ in sup norm on $F.$

\medskip

Let $C:=\{T;T \geq 0\;{\mbox{bidimension}}\; (1,1), \|T\|=1,
T \wedge \gamma=0, {\mbox{Supp}}(T)\subset K\}.$
This is a closed compact set since $\gamma$ is continuous. Elements
$T$ of $C$ act on $F$ since $T \wedge \gamma=0.$
If $C \cap
F^\perp_0=\emptyset,$ then by Hahn-Banach they can be
separated strongly with an element of $F_0$.
Then there is an element $\phi,$ $\phi$
is a smooth $(1,1)$ form on $\PP^2,$
$\phi\in F_0^{\perp \perp}$ and
$<T,\phi>\geq \delta >0\; \forall\; T\in C.$
There is a continuous function on $K$, strictly
subharmonic on leaves. There is a point $z_0$ on $K$ where the maximum is
reached. 
Since $E$ is locally pluripolar we 
easily get a contradiction as in ([BS],p.37).
Hence there is a current $T$ in $C$ such that $<T,i\partial_b
\overline{\partial}_bf>=0$ for all $f \in {\cal C}^2$
along leaves and with support in $K \setminus E$ and $i\partial
\overline{\partial}T=0.$\\


\CQFD\\

\begin{theo}
Let $(K,L,E)$ be a laminated set with singularities
 in a complex surface $M.$ Suppose there exists a smooth function
$\sigma:M \rightarrow \RR$ such that $\sigma$ is strictly
plurisubharmonic in a neighborhood of $E.$ Suppose there
exists a fuction $\rho:M \rightarrow \RR \cup \{-\infty\},$
$E=\{\rho=-\infty\}$, $\rho$ is plurisubharmonic in a neighborhood
 of $E,$ $\rho \in 
{\mathcal C}^\infty(M \setminus E).$ Then there exists
a positive harmonic current $T$ on $K$ which on $K \setminus E$
has the form
$$
T=\int H^\alpha [\Delta_\alpha]d\mu(\alpha)
$$
\noindent in flow boxes, where $\mu$ is a measure on leaves, $H^\alpha$
is a local Borel measurable function on $K$ which is
uniformly bounded below and above and is harmonic on each
local leaf $\Delta_\alpha.$ If $\Lambda_2(E)=0,$ then $T$ has no mass
on $E.$
\end{theo}

{\bf Proof:} Let $L$ denote the space of bounded continuous 
$(1,1)$ forms along leaves on $K \setminus E.$ We us the supnorm topology
on $L,$ so $L$ is a complex Banach space. Let $L_1\subset L$
denote the subspace of restrictions of forms $i\partial \overline{\partial} f$
where $f \in {\mathcal C}^\infty(M).$ Set $L_1'=\overline{L_1}.$

\begin{lem}
Let $\lambda \in L, \lambda>0 \; {\mbox{on}}\; on K \setminus E.$ Then
$\lambda$ is not in $L_1'.$
\end{lem}

{\bf Proof:}
If $r>0$ is large enough, then
$$
i\partial \overline{\partial}[\sigma+\rho]+r\lambda>0
\; {\mbox{on}}\; K \setminus E\; {\mbox{along leaves}}.
$$

If $\lambda \in L_1'$, there exists a real valued $f\in {\mathcal C}^\infty(M)$
so that 

$$
i\partial \overline{\partial}[\sigma+\rho]+r i \partial_b
\overline{\partial}_b f>0$$

\noindent on $K \setminus E.$ Then $g:=\sigma+\rho+rf$ is strictly subharmonic
along leaves, $g_{|E} \equiv -\infty.$ This contradicts the maximum principle
at maximum points of $g$ on $K.$\\

\CQFD\\

Pick a continuous linear functional $\Phi:L \rightarrow \CC$ so that
$\Phi_{|L_1'} \equiv 0, $ $\Phi(\lambda)>0$ for a fixed $\lambda
\in L,$ $\lambda>0$ on $K \setminus E$ ([]).

\medskip



\section{Finite transverse energy}

\subsection{content}
Theorem Define $T\wedge T$ pointwise. $T \wedge T$ supported on
singular points.\\

Note: When we consider intersection points of $L,L_\epsilon$,
then move but not to the original coordinate chart? (They do, but can one
make some estimates anyways??\\

Note: Duval has proved the Hurder theorem for finite transverse energy,
closed currents.

Formula for solving $\partial \overline{\partial}$
Levine formula, in Shabat Value distribution theory
in several variables. 
$$
\Lambda^\ell_w=
\log \frac{|\zeta|^2|w|^2}{|\zeta\wedge w|^2}
\sum_{j=0}^{\ell-1}
(d_\zeta d^c_\zeta \log |\zeta\wedge w|^2)^j\wedge \omega^{\ell-1-j}$$

$$
dd^c \Lambda^\ell_w=\omega^\ell-[w]$$

$[w]$ plane of codimension $\ell.$ (This works perhaps for $(p,p)$
currents, but that is not what we want...)

\subsection{More}

Suppose that $T$ is a harmonic current with a harmonic tranverse
measure on a laminated compact in $\mathbb P^2.$ So
$T=\int H^\alpha[\Delta_\alpha] d\mu(\alpha)$ in local flow boxes.
In this case we introduce another notion of energy which we call
transverse finite energy. 


Set $\tilde{T}_k:= \frac{1}{k} \log \frac{1}{|z|} [\Delta].$
We define $T_k:= \Phi_* (\tilde{T}_k \chi_{|x|<1-\frac{1}{R^{k+1}}}).$
We estimate $T_k$ in each flow box $B_i$ from above and below. 

For $T_k^{i+}$ we include the complete graph for each piece
of graph in $\Phi(\{|x|<1-\frac{1}{R^{k+1}}\})$ and for $T_k^{i-}$
we only include graphs from $T_k$ which are completely in the flow box.
Note that $T_k^{i+}$ and $T_k^{i-}$ are pluriharmonic on
$B_i$ and are given by transverse measures and harmonic
functions along leaves. Note also that the mass of
$T_k^{i+}-T_k$ and $T_k-T_k^{i-}$ are at most on the order of
$\frac{1}{k}.$ Note that the mass of $T_k$ is on the order of $1$
in each flow box. It follows that any weak limit of $\{T_k\}$
can be obtained in any flow box by taking the
weak limit of either $T_k^{i+}$ or $T_k^{i-}$ and hence have the
description as transverse measure with harmonic functions on leaves.



{\bf Proof:}
The only missing step is to prove the more precise estimate that $\|\partial
\overline{\partial} T_k\| \leq C/k.$
Fix a flow box $B_i$. 
Let $\{U^i_n\}$ denote the intersections of ${\cal L}$ with the flow box.
These are discs over $|z_i|<1.$

\medskip

Let $\{V^i_{n,j}\}_j \subset \Delta$ denote the connected components
of $\Phi^{-1}(U^i_n).$ We reorder the $\{V^i_{n,j}\}$ as
$\{V^i_{r,\ell}\}$ where
$$
V^i_{r,\ell} \cap \Delta(1-\frac{1}{R^{r+1}})\neq \emptyset,
V^i_{r,\ell} \cap \Delta(1-\frac{1}{R^{r}})=\emptyset.
$$

We push forward the current $\tilde{T}_k$ to $B_i$. The push forward
 is only nonzero
for discs $U^i_{r \ell}=\Phi(V^i_{r,\ell})\subset B_i$ with $r \leq k.$ 

\medskip

Suppose at first that $r<k.$ Then the push forward on $U^i_{r, \ell}$ from
$V^i_{r, \ell}$ is about
$\frac{1}{k} \frac{1}{R^r}[U^i_{r, \ell}].$ Hence the total mass
of $T_k$ from these contributions is about $\frac{1}{k}.$ We notice also that
the push forward is harmonic on these contributions.
Hence they don't contribute
to $\partial \overline{\partial} T_k.$ 

\medskip

Suppose next that $r=k.$ 
Notice that some of the $V^i_{k, \ell}$ might cut the boundary 
$|x|=1-1/R^{k+1}.$
But still the total mass of $T_k$ on these discs
is at most $1/k.$ It follows that the mass of $T_k$ is about $1$
as claimed. We also need to estimate the mass of $\partial
 \overline{\partial}T_k.$ Fix a $U^i_{k,\ell}$ given
by $|z_i|<1.$ Moreover say that $V^i_{k,\ell}=\{|x-(1-1/R^k)|<\delta/R^k\}$
and that $\Phi(z_i)=x=1-1/R^k+\delta z_i/R^k.$
Then the current $T_k$ has roughly the formula
$T_k= \frac{1}{k} \Re \frac{\delta z_i}{R^k} \chi_{\Re z_i<0} [\Delta]$
on $U^i_{k,\ell}.$ We calculate $\partial \overline{\partial}T_k$
on the interior of $\Delta.$ The problem is only the $\Im z_i$ axis.
However, one calculates that $|(\partial \overline{\partial}T_k)(\phi_{00})|
=|T_k(\partial \overline{\partial}\phi| \leq C\frac{\delta}{kR^k}
\|\phi\|_\infty$ from which the last statment follows.\\

\CQFD\\

\subsection{Laminations given by polynomial forms}

Suppose $\lambda=P(z,w)dz+Q(z,w)dw$ is a 1 form on $\mathbb P^2$
which doesn't vanish on a laminated compact $K$ (and defines
the leaves). We then get a holomorphic vector field on the leaves,
$X=-Q(z,w) \frac{\partial}{\partial z}+ 
P(z,w) \frac{\partial}{\partial w}$ which has a pole at infinity
of order $d$ at each point where a leaf intersects the line at infinity
transversally. 

\begin{lem}
Suppose $\Phi:\Delta\rightarrow L$ is the universal
covering of a leaf. Then $X$ lifts to a vector field
$\psi(\tau) \frac{\partial}{\partial \tau}$ where $\psi$
has poles at centers $z_j$ of discs where the leaves cross the
line at infinity. Moreover the size $|\psi| \sim (1-|\tau|)$ outside
these discs.
\end{lem}

{\bf Proof:} Clear.\\

\CQFD\\

We can write the holomorphic function $\phi(\tau)=1/\psi$
as

$$
\phi:= \Pi_{k=1}^\infty \Pi_{1-1/R^k}<|z_j|\leq 1-1/R^{k+1}
\left( \frac{1-\tau/z_j}{1-\tau \overline{z}_j}\right)^d.
$$

which converges.



Let $M$ be a complex 
Hermitian manifold of dimension $m.$ Let $X$ be a compact
set in $M.$ Let $E$ be a closed locally strictly pluripolar set such that
$\overline{X\setminus E}=X.$ Recall that $E$ is locally complete
pluripolar if for every $p\in E$ there is a neighborhood
$U(p)$ and a plurisubharmonic function $u$ on $U(p)$
so that $E \cap U(p)=\{u=-\infty\}.$ 
We assume that $X \setminus E$ is a union of
disjoint complex curves, i.e. immersed Riemann surfaces. We assume that
in local flow boxes the leaves are graphs varying continuously,
but there are weaker notions.
We call a maximal
complex curve $L$ a leaf and define $\cal{L}$ as the union of leaves.
We will call such a triple $(X,{\cal L},E)$ 
laminated
with singularities and we call $E$ the singular set.


\begin{remark}
Let  ${\cal L}_i$ be laminations of compact subsets $K_i \subset \PP^2,$
$i=1,2.$ If ${\mathcal L}_1$ and  ${\mathcal L}_2$ have a common leaf,
they have a common minimal set  ${\mathcal L}_0$. If $p \in
{\mathcal L}_0$ and $L$ is the local leaf through $p,$ then if
$\Phi$ is any automorphism of $\mathbb P^2$ which fixes $p,$
$\Phi'(T_pL) \neq T_pL,$ then  ${\mathcal L}_0$ and  $\Phi({\mathcal L}_0)$
have no common leaf since the leaves are dense.
\end{remark}

\begin{remark}
The above results can be extended to homogeneous K\"{a}hler
surfaces $\Sigma$ with $H^{(0,1)(\Sigma)}=0.$.\\
\end{remark}

\begin{coro}
Let $\Phi:\Delta \rightarrow L$ be a covering of a leaf.
Fix $R>>1$. Set $A_k:=\{1-\frac{1}{R^k}<|z|<1-\frac{1}{R^{k+1}}\}.$
Let $A_k^i:=\Phi(A_k) \cap B_i, \partial A_{ki}:= \Phi(
\{|z|=1-\frac{1}{R^k}\})$
Then the number of whole graphs in $A_k^i$ is on the order of $R^k$
and the number of partially covered graphs is at most on the order
of $NR^k.$ Also the length of $\partial A_{ki}$ is at most on the
order of $NR^k.$
\end{coro}

\begin{theo}
Suppose that $\Phi:\Delta \rightarrow {\cal L} $ is a covering
of a leaf of a lamination. Suppose that no leaf is
a holomorphic image of $\CC.$ Let $\tilde{T}_k:= \frac{1}{k}
\log \frac{1-\frac{1}{R^k}}{|x|}\chi_{\{|x|<1-\frac{1}{R^k}\}}[\Delta].$
Let $T_k:=\Phi_*(\tilde{T}_k).$ Then $\|T_k\| \sim 1$ and $\|\partial
\overline{\partial} T_k\| \leq C/k.$
Any limit point is a positive harmonic current given by transverse
measures and harmonic functions on leaves.
\end{theo}

\end{document}

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