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\begin{document}

\title{Remarks on endomorphisms and rational points}

\author{E. Amerik, F. Bogomolov, M. Rovinsky}

\date{\today}
\maketitle

WORK IN PROGRESS!!! VERY PRELIMINARY PARTIAL VERSION

\section{Invariant neighbourhoods}

Let $X$ be a smooth projective variety of dimension $n$ and let 
$f:X\dasharrow X$ be a
rational self-map, both defined over a "sufficiently large" number field $K$.  
We assume that $f$ has a fixed point $q\in X(K)$.
This assumption is not restrictive if, for example, $f$ is a regular polarized
(that is, such that $f^*L=L^{\otimes k}$ for a certain ample line bundle 
$L$ and a 
positive integer $k$) endomorphism: indeed, in this case the set 
of periodic points in
$X(\bar \Q)$ is even Zariski-dense \cite{F}, so replacing $f$
by a power and taking a finite extension of $K$ if necessary, 
we find a fixed point. 

Our starting point is that, for
a suitable prime ideal $\p\subset {\cal O}_K$, 
we can find a "$\p$-adic neighbourhood" $q\in O_{\p,q}\subset X(K_{\p})$,
on which $f$ is defined and which is $f$-invariant. 

More precisely, choose an affine neighbourhood $q\in U\subset X$,
which is the domain of the definition of a set of local coordinates
$x_1, \dots x_n$ (so that the $x_i$ define an \'etale (AT $q$???) map from $U$
to the affine space). Then choose a model ${\cal X}$ over ${\rm Spec}(A)$
where $A={\cal O}_K[1/N]$ is a suitable localization of ${\cal O}_K$, so that $\cal X$
has good reduction everywhere. Let ${\cal U}\subset {\cal X}$ be the 
corresponding open subset of ${\cal X}$. We have 
$${\cal O}({\cal U})=A[x_1,\dots x_n,x_{n+1}\dots x_m]/I$$ 
for some regular functions $x_{n+1},\dots x_m$ integral over
$A[x_1,\dots x_n]$. By Hensel's lemma, we can write $x_{n+1},\dots x_m$ 
as power series in $x_1,\dots x_n$, with coefficients in some further
localisation $A[1/M]$:
$${\cal O}({\cal U})\subset A[1/M][[x_1,\dots, x_n]].$$ 

The functions $f^*x_i$, $1\leq i\leq n$, are power series in $x_i$
with coefficients in $K$, since $f^*$ defines an endomorphism of
the ring ${\cal O}_{q,X}$ and of its completion. We have the following



\begin{lemm} \label{fg} Let $k$ be a field of characteristic zero and let
$f\in k[[x_1,\dots x_n]]$ be a function algebraic over $k(x_1,\dots,x_d)$. 
Then 
$f\in A[[x_1,\dots x_n]]$, where $A$ is a finitely generated
$\Z$-algebra.
\end{lemm}

(THE PROOF COPIED FROM YOUR FILE. It is OK up to some details which
should be verified and possibly corrected)

\smallskip

{\it Proof.} Let $F$ be a minimal polynomial of $f$ over $k[x_1,\dots,x_d]$,
so $F(f)=0$ and $F'(f)\neq 0$. Then $F'(f)\in\mathfrak{m}^s\smallsetminus
\mathfrak{m}^{s+1}$ for some $s\ge 0$, where $\mathfrak{m}$
is the maximal ideal in $k[[x_1,\dots,x_d]]$.

Denote by $f_n$ the only polynomial of degree $<n$ congruent to $f$
modulo $\mathfrak{m}^n$. For a polynomial $\Phi$ in $x$ and an integer
$m$ denote by $\Phi_{(m)}$ the homogeneous part of $\Phi$ of degree $m$.
Clearly, $F'(f_n)_{(m)}$ is independent of $n$ for $n>m$.

We are going to show that the ${\mathbb Z}$-subalgebra in $k$ generated
by the coefficients of $f$ (equivalently, by the coefficients of the
homogeneous components of $f$ of all degrees $n>s$) is generated,
in fact, by coefficients of $F$ (as polynomial in $d+1$ variables),
by coefficients of $f_{s+1}$ and by the inverse %the $f,x_1,\dots,x_d$
of a polynomial in coefficients of $F$ and in coefficients of $f_{s+1}$.

This is done by induction on degree $n>s$: by definition,
$F(f_n)\in\mathfrak{m}^n$ and we have to find (assuming that it exists!)
a homogeneous polynomial $\Delta$ of degree $n$ such that
$F(f_n+\Delta)\in\mathfrak{m}^{n+1}$.
One has $F(f_n+\Delta)\equiv F(f_n)+F'(f_n)\Delta\pmod{\Delta^2}$,
so the condition is $F(f_n)_{(n+s)}+F'(f_n)_{(s)}\Delta=0$.
This is a linear system with polynomial coefficients in coefficients
of $F$ and in coefficients of $f_{s+1}$. As $\Delta$ is a (unique!)
solution of this linear system, one can talk about the {\sl determinant}
of this linear system, which is non-zero and denoted by $D$. Then
the coefficients of $\Delta$ are polynomials over ${\mathbb Z}$ in
coefficients of $F$, coefficients of $f_{s+1}$ and in $D^{-1}$. So the
lemma is proved.

\medskip

Therefore, for almost all primes $\mathfrak p\subset {\cal O}_K$,
the coefficients of the power series $f^*x_1, \dots f^*x_n$ are 
integral in $K_{\p}$. We choose a $\p$ not dividing $N$ and $M$ and such that 
this last condition holds. Obviously, 
$${\cal O}({\cal U})\subset {\cal O}_{\p}[[x_1,\dots, x_n]]. \ \ (\ast)$$ 
     

Define the $\p$-adic neighbourhood $O_{{\p}, q}$ of the point $q$ as follows:
$$O_{{\p}, q}=\{t\in U(K_{\p})|g(t)\equiv g(q)\ mod\ \p\ for\ g
\in {\cal O}(U)\}.$$

(THE USE OF $U$ IS NOT A PROBLEM? It is arbitrary,
not $f$-invariant, etc)

By a subneighbourhood $O_{{\p}, q, n}\subset O_{{\p}, q}$, we shall mean,
throughout the paper, the
subset of points of $O_{{\p}, q}$ such that
the values of regular functions at those points are congruent to the values
at $q$ modulo $\p^n$.

The following properties are clear from the definition, using the observation
($\ast$):


\begin{prop} \label{nbhd}  (1) The functions $x_1, \dots x_n$ give a bijection
between $O_{{\p}, q}$ and the n-th cartesian power of $\p$.

(2) $f(O_{{\p}, q})\subset O_{{\p}, q}$.

(3) $Indet(f)$ does not intersect $O_{{\p}, q}$.

(4) The $\bar Q$-points are dense in $O_{{\p}, q}$ and in $O_{{\p}, q, n}$.

\end{prop}

 
(A FEW WORDS OF PROOF??)


Let $\lambda_1, \dots, \lambda_n$ be the eigenvalues of the tangent map
$Df_q$. The following is a consequence of the $p$-adic versions of 
several well-known results in
dynamics and number theory:


\begin{prop}\label{linear-gen}  Assume that $\lambda_1, \dots, \lambda_n$ are multiplicatively
independent. Then in some subneighbourhood $O_{{\p}, q, n}$ of $O_{{\p},q}$, the map $f$ is
equivalent to its linear part $\Lambda$.

\end{prop}



{\it Proof}: Note that the eigenvalues $\lambda_i$ are algebraic numbers. 
It is well-known that in absence of relations 
$$\lambda_1^{m_1}...\lambda_n^{m_n}=\lambda_j,\ 1\leq j\leq n,\ m=\sum m_i\geq 2,
\ m_i\geq 0$$ ("resonances"), there is a unique
formal linearization of $f$, obtained by formally solving the 
equation $f(\phi(x))=\phi(\Lambda(x))$; the expressions $\lambda_1^{m_1}...\lambda_n^{m_n}-\lambda_j$
appear in the denominators of the coefficients of $\phi$ (see for example \cite{Arn}). 
The problem
is of course whether $\phi$ has non-zero radius of convergence.
By Siegel's theorem (see \cite{HY}) for its $p$-adic version) this is
the case as soon as the numbers $\lambda_i$ satisfy the diophantine
condition $$|\lambda_1^{m_1}...\lambda_n^{m_n}-\lambda_j|_p>Cm^{-\alpha}$$
for some $C, \alpha$. By \cite{Yu} , this condition is always satisfied by algebraic numbers.



\begin{coro} \label{indep-dense} If $\lambda_1, \dots, \lambda_n$ are multiplicatively independent,
the rational points on $X$ are potentially dense.
\end{coro}

{\it Proof:} since algebraic points are dense in $O_{{\p}, q, n}$, we can find a point
$x\in X(\bar Q)$ which is contained in $O_{{\p}, q, n}$, away from the coordinate
hyperplanes in the local coordinates linearizing $f$. 

(PLEASE WRITE A KIND OF
A PROOF: since everithing is analytic not algebraic we need some words)



\section{Variety of lines of the cubic fourfold}

Now the difficulty is that it can be hard to find an interesting
example such that the eigenvalues of the tangent map at some
fixed point are multiplicatively independent. For instance if 
$f$ is an automorphism and $X$ is a projective $K3$ surface, or, more
 generally,
an irreducible holomorphic symplectic variety,
then the product of the eigenvalues is always a root of unity, as noticed
in \cite{Bv}. 

So first of all the map $f$ may be only "partially linearized" in the 
neighbourhood of $q$, 
in some sense, and after such a "partial linearization",
the orbit of a general algebraic point may be contained in a relatively small 
analytic subvariety of the neighbourhood (of course this subvariety does not have
to be algebraic, but it is unclear how to prove that it actually is not). 
Nevertheless, with some
additional geometric information, this partial linearization can 
still be used to prove potential density.

In the rest of this note, we illustrate this by giving a simplified
proof of the potential density of the variety of lines of a cubic fourfold, 
which is the main result of \cite{AV}. The proof uses several ideas
from \cite{AV}, but we think that certain aspects become more
transparent thanks to the introduction of our dynamical point of view and
the use of $\p$-adic neighbourhoods. 

We recall the setting of \cite{AV} (the facts listed below are taken from 
\cite{V1} and \cite{A}).
Let $V$ be a general smooth cubic in $\P^5$ and let
$X\subset G(1,5)$ be the variety of lines on $V$. This is an irreducible
holomorphic symplectic fourfold: $H^{2,0}(X)$ is generated by a nowhere vanishing
form $\sigma$. 
For $l\subset V$ general, there is a unique plane $P$ tangent to $X$
along $l$ (consider the Gauss map, it sends $l$ to a conic in the dual
projective space). The map $f$ maps $l$ to the residual line $l'$. It 
multiplies the form $\sigma$ by $-2$; in particular, its degree is 16. 
The indeterminacy locus $S$ consists of points such that the image of the 
corresponding
line by the Gauss map is a line (and the mapping is 2:1). This is a smooth 
surface of general type, resolved by a single blow-up. For a general $X$,
the Picard group is cyclic and thus the Hodge structure on $H^2(X)^{prim}$
is irreducible (thanks to $h^{2,0}(X)=1$); the space of algebraic cycles
is generated by $H^2=c_1^2(U^*)$ and $\Delta=c_2(U^*)$, where $U$ is the restriction
of $U_{G(1,5)}$, the
universal rank-two bundle on $G(1,5)$. By Terasoma's theorem \cite{T}, these
conditions are satisfied by a ``sufficiently general'' $X$ defined over a number field, in fact
even over $\Q$; ``sufficiently general'' meaning ``outside of a thin subset in
the parameter space''.
One computes that the cohomology class of $S$ is $5(H^2-\Delta)$ to conclude 
that
$S$ is irreducible and non-isotropic with respect to $\sigma$.

\subsection{Fixed points and linearization}

The fixed point set $F$ of our rational self-map $f:X\dasharrow X$ is the set of 
points such that along the corresponding
line $l$, there is a tritangent plane to $V$. Strictly speaking, this is
the closure of the fixed point set, since some of such points are in
the indeterminacy locus; but for simplicity we shall use the term 
``fixed point set'' as far as there is no danger of confusion.

\begin{prop}\label{fixed} The fixed point set $F$ of $f$ is an isotropic surface
of general type.
\end{prop}

{\it Proof:} It is clear from $f^*\sigma=-2\sigma$ that $F$ is isotropic.
Let $I\subset G(1,5)\times G(2,5)$ with projections
$p_1, p_2$ be the incidence
variety $\{(l,P)| l\subset P\}$ and let ${\cal F}\subset I\times 
\P H^0({\cal O}_{\P^5}(3))$ denote the 
variety of triples $\{(l,P,V)| V\cup P=3l\}$. This is a projective bundle
over $I$, so ${\cal F}$ is smooth and thus its fiber $F'_V$ over a general
$V\in \P H^0({\cal O}_{\P^5}(3))$ is also smooth. This fiber clearly
projects generically one-to-one on the corresponding $F=F_V$, since
along a general line $l\subset V$ there is only one tangent plane, 
and a fortiori only one
tritangent plane if any; so $F'=F'_V$ is a desingularization of $F$. 
Since $dim(I)=11$ and since intersecting the plane $P$ along the triple
line $l$ imposes 9 conditions on a cubic $V$, we conclude that 
$F'$ and $F$ are surfaces.

To compute the canonical class, remark that $F'$ is the zero locus of a
section of a globally generated vector bundle on $I$. This vector bundle
is the quotient of $p_2^*S^3U_{G(2,5)}^*$ (where $U_{G(2,5)}$ denotes the tautological 
subbundle on $G(2,5)$) by a line subbundle ${\cal L}_3$ whose fiber at $(l,P)$ is the
space of degree 3 homogeneous polynomials on $P$ with zero locus $l$.
One computes that the class of ${\cal L}_3$ is three times the difference
of the inverse images of the Pl\"ucker hyperplane classes on $G(2,5)$ and $G(1,5)$, and it follows that 
the canonical class of $F$ is $p_2^*(3c_1(U^*))$, which is ample 
(we omit the details since an analogous
computation is given in \cite{V}, and a more detailed version of it 
in \cite{P}).

\begin{rema}\label{fixed-indet} Since $F$ is isotropic and $S$ is not,
$S$ cannot coincide with a component of $F$. In fact, dimension count
shows that $F\cap S$ is a curve.
\end{rema}
 




\begin{prop}\label{eigenvalues} For a general point $q\in F$, the tangent 
map
$Df_q$ is diagonalized with eigenvalues $1, 1, -2, -2$.
\end{prop}



{\it Proof:} This follows from the fact that $f^*\sigma = -2\sigma$.
and the fact that the map is the identity on the lagrangian plane 
$T_pF\subset T_pX$. Let
$e_1,e_2,e_3,e_4$ be the Jordan basis with $e_1, e_2\in T_pF$.
There is no Jordan cell corresponding to the eigenvalue 1, since
in this case $e_4$ would be an eigenvector with eigenvalue 4,
but then $\sigma(e_1,e_4)=\sigma(e_2,e_4)=\sigma(e_3,e_4)=0$, contradicting 
the fact
that $\sigma$ is non-degenerate.
By the same reason, the eigenvalues at $e_3$ and $e_4$ are both equal to
$\pm 2$. 
Suppose that $Df_q$ is not diagonalized, so sends $e_3$ to $\pm 2e_3$ and 
$e_4$ to $e_3\pm 2e_4$. In both cases $\sigma(e_3, e_4)=0$. If $e_3$ goes to
$2e_3$, we immediately see that $e_3\in Ker(\sigma)$, a contradiction.
Finally, if $Df_q(e_3)=-2e_3$ and $Df_q(e_4)=e_3-2e_4$, 
we have 
$$-2\sigma(e_1,e_4)=\sigma(e_1,e_3)-2\sigma(e_1,e_4),$$
so that $\sigma(e_1,e_3)=0$, but by the same reason $\sigma(e_2,e_3)=0$,
again a contradiction to non-degeneracy of $\sigma$.

\begin{prop}\label{linear} 
(1) Let $q$ be a general fixed point of $f$ and let $O_{\p,q}$ be its
$\p$-adic neighbourhood for a suitable $\p$, as in the previous section.
We identify $O_{\p,q}$ with ${\cal O}_{\p}^4$ (so that $q$ 
becomes a $0$). There exists a formal power series $h=h_q$ in two 
variables $(t_1,t_2)=t$ such that $h(-2t)=f\circ h(t)$. This series
is determined uniquely by its linear part 
and converges
on a certain neighbourhood of zero $\p^n\times \p^n$ for some $n$ 
(which can be chosen depending only 
on the prime $\p$). Locally, $h_q$ can be chosen to 
depend analytically on $q$. 

(2) In the complex setting, the analogous statements are true. Moreover,
the maps $h_q$ extend to global meromorphic maps from $\C^2$ to $X$.

\end{prop}

{\it Proof:} 1) We take the coordinates in $O$ in which 
$$Df_q=\left(
\begin{array}{cccc}
1&0&0&0\\
0&1&0&0\\
0&0&-2&0\\
0&0&0&-2
\end{array}
\right)$$
and fix the linear part $h^{(1)}(t_1,t_2)=(0,0,t_1,t_2)$.
The existence and uniqueness of the formal solution is classical
and goes in the same way as for the Poincare-Dulac normal form 
(see for example \cite{Arn}): finding 
$h^{(2)}, h^{(3)}, \dots$ is linear algebra involving division by
$(-2)^2-1$, $(-2)^2+2, \dots, (-2)^m-1,$ $(-2)^m+2$, $m\in \N$.
Since none of those is equal to zero ("no resonances"), this works.
To begin with, taking the second order terms in $h(-2t)=f\circ h(t)$
gives
$$h^{(2)}(-2t)=Df_q\cdot h(t) + f^{(2)}(0,0,t_1,t_2)$$ and thus
the coefficients of $h^{(2)}$ are obtained from those of $f^{(2)}$
by dividing by $(-2)^2-1$ (for the first two components $h_1, h_2$) or by
$(-2)^2+2$ (for the last two).
In general, comparing the terms of order $m$, we get that
$((-2)^m-1)h_i^{(m)}(t),\ i=1,2$ (resp. $((-2)^m+2)h_i^{(m)}(t),\ i=3,4$)
are sums of terms of type
$$f^{(l)}(h^{(i_1)}, h^{(i_2)} \dots, h^{(i_l)},$$
where $i_1+i_2+\dots +i_l=m$. Recall that we may assume that the
coefficients of $f$ are integers by \ref{fg}. Thus, the denominators of the 
coefficients of $h_i^{(m)}$ are products of numbers of the form $(-2)^j-1$
or $(-2)^j+2$, where $2\leq j\leq m$.

The following claim is elementary (induction by $m$):

\smallskip

{\it Claim:} Each denominator is a product of at most $m-1$ factors,
and the exponents $j_1\geq j_2\geq \dots \geq j_{m-r}$ satisfy
$j_{k}\leq m-k+1$.

\smallskip

To see that the radius of convergence of our formal power series
is positive, we must estimate the $p$-adic order of the denominators 
and conclude that it is at most linear in $m$. Suppose for simplicity
that all our factors are of the form $(-2)^j-1$. Since 
$\Z_p^*\cong \F_p^*\times \Z_p$, we have
$$ord_p((-2)^j-1)=ord_p(ja), j\equiv 0 (s)$$
and equal to zero otherwise; here $-2$ corresponds to $(x,a)$ under the
above isomorphism and $s$ is the order of $x$ in $\F_p^*$.
Thus a very rough estimate gives that the order of our denominator 
is at most $$\sum_{j=2}^m ord_p(ja)\leq m\cdot ord_p(a)+ord_p(m!)\leq m\log a+
\frac{m}{p-1}.$$
So the series converges on $\p^n\times \p^n$ as soon as 
$n\geq \log a+\frac{1}{p-1}+1.$ 

In the general case with some factors of the form $(-2)^j+2$, it suffices to double
our estimate for $n$.

2) In the complex case, the convergence of the power series on some
neighbourhood $U$ of zero in $\C^2$ follows from $|-2|>1$ as in the classical
Poincare
theorem. To extend the map $h$ to $\C^2$, set $$h(x)=f^k(h((-2)^rx)),$$
where $(-2)^rx\in U$; one checks that this is independent of choices.

\medskip

We immediately get the following corollary (which follows from the results
of \cite{AV}, but for which there was as yet no elementary proof):

\begin{coro}\label{nonpreppoints}
There exist points in $X(\bar \Q)$ which are not preperiodic for $f$.
\end{coro} 

{\it Proof:} Indeed, $\bar Q$-points are dense in $O$. Take one in a suitable 
invariant subneighbourhood and use the linearization given by the proposition
above.

\medskip

\begin{rema}\label{heights}
If f were regular, this would follow from the theory of canonical heights;
but this theory does not seem to work sufficiently well for polarized 
rational self-maps.
\end{rema}

\subsection{Non-preperiodicity of certain surfaces}

The starting point of \cite{AV} was the observation that $X$ is covered
by a two-parameter family $\Sigma_b, b\in B$  of birationally abelian 
surfaces, namely,
surfaces parametrizing lines contained in a hyperplane section of $V$
with 3 double points. On a general $X$, a general such surface has cyclic 
Neron-Severi group (\cite{AV}, ??); moreover, many of those surfaces $\Sigma$
defined over a number field have the same property, as shown by an argument
similar to that of Terasoma \cite{T}. In fact, given a general $X$, 
the set of such surfaces on $X$ whose Neron-Severi group is not cyclic,
is {\it thin}.

In \cite{AV}, it is shown that the iterations of a suitable $\Sigma$
defined over a number field and with cyclic Neron-Severi group is 
Zariski-dense. The first step is to prove its non-preperiodicity, that is,
the fact that the number of $f^k(\Sigma)$, $k\in \N$, is infinite.
Already at this stage the proof is highly non-trivial, using the $l$-adic
Abel-Jacobi invariant in the continuous \'etale cohomology.

In this subsection, we give an elementary proof of the non-preperiodicity
of a suitable $\Sigma$,
which is based on \ref{linear}. Moreover, this works without an assumption
on its N\'eron-Severi group.

\begin{lemm}\label{noninv} The surface $\Sigma$ is not invariant by $f$.
\end{lemm}

{\it Proof:} The surface $\Sigma$ is the variety of lines contained
in the intersection $Y=V\cap H$, where $H$ is a hyperplane in $\P^5$
tangent to $V$ at exactly three points. For a general line $l$
corresponding to a point of $\Sigma$, there is a unique plane $P$
tangent to $V$ along $l$, and the map $f$ sends $l$ to the residual
line $l'$. If $\Sigma$ is invariant, $l'$ and therefore $P$ lie in $H$,
and $P$ is tangent to $Y$ along $l$. But this means that $l$ is
"of the second type" on $Y$ in the sense of Clemens-Griffiths
(i.e. the Gauss map of $Y\subset H=\P^4$ sends $l$ $2:1$ to a line in
$(\P^4)^*$), see \cite{CG}. At the same time it follows from the results
of \cite{CG} that a general line on a cubic threefold with double
points is "of the first type" (mapped bijectively onto a conic by the
Gauss map), a contradiction.

\medskip

Now let us work in the $p$-adic setting.

Let $x\in \Sigma(K)$ be a point of the p-adic neighbourhood.

\begin{prop}\label{containlines} The Zariski closure $D$ of the set of iterates
of $\Sigma$ contains the image of the line $Ox$. 
\end{prop}

Since in the complex situation, everything is given by the same power
series, this is also true over $\C$. 
Note that by \ref{noninv}, $\Sigma$ cannot coincide with a leaf of our
local fibration from \ref{linear}. Suppose that $\Sigma$ is preperiodic,
that is, $D$ is a finite union of surfaces; then from the form
of $f$ in \ref{linear} it is clear that each of them,
in particular $\Sigma$ itself,
contains the germ of the line $Ox$. Now there are two possible cases:
either the Zariski closure of this germ is the whole of $\Sigma$ and
then $\Sigma$ is invariant, an immediate contradiction with \ref{noninv};
or the Zariski closure is a proper subvariety. In this case we remark 
that extending the field $K$ if necessarily, we can construct as many
of such subvarieties as we wish (since there is one through any $K$-point
of $\Sigma$ in our $p$-adic neighbourhood). Since these are in the
intersection $\Sigma\cap f(\Sigma)$, this again means that $\Sigma$ must 
be invariant.   

To sum up, we have the following

\begin{theo}\label{nonprep} The Zariski closure $D$ is of dimension at least
three. If it is of dimension three, this is an irreducible divisor which 
either contains the surface of fixed points $F$, or has a curve in common
with $F$. In this last case, $D$ contains correspondent "leaves" 
(images of $\C^2$ from \ref{linear}) through the points of this curve.
\end{theo}







 





\subsection{Potential density}

In this subsection, we exclude the case when $D$ is a divisor.

Our proof is a case-by-case analysis on the Kodaira dimension of $D$. 
In \cite{AV} , we already have simple geometric arguments
ruling out the cases of $\kappa(D)=-\infty$ and $\kappa(D)=0$.
The case $\kappa(D)=-\infty$ is especially simple since then the holomorphic 
2-form
would be coming from the rational quotient of $D$, but $\Sigma$ obviously
must dominate the rational quotient and this cannot be isotropic. The
case $\kappa(D)=0$ is less easy and uses the fact that $Pic(X)=\Z$ or,
equivalently, that the Hodge structure $H^2_{prim}(X)$ is irreducible
of rank 22. Namely, an argument using Minimal Model theory and the
existence of an holomorphic 2-form on $D$ gives that
$D$ must be rationally dominated by an abelian threefold or by a product of
a K3 surface with an elliptic curve. But the second
transcendental Betti number of those varieties cannot exceed 21, which
contradicts the fact that $D$ carries an
irreducible Hodge substructure of rank 22; see \cite{AV} for details.

Let us deal with the case $\kappa(D)=2$. We need the following lemma:

\begin{lemm}\label{orderthree} On a general $X$, the points of order 3
with respect to $f$ form a curve.
\end{lemm} 

{\it Proof:} Let $l_1$ be (a line corresponding to) such a point, $l_2=f(l_1)$,
$l_3=f^2(l_1)$, so that $f(l_3)=l_1$. There are thus planes $P_1, P_2, P_3$,
such that $P_1$ is tangent to $V$ along $l_2$ and contains $l_3$, etc.
Clearly, $P_1\neq P_2\neq P_3$. The span of the planes $P_j$ is a projective
3-space $Q$. Let us denote the two-dimensional cubic, intersection of $V$ 
and $Q$, by $W$. We can choose the coordinates $(x:y:z:t)$ on $Q$ such that
$l_1$ is given by $y=z=0$, etc. Then the intersection of $W$ and $P_1$ is
given by the equation $z^2y=0$, etc. The only other monomial from the
equation of $W$, up to a constant, can be $xyz$, since it has to be 
divisible by the three
coordinates. Therefore $W$ is a cone over the plane cubic given by the
equation $ax^2y+by^2z+cy^2z+dxyz=0$ in the plane at infinity.
Now a standard dimension count shows that a general cubic admits a
one-parameter family of two-dimensional linear sections which are cones. 
Each cone on $V$ gives rise to a plane cubic on $X$. This cubic is invariant
under $f$, and $f$ acts by multiplication by $-2$ (for a suitable choice
of zero). The points of order 3
with respect to $f$ lie on such cubic and are their points of $9$-torsion.

\begin{rema}\label{orderthree-indet} In fact the lemma says slightly more:
it applies to the indeterminacy points which are "of order three in the 
generalized sense",
that is, points appearing if one replaces the condition 
"$f(l_1)=l_2$, $f(l_2)=l_3$, $f(l_3)=l_1$" by "$l_2\in f(l_1)$, etc."; here
by $f(l_1)$ we mean the rational curve which is the image of $l_1$ by 
the correspondence which is the graph of $f$ (equivalently, $l_2\in f(l_1)$
says that for some plane $P_3$ 
tangent to $V$ along
$l_1$, the residual line in $P_3\cap V$ is $l_2$).
\end{rema}     

\medskip    

By blowing-up $\tilde{D}$, we may assume that the Iitaka fibration 
$\tilde{D}\to B$ is regular. Its general fiber is
an elliptic curve. By \cite{NZ}, the rational self-map $f$ descends to $B$
and induces a transformation of finite order, so the elliptic curves
are invariant by a power of $f$. From proposition \ref{linear}, we
obtain that they are in fact invariant by $f$ itself: indeed, locally
in a neighbourhood of a fixed point, the curves invariant by $f$ are
the same as the curves invariant by its power. On a general elliptic curve, 
there is
a finite (non-zero) number of points of order three, since $f$ acts as
multiplication by $-2$. We have two possibilities:

\smallskip

1) These are mapped to points of order three
(in the "generalized sense" as in the \ref{orderthree-indet}) on $X$
(or the surface they form is contracted to any other curve on $X$).
Then any preimage of our surface by an iteration of $f$ is contracted as 
well, but since
there are infinitely many of them, this is impossible.

2) This surface dominates a component of the surface of fixed points of $f$.
 In this case, several points of
order three must collapse to the same fixed point $p$. But then the
resulting branches of each elliptic curve near the generic fixed point
are interchanged by $f$, which contradicts the local description
of $f$ in \ref{linear}.

This rules out the possibility $\kappa(D)=2$.

Finally, let us consider the case $\kappa(D)=1$. The Iitaka fibration
$D\to C$  maps $D$ to a curve $C$ and the general fiber $U$ is of Kodaira
dimension 0. As before, by \cite{NZ} $f$ induces a finite order
automorphism on $C$, and one deduces from \ref{linear} that this
is in fact the identity. We have two possible cases:

\medskip

{\it Case 1: $U$ is not isotropic with respect to the holomorphic 2-form $\sigma$.} We use the idea from \cite{AV} as in the case $\kappa(D)=0$. 
Namely, since $X$ is generic, the Hodge structure $H^2_{prim}(X, \Q)$ is simple.Since the 
restriction of $\sigma$ to $D$ is non-zero, $H^2(\tilde{D},\Q)$ carries
a simple Hodge substructure of rang 22. Since $U$ is non-isotropic, the
same is true for $U$, but a surface of Kodaira dimension zero never
satisfies this property.

\medskip

{\it Case 2: $U$ is isotropic with respect to $\sigma$.} The kernel of the
restriction of $\sigma$ to $D$ gives a locally free subsheaf of rank one
in the tangent bundle $T_D$, which is in fact a subsheaf of $T_U$ since
$U$ is isotropic. There is thus a foliation in curves on $U$, and this
foliation has infinitely many algebraic leaves (these are intersections
of $U$ with the iterates of our original surface $\Sigma$). By Jouanolou's
theorem, this is a fibration. In other words, $D$ is fibered over a 
surface $T$ in integral curves of the kernel of $\sigma_D$, and $U$ project
to curves. 
These cannot be rational curves since the surface $T$ is not uniruled
(indeed, the form $\sigma_D$ must be a lift of a holomorphic 2-form on $T$).
Therefore these are elliptic curves, and since $\kappa_U=0$, so are
the fibers of $\pi: D\to S$.

Recall from \ref{linear} that either $D$ contains $F$, or it contains a
curve on $F$; and in this last case, locally near generic such point,
$D$ is a fibration in (isotropic) two-dimensional disks over a curve; 
in particular,
such a point is a smooth point of $D$.  If $D$ contains $F$, this
is a contradiction with \ref{fixed}: indeed, $F$ must be dominated by a union
of fibers of $\pi$. If $D$ contains a curve on $F$, then we look at the
generic "leave" (image of $\C^2$) at this point $p$. Its intersection with
the image of $U$ is an invariant curve, that is, the image of a line through the origin.
And this must be an integral curve of the kernel of $\sigma_D$ and $U$
varies in a family, this implies that $\sigma_D$ is zero at $p$,
a contradiction since $\sigma$ is non-degenerate.




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\end{thebibliography}

\end{document}
