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\centerline{\hugebf The Algebraic Dimension of}\medskip \centerline{\hugebf
Compact Complex Threefolds}\medskip
\centerline{\hugebf with vanishing second Betti Number}\medskip \vskip1cm
{\centerline{{\twelverm Fr\'ed\'eric Campana, Jean-Pierre Demailly and
Thomas Peternell}}}
\vskip.5cm \vskip.5cm

\section{Abstract}

%\bn This note investigates compact complex manifolds $X$ of dimension
%3 with second Betti number $b_2(X) = 0.$ If $X$ admits a non-constant
%meromorphic function, then we prove that either $b_1(X) = 1$ and
%$b_3(X) = 0$ or that $b_1(X) = 0$ and $b_3(X) = 2.$ In particular
%$c_3(X) = 0.$ As a consequence a compact complex threefold
%homeomorphic to the 6-sphere $S^6$ cannot admit a non-constant
%meromorphic function. Furthermore we investigate the structure of
%threefolds with $b_2(X) = 0$ having a holomorphic algebraic reduction
%$X \la \bP _1.$ 

\bn This note investigates compact complex manifolds $X$ of dimension
$3$ with second Betti number $b_2(X) = 0.$ If $X$ admits a non-constant
meromorphic function, then we prove that either $b_1(X) = 1$ and
$b_3(X) = 0$, or $b_1(X) = 0$ and $b_3(X) = 2.$ The main idea
is to show that $c_3(X) = 0$ by means of a vanishing theorem for generic 
line bundles over~$X$. As a consequence, a compact complex 
threefold homeomorphic to the $6$-sphere $S^6$ cannot admit a non-constant
meromorphic function. Furthermore, we investigate the structure of
threefolds with $b_2(X) = 0$ and algebraic dimension $1$, in the
case when the algebraic reduction $X\dasharrow\bP_1$ is holomorphic. 

\sn {\it Keywords}~ Algebraic dimension, algebraic
reduction, generic vanishing theorem, topological Euler characteristic 

\sn {\it AMS classification}~ 32J17

\vskip.5cm


\section{Introduction}

\bn In this note we shall investigate compact complex manifolds of
dimension three and second Betti number $b_2(X) = 0$. Such a manifold
cannot be algebraic or K\"ahler.
Therefore we will be interested in the algebraic dimension $a(X)$ which is
by definition the transcendence degree of the field of meromorphic
functions over the
field of complex numbers. Note that $a(X) > 0$ if and only if $X$ admits a
non-constant meromorphic function. The topological Euler characteristic
will be denoted $\chi_\top(X) $ which is also the third Chern class
$c_3(X)$ by a theorem of Hopf. Our main result is

\bn {\bf Theorem}
\sn {\it Let $X$ be a compact 3-dimensional complex manifold with
$b_2(X)=0$ and $a(X)>0$. Then
$$
c_3(X)=\chi_\top(X)=2-2b_1(X)-b_3(X)=0,
$$
i.e.\ we either have $b_1(X) = 0$, $b_3(X) = 2$ or $b_1(X)=1$, $b_3(X) = 0$.}

\bn Notice that if $a(X) = 3$, i.e.\ $X$ is Moishezon, then we have $b_2(X)
>0$, but examples of compact threefolds $X$ with $a(X) = 1$ or 2 and with
the above Betti numbers exist (see sect.~2).

\bn The following corollary was actually our motivation for the Theorem:

\bn {\bf Corollary}
\sn {\it Let $X$ be a compact complex manifold homeomorphic to the
6-dimensional sphere $S^6$. Then $a(X) = 0$. }

\bn In other words, $S^6$ does not admit a complex structure with a
non-constant meromorphic function.

\bn Our Main Theorem is an immediate consequence of the following more general

\bn {\bf Theorem} \sn {\it Let $X$ be a compact 3-dimensional complex
manifold with $b_2(X) = 0$ and $a(X) > 0$. Let $B$ be a vector bundle on
$X$. Then $H^{i}(X,B \otimes \cM) = 0$ for $i \geq 0$ and generic
$\cM\in\Pic^0(X)$, in particular $\chi(X,B \otimes \cM) = 0$ for {\bf all}
$\cM \in\Pic^0(X)$.}

\sn In the last section we study more closely the structure of threefolds
$X$ with $b_2(X) = 0$ and algebraic dimension 1 whose algebraic reduction
is holomorphic.
We show e.g.\ that smooth fibers can only be Inoue surfaces, Hopf surface
with algebraic dimension 0 or tori.

\sn Finally we would like to thank the referee for suggestions of
improvements in the exposition.

\bn \bn
\section{1. Preliminaries and Criteria for the vanishing of $H^0$.} \bn
{\bf 1.0 Notations}

\sn (1) Let $X$ be a compact complex manifold, always assumed to be
connected. The algebraic dimension, denoted $a(X)$, is the transcendence
degree of the field of meromorphic functions over $\bC$. \sn (2) $b_i(X) =
\dim H^{i}(X,\bR)$ denotes the $i$-th Betti number of $X$. \sn (3) If $G$
is a finitely generated abelian group, then $\rk G$ will denote its rank
(over $\bZ)$. \sn (4) If $X$ is a compact space, then $h^q(X,\cF)$ denotes
the dimension of $H^q(X,\cF)$.

\bn {\bf 1.1 Proposition}
\sn {\it Let $Y$ be a connected compact complex space (not necessarily
reduced), every component $Y_i$ of $Y$ being of positive dimension. Let $D$
be an effective Cartier divisor on $Y$ such that $D_{\vert Y_i}\ne 0$ for
all $i$. Let $\cF$ be a locally free sheaf on $Y$. Then there exists $k_0
\in \bN$ such that $$
H^0(Y,\cF \otimes \cO_Y(-kD)) = 0
$$
for $k \geq k_0$.}

\bn {\bf Proof.} We have natural inclusions $$ H^0(Y,\cF \otimes
\cO_Y(-(k+1)D)) \subset H^0(Y,\cF \otimes \cO_Y(-kD)). $$ Take $k_0$ such
that this sequence is stationary for $k \geq k_0$. Then $s$ has to vanish
at any order along $D_{\vert Y_i}$ for every $i$ ($s$ can be thought of
locally as a tuple of holomorphic functions), hence $s_{\vert Y_i}= 0$, and
$s = 0$.

\bn {\bf 1.2 Corollary}
\sn {\it Let $S$ be a smooth compact complex surface containing an
effective divisor $C$ such that $c_1(\cO_S(C)) = 0$. Let $B$ be a vector
bundle on~$S$. Then for a generic $\cL \in \Pic^0(X)$ we have $$
H^0(S,B \otimes \cL) = H^2(S,B \otimes \cL) = 0. $$ In particular
$\chi(S,B) = \chi(S,B \otimes \cL) = -h^1(S,B \otimes \cL) \leq 0$.}

\bn {\bf Proof.} The vanishing $H^0(S,B \otimes \cO_S(-kC)) = 0$ for large
$k$ follows from (1.1). Since $\cO_S(kC)$ is topologically trivial for
large suitable $k$, the required $H^0$-vanishing follows from
semi-continuity. The $H^2$-vanishing follows by applying the previous
arguments
to $B^* \otimes K_S$ and Serre duality.

\bn {\bf 1.3 Corollary}
\sn {\it Let $X$ be a smooth compact threefold with $b_2(X) = 0$ carrying
an effective divisor $D$. Then $H^0(X,B \otimes \cL) = 0$ for generic $\cL$
in $\Pic^0(X)$ and every vector bundle $B$ on~$X$. }

\bn {\bf Proof.} The assumption $b_2(X) = 0$ means that $H^2(X,\bZ)$ is
finite, hence there exists an integer $m > 0$ such that $c_1(\cO_X(mD)) =
0$ in $H^2(X,\bZ)$. We can apply (1.1) to obtain $$ H^0(X,B \otimes
\cO_X(-kmD)) = 0
$$
for $k$ large. Now we conclude again by semi-continuity.

\bn The next lemma is well-known; we include it for the convenience of the
reader.

\bn {\bf 1.4 Lemma}
\sn {\it Let $X$ be a compact manifold of dimension $n$ with $a(X) = n$.
Then $b_2(X) > 0$.}

\bn {\bf Proof.} Choose a birational morphism $\pi : \hat X \La X$ such
that $\hat X$ is a projective manifold. Take a general very ample divisor
$\hat D$ on $\hat X$ and a general curve $\hat C \in \hat X$. Let $ D = \pi
(\hat D)$ and $C = \pi(\hat C)$. Then $D$ meets $C$ in finitely many
points, hence $D \cdot C > 0$, in particular $c_1(\cO_X(D)) $ is not
torsion in $H^2(X,\bZ)$.

\bn {\bf 1.5 Lemma}
\sn {\it Let $X$ be a smooth compact threefold and $f : X \La C$ be a
surjective holomorphic map to a smooth curve $C$. Let $\cF$ be a locally
free sheaf on~$X$. Then $R^{i}f_*(\cF)$ is locally free for all~$i$.}

\bn {\bf Proof.} (a) Note that local freeness is equivalent to torsion
freeness, since $\dim C = 1$. Hence the claim is clear for $i= 0$.

\sn (b) Next we treat the case $ i = 2$. We shall use relative duality (see
[RRV71], [We85]); it states in our special situation ($f $ is flat with
even Gorenstein fibers) that if $R^jf_*(\cG)$ is locally free for a given
locally free sheaf $\cG$ and fixed $j$, then $$ R^{2-j}f_*(\cG^* \otimes
\omega_{X/C}) \simeq R^jf_*(\cG)^*, $$ in particular $R^{2-j}f_*(\cG^*
\otimes \omega_{X/C}) $ is locally free. Here $\omega_{X/C} = \omega_X
\otimes f^*(\omega_C^*) $ is the relative dualizing sheaf. Applying this to
$j = 0$ and $\cG = \cF^* \otimes \omega^*_{X/C}$ our claim for $i = 2$
follows.

\sn (c) Finally we prove the freeness of $R^1f_*(\cF)$. By a standard
theorem of Grauert it is sufficient that $h^1(X_y,\cF_{\vert X_y})$ is
constant,
$X_y$ the analytic fiber over $y \in C$. By flatness, $\chi(X_y,\cF_{\vert
X_y})$ is constant, hence it is sufficient that $h^{j}(X_y,\cF_{\vert
X_y})$ is constant for $j=0$ and $j=2$. By the vanishing $R^3f_*(\cF) = 0$,
we have (see e.g.\ [BaSt76])
$$
R^2f_*(\cF)_{\vert \{ y\}} \simeq H^2(X_y, \cF_{\vert X_y}). $$ Therefore
$h^2(X_y,\cF_{\vert X_y})$ is constant by (b). Finally $$ h^{0}(X_y,
\cF_{\vert X_y}) = h^2(X_y,\cF^*_{\vert X_y}\otimes \omega_{X_y}) =
h^2(X_y, (\cF^* \otimes \omega_X)_{\vert X_y}) $$ is constant by applying
the same argument to $\cF^* \otimes \omega_X$.

\bn \bn
\section{2. The Main Theorem.}
\bn In this section we prove the main result of this note:

\bn {\bf 2.1 Theorem}
\sn {\it Let $X$ be a 3-dimensional compact complex manifold with $b_2(X) =
0$ and $a(X) > 0$. Let $B$ be a vector bundle on $X$. Then {\item {\rm (1)}
$H^{i}(X,B \otimes \cM) = 0$ for $i \geq 0$ and $\cM \in \Pic^0(X)$
generic.
\item {\rm (2)} $\chi (X,B \otimes \cM) = 0$ for all $\cM \in \Pic^0(X)$.
\item {\rm (3)} $c_3(X) = 0$, i.e.\ either $b_1(X) = 0$ and $b_3(X) = 2$ or
$b_1(X) = 1$ and $b_3(X) = 0$. }}

\bn {\bf Proof.} First notice that (2) and (3) follow from (1). In fact, by
(1) we have
$$
\chi(X,B \otimes \cM) = 0
$$
for generic $\cM$ and thus the same holds for all $\cM$ by Riemann-Roch and
the equality $c_j(B) = c_j(B \otimes \cM)$. For (3), we apply (2) to $B =
T_X$ and get
$$
\chi(X,T_X) = 0.
$$
Now, since $H^2(X,\bR) = H^4(X,\bR) = 0$, we have $c_1(X) = c_2(X) = 0$,
hence $$
0 = \chi(X,T_X) = {1 \over 2} c_3(X)
$$
by Riemann-Roch.

\sn So it suffices to prove (1).
Moreover by Serre duality we only need to prove the vanishing for $i = 0$
and $i = 2$.

\sn In case $i = 0$ we observe that there are non-zero effective divisors
on $X$ (since $a(X) > 0$) and we can apply (1.3) to get the claim.

\sn So let $i = 2$. Let $g: X \dasharrow \bP _1$ be a non-constant
meromorphic function. Let $\sigma : \hat X \la X$ be a resolution of the
indeterminacies of $g$ and let $f : \hat X \la C$ be the Stein
factorisation of the holomorphic map $\sigma \circ g$. Fix an ample divisor
$A$ on $C$ and let $\cL$ be the line bundle on $X$ determined by $$
f^*(A) = \sigma^*(\cL) \otimes \cO_{\hat X}(-E) \leqno (a) $$ with a
suitable effective divisor $E$ supported on the exceptional set of
$\sigma$. We need to exhibit a line bundle $\cM \in \Pic^0(X)$ with $$
H^2(X,B \otimes \cM) = 0.
$$
We shall distinguish two cases according to whether the indeterminacy locus
of $g$ is empty or not.
\sn We start treating the case that $g$ is not holomorphic. First note that
the canonical map
$$
H^2(X,B \otimes \cM) \la H^2(\hat X, \sigma^*(B \otimes \cM)) $$ is
injective. This is obvious from the Leray spectral sequence. Hence it is
sufficient to show
$$
H^2(\hat X, \sigma^*(B \otimes \cM)) = 0. \leqno (*) $$ Actually for $(*)$
we only need
$$
H^2(\hat X, \sigma^*(B \otimes \cM)(-tE)) = 0 \leqno (**) $$ for some $t
\geq 0$. To verify that $(**)$ implies $(*)$, consider the exact sequence
$$
H^2(\hat X, \sigma^*(B \otimes \cM)(-tE)) \la H^2(\hat X, \sigma^*(B
\otimes \cM)) \la H^2(tE, \sigma^*(B \otimes \cM)) $$ and note that
$H^2(tE, \sigma^*(B \otimes \cM)) = 0$. This last vanishing is seen as
follows: let $A_t$ be the complex subspace of $X$ defined by the ideal
sheaf $\sigma_*(-tE)$, then $$
H^2(tE, \sigma^*(B \otimes \cM)) = H^2(A_t,B \otimes \cM) = 0 $$ since
$\dim A_t = 1$.
\sn We make the ansatz $\cM = \cL^{t+k}$ with $t$ and $k$ to be determined;
of course we need to prove the vanishing only for one $\cM$ by
semi-continuity. Using the Leray spectral sequence for $f : \hat X \la C$,
$(**)$ comes down to
$$
H^q(C,R^pf_*(\sigma^*(B \otimes \cL^k)) \otimes tA) = 0 \leqno (*{*}*) $$
for $p + q = 2$, and large $t,k$. For $q = 2$, $(*{*}*)$ is obvious and for
$q = 1$ it follows from Serre's vanishing theorem for $t \gg 0$. So let $q
= 0$. We need to see that $$
\cF = R^2f_*(\sigma^*(B \otimes \cL^k)) = 0. $$ Since $\cF$ is locally free
by (1.5), it suffices that $\cF_{\vert F}= 0$ for the general fiber $F$ of
$f$ and large $k$. But this follows from (1.1), the effective divisor
$E_{\vert F}$ being non-zero: $$
H^2(F, \sigma^*(B \otimes \cL^k)) \simeq H^0(F, \sigma^*(B^*) \otimes K_F
\otimes \cO_F(-kE)) = 0.
$$
\sn If $g$ is holomorphic, i.e.\ we may take $\sigma = id$, so that $f =
g$, this argument does not work since $E = 0$. Here we have to replace
$\cL^{t+k}$
by a different line bundle. First note that $$ H^0(C,R^2f_*(B) \otimes
(-tA)) = 0 \leqno (+) $$ for $t$ sufficiently large. We claim that this
implies $$ H^0(C,R^2f_*(B \otimes \cM)) = 0 \leqno (++) $$ for general $\cM
\in \Pic^0(X)$.
Let $W = H^1(X,\cO_X)$. Then every element in $W$ is represented as a
topologically trivial line bundle.

Consider locally the universal bundle
$\hat {\cM}$ on $X \times W$. Let $F = f \times \id : X \times W \la C
\times W$ and $\hat B = \pr_X^*(B)$. The coherent sheaf $R^2F_*(\hat B
\otimes \hat {\cM}) $ satisfies $$
R^2F_*(\hat B \otimes \hat {\cM})_{\vert C} \times \{t\} \simeq R^2f_*(B
\otimes \hat {\cM}_t),
$$
where $\hat {\cM}_t$ is the line bundle corresponding to $t \in W$. Choose
$m \gg 0$ and $t_0$ such that $f^*(A^{-m}) = \hat \cM_{t_0}$. This is
possible since $b_2(X) = 0$. By (+) we have $$ H^0(C,R^2f_*(B \otimes \hat
{\cM}_{t_0})) = 0. $$ Hence it is sufficient to show that $R^jF_*(\hat B
\otimes \hat {\cM}) $ is flat with respect to the projection $ q :C \times
W \La W$, over a Zariski open set of $W$, then the usual semi-continuity
theorem gives the claim (2). Now $R^2f_*(B \otimes \hat {\cM}_t) $ is
locally free on $C = C \times t$ for every $t$ by (1.5), hence it is clear
that there is a Zariski open set $U \subset W$ such that $R^2F_*(\hat B
\otimes \hat {\cM})$ has constant rank over $U$, hence is locally free over
$U$ (observe just that the set where the rank of a coherent sheaf is not
minimal is analytic). This proves (++). \sn On the other hand we have for
$m \gg 0 $ by Serre's vanishing theorem 
$$
H^1(C, R^1f_*(B) \otimes A^m) = 0 .
$$
In the same way as above we conclude that
$$ 
H^1(C, R^1f_*(B \otimes {\cM})) = 0
$$
for general $\cM \in W$.

\sn By the Leray spectral sequence we therefore again obtain $H^2(X,B
\otimes \cM) = 0$ for general~$\cM$. This finishes the proof of the
theorem.

\bn {\bf 2.2 Corollary}
\sn {\it Let $X$ be a compact complex threefold homeomorphic to the sphere
$S^6$. Then every meromorphic function
on $X$ is constant.}

\bn {\bf Proof. } Note that $c_3(X) = \chi_\top(S^6) = 2$ and apply (2.1).


\bn Next we give examples of threefolds with $a(X) > 0$, $b_2(X) = 0$ and
$c_3(X) = 0$ so that the Main Theorem (2.1) is sharp.

\bn {\bf 2.3 Example}
\sn The so-called Calabi-Eckmann threefolds are compact threefolds
homeomorphic to\break \hbox{$S^3 \times S^3$}, see [Ue75]. They can be
realized as elliptic fiber bundles over $\bP_1 \times \bP_1$. Hence $a(X) =
2$, $b_1(X) = b_2(X) = 0$ and $b_3(X) = 2$.

\sn We now show that Calabi-Eckmann manifolds can be deformed to achieve
$a(X) = 1 $ or $a(X) = 0$ and $b_1 = b_2 = 0$, $b_3 = 2$. We choose
positive real numbers $a,b,c$ and let $B = \bC^2 \setminus \{(0,0)\}$. We
define the following action of $\bC$ on $B \times B :$ $$ (t,x,y,u,v)
\mapsto (\exp(t)x,\exp(at)y,\exp(ibt)u,\exp(ict)v). $$ One checks easily
that this action is holomorphic, free and almost proper so that the
quotient $X$ exists and is a compact manifold. If $a = 1$ and $b = c$, then
$X$ is a Calabi-Eckmann manifold. If however $a \not \in \bQ$ and $b = c$
resp.\ $ a \not \in \bQ$ and ${b\over c} \not \in \bQ$, then $a(X) = 1$
resp.\ $a(X) = 0$.

\bn {\bf 2.4 Example}
\sn Hopf threefolds of the form
$$
\bC^3 \setminus \{(0,0,0)\} /\bZ ,
$$
with the action of $\bZ\simeq\{\lambda^k\,;\,k\in\bZ\}$ being defined by $$
\lambda (x,y,z) = (\alpha x, \beta y, \gamma z),\qquad 0 < \vert \alpha
\vert, \vert \beta \vert, \vert \gamma \vert < 1 $$ are homeomorphic to
$S^1 \times S^5$. They have $a(X) = 0, 1$ or $2$ and $b_1(X) = 1$, while
$b_2(X) = b_3(X) = 0$. This realizes the other possibility for the pair
$(b_1,b_3)$ when $b_2 = 0$ and $a(X) > 0$, as stated in the Main Theorem.

\bn Notice that the algebraic reduction is holomorphic in (2.3) but it is
not holomorphic in (2.4) if $a(X) = 1$.

\bn {\bf 2.5 Example }
\sn We finally give other examples of compact threefolds $X$ with $a(X)=0$
and $b_1 = b_2 = 0$, $b_3 = 2$. Let
$\Gamma \subset \Sl(2,\bC)$ be a torsion free cocompact lattice in such a
way that the quotient $X := \Sl(2,\bC)/\Gamma$ has $b_1(X) = 0$.

\sn This last condition is not automatic. Let $Y = \SU(2)\backslash
\Sl(2,\bC)/\Gamma;$ then $Y$ is a compact differentiable manifold admitting
a
differentiable fibration $\pi: X \La Y$ with $S^3$ is fiber. Since $b_1(X)
= 0$, we also have $b_1(Y) = 0$, hence $b_2(Y) = 0$ by Poincar\'e duality.
Now the
Leray spectral sequence immediately gives $b_2(X) = 0$, the fibers of $\pi$
being 3-spheres. $b_3(X) = 2$ is again clear from the Leray spectral
sequence.
Finally the fact that $X$ does not carry any non-constant meromorphic
function results from [HM83] from which we even deduce that $X$ does not
carry any
hypersurface (as $X$ is homogeneous).

\bn \bn \section{3. On the finer structure of threefolds with}
\sn\section{\phantom{5. {}}$b_2(X)=0$ and $a(X)=1$.}

%\bn \bn In this section we investigate more closely threefolds $X$ with
%$b_2(X) = 0$ and holomorphic algebraic reduction $f : X \La V$ to a curve
%$V$. By construction of the algebraic reduction, $V$ is normal, thus
%smooth, and the fibers of $f$ are connected. From the Leray spectral
%sequence we have $b_1(X) \geq b_1(V)$, hence 2.1~(3) implies that $b_1(V) =
%0$, so that $V$ is rational. The key to our investigations~is

\bn \bn In this section we investigate more closely threefolds $X$ with
$b_2(X) = 0$ and algebraic dimension $1$. By construction, the algebraic 
reduction $f : X \dasharrow V$ is a meromorphic map to a normal algebraic
space, hence $V$ is a nonsingular curve. We claim that $V$ must be 
rational. In fact, we have $b_1(X)\leq 1$ by 2.1$\,$(3); on the other
hand, for every holomorphic $1$-form $u$ on $V$, then $f^\star u$ extends
to a $d$-closed holomorphic $1$-form on $X$, thus $b_1(X) \geq 2\,h^{1,0}(V)$
and $h^{1,0}(V)$ must be zero. In the rest of this section, we restrict
ourselves to the case when the algebraic reduction $f:X\to V$ is 
{\it holomorphic}. The key to our investigations~is

\bn {\bf 3.1 Theorem}
\sn {\it Let $F$ be a general smooth fiber of $f$. Then the restriction $r
: H^1(X,\cO_X) \La H^1(F,\cO_F) $ is surjective.}

\bn We need some preparations for the proof of (3.1). Let $\Delta \subset
V$ be a finite {\bf non empty } set such that $A = f^{-1}(\Delta) \subset
X$ contains all singular fibers of $f$. Let $V'= V \setminus \Delta$, and
$X' = f^{-1}(V')$ so that $f' = f_{\vert X'}$ is a smooth fibration. Let
$D_i$, $1\leq i \leq r$ be the irreducible components of $A$ and let $s
=\card\Delta$ be the number of connected components of $A$. Furthermore we
set $t =b_1(F)$ where $F$ is the general smooth fiber of $f$. For a
non-compact space $Z$ we let $b_i(Z) = \dim H_i(Z,\bR)$, whatever this
dimension is. We prepare the proof of (3.1) by three lemmas.

\bn {\bf 3.2 Lemma}
\sn {\it
\item{\rm(1)} The natural exact sequence of groups $$ 1 = \pi_2(V') \La
\pi_1(F) \La \pi_1(X') \La \pi_1(V') \La 1 $$ is exact and
$($non-canonically$)$ split. \item{\rm(2)} $b_1(X') = b_1(V') + b_1(F) =
s-1 + t$. \item{\rm(3)} $r = s-1 + t - b_1(X)$.}

\bn {\bf Proof.} (1) Since $V'$ is a non-compact Riemann surface,
$\pi_1(V')$ is a free group of $s-1$ generators and since $V'$ is
uniformized by either $\bC$ or by the unit disc, we have $\pi_2(V') = 1$.
Hence the exact homotopy sequence of the fibration $f': X' \La V'$ gives
the exact sequence of groups stated in (1). Since $\pi_1(V')$ is a free
group, the sequence splits.

\sn (2) From (1) we deduce that
$$
H_1(X',\bZ) \simeq H_1(V',\bZ) \oplus H_1(F,\bZ). $$ Moreover $f_* :
\pi_1(X) \La \pi_1(V)$ is surjective since the fibers of $f$ are connected.
Hence (2) follows.

\sn (3) The cohomology sequence with rational coefficients of the pair
$(X,A)$ gives
$$
0 = H^4(X) \La H^4(A) \La H^5(X,A) \La H^5(X) \La H^5(A) = 0. $$ By duality
we have $H^5(X,A) \simeq H_1(X')$ and $H^5(X) \simeq H_1(X)$. Hence $r =
\dim H^4(A) = b_1(X') - b_1(X) = s-1+t - b_1(X)$ by (2), as claimed.

\bn Now choose an integer $m > 0$ such that $c_1(\cO_X(mD_i)) = 0$ in
$H^2(X,\bZ)$ for all $1 \leq i \leq r$. Let $\bigoplus \bZ[mD_i] \simeq
\bZ^r$ be the free abelian group generated by $mD_i$, $1\leq i\leq r$, and
let $\phi:\bigoplus \bZ[mD_i]\La \Pic^0(X)$ be given by sending $D = \sum
_i a_imD_i$ to $\cO_X(D)$.

\bn {\bf 3.3 Lemma}
\sn {\it Let $K = \Ker \phi$ and $I = \Im \phi$. Then $\rk K \leq s-1 $ and
$\rk I \geq r-s+1$. }

\bn {\bf Proof.} The kernel $K$ consists of all divisors $D$ such that
$\cO_X(D)\simeq\cO_X$, i.e.\ such that $D$ is the divisor of a global
meromorphic function $h$ on $X$. As $f:X\to V$ is the algebraic reduction
of~$X$, there must exist a meromorphic function $\tilde h$ on $V$ such that
$h=\tilde h\circ f$. Now, the divisor $\tilde D$ of $\tilde h$ has degree
$0$ and support in $\Delta=\{x_1,\ldots,x_s\}$. This implies that $K$ is
contained in $f^*(\Pic^0(\Delta))$. As $\Pic^0(\Delta)\simeq \bZ^{s-1}$,
the claim follows.

\bn The last ingredient in the proof of (3.1) is provided by

\bn {\bf 3.4 Lemma}
\sn {\it The restriction map $\alpha: H^1(X,\cO_X) \La H^1(X',\cO_{X'})$ is
injective.}

\bn {\bf Proof.} The Leray spectral sequences of the fibrations $f:X\to V$
and $f':X'\to V'$ yield a commutative diagram { \def\quad{\kern3pt}
$$
\matrix{
0 &= &H^1(V,\cO_V) &\buildrel {\textstyle f^*} \over \La & H^1(X,\cO_X)
&\buildrel {\textstyle\simeq} \over \La & H^0(V,R^1f_*\cO_X) &\La&
H^2(V,\cO_V)&=0 \cr
\noalign{\vskip-4pt}\cr
&&\Big\downarrow && ~~\Big\downarrow\alpha && ~~\Big\downarrow\tilde\alpha
&& \Big\downarrow \cr
\noalign{\vskip-4pt}\cr
0 &= &H^1(V',\cO_{V'}) &\La & H^1(X',\cO_{X'}) &\buildrel
{\textstyle\simeq} \over \La & H^0(V',R^1f'_*\cO_{X'}) &\La&
H^2(V',\cO_{V'})&=0,\cr}
$$
}
and since $R^1f_*\cO_X$ is locally free by (1.5), we conclude that
$\tilde\alpha$ is injective, QED.

\bn We are now able to finish the proof of (3.1).

\sn Consider again $I \subset \Pic^0(X)$, the image of $ \phi :\bigoplus
\bZ[mD_i]\La\Pic^0(X)$, and let $\tilde I$ be the inverse image of $I$
under the natural map $H^1(X,\cO_X) \La \Pic^0(X)$. We have a commutative
diagram
{
\def\quad{\kern3pt}
$$
\matrix{
&&H^1(X,\bZ) &\La & H^1(X',\bZ) & \buildrel{\textstyle\beta^\bZ}\over\La &
H^0(V',R^1f'_*\bZ) &\buildrel{\textstyle\gamma^\bZ}\over\La& H^1(F,\bZ) \cr
\noalign{\vskip-4pt}\cr
&&\Big\downarrow && ~~\Big\downarrow\theta &&\Big\downarrow&&
\Big\downarrow\cr \noalign{\vskip-4pt}\cr
\tilde I &\subset &H^1(X,\cO_X) &\buildrel {\textstyle \alpha} \over \La &
H^1(X',\cO_{X'}) &\buildrel{\textstyle\beta}\over
{\mathop{\La}\limits_{\textstyle\simeq}} & H^0(V',R^1f'_*\cO_{X'})
&\buildrel{\textstyle\gamma}\over\La& H^1(F,\cO_F) \cr
\noalign{\vskip-4pt}\cr
\Big\downarrow&&\Big\downarrow && \Big\downarrow && \cr
\noalign{\vskip-4pt}\cr I&\subset &H^1(X,\cO_X^*)
&\buildrel{\textstyle\hat\alpha}\over\La & H^1(X',\cO_{X'}^*)&&&\cr}
$$
}where the two vertical sequences are exponential exact sequences, $\alpha$
and $\hat \alpha$ are restriction maps from $X$ to~$X'$, $\beta$ and
$\beta^\bZ$ arise
from the spectral sequence, and $\gamma$, $\gamma^\bZ$ are restriction maps
to a generic fiber~$F$. We get $$ \rk\tilde I=\rk I+\rk H^1(X,\bZ)\ge
r-s+1+b_1(X)=t $$ by 3.3 and 3.2~(3).
Now $\hat\alpha(I)=0$, since $\hat\alpha(\cO_X(mD_i))
=\cO_{X'}(mD_i))\simeq\cO_{X'}$ for all~$i$. It follows that $\alpha(\tilde
I)$ is contained in the image of $\theta$, thus $\theta^{-1}(\alpha(\tilde
I))\subset H^1(X',\bZ)$ has rank
$$
\rk(\theta^{-1}(\alpha(\tilde I)))\geq \rk\alpha(\tilde I)\geq\rk\tilde
I\ge t $$
thanks to the injectivity of~$\alpha$. Moreover, $R^1f'_*\bZ$ is a locally
constant system of rank $\rk H^1(F,\bZ)=t$, hence $H^0(V',R^1f'_*\bZ)$ has
rank at most $t$.
On the other hand, as $\beta$ is an isomorphism and $\rk\alpha(\tilde
I)\geq t$, we
see that $\beta^\bZ(\theta^{-1}(\alpha(\tilde I)))$ has rank at least~$t$.
Therefore, $\gamma^\bZ\circ\beta^\bZ(\theta^{-1}(\alpha(\tilde I)))$ is of
finite index in $H^1(F,\bZ)$. Since $H^1(F,\cO_F)$ is the complex linear
span of the image of $H^1(F,\bZ)$, we see that
$\gamma\circ\beta\circ\alpha(\tilde I)$ also generates $H^1(F,\cO_F)$. In
particular, the restriction map $H^1(X,\cO_X)\to H^1(F,\cO_F)$ must be
surjective. This concludes the proof of~(3.1).

\bn {\bf (3.5)} We now study the structure of the smooth fibers $F$ of $f$.
The exact sequence
$$
0 \La T_F \La (T_X)_{|F} \La \cO(F)_{|F} \La 0 $$ and the equality
$c_1(\cO(F))=0$ in $H^2(X,\bR)$ imply $$ c_1(F) = c_1(X)_{\vert F}= 0,
\qquad c_2(F) = c_2(X)_{\vert F} = 0 $$ in particular we also have
$\chi(F,\cO_F)=0$). We conclude from the classification of surfaces that
$F$ is one of the following: a Hopf surface, an Inoue surface, a Kodaira
surface (primary or secondary), a torus or a hyperelliptic surface (see
e.g.\ [BPV84]). By [Ka69] however, $F$ cannot be hyperelliptic. The reason
is the existence of a relative Albanese reduction in that case.

\bn {\bf 3.6 Proposition}
\sn {\it The general fiber $F$ of $f$ cannot be a Kodaira surface nor a
Hopf surface with algebraic dimension $1$. }

\bn {\bf Proof.} Let $F_0$ be a fixed smooth fiber and assume that $F_0$ is
a Kodaira surface or a Hopf surface with $a(F_0) = 1$. Let $g_0 : F_0 \La
C_0$ be ``the'' algebraic reduction which is an elliptic fiber bundle. Let
$\cL_0 = g_0^*({\cal G}_0)$ with ${\cal G}_0$ very ample on $C_0$.

\sn (1) There exists a line bundle $\tilde \cL$ on $X$ with $\tilde
\cL_{\vert F_0} = \cL_0$.

\sn Proof. By passing to some power $\cL_0^m$ if necessary, we have
$c_1(\cL_0) = 0$ in $H^2(X,\bZ)$. Let $$ \lambda_1 : H^1(X,\cO_X) \La
\Pic(X)
$$
and
$$
\lambda_2 : H^1(F_0, \cO_{F_0}) \La \Pic(F_0) $$ be the canonical maps, and
let $r : H^1(X,\cO_X) \La H^1(F_0,\cO_{F_0})$ be the restriction map.
Choose $\alpha \in H^1(F_0,\cO_{F_0})$ with $\lambda_2(\alpha) = \L_0$.
Since $r$ is surjective by (3.2), we find $\beta \in H^1(X,\cO_X)$ with
$r(\beta) = \alpha$. Now let $\tilde \cL = \lambda_1(\beta)$.

\sn (2) Let $F$ be any smooth fiber of $f$. Then $\kappa(\tilde \cL_{\vert
F}) = 1$. In fact, it follows from the local freeness of $R^jf_*(\tilde
\cL)$ stated in Lemma 1.5 that $$
f_*(\tilde \cL^{\mu})_{\vert\{y\}} \simeq H^0(F,\tilde \cL^{\mu}), $$ where
$F = f^{-1}(y)$,
see [BaSt76, chap.3, 3.10].

\sn (3) From the generically surjective morphism $$ f^*f_*(\tilde \cL^m)
\La \tilde \cL^m,
$$
($m \gg 0)$, we obtain a meromorphic map $$ g: X \dasharrow \bP(f_*(\tilde
\cL^m)),
$$
which, restricted to $F$ is holomorphic and just gives the algebraic
reduction of $F$. Let $Z$ be the closure of the image of $g$. Then $f$
factors via the meromorphic map $ h_1 : X \dasharrow Z$ and the holomorphic
map $h_2 : Z \La V$. Now $h_2$ is the restriction of the canonical
projection $\bP(f_*(\tilde \cL^m)) \La V$, therefore $h_2$ is a projective
morphism and $Z$ is projective. Hence \hbox{$a(X) \geq 2$,} contradiction.

\bn From (3.6) it follows that $F$ can only be an Inoue surface, a Hopf
surface without meromorphic functions or a torus. In order to exclude by a
similar method as in (3.6) also tori of algebraic dimension 1, we would
need the existence of a relative algebraic reduction (the analogue of $h_2
: X \dasharrow Z$) in that case, too.

\sn We now look more closely to the structure of $f$.

\bn {\bf 3.7 Proposition}
\sn {\it Assume that $F$ is not a torus. Then \item {\rm(1)} $R^1f_*(\cO_X) = \cO_V $
\item {\rm(2)} $R^2f_*(\cO_X) = 0 $
\item {\rm(3)} $\dim H^1(X,\cO_X) = 1$
\item {\rm(4)} $H^2(X,\cO_X) = H^3(X,\cO_X) = 0$. }

\bn {\bf Proof.} (2) Since $H^2(F,\cO_F) = 0$, the sheaf $R^2f_*(\cO_X) $
is torsion, hence identically zero by 1.5. Then $H^3(X,\cO_X) = 0$ is
immediate from the Leray spectral sequence.

\sn (1) Since $h^1(F,\cO_F) = 1, R^1f_*(\cO_X)$ is a line bundle on $V$.
Let $$ d = \deg R^1f_*(\cO_X).
$$
Then Riemann-Roch gives $\chi(R^1f_*(\cO_X)) = d+1. $ On the other hand the
Leray spectral sequence together with $H^3(X,\cO_X) = 0$ and (2) yields $$
\chi(R^1f_*(\cO_X)) = h^1(\cO_X) - h^2(\cO_X) = - \chi(\cO_X) + 1. $$ We
conclude $d = - \chi(\cO_X) = 0$. This proves (1). Now (3) and the second
part of (4) are obvious.

\bn {\bf 3.8 Remark} \ In case $F$ is a torus, $R^1f_*(\cO_X)$ is a rank 2
bundle and $R^2f_*(\cO_X)$ is a line bundle. Using Theorem~3.1 it is easy
to see that
\item {(1)} $R^1f_*(\cO_X) = \cO(a) \oplus \cO(b) $ with $a,b \geq 0$ \item
{(2)} $R^2f_*(\cO_X) = \cO(a+b)$.

\sn Note that (2) gives dually $f_*(\omega_{X\vert V}) = \cO(-a-b)$.
Usually one expects the degree of $f_*(\omega_{X \vert V})$ to be
semi-positive, but here we are in a highly non-K\"ahler situation where it
might happen that the above degree is negative, see [Ue87].

\bn {\bf 3.9 Proposition}
\sn {\it Assume that $F$ is not a torus. Then $H^0(X,\Omega^{i}_X) = 0$, $1
\leq i \leq 3$.}

\bn {\bf Proof.} For $i = 3$ the claim follows already from 3.7~(4) and
Serre duality.

\sn (1) First we treat the case $i = 1$. Let $\omega$ be a holomorphic
$1$-form. Let $j : F \La X$ be the inclusion. Then $j^*(\omega) = 0$, hence
at least locally near $F$ we have $\omega = f^*(\eta)$, hence $d\omega = 0$
near $F$ and therefore the holomorphic $2$-form $d\omega$ is identically
zero on~$X$. Now the space of closed holomorphic $1$-forms can be
identified with $H^0(X,d\cO_X)$ and, as it is well known (see e.g.\
[Ue75]), we have the inequality
$$
2 h^0(X,d\cO_X) \leq b_1(X).
$$
The inequality $b_1(X) \leq 1 $ implies $h^0(X,\Omega^1_X) =
h^0(X,d\cO_X)=0$, as desired.

\sn (2) In case $i = 2$, we again have $j^*(\omega) = 0$. Let $U$ be a
small open set in $V$ such that $f_{\vert f^{-1}(U)}$ is smooth. Let $z$ be
a coordinate on $U$ and $h = f^*(z)$. Then we conclude that $$
\omega_{\vert f^{-1}(U)} = dh \wedge\alpha $$ with some relative
holomorphic $1$-form $\alpha\in H^0(f^{-1}(U), \Omega^1_{X/V})$. Now again
$j^*(\alpha) = 0$ and therefore $\alpha = 0$, $\omega = 0$.

\bn {\bf 3.10 Corollary } \sn {\it Assume that $F$ is not a torus. Then
either $f_*(\Omega^1_{X/V}) = 0$  or there exists some $x\in V$ such that
$f_*(\Omega^1_{X/V}) = \bC_x$, i.e.\ a sheaf
supported on $x$ with a 1-dimensional stalk at $x$. In particular $f$ has
at most one singular fiber and such a fiber is normal with exactly one
singularity of embedding dimension $3$.}

\bn {\bf Proof.} Consider the exact sequence $$
0 \La f^*(\Omega^1_V) \La \Omega^1_X \La \Omega^1_{X/V} \La 0.$$
Since $F$ has no holomorphic 1-forms, $f_*(\smash{\Omega^1_{X/V}})$ is a
torsion sheaf on the curve $V$. The corollary will follow if we check that
$h^0(V,f_*(\Omega^1_{X/V}))=h^0(X,\Omega^1_{X/V}) \leq 1$. Now, observe the
following facts.
{\parindent=6mm \smallskip
\item {(1)} $ H^0(X,\Omega^1_X ) = 0$, by (3.9); \smallskip \item {(2)}
$H^1(X,f^*(\Omega^1_V)) = H^1(V,\Omega^1_V)$\hfil\break by Leray's spectral
sequence
and the equalities $R^if_*(f^*\Omega^1_V)=
R^if_*(\cO_X)\otimes\Omega^1_V=\Omega^1_V$, $i=0,1$ (cf.\ 3.7~(1));
\smallskip \item {(3)} $\dim H^1(V,\Omega^1_V) = 1.$ \par}

\sn Then, taking cohomology groups in the first exact sequence, we get the
desired inequality
$$
h^0(X,\Omega^1_{X/V}) \leq h^1(V,\Omega^1_V) = 1. $$

\bn We can say something more about the structure of the singular fibers of $f$.

\bn {\bf 3.11 Proposition}
\sn {\it Assume that $F$ is not a torus. Let $A$ be a union of fibers
containing all singular fibers of $f$. Let $s =\card(f(A))$ and $r$ the
number of irreducible componentsof $A$. Then $r = s$, i.e.\ all fibers of
$f$ are irreducible and $b_1(X) = 0$. }

\bn {\bf Proof}. Since $F$ is an Inoue surface or a Hopf surface, we have
$b_1(F) = 1$, thus 3.2~(3) implies $r=s-b_1(X)$. As $r\geq s$, we must have
$r=s$ and $b_1(X)=0$.

\bn {\bf 3.12 Remark} In case $F$ is a torus, 3.2~(3) implies $r=s+3-b_1(X)
\geq s+2$. It seems rather reasonable to expect that tori actually cannot
appear as fibers of $f$. Observe that $f$ must have a singular fiber in
this case because of $r>s$. So a study of the singular fibers is needed to
exclude tori as fibers of $f$. However there is a significant difference :
the case of tori is one (in fact the only one) where $C_{3,1}$ might fail,
see [Ue87].

\bn {\bf 3.13 Proposition}
\sn {\it Assume that $F$ is not a torus. Then $h^{1,1} = h^{1,2} = h^{2,1}
= 1 $ $($so that we know all Hodge numbers of $X)$.}

\bn {\bf Proof.}(1) $h^{1,2} = h^{2,1}$ is of course Serre duality.

\sn (2) By (3.9) and $\chi(X,\Omega_X^1) = 0$ it suffices to see $h^{1,3} =
0$ in order to get $h^{1,1} = h^{1,2}$. But this follows again from Serre
duality and the equality $h^{2,0}=h^0(X,\Omega^2_X)=0$.

\sn (3) From the exact sequence
$$
0 \La f^*(\Omega^1_V) \La \Omega^1_X \La \Omega^1_{X/V} \La 0 \eqno (S) $$
we deduce that it suffices to show
\smallskip
\item {(a)} $h^2(X,f^*(\Omega^1_V)) = 1$ \smallskip \item {(b)}
$h^2(X,\Omega^1_{X/V}) = 0$ \sn in order to get $h^{1,2} \leq 1$.

\sn (a) By the Leray spectral sequence and (3.7) we have
$h^2(X,f^*(\Omega_V^1)) = h^1(V,\Omega^1_V) = 1$.

\sn (b) Again we argue by the Leray spectral sequence. Since
$R^1f_*(\Omega^1_{X/V})$ is a torsion sheaf, we need only to show that $$
R^2f_*(\Omega^1_{X/V}) = 0.
$$
In fact, taking the direct image $f_*$ of (S), we see that
$R^2f_*(\Omega^1_{X/V})$ is a quotient of $R^2f_*(\Omega^1_X)$ which is $0$
by the equality $H^2(F,\Omega^1_F) = H^0(F,\Omega^1_F) = 0$ and by (1.5).

\sn (4) We finally show $h^{1,1} \ne 0$ to conclude the proof. Let
$(E^{p,q}_r)$ be the Fr\"olicher spectral sequence on $X$. Since $b_1(X) =
0$ by (3.11), we get $E_{\infty}^{0,1} = 0$. Hence $E_2^{0,1} = 0$. On the
other hand
$$
E_2^{0,1} = \Ker\partial : E_1^{0,1} \La E_1^{1,1}. $$ Since $E_1^{p,q} =
H^{p,q}(X)$, we conclude that $H^1(X,\cO_X)$ injects into $H^{1,1}(X)$. So
by (3.7) $H^{1,1}(X) \ne 0$.

\bn We finally collect all our knowledge in the case the general fiber of
$f$ is not a torus.

\bn {\bf 3.14 Theorem}
\sn {\it Let $X$ be a smooth compact threefold with $b_2(X) = 0$ and
holomorphic algebraic reduction $f: X \La V$ to the smooth curve $V$.
Assume that the general smooth fiber is not a torus. Then: \item{\rm(1)}
$b_1(X) = 0$, $b_3(X) = 2$. \item{\rm(2)} Any smooth fiber of $f$ is a Hopf
surface without meromorphic functions or an Inoue surface. \item{\rm(3)}
The Hodge numbers of $X$ are as follows: $h^{1,0}= 0$, $h^{0,1} = 0$,
$h^{2,0} = 0$, $h^{1,1} = 1$, $h^{0,2} = 0$, $h^{3,0} = 0$, $h^{2,1} = 1$
$($the others are determined by these via Serre duality$)$. \item{\rm(4)}
All fibers of $f$ are irreducible. There is at most one normal singular
fiber. \vskip30pt}

\section{References}
\bn
\parindent=21pt

\itemitem{[BaSt76]} Banica, C.; Stanasila, O.: Algebraic methods in the
global theory of complex spaces. Wiley 1976

\itemitem{[BPV84]} Barth, W.; Peters, C.; Van de Ven, A.: Compact complex
surfaces. Erg.\ d.\ Math.\ (3),
Band 4, Springer 1984

\itemitem{[HM83]} Huckleberry, A.T.; Margulis, G.: Invariant analytic
hypersurfaces. Inv.\ Math.\ {\bf 71}, 235-240 (1983)

\itemitem{[Ka69]} Kawai, S.: On compact complex analytic manifolds of
complex dimension 3, II. J.\ Math.\ Soc.\ Japan {\bf 21}, 604-616 (1969)

\itemitem{[RRV71]} Ramis, J.; Ruget, G.; Verdier, J.-L.: Dualit\'e relative
en g\'eom\'etrie analytique complexe. Inv.\ Math.\ {\bf 13}, 261-283 (1971)



\itemitem{[Ue75]} Ueno, K.: Classification theory of algebraic varieties
and compact complex spaces. Lecture Notes in Math.\ 439, Springer 1975

\itemitem{[Ue87]} Ueno, K.: On compact analytic threefolds with non-trivial
Albanese tori. Math.\ Ann.\ {\bf 278}, 41-70 (1987)

\itemitem{[We85]} Wehler, J.: Der relative Dualit\"atssatz f\"ur
Cohen-Macaulay R\"aume. Schriftenreihe des Math.\ Instituts der Univ.\
M\"unster, 2.~Serie, Heft {\bf 35} (1985)

\bn \bn Fr\'ed\'eric Campana, D\'epartement de math\'ematiques,
Universit\'e de Nancy, BP 239, 54506 Vandoeuvre les Nancy, France

\bn Jean-Pierre Demailly, Universit\'e de Grenoble~I, Institut Fourier, BP
74, U.M.R. 5582 du C.N.R.S., 38402 Saint-Martin d'Heres, France

\bn
Thomas Peternell, Mathematisches Institut, Universit\"at Bayreuth, 95440
Bayreuth, \hbox{Germany}

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