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% |Theta divisors and the Frobenius morphism |
% |Notes of a conference given in Luminy on Fri Dec 04 1998|
% |By David A. Madore |
% |This version's date: Wed Apr 21 12:30:03 CEST 1999 |
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\begin{document}
\title{Theta divisors and the Frobenius morphism}
\author{David A. Madore}
%\date{9th January 1999}
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\begin{abstract}
We introduce theta divisors for vector bundles and relate them to the
ordinariness of curves in characteristic $p>0$. We prove, following
M. Raynaud, that the sheaf of locally exact differentials in
characteristic $p>0$ has a theta divisor, and that the generic curve
in (any) genus $g\geq 2$ and (any) characteristic $p>0$ has a cover
that is not ordinary (and which we explicitely construct).
\end{abstract}
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\section{Theta divisors for vector bundles}
\par
Let $k$ be an algebraically closed field and $X$ a smooth proper connected
curve over $\Spec k$ having genus $g$. We assume throughout that
$g\geq 2$.
\par
If $E$ is a vector bundle (i.e. a locally free invertible sheaf) of rank
$r$ and degree $d$ over $X$, we define its {\em slope\/} to be
$\lambda=d/r$. The Riemann-Roch formula gives the Euler-Poincaré
characteristic of $E$:
$$
\chi(X,E)=h^0(X,E)-h^1(X,E)=r(\lambda-(g-1))
$$
\par
In particular for $\lambda=g-1$ (the {\em critical slope\/}) we have
$\chi(X,E)=0$; moreover, it is still true for any invertible sheaf
$L$ of degree $0$ over $X$ that $\chi(X,E\otimes L)=0$, in other
words, $h^0(X,E\otimes L)=h^1(X,E\otimes L)$. Under those
circumstances, it is natural to ask the following question:
\begin{que}
Suppose $E$ has critical slope. Then for which invertible sheaves $L$
of degree $0$ (if any) is it true that $h^0(X,E\otimes L)=0$ (and
consequently also $h^1(X,E\otimes L)=0$)? Is this true for some $L$,
for many $L$, or for none?
\label{MainQuestion}
\end{que}
\par
We start with a necessary condition. Suppose there were some
subbundle $F\rightarrowtail E$ having slope
$\lambda(F)>\lambda(E)=g-1$. Then we would have $\chi(X,F\otimes
L)>0$, hence $h^0(X,F\otimes L)>0$. Now $H^0(X,F\otimes
L)\rightarrowtail H^0(X,E\otimes L)$, so this implies $h^0(X,E\otimes
L)>0$. So if we are to have $h^0(X,E\otimes L)>0$ for some $L$, this
must not happen, and we say that $E$ is {\em semi-stable\/}:
\begin{defn}
A vector bundle $E$ over $X$ is said to be {\em stable\/} (resp. {\em
semi-stable\/}) iff for every sub-vector-bundle $F$ of $E$ (other than
$0$ and $E$) we have $\lambda(F)<\lambda(E)$
(resp. $\lambda(F)\leq\lambda(E)$).
\end{defn}
\par
Another remark we can make bearing some relation with
question (\ref{MainQuestion}) is that, by the semicontinuity theorem,
if we let $L$ vary on the jacobian of $X$, the functions
$h^0(X,E\otimes L)$ and $h^1(X,E\otimes L)$ are upper semicontinuous.
This means that they increase on closed sets. In particular, if
$h^0(X,E\otimes L)=0$ (the smallest possible value) for some $L$, then
this is true in a whole neighborhood of $L$, that is, for almost all
$L$. We then say that this holds for a {\em general\/} invertible
sheaf of degree $0$ and we write $h^0(X,E\otimes L_\gen)=0$.
\par
Now introduce the jacobian variety $J$ of $X$ and let $\mathscr{L}$ be
the (some) Poincaré sheaf (universal invertible sheaf of degree $0$)
on $X\times_{\Spec k}J$. We aim to use $\mathscr{L}$ to let $L$ vary
and provide universal analogues for our formulæ. Let
$$
\xymatrix{
X\times_{\Spec k}J \ar[d]^f \cr
J}
$$
be the second projection.
\par
Consider the sheaf $E\otimes\mathscr{L}$ (by this we mean the twist by
$\mathscr{L}$ of the pullback of $E$ to $X\times_{\Spec k}J$). Our
interest is mainly in the higher direct image
$Rf_*(E\otimes\mathscr{L})$, which incorporates information about
$H^i(X,E\otimes L)$ (and much more).
\par
To be precise, we know that there exists a complex
$0\to M^0\buildrel u\over\longrightarrow M^1\to 0$
of vector bundles on $J$ that universally computes the
$R^i f_*(E\otimes\mathscr{L})$, in the sense that the $i$-th
cohomology group ($i=0,1$) of the complex is $R^i
f_*(E\otimes\mathscr{L})$ and that this remains true after any base
change $J'\to J$. (One particularly important such base change, of
course, is the embedding of a closed point $\{L\}$ in $J$.)
\par
Now $M^0$ and $M^1$ have the same rank, say $s$ (because the
Euler-Poincaré characteristic of $E$ is $0$). So we can consider the
determinant of $u$, $\medwedge^s M^0\buildrel\det u\over\longrightarrow
\medwedge^s M^1$, or rather
$$
{\cal O}_J\buildrel\det u\over\longrightarrow
\medwedge^s M^1\otimes\big(\medwedge^s M^0\big)^{\otimes-1}
$$
\par
Exactly one of the following two things happens:
\begin{itemize}
\item
{\em Either\/} the determinant $\det u$ is zero (identically). In
this case, $u$ is nowhere (i.e. on no fiber) invertible, it always
has a kernel and a cokernel: we have $h^0(X,E\otimes L)>0$ (and of
course $h^1(X,E\otimes L)>0$) for all $L$.
\item
{\em Or\/} $\det u$ is a nonzero section of the invertible sheaf
$\medwedge^s M^1\otimes\big(\medwedge^s M^0\big)^{\otimes-1}$
on $J$ and it defines a positive divisor $\theta_E$ on $J$, whose
support is precisely the locus of $L$ such that $h^0(X,E\otimes
L)>0$. We then have $h^0(X,E\otimes L_\gen)=0$.
\end{itemize}
\par
In other words, precisely in the case where $h^0(X,E\otimes L_\gen)=0$
we can define a positive divisor $\theta_E$ on $J$ which tells us
``where the bundle $E$ has cohomology''. We call this divisor the
{\em theta divisor\/} of the vector bundle $E$. And we will use the
expression ``to admit a theta divisor'' as synonymous for
$h^0(X,E\otimes L_\gen)=0$. For example, a vector bundle that admits
a theta divisor is semi-stable (but the converse is not true,
cf. \cite{RaynaudOld}).
\par
%
%
%
\section{Enters the Frobenius morphism}
\par
We now assume that the base field $k$ has characteristic $p>0$. We
then have a relative Frobenius morphism
$$
\xymatrix@=1em{
X\ar[rr]^\pi\ar[dr]&&X_1\ar[dl]\cr
&\Spec k&
}
$$
which is obtained by factoring the absolute Frobenius morphism through
the pullback to $X$ of the Frobenius on $k$ (more descriptively, $\pi$
has the effect, in projective space, of raising the coordinates to the
$p$-th power, while $X_1$ is the curve obtained by raising to the
$p$-th powers the coefficients in the equations defining $X$).
\par
The curve $X_1$ has the same genus $g$ as $X$. The morphism $\pi$ is
flat, finite and purely inseparable of degree $p$. From it we deduce
a morphism ${\cal O}_{X_1}\to\pi_*{\cal O}_X$ (of
${\cal O}_{X_1}$-modules), which is mono because $\pi$ is surjective.
Call $B_1$ the cokernel, so that we have the following short exact
sequence:
\begin{equation}
0\to{\cal O}_{X_1}\to\pi_*{\cal O}_X\to B_1\to 0
\label{ShortExact}
\end{equation}
Now the sheaf $B_1$ can be viewed in a different way: if we call $d$
the differential ${\cal O}_X\to\Omega^1_X$ (between sheaves of
$\mathbb{Z}$-modules) then $\pi_*(d)$ is ${\cal O}_{X_1}$-linear and
has ${\cal O}_{X_1}$ as kernel. Consequently, $B_1$ can also be seen
as the image of $\pi_*(d)$, hence its name of {\em sheaf of locally
exact differentials\/}. As a subsheaf of the locally free sheaf
$\pi_*\Omega^1_X$, it is itself a vector bundle.
\par
In the short exact sequence (\ref{ShortExact}) above, the vector
bundles ${\cal O}_{X_1}$ and $\pi_*{\cal O}_X$ have rank $1$ and $p$
respectively, so that $B_1$ has rank $p-1$. On the other hand, since
${\cal O}_{X_1}$ and $\pi_*{\cal O}_X$ each have Euler-Poincaré
characteristic $g-1$, we have $\chi(X_1,B_1)=0$, or in other words,
$\lambda(B_1)=g-1$ (the critical slope), and what we have said in the
previous section applies to the sheaf $B_1$.
\par
More precisely, we have the following long exact sequence in
cohomology, derived from (\ref{ShortExact}):
$$
\begin{array}{ccccccccc}
&&k&\mkern-30mu\widetilde{\relbar\joinrel\relbar\joinrel\relbar
\joinrel\relbar\joinrel\longrightarrow}\mkern-30mu&k&&&&\cr&&\|&&\|&&&&\cr
0&\to&H^0(X_1,{\cal O}_{X_1})&\to&H^0(X,{\cal O}_X)&\to&H^0(X_1,B_1)&\to&\cr
\noalign{\medskip}
&\to&H^1(X_1,{\cal O}_{X_1})&\to&H^1(X,{\cal O}_X)&\to&H^1(X_1,B_1)&\to&0\cr
\end{array}
$$
\par
Here the first arrow is an isomorphism as shown. Consequently, the
arrow $H^1(X_1,{\cal O}_{X_1})\to H^1(X,{\cal O}_X)$ is also an
isomorphism if and only if $h^0(X_1,B_1)=0$, or, what amounts to the
same, $h^1(X_1,B_1)=0$. This is again the same as saying that $B_1$
has a theta divisor (something which we will see is always true) and
that it does not go through the origin.
\par
\begin{defn}
When the equivalent conditions mentioned in the previous paragraph are
satisfied, we say that the curve $X$ is {\em ordinary\/}.
\end{defn}
\par
%
%
%
\section{The sheaf of locally exact differentials has a theta divisor}
\par
In this section we prove the following result due to
M. Raynaud (\cite{RaynaudOld}):
\begin{thm}
If $X$ is a smooth projective connected curve over an algebraically
closed field $k$ of characteristic $p>0$ and $B_1$ is the sheaf of
locally exact differentials on $X_1$, as introduced above, then we
have $h^0(X_1,B_1\otimes L_\gen)=0$, i.e. the vector bundle $B_1$
admits a theta divisor (in particular, it is semi-stable).
\end{thm}
\par
Thus we can state the fact that a curve is ordinary simply by saying
that the theta divisor of $B_1$ does not go through the origin.
\par
To start with, introduce the jacobians $J$ and $J_1$ of $X$ and $X_1$
respectively. Then $J_1$ is the Frobenius image of $J$, and we have a
relative
Frobenius morphism $F\colon J\to J_1$ that is purely inseparable of
degree $p^g$; it corresponds to taking the norm on invertible sheaves
of degree $0$ --- or, if we prefer using points, it takes
${\cal O}_X(\Sigma n_i x_i)$ to ${\cal O}_{X_1}(\Sigma n_i\pi(x_i))$.
On the other hand, we also have the {\em Verschiebung\/} morphism in
the other direction $V\colon J_1\to J$, which corresponds to pulling
back by $\pi$ --- or again, it takes ${\cal O}_{X_1}(\Sigma n_i x_i)$
to ${\cal O}_X(\Sigma pn_i \pi^{-1}(x_i))$. The Verschiebung map also
has degree $p^g$. The composite of the Verschiebung and Frobenius
morphisms, in any direction, is the raising to the $p$-th power.
\par
We will show something more precise than just saying that $B_1$ has a
theta divisor: we will actually show that this theta divisor does not
contain all of $\ker V$ in the neighborhood of $0$. However, we will
see from actual equations that it ``almost'' does.
\par
If $L_1$ is an invertible sheaf of degree $0$ on $X_1$ (that is, a
$k$-point of $J_1$), the short exact sequence (\ref{ShortExact})
becomes, after tensoring by $L_1$:
\begin{equation}
0\to L_1\to\pi_*\pi^* L_1\to B_1\otimes L_1\to 0
\label{ShortExactTwisted}
\end{equation}
Now let $\mathscr{L}_1$ be the Poincaré bundle on
$X_1\times_{\Spec k} J_1$. The universal analogue
of (\ref{ShortExactTwisted}) above is
$$
0\to\mathscr{L}_1\to
(\pi\times 1_{J_1})_*(\pi\times 1_{J_1})^*\mathscr{L}_1\to
B_1\otimes\mathscr{L}_1\to 0
$$
But by the definition of the Verschiebung, the sheaf
$(\pi\times 1_{J_1})^*\mathscr{L}_1$ is also
$(1_X\times V)^*\mathscr{L}$ so that the exact sequence can be written
as
$$
0\to\mathscr{L}_1\to
(\pi\times 1_{J_1})_*(1_X\times V)^*\mathscr{L}\to
B_1\otimes\mathscr{L}_1\to 0
$$
\par
We now introduce projections as designated on the following diagram:
\begin{equation}
\xymatrix{
X\times J\ar[d]_f&%
X\times J_1\ar[l]_{1_X\times V}%
\ar[r]^{\pi\times 1_{J_1}}\ar[d]_g\ar@{}[dl]|{\square}%
&X_1\times J_1\ar[dl]^{f_1}\cr
J&J_1\ar[l]_V&
}
\label{Diagram}
\end{equation}
\par
Now we want to calculate the $R(f_1)_*$ of this. For one thing,
looking at the diagram (\ref{Diagram}) above, we see that
$R(f_1)_*(\pi\times 1_{J_1})_*(1_X\times V)^*\mathscr{L}$ is
$Rg_*(1_X\times V)^*\mathscr{L}$, and by base change (note that the
morphism $V$ is flat), this is $V^* Rf_*\mathscr{L}$. Thus we have
the following distinguished triangle, in the derived category of the
category of sheaves on $J_1$:
\begin{equation}
\xymatrix@=.8em{
&R(f_1)_*(B_1\otimes\mathscr{L}_1)\ar[ddl]_{\relax +1}&\cr
&&\cr
R(f_1)_*\mathscr{L}_1\ar[rr]^a&&V^* Rf_* \mathscr{L}\ar[luu]
}
\label{Triangle}
\end{equation}
And the corresponding long exact sequence of cohomology is
$$
0\to(f_1)_*(B_1\otimes\mathscr{L}_1)\to
R^1(f_1)_*\mathscr{L}_1\buildrel a\over\to V^*R^1f_*\mathscr{L}\to
R^1(f_1)_*(B_1\otimes\mathscr{L}_1)\to 0
$$
(the two first terms cancel). We want to show that $a$ is generically
invertible.
\par
To start with, consider a minimal resolution of $Rf_*\mathscr{L}$ in
the neighborhood of the origin. It has the form
$$
{\cal O}_{J,0}\buildrel u'\over\to{\cal O}_{J,0}^g
$$
where $u'(1)=(x_1,\ldots,x_g)$ is a system of parameters around $0$.
Indeed, this last statement is the same as saying, if $u$ is the
transpose of $u'$, that the image of $u$ is the maximal ideal
of the regular local ring ${\cal O}_{J,0}$, and this is easy because
$\{0\}$ is the largest closed subscheme of $\Spec{\cal O}_{J,0}$ on
which $\mathscr{L}$ is trivial.
\par
Now apply what we have just proven to $J_1$ on the one hand, and to
$J$ on the other, but pulling back by $V$, we find the following
resolution for the arrow $a$ in triangle (\ref{Triangle}):
\par
\begin{equation}
\xymatrix{
R\ar[r]^{a_0}\ar[d]_{u'}&R\ar[d]^{v'}\cr
R^g\ar[r]^{a_1}&R^g
}
\label{ResolOfA}
\end{equation}
where we have written $R={\cal O}_{J_1,0}$, and where
$u'(1)=(x_1,\ldots,x_g)$ is a system of parameters of $J_1$ around $0$
and $v'(1)=(y_1,\ldots,y_g)$ is a regular sequence that gives an
equation of $\ker V$ around $0$. Now of course $a_0$ is just an
element of $R$, and it is invertible because modulo the maximal ideal
of $R$ (that is, {\em at\/} the origin) the arrow $a$ is just the
identity on $k$. So we can assume that $a_0$ is the identity. What
we want to prove is that $\det a_1$ is not zero (of course, it is
invertible precisely when the curve is ordinary).
\par
Consider the diagram (\ref{ResolOfA}) and its transpose (i.e. its
image by the functor $\Hom_R(\cdot,R)$), and complete them both by
adding the Koszul complex on either column. That is, consider the
diagrams:
$$
\xymatrix{
0\ar[d]&0\ar[d]\cr
R\ar[r]^{1}\ar[d]_{u'}&R\ar[d]^{v'}\cr
R^g\ar[r]^{a_1}\ar@{.}[d]&R^g\ar@{.}[d]\cr
{\wedge^g R^g}\ar[r]^{\det a_1}\ar[d]&{\wedge^g R^g}\ar[d]\cr
M'\ar[r]^{h'}\ar[d]&N'\ar[d]\cr
0&0\cr
}
\hbox{ and }
\xymatrix{
0\ar[d]&0\ar[d]\cr
{\wedge^g R^g}\ar[r]^{\det a_1}\ar@{.}[d]&{\wedge^g R^g}\ar@{.}[d]\cr
R^g\ar[r]^{a_1^\vee}\ar[d]_{v}&R^g\ar[d]^{u}\cr
R\ar[r]^{1}\ar[d]&R\ar[d]\cr
N\ar[r]^{h}\ar[d]&M\ar[d]\cr
0&0\cr
}
$$
in which we have written $u$, $v$ and $a_1^\vee$ for the transposes of
$u'$, $v'$ and $a_1$ respectively and $M$ and $N$ for the cokernels of
$u$ and $v$ respectively. Since $u'(1)$ and $v'(1)$ are regular
sequences, $M$ and $N$ are modules of finite length, and the Koszul
complex is a resolution of them: the columns of both diagram are
exact. We have $M'=\Ext^g_R(M,R)$ and $N'=\Ext^g_R(N,R)$ (since we
have taken a resolution, transposed it, and shifted in $g$ degrees).
But since the Koszul complex is autodual (that is, the left column of
the right diagram is the same as the right column of the left diagram,
and {\it vice versa\/}), $M$ and $M'$ are the same and so are $N$ and
$N'$. Finally, it is known that ($R$ being a regular local ring) the
functor $\Ext^g_R(\cdot,R)$ is dualizing on modules of finite length.
Now $h$ is surjective as is seen on the diagram on the right, so that
its image $h'$ by the functor in question is injective. Hence $\det
a_1$ is nonzero, what we wanted.
\par
We can be more precise than this. As we have seen, $M$ is isomorphic
to $k$, and $N$ to the local ring of $\ker V$ at $0$: the support of
$\theta_{B_1}$ swallows everything in $\ker V$ around the origin but
just one $k$. (Incidentally, $X$ is ordinary if and only if the
support of $\theta_{B_1}$ does not contain the origin, so we recover
the known fact that $X$ is ordinary if and only if the local ring of
$\ker V$ at the origin is $k$, i.e. $V$ is étale.)
\par
%
%
%
\section{Constructing a non ordinary cover}
\par
We now present another result of M. Raynaud's (\cite{RaynaudNew}),
namely the fact that a finite étale cover of an ordinary curve is not
necessarily ordinary, even when the base curve is generic.
In fact, we obtain a cover $Y\twoheadrightarrow
X$ such that the image of the map $J(X_1)\rightarrowtail J(Y_1)$ on
the jacobians is completely contained in the support of the theta
divisor of $B_1$ on $Y_1$ --- and in particular $0$ is, so that $Y$ is
not ordinary. The construction is sufficiently general to apply to
the generic curve (for a given genus $g\geq 2$ and characteristic
$p$). We also get estimations on the Galois group of $Y$ over $X$;
a theorem of Nakajima states that an abelian cover of the generic
curve is ordinary, so we have to work with non abelian groups if we
want a non ordinary cover --- however, we will see that a nilpotent
group can suffice.
\par
We start with a few generalities on representations of the fundamental
group of curves. We refer to \cite{RaynaudNew} for details. If
$\rho\colon\pi_1(X)\to GL(r,k)$ is a representation of the fundamental
group of $X$ in $k$-vector spaces of rank $r$, and $\rho$ has open
kernel (or, which amounts to the same, $\rho$ is continuous and has
finite image), then $\rho$ defines a locally constant étale sheaf in
$k$-vector spaces of rank $r$ on $X$, written $\mathbb{V}_\rho$ (very
succintly, $\mathbb{V}_\rho$ can be obtained as follows: find a Galois
cover $Y\twoheadrightarrow X$ whose Galois group factors through the
kernel of $\rho$, then make $\pi_1(X)/\ker\rho$ act on $Y\times k^r$
componentwise, and take the fixed points of that action). Tensoring
$\mathbb{V}_\rho$ by ${\cal O}_X$ gives a Zarisky sheaf $V_\rho$ which
is locally free of rank $r$ and has degree $0$, i.e. a vector bundle
of rank $r$ and slope $0$ on $X$. Among the functoriality properties
of $V_\rho$ cited in \cite{RaynaudNew}, we will need the fact that if
$Y\buildrel a\over\twoheadrightarrow X$ is finite étale and $\rho$ is
a representation of $\pi_1(Y)$ as above then $a_* V_\rho$ is precisely
$V_{\rho'}$, where $\rho'$ is the representation of $\pi_1(X)$ induced
by $\rho$.
\par
We say that a representation $\rho$ as above has a theta divisor
(respectively, is ordinary) if and only if the sheaf
$V_{1,\rho}\otimes B_1$ on $X_1$ has a theta divisor (respectively,
has a theta divisor that does not go through the origin), $V_{1,\rho}$
being the bundle $V_\rho$ as above constructed on $X_1$. Thus, we
have seen that the trivial representation has a theta divisor, and it
is ordinary precisely when the curve $X$ is ordinary. The existence
of a theta divisor for $B$ shows that if $L$ is a general invertible
sheaf of finite order $n$ prime to $p$ then the representation $\rho$
of rank $1$ associated to it is ordinary.
\par
\begin{thm}
Let $k$ be an algebraically closed field of characteristic $p>0$, and
let $X$ be the generic curve of any genus $g\geq 2$ over $\Spec k$.
Then there exists a Galois cover of $X$ with solvable Galois group
of order prime to $p$ that is not ordinary.
\end{thm}
\par
Let $X$ be as stated, and let $J$ be its jacobian. If we choose a
base point on $X$ then we get a map $S^{g-1}X\to J$ from the
$(g-1)$-th symmetric power of $X$, whose image defines a positive
divisor on $J$, called the {\em classical theta divisor\/}, and
written $\Theta$. Let $N={\cal O}_J(\Theta)$ be the invertible sheaf
defined by $\Theta$. We can assume that $N$ is symmetric (i.e. that
if $\iota\colon J\to J$ is the inverse map then $\iota^* N=N$) and we
will do so.
\par
Let $n$ be a positive integer that is prime to $p$, and denote by
$\alpha$ the multiplication by $n$ map on $J$, which is étale of
degree $n^{2g}$. Call $A$ the kernel of $\alpha$, the (étale) set of
points of $J$ whose order divides $n$. Because we have chosen $N$
symmetric, we have $\alpha^*N=N^{\otimes n^2}$.
\par
We recall (cf. \cite{MumfordEquations}) that the kernel
$H(N^{\otimes n})$ is the subgroup of closed $x$ in $J$ such that
$T_x^*N^{\otimes n}\cong N^{\otimes n}$ (where $T_x$ denotes
translation by $x$). This kernel is obviously $A$. Now in
$\cite{MumfordEquations}$, D. Mumford defines another, more
interesting, group associated to an invertible sheaf on an abelian
variety. In our case, it is the group
$$
\mathscr{G}(N^{\otimes n})=\{(x,\varphi)\mid
\varphi\colon N^{\otimes n}\tilde\to
T_x^* N^{\otimes n}\}
$$
with multiplication defined in the obvious way. There is a short
exact sequence
$$
1\to k^\times\to\mathscr{G}(N^{\otimes n})\to H(N^{\otimes n})\to 1
$$
and in fact $k^\times$ is precisely the center of
$\mathscr{G}(N^{\otimes n})$. The commutator of two elements of
$\mathscr{G}(N^{\otimes n})$ is an element of $k^\times$ and it
depends only on the class in $H(N^{\otimes n})$ of the two elements.
Thus, the commutator defines a skew-symmetric biadditive form
$\langle\cdot,\cdot\rangle\colon A\times A\to k^\times$. It is
moreover shown in \cite{MumfordEquations} that this form is non
degenerate.
\par
Let $B$ a maximal totally isotropic subgroup of $A$ for the form we
have just defined. So $B$ has order $n^g$, and $C=A/B$ has order
$n^g$. We factor $\alpha$ as follows
$$
\xymatrix{
J\ar[dr]^{\beta}\ar[dd]_{\alpha}&\cr
&J'=J/B\ar[dl]^{\gamma}\cr
J&
}
$$
where $\beta$ has kernel $B$ and $\gamma$ has kernel (identified with)
$C$.
\par
Because $B$ is isotropic, by a result in \cite{MumfordEquations}, the
sheaf $N^{\otimes n}$ descends to an invertible sheaf $M$ on $J'$, i.e. a sheaf
such that $N^{\otimes n}=\beta^* M$. And $M$ is a principal
polarization on $J'$. Now note that $\gamma^* N$ and $M^{\otimes n}$
have the same pullback (namely $N^{\otimes n^2}$) by $\beta$; if $n$
is odd we can choose $M$ to be symmetric, so that $\gamma^* N$ and
$M^{\otimes n}$ coincide. We will now suppose this to be the case.
\par
If $L$ is an invertible sheaf on $J$ that is algebraically equivalent
to $0$ (that is, a closed point of $J^\vee$) then we have
$\gamma_*(M\otimes\gamma^*L)\cong(\gamma_*M)\otimes L$, so that
$h^0(J',M\otimes\gamma^*L)=h^0(J,(\gamma_*M)\otimes L)$. Now the
point is that $M\otimes\gamma^*L$ is a principal polarization on $J$,
so this number is $1$. In particular $h^0(J,(\gamma_*M)\otimes L)>0$,
and this implies that for any invertible sheaf $L$ of degree $0$ on
$X$ we have $h^0(X,F\otimes L)>0$, where $F$ is the restriction of
$\gamma_*M$ to $X$. This is a good first step, but we need to twist
$F$ by an invertible sheaf having the right degree to compensate for
the slope of $F$ (since the sheaves $V_\rho$ have slope zero).
\par
We now calculate the slope of $F$. Its rank is $n^g$. Introduce the
curves $Y$ and $Z$ that are inverse image of $X$ by $\gamma$ and
$\alpha$ respectively, thus:
$$
\xymatrix{
Z\ar[dr]^{\beta}\ar[dd]_{\alpha}&\cr
&Y\ar[dl]^{\gamma}\cr
X&
}
$$
The degree of $N$ restricted to $X$ is well-known: it is $g$. Pulling
this back by $\alpha$, we see that the degree of $N^{\otimes n^2}$
restricted to $Z$ is $gn^{2g}$, and that of $N^{\otimes n}|Z$ is
$gn^{2g-1}$. Descending to $Y$, we see that the degree of $M|Y$ is
$gn^{g-1}$. So the slope of $F$ is finally $g/n$.
\par
Now assume that $g$ divides $n$, i.e. that $g/n=d$, the slope of $F$,
is an integer. The degree of $N|X$ is $g=nd$, so there exists an
invertible sheaf $P$ of degree $d$ on $X$ such that
$N|X=P^{\otimes n}$. Let $L'=(M|Y)\otimes\gamma^* P^{\otimes-1}$,
which is an invertible sheaf of degree zero. Its inverse image
$L''=\beta^*L'$ is such that $L^{\prime\prime\otimes n}$ is trivial on
$Z$, so that the order of $L''$ divides $n$ (in fact, it is exactly
$n$, but we won't need this). If $E=\gamma_*L'$ then $E=F\otimes
P^{\otimes-1}$, which is an invertible sheaf of degree $0$ on $X$ and
satisfies $h^0(X,E\otimes L_{d,\gen})>0$ for a general invertible
sheaf $L_{d,\gen}$ of degree $d$ on $X$.
\par
Now $L''$ is of order dividing $n$, so there is a cyclic covering of
degree $n$ $Z''\twoheadrightarrow Z$ which trivializes it. It is
$Z''$ that we will prove not to be ordinary (under certain numerical
conditions at least). The invertible sheaf $L'$ of degree $0$
corresponds to an abelian representation of $\pi_1(Y)$ that factors
trough $\pi_1(Z'')$, and when we induce that representation to
$\pi_1(X)$ we see that $E=\gamma_*L'$ is of the form $V_\rho$ for some
representation $\rho$ of $\pi_1(X)$ that factors through $\pi_1(Z'')$
(and which can actually be described: see \cite{RaynaudNew}).
\par
All these constructions were performed on $X$ and $J$. They could
equally well have been performed on $X_1$ and $J_1$. We now consider
$E$ as a sheaf of $X_1$. We have seen $h^0(X,E\otimes L_{d,\gen})>0$
and we wish to have $h^0(X,E\otimes B_1\otimes L_\gen)>0$. We are
therefore done if we can show that $B_1$ contains an invertible
subsheaf of degree $d$.
\par
But A. Hirschowitz claims in \cite{Hirschowitz} and proves in
\cite{HirschowitzWWW} that a general
bundle of rank $r_0$ and slope $\lambda_0$ contains a subbundle of
rank $r'$ and slope $\lambda'$ (the quotient having rank $r''=r_0-r'$
and slope $\lambda''$) if $\lambda''-\lambda'\geq g-1$. If we are
looking for $r'=1$ and $\lambda'=d$, with $r_0=p-1$ and
$\lambda_0=g-1$ (the numerical values of $B_1$), so $r''=p-2$ and
$\lambda''=[(g-1)(p-1)-d]/(p-2)$, this condition is satisfied iff
$(g-1-d)(p-1)/(p-2)\geq g-1$, that is iff $d\leq\frac{g-1}{p-1}$. By
deforming and specializing to $B_1$, we see that if this inequality is
satisfied then $B_1$ contains an invertible sheaf of degree $d$.
\par
Finally, we have shown that if $p$ and $g$ are such that there exists
a positive odd integer $n$, prime to $p$, dividing $g$, and satisfying
$\frac{g}{n}\leq\frac{g-1}{p-1}$ then the generic curve $X$ of genus
$g$ in characteristic $p$ has a covering that is not ordinary. This
is not always the case, but we can always reduce to that case by first
taking a cyclic cover $X'$ of degree $m$ prime to $p$ of $X$ ($X'$
then has genus $g'=1+m(g-1)$), and apply the result to $X'$. Here are
the details:
\begin{itemize}
\item
If $p$ is odd, take $m$ even, not multiple of $p$ and large enough so
that $g'=1+m(g-1)\geq p$.
\begin{itemize}
\item
If $p$ does not divide $g'$, then $n=g'$ works (it is odd because $m$
is even, it is prime to $p$, and $\frac{g'}{n}=1\leq\frac{g'-1}{p-1}$
because $g'\geq p$).
\item
If $p$ does divide $g'$ then we double $m$ and this is no longer the
case, so we are reduced to the previous point.
\end{itemize}
\item
If $p=2$, write $g=2^r s$ with $s$ odd.
\begin{itemize}
\item
If $s\geq 3$, take $m=1$, $n=s$. (Then $n$ is odd, and
$\frac{g}{n}=2^r\leq g-1$.)
\item
If $s=1$ then $g=2^r$.
\begin{itemize}
\item
If $r\geq 2$, take $m=3$, $n=g'/2=3\times 2^{r-1}-1$. (Then $n$ is
odd, and $\frac{g'}{n}=2\leq g'-1$.)
\item
If $r=1$ so $g=2$ and we take $m=5$, $g'=6$, $n=3$.
\end{itemize}
\end{itemize}
\end{itemize}
\par
Finally, we note that our final covering was constructed as a
composite $Z''\twoheadrightarrow Y\twoheadrightarrow
X'\twoheadrightarrow X$ of coverings all of which are abelian: it is
therefore solvable.
\par
%
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\begin{thebibliography}{RaynaudOld}
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\bibitem{RaynaudOld}
M. Raynaud, ``Sections des fibrés vectoriels sur une courbe'',
{\it Bull. Soc. math. France\/},
{\bf 110} (1982), p. 103--125.
%
\bibitem{RaynaudNew}
M. Raynaud, ``Revêtements des courbes en caractéristique $p>0$
et ordinarité'', forthcoming.
%
\bibitem{MumfordEquations}
D. Mumford, ``On the Equations Defining Abelian Varieties, I'',
{\it Inventiones mathematicæ\/},
{\bf 1} (1966), p. 287--354.
%
\bibitem{Hirschowitz}
A. Hirschowitz, ``Problèmes de Brill-Noether en rang supérieur'',
{\it C. R. Acad. Sci\/},
{\bf 307} (1988), p. 153--156.
%
\bibitem{HirschowitzWWW}
A. Hirschowitz, ``Problèmes de Brill-Noether en rang supérieur'',
{\chardef\asciitilde="7E % I HATE THIS
{\tt http://math.unice.fr/\asciitilde ah/Brill/},}
on the World Wide Web.
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\end{thebibliography}
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{\par\vskip.3truein\relax
\leftline{\hskip2truein\relax David Madore}
\leftline{\hskip2truein\relax c/o École Normale Supérieure de Paris}
\leftline{\hskip2truein\relax 45 rue d'Ulm}
\leftline{\hskip2truein\relax 75\thinspace 230 PARIS CEDEX 05}
\leftline{\hskip2truein\relax FRANCE}
\leftline{\hskip2truein\relax\tt david.madore@ens.fr}
\relax}
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\end{document}
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