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1998-05-11T05:39:17 | 9708 | alg-geom/9708002 | en | https://arxiv.org/abs/alg-geom/9708002 | [
"alg-geom",
"math.AG"
] | alg-geom/9708002 | James A. Carlson | James A. Carlson and Domingo Toledo | Discriminant Complements and Kernels of Monodromy Representations | 20 page dvi file available at
http://www.math.utah.edu/~carlson/eprints.html Minor changes for final
version to appear in Duke J. Math | null | null | null | null | We show that the kernel of the monodromy representation for hypersurfaces of
degree d and dimension n is large for d at least three with the exception of
the cases (d,n) = (3,0) and (3,1). For these the kernel is finite. By "large"
we mean a group that admits a homomorphism to a semisimple Lie group of
noncompact type with Zariski-dense image. By the Tits alternative a large group
contains a free subgroup of rank two.
| [
{
"version": "v1",
"created": "Fri, 1 Aug 1997 23:18:27 GMT"
},
{
"version": "v2",
"created": "Fri, 13 Feb 1998 16:48:02 GMT"
},
{
"version": "v3",
"created": "Mon, 11 May 1998 03:39:15 GMT"
}
] | 2008-02-03T00:00:00 | [
[
"Carlson",
"James A.",
""
],
[
"Toledo",
"Domingo",
""
]
] | alg-geom | \section{Introduction}
\secref{introsection}
A hypersurface of degree $d$ in a complex projective space
$\P^{n+1}$ is defined by an equation of the form
$$
F(x) = \sum a_L x^L = 0,
\eqn
\eqref{universalhypersurface}
$$
where $x^L = x_0^{L_0} \cdots x_{n+1}^{L_{n+1}}$ is a monomial of degree
$d$ and where the $a_L$ are arbitrary complex numbers, not all zero. Viewed as
an equation in both the $a$'s and the $x$'s, \eqrefer{universalhypersurface}
defines a hypersurface ${\bf X}$ in $\P^N\times\P^{n+1}$, where $N+1$ is the
dimension of the space of homogeneous polynomials of degree $d$ in $n+2$
variables, and where the projection
$p$ onto the first factor makes ${\bf X}$ into a family with fibers
$X_a = p^{-1}(a)$. This is the universal family of hypersurfaces of degree $d$
and dimension
$n$. Let $\Delta$ be the set of points $a$ in $\P^N$ such that the
corresponding fiber is singular. This is the {\sl discriminant locus}; it
is well-known to be irreducible and of codimension one. Our aim is to study
the fundamental group of its complement, which we write as
$$
\Phi = \pi_1(\P^N - \Delta).
$$
When we need to make precise statements we will sometimes write
$
\Phi_{d,n} = \pi_1(U_{d,n}, o),
$
where $d$ and $n$ are as above, $U_{d,n} = \P^N - \Delta$, and $o$ is a base
point.
The groups $\Phi$ are almost always nontrivial and in fact are almost
always {\sl large}. By this we mean that there is a homomorphism of $\Phi$
to a non-compact semi-simple real algebraic group which has Zariski-dense
image. Large groups are infinite, and, moreover, always contain a free group of
rank two. This follows from the Tits alternative
\cite{Tits}, which states that in characteristic zero a linear group either
has a solvable subgroup of finite index or contains a free group of rank
two.
To show that $\Phi = \Phi_{d,n}$ is large we consider the image
$\Gamma = \Gamma_{d,n}$ of the
monodromy representation
$$
\rho: \Phi \map G.
\eqn
\eqref{monodrep}
$$
Here and throughout this paper $G = G_{d,n}$ denotes the group of
automorphisms of the primitive cohomology $H^n(X_o,\R)_o$ which preserve the
cup product. When $n$ is odd the primitive cohomology is the
same as the cohomology, and when $n$ is even it is the orthogonal complement
of $h^{n/2}$, where $h$ is the hyperplane class. Thus $G$ is either a
symplectic or an orthogonal group, depending on the parity of $n$, and is an
almost simple real algebraic group.
About the image of the monodromy representation, much is known. Using
results of Ebeling \cite{Ebeling} and Janssen \cite{Janssen}, Beauville in \cite{BeauvilleLattice}
established the following:
\proclaim{Theorem}. Let $G_\Z$ be the subgroup of $G$ which preserves the
integral cohomology. Then the monodromy group $\Gamma_{d,n}$ is of finite index in
$G_\Z$. Thus it is an arithmetic subgroup.
\endproclaim
\procref{Beauvillethm}
\noindent
The result in \cite{BeauvilleLattice} is much more
precise: it identifies $\Gamma$ as a specific subgroup of finite
(and small) index in $G_\Z$. Now suppose that $d > 2$ and that $(d,n) \ne (3,2)$.
Then $G$ is noncompact, and the results of Borel \cite{BorelDT} and
Borel-Harish-Chandra \cite{BHC}
apply to show that
$\Gamma$ is (a) Zariski-dense and (b) a lattice. Thus (a) the smallest algebraic
subgroup of $G$ which contains $\Gamma$ is $G$ itself and (b) $G/\Gamma$ has
finite volume.
Consider now the kernel of the monodromy representation, which we denote by
$K$ and which fits in the exact sequence
$$
1 \map K \map \Phi \mapright{\rho} \Gamma \map 1.
\eqn
\eqref{KPhiGamma}
$$
The purpose of this paper is to show that in almost
all cases it is also large:
\proclaim{Theorem.} The kernel of the monodromy
representation \eqrefer{monodrep} is large if $d > 2$ and $(d,n) \ne (3,1),\
(3,0)$.
\endproclaim
\procref{maintheorem}
The theorem is sharp in the sense that the remaining groups are finite.
When $d = 2$, the case of quadrics, $\Phi$ is finite cyclic.
When $(d,n) = (3,0)$, the configuration space $U$ parametrizes
unordered sets of three distinct points in the projective line and so $\Phi$
is the braid group for three strands in the sphere. It has order
12 and can be faithfully represented by
symmetries of a regular hexagon.
When $(d,n) = (3,1)$ the configuration space $U$ parametrizes smooth cubic
plane curves and the above sequence can be written as
$$
1 \map K \map \Phi_{3,1} \mapright{\rho} SL(2,\Z) \map 1,
$$
where $K$ is the three-dimensional Heisenberg group over the field $\Z/3$, a
finite group of order 27. Moreover, $\Phi_{3,1}$ is a semi-direct product,
where $SL(2,\Z )$ acts on $K$ in the natural way. This result, due
to Dolgachev and Libgober \cite{DolgLib}, is to our knowledge the only one
which determines
the exact sequence \eqrefer{KPhiGamma} for hypersurfaces of positive
dimension and degree larger than two.
Note that in this case $\Phi$ is large but $K$ is finite.
Note also that there are two kinds of groups for which the natural monodromy
representation has finite image but large kernel. These are the braid
groups $\Phi_{d,0}$ for $d > 3$ and the group $\Phi_{3,2}$ for the space of
cubic surfaces. Thus all of them are large. For the braid groups
this result is classical, but for $\Phi_{3,2}$ it is new. Since $\Phi_{3,2}$
is large it is infinite, a fact which answers a question left open
by Libgober in \cite{Lib}.
Concerning the proof of Theorem \xref{maintheorem}, we would like to say first of
all that it depends, like anything else in this subject, on the Picard-Lefschetz
formulas. We illustrate their importance by sketching how they imply
the non-triviality of the monodromy representation
\eqrefer{monodrep}. Consider a smooth point
$c$ of the discriminant locus. For these $X_c$ has a
exactly one node: an isolated singularity defined in suitable local coordinates
by a nondegenerate sum of squares. Consider also a loop $\gamma = \gamma_c$
defined by following a path $\alpha$ from the base point to the edge of a
complex disk normal to $\Delta$ and centered at $c$, traveling once around the
circle bounding this disk, and then returning to the base point along
$\alpha$ reversed. By analogy with the case of knots, we
call these loops (and also their homotopy classes) the {\sl meridians} of
$\Delta$. Then $T = \rho(\gamma)$ is a {\sl Picard-Lefschetz} transformation,
given by the formula
$$
T(x) = x \pm (x,\delta) \delta .
\eqn
\eqref{plformula}
$$
Here $(x,y)$ is the cup product and $\delta$ is the {\sl vanishing cycle}
associated to $\gamma$. When $n$ is odd, $(\delta,\delta) = 0$ and the sign
in \eqrefer{plformula} is $-$. When $n$ is even and $(\delta,\delta) = \pm
2$, the sign in \eqrefer{plformula} is $\mp$ (see \cite{DeWeOne}, paragraph
4.1). Thus when $n$ is even $\delta$ is automatically nonhomologous to zero, and so
$T$ must be nontrivial. Since vanishing cycles exist whenever the hypersurface
$X_o$ can degenerate to a variety with a node, we conclude that $\rho$ is
nontrivial for $n$ even and $d > 1$. Slightly less elementary arguments show that
the homology class of the vanishing cycle, and hence the monodromy representation,
is nontrivial for all $d > 1$ except for the case $(d,n) = (2,1)$.
The proofs of theorem \xref{Beauvillethm}, an earlier result of
Deligne asserting the Zariski density of $\Gamma_{d,n}$, and the main result of
this paper are based on the Picard-Lefschetz formulas
\eqrefer{plformula}. Our proof begins with the construction of a universal family
of cyclic covers of $\P^{n+1}$ branched along the hypersurfaces
$X$. From it we define a second monodromy representation $\bar \rho'$ of
$\Phi$. Suitable versions of the Picard-Lefschetz formulas and
Deligne's theorem apply to show that $\bar \rho'$ has Zariski-dense
image. Finally, we apply Margulis' super-rigidity theorem to show that
$\bar\rho'(K)$, where $K$ is the kernel of the natural monodromy representation, is
Zariski-dense. Thus $K$ is large.
We mention the paper
\cite{Mag} as an example of the use of an associated family of
cyclic covers to
construct representations (in this case for the braid
groups of the sphere). We also note the related results of the
article \cite{DOZ} which we learned of while preparing the final version of this
manuscript. The main theorem is that the complement of the dual $\widehat
C$ of an immersed curve $C$ of genus at least one, or of an immersed rational
curve of degee at least four, is {\sl big} in the sense that it contains a free
group of rank two. When $C$ is smooth, imbedded, and of even degree at least
four this follows from a construction of Griffiths \cite{GriffHyperbolic}:
consider the family of hyperelliptic curves obtained as double covers of a line
$L$ not tangent to $C$ which is branched at the points $L\cap C$. It defines
a monodromy representation of $\Phi = \pi_1(\widehat{\P}^2 - \widehat{C})$ with
Zariski-dense image. Consequently $\Phi$ is large, and, {\sl a fortiori},
big. Such constructions have inspired the present paper. By using cyclic
covers of higher degree one can treat the case of odd degree greater than
four in the same way.
The authors would like to thank Herb Clemens and Carlos Simpson for
very helpful discussions.
\section{Outline of the proof}
\secref{outlinesection}
As noted above, the proof of the main theorem is based on the construction of an
auxiliary representation $\rho'$ defined via a family of cyclic covers $Y$ of
$\P^{n+1}$ branched along the hypersurfaces $X$. To describe it, let
$k$ be a divisor of $d$ and consider the equation
$$
F(a,x) = y^k + \sum a_L x^L = 0,
\eqn
\eqref{universalcyclic}
$$
which for the moment we view as defining
a set $\widehat {\bf Y}$ in $(\C^{N+1} - \set{0})\times \C^{n+3}$ with coordinates
$a_L$ for $\C^{N+1}$ and coordinates $x_0,\cdots ,x_{n+2}$ and $y$ for
$\C^{n+3}$. Construct an action of
$\C^*$ on it by multiplying the coordinates $x_i$ by $t$ and by multiplying
$y$ by $t^{d/k}$. View the quotient ${\bf Y}$ in
$(\C^{N+1} - \set{0})\times \P^{n+2}$, where we use $\P^{n+2}$ to denote
the weighted projective space for which the $x_i$ have weight one and for which $y$
has weight $d/k$.
The resulting universal family of cyclic covers ${\bf Y}$ is defined on
$\C^{N+1} - \set{0}$ and has smooth fibers over $\widetilde U = \C^{N+1} -
\widetilde
\Delta$, where $\widetilde \Delta$ is the pre-image of $\Delta$. Since $\C^{N+1} -
\set{0}$ is a principal
$\C^*$ bundle over $\P^N$, the same holds over $\widetilde U$ and $\widetilde
\Delta$. It follows that one has a central extension
$$
0 \map \Z \map \widetilde \Phi \map \Phi \map 1,
$$
where $\widetilde \Phi = \pi_1(\widetilde U)$. We introduce
$\widetilde U$ and $\widetilde\Phi$ purely for the technical reason that the
universal family of cyclic branched covers need not be defined over $U$ itself.
The family ${\bf Y} |\widetilde U$ has a monodromy representation which
we denote by
$\tilde\rho$ and which takes values in a real algebraic group $\widetilde G$ of
automorphisms of $H^{n+1}(Y_{\tilde o},\C )$ which commute with the cyclic
group of covering transformations (and which preserve the hyperplane class and
the cup product). Here $\tilde o$ is a base point in $\widetilde U$ which lies
above the previously chosen base point $o$ of $U$, and $Y_{\tilde o}$ denotes the
$k$-fold cyclic cover of $\P^{n+1}$ branched over $X_o$.
The group $\widetilde G$ is semisimple but in general has more than one
simple factor. Let $G'$ be one of these and let
$$
\rho ' :\widetilde\Phi\map G',
$$
denote the composition of $\tilde\rho$ with the projection to $G'$. Then
we must establish the following:
\proclaim{Technical point}. The factor $G'$ can be chosen to be a non-compact
almost simple real algebraic group. The image of
$\rho '$ is Zariski-dense in
$G'$.
\endproclaim
\procref{technicalpoint}
Suppose that this is true. Then we can argue as follows. First, the group
of matrices which commute with
$\rho'(\Z)$ contains a Zariski-dense group. Consequently $\rho'(\Z)$ lies in the
center of
$G'$. Therefore there is a quotient representation
$$
\bar \rho': \Phi\map \bar G',
$$
where $\bar G' $ is the adjoint group of $G'$ (that is, $G'$ modulo its center).
Moreover, the representation
$\bar\rho'$ also has Zariski-dense image.
Now consider our original representation \eqrefer{monodrep}. Replacing
$\Phi$ by a normal subgroup of finite index we may assume that the image of
$\rho$ lies in the identity component of $G$ in the analytic topology and
that the image of $\bar\rho '$ lies in the identity component of $\bar G'$
in the Zariski topology. Let $\bar G$ denote the identity component (in the
analytic topology) of $G$ modulo its center, and let
$\bar\rho :\Phi\map \bar
G$ denote the resulting representation. We still have that
$\bar\rho (\Phi)$ is a lattice in
$\bar G$ and that $\bar \rho '(\Phi)$ is Zariski-dense in $\bar G'$.
Now let $\bar K$ be the kernel of $\bar\rho$, and let $L$ be the
Zariski-closure of $\bar\rho'(\bar K)$. Since $\bar K$ is normal in $\Phi$
and $\bar\rho'(\Phi)$ is Zariski-dense in $\bar G'$, $L$ is normal in $\bar
G'$. Since $\bar G'$ is a {\sl simple} algebraic group, either $L = \bar G'$
or $L = \set{1}$. If the first of the two alternatives holds, then
$\bar\rho'(K)$ is Zariski dense, and so $K$ is large. This is because $K$ has
finite index in $\bar K$ and so $\bar\rho'(K)$ and $\bar\rho'(\bar K)$ have
the same Zariski closure.
We now show that the second alternative leads to a contradiction, from which it
follows that $K$ must be large. Indeed, if $\bar \rho'(\bar K) =
\set{1}$, then the expression $\bar \rho'\circ\bar \rho^{-1}$
defines a homomorphism from the lattice $\bar \rho(\Phi)$ in $\bar G$ to the
Zariski-dense subgroup $\bar \rho'(\Phi)$ in
$\bar G'$. If the real rank of $\bar G$ is at least two, the Margulis
rigidity theorem
\cite{MargulisRigidity}, \cite{Zimmer} Theorem 5.1.2, applies to give an
extension of $\bar \rho'$ to a homomorphism of $\bar G$ to
$\bar G'$. Since $\bar\rho'(\Phi)$ is Zariski-dense, the extension is surjective.
Since $\bar G$ is simple, it is an isomorphism. Thus the complexified lie algebras
$\g_\C, \g'_\C$ must be isomorphic. However, one easily shows that $\g_\C
\not\cong \g'_\C$, and this contradiction completes the proof.
We carry out the details separately in two cases. First, for the simpler case
where $d$ is even and $(d,n)\ne (4,1)$, we use double covers ($k = 2$). Then
$G'$ is the full group of automorphisms of the
primitive (or anti-invariant) part of $H^{n+1}(Y,\R)$ and so is again an
orthogonal or symplectic group. The technical point \xref{technicalpoint}
follows from a density result of Deligne that we recall in section
\xref{zardensitysection}. Deligne's result gives an alternative between
Zariski
density and finite image, and the possibility of finite image is excluded in
section
\xref{ratdiffsection}. Finally the Lie algebras $\g_\C$ and
$\g'_\C$ are not isomorphic, since when one of them is symplectic (type
$C_\ell$), the other is orthogonal (type $B_\ell$ or $D_\ell$). By lemma
\xref{rankboundslemma} the rank $\ell$ is at least three, so there are no
accidental isomorphisms, e.g., $B_2 \cong C_2$.
For the remaining cases, namely $d$ odd or $(d,n) = (4,1)$ we use $d$-fold
covers, i.e., $k = d$. For these we must identify the group $\widetilde G$
of automorphisms of $H^{n+1} (Y,\R)_0$ which preserve the cup product and
which commute with the cyclic automorphism $\sigma$. This is the natural
group in which the monodromy representation $\tilde \rho$ takes its values.
Now a linear map commutes with $\sigma$ if and only if it preserves the
eigenspace decomposition of $\sigma$, which we write as
$$
H^{n+1} (Y,\C)_0 = \bigoplus_{\mu \ne 1} H(\mu).
$$
As noted in \eqrefer{eigenspacedimform}, the dimension of $H(\mu)$
is independent of $\mu$. Now let $\widetilde G(\mu)$ be subgroup of $\widetilde G$ which acts by
the identity on $H(\lambda)$ for $\lambda \ne \mu,\; \bar\mu$. It can be viewed as a
group of transformations of $H(\mu) + H(\bar\mu)$. Thus there is a
decomposition
$$
\widetilde G
=
\prod_{ \mu \in S} \widetilde G(\mu),
\eqn
\eqref{tildeGdecomp}
$$
where
$$
S = \sett{ \mu }{ \mu^k =1,\ \mu \ne 1,\ \Im \mu \ge 0 }.
$$
When $\mu$ is non-real, $\widetilde G(\mu)$ can be identified
via the projection $H(\mu) \oplus H(\bar\mu) \map H(\mu)$
with the group of transformations of $H(\mu)$ which are unitary with respect
to the hermitian form $h(x,y) = i^{n+1}(x,\bar y)$, where $(x,y)$ is the cup
product. This form may be (and usually is) indefinite. When
$\mu = -1$, $\widetilde G(\mu)$ is the group of transformations of $H(-1)$
which preserve the cup product. It is therefore an orthogonal or symplectic group.
We will show that at least one of the components $\widetilde\rho_\mu(\Phi)\subset
\widetilde G(\mu)$ is Zariski-dense, and we will take $G' = \widetilde G(\mu)$.
The necessary Zariski density result, which is a straightforward adaptation of
Deligne's, is proved in section \xref{unitarydensitysection} after some preliminary
work on complex reflections in section \xref{complexreflectionsection}. Again,
the possibility of finite image has to be excluded, and the argument for this is
in section \xref{cycliccoversection}. Finally, to prove that
$\g_\C$ and $\g'_\C$ are not isomorphic one observes that
$\g_\C$ is of type $B_\ell, C_\ell$ or $D_\ell$ while $\g'_\C$ is of type
$A_\ell$ (since
$G'$ is of type $SU(r,s)$. One only needs to avoid the isomorphism $D_3\cong
A_3$, which follows from the lower bound of the rank of $\g_\C$ in lemma
\xref{rankboundslemma}.
In order to apply Margulis' theorem we also need to verify that the real
rank of $G$ is at least two. This is done in section \xref{rankbounds}.
To summarize, we have established the following general criterion, and our proof of
Theorem \xref{maintheorem} is an application of it.
\proclaim{Criterion.} The kernel $K$ of a linear representation $\rho: \Phi \map
G$ is large if
\list
\i $\rho(\Phi) \subset G$ is a lattice in a simple Lie group $G$ of
real rank at least two.
\i There exist a non-compact, almost simple real algebraic group $G'$, a central
extension $\widetilde\Phi$ of $\Phi$ and a linear
representation
$\rho':
\widetilde
\Phi
\map G'$ with Zariski-dense image.
\i $G$ and $G'$ are not locally isomorphic.
\endlist
\endproclaim
\procref{largekernelcriterion}
\noindent
An immediate consequence is the following:
\proclaim{Corollary.} Let $\Phi$ be a group which admits a representation
$\rho: \Phi \map G$
to a simple Lie group of real rank greater than 1 with image a lattice.
Suppose further that there exist an almost simple real algebraic group $G'$,
a central extension $\widetilde \Phi$ of $\Phi$, and a representation $\rho':
\widetilde\Phi \map G'$ with Zariski-dense image.
Suppose in addition that $G$ and $G'$ are not locally isomorphic. Then $\Phi$ is not
isomorphic to a lattice in any simple Lie group of real rank greater than 1.
\endproclaim
\noindent{\bf Proof:\ } Suppose that $\tau: \Phi \map \Sigma$ is an isomorphism of $\Phi$
with a lattice $\Sigma$ in a Lie group $H$ of real rank greater than one.
If $H$ is not locally isomorphic to $G'$, then apply the criterion with $\tau$ in
place of $\rho$ to conclude that $\tau$ has large kernel, hence cannot be an
isomorphism. Suppose next that $H$ is locally isomorphic to $G'$. Apply
the criterion with $\tau$ in place of $\rho$ and with $\rho$ in place of $\rho'$
to conclude as before that the kernel of $\tau$ is large.
For most families of hypersurfaces the natural monodromy representation and the
representation for the associated family of cyclic covers satisfy the hypotheses of
the corollary to give the following:
\proclaim{Theorem.} If $d > 2$, $n>0$, and $(d,n) \ne (3,1),\; (3,2)$, the group
$\Phi_{d,n}$ is not isomorphic to a lattice in a simple Lie group of real rank
greater than one.
\endproclaim
It seems reasonable that the preceding theorem holds with ``semisimple''
in place of ``simple.'' However, we are unable show that this is the case.
Indeed, our results so far are compatible with an isomorphism $\Phi \cong
\Gamma\times\Gamma'$. We can exclude this in certain cases (see section
\xref{remarkssection}), but not for an arbitrary subgroup of finite index, which
is what one expects.
\proclaim{Remarks.}
\procref{remarkMeridiansInfiniteOrder}
\rm
(a) Suppose that $d \ge 3$ and let $\gamma$ be a meridian of $\Phi_{d,n}$.
When $n$ is odd, $\rho(\gamma)$ is a nontrivial symplectic transvection. Since
it is of infinite order, so is the meridian $\gamma$. When $n$ is even,
$\rho(\gamma)$ is a reflection, hence of order two. Now suppose that $d$
is even and consider the monodromy representation of the central extension $\widetilde
\Phi$ constructed from double covers. Let $\tilde\gamma$ be a lift of
$\gamma$ to an element of $\widetilde\Phi$. Then
$\rho'(\tilde\gamma)$ is a nontrivial symplectic transvection, no power of which is central. Thus
$\bar\rho'(\gamma)$ is of infinite order, and, once again, we conclude that
$\gamma$ is of infinite order.
(b) M. Kontsevich informs us that he can prove that for any $d>2$ (and at
least for $n=2$) the local monodromy corresponding to a meridian is of
infinite order in the group of connected components of the symplectomorphism
group of $X_o$. This implies that the meridians are of infinite order for all
$d>2$, not necessarily even as above. The symplectic nature of the monodromy
for a meridian (for $n=2$) is studied in great detail by P. Seidel in his
thesis \cite{Seidel}.
(c) For the case of double covers the image $\Gamma'$ of the fundamental group
under the second monodromy representation $\rho'(\widetilde\Phi)$ is a
lattice. This follows from the argument given by Beauville to prove theorem
\xref{Beauvillethm}. It is enough to be able to degenerate the branch locus
$X$ to a variety which has an isolated singularity of the form $x^3 + y^3 + z^4
+ \hbox{a sum of squares} = 0$. Then the roles of the kernels $K$ and $K'$ are
symmetric and one concludes that $K'$ is also large.
\endproclaim
\section{Zariski Density}
\secref{zardensitysection}
The question of Zariski-density for monodromy groups of Lefschetz
pencils was settled by Deligne in \cite{DeWeOne} and \cite{DeWeTwo}.
We review these results here in a form convenient for the proof of the main
theorem in the case of even degree and also for the proof of a
density theorem for unitary groups (section
\xref{unitarydensitysection}). To begin, we have the following
purely group-theoretic fact: \cite{DeWeTwo}(4.4):
\proclaim{Theorem. (Deligne)} Let $V$ be a vector space (over $\C$)
with a non-degenerate bilinear form $(\ ,\ )$ which is either symmetric or
skew-symmetric. Let
$\Gamma$ be a group of linear transformations of $V$ which preserves the
bilinear form. Assume the existence of a subset $E\subset V$ such that
$\Gamma$ is generated by the Picard-Lefschetz transformations \eqrefer{plformula}
with $\delta\in E$.
Suppose that $E$ consists of a single $\Gamma$-orbit
and that it spans $V$. Then $\Gamma$ is either finite or Zariski-dense.
\endproclaim
\procref{delignedensity}
To apply this theorem in a geometric setting, consider a family of
$n$-dimensional varieties $p: {\bf X} \map S$ with discriminant locus $\Delta$ and
monodromy representation
$\rho:\pi_1(S - \Delta) \map \hbox{Aut}(H^n(X_o))$. Assume that $S$ is either
$\C^{N+1} - \{ 0\} , N\ge 1$ or $\P^N$, so that $S$ is simply connected and
hence that
$\pi_1 (S-\Delta)$ is generated by {\sl meridians} (cf. \S
\xref{introsection} for the definition). Assume also that for
each meridian there is a class $\delta\in H^n (X_o )$ such that the
corresponding monodromy transformation is given by the Picard-Lefschetz
formula
\eqrefer{plformula}. Let $E$ denote the set of these classes (called the
{\sl vanishing cycles}). Let $V^n(X_o)\subset H^n (X_o)$ be the span of
$E$, called the {\sl vanishing cohomology}.
A cycle orthogonal to $V = V^n(X_o)$ is invariant under all Picard-Lefschetz
transformations, hence is invariant under the action of monodromy.
Consequently its orthogonal complement $V\perp$ is the space of invariant
cycles. The image of $H^n({\bf X})$ in $H^n(X_o)$ also consists
of invariant cycles. By theorem 4.1.1 (or corollary (4.1.2)) of
\cite{DeHodgeTwo}, this inclusion is an equality. One concludes that $V\perp$
is the same as the image of $H^n({\bf X})$, which is a sub-Hodge structure, and so
the bilinear form restricted to it is nondegenerate. Therefore the
bilinear form restricted to $V = V^n(X_o)$ is also nondegenerate. Consequently
$V^n(X_o)$ is an orthogonal or symplectic space, and the monodromy group acts
on $V^n(X_o)$ by orthogonal or symplectic transformations.
When the discriminant locus is irreducible the argument of Zariski
\cite{Zar} or \cite{DeWeOne}, paragraph preceding Corollary 5.5, shows that the
meridians of $\pi_1(S - \Delta)$ are mutually conjugate. Writing down a
conjugacy $\gamma' = \kappa^{-1}\gamma\kappa$ and applying it to
\eqrefer{plformula}, one concludes that $\delta' =
\rho(\kappa^{-1})(\delta)$. Thus the vanishing cycles constitute a single
orbit. To summarize, we have the following, (c.f.
\cite{DeWeOne}, Proposition 5.3, Theorem 5.4, and \cite{DeWeTwo}, Lemma 4.4.2):
\proclaim{Theorem.} Let ${\bf X} \map S$, with $S = \C^{N+1} - \{ 0\}$ or $\P^N$
and $N \ge 1$, be a
family with irreducible discriminant locus and such that the monodromy
transformations of meridians are Picard-Lefschetz transformations. Then
the monodromy group is either finite or is a Zariski-dense subgroup of the
(orthogonal or symplectic) group of automorphisms of the vanishing cohomology.
\endproclaim
To decide which of the two alternatives holds, consider the period mapping
$$
f : U \map D/\Gamma,
$$
where $D$ is the space \cite{GriffPerDom} which classifies the Hodge
structures $V^n(X_a)$ and where $\Gamma$ is the monodromy group. Then one
has the following well-known principle:
\proclaim{Lemma.} If the monodromy group is finite, then the period map is
constant.
\endproclaim
\procref{finiteimagelemma}
\noindent{\bf Proof:\ } Let $f$ be the period map and suppose that the monodromy representation
is finite. Then there is an unramified cover $\widetilde S$ of the domain of
$f$ for which the monodromy representation is trivial. Consequently there is
lift
$\tilde f$ to $\widetilde S$ which takes values in the period domain $D$. Let
$\bar S$ be a smooth compactification of
$\widetilde S$. Since $D$ acts like a bounded domain for horizontal holomorphic
maps,
$\tilde f$ extends to a holomorphic map of $\bar S$ to $D$.
Any such map with compact domain is constant \cite{GS}.
As a consequence of the previous lemma and theorem, we have a practical
density criterion:
\proclaim{Theorem.} Let ${\bf X}$ be a family of varieties over $\C^{N+1} - \{ 0\}$
or
$\P^N$, $N\ge 1$, whose monodromy group is generated by Picard-Lefschetz
transformations
\eqrefer{plformula},
which has irreducible discriminant locus, and whose period map has nonzero
derivative at one point. Then the monodromy group is Zariski-dense in the
(orthogonal or symplectic) automorphism group of the vanishing cohomology.
\endproclaim
\procref{practicaldensitycriterion}
Irreducibility of the discriminant locus for hypersurfaces is well known, and
can be proved as follows.
Consider the Veronese imbedding
$v$ of $\P^{n+1}$ in $\P^N$. This is the map which sends the homogeneous
coordinate vector $[ x_0 \commadots x_{n+1} ]$ to $[ x^{M_0} \commadots x^{M_N}
]$ where the $x^{M_i}$ are an ordered basis for the monomials of degree $d$ in
the $x_i$. If $H$ is a hyperplane in $\P^N$, then $v^{-1}(H)$ is a hypersurface
of degree $d$ in $\P^{n+1}$. All hypersurfaces are obtained in this way,
so the dual projective space $\widehat \P^N$ parametrizes the universal
family. A hypersurface is singular if and only if $H$ is tangent to
the Veronese manifold $\VV = v(\P^{n+1})$. Thus the discriminant
is the variety $\widehat \VV$ dual to $\VV$. Since the variety dual to an
irreducible variety is also irreducible, it follows that the discriminant is
irreducible.
Finally, we observe that in the situations considered in this paper, vanishing
cohomology and primitive cohomology coincide. This can easily be checked by
computing the invariant cohomology using a suitable compactification and appealing
to (4.1.1) of \cite{DeHodgeTwo}. Since this is not essential to our
arguments we omit further details.
\section{Rational differentials and the Griffiths residue calculus}
\secref{ratdiffsection}
Griffiths' local Torelli theorem \cite{GriffPerRat} tells us
that the period map for hypersurfaces of degree $d$ and dimension $n$ is
is nontrivial for $d > 2$ and $n > 1$ with the exception of the case $(d,n) = (3,2)$.
In fact, it says more: the kernel of the differential is the tangent space
to the orbit of the natural action of the projective linear group. The proof
is based on the residue calculus for rational differential forms and some
simple commutative algebra (Macaulay's theorem).
What we require here is a weak (but sharp) version of Griffiths' result for
the variations of Hodge structures defined by families of cyclic covers of
hypersurfaces. For double covers this is straightforward, since such covers
can be viewed as hypersurfaces in a weighted projective space \cite{Dolgachev}.
For higher cyclic covers the variations of Hodge structure are complex, and in
general the symmetry of Hodge numbers, $h^{p,q} = h^{q,p}$ is broken. Nonetheless,
the residue calculus still gives the needed result. Since this last part
is nonstandard, we sketch recall the basics of the residue calculus, how
it applies to the case of double covers, and how it extends to the case of
higher cyclic covers.
To begin, consider weighted projective space $\P^{n+1}$ where the weights of
$x_i$ are $w_i$. Fix a weighted homogeneous polynomial $P(x)$ and let $X$
be the variety which it defines. We assume that it is smooth. Now take a
meromorphic differential $\nu$ on $\P^{n+1}$ which has
a pole of order $q+1$ on $X$.
Its residue is the cohomology class on $X$ defined by the formula
$$
\int_\gamma \hbox{res}\, \nu = { 1 \over 2 \pi } \int_{\partial T(\gamma)} \nu,
$$
where $T(\gamma)$ is a tubular neighborhood of an $n$-cycle $\gamma$. The
integrand can be written as
$$
\nu(A,P,q) = { A\Omega \over P^{q+1} }. \eqn\eqref{ratdiff}
$$
where
$$
\Omega = \sum (-1)^i\;w_i x_i \;dx_0 \wedges \widehat{dx_i} \wedges dx_{n+1}.
$$
The ``volume form'' $\Omega$ has weight $w_0 + \cdots + w_{n+1}$ and the degree of $A$,
which we write as $a(q)$, is such that $\nu$ is of weight zero. The primitive
cohomology of $X$ is spanned by Poincar\'e residues of rational differentials,
and the space of residues with a pole of order $q+1$ is precisely
$F^{n-q}H^n_o(X)$, the $(n-q)$-th level of the Hodge filtration on the primitive cohomology. When
the numerator polynomial is a linear combination of the partial derivatives of $P$, the residue
is cohomologous in $\P^{n+1} - X$ to a differential with a pole of order one
lower. Let $J = (\partial P/\partial x_0 \commadots \partial P/\partial x_{n+1})$
be the Jacobian ideal and let $R = \C[x_0 \commadots x_{n+1}]/J$ be the quotient
ring, which we note is graded. Then the residue maps $R^{a(q)}$ to
$F^q/F^{q+1}$. By a theorem of Griffiths \cite{GriffPerRat}, this map is an isomorphism.
For a smooth variety the ``Jacobian ring'' $R$ is finite-dimensional, and so there is
a least integer
$$
t = (n+2)(d-2)
\eqn
\eqref{topJ}
$$
such that $R^i = 0$ for $i > t$. Moreover,
and $R^t$ is one-dimensional and the bilinear map
$$
R^i\times R^{t-i} \map R^t \cong \C.
$$
is a perfect pairing (Macaulay's theorem). When $R^i$ and $R^{t-i}$
correspond to graded quotients of the Hodge filtration, the pairing corresponds
to the cup product \cite{CG}.
The derivative of the period map is given by formal differentiation
of the expressions (\xref{ratdiff}). Thus, if $P_t = P + tQ + \cdots$
represents a family of hypersurfaces and $\omega
= \hbox{res}\,(A\Omega/F^\ell)$ represents a family of cohomology classes
on them, then
$$
{ d \over dt} \hbox{res}\,{ A\Omega \over P^{q+1} } = -(q+1) \hbox{res}\, { QA\Omega \over P^{q+2}
}.
$$
To show that the derivative of the period map is nonzero,
it suffices to exhibit an $A$ and a $Q$ which are nonzero
in $R$ and such that the product $QA$ is also nonzero.
Here we implicitly use
the identification
$
T \cong R^d
$
of tangent vectors to the moduli space with the component
of the Jacobian ring in degree $d$. Thus the natural components
of the differential of the period map,
$$
T \map \Hom(H^{p,q}(X),H^{p-1,q+1}(X)),
$$
can be identified with the multiplication homomorphism
$$
R^d \map \Hom(R^a,R^{a+d}),
$$
where $a$ is the degree of the numerator polynomial used in
the residues of the forms (\xref{ratdiff}). All of these results,
discovered first by Griffiths in the case of hypersurfaces,
hold for weighted hypersurfaces by the results described in
\cite{Dolgachev} and \cite{Tu}.
Consider now a double cover $Y$ of a hypersurface $X$ of even degree $d$.
If $X$ is defined by $P(x_0 \commadots x_n) = 0$ then $Y$ is defined by
$y^2 +P(x_0 \commadots x_n) = 0$, where $y$ has weight $d/2$ and where the $x$'s
have weight one. This last equation is homogeneous of degree $d$ with respect
to the given weighting, and $\Omega$ has weight $d/2 + n + 2$. Thus
$\nu(A,y^2 + P,q)$ is of weight zero if
$a(q) = (q + 1/2)d - (n+2)$. Since $y$ is in the Jacobian ideal,
we may choose $A$ to be a polynomial in the $x$'s, and we may consider
it modulo the Jacobian ideal of $P$. Thus the classical considerations
of the residue calculus apply. If we choose $a(q)$ maximal subject to
the constraints $p > q$ and $a \ge 0$ then
$$
q = \left\{ { n + 1 \over 2 } \right\},
$$
where $\set{ x }$ is the greatest integer {\sl strictly less}
than $x$. Both conditions are satisfied for $d \ge 4$ except
that for $n = 1$ we require $d \ge 6$. Thus we have excluded the case $(d,n)
= (4,1)$ in which the resulting double cover is rational and the period map is
constant.
Now let $A$ be a polynomial of degree $a$ which is
nonzero modulo the Jacobian ideal. We must exhibit a polynomial $Q$ of degree
$d$ such that $AQ$ nonzero modulo $J$. By Macaulay's theorem there is a polynomial
$B$ such that $AB$ is congruent to a generator of $R^t$, hence satisfies
$AB \not\equiv 0 \hbox{ mod $J$}$. Write $B$ as a linear combination of monomials
$B_i$ and observe that there is an $i$ such that $AB_i \not\equiv 0$. If $B_i$
is of degree at least $d$, we can factor it as $QB_i'$ with $Q$ of degree $d$.
Since $AQB_i' \not\equiv 0$, $AQ \not\equiv 0$, as required.
The condition that $B$ have degree at least $d$ reads $a + d \le t$. Using
the formulas \eqrefer{topJ} for $t$ and the optimal choice for $a$, we see that this
inequality is satisfied for the range of $d$ and $n$ considered. This computation
completes the proof of the main theorem in the case $d$ even, $d \ge 4$,
except for the case $(d,n) = (4,1)$.
\section{Rational differentials for higher cyclic covers }
\secref{cycliccoversection}
To complete the proof of the main theorem we must consider
arbitrary cyclic covers of $\P^{n+1}$ branched along a smooth hypersurface
of degree $d$. Since the fundamental group of the complement of $X$
is cyclic of order $d$, the number of sheets $k$ must be a divisor of
$d$. As mentioned in the outline of the proof, there is an automorphism
$\sigma$ of order $k$ which operates on the universal family ${\bf Y}$ of
such covers. Consequently the local system $\H$ of vanishing cohomology
(cf. \S \xref{zardensitysection})
splits over $\C$ into eigensystems $\H(\mu)$, where $\mu \ne 1$
is a $k$-th root of unity. Therefore the monodromy representation,
which we now denote by $\rho$, splits as a sum of representations $\rho_\mu$
with values in the groups $\widetilde G(\mu)$ introduced in \eqrefer{tildeGdecomp}.
As noted there we can view $\rho_\mu$ as taking values in a group of linear
automorphisms of $H(\mu)$. This group is unitary for the
hermitian form $h(x,y) = i^{n+1}(x,\bar y)$ if $\mu$ is
non-real, and that is the case that we will consider here.
Although the decomposition of $\H$ is over the complex numbers, important
Hodge-theoretic data survive. The hermitian form $h(x,y)$ is
nondegenerate and there is an induced Hodge decomposition, although
$h^{p,q}(\mu) = h^{q,p}(\mu)$ may not hold. However, Griffiths'
infinitesimal period relation,
$$
{ d \over dt } F^p(\mu) \subset F^{p-1}(\mu)
$$
remains true. Thus each $\H(\mu)$ is a {\sl complex variation of Hodge
structure}, c.f. \cite{DeMo}, \cite{SimpsonHiggs}. The
associated period domains are homogeneous for the groups
$\widetilde G(\mu)$.
To extend the arguments given above to the unitary representations
$\rho_\mu$ we must extend Deligne's density theorem to this case.
The essential point is that the monodromy groups $\Gamma(\mu)$ are
generated not by Picard-Lefschetz transformations, but by their unitary
analogue, which is a {\sl complex reflection} \cite{Pham},
\cite{Givental}, \cite{Mostow}. These are linear maps of the form
$$
T(x) = x \pm (\lambda - 1)h(x,\delta)\delta,
$$
where $h$ is the hermitian inner product defined above, $h(\delta,\delta) = \pm
1$, where
$\pm$ is the same sign as that of $h(\delta ,\delta )$, and where
$\lambda \ne 1$ is a root of unity. The vector $\delta$ is an eigenvector of $T$
with eigenvalue $\lambda$ and $T$ acts by the identity on the hyperplane
perpendicular to $\delta$. It turns out that the eigenvalue $\lambda$ of $T$
is, up to a fixed sign that depends only on the dimension of $Y$, equal
to the eigenvalue $\mu$ of $\sigma$.
In section \xref{unitarydensitysection} we will prove an analogue of Deligne's
theorem \eqrefer{delignedensity} for groups of complex reflections.
It gives the usual dichotomy: either the monodromy group is finite, or it is
Zariski-dense. In section \xref{complexreflectionsection} we will show that the
monodromy groups $\Gamma(\mu)$ are indeed generated by complex reflections.
It remains to show that the derivative of the period map for the complex
variations of Hodge structure $\H(\mu)$ are nonzero given appropriate
conditions on $d$, $k$, $n$, and $\mu$.
For the computation fix $\zeta = e^{2\pi i/k}$ as a primitive $k$-th
{\sl root of unity} and let the cyclic action on the universal family
\eqrefer{universalcyclic} be given by $y\circ \sigma = \zeta y$.
Then the ``volume form'' $\Omega(x,y)$ is an eigenvector
with eigenvalue $\zeta$ and the rational differential
$$
{ y^{i-1} A(x) \Omega(x,y) \over ( y^k + P(x) )^{q+1} }
\eqn
\eqref{iratdiff}
$$
has eigenvalue $\mu = \zeta^i$, as does its residue.
Thus we will sometimes write $\H(i)$ for
$\H(\zeta^i)$ and will use the corresponding notations
$\widetilde G(i)$, $\tilde\rho_i$, etc. Residues with numerator
$y^{i-1}A(x)$ and denominator $(y^k + P(x))^{q+1}$ span the spaces $H^{p,q}_0(i)$, where
$i$ ranges from 1 to $k-1$. Moreover, the corresponding space of numerator
polynomials, taken modulo the Jacobian ideal of $P$, is isomorphic via the residue
map to $H^{p,q}_0(i)$. Since $P$ varies by addition of a polynomial in the $x$'s,
the standard unweighted theory applies to computation of the derivative map.
Let us illustrate the relevant techniques by computing the Hodge numbers and
period map for triple covers of $\P^3$ branched along a smooth cubic surface.
(This period map is studied in more detail in \cite{ACT}.)
A triple cover of the kind considered is a cubic hypersurface in $\P^4$, and
the usual computations with rational differentials show that $h^{3,0} = 0$,
$h^{2,1} = 5$. The eigenspace
$H^{2,1}(i)$ is spanned by residues of differentials with numerator
$ A(x)\Omega(x,y) $ and denominator $(y^3 + P(x))^2$. Since the degree of
$\Omega(x_0,x_1,x_2,x_3, y)$ is 5, $A$ is must be linear in the variables $x_i$. Thus
$h^{2,1}(1) = 4$. The space
$H^{1,2}(1)$ is spanned by residues of differentials with numerator
$ A(x)\Omega(x,y) $ and denominator $(y^3 + P(x))^3$.
Thus the numerator is of degree four, but must be viewed modulo
the Jacobian ideal. For dimension counts it is enough
to consider the Fermat cubic, whose Jacobian ideal is generated
by squares of variables. The only square-free quartic in four variables
is $x_0x_1x_2x_3$, so $h^{1,2}(1) = 1$. Similar computations show
that the remaining Hodge numbers for $H^3(1)$ are zero and
yield in addition the numbers for $H^3(2)$. One can
also argue that $H^3(1)\oplus H^3(2)$ is defined over $\R$, since the
eigenvalues are conjugate. A Hodge structure defined over $\R$
satisfies $h^{p,q} = h^{q,p}$. From this one deduces that
$h^{2,1}(2) = 1$, $h^{1,2}(2) = 4$. Since there is just one conjugate
pair of eigenvalues of $\sigma$, there is just one component in the
decomposition \eqrefer{tildeGdecomp}, $\widetilde G = \widetilde G(\zeta)$,
and this group is isomorphic to $U(1,4)$. Since the coefficients
of the monodromy matrices lie in the ring $\Z[\zeta]$, where
$\zeta$ is a primitive cube root of unity, the representation $\tilde\rho$
takes values in a discrete subgroup of $\widetilde G$. Therefore the complex variation
of Hodge structures define period mappings
$$
p : U_{3,2} \map B_4/\Gamma',
$$
where $B_4$ is the unit ball in complex 4-space and $\Gamma'$ a
discrete group acting on it.
To show that the period
map $p_i$ is nonconstant it suffices to show that its differential is
nonzero at a single point. We do this for the Fermat variety. A basis
for $H^{2,1}(1)$ is given by the linear forms $x_i$,
and a basis for $H^{1,2}$ is given by their product
$x_0x_1x_2x_3$. Let $m_i$ be the product of all the $x_k$
except $x_i$. These forms constitute a basis for the tangent
space to moduli. Since $m_ix_i = x_0x_1x_2x_3$, multiplication
by $m_i$ defines a nonzero homomorphism from $H^{2,1}(1)$
to $H^{1,2}(1)$. Thus the differential of the period map
is nonzero at the Fermat. In fact it is of rank four, since
the homomorphisms defined by the $m_i$ are linearly independent. Similar
considerations show that the period map for $\H(2)$ is of rank
four. The relevant bases are $\set{ y }$ for $H^{2,1}(2)$
and $\set{ ym_0, ym_1, ym_2, ym_3 }$ for $H^{1,2}(2)$.
For the general case it will be enough to establish the following.
\proclaim{Proposition.} Let ${\bf Y}$ be the universal family of $d$-sheeted
covers of $\P^{n+1}$ branched over smooth hypersurfaces of degree $d$.
The derivative of the period map for $\H^{n+1}(1)$ is nontrivial
if $n \ge 2$ and $d \ge 3$ or if $n = 1$ and $d \ge 4$.
\endproclaim
\procref{PropositionDerivNonTrivialNgeTwo}
\noindent{\bf Proof:\ } Elements of
$H^{p,q}(1)$ with $p+q = n+1$ are given by rational differential
forms with numerator $ A(x) \Omega(x,y) $ and denominator $( y^d + P(x) )^{q+1}$.
The numerator must have degree $ a = (q + 1)d - (n+3) $.
As before choose $q$ so that $a$ is maximized subject to the constraints
$p > q$ and $a \ge 0$. Then $q = \set{ {n / 2} + { 1 / d } }$.
If $n \ge 2$ and $d \ge 3$ or if $n = 1$ and $d \ge 4$, then $a \ge 0$.
Thus numerator polynomials $A(x)$ which are nonzero modulo the Jacobian ideal
exist. One establishes the existence of a polynomial $Q(x)$ of degree $d$
such that $QA$ is nonzero modulo the Jacobian ideal using the same argument
as in the case of double covers.
A different component of the period map is required if the branch locus
is a finite set of points, which is the case for the braid group of $\P^1$:
\proclaim{Proposition.} For $n=0$ the period map for $\H^1(i)$ is
non-constant if $d \ge 4$ and $i \ge 2$.
\endproclaim
\noindent{\bf Proof:\ } An element of $H^{1,0}(i)$ is the residue of a rational differential
with numerator $ y^{i-1}A(x_0,x_1)\Omega $ and denominator $ y^d + P(x_0,x_1) $.
The degree of $A$ is $a = d - 2 - i$. The top degree for the Jacobian
ideal is $2d-4$. Thus we require $a + d \le 2d - 4$, which is satisfied
if $i \ge 2$. Since $a \ge 0$, one must also require $d \ge 4$.
We observe that the local systems which occur
as constituents for $k$-sheeted covers, where $k$ divides $d$,
also occur as constituents of $d$-sheeted covers.
\proclaim{Remark.} Let $\H({k,\mu})$ be the complex variation
of Hodge structure associated to a $k$-sheeted cyclic cover
of $\P^{n+1}$ branched along a hypersurface of degree $d$,
belonging to the eigenvalue $\mu$, where $k$ is a divisor
of $d$. Then $\H({k,\mu})$ is isomorphic to $\H({d,\mu})$.
\endproclaim
\noindent{\bf Proof:\ } Consider the substitution $y = z^{d/k}$ which effects the transformation
$$
{ y^i A(x) \Omega(x,y) \over (y^k + P(x) )^{q+1} }
\mapsto
{ (d / k) }{ z^{(i+1)(d/k) -1} A(x) \Omega(x,z) \over
( z^d + P(x) )^{q+1} } .
$$
These differentials are eigenvectors with the same eigenvalue. The
map which sends residues of the first kind of rational differential
to residues of the second defines the required isomorphism.
\section{Complex Reflections}
\secref{complexreflectionsection}
We now review some known facts on how complex reflections arise for
degenerations of cyclic covers. When the branch locus acquires
a node, the local equation is
$$
y^k + x_1^2 + \cdots + x_{n+1}^2 = t,
\eqn
\eqref{kdoublept}
$$
which is a special case of the situation studied by Pham in \cite{Pham},
where the left-hand side is a sum of powers. Our
discussion is based on Chapter 9 of \cite{Milnor} and Chapter 2 of \cite{Arnold}.
Consider first the case $y^k = t$. It is a family of
zero-dimensional varieties $\set{ \xi_1(t) \commadots \xi_k(t)}$ whose vanishing
cycles are successive differences of roots,
$$
\xi_1 - \xi_2, \
\ldots,\
\xi_{k-1} - \xi_k,
\eqn
\eqref{stdvanishingbasis}
$$
and whose monodromy is given by cyclically shifting indices to the right:
$$
T( \xi_i - \xi_{i+1} ) = \xi_{i+1} - \xi_{i+2},
$$
where $i$ is taken modulo $k$. Thus $T$ acts on the $(k-1)$-dimensional
space of vanishing cycles as a transformation of order $k$. Over the complex
numbers it is diagonalizable, and the eigenvalues are the $k$-th roots of unity
$\mu \ne 1$. Note that $T = \sigma_0$ where $\sigma_0$ is the generator for the
automorphism group of the cyclic cover $y^k = t$ given by $y\map \zeta y$, where
$\zeta = e^{2\pi i/k}$ is our chosen primitive $k$-th root of unity.
The intersection product $B$ defines a possibly degenerate bilinear
form on the space of vanishing cycles. For the singularity $y^k = t$
it is $(\xi_i,\xi_j) = \delta_{ij}$, so relative
to the basis \eqrefer{stdvanishingbasis} it is the negative of the matrix for
the Dynkin diagram $A_{k-1}$ --- the positive-definite matrix with two's along
the diagonal, one's immediately above and below the diagonal, and zeroes elsewhere.
Now suppose that $f(x) = t$ and $g(y) = t$ are families which
acquire an isolated singularity at $t = 0$. Then $f(x) + g(y) = t$
is a family of the same kind; we denote it by $f \oplus g$. The theorem of
Sebastiani and Thom \cite{ST}, or
\cite{Arnold}, cf. Theorem 2.1.3, asserts that vanishing cycles for the
sum of two singularities are given as the join of vanishing cycles for $f$ and
$g$. Thus, if $a$ and $b$ are vanishing cycles of dimensions $m$ and $n$, then
the join
$a*b$ is a vanishing cycle of dimension $m+n+1$, and, moreover,
the monodromy acts by
$
T(a * b) = T(a)*T(b).
$
{}From an algebraic standpoint the join is a tensor product, so one can write
$V(f\oplus g) = V(f)\otimes V(g)$ where $V(f)$ is the space of vanishing
cycles for $f$, and one can write the monodromy operator as
$
T_{f \oplus g} = T_f\otimes T_g.
$
The {\sl suspension} of a singularity $f(x) = t$ is by definition the
singularity $y^2 + f(x) = t$ obtained by adding a single square. If $a$ is
a vanishing cycle for $f$ then $(y_0 - y_1)\otimes a$ is a vanishing
cycle for the suspension, and the suspended monodromy is given by
$$
T( (y_0 - y_1)\otimes a ) = - (y_0 - y_1)\otimes T(a) .
$$
In particular, the local monodromy of a singularity
and its double suspension are isomorphic.
The intersection matrix $B'$ of
a suspended singularity (relative to the same canonical basis) is a function of
the intersection matrix $B$ for the given singularity, cf. Theorem 2.14 of
\cite{Arnold}. When the bilinear form for $B$ is symmetric, the rule for producing $B'$ from $B$
is: make the diagonal entries zero and change the sign of the above-diagonal entries. When $B'$
has an even number of rows of columns, the determinant is one, and when the number
of rows and columns is odd, it is zero. Thus the intersection matrix
for $x^2 + y^k = t$ is nondegenerate if and only if $k$ is odd. In addition,
the intersection matrix of a double suspension is the negative of the given
matrix. Thus the matrix of any suspension of
$y^k = t$ is determined. It is nondegenerate if the dimension of the
cyclic cover \eqrefer{kdoublept} is even or if the dimension is odd and $k$ is
also odd. Otherwise it is degenerate.
It follows from our discussion that the space of vanishing cycles
$V$ for the singularity \eqrefer{kdoublept} is $(k-1)$-dimensional
and that the local monodromy transformation is
$
T = \sigma_0\otimes(-1)\otimes\cdots\otimes(-1)
$
where $\sigma_0$ is the covering automorphism $y\map \zeta y$ for $y^k = t$. Thus
$T$ is a cyclic transformation of order $k$ or $2k$, depending on whether
the dimension of the cyclic cover is even or odd. In any case, $T$ is
diagonalizable with eigenvectors $\eta_i$ and eigenvalues $\lambda_i$,
where $\lambda_i = \pm \mu_i$ with $\mu_i = \zeta^i$ where $\zeta$ is our fixed
primitive
$k$-th root of unity and $i = 1,\cdots , k-1$.
Note that the cyclic automorphism $\sigma$ of the universal family
\eqrefer{universalcyclic}, given by
$y\mapsto\zeta y$
acts as $\sigma_0\otimes(+1)\otimes\cdots\otimes(+1)$ on the vanishing homology
of \eqrefer{kdoublept}. Thus the eigenspaces of $\sigma$ and $T$ coincide, and
their respective eigenvalues differ by the fixed sign $(-1)^{n+1}$. Since the
eigenvalues
$\mu_i$ are distinct, the eigenvectors $\eta_i$ are orthogonal with respect
to the hermitian form. Thus $h(\eta_i,\eta_i) \ne 0$. Moreover the sign of
$h(\eta_i,\eta_i)$ depends only on the index $i$, globally determined on
\eqrefer{universalcyclic}, independently of the particular smooth point on
the discriminant locus whose choice is implicit in \eqrefer{kdoublept}. We
conclude that on the space of vanishing cycles,
$$
T(x) = \sum_{i = 1}^{k-1} \lambda_i{h(x,\eta_i) \over
h(\eta_i,\eta_i)}\eta_i,
\eqn
\eqref{TVcxreflectionformula}
$$
where $\lambda_i = (-1)^{n+1} \mu_i$.
Now consider a cycle $x$ in $H^{n+1}(Y_{\tilde o})$, and {\sl suppose that}
$k$ {\sl is odd}. Then the intersection form on the space $V$ of local vanishing
cycles
for the degeneration \eqrefer{kdoublept} is {\sl nondegenerate}. Consequently
$H^{n+1}(Y_{\tilde o})$ splits orthogonally as
$V
\oplus V\perp$. The action on $H^{n+1}(Y_{\tilde o})$ of the monodromy
transformation $T$ for the meridian corresponding to the degeneration
\eqrefer{kdoublept}
is given by \eqrefer{TVcxreflectionformula} on $V$ and
by the identity on $V\perp$. Thus it is given for arbitrary $x$ by the formula
$$
T(x) = x + \sum_{i=1}^{k-1} (\lambda_i-1){h(x,\eta_i) \over
h(\eta_i,\eta_i)}\eta_i .
\eqn
\eqref{Tcxreflectionformula}
$$
Finally, for each $i = 1,\cdots ,k-1$ we can normalize the eigenvector $\eta_i$
to an eigenvector $\delta_i$ satisfying $h(\delta_i , \delta_i ) = \epsilon_i =
\pm 1$. To summarize, we have proved the following:
\proclaim{Proposition.} Consider the family \eqrefer{universalcyclic} of $k$-fold
cyclic covers of $\P^{n+1}$ branched over a smooth hypersurface of degree $d$,
where both $k$ and $d$ are odd. Let $T$ be the monodromy corresponding to
a generic degeneration of the branch locus, as in \eqrefer{kdoublept}.
Then $T$ acts on the $i$-th eigenspace of the cyclic automorphism
$\sigma$ (defined by $y\mapsto\zeta y$ in \eqrefer{universalcyclic}) by a
complex reflection with eigenvalue
$\lambda_i = (-1)^{n+1} \zeta^i$. Thus
$$
T(x) = x + \epsilon_i (\lambda_i - 1) h(x,\delta_i)\delta_i
$$
holds for all $x\in\H (i)$.
\endproclaim
\proclaim{Remark.} \rm In remark \xref{remarkMeridiansInfiniteOrder}.a we observed that
the meridians of $\Phi_{d,n}$ are of infinite order for $n$ odd and for
$n$ even, $d \ge 4$ even. Consider now the case in which $n$ is even and $d$ is odd,
let $\zeta = \exp(2\pi i/d)$, and let $\bar\rho'$ be the
corresponding representation, in which meridians of $\widetilde \Delta$
correspond to complex reflections of order $2d$. These complex reflections
and their powers different from the identity are non-central if the $\zeta$ eigenspace
has dimension at least two, which is always the case for $d \ge 3$, $n \ge 2$.
Thus $\bar\rho'(\gamma)$ has order $2d$. By this simple argument
we conclude that in the stated range of $(n,d)$, meridians always have order greater than two.
However, our argument does not give the stronger result \xref{remarkMeridiansInfiniteOrder}.b
asserted by Kontsevich.
\endproclaim
\section{Density of unitary monodromy groups}
\secref{unitarydensitysection}
We now show how the argument Deligne used in \cite{DeWeTwo}, section 4.4,
to prove Theorem \xref{delignedensity} can
be adapted to establish a density theorem for groups generated
by complex reflections on a space $\C(p,q)$ endowed with
a hermitian form $h$ of signature $(p,q)$. If $A$ is a subset of $\C(p,q)$ or
of $U(p,q)$, we use $PA$ to denote its projection in $\P(\C(p,q))$ or $PU(p,q)$.
\proclaim{Theorem.} Let $\epsilon = \pm 1$ be fixed, and let $\Delta$ be a set of
vectors in a hermitian space
$\C(p,q)$ which lie in the unit quadric $h(\delta,\delta) = \epsilon$.
Fix a root of unity $\lambda \ne \pm 1$ and let $\Gamma$ be the
subgroup of $U(p,q)$ generated by the complex reflections
$s_\delta(x) = x + \epsilon ( \lambda - 1 )h(x,\delta)\delta$ for all $\delta$ in
$\Delta$. Suppose that $p+q >1$, that $\Delta$ consists of a single
$\Gamma$-orbit, and that $\Delta$ spans $\C(p,q)$. Then either $\Gamma$ is
finite or $P\Gamma$ Zariski-dense in $PU(p,q)$.
\endproclaim
\procref{udensitytheo}
Let $\bar\Gamma$ be the Zariski closure of a subgroup $\Gamma$
of $U(p,q)$ which contains the $\lambda$-reflections
for all vectors $\delta$ in a set $\Delta$. Then
$\bar\Gamma$ also contains the $\lambda$-reflections for
the set $R = \bar\Gamma\Delta$. Indeed, if $g$ is an element
of $\bar\Gamma$, then
$$
g^{-1}s_\delta g = s_{g^{-1}(\delta)}.
\eqn\eqref{reflectionconjugacy}
$$
Thus it is enough to establish the following result in order
to prove our density theorem:
\proclaim{Theorem.} Let $\epsilon = \pm 1$ be fixed, and let
$R$ be a set of vectors in a hermitian space
$\C(p,q)$ which lie in the unit quadric $h(\delta,\delta) = \epsilon$.
Fix a root of unity $\lambda \ne \pm 1$ and let $M$ be the smallest algebraic
subgroup of $U(p,q)$ which contains the complex reflections
$s_\delta(x) = x + \epsilon ( \lambda - 1 )h(x,\delta)\delta$
for all $\delta$ in $R$.
Suppose that $p+q >1$, that $R$ consists of a single $M$-orbit,
and that $R$ spans $\C(p,q)$. Then either $M$ is finite or
$PM = PU(p,q)$.
\endproclaim
\procref{udensityprop}
We begin with a special case of the theorem for groups
generated by a pair of complex reflections.
\proclaim{Lemma.} Let $\lambda \ne \pm 1$ be a root of unity, and let
$U$ be the unitary group of a nondegenerate
hermitian form on $\C^2$. Let $\delta_1$ and $\delta_2$ be
independent vectors with nonzero inner product, and let
$\Gamma$ be the group generated by complex reflections with common
eigenvalue $\lambda$. Then either $\Gamma$ is finite or its
image in the projective unitary group is Zariski-dense.
In the positive-definite case $\Gamma$ is finite if and only if
the inner products $(\delta_1,\delta_2)$ lie in a
fixed finite set $S$ which depends only on $\lambda$ and $h$. In the indefinite
case
$\Gamma$ is never finite.
\endproclaim
\procref{Ulemma}
We treat the definite case first. To begin, note
that the group $U$ acts on the Riemann sphere $\P^1$ via the natural map
$U \map PU$, where $PU$ is the projectivized unitary group. Let $PR$ be
the image of $R \subset \C^2$ in $\P^1$. Since $\lambda$ is a root of
unity, the projection $P\Gamma$ is a
finite group if and only if $\Gamma$ is. The finite subgroups of rotations
of the sphere are well known. There are two infinite series: the cyclic
groups, where the vectors $\delta$ are all proportional, and the dihedral
groups where $\lambda = -1$. There are three additional groups, given
by the symmetries of the five platonic solids, and $S$ is the set of possible
values of $h(\delta_1,\delta_2)$ that can arise for these three groups.
We suppose that
$(\delta_1,\delta_2)$ lies outside $S$, so that $P\Gamma$ is infinite.
Then its Zariski closure $PM$ is either $PU$ or a group whose identity
component is a circle. In this case $PR$ contains a great circle $\alpha$.
However, $PR$ is stable under the action of $PM$, hence under the rotations
corresponding to axes in $PR$. Since $\lambda \ne \pm 1$, the orbit
$PR$ contains additional great circles which meet $\alpha$ in an angle $0 < \phi
\le \pi/2$. The union of these, one for each point of the given
circle, forms a band about the equator, hence has nonempty interior. Such a
set is Zariski-dense in the Riemann sphere viewed as a real algebraic
variety. Since
$PR$ is a closed real algebraic set, $PR = S^2$. Since
$PR \cong PM/H$, where $H$ is the isotropy group of a point on the sphere,
$PM = PU$.
In the case of an indefinite hermitian form, the group
$U = U(1,1)$, acts on the hyperbolic plane via the projection
to $PU$, and $P\Gamma$
is a group generated by a pair of elliptic elements of
equal order but with distinct fixed points. One elliptic
element moves the fixed point of the other, and so their commutator $\gamma$
is hyperbolic (c.f. Theorem
7.39.2 of \cite{Beardon}). The Zariski closure of the cyclic group
$\set{\gamma^n}$ is a one-parameter subgroup of $PU$.
Consequently the orbit $PR$ contains a geodesic $\alpha$ through one of the
elliptic fixed points. By \eqrefer{reflectionconjugacy} the other points of
$\alpha$ are
fixed points of other elliptic transformations in $PM$. Now the orbit $PR$
contains the image of $\alpha$ under each of these transformations,
and so $PR$ contains an open set of the hyperbolic plane. This implies that
either $PM = PU$ or $PM$ is contained in a parabolic subgroup. Since $PM$
contains non-trivial elliptic elements that last possibility cannot occur,
and so $PM = PU$.
Next we show that if the set $R$ which defines the reflections
is large, then so is the group containing those reflections.
\proclaim{Lemma.} Fix a root of unity $\lambda\ne\pm 1$ and $\epsilon = \pm 1$.
Let $R$ be a semi-algebraic subset of the unit quadric
$h(\delta,\delta) = \epsilon$.
Let $M$ be the smallest algebraic subgroup of $U(p,q)$
containing the complex reflections $s_\delta(x) = x + \epsilon (\lambda - 1)
h(x,\delta)\delta$, $\delta\in R$. If $p+q >1$ and if $PR$
is Zariski-dense in $\P(\C(p,q))$,
then
$M =PU(p,q)$.
\endproclaim
The proof is by induction on $n = p+q$. For $n = 2$ the result
follows from the proof of lemma \xref{Ulemma}. Let $n>2$ and assume
$p\le q$. Then $q\ge 2$. Fix a codimension two subspace of $\C (p,q)$ of
signature $(p,q-2)$ and let $W_t$ be the pencil of hyperplanes of $\C (p,q)$
containing this codimension two subspace. Then the restriction of $h$ to each
$W_t$ is a non-degenerate form of signature $(p,q-1)$.
Consider a subgroup $M$ of $U(p,q)$ which satisfies
the hypotheses of the lemma, and let $R_t = R\cap W_t$. Since $PR$, respectively
$PR_t$ is semi-algebraic in $\P (\C (p,q))$, respectively in $PW_t$, it is
Zariski dense if and only if it has non-empty interior in the analytic topology.
Thus $R$ has non-empty interior in $\P(\C(p,q))$, and so for dimension reasons
$PR_t$ has non-empty interior in $PW_t$ for generic $t$. Thus $PR_t$ is
Zariski dense in $PW_t$ for generic $t$.
Fix one such value of $t$, let $W = W_t$ and let $M'(R\cap W)\subset M$ denote the
smallest algebraic subgroup of $M$ containing $R\cap W$. Let $M(R\cap W)$ denote
the set of restrictions of elements of $M'(R\cap W)$ to $W$. Then $R\cap W$ and
$M(R\cap W)$ satisfy the induction hypothesis, thus $PM(R\cap W) = PU(W)$.
Now the orthogonal
complement of $W$ is a Zariski closed set, as is $W \cup W\perp$.
Since $R$ is Zariski-dense there is a $\delta$ in $R - W$
and a $\delta'$ in $W$ such that $h(\delta,\delta') \ne 0$.
Consider the function $f_\delta(x) = h(x,\delta')$. If it is constant
on the Zariski closure $C$ of $R\cap W$, then the derivative
$df_\delta$ vanishes on $C$. Therefore $C$ lies in the intersection
of the hyperplane $df(x) = 0$ with $W$, which is a proper
algebraic subset of $W$. Consequently $R \cap W$ is not Zariski-dense,
a contradiction. Thus $f_\delta$ is nonconstant and so we can choose
$\delta$ in $R \cap W$ such that $h(\delta',\delta)$ lies outside
the fixed set $S$. Then lemma \xref{Ulemma} implies that the unitary group of
the plane $F$ spanned by $\delta$ and $\delta'$ is contained in $M$. But $U(W)$
and $U(F)$ generate $U(p,q)$ and the proof of the lemma is complete.
To complete the proof of Theorem (\xref{udensityprop})
we must show that either $R$ is sufficiently large or that $M$
is finite. Observe that since $R$ is an $M$-orbit, it is a semi-algebraic set.
Let $W$ be a subspace of $\C(p,q)$
which is maximal with respect to the property ``$W \cap R$
is Zariski-dense in the unit quadric of $W$.'' Our aim is to show
that either $W = \C(p,q)$ or that $M$ is finite. Consider
first the case $W = 0$. Then the inner products
$h(\delta,\delta')$ for any pair of elements in $R$ lie in the
fixed finite set $S$ of lemma \xref{Ulemma}. Now let $\delta_1 \commadots
\delta_n$ be a basis of $\C(p,q)$ whose elements are chosen from $R$.
Then the inner products $h(\delta,\delta_i)$ lie in $S$
for all $\delta$ and $i$. Consequently $R$ is a finite set
and $M$, which is faithfully represented as a group of permutations
on $R$, is finite as well.
Henceforth we assume that $W$ is nonzero. If it is not maximal
there is a vector $\delta$ in $R - W$ and we may consider
the function $f_\delta(x) = h(x,\delta)$ on the set $R \cap W$. If $f_\delta$
is identically zero for all $\delta$ in $R - W$, then $R \subset W \cup W\perp$.
Therefore $\C(p,q) = W + W\perp$, from which one concludes that
$W = W \oplus W\perp$ and so $M$ is a subgroup of $U(W)\times
U(W\perp)$. But $R$ consists of a single $M$-orbit and contains a point of $W$,
which implies that $R \subset W$, a contradiction.
We can now assume that there is a $\delta \in R - W$ such that
the function $f_\delta$ is not identically zero.
If one of these functions is not locally
constant, then it must take values outside the set $S$. Then
the inner product $(x,\delta)$ lies outside $S$ for an open dense
set of $x$ in $R \cap W$. For each such $x$, $R$ is dense in
the span of $x$ and $\delta$. We conclude that $R$ is dense in
$W + \C\delta$. Thus $W$ is not maximal, a contradiction.
At this point we are reduced to the case in which all the functions
$f_\delta$ are locally constant, with at least one which is not
identically zero. To say that $f_\delta$ is locally constant
on a dense subset of the unit quadric in $W$ is to say that its
derivative is zero on that quadric. Equivalently, tangent spaces
to the quadric are contained in the kernel of $df_\delta$,
that is, in the hyperplane $\delta\perp$. But if all tangent spaces
to the quadric are contained in that hyperplane, then so is the quadric
itself. Then the function in question is identically zero, contrary
to hypothesis. The proof is now complete.
To apply the density theorem we need to show that the
``complex vanishing cycles'' contain a basis for the vanishing cohomology and
form
a single orbit. These cycles are by
definition the eigencomponents of ordinary vanishing cycles.
Consider now a generalized Picard-Lefschetz transformation given by
\eqrefer{Tcxreflectionformula}. It can be rewritten as
$$
\rho(\gamma)(x) =
x + \sum \epsilon_i(\lambda_i-1) h(x,\delta_i) \delta_i,
$$
where the $\delta_i$ are complex vanishing cycles and the $\lambda_i$
are suitable complex numbers. Let
$$
\rho(\gamma')(x) = x + \sum\epsilon_i (\lambda_i-1) h(x,\delta'_i) \delta'_i
$$
be another generalized Picard-Lefschetz tranformation. If
$\gamma' = \kappa^{-1}\gamma\kappa$ then the two preceding equations
yield
$$
\sum \epsilon_i(\lambda_i - 1) h(x,\delta'_i)
\delta'_i
=
\sum \epsilon_i(\lambda_i - 1 ) h(\kappa.x,\delta_i)
\kappa^{-1}.\delta_i ,
$$
where $\kappa.x$ stands for $\rho(\kappa)(x)$.
Comparing eigencomponents on each side we find
$$
\delta'_i = \kappa^{-1}.\delta_i,
$$
as required. By the same argument as used in
\S \xref{zardensitysection}, one sees that the complex vanishing cycles
span $H(i)$.
\section{Bounds on the real and complex rank}
\secref{rankbounds}
In this section we derive lower bounds for the complex and real ranks of the
groups
$G_{d,n}$ of automorphisms of the primitive cohomology
$H^n_o(X_{d,n},\R)$ where
$X_{d,n}$ is a hypersurface of degree $d$ and dimension $n$.
Recall that for a field $k$,
the $k$-rank is the dimension of the largest subgroup that
can be diagonalized over $k$. These bounds
complete the outline of proof. We also show that all the eigenspaces
of the cyclic automorphism $\sigma$ have the same dimension.
The main result is the following:
\proclaim{Lemma}. The complex rank of $G_{d,n}$ is at least
five
for $d \ge 3$, $n \ge 1$, with the exception of $(d,n) = (3,1)$, for which it is
one, and $(d,n) = (4,1), (3,2)$ for which it is three. Under the same
conditions the real rank is at least two with the exception of the cases $(d,n)
= (3,1),\ (3,2)$ for which the real ranks are one and zero, respectively.
\endproclaim
\procref{rankboundslemma}
To prove the first assertion we note that the complex rank is given by
$\rank_\C G_{d,n} = [ B_{d,n} / 2 ]$
where $[x]$ is the greatest integer in $x$ and where
$ B_{d,n} = \dim H^n_o(X_{d,n})$ is the
primitive middle Betti number. To compute it we compute the Euler
characteristic $\chi_{d,n}$ recursively using the fact that a $d$-fold cyclic
cover of $\P^{n}$ branched along a hypersurface of degree $d$ is a hypersurface
of degree $d$ in $\P^{n+1}$. Thus, mimicking the proof of Hurwitz's formula for
Riemann surfaces, we have
$$
\chi_{d,n} = d\,\chi(\P^n - B) + \chi(B) = d(n+1) + (1-d)\chi_{d,n-1} .
$$
Since $\chi_{d,0} = d$, the Euler characteristics of all hypersurfaces are
determined. Rewriting this recursion relation in terms of the $n$-th
primitive Betti number we obtain
$$
B_{d,n} = (d-1)\left(B_{d,n-1} + (-1)^n\right),
\eqn
\eqref{bettinumberrecursion}
$$
{}From it we deduce an expression in closed form:
$$
B_{d,n} = (d-1)^n \,(d-2) + { ( d-1 )^n - (-1)^n \over d } + (-1)^n .
\eqn
\eqref{bettinumberformul}
$$
{}The preceding two formulas
imply that $B_{d,n}$ is an increasing
function of $n$ and of $d$. Now assume $d \ge 3$, $n \ge 1$. Then $d+n\ge 4$.
If $d+n \le 6$, then $(d,n) = (3,3), (4,2), (5,1)$ and $B_{3,3} = 10 , B_{4,2} =
21, B_{5,1} = 12$. Thus $B_{d,n} \ge 10$ except when $d+n = 4$
or $5$. These are the cases $(d,n) = (3,1), (3,2), (4,1)$ where $B_{d,n} =
2,6,6$ respectively. The inequalities on the complex rank are now established.
Let us now turn to the proof of the second assertion of the lemma. For $n$ odd
the group $G_{d,n}$ is a real symplectic group. Its real and complex ranks
are the same, and so the bound follows from the first assertion. For $n$ even
the group
$G_{d,n}$ is the orthogonal group of the cup product on the primitive
cohomology. This bilinear form has signature
$(r,s)$, and the real rank of $G$ is the minimum of $r$ and $s$. The signature is
computed from the Hodge decomposition:
$r$, the number of positive eigenvalues, is the sum of the
$h^{p,q}$ for $p$ even, while $s$ is the sum for $p$ odd. According to the
first inequality of lemma \xref{hodgeinequalities}, the Hodge numbers
$h^{p,q}(d,n)$ of $X_{d,n}$ satisfy
$h^{p,q}(d+1,n) > h^{p,q}(d,n)$. Thus the real rank is an increasing function of
the degree. Consequently it is enough to show that it is at least two for
quartic surfaces and for cubic hypersurfaces of dimension four or more. For
quartic hypersurfaces $h^{2,0} = 1$ and $h^{1,1} = 19$, so $(r,s) = (2,19)$.
For cubic hypersurfaces there is a greatest integer
$p \le n$ such that $h^{p,q} \ne 0$, where $p+q = n$. We will compute this
``first'' Hodge number and see that under the hypotheses of the lemma, $p > q$.
Since $n$ is even,
$h^{p,q}$ and $h^{q,p}$ have the same parity. Thus one of $r$, $s$ is at least
two. According to the second inequality of lemma
\xref{hodgeinequalities}, $h^{p-1,q+1}(d,n) > h^{p,q}(d,n)$ if $p > q$. Thus
$h^{p-1,q+1}(d,n) > h^{p,q}(d,n) > 0$. We conclude that the other component of
the signature, $s$ or $r$, must be at least two. For the Hodge numbers of cubic
hypersurfaces of dimension $n = 3k + r$ where $r = 0$, 1, or 2, one uses the
calculus of \cite{GriffPerRat} to show the following:
(a) if $n \equiv 0 \hbox{ mod } 3$ then the first
Hodge number is $h^{2k,k} = n+2$,
(b) if $n \equiv 1 \hbox{ mod } 3$ then it is $h^{2k+1,k} = 1$,
(c) if $n \equiv 2 \hbox{ mod } 3$ then it is $h^{2k+1,k+1} = (n+1)(n+2)/2$.
When $k > 0$ these Hodge numbers satisfy $p > q$, and so
the proof of the lemma is complete.
\proclaim{Lemma}
Let $h^{p,q}(d,n)$ be the dimension of $H^{p,q}_o(X_{d,n})$. Then
the inequalities below hold:
$$
\eqalign{
& h^{p,q}(d+1,n) > h^{p,q}(d,n) \cr
& h^{p,q}(d,n) > h^{p+1,q-1}(d,n) \hbox{ if $p \ge q$} \cr
}
$$
\endproclaim
\procref{hodgeinequalities}
\noindent{\bf Proof:\ }{} It is enough to prove the inequalities when $X_{d,n}$
is the Fermat hypersurface defined by
$F_d(x) = x_0^d + \cdots + x_{n+1}^d = 0$. Because of the
symmetry $h^{p,q} = h^{q,p}$, it is also enough to prove
the inequalities for $p \ge q$. To this end recall that
$h^{p,q} = \dim R^a$,
where $R$ is the Jacobian ring for $F_d$ and where $a = (q+1)d - (n+2)$
is the degree of the adjoint polynomial in the numerator
of the expression
$$
\hbox{res}\, { A \Omega \over F_d^{q+1} }.
$$
Now there is a map $\mu: R^{a(q,d)}(F_d) \map R^{a(q,d+1)}(F_{d+1})$ defined
by $\mu(P) = (x_0 \cdots x_q) P$. This makes sense because
$q \le n$. We claim that that resulting map from $H^{p,q}(X_{d,n})$ to
$H^{p,q}(X_{d+1,n})$ is injective but not surjective.
To prove the claim, observe that the Jacobian ideal is generated by
the powers $x_i^{d-1}$ and so
has a vector space basis consisting of monomials $x^M$. The same
is true of the quotient ring $R(F_d)$.
Indeed, a basis is given by (the classes of)
those monomials not divisible by $x_i^{d-1}$
for any $i$. Now consider a polynomial which represents
an element of the kernel of $\mu$.
It can be be chosen to be a linear combination of monomials $x^M$ which are not
divisible by $x_i^{d-1}$ for any $i$. Its image is represented by a linear
combination of monomials $(x_0 \cdots x_q)x^M$. Each of these is divisible by some
$x_i^d$. Thus either $x^M$ is divisible by $x_i^d$, $i > q$, a contradiction,
or by $x_i^{d-1}$, $i \le q$, also a contradiction. Thus injectivity part the claim
is established.
For the surjectivity part note that image of the map $\mu$ has a basis of monomials
$x^M$ which are divisible by $x_i$ for $i = 0 \commadots q$. Thus, to show that
$\mu$ is not surjective it suffices to show that there is a monomial for
$R^{a(q,d+1)}(F_{d+1})$ that is not divisible by $x_0$. Such a monomial has the
form
$x_1^{M_1} \cdots x_{n+2}^{M_{n+2}}$ where $M_i \le d-1$. It
exists if $a(q,d+1) \le (n+1)(d-1)$. The largest relevant values of
$q$ and $a(q,d+1)$ are $n/2$ and $(n/2 + 1)d - (n+2)$. For these the
preceding inequality holds and so the first inequality
of the lemma holds strictly.
For the second inequality we use the fact that basis elements for the
Jacobian ring of $F_d$ correspond to lattice points of the cube in $(n+2)$-space
defined by the inequalities $0 \le m_i \le d-2$. A basis for $R^a$
corresponds to the set of lattice points which lie on the convex subset $C(a)$
of the cube obtained by slicing it with the hyperplane $m_0 + \cdots + m_{n+1} =
a$. The volume of $C(a)$ is a strictly increasing function of $a$ for
$0 \le a \le t/2$, where $t = (n+2)(d-2)$. For $t/2 \le a \le t$ the volume
function $V(a)$ is strictly decreasing, and in general its graph is symmetric around
$a = t/2$. Let $L(a)$ be the number of lattice points in $C(a)$. If $L(a)$ satisfies
the same monotonicity properties as does $V(a)$, then the second inequality
follows. To show this, we prove the following result.
\proclaim{Lemma.} Let $L_{d,n}(k)$ be the number of points in the set
$\LL_{d,n}(k) = \sett{ x \in \Z^n }{ 0 \le x_i \le d,\ x_1 + \cdots + x_n = k }$.
Assume that $n > 1$. Then $L_{d,n}(k)$ is a strictly increasing function of $k$ for $k < dn/2$
and is symmetric around $k = dn/2$.
\endproclaim
\noindent{\bf Proof:\ }{} Symmetry follows from the bijection
$\LL_{d,n}(k) \map \LL_{d,n}(dn - k)$ given by $x \mapsto \delta - x$
where $\delta = (d \commadots d)$. We shall say that these two sets
are dual to eachother. For the inequality we argue by induction, noting
first that
$L_{d,2}(k) = k + 1$ for $k \le d$. Now observe that $\LL_{d,n}(k)$
can be written as a disjoint union of sets
$S_i = \sett{ x \in \LL_{d,n}(i) }{ x_n = k-i }$ where $i$ ranges from
$k-d$ to $k$. Thus
$$
L_{d,n}(k) = \sum_{i = k-d}^k L_{d,n-1}(i) .
$$
Consequently
$$
L_{d,n}(k+1) - L_{d,n}(k) = L_{d,n-1}(k+1) - L_{d,n-1}(k-d) .
$$
By the induction hypothesis the right-hand side is positive if
$k-d < (n-1)d/2$ and if $k+1$ is not greater than the index dual to $k-d$,
namely $(n-1)d - (k-d)$. Thus we require also that $k+1 \le (n-1)d - (k-d)$.
Both inequalities hold if $k < nd/2$, which is what we assume. Thus the proof
is complete.
\subheading{Dimension of the eigenspaces.}
We close this section by noting that the eigenspaces $H^n(X)(\lambda)$
for $\lambda\ne 1$ all have the same dimension, explaining why the
primitive middle Betti number is divisible by $d-1$, where $d$ is the
degree. Indeed, we have the following,
$$
\dim H^n(X,\C)(\lambda)
= \dim H^n(X,\C)(\mu)
= \dim H^n(\P^n - B,\C) + (-1)^n .
\eqn
\eqref{eigenspacedimform}
$$
When the degree is prime there is a short proof: consider the field $k
= \Q[\omega]$ where $\omega$ is a primitive $d$-th root of unity and
observe that its Galois group permutes the factors $H^n(X,k)(\lambda)$
for $\lambda \ne 1$. For the general case let $p: X \map \P^n$ be the
projection and note that $H^n(X,\C) = H^n(\P^n, p_*\C)$. The group of
$d$-th roots of unity acts on $p_*\C$ and decomposes it into
eigensheaves $\C_\lambda$, where $\lambda^d = 1$. Thus the
$\lambda$-th eigenspace of $H^n(X,\C)$ can be identified with
$H^n(\P^n,\C_\lambda)$. The component for $\lambda = 1$ is
one-dimensional and is spanned by the hyperplane class. For $\lambda
\ne 1$ the sheaf $\C_\lambda$ is isomorphic to the extension by zero
of its restriction to $\P^n - B$. Thus the eigenspace can be
identified with $H^n(\P^n - B, \C_\lambda)$. By the argument of
lecture 8 in
\cite{CKM} used in the proof of vanishing theorems, the groups $H^i(\P^n - B,
\C_\lambda)$ vanish for
$i \ne n$, $\lambda \ne 1$. Thus $\dim H^n(\P^n - B,\C_\lambda) =
(-1)^n \chi(\lambda)$, where $ \chi(\lambda)$ is the Euler
characteristic of $\C_\lambda$. Fix a suitable open tubular
neighborhood $U$ of $B$ and a good finite cell decomposition $K$ of
$\P^n - U$. Then $\chi(\lambda)$ is the Euler characteristic of the
complex of $\C_\lambda$-valued cochains on $K$, which depends only on
the number of cells in each dimension, not on $\lambda$. This
establishes the first equality above. For the second use
$\chi(\lambda) = \chi(1)$ and the vanishing of $H^i(\P^n - B,\C)$ for
$i \ne n, 0$.
\section{Remarks and open questions}
\secref{remarkssection}
We close with some remarks on (a) the possiblity of an isomorphism $\Phi \cong
\Gamma\times\Gamma'$, (b) the impossibility of producing additional
representations by iterating the suspension (globally), and (c) generalizations of
the main theorem.
\subheading{(A) Products}
So far everything that has been said is consistent with
an isomorphism between $\Phi$ and the product
$\Gamma\times\Gamma'$, where $\Gamma'$ is the monodromy group
$\bar \rho'(\Phi)$. This, however, is not
the case, at least for surfaces, for we can show that {\sl if $k$ is a divisor of $d$ and $d$ is
odd, then $\Phi_{d,2}$ and $\Gamma\times\Gamma'$ are not isomorphic}.
The argument is based on the fact that the abelianization
of $\Phi$ is a cyclic group of order equal to the
degree of the discriminant, which we denote by $r$.
This is because (a) the generators $g_1 \commadots g_r$ of $\Phi$ are mutually
conjugate, hence equal in the abelianization, (b) $g_1 \cdots g_r = 1$, (c)
the additional relations are trivial when abelianized. See \cite{Zar}.
For the last point note that $\Phi$ is also the fundamental group
of the complement of a generic plane section $\Delta'$
of $\Delta$. This complement has nodes and cusps as its
only singularities. The nodes yield relations of the
form $gg' = g'g$ where $g$ and $g'$ are conjugates of the
given generators. The cusps yield braid relations
$gg'g = g'gg'$. Both are trivial in the abelianization.
Thus the abelianization is generated by a single
element with relation $g^r = 1$. The degree of
the discriminant is given in \cite{DolgLib}, page 6, line 2:
$$
r = \hbox{deg}(\Delta) = 4(d-1)^3.
$$
If $\Phi$ is isomorphic to the Cartesian product,
then there is a corresponding isomorphism of abelianizations.
Let us therefore compute what we can of
the abelianizations of $\Gamma$ and $\Gamma'$.
For $\Gamma$ we note that the generators are the elements
$g_i$ as above satisfying additional relations which include
$g_i^2 = 1$. Therefore $\Gamma$ abelianized is a quotient of $\Z/2$.
Consider next the case of $\Gamma'$ for cyclic covers of
degree $k$. Then $\Gamma'$
is a product of groups
$\Gamma'(i)$ for $i = 1 \commadots k-1$. Generators
and relations are as in the previous case except that
among the additional relations are $g_i^{2k} = 1$ instead of $g_i^2 = 1$.
Therefore the abelianization is a quotient of $\Z/{2k}$.
Consequently the abelianization of the
product $\Gamma\times \Gamma'$ is a quotient
of the product of $\Z/2$ with a product of $\Z/2k$'s.
But the largest the order of an element in such
a quotient can be is $2k$, which is always
less than the degree of the discriminant,
provided that $d > 2$, which is the case.
\subheading{(B) Suspensions}
Since $\Phi$ is not in general isomorphic to
$\Gamma\times\Gamma'$ it is natural to ask
whether there are further representations
with large kernels. One potential
construction of new representations is
given by iterating the suspension. By this we mean that
we take repeated double covers. Unfortunately,
this produces nothing new, since it turns
out that the global suspension is periodic of period two.
To make a precise statement, let $P(x)$ be a polynomial
of degree $2d$ which defines
a smooth hypersurface $X$ in $\P^n$.
Let $X(2)$ be the hypersurface defined by
$$
P(x) + y_1^2 + y_2^2
$$
in a weighted projective space $\P^{n+2}$
where the $x_i$ have weight one and the
$y_i$ have weight $d$. {\sl Then there is an
isomorphism
$$
H^n_o(X)\otimes T \map H^{n+2}_o(X(2)),
$$
where $T$ is a trivial Hodge structure
of dimension one and type $(1,1)$ and
where the subscript denotes primitive cohomology}.
For the proof we note that the map
$$
{ A\Omega(x) \over P^{q+1} }
\mapsto
{ A\Omega(x, y_1, y_2) \over ( y_1^2 + y_2^2 +P )^{q+2} }
$$
is well-defined and via the residue provides an isomorphism
compatible with the Hodge filtrations which is defined over the
complex numbers. However, it can be defined geometrically
and so is defined over the integers. To see why, consider
first the trivial case $g(x) = f(x) + y_1^2 + y_2^2 = 0$ in affine
coordinates, where $x$ is a scalar variable and $f$ has degree
$2d$. Thus $f(x) = 0$ defines a finite point set, and
$g(x) = 0$ is its double suspension. Let $p$ be one point
of the given finite set. Then $f(p) = 0$, so the locus
$\sett{ (p,y_1,y_2) }{ y_1^2 + y_2^2 = 0 }$ lies on the double
suspension. This locus is a pair of lines meeting in a point, and
the statement remains true in projective coordinates. Thus we
may associate to $p$ a difference of lines $\ell_p - \ell'_p$.
This map induces an isomorphism $H_0(X,\Z) \map H_2(X(2),\Z)$
which is in fact a morphism of Hodge structures. For the general
case we parametrize the construction just made. The map in cohomology
which corresponds to the previous construction is the dual of the
inverse of the map in homology.
\subheading{(C) Generalizations}
The main theorem \xref{maintheorem} can be generalized in a number of
ways. First, using the techniques of \cite{Tu}, it is certainly
possible to get sharp results for various kinds of weighted
hypersurfaces, just as we have obtained sharp results for standard
hypersurfaces. Second, one can prove a quite general (but not
sharp) result that reflects the fairly weak hypothesis of criterion
\xref{largekernelcriterion}:
\proclaim{Theorem} Let $L$ be a positive line bundle on a projective
algebraic manifold $M$ of dimension at least three. Let $P$ be the
projectivization of the space of sections of $L^d$, and let $\Delta$
be the discrimant locus defined by sections of $L^d$ whose zero set $Z$
is singular. Then for $d$ sufficiently large the kernel of
the monodromy representation of $\Phi = \pi_1(P - \Delta)$ is large
and its image is a lattice.
\endproclaim
The monodromy representation has the primitive cohomology
$$
H^{m-1}(Z)_0 = \mathop{kernel}{\left[ H^{m-1}(Z) \mapright{Gysin} H^{m+1}(M)
\right]}
$$
as underlying vector space. The results needed for the proof are all in
the literature. First, note that the condition that a section $s$ of
$L^d$ have a singularity of type ``$x^3 + y^3 + z^4 + \hbox{sum of
squares}$'' at a given point is set of linear conditions and so can be
satisfied for $d$ sufficiently large. Consequently by the
Beauville-Ebeling-Janssen argument, the image of the natural monodromy
representation is a lattice. Second, by the results of Green
\cite{GreenPM}, the local Torelli theorem for cyclic covers holds for $d$
sufficiently large, so some component of the second monodromy
representation has nonzero differential. The standard argument used
just following Theorem
\xref{practicaldensitycriterion} proves that the discriminant locus is
irreducible, and so Theorem
\xref{udensityprop} applies to give Zariski-density for the
second monodromy representation. Finally, the Hodge numbers, like the
standard case of projective hypersurfaces, are polynomials in $d$ with
positive leading coefficient and of degree equal to the dimension of
$M$. Consequently they are large for $d$ large, and therefore both the
real and complex rank of the relevant algebraic groups can be assumed
sufficiently large by taking $d$ large enough. Thus the hypotheses of
criterion \xref{largekernelcriterion} are satisfied.
For a quick proof of the statement on the behavior of the Hodge numbers,
consider first the Poincar\'e residue sequence
$$
0 \map \Omega^m_M \map \Omega^m_M(L^d) \map \Omega^{m-1}_Z \map 0,
$$
where $Z$ is a smooth divisor of $L^d$ and $m$ is the dimension of $M$.
From the Kodaira vanishing theorem we have
$$
H^0(\Omega^{m-1}_Z)_0
\cong
\mathop{cokernel}{ \left[
H^0(\Omega^m_M) \map H^0(\Omega^m_M(L^d))
\right] }.
$$
By the Riemann-Roch theorem the dimension of the right-most term
is a polynomial with leading coefficient $Cd^m$, while
the dimension of the middle term is constant as a
function of $d$. Therefore the Hodge number in question
is a polynomial in $d$ of the required form.
For the other Hodge numbers we use the identification
$$
H^q(\Omega^p_Z)_0
\cong
\mathop{cokernel}{ \left[ H^0(\Omega^{m-1}_Z\otimes\Theta_M\otimes
N_Z^{q-1} )
\map
H^0(\Omega^{m-1}_Z\otimes N_Z^q) \right] },
$$
where $N$ is the normal bundle of $Z$ in $M$, where $\Theta_M$
is the holomorphic tangent bundle of $M$, and where $p+q=m-1$. See Proposition 6.2,
\cite{JC}, a consequence of Green's Koszul cohomology formula (Theorem
4.f.1 in \cite{GreenKC}) for $d$ sufficiently large. Now tensor the
Poincar\'e residue sequence with $L^{qd}$ to get
$$
0
\map
\Omega^m_M(L^{qd})
\map
\Omega^m_M(L^{(q+1)d})
\map
\Omega^{m-1}_Z \otimes N_Z^q
\map
0 .
$$
From the Kodaira vanishing theorem and the Riemann-Roch formula one
finds that the dimension of the right-hand part of the cokernel formula
is a polynomial with leading term $C((q+1)d)^m$, where
$C = c_1(L)^m/m!$. A similar argument shows that the dimension of
the left-hand part is a polynomial with leading coefficient
$C(qd)^m$. Thus the leading term of $\dim H^q(\Omega^p)_0$ is bounded
below by a positive constant depending on $L$, $q$, and $M$, times $d^m$.
\subheading{(D) Questions.}
We close with some open questions. The main
problem is, of course, to understand the nature of
the groups $\Phi_{d,n}$. Are they linear?
Are they residually finite? It
seems reasonable to conjecture that in general
they are not linear groups, and, in particular,
are not lattices in Lie groups. We settle this last question
for $\Phi_{3,2}$ in the note \cite{ACT}.
The structure of $\Phi_{d,n}$ is
closely related to the structure of the
kernel $K$ of the natural monodromy
representation. For $n=0$, $K$ is the pure $d$-strand
braid group the sphere and so is finitely
generated. For $(d,n) = (3,2)$, the case of cubic
surfaces, $K$ is not finitely generated (see \cite{ACT}).
It is therefore natural to ask when $K$ is finitely generated
and when it is not.
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\endbibliography
\bigskip
\begingroup
\obeylines
\parskip=0pt
\parindent1cm
\baselineskip=10pt
\def\vskip7pt{\vskip7pt}
Department of Mathematics
University of Utah
Salt Lake City, Utah 84112
\vskip7pt
[email protected] $\qquad$ [email protected]
\vskip7pt
http://www.math.utah.edu/$\sim$carlson
http://xxx.lanl.gov --- alg-geom/9708002
To appear in Duke J. Math.
\endgroup
\enddoc
\end
|
1997-08-14T10:59:42 | 9708 | alg-geom/9708012 | en | https://arxiv.org/abs/alg-geom/9708012 | [
"alg-geom",
"math.AG"
] | alg-geom/9708012 | Lothar Goettsche | Barbara Fantechi, Lothar G\"ottsche, Duco van Straten | Euler number of the compactified Jacobian and multiplicity of rational
curves | LaTeX, 16 pages with 1 figure | null | null | null | null | We show that the Euler number of the compactified Jacobian of a rational
curve $C$ with locally planar singularities is equal to the multiplicity of the
$\delta$-constant stratum in the base of a semi-universal deformation of $C$.
In particular, the multiplicity assigned by Yau, Zaslow and Beauville to a
rational curve on a K3 surface $S$ coincides with the multiplicity of the
normalisation map in the moduli space of stable maps to $S$.
| [
{
"version": "v1",
"created": "Thu, 14 Aug 1997 08:59:50 GMT"
}
] | 2008-02-03T00:00:00 | [
[
"Fantechi",
"Barbara",
""
],
[
"Göttsche",
"Lothar",
""
],
[
"van Straten",
"Duco",
""
]
] | alg-geom | \section{Introduction}
Let $C$ be a reduced and irreducible projective curve with singular set
$\Sigma \subset C$ and let $n: \widetilde{C} \longrightarrow C$ be
its normalisation. The generalised Jacobian $JC$ of $C$ is an extension of
$J\widetilde{C}$ by an affine commutative group of dimension
$$\delta:=\dim H^0(n_*({\cal O}_{\widetilde{C}})/{\cal O}_C)=\sum_{p \in \Sigma}
\delta(C,p)$$
so that $\dim JC=\dim J\widetilde{C} +\delta=g(\widetilde{C})+\delta$
is equal to
the arithmetic genus $g_a(C)$ of $C$. The non-compact space $JC$
is naturally an
open subset of the {\em compactified Jacobian} ${\overline{J}} C$ of $C$,
whose points
correspond to isomorphism classes of rank one torsion free sheaves
${\cal F}$ of degree
zero (i.e. $\chi({\cal F})=1-g_a(C)$) on $C$. The space ${\overline{J}} C$ is irreducible
if and only
if $C$ has planar singularities; then ${\overline{J}} C$ is in fact a
compactification of
$JC$, i.e., $JC$ is dense in ${\overline{J}} C$. If moreover $C$ is rational
and unibranch,
then ${\overline{J}} C$ is topologically the product of compact spaces
$M(C,p)$ for every
$p\in \Sigma$. The space $M(C,p)$ only depends on the
analytic singularity
$(C,p)$; it can be defined as ${\overline{J}} D$ for any rational curve $D$
having $(C,p)$ as
unique singularity.\\
Let $B=B(C,p)$ be the base of a semi-universal deformation of
the singularity
$(C,p)$. Inside $B$ let $B^\delta=B^\delta(C,p)$ be
the locus of points for which $\delta$ remains constant. This
means that
$$t \in B^\delta \Leftrightarrow \sum_{p \in C_t} \delta(C_t,p)
=\delta(C).$$
The codimension of $B^\delta$ is $\delta(C,p)$; its multiplicity
$m(C,p)$ at
$[(C,p)]$ is by definition equal to the
number of intersection points with a generic $\delta$-dimensional
smooth subspace
of $B$. The $\delta$-constant stratum can be defined in a similar
way for a
semi-universal deformation of a projective curve with only planar
singularities.
In this paper we show the following theorem.
{\bf Theorem 1.}
{\sl Let $(C,p)$ be a reduced plane curve singularity. Then the
Euler number of
$M(C,p)$ is
equal to the multiplicity of the $\delta$-constant stratum:
$$e(M(C,p))=m(C,p).$$
Let $C$ be a projective, reduced rational curve with only planar
singularities.
Then $e({\overline{J}} C)= m(C)$, the multiplicity of the $\delta$-constant
stratum
$B^\delta$ at $0$.}
Note that this gives an independent proof of the following result
of Beauville:
Let $C$ be an irreducible and reduced rational curve with planar
singularities.
Then $e({\overline{J}} C)$
can be written as a product over the singularities of $C$ of a
number only
depending on the type of the singularity, and it is the same
for $C$ and its
minimal unibranch partial normalisation.
Theorem 1 has an application in the following situation.
Let $X$ be a (smooth) $K3$ surface with a complete
(hence $g$-dimensional) linear
system of curves of genus $g$.
Under the assumption that all curves in the system are
irreducible and reduced,
it was shown in \cite{Y-Z} and \cite{B} that the ``number''
$n(g)$ of rational
curves
occuring in the linear system, is equal to the $g^{\rm th}$
coefficient of the
$24^{\rm th}$ power of
the partition function, i.e:
$$\sum_{g \ge 0}n(g)q^g=\frac{q}{\Delta(q)}$$
where $\Delta(q)=q\prod_{n \ge 1}(1-q^n)^{24}$.
In this counting, a rational curve $C$ in the linear
system contributes
$e({\overline{J}} C)$ to $n(g)$:
$$n(g)=\sum_{C} e({\overline{J}} C).$$
If $C$ is a rational curve with only nodes as singularities,
then $e({\overline{J}} C) =1$,
so that $e( {\overline{J}} C)$
seems to be a reasonable notion of multiplicity.
Theorem 1 implies that $e({\overline{J}} C)$
is always positive, and in principle allows an
explicit computation of it (see
section {G}).\\
In fact, we prove a more precise statement. For any
projective scheme $Y$ and
$d \in H_2(Y,{\bf Z})$ let $M_{0,0}(Y, d)$ be the
moduli space of
genus zero stable maps $f: {\bf P}^1 \longrightarrow Y$ with
$f_*([{\bf P}^1])=d$.
Under the above assumptions on the K3 surface $X$
and the linear system
corresponding to
$d$, the space $M_{0,0}(X,d)$ is a zero-dimensional scheme.
If $C \stackrel{i}{\hookrightarrow} X$ is a rational
curve in $X$ (always assumed
to be irreducible and reduced), $n:{\bf P}^1 \longrightarrow C$ its
normalisation, then $f=i
\circ n:{\bf P}^1 \longrightarrow X$ is a point of
$M_{0,0}(X,d)$. The moduli space
$M_{0,0}(X,d)$ contains naturally as a closed subscheme
$M_{0,0}(C,[C])$, the submoduli space of maps whose scheme
theoretic image
is $C$; the latter scheme is of course defined for any
projective reduced curve
$C$, and it is zero-dimensional if the curve is rational.
More generally, $M_{g,0}(C,[C])$ is zero dimensional, where
$g$ denotes the genus of the normalisation of $C$. The following
theorem gives another
interpretation of $e({\overline{J}} C)$ in terms of the length of such
zero-dimensional
schemes.\\
{\bf Theorem 2.}
{\sl Let $C$ be a reduced, irreducible projective curve with
only planar
singularities, and let $g$ be the genus of its normalisation.
Then $m(C)=l(M_{g,0}(C,[C]))$. If moreover $C$ is rational and
contained in a
smooth $K3$ surface $X$, then $e({\overline{J}} C)=
l(M_{0,0}(X,d),f)$ (length of the zero-dimensional component
supported at $f$).}
We now sketch briefly the idea of the proof of Theorem 1.
Let ${\cal C}\to B$ be a semi-universal family of deformations
of a curve $C$
with planar singularities. We prove that the relative
compactified Jacobian $\bar
J {\cal C}$ is smooth; moreover, given any deformation ${\cal C}'\to S$
of $C$ with a smooth
base, $\bar J{\cal C}'$ is smooth if and only if the image of $TS$
is transversal in
$TB$ to the $\delta$-codimensional vector space $V$, the
support of the tangent
cone to the $\delta$-constant stratum $B^\delta$.
Assume now $C$ is rational and has $p$ as unique singularity.
We have to show that
$e(\bar JC)=m(C,p)$. Choose a one-parameter family $W_t$
of smooth
$\delta$-dimensional subspaces of $B$ such that $0\in W_0$,
$T_{W_0,0}\cap
V=\{0\}$, and for general $t$ the intersection
$W_t\cap B^\delta$ is a set of
$m(C,p)$ distinct points corresponding to nodal
curves.
Let ${\cal C}_t\to W_t$ be the induced families. Then
$\bar J{\cal C}_t$ is a family of
smooth compact varieties, hence $e(\bar J{\cal C}_t)$
does not depend on $t$. Arguing as in
\cite{Y-Z} and \cite{B}, we prove that
$e(\bar J{\cal C}_0)=e({\overline{J}} C)$, while $e({\overline{J}} {\cal C}_t)=m(C,p)$
for $t$ general.
{\bf Conventions.} In this paper we will always work
over the complex numbers, and
open will mean open in the strong (euclidean) topology
(unless of course we specify Zariski open).
{\bf Preliminaries.} We will use the language of
deformation functors; we recall a
few facts about them for the reader's convenience.
A deformation functor $D$ will always be a covariant
functor from local artinian
${\bf C}$-algebras to sets, satisfying Schlessinger's
conditions $(H1)$, $(H2)$,
$(H3)$, hence admitting a hull (see \cite{Sch}).
In particular $D$ admits a
finite-dimensional tangent space, which we denote by
$TD$, functorial in $D$. A
functor is smooth if its hull is. The dimension of
the functor will be equal to
the dimension of the hull. We will need the following
elementary result.
{\bf Lemma.} {\sl Let $X\to Y$ and $Z\to Y$ be morphisms
of smooth deformation
functors. Then $X\times_YZ$ is smooth of dimension
$\dim X+\dim Z-\dim Y$ if and
only if the images of $TX$ and $TZ$ span $TY$.}\\
{\bf Proof.} Base change considerations reduce the
problem to the case of
prorepresentable functors, where it is obvious.\hfill $\Diamond$
It would be possible to replace deformation functors
with contravariant functors
on the category of germs of complex spaces, and the
hull with the base of a
semi-universal family of deformations.
The two viewpoints correspond to working with formal
versus convergent power
series.
{\bf Acknowledgements.} This paper was written at the
Mittag-Leffler Institute in
Stockholm, during a special year on Enumerative Geometry.
The authors are grateful
for the support received and for making our
collaboration possible.
The first author is a member of GNSAGA of CNR.
\section{A. Deformations of curves and sheaves.}
Let $C$ be a reduced projective curve, with
singular set $\Sigma$.
Any deformation ${\cal C} \longrightarrow S$ of $C$ over a
base $S$ induces a deformation of its
singularities. More
precisely, one can introduce the {\em functor
of local deformations} by
letting
$D^{loc}(C)(T)$ be the set of isomorphism
classes of data $(U_i,U_i^T)_{i\in
I}$, where $(U_i)_{i\in I}$ is an affine open
cover of $C$ and,
for each $i$, $U_i^T$ is a deformation of $U_i$
over $T$; we require that the
induced deformations of $U_{ij}:=U_i\cap U_j$
be the same.
There is a natural
transformation of functors $\hbox{\sl loc}: D(C) \longrightarrow D^{loc}(C)$;
the induced map of
tangent spaces can be identified with the edge homomorphism
$${\bf T}_C^1 \longrightarrow H^0({\cal T}_C^1)$$
of the local-to-global spectral sequence for the ${\cal T}^i$.
The kernel of this map is $H^1(\Theta_C)$, the cokernel
injects in
$H^2(\Theta_C)$ which is zero. The obstruction space
${\bf T}_C^2$ sits
in an exact sequence
$$0 \longrightarrow H^1({\cal T}_C^1) \longrightarrow {\bf T}_C^2 \longrightarrow H^0({\cal T}_C^2)
\longrightarrow 0.$$
As $C$ is reduced, ${\cal T}_C^1$ is supported on a finite
set of points, hence
$H^1({\cal T}_C^1)=0$. If $C$ has locally complete
intersection singularities, then also ${\cal T}_C^2 =0$, so that
in that case ${\bf T}_C^2=0$. Hence in such a situation, and
in particular when
$C$ is a reduced curve with only planar singularities, the
functors $D(C)$ and $D^{loc}(C)$ are smooth and $\hbox{\sl loc}$
is a smooth map.
Let ${\cal F}$ be a torsion free coherent sheaf on $C$. Analogously
, we denote by
$D(C,{\cal F})$ the functor of deformations of the pair, and
define the functor of
local deformations by letting $D^{loc}(C,{\cal F})(T)$ be the
set of isomorphism
classes of data $(U_i,U_i^T,F_i^T)_{i\in I}$ where
$(U_i)_{i\in I}$ is an affine
open cover of $C$, and for each $i$, $(U_i^T,F_i^T)$
is a $T$-deformation
of $(U_i,{\cal F}|_{U_i})$ such that the induced deformations
on $U_{ij}$ are the same.
Again we have a localisation map
$D(C,{\cal F})\to D^{loc}(C,{\cal F})$.
The four functors introduced sit in a natural commutative diagram
$$
\begin{array}{ccc}
D(C,{\cal F}) & \longrightarrow & D^{loc}(C,{\cal F})\\
\downarrow&&\downarrow\\
D(C) & \longrightarrow & D^{loc}(C)\\
\end{array}
$$
with horizontally localisation maps and vertically forget maps.
Note that this diagram in general is {\em not} cartesian.
{\bf Proposition A.1.} {\sl
The canonical map
$$ D(C,{\cal F}) \longrightarrow D(C) \times_{D^{loc}(C)}D^{loc}(C,{\cal F})$$
is smooth.}
{\bf Proof.}
We have to show the following: Let ${\cal F}_T$, $C_T$ be flat
deformations of
$C$ and ${\cal F}$ over $T$, $\xi_T \in D^{loc}(C,{\cal F})(T)$
the induced local
deformation. If we are given lifts $C_{T'}$ and $\xi_{T'}$
over a
small extension $T'$ of $T$, then we can lift ${\cal F}_T$ to
a deformation
${\cal F}_{T'}$ of ${\cal F}$ over $C_{T'}$ inducing $\xi_{T'}$.
This can be done
as follows: choose an affine open cover $U_i$ of $C$
such that $\xi_{T'}$
is defined by coherent sheaves $F_i'$ on the induced
cover $U_{i,T'}$ of $C_{T'}$.
Assume also that $U_{ij}:=U_i\cap U_j$ is smooth for every $i\ne j$.
Let $F_i$ be the restriction of $F_i'$ to $U_{i,T}$.
The fact that ${\cal F}$ induces
$\xi_T$ means that we can find isomorphisms
$\phi_i:{\cal F}_T|_{U_{i,T}}\to F_i$.
The $\phi_i$ induce isomorphisms $\phi_{ij}:F_i\to F_j$
over $U_{ij,T}$,
satisfying the cocycle condition. What we need to prove
is that the $\phi_{ij}$
can be lifted to $\phi_{ij}':F'_i\to F'_j$, again
satisfying the cocycle
condition; then the $\phi_{ij}'$ can be used to glue
together the $F_i'$'s to
a coherent sheaf ${\cal F}_{T'}$ as required. But on $U_{ij}$
all the sheaves under
consideration are line bundles, hence the obstruction to
the existence of such a
lifting is an element in $H^2(C_,{\cal O}_{C})$, which is
zero as $C$ has dimension
$1$. \hfill $\Diamond$
If $R$ is a ring and $M$ is an $R$-module, we denote by $D(R)$,
respectively $D(R,M)$ the corresponding deformation functors.
{\bf Lemma A.2}. {\sl Let $C$ be a reduced projective curve, ${\cal F}$
a torsion free
module on $C$. Let $\Sigma$ denote the singular locus.
Then the natural morphisms of functors $$
D^{loc}(C)\to \prod_{p\in \Sigma}D({\cal O}_{C,p})\qquad
\hbox{\sl and}\qquad
D^{loc}(C,{\cal F})\to \prod_{p\in \Sigma}D({\cal O}_{C,p},{\cal F}_p)$$
are isomorphisms.}
{\bf Proof.} Both morphisms are clearly injective. On the other
hand,
surjectivity is obvious since on the smooth open locus, every
infinitesimal
deformation is locally trivial and every torsion free sheaf is
locally free.
\hfill $\Diamond$
{\bf Proposition A.3}. {\sl Let $P$ be a regular local ring of
dimension $2$,
$f\in P$ a nonzero element, and $R=P/(f)$; assume that $R$ is
reduced. Let $M$ be
a finitely generated, torsion free
$R$-module of rank $1$. Then $D(R,M)$ is a smooth functor.}
{\bf Proof.}
As it is torsion free, the module $M$ has depth $1$. By the
Auslander-Buchsbaum theorem (see e.g. \cite{Ma}), $M$ has a
free resolution of length $1$ as a $P$-module, so is
represented as the cokernel
of some
$n \times n$ matrix $A$ with entries from $P$:
$$ 0 \longrightarrow P^n \stackrel{A}{\longrightarrow}P^n \longrightarrow M \longrightarrow 0$$
As $M$ is an $R$-module of rank $1$, the determinant ideal
$(det(A))$ is equal
to $(f)$.\\
Any flat deformation $M_T$ of $M$ over $T$ (as $P$-module)
is obtained by
deforming the matrix
$A$ to a matrix $A_T$ with entries from
$P_T:=T \otimes_{{\bf C}} P$, so that
$M_T$ has a presentation
$$0 \longrightarrow P_T^n \stackrel{A_T}{\longrightarrow}P_T^n \longrightarrow M_T \longrightarrow 0.$$
There is a unique deformation $R_T$ of $R$ over $T$
such that $M_T$ is a flat $R_T$-module, given by the ideal
$(det(A_T))$.
It follows that the natural transformation
$$D(A) \longrightarrow D(R,M)\qquad A_T \mapsto
(P_T/det(A_T),Coker(A_T))$$
is {\em smooth}.
As $D(A)$, the functor of deformations of the matrix $A$,
is clearly smooth,
the functor $D(R,M)$ is also smooth. \hfill $\Diamond$
Note that in the assumption of A.3, although both functors
$D(R,M)$ and $D(R)$ are
smooth, the forgetful morphism $D(R,M)\to D(R)$ is not smooth
in general.
{\bf Remark A.4.} Let $R$ be a one-dimensional local
${\bf C}$-algebra, and let $M$ be a finitely generated torsion free
$R$-module. Let $\hat R$ be the completion of $R$, and
$\hat M=M\otimes_R\hat R$.
The natural morphism $D(R,M)\to D(\hat R, \hat M)$
is smooth and induces an isomorphism on tangent spaces,
and the same is true for
$D(R)\to D(\hat R)$. In fact it is easy to see that
the induced morphisms of
tangent and obstruction spaces are isomorphisms.
\section{B. Relative compactified Jacobians.}
For any flat projective family of curves ${\cal C}\to S$
we let ${\overline{J}} {\cal C}\to S$
be the
relative compactified Jacobian (see \cite{R}).
For every closed point $s \in S$
the fiber over $s$ of ${\overline{J}} {\cal C}$ is canonically
isomorphic to the compactified
Jacobian ${\overline{J}} C_s$;
in particular, its points correspond to isomorphism
classes of torsion free rank
$1$ degree zero sheaves on $C_s$.
Fix a point ${\cal F} \in {\overline{J}} {\cal C}$ over $s\in S$, and
denote again by $({\overline{J}}{\cal C},{\cal F})$ and $(S,s)$
the deformation functors induced by the respective
germs of complex spaces.
Let $C=C_s$. Remark that if ${\cal C} \longrightarrow S$ is a
semi-universal family of
deformations of $C$, then we have an isomorphism
of functors
$$ ({\overline{J}}{\cal C}, {\cal F}) \simeq D(C,{\cal F}) $$
For a general flat family one has a natural
commutative diagram
$$
\begin{array}{ccc}
({\overline{J}}{\cal C},{\cal F}) & \longrightarrow & D^{loc}(C,{\cal F})\\
\downarrow&&\downarrow\\
(S,s) & \longrightarrow & D^{loc}(C)\\
\end{array}
$$
and analogous to {\bf A.1.} one has:
{\bf Proposition B.1.} {\sl
The canonical map
$$ ({\overline{J}}{\cal C},{\cal F}) \longrightarrow (S,s) \times_{D^{loc}(C)}
D^{loc}(C,{\cal F})$$
is smooth.}
We omit the proof, which is almost identical to
that of {\bf A.1.}
{\bf Corollary B.2.} {\sl
Let $C$ be a reduced curve with only plane curve
singularities.
If ${\cal C}\to S$ is a versal family of deformations
of $C$, then ${\overline{J}} {\cal C}$ is smooth along ${\overline{J}} C$, and
${\overline{J}} C$ has local complete intersection singularities.}
{\bf Proof.} The family is versal if and only if the
natural map $S\to D(C)$ is
smooth. This in turn implies that $S\to D^{loc}(C)$
is smooth, hence the first
claim follows from Proposition B.1. On the other hand,
all fibres of ${\overline{J}} {\cal C}\to
S$ have the same dimension $g_a(C)$, therefore each of
them has local complete
intersection singularities. \hfill $\Diamond$
{\bf Corollary B.3.} {\sl With the same assumptions as
B.2, let ${\cal C}'\to S'$ be
any deformation of $C$ with smooth base $S'$. Let ${\cal F}$
be a torsion free rank $1$
degree zero coherent sheaf on $C$. Then the relative
compactified Jacobian ${\overline{J}}
{\cal C}'$ is smooth at $[{\cal F}]$ if and only if the image
of $TS'$ in $TD^{loc}(C)$
is transversal to the image of $TD^{loc}(C,{\cal F})$.}
{\bf Proof.} We keep the notation of B.2. The dimension
of $\bar J{\cal C}'$ is equal
to $\dim S'+g_a(C)$.
As ${\overline{J}} {\cal C}'$ is equal to the fibred product of
${\overline{J}} {\cal C}$ and $S'$ over $S$,
it follows that ${\overline{J}} {\cal C}'$ is smooth at $[{\cal F}]$
if and only if the image
of $TS'$ in $TS$ is transversal to that of
$T({\overline{J}} {\cal C}, {\cal F})$. Proposition B.1
implies that the image of $T({\overline{J}} {\cal C}, {\cal F})$
is the inverse image of the image of
$TD^{loc}(C,{\cal F})$ in $TD^{loc}(C)$.\hfill $\Diamond$
\section{C. The canonical sub-space $V$}
Let $C$ be a reduced curve with only planar
singularities.
In this section we study the map
$$D^{loc}(C,{\cal F})\to D^{loc}(C)$$
at the level of tangent spaces. As both functors
are products corresponding to the
singularities of $C$ (Lemma A.2) and the tangent
spaces only depend on the formal
structure of the singularity (Remark A.4), it suffices
to analyse what happens for
$$
D(R,M)\to D(R)$$
where $P={\bf C}[[x,y]]$, $R=P/(f)$, $f$ a non-zero
element of the maximal ideal
such that $R$ is reduced, and $M$ a torsion free rank one
$R$-module given by a presentation
$$ 0 \longrightarrow P^n \stackrel{A}{\longrightarrow}P^n \longrightarrow M \longrightarrow 0.$$
{\bf Proposition C.1.} {\sl The image of the map $
TD(R,M)\to TD(R)$
is the image of the first Fitting ideal $F_1(M)$ in the
quotient ring $TD(R)=P/(f,\partial_xf,\partial_yf)$.}\\
{\bf Proof.}
Let $E_{i,j}$ be the $n \times n$ matrix which has entry
$(i,j)$ equal to $1$ and
all other entries equal to zero. If $\epsilon^2=0$, then
$\det(A+\epsilon.E_{i,j})=\det(A)+\epsilon \wedge^{n-1}(A)_{i,j}$,
therefore
we see that by perturbing the matrix $A$ to first order, we
generate precisely the
ideal
of $(n-1) \times (n-1)$ minors of the matrix $A$ as first
order perturbations of
$f$.
This is by definition the first Fitting ideal of $F_1(M)$.
\hfill $\Diamond$
Another description of the ideal $F_1(M)$ is the following
{\bf Proposition C.2.} {\sl $F_1(M)$ is the set of elements
$r\in R$ such that
$r=\varphi(m)$ for some $m\in M$, $\varphi\in Hom_R(M,R)$.
}\\
{\bf Proof.}
As $M$ is maximal Cohen-Macaulay, a resolution of $M$ as an
$R$-module will be
$2$-periodic of the form $$
\dots \longrightarrow R^n \stackrel{\bar B}{\longrightarrow} R^n \stackrel{\bar A}
{\longrightarrow} R^n \longrightarrow M \longrightarrow
0$$
for some $n\times n$ matrix with $P$-coefficients $B$ with
the property that $$
AB=BA=f{\bf 1}$$
and $\bar A$, $\bar B$ are the induced matrices with $R$
coefficients
(see \cite{E} or \cite{Yo}).
From the $2$-periodicity it follows that there is an exact
sequence $$
0\longrightarrow M \longrightarrow R^n \stackrel{\bar A}{\longrightarrow} R^n \longrightarrow M\longrightarrow 0,$$
where $M=\ker A=\hbox{\rm im}\, B$. We split this sequence into $$
\begin{array}{c}
0\longrightarrow M\longrightarrow R^n \longrightarrow N\longrightarrow 0\\
0\longrightarrow N \longrightarrow R^n \longrightarrow M \longrightarrow 0.\end{array}$$
As $N$ is also torsion free and $R$ is Gorenstein,
$Ext^1_R(N,R)=0$ by local
duality. Hence we see from the first sequence that
the map $Hom_R(R^n,R)\longrightarrow
Hom_R(M,R)$ is surjective.
From this it follows that the ideal obtained by evaluating all
homomorphisms $\phi\in Hom_R(M,R)$ on all elements of $M$
is the same as the ideal generated by the entries of the
matrix $\bar B$.
As $M$ has rank $1$, it follows that $det(A)=f$,
and hence the matrix $B$ is the Cramer matrix
$(\Lambda^{n-1}A)^{tr}$ of $A$. The claim follows.
\hfill $\Diamond$
Locally, the normalisation $\widetilde C \longrightarrow C$ corresponds
to the inclusion of $R$ in its integral closure $\overline R$
$$R \hookrightarrow \overline R.$$
Recall that the conductor is the ideal $I=Hom_R(\overline R,R)$.
One has
$$I\subset R\subset \overline R$$
and
$dim(R/I)=dim(\overline R/R)=\delta(C,p)$.
As an important corollary of {Proposition C.2} we have
{\bf Corollary C.3.}
$F_1(M)\supset I$.\\
{\bf Proof.} Write $\overline R=\oplus \overline R_i$, with
$\overline R_i$ a
domain isomorphic to ${\bf C}[[t]]$. Let $Q(\overline R_i)$
be the quotient field
of $\overline R_i$, and let $Q(R)=\oplus Q(\overline R_i)$ be
the total quotient ring of $R$.
As $M$ has rank $1$, $M\otimes_RQ(R)$ is isomorphic to $Q(R)$;
as it is
torsion-free, the natural map $M\to M\otimes_RQ(R)$ is
injective. Hence up to
isomorphism we can assume that $M$ is a submodule of $Q(R)$.
Let $m\in M$ be an
element of minimal valuation
(it exists as $M$ is finitely generated). Then multiplication
by $m^{-1}$, an
isomorphism of $Q(R)$ as an $R$-module, sends $M$ to a submodule
of $\overline R$
containing $1$.
So we can assume that $R\subset M\subset \overline R$.
Let
$c$ be any element of $I$.
Multiplication by $c$ defines a homomorpism
$\phi\in Hom_R(M,R)$ with
$\phi(1)=c$ (note that $1\in R\subset M$). Hence
$$\bigl\{\phi(m)\bigm| m\in M, \ \phi\in Hom(M,R)\bigr\}\supset I.$$
\hfill $\Diamond$
{\bf Remark C.4.} From the above description one also sees
that $F_1(\overline R)=I$. Hence the differential of the map
$D(R,M)\to
D(R)$ has minimal rank for $M=\overline R$.
Let $C$ be a reduced projective curve with only planar
singularities, $\Sigma$ its
singular locus. For $p\in \Sigma$, let $V_p$ be the
subspace of codimension
$\delta(C,p)$ in
$TD(C,p)$ generated by the conductor, and put
$$V^{loc}=\prod_{p \in \Sigma} V_p \subset
TD^{loc}(C) =\prod_{p \in
\Sigma}TD(C,p).$$
Let $V$ be the inverse image of $V^{loc}$ in $TD(C)$; note
that $V$ is a linear
subspace of codimension $\delta(C)$.
If $B$ is the base space of a semi-universal family of deformations
of $C$, then $TB$ is identified with $TD(C)$.
{\bf Proposition C.5.} {\sl Let ${\cal C}\to B$ be a semi-universal
family of
deformations of $C$. Then for any ${\cal F}\in \bar JC$ the image of
the tangent map
${\overline{J}} {\cal C}\to B$ at ${\cal F}$ contains the subspace $V$, and there
exists at least
one such ${\cal F}$ for which the image is exactly $V$. }
\\
{\bf Proof.} The first statement follows immediately from
Proposition C.1 and
Corollary C.3, by applying Proposition B.1 and Lemma A.2.
The second statement
follows in the same way from Remark C.4; e.g., we can take
${\cal F}=n_*({\cal O}_{\tilde
C})$, where $n:\tilde C\to C$ is the normalisation map.
\hfill $\Diamond$
\section{D. The $\delta$-constant stratum.}
Let $C$ be a reduced curve with only planar singularities.
We denote by $B$ an
appropriate representative of the semi-universal
deformation of $C$. The stratum $B^\delta$ is defined as
the set of points where
the geometric genus of the fibres is constant.
This amounts to saying that
$$\sum_{x\in C_t} \delta(C_t,x)$$
is constant for $t\in B^\delta$
and equal to $g_a(C)-g(\widetilde C)$, hence the name.
The analytic set $B^\delta$ (we give it the reduced induced
structure) is very
singular in general, but its properties
can be related directly to the local $\delta$-constant strata
$$B^\delta(C,p).$$
To be more precise, $B^\delta$ is the pullback of
$B^{\delta,loc}=\prod B^\delta(C,p)$ under the smooth map
$B\longrightarrow B^{loc}$.
So let $(C,p)\subset ({\bf C}^2,0)$ be a reduced plane curve
singularity, with normalisation
$$(\widetilde C,q)\stackrel{n}{\longrightarrow}(C,p),\qquad q=n^{-1}(p).$$
Note that in general $q$ will be a finite set of distinct points,
one for each
branch of $C$ at $p$.
We denote for brevity by $D(n)$ the functor of deformations of
$n:(\widetilde
C,q)\to (C,p)$ (that is, we are allowed to deform $C$ and
$\tilde C$ as well as
the map).
{\bf Lemma D.1.} {\sl $D(n)$ is smooth.}\\
{\bf Proof.} The morphism $D((C,p)\to ({\bf C}^2,0))
\longrightarrow D(n)$ (given by taking the image of the deformation of the map)
is smooth.
Hence it is enough to verify that $D((C,p)\to ({\bf C}^2,0)$ is smooth,
and this is obvious.\hfill $\Diamond$
{{\bf Theorem D.2.} (\cite{T}, \cite{D-H}).}
{\sl Let $(C,p)\subset ({\bf C}^2,0)$ be a reduced plane curve
singularity,
$n:(\widetilde C,q) \longrightarrow(C,p) $ its normalisation.
Let $B(C,p)$ be a semi-universal family for $D(C,p)$ and
$$B^{\delta}(C,p)\subset B(C,p)$$
the $\delta$-constant stratum. Then one has:\\
{(1)}
The normalisation $\widetilde B^{\delta}(C,p)$ of
$B^\delta(C,p)$ is a smooth space. \\
{(2)} The pullback of the semi-universal family to
$\tilde B^\delta$ admits a
simultaneous resolution of singularities. This makes
$\widetilde B^{\delta}(C,p)$ into a semi-universal
family for $D(n)$.\\
{(3)}
The codimension of $B^\delta\subset B$ is $\delta(C,p)$.
Over the generic point $p\in B^{\delta}$, the curve $C_p$
has precisely
$\delta(C,p)$ double points as its only singularities.\\
(4) The tangent cone to the $\delta$-constant stratum is
supported on $V_p$, the
vector subspace generated by the conductor ideal. \hfill $\Diamond$}
The second half of (2) is in fact not explicitly stated
in either of \cite{T},
\cite{D-H}; however it follows easily from Lemma D.1. A
similar argument is
presented in the proof of Proposition F.2, so we don't
repeat it here.
\section{E. Proof of Theorem 1.}
Let $C$ be a reduced projective rational curve with only planar
singularities.
We want to show that $e(\bar JC)=m(C)$. In particular let
$(C,p)$ be a reduced
plane curve singularity. Let $C$
be a projective rational curve that has $(C,p)$ as its only
singular point.
Then it follows that $e(\bar JC)=m(C,p)$.
Let $\Phi:{\cal C}\longrightarrow B$ be a semi-universal family of
deformations of $C$;
we denote its fibres by $C_s=\Phi^{-1}(s)$, $C_0=C$.
Let $\pi:{\overline{J}}{\cal C} \longrightarrow B$ be the corresponding family of compactified
Jacobians.
We always assume that we have chosen discs as representatives
for the corresponding germs.
We may also assume that the induced morphism
$j:B\longrightarrow B^{loc}$ is smooth and has contractible fibres.
We choose a section $\sigma:B^{loc}\longrightarrow B$ of $j$ with $\sigma(0)=0$.
We will denote $\overline B:=\sigma(B^{loc})$,
$\overline B^\delta:=\sigma(B^\delta)$ and $\overline V:=\sigma(V)$.
Let $(W,0)\subset (\overline B,0)$ be a smooth subspace of dimension
$\delta+1$ containing the point $(0,0)$ together with a smooth map
$\lambda:(W,0)\longrightarrow (T,0)$ to a disc $(T,0)\subset ({\bf C},0)$.
$W$ is a one parameter family of $\delta$-dimensional subspaces
$W_t=\lambda^{-1}(t)\subset \overline B$.
We require in addition that $W_0$ is transverse to $V$.
\bigskip
\epsfxsize 4cm
\epsfysize 4cm
\centerline{\epsfbox{del.eps}}
\bigskip
By {Theorem D.2} we can choose $W$ in such a way that
for $t\ne 0$ the fibre $W_t$ intersects $\overline B^\delta$ in
$mult(B^{\delta})$ points, and for $s\in W_t\cap \overline B^\delta$ the
corresponding curve $C_s$ has precisely $\delta$ nodes as singularities.
For $s\in W_t \setminus \overline B^\delta$ the curve $C_s$ will have
positive genus. Let $\bar \Delta\subset B$ be a closed disc, and let $Z
=W\cap \bar\Delta$.
We define the family $\rho:{\overline{J}}{\cal C}_Z\longrightarrow T$ by pullback:
\def\mapd#1{\Big\downarrow
\rlap{$\vcenter{\hbox{$\scriptstyle#1$}}$}}
$$\begin{array}{ccccc}
&&{\overline{J}}{\cal C}_Z&\longrightarrow&{\overline{J}}{\cal C}\\
&\stackrel{\rho}{\swarrow}&\mapd{\pi}&&\mapd{\pi}\\
T&\stackrel{\lambda}{\longleftarrow}&Z&\longrightarrow&B
\end{array}
$$
As we have chosen $W_0$ to be transversal to $V$, Proposition C.5
implies that $\rho$ is smooth along $\pi^{-1}(0)$; by making
$\bar\Delta$ and $T$
smaller we can assume
that $\rho$ is smooth. As $\rho$ is also proper,
all the fibres $\rho^{-1}(t)$ are diffeomorphic, in particular
they all have the
same Euler number.
The space $\rho^{-1}(t)$ is the union, for $s\in W_t$, of
$\bar JC_s$. We know
that if $C_s$ has positive geometric genus, then $e(\bar JC_s)$
is zero; arguing
as in \cite{B}, we obtain that $$
e(\rho^{-1}(t))=\sum_{s\in W_t\cap \bar B^\delta}e(\bar JC_s)$$
(note that if $s\in W_t$, then $C_s$ is rational if and only
if $s\in \bar
B^\delta$).\\
The intersection of $W_0\subset \overline B$ with
$\overline B^\delta$
consists only of the point $0$ corresponding to the curve $C$.
Therefore $e(\rho^{-1}(0))=e({\overline{J}} C)$.\\
On the other hand, for $t\ne 0$, $W_t$ intersects
$\overline B^{\delta}$ in $mult( B^\delta)$ points and for $s\in
\overline B^{\delta}\cap W_t$ the curve $C_s$ has precisely
$\delta$
nodes as singularities.
As for a nodal rational curve $C_s$, the Euler number
$e({\overline{J}} C_s)$ is equal to $1$, we obtain
$$e(\rho^{-1}(t))=
\sum_{s\in W_t}e({\overline{J}} C_s)=\sum_{s\in W_t\cap \overline B^{\delta}}1=
mult(B^\delta).$$
So we get
$$e({\overline{J}} C)=e(\rho^{-1}(0))=e(\rho^{-1}(t))=mult(B^\delta).$$
\hfill$\Diamond$
\section{F. The invariant as length of moduli of stable maps}
Let $C$ be a reduced projective curve with only plane curve
singularities; let
$n:\tilde C\to C$ be its normalisation, and $g$ the genus of
$\tilde C$.
Let $m(C)=\prod m(C,p)$. The scheme $\overline M_{g,0}(C,[C])$
parametrizing
stable birational maps from a genus $g$ curve to $C$ contains
only one point,
namely the normalisation of $C$.
The aim of this section is to prove that its length is equal
to $m(C)$.
Note that if $C$ is an isolated rational curve inside a
smooth manifold $Y$,
$\overline M_{g,0}(C,[C])$ is naturally a closed subscheme
of $\overline
M_{g,0}(Y,[C])$; in particular, $m(C)$ is a lower bound
for the length of the
corresponding component of $M_{g,0}(Y,[C])$ (in case this
scheme also has
dimension zero).
Denote by $D(n)$ the deformation functor of the triple
$(n:\widetilde C\to C)$,
and by $D^{loc}(n)$ the corresponding local deformation
functor. As before
$D^{loc}(n)$ is the product over the singular points $p$
of $C$ of $D(n,p)$,
the deformation functor of the triple $n:(\tilde C,
n^{-1}(p))\to (C,p)$.
If $(C,p)$ is the germ of a planar reduced curve singularity,
then $D(n,p)$ is
a smooth functor (see section D).
{\bf Lemma F.1.} {\sl
The natural morphism of functors $D(n)\to
D^{loc}(n)\times_{D^{loc}(C)}D(C)$ is an isomorphism.}
{\bf Proof.} Let $C_T$ be an infinitesimal deformation of $C$,
and let $U_i$ be an
open cover of $C$ such that $U_{ij}$ is smooth for each $i\ne j$.
Let $V_i=n^{-1}(U_i)$. Let $U_{i,T}$ be the deformation of $U_i$
induced by $C_T$,
and assume we are given a deformation $n_{i,T}:V_{i,T}\to U_{i,T}$
of
$n_i:=n|_{V_i}$. Then to lift $(C_T,n_{i,T})$ to a deformation
of $n$ we must
choose gluing isomorphisms $\psi_{ij}:V_{ij,T}\to V_{ji,T}$
satisfying the cocycle
condition and compatible with the other data, namely the maps
$n_{i,T}$ and the
gluing isomorphisms $\phi_{ij}:U_{ij,T}\to U_{ji,T}$ induced
by $C_T$. But
$U_{ij}$ is smooth, so that $n|_{V_{ij}}$ is an isomorphism
for each $i\ne j$;
hence the $\psi_{ij}$ are univocally determined by the
$\phi_{ij}$ and
automatically satisfy the cocycle condition.
\hfill $\Diamond$
Let us now denote by $B(\cdot)$ the germ of complex
space being a hull for the
functor $D(\cdot)$. Note that Lemma F.1 implies
that there is a cartesian
diagram $$
\cdia{B(n)}{B(C)}{B^{loc}(n)}{B^{loc}(C).}$$
{\bf Proposition F.2.} {\sl Let $C$ be a reduced projective
curve with planar
singularities, $n:\tilde C\to C$ be the normalisation,
$g=g(\tilde C)$.
Let $\pi:{\cal C}\to B(C)$ be a semi-universal deformation
of $C$.
Denote by $M= M_{g,0}({\cal C},[C])$; then $M$ is smooth at
$n$, and the natural
map $M\to B^\delta:=B^\delta(C)$ is the normalisation map.}
{\bf Proof.} Write $M$ for the germ of $M$ at $n$.
As the domain of $n$ is a smooth curve, the same is true
for all stable maps in a
neighborhood of $n$. Hence $M$ is isomorphic to $B(n)$.
By Lemma F.1, together
with Lemma D.1, we deduce that $B(n)$ is smooth. By the
definition of $B^\delta$
the natural map $M\to B(C)$ factors via $B^\delta$, hence,
as $M$ is smooth, via
its normalisation $\tilde B^\delta$. On the other hand, we
know that the family
$\tilde {\cal C}\to \tilde B^\delta$ gotten by pullback
admits a very weak
simultaneous resolution of singularities \cite{T}, inducing a morphism
$\tilde B^\delta\to M$. It is easy to check pointwise that
these two morphisms are
inverse to each other (both $\tilde B^\delta$ and $M$
just parametrize the normalisation maps of the fibres of $\pi$).
As
both $\tilde B^\delta$ and $M$ are smooth, a bijective morphism
must be an
isomorphism. \hfill $\Diamond$
{\bf Proof of Theorem 2}. The scheme $M_{g,0}(C,[C])$ is
the fibre over the point
$[C]$ of the morphism $\tilde B^\delta\to B^\delta$; this
is the multiplicity of
$B^\delta$ at $[C]$, as $\tilde B^\delta$ is smooth. This
proves the first
equality.
Let now $X$ be a smooth projective surface, $C\subset X$ a
reduced irreducible
curve, $n:\tilde C\to C$ the normalisation, $g=g(\tilde C)$.
Assume that
$n$ is an isolated point of $\overline{M}_{g,0}(X,[C])$, and
let $M_n$ be the
connected component of $n$. $M_n$ contains $M_{g,0}(C,[C])$
as a closed subscheme,
so we always have an inequality $$
l(M_n)\ge l(M_{g,0}(C,[C]))=m(C).$$
This inequality is an equality if and only the natural morphism
$M_n\to Hilb(X)$
sending each map to its image factors scheme theoretically
(and not only
set-theoretically) via $C$.
Hence to complete the proof of Theorem 2, it is enough to
show that this is the
case if $C$ is rational and $X$ is a $K3$ surface. Let $S$
be the complete linear
system defined by $C$ on $X$, and let ${\cal C}\to S$ be the
universal curve.
It is known that $\bar J{\cal C}$ is smooth, see \cite{Mu}; but
this means precisely that $S$ maps transverse
to the $\delta$-constant stratum in $B(C)$, and we are done
in view of Corollary
B.3.\hfill $\Diamond$
\section{G. Examples.}
\def\ell{\ell}
\defM{M}
\def\alpha{\alpha}
\def{\bf A}{{\bf A}}
\def{\bf P}{{\bf P}}
{\bf Example 1 (Beauville):} {\sl Let $(C,o)$ be the
singularity of equation
$x^q=y^p$, with $p<q$ and $(p,q)=1$. Then $$m(C,o)=
{{1}\over{p+q}}{p+q\choose p}.$$}
{\bf Proof.} We write for simplicity $\overline M(X,\beta)$
instead of $\overline
M_{0,0}(X,\beta)$; if $X$ is a curve and $\beta=[X]$ we omit
it. Let $C$ be the
plane curve of equation $y^pz^{q-p}=x^q$. $C$ is a rational
curve with two
singular points, $o=(0,0,1)$ and $\infty=(1,0,0)$.
Let $\alpha:C'\to C$ be the partial normalisation of $C$ at
$\infty$.
By Theorem 2, it is enough to prove that
$$l(\overline M(C'))={{1}\over{p+q}}{p+q\choose p}=:N(p,q).$$
The natural map $\overline M(C')\to \overline M(C)$ given by
$\mu\mapsto \alpha\circ
\mu$ is a closed embedding, and the closed subscheme
$\overline M(C')$ is
identified by requiring the deformation of the normalisation
morphism to be
locally trivial near $\infty$.
On the other hand $\overline M(C)$ is naturally a closed subset
of $\overline
M({\bf P}^2,q\ell)$, where $\ell$ is the class of a line.
Let $n:{\bf P}^1\to C$ be the normalisation map, and choose coordinates
on ${\bf P}^1$ such
that $n(s,t)=(t^ps^{q-p},t^q,s^q)$. A morphism in $\overline
M({\bf P}^2,q\ell)$ near
$n$ has equations $$(t^ps^{q-p}+x,t^q+y,s^q+z),$$ for suitable
homogeneous
polynomials $x,y,z$ of degree $q$.
We impose the conditions that the image of the map be
contained in $C$
and that the deformation be locally trivial at $\infty$.
Then we eliminate the indeterminacy generated by
a reparametrization of ${\bf P}^1$ and a rescaling of the
coordinates on ${\bf P}^2$.
We get that all deformations of $n$ in $\overline M(C)$
must be (in affine
coordinates where $z=1$) of the form $$
t\mapsto (t^p+\textstyle\sum\limits_{i=0}^p x_i t^i,t^q+
\textstyle\sum\limits_{i=0}^qy_it^i).$$
Hence we are now left with the following problem:
compute the length
of the ${\bf C}$--algebra with generators
$x_0,\ldots,x_{p-2},y_0,\ldots,y_{q-2}$
and relations given by the coefficients of the polynomial
$f^q-g^p$, where
$f=t^p+\sum x_it^i$ and $g=t^q+\sum y_it^i$.
It is easy to check that the equation $f^q=g^p$ is
equivalent to $qf'g=pg'f$
by taking $d/dt\circ\log$ on both sides. The $t$-degree
of $qf'g-pg'f$ is $p+q-1$,
however we only get $p+q-2$ equations as the coefficients
of $t^{p+q-1}$ and
$t^{p+q-2}$ are zero anyway.
Moreover, if we consider the variables $x_i$ (resp.\ $y_i$)
as having degree
$p-i$ (resp.\ $q-i$), the equations we obtain are homogeneous of
degree $2,\ldots,p+q-1$.
Now we recall the weighted B\'ezout theorem, which says that
if we have a
zero-dimensional algebra given by $N$ homogeneous equations of
degrees $e_j$ in $N$ weighted variables of degree $d_j$, then
the length of the
algebra is $\prod e_j/\prod d_j$.
Applying the formula in our case, with $N=p+q-2$,
$(d_j)=(2,3,\ldots,p,2,3,\ldots,q)$ and
$e_j=(2,3,\ldots,p+q-1)$ gives
$$
N(p,q)={\prod e_j\over \prod d_j}={(p+q-1)!\over p!q!}
={1\over p+q}{p+q\choose
p}.$$
{\bf Example 2.} We would like to outline an algorithm for
the computation of
$m(C,p)$ for a planar, reduced and irreducible curve
singularity $(C,p)$.
Assume we know how to realize $(C,p)$ as singularity
of a rational curve. It is
then easy to realize it as singularity of a plane
rational curve $C$, whose other
singularities are only nodes. Let $d$ be the degree
of the curve, $F(x,y,z)=0$ its
equation, and $\bar n=(\bar x,\bar y,\bar z)$ an explicit
normalisation given by
homogeneous polynomials of degree $d$ in $s,t$. Assume
without loss of generality
that $\bar z$ contains the monomial $s^d$ with nonzero
coefficient.
Then we can describe the scheme $M_{0,0}(C,[C])$ explicitly
as follows.
Choose three points $p_i$ ($i=1,2,3$) in ${\bf P}^1$ mapping via
$n$ to smooth points
of $C$; let $L_i\subset {\bf P}^2$ be a line transversal to $C$
at $n(p_i)$.
Choose variables $x_i$, $y_i$ and $z_i$ for $i=0,\ldots d$,
and
let $x$ be the polynomial $\bar x+\sum_i x_i s^it^{d-i}$;
define $y$ and $z$ in a
similar way.
Then $M_{0,0}(C,[C])$ is naturally isomorphic the
subscheme of $
\hbox{\sl Spec}\, {\bf C} [x_i,y_i,z_i]$ defined
by the equations
$$\begin{array}{c}
z_d=0,\\
(x,y,z)(p_i)\in L_i \qquad i=1,2,3,\\
F(x,y,z)=0.
\end{array}$$
In fact, all deformations of $\bar n$ are again morphisms
of degree $d$ from
${\bf P}^1$ to ${\bf P}^2$, hence are given by polynomials of
degree $d$.
The first four equations, defining a linear subspace,
correspond to choosing local coordinates near $\bar n$
on $M_{0,0}({\bf P}^2,d)$; the
last one, which is a system of
$d^2$ equations, imposes the condition that the
scheme-theoretic image of the
morphism be contained in
$C$.\\
|
1997-12-16T01:08:53 | 9708 | alg-geom/9708007 | en | https://arxiv.org/abs/alg-geom/9708007 | [
"alg-geom",
"math.AG"
] | alg-geom/9708007 | Yuri G. Zarhin | Yuri G. Zarhin | Torsion of abelian varieties, Weil classes and cyclotomic extensions | LaTeX 2e 17 pages | null | null | null | null | Let $K$ be a field finitely generated over the field of rational numbers,
$K(c)$ the extension of $K$ obtained by adjoining all roots of unity, $L$ an
infinite Galois extension of $K$, $X$ an abelian variety defined over $K$. We
prove that under certain conditions on $X$ and $K$ the existence of infinitely
many L-rational points of finite order on $X$ implies that the intersection of
$L$ and $K(c)$ has infinite degree over $K$.
| [
{
"version": "v1",
"created": "Mon, 4 Aug 1997 23:46:42 GMT"
},
{
"version": "v2",
"created": "Wed, 27 Aug 1997 19:17:28 GMT"
},
{
"version": "v3",
"created": "Wed, 3 Sep 1997 15:12:11 GMT"
},
{
"version": "v4",
"created": "Tue, 9 Sep 1997 16:53:41 GMT"
},
{
"version": "v5",
"created": "Tue, 16 Dec 1997 00:08:53 GMT"
}
] | 2008-02-03T00:00:00 | [
[
"Zarhin",
"Yuri G.",
""
]
] | alg-geom | \section{Main construction}
Let $F$ be the center of $\mathrm{End}_K(X)\otimes{\mathbf Q}$, $R_F=F\bigcap \mathrm{End}_K(X)$ the center of $\mathrm{End}_K(X)$. We put
$$V_{{\mathbf Z}}=V_{{\mathbf Z}}(X)=H_1(X({\mathbf C}),{\mathbf Z}), \quad V=V(X)=H_1(X({\mathbf C}),{\mathbf Q})= V_{{\mathbf Z}}\otimes{\mathbf Q}.$$
For each nonnegative integer $m$ one may naturally identify the $m$th rational
cohomology group $H^m(X({\mathbf C}),{\mathbf Q})$ of $X({\mathbf C})$ with $\mathrm{Hom}_{{\mathbf Q}}(\Lambda^m_{{\mathbf Q}}(V(X),{\mathbf Q})$.
For each prime $\ell$ there are natural identifications
$$X_{\ell}=V_{{\mathbf Z}}/\ell V_{Z}, T_{\ell}(X)=V_{Z}\otimes{\mathbf Z}_{\ell}, V_{\ell}(X)=V(X)\otimes_{{\mathbf Q}}{\mathbf Q}_{\ell}=V_{{\mathbf Z}}\otimes{\mathbf Q}_{\ell}.$$
There is a natural Galois action
$$\rho_{\ell}=\rho_{\ell,X}:G(K)\to \mathrm{Aut}_{{\mathbf Z}_{\ell}}(T_{\ell}(X)) \subset \mathrm{Aut}_{{\mathbf Q}_{\ell}}(V_{\ell}(X)),$$
induced by the Galois action on the torsion points of $X$ \cite{Serre}.
One may naturally identify the $m$th $\ell-$adic cohomology group $H^m(X_a,{\mathbf Q}_{\ell})$ of $X_a=X\times K(a)$ with
$$\mathrm{Hom}_{{\mathbf Q}_{\ell}}(\Lambda^m_{{\mathbf Q}_{\ell}}(V_{\ell}(X),{\mathbf Q}_{\ell})=\mathrm{Hom}_{{\mathbf Q}}(\Lambda^m_{{\mathbf Q}}(V(X),{\mathbf Q}))\otimes_{{\mathbf Q}}{\mathbf Q}_{\ell}).$$
This identification is an isomorphism of the Galois modules.
Assume now that $F$ is a number field, i.e., $X$ is either simple or is
isogenous over $K$ to a self-product of
a
simple abelian variety. Let $O_F$
be the ring of integers in $F$. It
is well-known that $R_F$ is a subgroup of finite index in $O_F$. Recall that
for each prime $\ell$ there is a splitting $F\otimes_{{\mathbf Q}}{\mathbf Q}_{\ell}=\oplus F_{\lambda}$
where $\lambda$ runs through the set of prime ideals dividing $\ell$ in $O_F$
and $F_{\lambda}$ is the
completion of $F$ with respect to $\lambda-$adic topology. There is a natural splitting
$V_{\ell}(X)=\oplus V_{\lambda}(X)$
where
$$V_{\lambda}(X)=F_{\lambda} V_{\ell}(X) =V(X)\otimes_F F_{\lambda}.$$
It is well-known that all $V_{\lambda}(X)$ are $G(K)-$invariant $F_{\lambda}-$vector
spaces of dimension
$2\mathrm{dim}(X)/[F:{\mathbf Q}]$. We write $\rho_{\lambda,X}$ for the corresponding
$\lambda-$adic representation
$$\rho_{\lambda,X}:G(K) \to\mathrm{Aut}_{F_{\lambda}}V_{\lambda}(X)$$
of $G(K)$ \cite{Serre},\cite{RibetA}.
Similarly, for all but finitely many $\ell$
$$R_F/\ell R_F=O_F/\ell O_F = \oplus_{\lambda\mid\ell} O_F/\lambda$$
is a direct sum of finite fields $O_F/\lambda$ of characteristic $\ell$. Also, $X_{\ell}=V_{{\mathbf Z}}/\ell V_{{\mathbf Z}}$ is a free $R_F/\ell R_F=O_F/\ell O_F-$module of rank $2\mathrm{dim}(X)/[F:{\mathbf Q}]$ and there is a natural splitting
$$X_{\ell}=V_{{\mathbf Z}}/\ell V_{{\mathbf Z}}=\oplus_{\lambda\mid\ell} X_{\lambda}$$
where $X_{\lambda}=(O_F/\lambda) \cdot X_{\ell}.$
Clearly, each $X_{\lambda}$ is a $G(K)-$invariant $O_F/\lambda-$vector space of dimension $2\mathrm{dim}(X)/[F:{\mathbf Q}]$.
We write $\bar{\rho}_{\lambda,X}$ for the corresponding modular
representation
$$\bar{\rho}_{\lambda,X}:G(K) \to\mathrm{Aut}_{O_F/\lambda}X_{\lambda}$$
of $G(K)$ \cite{RibetA}.
Let $d$ be a positive integer and assume that there exists a non-zero
$2d-$linear form $\psi \in \mathrm{Hom}_{{\mathbf Q}}(\otimes^{2d}_{{\mathbf Q}} V(X),{\mathbf Q})$, enjoying the following properties.
\begin{enumerate}
\item For all $f\in F; v_1, \ldots v_{2d}\in V(X)$
$$\psi(f v_1,v_2,\ldots ,v_{2d})=\psi(v_1,fv_2,\ldots ,v_{2d})=\cdots =
\psi(v_1,v_2,\ldots ,fv_{2d}).$$
\item
For any prime $\ell$ let us extend $\psi$ by ${\mathbf Q}_{\ell}-$linearity to the
non-zero multilinear form
$\psi_{\ell} \in \mathrm{Hom}_{{\mathbf Q}_{\ell}}(\otimes^{2d}_{{\mathbf Q}_{\ell}} V_{\ell}(X),{\mathbf Q}_{\ell})$.
Then for all $\sigma \in G(K); v_1, \ldots v_{2d}\in V_{\ell}(X)$
$$\psi_{\ell} (\sigma(v_1),\sigma(v_2),\ldots,\sigma(v_{2d}))=
\chi_{\ell}^d(\sigma)\psi_{\ell}(v_1,v_2,\ldots ,v_{2d}).$$
\end{enumerate}
We call such a form {\sl admissible} or $d-${\sl admissible}.
\vskip .5cm
{\bf Example.} Let us assume that $F$ is a {\sl totally real} number field. If
$\mathcal L$ is an invertible sheaf on $X$ defined over $K$ and algebraically
non-equivalent to zero then one may associate to $\mathcal L$ its first Chern
class
$$c_1({\mathcal L})\in H^2(X({\mathbf C}),{\mathbf Q})=\mathrm{Hom}_{{\mathbf Q}}(\Lambda^2_{{\mathbf Q}}(V(X),{\mathbf Q}).$$
The well-known properties of Rosati involutions and Weil pairings imply
that $c_1({\mathcal L})$ is $1-$admissible (see p.~237 of \cite{MumfordAV}, especially the last sentence and Section 2 of \cite{SZMathZ}).
\vskip .5cm
There exists a unique $F-2d-$linear form
$\psi_F\in \mathrm{Hom}_F(\otimes^{2d}_F V(X),F)$ such that
$$\psi={\mathbf{Tr}}_{F/{\mathbf Q}}(\psi_F).$$
Multiplying $\psi$ by a sufficiently divisible positive integer, we may and
will assume that the restriction of $\psi_F$ to
$V_{{\mathbf Z}}\times \cdots V_{{\mathbf Z}}$ takes on values in $R_F$. Let
$\mathrm{Im}(\psi_F)$ be the additive subgroup of $R_F$ generated by values of
$\psi_F$ on $V_{{\mathbf Z}}\times \cdots V_{{\mathbf Z}}$ takes on values in $R_F$. Let
$\mathrm{Im}(\psi_F)$ be the additive subgroup of $R_F$ generated by values of
$\psi_F$ on $V_{{\mathbf Z}}\times\cdots V_{{\mathbf Z}}$. Clearly, $\mathrm{Im}(R_F)$ is a
subgroup of finite index in $R_F$ that is an ideal.
Notice that for all but finitely many primes $\ell$
$$O_F=R_F/\ell R_F, \mathrm{Im}(\psi_F)=R_F/\ell R_F.$$
Let us extend $\psi_F$ by $F_{\lambda}-$linearity to the
{\sl non-zero} multilinear form
$$\psi_{F,\lambda} \in \mathrm{Hom}_{F_{\ell}}(\otimes^{2d}_{F_{\lambda}} V_{\lambda}(X),F_{\lambda}).$$
Then
$$\psi_{F,\lambda}(\sigma(v_1),\sigma(v_2),\ldots,\sigma(v_{2d}))=
\chi_{\ell}^d(\sigma)\psi_{F,\lambda}(v_1,v_2,\ldots ,v_{2d})$$
for all $\sigma \in G(K); v_1, \ldots v_{2d}\in V_{\lambda}(X)$.
Similarly, for all but finitely many $\ell$ the form $\psi_F$ induces a
non-zero multilinear form
$$\psi_{F}^{(\ell)} \in \mathrm{Hom}_{R_F/\ell R_F}(\otimes^{2d}_{R_F/\ell R_F} X_{\ell},R_F/\ell R_F)$$
enjoying the following properties:
\begin{itemize}
\item
The subgroup of $R_F/\ell R_F$ generated by all the values of $\psi_{F}^{(\ell)}$ coincides with $R_F/\ell R_F$;
\item For all $\sigma \in G(K); v_1, \ldots v_{2d}\in X_{\ell}$
$$\psi_{F}^{(\ell)}(\sigma(v_1),\sigma(v_2),\ldots,\sigma(v_{2d}))=
\bar{\chi}_{\ell}^d(\sigma)\psi_{F}^{(\ell)}(v_1,v_2,\ldots ,v_{2d}).$$
\end{itemize}
This implies that for all but finitely many $\ell$ the restriction of $\psi_{F}^{\ell}$
to $X_{\lambda}$ defines a {\sl non-zero} multilinear form
$$\psi_{F}^{(\lambda)} \in \mathrm{Hom}_{O_F/\lambda}(\otimes^{2d}_{O_F/\lambda} X_{\ell},O_F/\lambda)$$
enjoying the following property:
$$\psi_{F}^{(\lambda)}(\sigma(v_1),\sigma(v_2),\ldots,\sigma(v_{2d}))=
\bar{\chi}_{\ell}^d(\sigma)\psi_{F}^{(\lambda)}(v_1,v_2,\ldots ,v_{2d})$$
for all $\sigma \in G(K); v_1, \ldots v_{2d}\in X_{\lambda}$.
\begin{rem} Using the K\"unneth formula for $X_a^{2d}$, one may view $\psi_{\ell}$ as a
Tate cohomology class on $X_a^{2d}$. If $\psi$ is skew-symmetric then $\psi_{\ell}$ is a Tate
cohomology class on $X_a$.
\end{rem}
\begin{thm} Assume that the center $F$ of $\mathrm{End}^0 X$
is a field
and there is a $d-$admissible form
$\psi$ on $X$. Let $\ell$ be a prime and assume that the
$\ell-$ torsion
in $X(L)$ is infinite. If $L^{(\ell)}$ is the intersection of
$L$ and $K(\ell)$
then the field extension $K(\ell)/L^{(\ell)}$ has finite degree dividing $(d,\ell-1)$ if $\ell$ is odd and dividing $2$ if $\ell=2$ respectively. In addition, $L$ contains ${\mathbf Q}(\ell)'$.
\end{thm}
\begin{proof} As explained in (\cite{ZarhinMA}, 0.8, 0.11) the assumption
that the $\ell-$torsion in $X(L)$ is infinite means that there exists a place
$\lambda$ of F, dividing $\ell$
such that the Galois group $G(L)$ of $L$ acts trivially on $V_{\lambda}(X)$. Since $\psi_{F,\lambda}$ is not identically zero, we conclude that
$$\chi_{\ell}^d(\sigma)=1 \quad \forall \sigma \in G(L) \subset G(K).$$
We write $G'$ for the kernel of $\chi_{\ell}^d$. We have
$G(L)\subset G'\subset G(K)$.
Recall that the kernel of $\chi_{\ell}:G(K) \to {\mathbf Z}_\ell^*$ coincides with the Galois group
$G(K(\ell))$ of $K(\ell)$ and $\chi_{\ell}$ identifies $\mathrm{Gal}(K(\ell)/K)$ with a subgroup of ${\mathbf Z}_\ell^*=\mathrm{Gal}({\mathbf Q}(\ell)/{\mathbf Q})$. Since the torsion subgroup of ${\mathbf Z}_\ell^*$ is the cyclic group $\mu({\mathbf Z}_{\ell})$ of order $\ell-1$ if $\ell$ is odd and of order $2$ if $\ell=2$, $G'$ coincides with the kernel of $(\chi_{\ell})^{d'}$ with $d'=(d,\ell-1)$ if $\ell$ is odd and $d'=(d,2)$ if $\ell=2$ respectively. This implies that the field $K'=K(a)^{G'}$ of $G'-$invariants is a subfield of $K(\ell)$ and $[K(\ell):K']$ divides $d'$, since $\chi_{\ell}$ establishes an isomorphism between $\mathrm{Gal}(K(\ell)/K')$ and
$$\{s \in \mathrm{Im}(\chi_{\ell})\subset {\mathbf Z}_{\ell}^*\mid s^{d'}=1\}
\subset \{s \in \mu({\mathbf Z}_{\ell})\mid s^{d'}=1\}.$$
Now it is clear that $K'\subset L$, since $G(L) \subset G'=G(K')$.
It is also clear that $K(\ell)/K'$ is a cyclic extension of degree dividing $d'$.
In order to prove the last assertion of the theorem, notice that
$\mathrm{Gal}(K(\ell)/K) \subset \mathrm{Gal}({\mathbf Q}(\ell)/{\mathbf Q})={\mathbf Z}_{\ell}^*$ and the finite subgroup $\mathrm{Gal}(K(\ell)/K')$ of $\mathrm{Gal}(K(\ell)/K)$ sits in $\mu({\mathbf Z}_{\ell})\subset{\mathbf Z}_{\ell}^*$. Since $\mu({\mathbf Z}_{\ell})=\mathrm{Gal}({\mathbf Q}(\ell)/{\mathbf Q}(\ell)')$, ${\mathbf Q}(\ell)'\subset K'$. Since $K'\subset L$, ${\mathbf Q}(\ell)'\subset L$.
\end{proof}
\begin{thm} Assume that the center $F$ of $\mathrm{End}^0 X$
is a field and there is a $d-$ admissible form $\psi$ on
$X$. Let $S$ be an infinite set of primes $\ell$ such that
for all but finitely many $\ell\in S$ the $\ell-$torsion in
$X(L)$ is not
zero. Then for all but finitely many $\ell\in S$ the field extension
$K(\mu_{\ell})/K(\mu_{\ell})\bigcap L$ has degree dividing $(d,\ell-1)$.
\end{thm}
\begin{proof}
For all but finitely
many $\ell$ the $G(K)-$module
$X_{\ell}$ is semisimple and the centralizer of $G(K)$ in $\mathrm{End}(X_{\ell})$
coincides with
$\mathrm{End}_K(X)\otimes {\mathbf Z}/\ell{\mathbf Z}$. This assertion was proven in
\cite{ZarhinInv} for number fields $K$; the proof is based on results of Faltings
\cite{Faltings1}. (See \cite{MW} for an
effective version.) However, the same proof works for arbitrary finitely
generated fields $K$, if one uses results of \cite{Faltings2},
generalizing the results of \cite{Faltings1}.
Clearly, for all but finitely many $\ell$ the center of $\mathrm{End}_K(X)\otimes {\mathbf Z}/\ell{\mathbf Z}$
coincides
with $R_F/\ell R_F=O_F/\ell O_F$. Applying Theorem 5f of \cite{ZarhinDuke}
to $G=G(K), G'=G(L), H=X_{\ell}, D=\mathrm{End}_K(X)\otimes {\mathbf Z}/\ell{\mathbf Z},
R=F_F/\ell R_F$, we conclude that for all but finitely many $\ell \in S$ there exists
$\lambda\mid\ell$ such that $G(L)$ acts trivially on $X_{\lambda}$. Using the Galois
equivariance of the non-zero form $\psi_{F}^{(\lambda)}$, we conclude that
for all but finitely many $\ell\in S$ the character
$\bar{\chi}_{\ell}^d$ kills $G(L)$. We write $G'$ for the kernel of $\bar{\chi}_{\ell}^d$. We have
$G(L)\subset G'\subset G(K)$.
Recall that the kernel of $\bar{\chi}_{\ell}:G(K) \to ({\mathbf Z}/\ell {\mathbf Z})^*$ coincides with
$G(K(\mu_{\ell}))$ and
$({\mathbf Z}/\ell {\mathbf Z})^*$ is a cyclic group of order $\ell-1$. This implies that
the field $K'=K(a)^{G'}$ of
$G'-$invariants is a subfield of $K(\mu_{\ell})$ and $[K(\mu_{\ell}):K']$
divides $(\ell-1,d)$, since
$\bar{\chi}_{\ell}$ establishes an isomorphism between $\mathrm{Gal}(K(\mu_{\ell})/K')$ and $\{s \in \mathrm{Im}(\bar{\chi}_{\ell})\subset ({\mathbf Z}/\ell {\mathbf Z})^*\mid s^d=1\}$. One has only to notice that $K'\subset L$, since $G(L) \subset G'=G(K')$.
\end{proof}
\begin{cor} Assume that the torsion subgroup of $X(L)$
is infinite. Then the intersection of $L$ and $K(c)$ has infinite degree over $K$.
\end{cor}
\begin{proof} Indeed, either there is a prime $\ell$ such that the $\ell-$torsion in $X(L)$ is infinite or for infinitely many primes $\ell$ the $\ell-torsion$ in $X(L)$ is not zero. Now, one has only to apply the previous two theorems.
\end{proof}
\section{Weil classes and admissible forms}
Suppose $A$ is an abelian variety defined over $K$, $k$ is a CM-field,
$\iota : k \hookrightarrow \mathrm{End}_K^0(A)$
is an embedding, and $C$ is an algebraically
closed field containing $K$ (for instance, $C={\mathbf C}$ or $C=\bar{{\mathbf Q}}$). Let $\mathrm{Lie}(A)$ be the tangent space of $A$ at the origin,
an $K$-vector space. If $\sigma$ is an embedding of $k$ into $C$, let
$$n_\sigma = \mathrm{dim}_C\{t \in \mathrm{Lie}(A)\otimes_K C :
\iota(\alpha)t = \sigma(\alpha)t {\text{ for all }} \alpha \in k\}.$$
Write ${\bar \sigma}$ for the composition of $\sigma$ with the involution
complex conjugation of $k$.
Recall that a triple $(A,k,\iota)$ is {\em of Weil type} if $A$ is an abelian
variety over an algebraically closed field $C$ of
characteristic zero, $k$ is a CM-field, and
$\iota : k \hookrightarrow \mathrm{End}^0(A)$ is an embedding, such that
$n_\sigma = n_{\bar \sigma}$ for all embeddings $\sigma$ of $k$ into $C$.
It is known \cite{SZMathZ} that $(A,k,\iota)$ is of Weil type if and only
if $\iota$ makes
$\mathrm{Lie}(A) \otimes_K C$ into a free $k \otimes_{\mathbf Q} C$-module
(see p.~525 of \cite{Ribet} for the case where $k$ is an imaginary quadratic field).
Now, assume that $A=X$ and the image $\iota(k)$ contains the center $F$ of
$\mathrm{End}_K(X)\otimes{\mathbf Q}$ (for instance, $F=k$). Notice that in the case of Weil type the
degree $[k:{\mathbf Q}]$ divides $\mathrm{dim}(A)$. In particular, $\mathrm{dim}(A)$ is even.
Our goal is to construct an admissible form, using a triple $(A,k,\iota)$ of Weil type.
Recall that the degree $[k:{\mathbf Q}]$ divides $\mathrm{dim}(A)$, put $d=\mathrm{dim}(X)/[k:{\mathbf Q}]$
and consider the space of Weil classes (\cite{WeilHodge}, \cite{Deligne}, \cite{SZMathZ})
$$W_{k,X}=\mathrm{Hom}_k(\Lambda_k^{2d} V(X),{\mathbf Q}(d)) \hookrightarrow \mathrm{Hom}_{{\mathbf Q}}(\Lambda_{{\mathbf Q}}^{2d}V(X),{\mathbf Q}(d))=H^{2d}(X({\mathbf C}),{\mathbf Q})(d).$$
Clearly, $W_{k,X}$ carries a natural structure of one-dimensional $k-$vector
space. If fix an isomorphism of one-dimensional ${\mathbf Q}-$vector spaces ${\mathbf Q} \cong
{\mathbf Q}(2d)$ then one may naturally
identify $\mathrm{Hom}_{{\mathbf Q}}(\Lambda_{{\mathbf Q}}^{2d}V(X),{\mathbf Q}(d))$ with $\mathrm{Hom}_{{\mathbf Q}}(\Lambda_{{\mathbf Q}}^{2d}V(X),{\mathbf Q})$
and $W_{k,X}$ can
be described as the space of all $2d-$linear skew-symmetric form
$\psi \in \mathrm{Hom}_{{\mathbf Q}}(\Lambda^{2d}_{{\mathbf Q}} V,{\mathbf Q})$ with
$$\psi(f v_1,v_2,\ldots ,v_{2d})=\psi(v_1,fv_2,\ldots ,v_{2d})=\cdots =
\psi(v_1,v_2,\ldots ,fv_{2d})$$
for all $f\in F; v_1, \ldots v_{2d}\in V(X)$.
Since $(X,k.\iota)$ is of Weil type, all elements of $W_k$ are Hodge classes by Proposition 4.4 of \cite{Deligne}. Therefore, by Theorem 2.11 of \cite{Deligne} they must be also {\sl absolute Hodge cycles}; cf. \cite{Deligne}.
\begin{lem} Let $\mu_k$ be the finite multiplicative group of all roots of
unity in $k$. There is a continuous
character $\chi_{X,k}:G(K) \to \mu_k \subset k^*,$
enjoying the following properties:
For each prime $\ell$ the subgroup
$$W_k \subset W_{k,X}\otimes_{{\mathbf Q}}{\mathbf Q}_{\ell}\subset H^{2d}(X({\mathbf C}),{\mathbf Q})(d)\otimes_{{\mathbf Q}}{\mathbf Q}_{\ell}
=H^{2d}(X_a,{\mathbf Q}_{\ell})(d)$$
is $G(K)-$stable and the action of $G(K)$ on $W_k$
is defined via the character
$$\chi_{X,k}:G(K) \to \mu_k \subset k^* =\mathrm{Aut}_k(W_{k,X}).$$
\end{lem}
\begin{proof} Since all elements of $k$ are endomorphisms of $X$ defined over $K$,
it follows easily that $W_{k,X}\otimes_{{\mathbf Q}}{\mathbf Q}_{\ell}$ is $G(K)-$stable and $G(K)$ acts
on $W_{k,X}\otimes_{{\mathbf Q}}{\mathbf Q}_{\ell}$ via a certain character
$\chi_{X,k,\ell}:G(K) \to [k\otimes_{{\mathbf Q}}{\mathbf Q}_{\ell}]^*=
\Pi_{\lambda\mid \ell}k_{\lambda}^*.$
Let us consider the ${\mathbf Q}-$vector subspace
$$C^d_{\mathrm{AH}}(X)\subset H^{2d}(X({\mathbf C}),{\mathbf Q})(d)\subset H^{2d}(X_a,{\mathbf Q}_{\ell})(d)$$
of absolute Hodge classes. Then $C^d_{\mathrm{AH}}(X)$ is $G(K)-$stable and the action of $G(K)$ on $C^d_{\mathrm{AH}}(X)$ does not depend on $\ell$ and factors through a finite quotient; cf. \cite{Deligne}, Prop. 2.9b. Since $W_{k,X}$ consists of Hodge classes and $X$ is an abelian variety, all Weil classes are absolute Hodge classes, i.e,
$W_{k,X}\subset C^d_{\mathrm{AH}}(X),$
\cite{Deligne}, Th. 2.11. This implies easily that
the subgroup $\mathrm{Im}(\chi_{X,k,\ell})$ is finite and contained in $k^*$, since the
intersection of $W_{k,X}\otimes_{{\mathbf Q}}{\mathbf Q}_{\ell}$ and $C^d_{\mathrm{AH}}(X)$ coincides with
$W_{k,X}$. (In fact, $W_{k,X}$ coincides even with the intersection of $W_{k,X}\otimes_{{\mathbf Q}}{\mathbf Q}_{\ell}$ and $H^{2d}(X({\mathbf C}),{\mathbf Q})(d)$.) This implies also that $\chi_{X,k,\ell}$ does not depend on the choice of $\ell$. So, we may view $\chi_{X,k,\ell}$ as the continuous homomorphism
$$\chi_{X,k}:=\chi_{X,k,\ell}:G(K) \to \mu_k \subset k^*,$$
which does not depend on the choice of $\ell$.
\end{proof}
Let $r$ be the order of the finite group $\mathrm{Im}(\chi_{X,k})$. Clearly, $r$ divides the order of $\mu_k$. Let us put $Y=X^r$ and consider the K\"unneth chunk
$$H^{2d}(X({\mathbf C}),{\mathbf Q})(d)^{\otimes r} \subset H^{2dr}(X({\mathbf C})^r,{\mathbf Q})(dr)=H^{2dr}(Y({\mathbf C}),{\mathbf Q})(dr)$$
of the $2dr$th rational cohomology group of $Y$. One may easily check that the tensor power
$$W_{k,X}^{\otimes r}\subset H^{2d}(X({\mathbf C}),{\mathbf Q})(d)^{\otimes r} \subset H^{2dr}(X({\mathbf C})^r,{\mathbf Q})(dr)=H^{2dr}(Y({\mathbf C}),{\mathbf Q})(dr)$$
coincides with the space $W_{k,Y}$ of Weil classes on $Y$ attached to the ``diagonal" embedding
$$k \to \mathrm{End}^0(X) \subset \mathrm{End}^0(X^r)=\mathrm{End}^0(Y).$$
Since the centers of $\mathrm{End}^0(X)$ and $\mathrm{End}^0(X^r)$ coincide, the image of $k$ in $\mathrm{End}^0(Y)$ does contain the center of $\mathrm{End}^0(Y)$.
One may easily check that $G(K)$ acts on $W_{k,Y}=W_{k,X}^{\otimes r}$ via the character $\chi_{X,k}^r$, which is trivial, i.e., $W_{k,Y}$ consists of $G(K)-$invariants.
Let us fix an isomorphism of one-dimensional ${\mathbf Q}-$vector spaces ${\mathbf Q} \cong {\mathbf Q}(2dr)$ and
choose a {\sl non-zero} element
$$\psi \in W_{k,Y} \subset H^{2dr}(Y,{\mathbf Q})(dr)=\mathrm{Hom}_{{\mathbf Q}}(\Lambda^{2dr}_{{\mathbf Q}}(V(Y),{\mathbf Q}).$$
Then a skew-symmetric $2dr-$linear form $\psi$ is admissible.
Applying to $\psi$ the theorems of the previous section, we obtain the following
statement, which
implies the case 4 (in the hypothesis (H)) of Theorem \ref{Theorem 1}.
\begin{thm} Assume that the center $F$ of $\mathrm{End}^0 X$
is a CM-field and $(X,F, \mathrm{id})$ is of Weil type. Let us put
$d= \#(\mu_F) \times 2 \mathrm{dim}(X)/[F:{\mathbf Q}] \in {\mathbf Z}_{+}.$
Let $L$ be an infinite Galois extension of $K$.
\begin{enumerate}
\item Let $\ell$ be a prime such that the $\ell-$torsion
in $X(L)$ is infinite. Let $L^{(\ell)}$ be the intersection of
$L$ and $K(\ell)$.
Then the field extension $K(\ell)/L^{(\ell)}$ has finite degree dividing $(d,\ell-1)$ if $\ell$ is odd and dividing $2$ if $\ell=2$ respectively. In addition, $L$ contains ${\mathbf Q}(\ell)'$.
\item Let $S$ be the set of primes $\ell$ such that
$X(L)$ contains a point of order $\ell$ and assume that $S$
is infinite. Then for all but finitely many $\ell\in S$ the field
extension
$K(\mu_{\ell})/K(\mu_{\ell})\bigcap L$ has degree dividing $(d,\ell-1)$.
\end{enumerate}
\end{thm}
\begin{rem} Since $[F:{\mathbf Q}]$ divides $2\mathrm{dim}(X)=2g$, one may easily find an explicit
positive integer $M=M(g)$, depending only on $g$ and divisible by
$\#(\mu_F) \times 2 \mathrm{dim}(X)/[F:{\mathbf Q}]$
\end{rem}
\section{Proof of Theorem \ref{Theorem 1}}
We may and will assume that $X$ is $K-$simple. Then the center $F$ of $\mathrm{End}^0
X$ is either a totally real number field or a CM-field. If $F$ is totally
real then the assertion of
Theorem \ref{Theorem 1} is proven in \cite{ZarhinDuke} with $N=1$. So, further
we assume that $F$ is a
CM-field. We also know that the assertion of Theorem \ref{Theorem 1} is true
when $(X,F)$ is of Weil type
(Case 4 of Hypothesis (H)).
\subsection{ Cases 1 and 3 of Hypothesis (H)} Enlarging $K$ if necessary, we may and will assume
that $X$ is absolutely simple
and has semistable reduction. Then, the results of \cite{SZ} imply that in
both cases $\mathrm{Hdg}_X$ is semisimple.
This means that $(X,F,\mathrm{id})$ is of Weil type (cf. for instance \cite{SZ}).
Now, one has only to apply the
result of the previous section with $d=\#(\mu_F) \times 2\mathrm{dim}(X)/[F:{\mathbf Q}]$
and get the assertion of Theorem \ref{Theorem 1} with $N=M(g)$.
\subsection{Case 2 of Hypothesis (H)} We know that the assertion of the theorem is true if $(X,F,\mathrm{id})$ is of Weil type. So, we may assume that $(X,F,\mathrm{id})$ is not of Weil type.
Let us consider the trace map
$$\mathbf{Tr}_{\mathrm{Lie}(X)}: F \subset \mathrm{End}^0(X)\hookrightarrow\mathrm{End}_K(\mathrm{Lie}(X)) \to K\subset {\mathbf C}.$$
Our assumption means that the image $\mathbf{Tr}_{\mathrm{Lie}(X)}(F)$ is not contained in ${\mathbf R}$. On the other hand, let us fix an embedding of $F$ into ${\mathbf C}$ and let $L$ be the normal closure of $F$ into ${\mathbf C}$. Clearly, $L$ is a CM-field, containing $\mathbf{Tr}_{\mathrm{Lie}(X)}(F)$. Since $\mathbf{Tr}_{\mathrm{Lie}(X)}(F)\subset K$, the intersection $L\bigcap K$ contains an element, which is not totally real. Since any subfield of a CM-field is either totally real or CM, the field $L\bigcap K$ is a CM-subfield of $K$.
\begin{rem} If $K$ is a number field not containing a CM-field, one may give another proof, using theory of abelian $\lambda-$adic representations
\cite{Serre} instead of Weil/Hodge classes. The crucial point is that in this case the Serre's tori $T_{\mathfrak m}$ are isomorphic to the multiplicative group ${\mathbf G}_m$ \cite{Serre}, Sect. 3.4.
\end{rem}
\begin{cor} Let $X$ be a $K-$simple abelian variety of odd dimension. Assume that $K$ does not contain a CM-subfield (e.g., $K\subset {\mathbf R}$). If $X(L)$ contains infinitely many points of finite order then $L$ contains infinitely many roots of unity.
\end{cor}
\begin{proof} In the case of the totally real center $F$ this assertion is proven in (\cite{ZarhinDuke}, Th.6, p. 142) without restrictions on the dimension. So, in order to prove Corollary, it suffices to check that $F$ is not a CM-field.
Assume that $F$ is a CM-field. Since $\mathrm{dim}(X)$ is odd, $(X,F,\mathrm{id})$ is not
of Weil type. Now, the arguments, used in the proof of
Case 2 imply that $K$ contains a CM-subfield. This gives us a desired contradiction.
\end{proof}
\begin{rem} The assertion of Corollary cannot be extended to the even-dimensional
case. In Section \ref{roots} we give an explicit counterexample.
\end{rem}
\begin{rem} Let $X$ be a $g-$dimensional abelian variety that is not necessarily $K-$simple and let $F$ be the center of $\mathrm{End}^0(X)$. Assume that
$$\mathbf{Tr}_{\mathrm{Lie}(X)}(F)\subset{\mathbf R}.$$
Then the assertion of Theorem \ref{Theorem 1} holds true for $X$.
Indeed, if $Y$ is a $K-$simple abelian subvariety of $X$ and $F_Y$ is the center of $\mathrm{End}^0(Y)$ then one may easily check that either $F_Y$ is a totally real number field or $(Y,F_Y,\mathrm{id})$ is of Weil type.
\end{rem}
\section{Example}
\label{roots}
In this section we construct an abelian surface $X$ over ${\mathbf Q}$ and a Galois
extension $L$ of ${\mathbf Q}$ such that $L$ contains only
finitely many roots of unity but $X(L)$ contains infinitely
many points of finite order.
Of course, the intersection of $L$ and ${\mathbf Q}(c)$ is of infinite degree over ${\mathbf Q}$.
\subsection{} Let $E$ be an elliptic curve over ${\mathbf Q}$ without complex multiplication
(e. g., $j(E)$ is not an
integer). Let us put
$$Y=\{(e_1,e_2,e_3) \in E^3\mid e_1+e_2+e_3=0\}.$$
Clearly, $Y$ is an abelian surface over ${\mathbf Q}$ isomorphic to $E^2$.
Denote by $s$ an automorphism of $Y$ induced by the cyclic permutation of factors of $E^3$, i.e.,
$$s(e_1,e_2,e_3)=(e_3,e_1,e_2) \quad \forall\ (e_1,e_2,e_3) \in Y.$$
Let $C$ be the cyclic subgroup in $\mathrm{Aut}(X)$ of order $3$ generated by $s$.
By a theorem of Serre \cite{Serre2} the homomorphism
$$\rho_{\ell,E}: G({\mathbf Q}) \to \mathrm{Aut}_{{\mathbf Z}_{\ell}}(T_{\ell}(E)) \cong \mathrm{GL}(2,{\mathbf Z}_{\ell})$$
is {\sl surjective} for all but finitely many $\ell$.
Notice, that the composition
$$\mathrm{det} \rho_{\ell,E}: G({\mathbf Q}) \to \mathrm{GL}(2,{\mathbf Z}_{\ell}) \to {\mathbf Z}_{\ell}^*$$
coincides with $\chi_{\ell}: G(K)\to {\mathbf Z}_{\ell}^*$ \cite{Serre2}.
In particular, if ${\mathbf Q}(E(\ell^{\infty}))$ is the field of definition of all
points on $E$ of $\ell$-power order
then ${\mathbf Q}(E(\ell^{\infty}))/{\mathbf Q}$ is the Galois extension with the Galois group
$\mathrm{GL}(2,{\mathbf Z}_{\ell})$. In addition, the cyclotomic
field ${\mathbf Q}(\ell)$ is the {\sl maximal abelian} subextension of ${\mathbf Q}(E(\ell^{\infty}))$ and the subgroup
$\mathrm{Gal}({\mathbf Q}(E(\ell^{\infty}))/{\mathbf Q}(\ell)) \subset \mathrm{Gal}({\mathbf Q}(E(\ell^{\infty}))/{\mathbf Q})$ coincides with $\mathrm{SL}(2,{\mathbf Z}_{\ell})$.
Let us fix such an $\ell$, assuming in addition that $\ell-1$ is divisible
by $3$ but not divisible by $9$.
Let $\mu_{3,\ell}$ be the group of cubic roots of unity in ${\mathbf Z}_{\ell}^*$.
Then there exists a continuous
surjective homomorphism
$\mathrm{pr}_3: {\mathbf Z}_{\ell}^* \to \mu_{3,\ell},$
coinciding with the identity map on $\mu_{3,\ell}$.
These properties determine $\mathrm{pr}_3$ uniquely.
Let us define field $L$ as a subextension of ${\mathbf Q}(E(\ell^{\infty}))$ such that ${\mathbf Q}(E(\ell^{\infty}))/L$ is a cubic extension, whose Galois (sub)group coincides with
$$\mu_{3,\ell}\cdot\mathrm{id}=\{\gamma \cdot \mathrm{id}\mid \gamma \in \mu_{3,\ell}\} \subset \mathrm{GL}(2,{\mathbf Z}_{\ell}).$$
It follows immediately that $L$ is a Galois extension of ${\mathbf Q}$ and it does
not contain a primitive $\ell$th root of unity.
This implies that $1$ and $-1$ are the only roots of unity in $L$.
Let us choose a {\sl primitive} cubic root of unity $\gamma \in \mu_{3,\ell}$ and let
$\iota: \mu_{3,\ell} \to C$
be the isomorphism, which sends $\gamma$ to
$s$.
Now, let us define $X$ as the twist of $Y$ via the cubic character
$$\kappa:=\iota\mathrm{pr}_3 \chi_{\ell}
=\iota \mathrm{pr}_3 \mathrm{det} \rho_{\ell,E}
: G({\mathbf Q})\to \mu_{3,\ell} \to C \subset \mathrm{Aut}(Y).$$
The Galois module $T_{\ell}(X)$ is the twist of $T_{\ell}(E)^2$ via $\kappa$. Namely,
$$T_{\ell}(X)=\{(v_1,v_2,v_3)\in T_{\ell}(E)\oplus T_{\ell}(E)\oplus T_{\ell}(E)\mid v_1+v_2+v_3=0\}$$
as the ${\mathbf Z}_{\ell}-$module but
$$\rho_{\ell,X}(\sigma)(v_1,v_2,v_3)=\kappa(\sigma)(\rho_{\ell,E}(\sigma)(v_1),\rho_{\ell,X}(\sigma)(v_2),\rho_{\ell,X}(\sigma)(v_3))$$
for all $\sigma \in G({\mathbf Q})$. Now, we construct explicitly $G(L)-$invariant
elements of $T_{\ell}(X).$ Starting
with any $v \in T_{\ell}(E)$, put
$$w=(\gamma^{-1}v,\gamma v,v)=(\gamma^2 v,\gamma v, v)\in T_{\ell}(E)\oplus T_{\ell}(E)\oplus T_{\ell}(E).$$
Clearly,
$w \in T_{\ell}(X);\quad sw =\gamma w.$
Let us check that $w$ is $G(L)-$invariant. Clearly,
$$G(L)=\{\sigma\in G({\mathbf Q})\mid \rho_{\ell,E}(\sigma) \in \mu_{3,\ell}\cdot \mathrm{id}\}.$$
Let $\sigma \in G(L)$ with $\rho_{\ell,E}(\sigma)=\zeta\mathrm{id}, \quad \zeta \in
\mu_{3,\ell}.$ If $\zeta=1$ , i.e., $\rho_{\ell,E}(\sigma)=\mathrm{id}$ then all
elements of $V_{\ell}(X)$ are
$\sigma-$invariant. Since $\mu_{3,\ell}=\{1,\gamma,\gamma^{-1}\}$, we may
assume that $\zeta=\gamma$, i.e.,
$\rho_{\ell,E}(\sigma)=\gamma\cdot \mathrm{id}$
and therefore
$\mathrm{det} \rho_{\ell,E}(\sigma)=\gamma^2=\gamma^{-1}.$
Then
$$\rho_{\ell,X}(\sigma)(w)=$$
$$\iota(\mathrm{pr}_3(\mathrm{det} \rho_{\ell,E}(\sigma)))(\rho_{\ell,E}(\sigma)(\gamma^2
v),\rho_{\ell,E}(\sigma)(\gamma v), \rho_{\ell,E}(\sigma)( v))=\iota(\gamma^2)(\gamma w)=$$
$$s^2(\gamma w)=\gamma s^2 w=\gamma\gamma^{2} w=w.$$
This proves that $w$ is $G(L)-$invariant.
Now, I claim that $X(L)$ contains infinitely many points, whose order is
a power of $\ell$. Indeed, starting with a non-divisible element $v \in T_{\ell}(E)$
and identifying the group $X_{\ell^n}$ with the quotient
$T_{\ell}(X)/\ell^n T_{\ell}(X)$, we get a $L-$rational point
$(\gamma^2 v,\gamma v,v)\mod\ell^n T_{\ell}(X) \in T_{\ell}(X)/\ell^n
T_{\ell}(X)=X_{\ell^n}$
of order $\ell^n$.
\section{Another Example}
Let $K$ be an imaginary quadratic field with class number $1$ and let
$E$ be
an elliptic curve over ${\mathbf Q}$ such that $\mathrm{End}_K(E)=O_K$ is the ring of
integers in $K$. In this section we construct a Galois
extension $L$ of $K$ such that $E(L)$ contains
infinitely many points of finite order but the intersection of $L$ and
$K(c)$ is of finite degree over $K$ (even coincides with $K$).
We write $\iota:{\mathbf C} \to {\mathbf C}$ for the complex conjugation $z\mapsto
\bar{z}$. We write $R$ for $O_K$. Clearly, $\mathrm{End}_{{\mathbf Q}}(E)={\mathbf Z}\ne R$.
It follows easily that
$$\iota(ux)=\bar{u}(\iota(x))\quad \forall x \in E({\mathbf C}), u\in R.$$
Notice that $K$ is abelian over ${\mathbf Q}$. Since ${\mathbf Q}(c)={\mathbf Q}(ab)$, $K \subset
{\mathbf Q}(c)$ and therefore
$$K(c)={\mathbf Q}(c).$$
\subsection{}
Let $\ell$ be a prime number. We write $R_{\ell}$ for $R \otimes {\mathbf Z}_{\ell}$.
It is well-known that $T_{\ell}(E)$ is a free $R \otimes {\mathbf Z}_{\ell}$-module
of rank $1$ and therefore
$$\mathrm{End}_{R_{\ell}}(T_{\ell}(E))=R_{\ell},\quad \mathrm{Aut}_{R_{\ell}}(T_{\ell}(E))=R_{\ell}^*.$$
Let us consider the corresponding $\ell$-adic representation
$$\rho_{\ell,E}: G({\mathbf Q}) \to \mathrm{Aut}_{{\mathbf Z}_{\ell}}(T_{\ell}(E)) \cong \mathrm{GL}(2,{\mathbf Z}_{\ell}).$$
Clearly, $G_{\ell}:=\rho_{\ell,E}(G({\mathbf Q}))$ is not a subgroup of
$R_{\ell}^*=\mathrm{Aut}_{R_{\ell}}(T_{\ell}(E))$ but
$$H_{\ell}:=\rho_{\ell,E}(G(K))\subseteq R_{\ell}^*.$$
It is also known (\cite{Serre2}, Sect. 4.5) that
$$H_{\ell}=R_{\ell}^*$$
for all but finitely many primes $\ell$. Let us fix such an $\ell$, assuming
in addition that $\ell$ is unramified and splits in $K$.
This implies that $\ell={\mathfrak q}\bar{{\mathfrak q}}$ for some ${\mathfrak q}\in K$ and
$$O_K={\mathfrak q}\cdot O_K+\bar{{\mathfrak q}}\cdot O_K,\quad R_{\ell}=R_{{\mathfrak q}}\oplus R_{\bar{{\mathfrak q}}},
\quad R_{{\mathfrak q}}={\mathbf Z}_{\ell}, R_{\bar{{\mathfrak q}}}= {\mathbf Z}_{\ell},$$
$${\mathfrak q} R_{\ell}=\ell R_{{\mathfrak q}}\oplus R_{\bar{{\mathfrak q}}}=
\ell {\mathbf Z}_{\ell}\oplus{\mathbf Z}_{\ell}\subset {\mathbf Z}_{\ell}\oplus{\mathbf Z}_{\ell}= R_{\ell},$$
$$\bar{{\mathfrak q}} R_{\ell}= R_{{\mathfrak q}}\oplus \ell R_{\bar{{\mathfrak q}}}=
{\mathbf Z}_{\ell}\oplus\ell{\mathbf Z}_{\ell}\subset {\mathbf Z}_{\ell}\oplus{\mathbf Z}_{\ell}= R_{\ell},$$
$$R_{\ell}^*=R_{{\mathfrak q}}^*\times R_{\bar{{\mathfrak q}}}^*,\quad R_{{\mathfrak q}}^*={\mathbf Z}_{\ell}^*, R_{\bar{{\mathfrak q}}}^*= {\mathbf Z}_{\ell}^*.$$
We also have
$$T_{\ell}(E)=T_{{\mathfrak q}}(E)\oplus T_{\bar{{\mathfrak q}}}(E)$$
where
$$T_{{\mathfrak q}}(E):=R_{{\mathfrak q}}\cdot T_{\ell}(E),\quad T_{\bar{{\mathfrak q}}}(E):=R_{\bar{{\mathfrak q}}}\cdot T_{\ell}(E)$$
are free ${\mathbf Z}_{\ell}$-modules of rank $1$. This implies that for each positive integer $i$
$${\mathfrak q}^i T_{{\mathfrak q}}(E)=\ell^i T_{{\mathfrak q}}(E), \quad \bar{{\mathfrak q}}^i T_{{\mathfrak q}}(E)=T_{{\mathfrak q}}(E),$$
$$\bar{q}^i T_{\bar{{\mathfrak q}}}(E)=\ell^i T_{\bar{{\mathfrak q}}}(E),\quad
{\mathfrak q}^i T_{\bar{{\mathfrak q}}}(E)=T_{\bar{{\mathfrak q}}}(E)$$
and therefore
$$T_{\ell}(E)/\ell^i T_{\ell}(E)=
T_{{\mathfrak q}}(E)/\ell^i T_{{\mathfrak q}}(E)\oplus T_{\bar{{\mathfrak q}}}(E)/\ell^i T_{\bar{{\mathfrak q}}}(E)=
T_{{\mathfrak q}}(E)/{\mathfrak q}^i T_{{\mathfrak q}}(E)\oplus T_{\bar{{\mathfrak q}}}(E)/\bar{{\mathfrak q}}^i T_{\bar{{\mathfrak q}}}(E).$$
It follows easily that a point
$x \in E_{\ell^i}=T_{\ell}(E)/\ell^i T_{\ell}(E)$ satisfies ${\mathfrak q}^i x=0$ (respectively $\bar{{\mathfrak q}}^i x=0$) if and only if
$x \in T_{{\mathfrak q}}(E)/\ell^i T_{{\mathfrak q}}(E)$
(respectively $x \in T_{\bar{{\mathfrak q}}}(E)/\ell^i T_{\bar{{\mathfrak q}}}(E)$).
Let us put
$$\tau:=\rho_{\ell,E}(\iota) \in G_{\ell} \subset \mathrm{Aut}_{{\mathbf Z}_{\ell}}(T_{\ell}(E)).$$
Then $\tau^2=\mathrm{id}$ and
$$\tau( R_{{\mathfrak q}}^*\times\{1\}) \tau^{-1}=\{1\}\times R_{\bar{{\mathfrak q}}}^*\subset R_{\ell}^* ,\quad
\tau(\{1\}\times R_{\bar{{\mathfrak q}}}^*) \tau^{-1}=R_{{\mathfrak q}}^*\times\{1\})\subset R_{\ell}^*.$$
It is also clear that
$$\tau(T_{{\mathfrak q}}(E))=T_{\bar{{\mathfrak q}}}(E), \quad \tau(T_{\bar{{\mathfrak q}}}(E))=T_{{\mathfrak q}}(E).$$
Let us consider the field $K(E(\ell^{\infty}))$ of definition of all
points on $E$ of $\ell$-power order. It is the Galois extension of $K$ with the Galois group
$R_{\ell}^*=R_{{\mathfrak q}}^*\times R_{\bar{{\mathfrak q}}}^*.$
It is also normal over ${\mathbf Q}$ and $\mathrm{Gal}(K(E(\ell^{\infty}))/{\mathbf Q})=G_{\ell}$, since $E$ is defined over ${\mathbf Q}$ and $K$ is normal over ${\mathbf Q}$.
Let us define $L$ as a subextension of $K(E(\ell^{\infty}))/K$ such
that
$$\mathrm{Gal}(K(E(\ell^{\infty}))/L)=\{1\}\times R_{\bar{{\mathfrak q}}}^*\subset
R_{{\mathfrak q}}^*\times R_{\bar{{\mathfrak q}}}^*=R_{\ell}^*=\mathrm{Gal}(K(E(\ell^{\infty}))/K).$$
One may
easily check that $L$ coincides with the field $K(E({\mathfrak q}^{\infty}))$ of definition of all
torsion points on $E$ which are killed by a power of ${\mathfrak q}$. In
particular, $E(L)$ contains infinitely many points, whose order is a power
of $\ell$.
Let us consider the field $L'=\iota(L)$. Clearly, $K\subset L'\subset K(E(\ell^{\infty}))$
and $L'$ coincides with the field $K(E(\bar{{\mathfrak q}}^{\infty}))$
of definition of all torsion points on $E$ which are killed by a power of
$\bar{{\mathfrak q}}$.
It is also clear that
$$\mathrm{Gal}(K(E(\ell^{\infty}))/L)=\tau(\{1\}\times
R_{\bar{{\mathfrak q}}}^*)\tau^{-1}=R_{{\mathfrak q}}^*\times\{1\}\subset
R_{{\mathfrak q}}^*\times R_{\bar{{\mathfrak q}}}^*=R_{\ell}^*=\mathrm{Gal}(K(E(\ell^{\infty}).$$
Since the subgroups $\{1\}\times R_{\bar{{\mathfrak q}}}^*$ and
$R_{{\mathfrak q}}^*\times\{1\}$ generate the whole group $R_{\ell}^*=\mathrm{Gal}(K(E(\ell^{\infty}))/K)$,
$$L\bigcap \iota(L)=L\bigcap L'=K.$$
It follows that if $M/K$ is a subextension of $L/K$ such that $M$ is
normal over ${\mathbf Q}$ then $M=K$. Since $K(c)={\mathbf Q}(c)$,
$L\bigcap K(c)=L\bigcap {\mathbf Q}(c)$ is a subfield of
${\mathbf Q}(c)$ and therefore is normal (even abelian) over ${\mathbf Q}$. It follows that
$$L\bigcap K(c)=K.$$
\section{Abelian subextensions}
The following statement may be viewed as a variant of Theorem \ref{Theorem 1} for
arbitrary abelian varieties over number fields.
\begin{thm}
\label{abelian}
\sl Let $X$ be an abelian variety over a number field $K$. Then:
\begin{enumerate}
\item If for some prime $\ell$ the $\ell-$primary part of $\mathrm{TORS}(X(L))$
is infinite then $L$ contains an
abelian infinite subextension $E\subset L$ such that $\mathrm{Gal}(E/K)\cong {\mathbf Z}_{\ell}$ and $E/K$ is ramified only at divisors of $\ell$.
\item Let $P=P(X,L)$ be the set of primes $\ell$ such that $X(L)$ contains
a point of order $\ell$. If
$P$ is infinite then for all but finitely many primes $\ell \in P$ there
exist a finite subextension
$E^{(\ell)}\subset L$ such that $E^{(\ell)}/K$ is a ramified abelian extension
which is unramified outside
divisors of $\ell$.
In addition, the degree $[E^{(\ell)}:K]$ is prime to $\ell$ and degree
$[E^{(\ell)}:K]$ tends to infinity while $\ell$ tends to infinity.
\end{enumerate}
\end{thm}
\begin{cor}[Theorem of Bogomolov] If $\mathrm{TORS}(X(L))$ is infinite then $L$ contains an infinite abelian subextension of $K$.
\end{cor}
\begin{proof}[Proof of Theorem \ref{abelian}]
First, we may and will assume that $X$ is $K-$simple, i.e., the center
$F$ of the endomorphism algebra of $X$ is a number field.
Second, there is a positive integer $d$, enjoying the following property:
If $m$ is a positive integer such that $\varphi(m) \le 2g=2\mathrm{dim}(X)$ then
$d$ is divisible by $m$.
Third, let $\lambda$ be a prime ideal in $O_F$ dividing a prime number
$\ell$. Then, in the notations of Section 1 the following statement is
true.
\begin{lem}
\begin{enumerate}
\item
The composition
$$\pi_{\lambda}:=(\mathrm{det}_{F_{\lambda}} \rho_{\lambda,X})^d:G(K) \to
\mathrm{Aut}_{F_{\lambda}}V_{\lambda}(X) \to {F_{\lambda}}^* \to
{F_{\lambda}}^*$$
is an abelian representation of $G(K)$ unramified outside divisors of
$\ell$.
\item For all but finitely many $\lambda$ the composition
$$\bar{\pi}_{\lambda}:=(\mathrm{det}_{F_{\lambda}}\bar{\rho}_{\lambda,X})^d:G(K) \to
\mathrm{Aut}_{O_F/\lambda}X_{\lambda} \to (O_F/\lambda)^* \to
(O_F/\lambda)^*$$
is an abelian representation of $G(K)$ unramified outside divisors of
$\ell$.
\end{enumerate}
\end{lem}
We will prove Lemma at the end of this section. Now, let us finish the
proof of Theorem, assuming validity of Lemma.
First, notice that the ratio
$$e=2\mathrm{dim}(X)/[F:{\mathbf Q}]$$
is a positive integer. Second,
for all but finitely many primes $p$ there exists a finite
collection of {\sl Weil numbers}, i.e., certain algebraic integers $\{\alpha_1,\ldots
\alpha_{e}\} \subset F(a)$, enjoying the following properties:
\begin{itemize}
\item (Weil's condition) There is a positive integer $q>1$ such that $q$ is an integral power
of $p$ and all $\mid\alpha_i\mid^2=q$ for all embeddings
$F(a)\subset{\mathbf C}$.
\item
For all $\ell\ne p$ and $\lambda\mid\ell$ there is a subset
$S_{\lambda}\subset \{1,\ldots e\}$ such that
$(\prod_{i\in S_{\lambda}}\alpha_i)\in
O_F$ and the group $\mathrm{Im}(\pi_{\lambda})$ contains $\prod_{i\in
S_{\lambda}}\alpha_i$.
\item For all but finitely many $\lambda$ the subgroup
$\mathrm{Im}(\bar{\pi}_{\lambda})$
contains $(\prod_{i\in S_{\lambda}}\alpha_i)\mod \lambda \in (O_F/\lambda)^*$.
\end{itemize}
Indeed, let us choose a prime ideal $\mathbf v$ in the ring $O_K$ of all
algebraic integers in $K$ such that $X$ has good reduction at $\mathbf
v$. Let
$$Fr_{\mathbf v} \in \mathrm{Im}(\rho_{\lambda,X}) \subset \mathrm{Aut}_{F_{\lambda}}V_{\lambda}(X)$$
be {\sl Frobenius element} $Fr_{\mathbf v}$ at $\mathbf v$ (defined up to
conjugacy)\cite{Serre},\cite{RibetA}. Then the set of its eigenvalues belongs
to $F(a)$, does not depend on the choice of $\lambda$ and satisfies all the desired properties with $p$
the residual characteristic of $\mathbf v$ and $q=\#(O_K/{\mathbf v})$(\cite{Shimura},
Ch. 7, Prop. 7.21 and proof of Prop. 7.23).
\begin{proof}[Proof of assertion 1] We know that there exists $\lambda$ dividing
$\ell$ such that the subspace $V_{\lambda}(X)$ consists of $G(L)-$invariants. This means
that $G(L)$ lies in the kernel of $\pi_{\lambda}$. This implies that
the field $E'$ of $\ker(\pi_{\lambda})-$invariants is an abelian subextension
of L, unramified outside divisors of $\ell$ and $\mathrm{Gal}(E'/K)$ is
isomorphic to $\mathrm{Im}(\pi_{\lambda})$. Choosing a collection of Weil numbers
attached to prime $p\ne\ell$, we easily conclude that $\mathrm{Im}(\pi_{\lambda})$ is an
{\sl infinite} commutative $\ell-$adic Lie group \cite{Serre} and therefore,
there is a continuous quotient of $\mathrm{Im}(\pi_{\lambda})$, isomorphic to
${\mathbf Z}_{\ell}$. One has to take as $E$ the subextension of $E'$
corresponding to this quotient.
\end{proof}
\begin{proof}[Proof of assertion 2]
We know that for all but finitely many $\ell \in P$ there exists
$\lambda$ dividing $\ell$ such that $X_{\lambda}$ consists of $G(L)-$invariants.
This means that the field $E^{(\ell)}$ of
$\ker(\bar{\pi}_{\lambda})-$invariants is an abelian subextension
of L, unramified outside divisors of $\ell$ and $\mathrm{Gal}(E^{(\ell)}/K)$ is
isomorphic to $\mathrm{Im}(\pi_{\lambda})$. In order to prove that
$[E^{(\ell)}:K]$ tends to infinity, let us assume that there exist an
infinite subset $P'\subset P$ and a positive integer $D$ such that
$\#(\mathrm{Gal}(E^{(\ell)}/K))=[E^{(\ell)}:K]$ divides $D$ for all $\ell\in P'$. This means that
$$\bar{\pi}_{\lambda}^D: G(K)\to (O_F/\lambda)^*$$
is a trivial homomorphism for {\sl infinitely many} $\lambda$.
In order to get a contradiction, let us choose a collection of Weil
numbers $\{\alpha_1,\ldots
\alpha_{e}\}$ enjoying the properties described above.
Clearly. for any non-empty subset $ S \subset \{1,\ldots e\}$ the
product $\alpha_S:=\prod_{i\in S} \alpha_i$ is not a root of unity. In addition,
if $\alpha_S\in
O_F$ then there only finitely many $\lambda$ such that $\alpha_S^D-1$ is
an element of $\lambda$. Since there are only finitely many subsets of $\{1,\ldots2g\}$,
for all but finitely many $\lambda$ the group
$$\mathrm{Im}((\bar{\pi}_{\lambda})^D) \subset (O_F/\lambda)^*$$
contains an element of type $\alpha_S^D\mod \lambda$ different from 1.
This implies that $(\bar{\pi}_{\lambda})^D$ is a non-trivial
homomorphism for all but finitely many $\lambda$. This gives the
desired contradiction.
\end{proof}
\begin{proof}[Proof of Lemma]
Let $\mathbf v$ be a prime ideal in the ring $O_K$ of all
algebraic integers in $K$. We write $I_{\mathbf v} \subset G(K)$ for
the corresponding inertia subgroup defined up to conjugacy. Assume that the
residual characteristic of $\mathbf v$ is different from $\ell$. It is known
\cite{SGA} that for any $ \sigma \in I_{\mathbf v}$ there exists a
positive integer $m$ such that $\rho_{\ell,X}(\sigma)^m$ is an unipotent
operator in $V_{\ell}(X)$ and its characteristic polynomial has
coefficients in ${\mathbf Z}$. This implies that if $m$ is the smallest integer enjoying
this property then the characteristic polynomial is divisible by the
$m$th cyclotomic polynomial. This implies that $2g\ge \varphi(m)$ and
therefore $m$ divides $d$. Since $V_{\lambda}(X)$ is a Galois-invariant
subspace of $V_{\ell}(X)$ and (for all but finitely many $\ell$) $X_{\lambda}$
is a Galois-invariant
subspace of $T_{\ell}(X)/\ell T_{\ell}(X)$, a Galois automorphism
$\sigma^d$ acts as an unipotent operator in $V_{\lambda}(X)$ and (for
all but finitely many $\lambda$) in $X_{\lambda}$. One has only to
recall that the determinant of an unipotent operator is always $1$.
\end{proof}
\end{proof}
|
1997-08-22T16:19:27 | 9708 | alg-geom/9708020 | en | https://arxiv.org/abs/alg-geom/9708020 | [
"alg-geom",
"math.AC",
"math.AG"
] | alg-geom/9708020 | Gunnar Floystad | Gunnar Floystad | A property deducible from the generic initial ideal | Completely revised compared to earlier hardcopy versions. AMS-Latex
v1.2, 13 pages | Journal of Pure and Applied Algebra, 136 (1999), no.2, p.127-140 | 10.1016/S0022-4049(97)00165-5 | null | null | " Let $S_d$ be the vector space of monomials of degree $d$ in the variables\n$x_1, ..., x_s$. For a(...TRUNCATED) | [
{
"version": "v1",
"created": "Fri, 22 Aug 1997 14:19:15 GMT"
}
] | 2011-12-14T00:00:00 | [
[
"Floystad",
"Gunnar",
""
]
] | alg-geom | "\\section*{Introduction}\n\nDuring the recent years the generic initial ideal of a homogeneous idea(...TRUNCATED) |
1997-08-22T10:49:15 | 9708 | alg-geom/9708019 | en | https://arxiv.org/abs/alg-geom/9708019 | [
"alg-geom",
"math.AG"
] | alg-geom/9708019 | Alexander A. Voronov | Alexander A. Voronov (RIMS and M.I.T.) | Stability of the Rational Homotopy Type of Moduli Spaces | 7 pages, 1 figure | null | null | RIMS-1157 | null | " We show that for g > 2k+2 the k-rational homotopy type of the moduli space\nM_{g,n} of algebraic (...TRUNCATED) | [
{
"version": "v1",
"created": "Fri, 22 Aug 1997 08:49:13 GMT"
}
] | 2008-02-03T00:00:00 | [
[
"Voronov",
"Alexander A.",
"",
"RIMS and M.I.T."
]
] | alg-geom | "\\section*{Introduction}\n\nThe description of the algebraic topology of the moduli space\n$\\mgn{g(...TRUNCATED) |
1997-08-07T16:22:05 | 9708 | alg-geom/9708010 | en | https://arxiv.org/abs/alg-geom/9708010 | [
"alg-geom",
"math.AG",
"math.QA",
"q-alg"
] | alg-geom/9708010 | Carlos Simpson | Carlos Simpson (CNRS, Universit\'e Paul Sabatier, Toulouse, France) | Limits in $n$-categories | Approximately 90 pages | null | null | null | null | " We define notions of direct and inverse limits in an $n$-category. We prove\nthat the $n+1$-categ(...TRUNCATED) | [
{
"version": "v1",
"created": "Thu, 7 Aug 1997 16:31:55 GMT"
}
] | 2008-02-03T00:00:00 | [
[
"Simpson",
"Carlos",
"",
"CNRS, Université Paul Sabatier, Toulouse, France"
]
] | alg-geom | "\\section*{Limits in $n$-categories}\n\nCarlos Simpson\\newline\nCNRS, UMR 5580, Universit\\'e Paul(...TRUNCATED) |
1997-08-18T09:52:59 | 9708 | alg-geom/9708014 | en | https://arxiv.org/abs/alg-geom/9708014 | [
"alg-geom",
"math.AG"
] | alg-geom/9708014 | Leticia B. Paz | L. Brambila-Paz and H. Lange | A stratification of the moduli space of vector bundles on curves | Latex, Permanent e-mail L. Brambila-Paz: [email protected]
Classification: 14D, 14F | null | null | null | null | " Let $E$ be a vector bundle of rank $r\\geq 2$ on a smooth projective curve $C$\nof genus $g \\geq(...TRUNCATED) | [
{
"version": "v1",
"created": "Mon, 18 Aug 1997 07:52:26 GMT"
}
] | 2016-08-30T00:00:00 | [
[
"Brambila-Paz",
"L.",
""
],
[
"Lange",
"H.",
""
]
] | alg-geom | "\\section{The invariants ${ {}{\\mbox{\\euf s}_k}}(E)$}\n\n\nLet $C$ be a smooth projective curve o(...TRUNCATED) |
1998-08-05T18:28:10 | 9708 | alg-geom/9708011 | en | https://arxiv.org/abs/alg-geom/9708011 | [
"alg-geom",
"math.AG"
] | alg-geom/9708011 | Balazs Szendroi | Balazs Szendroi | Some finiteness results for Calabi-Yau threefolds | "15 pages LaTex, uses amstex, amscd. New title, paper completely\n rewritten, results same as in pr(...TRUNCATED) | null | null | null | null | " We investigate the moduli theory of Calabi--Yau threefolds, and using\nGriffiths' work on the per(...TRUNCATED) | [{"version":"v1","created":"Tue, 12 Aug 1997 15:15:06 GMT"},{"version":"v2","created":"Wed, 29 Oct 1(...TRUNCATED) | 2008-02-03T00:00:00 | [
[
"Szendroi",
"Balazs",
""
]
] | alg-geom | "\\section*{Introduction}\n\nIf $X$ is a smooth complex projective $n$-fold, \nHodge--Lefschetz theo(...TRUNCATED) |
1997-08-26T19:03:35 | 9708 | alg-geom/9708022 | en | https://arxiv.org/abs/alg-geom/9708022 | [
"alg-geom",
"math.AC",
"math.AG"
] | alg-geom/9708022 | Uwe Nagel | J. C. Migliore, U. Nagel, C. Peterson | Buchsbaum-Rim sheaves and their multiple sections | 27 pages, AMS-LaTeX | null | null | null | null | " This paper begins by introducing and characterizing Buchsbaum-Rim sheaves on\n$Z = \\Proj R$ wher(...TRUNCATED) | [
{
"version": "v1",
"created": "Tue, 26 Aug 1997 17:03:21 GMT"
}
] | 2008-02-03T00:00:00 | [
[
"Migliore",
"J. C.",
""
],
[
"Nagel",
"U.",
""
],
[
"Peterson",
"C.",
""
]
] | alg-geom | "\\section{Introduction}\n\nA fundamental method for constructing algebraic varieties is to\nconside(...TRUNCATED) |
2005-11-19T08:38:21 | 9708 | alg-geom/9708006 | en | https://arxiv.org/abs/alg-geom/9708006 | [
"alg-geom",
"math.AC",
"math.AG"
] | alg-geom/9708006 | Joseph Lipman | Leovigildo Alonso, Ana Jeremias, Joseph Lipman | Duality and flat base change on formal schemes | "89 pages. Change from published version: in section 2.5, about\n dualizing complexes on formal sch(...TRUNCATED) | Contemporary Math. 244 (1999), 3-90 | null | null | null | " We give several related versions of global Grothendieck Duality for unbounded\ncomplexes on noeth(...TRUNCATED) | [{"version":"v1","created":"Mon, 4 Aug 1997 17:48:14 GMT"},{"version":"v2","created":"Wed, 14 Oct 19(...TRUNCATED) | 2008-02-03T00:00:00 | [
[
"Alonso",
"Leovigildo",
""
],
[
"Jeremias",
"Ana",
""
],
[
"Lipman",
"Joseph",
""
]
] | alg-geom | "\\section{Preliminaries and main theorems.}\n\\label{S:prelim}\n\nFirst we need some notation and t(...TRUNCATED) |
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