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alg-geom/9708002

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https://arxiv.org/abs/alg-geom/9708002

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

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.

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. \bibliography \parskip=2pt \baselineskip=10.5pt \bi{ACT} D. Allcock, J. Carlson, and D. Toledo, Complex Hyperbolic Structures for Moduli of Cubic Surfaces, C. R. Acad. Sci. Paris {\bf 326}, ser I, pp 49-54 (1998) (alg-geom/970916) \bi{Arnold} V.I. Arnold, S.M. Gusein-Zade, and A.N. Varchenko, Singularities of Differentiable Maps, vol II, Birkhauser, Boston, 1988. \bi{Beardon} A.F. Beardon, The Geometry of Discrete Groups, Springer-Verlag (1983), pp 333. \bi{BeauvilleLattice} A. Beauville, Le groupe de monodromie d'hypersurfaces et d'intersections compl\`etes. Springer Lecture Notes in Mathematics, {\bf 1194} (1986) 8--18. \bi{BorelDT} A. Borel, Density and maximality of arithmetic groups, J. Reine Andgew. Math. {\bf 224} (1966) 78--89. \bi{BHC} A. Borel and Harish-Chandra, Arithmetic subgroups of algebraic groups, Ann. Math. (2) {\bf 75} (1962) 485--535. \bi{CKM} H. Clemens, J. Koll\'ar, and S. Mori, editors, Higher Dimensional Complex Geometry, Ast\'erisque {\bf 166} (1988). \bi{JC} J. Carlson, Hypersurface Variations are Maximal, II, Trans. Am. Math. 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Dolgachev, Weighted projective varieties, in Group Actions and Vector Fields, Proceedings 1981, Lecture Notes in Math. {\bf 956}, Springer-Verlag, New York (1982). \bi{DolgLib} I. Dolgachev and A. Libgober, On the fundamental group to the complement of a discriminant variety, Springer Lecture Notes in Mathematics {\bf 862} (1981), pp 1--25. \bi{Ebeling} W. Ebeling, An arithmetic characterisation of the symmetric monodromy groups of singularities, Invent. Math. {\bf 77} (1984) 85-99. \bi{Givental} A. B. Givental, Twisted Picard-Lefschetz Formulas, Funct. Analysis Appl. {\bf 22} (1988) 10-18. \bi{GreenKC} M. Green, Koszul cohomology and the geometry of projective varieties. J. Differential Geom. {\bf 19} (1984), no. 1, 125--171. \bi{GreenPM} M. Green, The period map for hypersurface sections of high degree of an arbitrary variety. Compositio Math. {\bf 55} (1985), no. 2, 135--156. \bi{GriffHyperbolic} P.A. Griffiths, oral communication 1969. \bi{GriffPerRat} P.A. Griffiths, On the periods of certain rational integrals: I and II, Ann. of Math. {\bf 90} (1969) 460-541. \bi{GriffPerDom} P.A. Griffiths, Periods of integrals of algebraic manifolds, III, Pub. Math. I.H.E.S. {\bf 38} (1970), 125--180. \bi{GS} P.A. Griffiths, and W. Schmid, Locally homogenous complex manifolds, Acta Math. {\bf 123} (1969), 253--302. \bi{Janssen} W. A. M. Janssen, Skew-symmetric vanishing lattices and their monodromy groups, Math. Annalen {\bf 266} (1983) 115-133, {\bf 272} (1985) 17-22. \bi{Lib} A. Libgober, On the fundamental group of the space of cubic surfaces, Math. Zeit. {\bf 162} (1978) 63--67. \bi{Mag} W. Magnus and A. Peluso, On a Theorem of V.I. Arnol'd, Comm. of Pure and App. Math. {\bf 22}, 683--692 (1969). \bi{MargulisRigidity} G.A. Margulis, Discrete groups of motions of manifolds of non-positive curvature, Amer. Math. Soc. Translations {\bf 109} (1977), 33-45. \bi{Milnor} J. Milnor, Singular Points of Complex Hypersurfaces, Annals of Mathematics Studies {\bf 61}, Princeton University Press (1968), pp 122. \bi{Mostow} G. D. Mostow, On a remarkable class of polyhedra in complex hyperbolic space, Pacific Journal of Mathematics {\bf 80} (1980), 171 -- 276. \bi{Pham} F. Pham, Formules de Picard-Lefschetz g\'en\'eralis\'ees et ramification des int\'egrales, Bull. Soc. Math. France {\bf 93}, 1965, 333-367. Reprinted in Homology and Feynman Integrals, Rudolph C. Hwa and Vidgor L. Teplitz, W. A. Benjamin 1966, pp. 331. \bi{Seidel} P. Seidel, Floer homology and the symplectic isotopy problem, thesis, Oxford University, 1997. \bi{ST} M. Sebastiani and R. Thom, Un r\'esultat sur la monodromie, Invent. Math. {\bf 13} (1971), 90--96. \bi{SimpsonHiggs} C. T. Simpson, Higgs bundles and local systems, Publ. Math. IHES {\bf 75} (1992), 5--95. \bi{Tits} J. Tits, Free subgroups in linear groups, J. Algebra {\bf 20}, 250--270 (1972). \bi{Tu} L. Tu, Macaulay's theorem and local Torelli for weighted hypersurfaces, Compositio Math. {\bf 60} (1986), 33--44. \bi{Zar} O. Zariski, On the problem of existence of algebraic functions of two variables possessing a given branch curve, Amer. J. Math. {\bf 51} (1929) 305--328. \bi{Zimmer} R. J. Zimmer, Ergodic Theory and Semisimple Groups, Birkh\"auser 1984, pp 209. \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

9708

alg-geom/9708012

en

https://arxiv.org/abs/alg-geom/9708012

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

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 $S$.

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 $C$.\\

9708

alg-geom/9708007

en

https://arxiv.org/abs/alg-geom/9708007

alg-geom/9708007

Yuri G. Zarhin

Yuri G. Zarhin

Torsion of abelian varieties, Weil classes and cyclotomic extensions

LaTeX 2e 17 pages

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 $K$.

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}

9708

alg-geom/9708020

en

https://arxiv.org/abs/alg-geom/9708020

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

" Let $S_d$ be the vector space of monomials of degree $d$ in the variables\n$x_1, ..., x_s$. For a(...TRUNCATED)

alg-geom

"\\section*{Introduction}\n\nDuring the recent years the generic initial ideal of a homogeneous idea(...TRUNCATED)

9708

alg-geom/9708019

en

https://arxiv.org/abs/alg-geom/9708019

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

RIMS-1157

" We show that for g > 2k+2 the k-rational homotopy type of the moduli space\nM_{g,n} of algebraic (...TRUNCATED)

alg-geom

"\\section*{Introduction}\n\nThe description of the algebraic topology of the moduli space\n$\\mgn{g(...TRUNCATED)

9708

alg-geom/9708010

en

https://arxiv.org/abs/alg-geom/9708010

alg-geom/9708010

Carlos Simpson

Carlos Simpson (CNRS, Universit\'e Paul Sabatier, Toulouse, France)

Limits in $n$-categories

Approximately 90 pages

" We define notions of direct and inverse limits in an $n$-category. We prove\nthat the $n+1$-categ(...TRUNCATED)

alg-geom

"\\section*{Limits in $n$-categories}\n\nCarlos Simpson\\newline\nCNRS, UMR 5580, Universit\\'e Paul(...TRUNCATED)

9708

alg-geom/9708014

en

https://arxiv.org/abs/alg-geom/9708014

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

" Let $E$ be a vector bundle of rank $r\\geq 2$ on a smooth projective curve $C$\nof genus $g \\geq(...TRUNCATED)

alg-geom

"\\section{The invariants ${ {}{\\mbox{\\euf s}_k}}(E)$}\n\n\nLet $C$ be a smooth projective curve o(...TRUNCATED)

9708

alg-geom/9708011

en

https://arxiv.org/abs/alg-geom/9708011

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)

" We investigate the moduli theory of Calabi--Yau threefolds, and using\nGriffiths' work on the per(...TRUNCATED)

alg-geom

"\\section*{Introduction}\n\nIf $X$ is a smooth complex projective $n$-fold, \nHodge--Lefschetz theo(...TRUNCATED)

9708

alg-geom/9708022

en

https://arxiv.org/abs/alg-geom/9708022

alg-geom/9708022

Uwe Nagel

J. C. Migliore, U. Nagel, C. Peterson

Buchsbaum-Rim sheaves and their multiple sections

27 pages, AMS-LaTeX

" This paper begins by introducing and characterizing Buchsbaum-Rim sheaves on\n$Z = \\Proj R$ wher(...TRUNCATED)

alg-geom

"\\section{Introduction}\n\nA fundamental method for constructing algebraic varieties is to\nconside(...TRUNCATED)

9708

alg-geom/9708006

en

https://arxiv.org/abs/alg-geom/9708006

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

" We give several related versions of global Grothendieck Duality for unbounded\ncomplexes on noeth(...TRUNCATED)

alg-geom

"\\section{Preliminaries and main theorems.}\n\\label{S:prelim}\n\nFirst we need some notation and t(...TRUNCATED)