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\datereceived{2014-01-02}
\dateaccepted{2014-05-16}
\dateepreuves{2014-05-26}

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\begin{document}
\frontmatter
\title[Some surfaces with maximal Picard number]{Some surfaces with maximal Picard number}
\author[\initial{A.} \lastname{Beauville}]{\firstname{Arnaud} \lastname{Beauville}}
\address{Laboratoire J.-A. Dieudonn�, UMR 7351 du CNRS, Universit� de Nice\\
Parc Valrose, F-06108 Nice cedex 2, France}
\email{arnaud.beauville@unice.fr}
\urladdr{http://math1.unice.fr/~beauvill/}

\begin{abstract}
For a smooth complex projective variety, the rank $\rho $ of the N�ron-Severi group is bounded by the Hodge number $h^{1,1}$. Varieties with $\rho =h^{1,1}$ have interesting properties, but are rather sparse, particularly in dimension $2$. We discuss in this note a number of examples, in particular those constructed from curves with special Jacobians.
\end{abstract}

\subjclass{14J05, 14C22, 14C25}

\keywords{Algebraic surfaces, Picard group, Picard number, curve correspondences, Jacobians}

\altkeywords{Surfaces alg�briques, groupe de Picard, nombre de Picard, correspondances de courbes, jacobiennes}

\alttitle{Quelques surfaces dont le nombre de Picard est maximal}

\begin{altabstract}
Le rang $\rho$ du groupe de N�ron-Severi d'une vari�t� projective lisse complexe est born� par le nombre de \hbox{Hodge $h^{1,1}$}. Les vari�t�s satisfaisant � $\rho =h^{1,1}$ ont des propri�t�s int�ressantes, mais sont assez rares, particuli�rement en dimension $2$. Dans cette note nous analysons un certain nombre d'exemples, notamment ceux construits � partir de courbes � jacobienne sp�ciale.
\end{altabstract}

\maketitle
\tableofcontents

\mainmatter

\section{Introduction}

The \emph{Picard number} of a smooth projective variety $X$ is the rank $\rho $ of the N�ron-Severi group -- that is, the group of classes of divisors in $H^2(X,\Z)$. It is bounded by the Hodge number $h^{1,1}:= \dim H^1(X,\Omega^1_X)$. We are interested here in varieties with maximal Picard number $\rho =h^{1,1}$. As we will see in \S 2, there are many examples of such varieties in dimension $\geq 3$, so we will focus on the case of surfaces.

Apart from the well understood case of K3 and abelian surfaces, the quantity of known examples is remarkably small. In \cite{P} Persson showed that some families of
double coverings of rational surfaces contain surfaces with maximal Picard number (see Section \ref{P} below); some scattered examples have appeared since then. We will review them in this note and examine in particular when the product of two curves has maximal Picard number -- this provides some examples, unfortunately also quite sparse.

\section{Generalities}
Let $X$ be a smooth projective variety over $ \C $. The N�ron-Severi group $ \NS(X) $ is the subgroup of algebraic classes in $ H^2(X,\Z) $; its rank $ \rho $ is the \emph{Picard number} of $X$. The natural map $ \NS(X) \otimes\C\to H^2(X,\C) $ is injective and its image is contained in~$ H^{1,1}$, hence $ \rho\leq h^{1,1}$.

\begin{prop}\label{triv}
The following conditions are equivalent:
\begin{enumeratei}
\item\label{triv-i}
$ \rho= h^{1,1}$;

\item\label{triv-ii}
The map $ \NS(X) \otimes\C\to H^{1,1}$ is bijective;

\item\label{triv-iii}
The subspace $ H^{1,1}$ of $ H^2(X,\C) $ is defined over $ \Q $.

\item\label{triv-iv}
The subspace $ H^{2,0}\oplus H^{0,2}$ of $ H^2(X,\C) $ is defined over $ \Q $.
\end{enumeratei}
\end{prop}

\begin{proof}
The equivalence of \eqref{triv-iii} and \eqref{triv-iv} follows from the fact that $ H^{2,0}\oplus H^{0,2}$ is the orthogonal of $ H^{1,1}$ for the scalar product on $ H^2(X,\C) $ associated to an ample class. The rest is clear.
\end{proof}

When $X$ satisfies these equivalent properties we will say for short that $X$ is
\emph{$\rho$\nobreakdash-maximal} (one finds the terms singular, exceptional or extremal in the literature).

\skippointrait
\begin{rems*}
\begin{enumerate}
\item
A variety with $ H^{2,0}=0 $ is $\rho$\nobreakdash-maximal. We will implicitly exclude this trivial case in the discussion below.

\item
Let $X$, $Y$ be two $\rho$\nobreakdash-maximal varieties, with $H^1(Y,\C)=0$. Then $X\times Y$ is $\rho$\nobreakdash-maximal. For instance $X\times \P^n$ is $\rho $\nobreakdash-maximal, and $Y\times C$ is $\rho$\nobreakdash-maximal for any curve~$C$.

\item
Let $Y$ be a submanifold of $X$; if $X$ is $\rho$\nobreakdash-maximal and the restriction map $ H^2(X,\C)\to H^2(Y,\C) $ is bijective, $Y$ is $\rho$\nobreakdash-maximal. By the Lefschetz theorem, the latter condition is realized if $Y$ is a complete intersection of smooth ample divisors in~$X$, of dimension $ \geq 3 $. Together with Remark 2, this gives many examples of $\rho$\nobreakdash-maximal varieties of dimension $ \geq 3 $; thus we will focus on finding $\rho$\nobreakdash-maximal \emph{surfaces}.
\end{enumerate}
\end{rems*}

\setcounter{rem}{3}

\begin{prop}\label{map}
Let $\pi: X\dasharrow Y$ be a rational map of smooth projective varieties.
\begin{enumeratea}
\item\label{mapa}
If $\pi^*:H^{2,0}(Y)\to H^{2,0}(X)$ is injective (in particular if $\pi$ is dominant), and
$X$ is $\rho$\nobreakdash-maximal, so is $Y$.

\item\label{mapb}
If $\pi^*:H^{2,0}(Y)\to H^{2,0}(X)$ is surjective and $Y$ is $\rho$\nobreakdash-maximal, so is $X$.
\end{enumeratea}
\end{prop}

Note that since $\pi$ is defined on an open subset $U\subset X$ with $\mathrm{codim}(X\smallsetminus U)\geq 2$, the pull back map $\pi^*: H^2(Y,\C)\to H^2(U,\C)\cong H^2(X,\C)$ is well defined.

\begin{proof}
Hironaka's theorem provides a diagram
\[
\xymatrix{& {\hat X} \ar[dl]_b \ar[dr]^{\hat \pi} & \\
X\ar@{-->}[rr]^\pi & & Y}
\]
where $ \hat{\pi}$ is a morphism, and $ b $ is a composition of blowing-ups with smooth centers. Then $b^*: H^{2,0}(X)\to H^{2,0}(\hat{X})$ is bijective, and
$\hat{ X}$ is $\rho$\nobreakdash-maximal if and only if $X$ is $\rho$\nobreakdash-maximal; so replacing $\pi$ by $ \hat{\pi}$ we may assume
that $\pi$ is a morphism.

\begin{enumeratea}
\item
Let $V:=(\pi^*)^{-1}(\NS(X)\otimes \Q)$. We have
\[
V\otimes_{\Q} \C= (\pi^*)^{-1}(\NS(X)\otimes \C)=(\pi^*)^{-1}(H^{1,1}(X))=H^{1,1}(Y)
\]
(the last equality holds because
$\pi^*$ is injective on $H^{2,0}(Y)$ and $H^{0,2}(Y)$), hence $Y$ is $\rho$\nobreakdash-maximal.

\item
Let $W$ be the $\Q$-vector subspace of $H^2(Y,\Q)$ such that
\[
W\otimes_{\Q}\C= H^{2,0}(Y)\oplus H^{0,2}(Y).
\]
Then $\pi^*W$ is a $\Q$-vector subspace of $H^2(X,\Q)$, and
\[
(\pi^*W)\otimes \C = \pi^*(W\otimes \C)=\pi^*H^{2,0}(Y)\oplus \pi^*H^{0,2}(Y)=H^{2,0}(X)\oplus H^{0,2}(X),
\]
so $X$ is $\rho$\nobreakdash-maximal.\qedhere
\end{enumeratea}
\end{proof}

\section{Abelian varieties}
There is a nice characterization of $\rho$\nobreakdash-maximal abelian varieties (\cite{K}, \cite{L}):

\begin{prop}\label{ab}
Let $ A $ be an abelian variety of dimension $ g $. We have
\[
\rk_{\Z} \End(A) \leq\nobreak 2g^2.
\]
The following conditions are equivalent:
\begin{enumeratei}
\item\label{ab-i}
$ A $ is $\rho$\nobreakdash-maximal;

\item\label{ab-ii}
$ \rk_{\Z} \End(A)=2g^2 $;

\item\label{ab-iii}
$ A $ is isogenous to $ E^g $, where $ E $ is an elliptic curve with complex multiplication.

\item\label{ab-iv}
$A$ is isomorphic to a product of mutually isogenous elliptic curves with complex multiplication.
\end{enumeratei}
\end{prop}

(The equivalence of \eqref{ab-i}, \eqref{ab-ii} and \eqref{ab-iii} follows easily from Lemma \ref{hom} below; the only delicate point is $\eqref{ab-iii}\Rightarrow \eqref{ab-iv}$, which we will not use.)

\medskip	
Coming back to the surface case,
suppose that our abelian variety $ A $ contains a surface $ S $ such that the restriction map $ H^{2,0}(A)\to H^{2,0}(S) $ is surjective. Then $ S$ is $\rho$\nobreakdash-maximal if $ A $ is $\rho$\nobreakdash-maximal (Proposition \ref{map}\eqref{mapb}). Unfortunately this situation seems to be rather rare. We will discuss below (Proposition \ref{c}) the case of $\mathrm{Sym}^2C$ for a curve $C$. Another interesting example is the \emph{Fano surface} $F_X$ parametrizing the lines contained in a smooth cubic threefold $X$, embedded in the intermediate Jacobian $JX$ \cite{CG}. There are some cases in which $JX$ is known to be $\rho$\nobreakdash-maximal:

\skippointrait
\begin{prop}\label{prop:4}
\begin{enumeratea}
\item\label{prop:4a}
For $ \lambda\in\C $, $ \lambda^3\neq 1 $, let $ X_{\lambda}$ (resp. $ E_{\lambda}$) be the cubic in $ \P^4 $ (resp. $\P^2$) defined by
\[
X_{\lambda}\colon X^3+Y^3+Z^3-3\lambda XYZ +T^3+U^3=0,\quad
E_{\lambda}\colon X^3+Y^3+Z^3-3\lambda XYZ =0.
\]
If $E_{\lambda}$ is isogenous to $E_0$, $JX_{\lambda}$ and $F_{X_{\lambda}}$ are $\rho$\nobreakdash-maximal. The set of $\lambda\in\C $ for which this happens is countably infinite.

\item\label{prop:4b}
Let $ X \subset \P^4$ be the Klein cubic threefold $\sum_{i\in \Z/5}X_i^2X_{i+1}=0 $. Then $JX$ and~$F_X$ are $\rho$\nobreakdash-maximal.
\end{enumeratea}
\end{prop}

\begin{proof}
Part \eqref{prop:4a} is due to Roulleau \cite{R}, who proves that $JX_{\lambda}$ (for any $ \lambda $) is isogenous to $ E_0^3\times E_{\lambda}^2 $. Since the family $(E_\lambda)_{\lambda \in \C}$ is not constant, there is a countably infinite set of $\lambda\in\C $ for which $E_{\lambda}$ is isogenous to $E_0$, hence $JX_{\lambda}$ and therefore $F_{X_{\lambda}}$ are $\rho$\nobreakdash-maximal.

\smallskip
Part \eqref{prop:4b} follows from a result of Adler \cite{A}, who proves that $JX$ is isogenous (actually isomorphic) to $ E^5 $, where $ E $ is the elliptic curve whose endomorphism ring is the ring of integers of $ \Q(\sqrt{-11}) $ (see also \cite{Rk} for a precise description of the group $ \NS(X) $).
\end{proof}

\section{Products of curves}\label{prod}

\begin{prop}\label{cc'}
Let $C,C'$ be two smooth projective curves, of genus $g$ and $g'$ respectively. The following conditions are equivalent:
\begin{enumeratei}
\item
The surface $C\times C'$ is $\rho$\nobreakdash-maximal;

\item
There exists an elliptic curve $E$ with complex multiplication such that $JC$ is isogenous to $E^g$ and $JC'$ to $E^{g'}$.
\end{enumeratei}
\end{prop}

\begin{proof}
Let $ p,p' $ be the projections from $ C\times C' $ to $ C $ and $C'$. We have
\begin{multline*}
H^{1,1}(C\times C')= p^*H^2(C,\C)\oplus p'^*H^2(C',\C)\oplus\left(p^*H^{1,0}(C) \otimes p'^*H^{0,1}(C')\right)\\
\oplus \big(p^*H^{0,1}(C) \otimes p'^*H^{1,0}(C') \big),
\end{multline*}
hence $ h^{1,1}(C\times C')=2gg'+2 $. On the other hand we have
\[
\NS(C\times C') = p^*\NS(C)\oplus p'^*\NS(C')\oplus \Hom (JC, JC')
\]
(\cite{BL}, Th.\ 11.5.1), hence $ C\times C' $ is $\rho$\nobreakdash-maximal if and only if $ \rk \Hom (JC, JC') = 2gg' $. Thus the Proposition follows from the following (well-known) lemma:

\begin{lem}\label{hom}
Let $A$ and $B$ be two abelian varieties, of dimension $a$ and $b$ respectively. The $\Z$-module $\Hom(A,B)$ has rank $\leq 2ab$; equality holds if and only if there exists an elliptic curve $E$ with complex multiplication such that $A$ is isogenous to $E^a$ and $B$ to~$E^b$.
\end{lem}

\begin{proof}
There exist simple abelian varieties $A_1,\ldots,A_s$, with distinct isogeny classes, and nonnegative integers $p_1,\ldots,p_s,q_1,\ldots,q_s$ such that $A$ is isogenous to $A_1^{p_1}\times \nobreak \ldots \times\nobreak A_s^{p_s}$ and $B$ to $A_1^{q_1}\times \ldots \times A_s^{q_s}$. Then
\[
\Hom(A,B)\otimes_{\Z}\Q\,\cong\, M_{p_1,q_1}(K_1)\times \ldots \times M_{p_s,q_s}(K_s),
\]
where $K_i$ is the (possibly skew) field $\End(A_i)\otimes_{\Z}\Q$. Put $a_i:=\dim A_i$. Since $K_i$ acts on $H^1(A_i,\Q)$ we have $\dim_{\Q}K_i\leq b_1(A_i)=2a_i$, hence
\[
\rk \Hom(A,B)\leq \sum_i 2p_iq_ia_i \leq 2 \Big(\sum p_ia_i \Big) \Big(\sum q_ia_i \Big)=2ab.
\]
The last inequality is strict unless $s=a_1=1$, in which case
the first one is strict unless $\dim_{\Q}K_1=2$. The lemma, and therefore the Proposition, follow.
\end{proof}
\skipqed
\end{proof}

The most interesting case occurs when $C=C'$. Then:
\begin{prop}\label{c}
Let $ C $ be a smooth projective curve. The following conditions are equivalent:
\begin{enumeratei}
\item\label{c-i}
The Jacobian $JC $ is $\rho$\nobreakdash-maximal;

\item\label{c-ii}
The surface $ C\times C $ is $\rho$\nobreakdash-maximal;

\item\label{c-iii}
The symmetric square $\mathrm{Sym}^2C$ is $\rho$\nobreakdash-maximal.
\end{enumeratei}
\end{prop}

\begin{proof}
The equivalence of \eqref{c-i} and \eqref{c-ii} follows from Proposition \ref{cc'}. The Abel-Jacobi map $\mathrm{Sym}^2C\to JC$ induces an isomorphism
\[
H^{2,0}(JC)\cong \wedge^2 H^{0}(C,K_C)\iso H^{2,0}(\mathrm{Sym}^2C),
\]
thus \eqref{c-i} and \eqref{c-iii} are equivalent by Proposition \ref{map}.
\end{proof}
	
When the equivalent conditions of Proposition \ref{c} hold, we will say that $ C $ has \emph{maximal correspondences} (the group $ \End(JC) $ is often called the group of divisorial correspondences of $ C $).

By Proposition \ref{ab} the Jacobian $JC$ is then isomorphic to a product of isogenous elliptic curves with complex multiplication.
Though
we know very few examples of such curves, we will give below some examples with $g=4$ or $10$.

For $ g=2 $ or $ 3 $, there is a countably infinite set of curves with maximal correspondences (\cite{HN}, \cite{H}). The point is that any indecomposable principally polarized abelian variety of dimension 2 or 3 is a Jacobian; thus it suffices to construct an indecomposable principal polarization on $ E^g $, where $ E $ is an elliptic curve with complex multiplication, and this is easily translated into a problem about hermitian forms of rank $ g $ on certain rings of quadratic integers.

This approach works only for $g=2$ or $3$; moreover it does not give an explicit description of the curves. Another method is by using automorphism groups, with the help of the following easy lemma:

\begin{lem}\label{irred}
Let $ G $ be a finite group of automorphisms of $ C $, and let $ H^0(C,K_C)=\oplus_{i\in I}V_i $ be a decomposition of the $ G $-module $ H^0(C,K_C) $ into irreducible representations. Assume that there exists an elliptic curve $ E $ and for each $ i\in I $, a nontrivial map $ \pi_i:C\to E $ such that $ \pi_i^*H^0(E,K_E)\subset V_i $. Then $JC $ is isogenous to $ E^g $.
\end{lem}
In particular if $H^0(C,K_C)$ is an irreducible $G$-module and $C$ admits a map onto an elliptic curve $E$, then $JC$ is isogenous to $ E^g $.

\begin{proof}
Let $ \eta $ be a generator of $ H^0(E,K_E) $. Let $ i\in I $; the forms
$ g^*\pi_i^*\eta $ for $ g\in G$ generate $V_i$, hence there
exists a subset $ A_i $ of $ G $ such that the forms $ g^*\pi_i^*\eta $ for $ g\in A_i $ form a basis of $ V_i $.

Put $ \Pi_i=(g\rond \pi_i)_{g\in A_i}:C\to E^{A_i}$, and $ \Pi=(\Pi_i)_{i\in I}: C\to E^g $. By construction $ \Pi^*:H^0(E^g,\Omega^1_{E^g})\to H^0(C,K_C) $ is an isomorphism. Therefore the map $JC\to E^g $ deduced from $ \Pi $ is an isogeny.
\end{proof}

In the examples which follow, and in the rest of the paper, we put $\omega:=e^{\sfrac{2\pi i}{3}}$.
\begin{ex} We consider the family $(C_t)$ of genus 2 curves given by $y^2=x^6+tx^3+1$, for $t\in \C\smallsetminus \{\pm 2\}$. It admits the automorphisms
\[
\tau:(x,y)\longmapsto \Big(\dfrac{1}{x}, \dfrac{y}{x^3} \Big)\quad\text{and}\quad \psi:(x,y)\longmapsto(\omega x,y).
\]

The forms $\sfrac{dx}{y}$ and $\sfrac{xdx}{y}$ are eigenvectors for $\psi $ and are exchanged (up to sign) by $\tau $; it follows that the action of the group generated by $\psi $ and $\tau $ on $H^0(C_t,K_{C_t})$ is irreducible.

Let $E_t$ be the elliptic curve defined by $v^2=(u+2)(u^3-3u+t)$; the curve $C_t$ maps onto $E_t$ by
\[
(x,y)\longmapsto\Big(x+\dfrac{1}{x}, \dfrac{y(x+1)}{x^2} \Big).
\]
By Lemma \ref{irred} $JC_t$ is isogenous to $E_t^2$.
Since the $j$-invariant of $E_t$ is a non-constant function of $t$, there is a countably infinite set of $t\in\C$ for which $E_t$ has complex multiplication, hence $C_t$ has maximal correspondences.
\end{ex}

\begin{ex}\label{ex2}
Let $ C $ be the genus 2 curve $ y^2=x(x^4-1) $; its automorphism group is a central extension of $\mathfrak{S}_4$ by the hyperelliptic involution $\sigma $ (\cite{BL}, 11.7); its action on $H^0(C,K_C)$ is irreducible.

Let $E$ be the elliptic curve $E\colon v^2=u(u+1)(u-2\alpha)$, with $\alpha =1-\sqrt{2}$.
The curve~$C$ maps to $E$ by
\[
(x,y)\longmapsto \Big(\dfrac{x^2+1}{x-1}, \dfrac{y(x-\alpha)}{(x-1)^2} \Big).
\]
The $j$-invariant of $E$ is $8000$, so $E$ is the elliptic curve $\C/\Z[\sqrt{-2}]$ (\cite{S}, Prop.\ 2.3.1).
\end{ex}

\begin{ex}[The $\mathfrak{S}_4$-invariant quartic curves]\label{fermat}
Consider the standard representation of $\mathfrak{S}_4$ on $\C^3$. It is convenient to view $\mathfrak{S}_4$ as the semi-direct product $(\Z/2)^2\rtimes\mathfrak{S}_3$, with $\mathfrak{S}_3$ (resp. $(\Z/2)^2$) acting on $\C^3$ by permutation (resp.\ change of sign) of the basis vectors. The quartic forms invariant under this representation form the pencil
\[
(C_t)_{t\in\P^1}\colon x^4+y^4+z^4+t(x^2y^2+y^2z^2+z^2x^2)=0.
\]
According to \cite{DK}, this pencil was known to Ciani. It contains the Fermat quartic $(t=0)$ and the Klein quartic $(t=\frac{3}{2}(1\pm i\sqrt{7}))$.

Let us take $t\notin\{2,-1,-2,\infty\}$; then
$C_t$ is smooth. The action of $\mathfrak{S}_4$ on $H^0(C_t,K)$, given by the standard representation, is irreducible. Moreover the involution $x\mapsto -x$ has $4$ fixed points, hence the quotient curve $E_t$ has genus~$1$. It is given by the degree~$4$ equation
\[
u^2+tu (y^2+z^2) +y^4+z^4+ty^2z^2=0
\]
in the weighted projective space $\P(2,1,1)$. Thus $E_t$ is a double covering of $\P^1$ branched along the zeroes of the polynomial $(t+2)(y^4+z^4)+2ty^2z^2$. The cross-ratio of these zeroes is $-(t+1)$, so $E_t$ is the elliptic curve $y^2= x(x-1)(x+t+1)$. By Lemma \ref{irred} $JC_t$ \emph{is isogenous to} $E_t^3$. For a countably infinite set of $t$ the curve $E_t$ has complex multiplication, thus $C_t$ has maximal correspondences.
For $t=0$ we recover the well known fact that the Jacobian of the Fermat quartic curve is isogenous to
$(\C/\Z[ i])^3$.
\end{ex}

\begin{ex}
Consider the genus 3 hyperelliptic curve $H\colon y^2=x(x^6+1)$. The space $H^0(H,K_H)$ is spanned by $\sfrac{dx}{y}$, $x\sfrac{dx}{y}$, $x^2\sfrac{dx}{y}$. This is a basis of eigenvectors for the automorphism $\tau:(x,y)\mapsto (\omega x,\omega^2 y)$. On the other hand the involution $\sigma:(x,y)\mapsto (\sfrac{1}{x},\sfrac{-y}{x^4})$ exchanges $\sfrac{dx}{y}$ and $x^2\sfrac{dx}{y}$, hence the summands of the decomposition
\[
H^0(H,K_H) = \Lr{\frac{dx}{y}, x^2\frac{dx}{y}}\oplus \Lr{x\frac{dx}{y}}
\]
are irreducible under the group $\mathfrak{S}_3$ generated by $\sigma $ and $\tau $.

Let $E_i$ be the elliptic curve $v^2=u^3+u$, with endomorphism ring $\Z[i]$. Consider the maps $f$ and $g$ from $H$ to $E_i$ given by
\[
f(x,y)=(x^2,xy) \qquad g(x,y)=\Big(\lambda^2 \big(x+\frac{1}{x} \big), \frac{\lambda^3y}{x^2} \Big)\quad\text{with } \lambda^{-4}=-3.
\]
We have
\[
f^*\dfrac{du}{v}=\dfrac{2xdx}{y}\quad\text{and}\quad g^*\dfrac{du}{v}=\lambda^{-1} (x^2-1)\dfrac{dx}{y}.
\]
Thus we can apply Lemma \ref{irred}, and we find that $JH$ \emph{is isogenous to $E_i^3$}.

Thus $JH$ is isogenous to the Jacobian of
the Fermat quartic $F_4$ (Example \ref{fermat}). In particular we see that the surface $H\times F_4$ is $\rho$\nobreakdash-maximal.

\end{ex}

	

We now arrive to our main example in higher genus. Recall that we put $\omega = e^{\sfrac{2\pi i}{3}}$.

\begin{prop}\label{C6}
The Fermat sextic curve $ C_6\colon X^6+Y^6+Z^6=0 $ has maximal correspondences. Its Jacobian $JC_6$ is isogenous to $E_\omega^{10}$, where $E_\omega $ is the elliptic curve $\C/\Z[\omega ]$.
\end{prop}

The first part can be deduced from the general recipe given by Shioda to compute the Picard number of $ C_d\times C_d $ for any $ d $ \cite{S2}. Let us give an elementary proof. Let $G:= T\rtimes \mathfrak{S}_3$, where $\mathfrak{S}_3$ acts on $\C^3$ by permutation of the coordinates and
$T$ is the group of diagonal matrices $t$ with $t^6=1$.
\[
\text{Let}\qquad\Omega =\frac{XdY-YdX}{Z^5}=\frac{Y dZ-ZdY}{X^5} = \frac{ZdX-XdZ}{Y^5}\,\in H^0(C,K_C(-3)).
\]
A basis of eigenvectors for the action of $T$ on $H^0(C_6,K)$ is given by the forms $X^aY^bZ^c\,\Omega $, with $a+b+c=3$; using the action of $\mathfrak{S}_3$ we get a decomposition into irreducible components:
\[
H^0(C_6,K)= V_{3,0,0}\oplus V_{2,1,0}\oplus V_{1,1,1},
\]
where $V_{\alpha,\beta,\gamma}$ is spanned by the forms $X^aY^bZ^c\,\Omega$ with $\{a,b,c\}=\{\alpha,\beta,\gamma \}$.

Let us use affine coordinates $x=\sfrac{X}{Z}$, $y=\sfrac{Y}{Z}$ on $C_6$. We consider the following maps from $C_6$ onto $E_\omega\colon v^2=u^3-1$:
\[
f(x,y)= (-x^2,y^3), \qquad g(x,y)=\Bigl(2^{-\sfrac{2}{3}}x^{-2}y^4,\, \frac{1}{2}(x^3-x^{-3})\Bigr) \,;
\]
and, using for $E_{\omega}$ the equation $\xi^3+\eta^3+1=0$, $h(x,y)=(x^2,y^2)$.

We have
\begin{align*}
f^*\frac{du}{v}&=-\frac{2xdx}{y^3}=-2 XY^2\,\Omega \in V_{2,1,0},\\
g^*\frac{du}{v}&=-2^{\sfrac{4}{3}}Y^3\,\Omega \in V_{3,0,0},\\
h^*\frac{d\xi}{\eta^2}&= 2XYZ\,\Omega \in V_{1,1,1},
\end{align*}
so the Proposition follows from Lemma \ref{irred}.\qed

\medskip
By Proposition \ref{map} every quotient of $ C_6 $ has again maximal correspondences. There are four such quotient which have genus $4$:
\begin{itemize}
\item
The quotient by an involution $\alpha \in T$, which we may take to be $ \alpha: (X,Y,Z)\mapsto (X,Y,-Z) $. The canonical model of $ C_6/\alpha $
is the image of $ C_6 $ by the map
\[
(X,Y,Z)\longmapsto (X^2,XY,Y^2, Z^2)\,;
\]
its equations in $ \P^3 $ are $ xz-y^2= x^3+z^3+t^3=0 $. Projecting onto the conic $xz-y^2=0$ realizes $ C_6/\alpha $ as the cyclic triple covering $v^3=u^6+1$ of $\P^1$.

\item
The quotient by an involution $\beta \in\mathfrak{S}_3$, say $\beta:(X,Y,Z)\mapsto (Y,X,Z)$. The canonical model of $ C_6/\beta $
is the image of $ C_6 $ by the map
\[
(X,Y,Z)\longmapsto ((X+Y)^2, Z(X+Y), Z^2,XY)\,;
\]
its equations are $xz-y^2=x(x-3t)^2+z^3-2t^3=0$.

Since the quadric containing their canonical model is singular, the two genus $4$ curves $C_6/\alpha $ and $C_6/\beta $ have a unique $g^1_3$. The associated triple covering $C_6/\alpha \to \P^1$ is cyclic, while the corresponding covering $C_6/\beta \to \P^1$
is not. Therefore the two curves are not isomorphic.

\item
The quotient by an element of order 3 of $T$ acting freely, say $ \gamma: (X,Y,Z)\mapsto (X,\omega Y,\omega^2Z) $. The canonical model of $ C_6/\gamma $ is the image of $ C_6 $ by the map
\[
(X,Y,Z)\mapsto (X^3,Y^3,Z^3,XYZ)\,;
\]
its equations are
$x^2+y^2+z^2=t^3-xyz=0$. Projecting onto the conic $x^2+y^2+z^2=0$ realizes $ C_6/\gamma $ as the cyclic triple covering $v^3=u(u^4-1)$ of $\P^1$; thus
$C_6/\gamma $ is not isomorphic to $C_6/\alpha $ or $C_6/\beta $.

\item
The quotient by an element of order 3 of $\mathfrak{S}_3$ acting freely, say $ \delta: (X,Y,Z)\mapsto (Y,Z,X) $. The canonical model of $ C_6/\delta $ is the image of $ C_6 $ by the map
\[
(X,Y,Z)\mapsto (X^3+Y^3+Z^3,XYZ,X^2Y+Y^2Z+Z^2X, XY^2+YZ^2+ZX^2).
\]
It is contained in the smooth quadric $ (x+y)^2+5y^2-2zt=0 $, so $ C_6/\delta $ is not isomorphic to any of the 3 previous curves.
\end{itemize}

Thus we have found four non-isomorphic curves of genus $4$ with Jacobian isogenous to $E_\omega^4$. The product of any two of these curves is a $\rho$\nobreakdash-maximal surface.

\begin{cor}\label{S6}
The Fermat sextic surface $ S_6\colon X^6+Y^6+Z^6+T^6=0$ is $\rho$\nobreakdash-maximal.
\end{cor}

\begin{proof}
This follows from Propositions \ref{C6}, \ref{map} and Shioda's trick: there exists a rational dominant map $ \pi: C_6\times C_6 \dasharrow S_6 $, given by
\[
\pi \bigl((X,Y,Z),(X',Y',Z')\bigr)= (XZ',YZ',iX'Z,iY'Z).\qedhere
\]
\end{proof}

\begin{rem}
Since the Fermat plane quartic has maximal correspondences (Example~\ref{ex2}), the same argument
gives the classical fact that the Fermat quartic surface is $\rho$\nobreakdash-maximal. It follows from the explicit formula for $\rho (S_d)$ given in \cite{Ao} that $S_d$ is $\rho$\nobreakdash-maximal (for $d\geq 4$) only for $d=4$ and $6$.
\end{rem}

Again every quotient of the Fermat sextic is $\rho$\nobreakdash-maximal. For instance, the quotient of $ S_6 $ by the automorphism $ (X,Y,Z,T)\mapsto (X,Y,Z,\omega T) $ is the double covering of $ \P^2 $ branched along $ C_6 $: it is a $\rho$\nobreakdash-maximal K3 surface. The quotient of $S_6$ by the involution $(X,Y,Z,T)\mapsto (X,Y,-Z,-T)$ is given in $\P^5$ by the equations
\[
y^2-xz = v^2-uw= x^3+z^3+u^3+w^3=0\,;
\]it is a complete intersection of degrees $(2,2,3)$, with 12 ordinary nodes.
Other quotients have $p_g $ equal to $2$, $3$, $4$ or $6$.

\section{Quotients of self-products of curves}\label{cub}
The method of the previous section may sometimes allow to prove that certain quotients of a product $C\times C$ have maximal Picard number.
Since we have very few examples we will refrain from giving a general statement and contend ourselves with one significant example.

Let $C$ be the curve in $\P^4$ defined by
\[
u^2=xy,\quad v^2=x^2-y^2,\quad w^2=x^2+y^2.
\]
It is isomorphic to the modular curve $X(8)$ \cite{FS}. Let $\Gamma\subset \mathrm{PGL(5,\C)}$ be the subgroup of diagonal elements changing an even number of signs of $u,v,w$; $\Gamma$ is isomorphic to $(\Z/2)^2$ and acts freely on $C$.

\skpt
\begin{prop}
\begin{enumeratea}
\item
$JC$ is isogenous to $E_i^3\times E^2_{\sqrt{-2}}$, where $E_\alpha =\C/\Z[\alpha ]$ for $\alpha =i$ or ${\sqrt{-2}}$.

\item
The surface $(C\times C)/\Gamma$ is $\rho$\nobreakdash-maximal.
\end{enumeratea}
\end{prop}

\skpt
\begin{proof}
\begin{enumeratea}
\item
The form $\Omega:=(xdy-ydx)/uvw$ generates $H^0(C,K_C(-1))$, and is $\Gamma$-invariant; thus multiplication by $\Omega $ induces a $\Gamma$-equivariant isomorphism
\[
H^0(\P^4,\O_{\P^4}(1))\iso H^0(C,K_C).
\]
Let $V$ and $L$ be the subspaces of $H^0(C,K_C)$ corresponding to $\lr{u,v,w}$ and $\lr{x,y}$. The projection $(u,v,w,x,y)\mapsto (u,v,w)$
maps $C$ onto the quartic curve \hbox{$F\colon 4u^4+v^4-w^4=0$}; the induced map $f:C\to F$
identifies $F$ with the quotient of~$C$ by the involution $(u,v,w,x,y)\mapsto(u,v,w,-x,-y)$, and we have $f^*H^0(F,K_F)=V$.

The quotient curve $H:=C/\Gamma$ is the genus 2 curve $z^2=t(t^4-1)$ \cite{Bt}. The pull-back of $H^0(H,K_H)$ is the subspace invariant under $\Gamma$, that is $L$. Thus $JC$ is isogenous to $JF\times JH$. From examples 1 and 2 of \S \ref{prod} we conclude that $JC$ is isogenous to $E_i^3\times E^2_{\sqrt{-2}}$.

\item
We have $\Gamma$-equivariant isomorphisms
\begin{align*}
H^{1,1}(C\times C)&=H^2(C,\C)\oplus H^2(C,\C)\oplus (H^{1,0}\boxtimes H^{0,1})\oplus (H^{0,1}\boxtimes H^{1,0})\\
&=\C^2\oplus \End(H^0(C,K_C))^{{\scriptscriptstyle \oplus}2}
\end{align*}
(where $\Gamma$ acts trivially on $\C^2$),
hence
\[
H^{1,1}((C\times C)/\Gamma)=\C^2\oplus \End_\Gamma(H^0(C,K_C))^{{\scriptscriptstyle \oplus}2}.
\]

As a $\Gamma$-module we have $H^0(C,K_C)=L\oplus V$, where $\Gamma$ acts trivially on $L$ and $V$ is the sum of the 3 nontrivial one-dimensional representations
of $\Gamma$. Thus\[
\End_\Gamma(H^0(C,K_C)) = \mathbb{M}_2(\C)\times \C^3.
\]
Similarly we have $\NS((C\times C)/\Gamma)\otimes \Q=\Q^2 \oplus (\End_\Gamma(JC)\otimes \Q)$ and
\[
\End_\Gamma(JC)\otimes \Q = (\End (JH) \otimes \Q) \,\times \, (\End_\Gamma(JF)\otimes \Q)^3 =\mathbb{M}_2(\Q(\sqrt{-2}))\times \Q(i)^3,
\]
hence the result.\qedhere
\end{enumeratea}
\end{proof}

\begin{cor}[\cite{ST}]\label{ST}
Let $\Sigma \subset \P^6$ be the \emph{surface of cuboids}, defined by
\[
t^2= x^2+y^2+z^2, \quad u^2= y^2+z^2, \quad v^2=x^2+z^2, \quad w^2= x^2+y^2.
\]
$\Sigma $ has $48$ ordinary nodes; its minimal desingularization $S$ is $\rho$\nobreakdash-maximal.\end{cor}
Indeed $\Sigma $ is a quotient of $(C\times C)/\Gamma$ \cite{Bt}.\qed

\smallskip	
(The result has been obtained first in \cite{ST} with a very different method.)
	
\section{Other examples}
\subsection{Elliptic modular surfaces}\label{mod}
Let $\Gamma$ be a finite index subgroup of $\mathrm{SL}_2(\Z)$ such that $-I\notin \Gamma$. The group $\mathrm{SL}_2(\Z)$ acts on the Poincar� upper half-plane $\mathbb{H}$; let $\Delta_\Gamma$ be the compactification of the Riemann surface
$\mathbb{H}/\Gamma$.
The universal elliptic curve over $\mathbb{H}$ descends to
$\mathbb{H}/\Gamma$, and extends to a smooth projective surface $B_\Gamma$ over $\Delta_\Gamma$, the \emph{elliptic modular surface} attached to $\Gamma$.
In \cite{S0} Shioda proves that $B_\Gamma$ \emph{is $\rho$\nobreakdash-maximal}.\footnote{ I am indebted to I.\,Dolgachev and B.\,Totaro for pointing out this reference.}

Now take $\Gamma=\Gamma(5)$, the kernel of the reduction map $\mathrm{SL}_2(\Z)\to \mathrm{SL}_2(\Z/5)$. In \cite{Li}
Livne constructed a $\Z/5$-covering $X\to B_{\Gamma(5)}$, branched along the sum of the 25 5-torsion sections of $B_{\Gamma(5)}$. The surface $X$ satisfies $c_1^2=3c_2\,(=225)$, hence it is a ball quotient and therefore rigid. By analyzing
the action of $\Z/5$ on $H^{1,1}(X)$ Livne shows that $H^{1,1}(X)$ is not defined over $\Q$, hence $X$ \emph{is not $\rho$\nobreakdash-maximal}. This seems to be the only known example
of a surface which cannot be deformed to a $\rho$\nobreakdash-maximal surface.
	
\subsection{Surfaces with $p_g=K^2=1$}

The minimal surfaces with $p_g=K^2=1$ have been studied by Catanese \cite{C} and Todorov \cite{T1}. Their canonical model
is a complete intersection of type $(6,6)$ in the weighted projective space $\P(1,2,2,3,3)$. The moduli space $\mathcal{M}$ is smooth of dimension 18.
\begin{prop}
The $\rho$\nobreakdash-maximal surfaces are dense in $\mathcal{M}$.
\end{prop}

\begin{proof}
We can replace $\mathcal{M}$ by the Zariski open subset $\mathcal{M}_{a}$ parametrizing surfaces with ample canonical bundle. Let $S\in\mathcal{M}_{a}$, and let
$f:\mathcal{S}\to (B,\mathrm{o})$ be a local versal deformation of $S$, so that $S\cong\mathcal{S}_{\mathrm{o}}$.
Let $L$ be the lattice $H^2(S,\Z)$, and $k\in L$ the class of~$K_S$. We may assume that $B$ is simply connected and fix an isomorphism of local systems $R^2f_*(\Z)\iso L_B$, compatible with the cup-product and mapping the canonical class $[K_{\mathcal{S}/B}]$ onto $k$. This induces for each $b\in B$ an isometry \hbox{$\varphi_b: H^2(\mathcal{S}_b,\C)\iso L_{\C}$}, which maps $H^{2,0}(\mathcal{S}_b)$ onto a line in $L_{\C}$; the corresponding point $\wp(b)$ of $\P(L_{\C})$ is the period of $\mathcal{S}_b$. It belongs to the complex manifold
\[
\Omega:= \big\{[x]\in \P(L_{\C})\mid x^2=0,\ x\cdot k=0,\ x\cdot\bar{x}>0 \big\}.
\]
Associating to $x\in\Omega $ the real 2-plane $P_x:=\lr{\mathrm{Re}(x), \mathrm{Im}(x)}\subset L_{\R}$ defines an isomorphism of $\Omega $ onto the Grassmannian of positive oriented 2-planes in $L_{\R}$.

The key point is that
the image of the \emph{period map} $\wp: B\to \Omega $ is open \cite{C}.
Thus we can find $b$ arbitrarily close to $\mathrm{o}$ such that the 2-plane $P_b$ is defined over $\Q$,
hence $H^{2,0}(\mathcal{S}_b)\oplus H^{0,2}(\mathcal{S}_b)=P_b\otimes_{\R}\C$ is defined over $\Q$.
\end{proof}

\begin{rem}
The proof applies to all surfaces with $p_g=1$ for which the image of the period map is open (for instance to K3 surfaces); unfortunately this seems to be a rather exceptional situation.
\end{rem}

\subsection{Todorov surfaces}
In \cite{T2} Todorov constructed a series of regular surfaces with $p_g=1$, $ 2\leq K^2\leq 8$, which provide counter-examples to the Torelli theorem. The construction is as follows:
let $ K \subset \P^3$ be a Kummer surface. We choose $ k $ double points of $ K $ in general position (this can be done with $ 0\leq k\leq 6$), and a general quadric $Q\subset \P^3 $ passing through these $ k $ points. The \emph{Todorov surface} $ S $ is the double covering of $ K $ branched along $ K\cap Q $ and the remaining $ 16-k $ double points. It is a minimal surface of general type with $ p_g=1 $, $ K^2= 8-k $, $q=0$. If moreover we choose $ K $ $\rho$\nobreakdash-maximal (that is, $ K=E^2/\{\pm 1\}$, where $ E $ is an elliptic curve with complex multiplication), then $ S $ is $\rho$\nobreakdash-maximal by Proposition \ref{map}\eqref{mapb}.

Note that by varying the quadric $ Q $ we get a continuous, non-constant family of $\rho$\nobreakdash-maximal surfaces.

\subsection{Double covers}\label{P}

In \cite{P} Persson constructs $\rho$\nobreakdash-maximal double covers of certain rational surfaces by allowing the branch curve to acquire some simple singularities (see also \cite{BE}). He applies this method to find $\rho$\nobreakdash-maximal surfaces in the following families:
\begin{itemize}
\item
Horikawa surfaces, that is, surfaces on the ``Noether line'' $K^2=2p_g-4$, for $p_g\not\equiv -1$ $(\mathrm{mod.}\ 6)$;

\item
Regular elliptic surfaces;

\item
Double coverings of $\P^2$.
\end{itemize}

In the latter case the double plane admits (many) rational singularities; it is unknown whether there exists a $\rho$\nobreakdash-maximal surface $S$ which is a double covering of $\P^2$ branched along a smooth curve of even degree $\geq 8$.

\subsection{Hypersurfaces and complete intersections}

Probably the most natural families to look at are smooth surfaces in $\P^3$, or more generally complete intersections.
Here we may ask for a smooth surface $S$, or for the minimal resolution of a surface with rational double points (or even any surface deformation equivalent to a complete intersection of given type). Here are the examples that we know of:
\begin{itemize}
\item
The quintic surface $x^3yz+y^3zt+z^3tx+t^3xy=0$ has four $A_9$ singularities; its minimal resolution is $\rho$\nobreakdash-maximal \cite{Sc}. It is not yet known whether there exists a
smooth $\rho$\nobreakdash-maximal quintic surface.

\item
The Fermat sextic is $\rho$\nobreakdash-maximal (\S \ref{prod}, Corollary \ref{S6}).

\item
The complete intersection $y^2-xz = v^2-uw= x^3+z^3+u^3+w^3=0$ of type $(2,2,3)$ in $\P^5$ has 12 nodes; its minimal desingularization is $\rho$\nobreakdash-maximal (end of \S \ref{prod}).

\item
The surface of cuboids is a complete intersection of type $(2,2,2,2)$ in $\P^6$ with $48$ nodes; its minimal desingularization is $\rho$\nobreakdash-maximal (\S \ref{cub}, Corollary \ref{ST}).
\end{itemize}
	
\section{The complex torus associated to a \texorpdfstring{$\rho$}{rho}-maximal variety}
For a $\rho$\nobreakdash-maximal variety $X$,
let $T_X$ be the $\Z$-module $H^2(X,\Z)/\NS(X)$. We have a decomposition
\[
T_X\otimes \C= H^{2,0}\oplus H^{0,2}
\]
defining a weight 1 Hodge structure on $T_X$, hence a complex torus $\mathcal{T}:=H^{0,2}/p_2(T_X)$, where $p_2:T_X\otimes \C\to H^{2,0}$ is the second projection. Via the isomorphism $H^{0,2}=\nobreak H^2(X,\O_X)$, $\mathcal{T}_X$ is identified with the cokernel of the natural map $H^2(X,\Z)\to H^2(X,\O_X)$.

The exponential exact sequence gives rise to an exact sequence
\[
0\to \NS(X)\to H^2(X,\Z)\to H^2(X,\O_X)\to H^2(X,\O_X^*)\qfl{\partial} H^3(X,\Z),
\]
hence to a short exact sequence
\[
0\to \mathcal{T}_X\to H^2(X,\O_X^*)\qfl{\partial} H^3(X,\Z),
\]
so that $\mathcal{T}_X$ appears as the ``continuous part'' of the group $H^2(X,\O_X^*)$.
	
\begin{ex}
Consider the elliptic modular surface $B_\Gamma$ of Section \ref{mod}. The space $H^0(B_\Gamma,K_{B_\Gamma})$ can be identified with the space of cusp forms of weight 3 for~$\Gamma$; then the torus $\mathcal{T}_{B_\Gamma}$ is the complex torus associated to this space by Shimura (see \cite{S0}).
\end{ex}
\begin{ex}
Let $X=C\times C'$, with $JC$ isogenous to $E^g$ and $JC'$ to $E^{g'}$ (Proposition~\ref{cc'}). The torus $\mathcal{T}_X$ is the cokernel of the map
\[
i\otimes i':H^1(C,\Z)\otimes H^1(C',\Z)\to H^1(C,\O_C)\otimes H^1(C',\O_{C'}),
\]
where $i$ and $i'$ are the embeddings
\[
H^1(C,\Z)\mono H^1(C,\O_C)\quad\text{and}\quad H^1(C',\Z)\mono H^1(C',\O_{C'}).
\]

We want to compute $\mathcal{T}_X$ up to isogeny, so we may replace the left hand side by a finite index sublattice. Thus, writing $E=\C/\Gamma$,
we may identify $i$ with the diagonal embedding $\Gamma^g\mono \C^g$, and similarly
for $i'$; therefore $i\otimes i'$ is the diagonal embedding of $(\Gamma\otimes \Gamma)^{gg'}$ in $\C^{gg'}$. Put $\Gamma=\Z+\Z\tau $;
the image $\Gamma'$ of $\Gamma\otimes \Gamma$ in $\C$ is spanned by $1,\tau,\tau^2$; since $E$ has complex multiplication, $\tau $ is a quadratic number, hence $\Gamma$ has finite index in $\Gamma'$. Finally we obtain that $\mathcal{T}_X$\emph{is isogenous to} $E^{gg'}$.

\end{ex}

For the surface $X=(C\times C)/\Gamma$ studied in \S \ref{cub} an analogous argument shows that~$\mathcal{T}_X$ is isogenous to $A=E_i^4\times E_{\sqrt{-2}}^3$. This is still an abelian variety of type CM, in the sense that $\End(A)\otimes \Q$ contains an �tale $\Q$-algebra of maximal dimension $2\dim(A)$. There seems to be no reason why this should hold in general. However it is true in the special case
$h^{2,0}=1$ (e.g. for holomorphic symplectic manifolds):

\begin{prop}
If $h^{2,0}(X)=1$, the torus $\mathcal{T}_X$ is an elliptic curve with complex multiplication.
\end{prop}

\begin{proof}
Let $T'_X$ be the pull back of $H^{2,0}+H^{0,2}$ in $H^2(X,\Z)$;
then $p_2(T'_X)$ is a sublattice of finite index in $p_2(T_X)$.
Choosing an ample class $h\in H^2(X,\Z)$ defines a
quadratic form on $H^2(X,\Z)$ which is positive definite on $T'_X$.
Replacing again $T'_X$ by a finite index sublattice we may assume that it admits an orthogonal basis $(e,f)$ with $e^2=a$, $f^2=b$. Then $H^{2,0}$ and $H^{0,2}$ are the two
isotropic lines of $T'_X\otimes \C$; they are spanned by the vectors $\omega =e+\tau f$ and $\bar{\omega}=e-\tau f$, with $\tau^2=-a/b$. We have $e=\frac{1}{2} (\omega +\bar{\omega}) $ and $f=\frac{1}{2\tau} (\omega -\bar{\omega}) $; therefore multiplication by $\frac{1}{2\tau} \bar{\omega}$ induces an isomorphism of $\C/(\Z+\Z \tau)$ onto $H^{0,2}/p_2(T'_X)$, hence $\mathcal{T}_X$ is isogenous to $\C/(\Z+\Z \tau)$ and
\[
\End(\mathcal{T}_X)\otimes \Q=\Q(\tau)=\Q\Big(\sqrt{-\mathrm{disc}(T'_X)}\,\Big).\qedhere
\]
\end{proof}

\section{Higher codimension cycles}

A natural generalization of the question considered here is to look for varieties~$X$ for which the group $ H^{2p}(X,\Z)_{\mathrm{alg}}$ of algebraic classes in $H^{2p}(X,\Z)$ has maximal rank~$h^{p,p}$. Very few nontrivial cases seem to be known. The following is essentially due to Shioda:
\begin{prop}
Let $F^n_d$ be the Fermat hypersurface of degree $d$ and even dimension $n=2\nu $. For $d=3,4$, the group $ H^{n}(F^n_d,\Z)_{\mathrm{alg}}$ has maximal rank $h^{\nu,\nu}$.
\end{prop}

\begin{proof}
According to \cite{S1} we have
\[
\rk H^{n}(F^n_3,\Z)_{\mathrm{alg}}=1+\frac{n!}{(\nu)!^2} \quad \text{and}
\quad \rk H^{n}(F^n_4,\Z)_{\mathrm{alg}}= \sum_{k=0}^{k=\nu +1} \dfrac{(n+2)!}{(k!)^2(n+2-2k)!}\, \cdot
\]
On the other hand, let $R^n_d:= \C[X_0,\ldots,X_{n+1}]/(X_0^{d-1},\ldots, X_{n+1}^{d-1})$ be the Jacobian ring of $F^n_d$; Griffiths theory \cite{G} provides an isomorphism of the primitive cohomology $H^{\nu,\nu}(F^n_d)_{\mathrm{o}}$ with the component of degree $(\nu +1)(d-2)$ of $R^n_d$. Since this ring is the tensor product of $(n+2)$ copies of $\C[T]/(T^{d-1})$, its Poincar� series $\sum_k \dim (R^n_d)_kT^k$ is $(1+T+\ldots +T^{d-2})^{n+2}$. Then an elementary computation gives the result.
\end{proof}

In the particular case of cubic fourfolds we have more examples:
\begin{prop}
Let $F$ be a cubic form in $3$ variables, such that the curve $F(x,y,z)=0$ in $\P^2$ is an elliptic curve with complex multiplication; let $X$ be the cubic fourfold defined by
$F(x,y,z)+F(u,v,w)=0$ in $\P^5$. The group $ H^4(X,\Z)_{\mathrm{alg}}$ has maximal rank $h^{2,2}(X)$.
\end{prop}

\begin{proof}
Let $u$ be the automorphism of $X$ defined by
\[
u(x,y,z;u,v,w)=(x,y,z;\omega u,\omega v,\omega w).
\]
We observe that $u$ acts trivially on the (one-dimensional) space $H^{3,1}(X)$. Indeed Griffiths theory \cite{G} provides a canonical isomorphism
\[
\mathrm{Res}: H^0(\P^5,K_{\P^5}(2X))\iso H^{3,1}(X)\,;
\]
the space $H^0(\P^5,K_{\P^5}(2X))$ is generated by the meromorphic form $\sfrac{\Omega}{G^2}$, with
\begin{align*}
\Omega &=xdy\wedge dz\wedge du \wedge dv \wedge dw - y dx \wedge dz\wedge du \wedge dv \wedge dw + \cdots,\\
G&=F(x,y,z)+F(u,v,w).
\end{align*}
The automorphism $u$ acts trivially on this form, and therefore on $H^{3,1}(X)$.

Let $F$ be the variety of lines contained in $X$. We recall from \cite{BD} that $F$ is a holomorphic symplectic fourfold, and that there is a natural isomorphism of Hodge structures $\alpha:H^4(X,\Z)\iso H^2(F,\Z)$. Therefore the automorphism $u_F$ of $F$ induced by $u$ is symplectic.
Let us describe its fixed locus.

The fixed locus of $u$ in $X$ is the union of the plane cubics $E$ given by $ x=y=z=0$ and $E'$ given by $ u=v=w=0$. A line in $X$ preserved by $u$ must have (at least) two~fixed points, hence must meet both $E$ and $E'$; conversely, any line joining a point of $E$ to a point of $E'$ is contained in $X$, and preserved by $u$. This identifies the fixed locus $A$ of $u_F$ to the abelian surface $E\times E'$. Since $u_F$ is symplectic $A$ is a symplectic submanifold, that is, the restriction map $H^{2,0}(F)\to H^{2,0}(A)$ is an isomorphism. By our hypothesis $A$ is $\rho$\nobreakdash-maximal, so $F$ is $\rho$\nobreakdash-maximal by Proposition~\ref{map}. Since $\alpha $ maps $H^4(X,\Z)_{\mathrm{alg}}$ onto $\NS(F)$ this implies the Proposition.
\end{proof}

\backmatter
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