The Tate Conjecture for Powers of Ordinary Cubic Fourfolds Over Finite Fields
Yuri G. Zarhin
Recently N. Levin proved the Tate conjecture for ordinary cubic fourfolds over finite fields. In this paper we prove the Tate conjecture for self-products of ordinary cubic fourfolds. Our proof is based on properties of so called polynomials of K3 type introduced by the author about a dozen years ago.
math0304014.tex


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\begin{document}
\title[Powers of ordinary cubic fourfolds]{The Tate Conjecture for Powers of Ordinary Cubic Fourfolds Over Finite Fields}
\author {Yuri G. Zarhin}
\address{The Pennsylvania State University, Department of Mathematics\\
\noindent 218 McAllister Building, University Park, PA 16802, USA}
%and
%\noindent Institute for Mathematical Problems in Biology \\
%\noindent Russian Academy of Sciences,  Pushchino, Moscow Region, 142292,
%Russia}
\email{zarhin@@math.psu.edu}
\date{}
\thanks{Partially supported by the NSF}
\begin{abstract}
Recently N. Levin proved the Tate conjecture for ordinary cubic
fourfolds over finite fields. In this paper we prove the Tate
conjecture for self-products of ordinary cubic fourfolds. Our
proof is based on properties of so called polynomials of K3 type
introduced by the author about a dozen years ago.
\end{abstract}
\maketitle


\section{Introduction.}
Let $X$ be a smooth projective variety over a finite field $k$ of
characteristic $p$. We write $X_a$ for $X \times k(a)$ where
$k(a)$ is an algebraic closure of $k$. For each non-negative
integer $m$ and each rational prime $l$ different from $p$ let us
consider the $2m$th twisted $\ell$-adic cohomology group
$H^{2m}(X_a,\Ql)(m)$ of $Y_a$. The Galois group $G(k)$ of $k$ acts
on  $H^{2m}(X_a,\Ql)(m)$ in a natural way. In \cite{T1} Tate
conjectured that the subspace fixed under the Galois action is
spanned by cohomology classes of codimension $m$ algebraic cycles
on $X$. The famous conjecture of Serre and Grothendieck
~\cite{T1,T2,Mz,S2} asserts that the action of $G(K)$ on
$H^{2m}(X_a,\Ql)(m)$ is semisimple, i.e., the Frobenius
automorphism acts on $H^{2m}(X_a,\Ql)(m)$ as a semisimple linear
operator.

 The Tate conjecture is known to be true in certain cases, e.g., Fermat varieties satisfying certain numerical
 conditions ~\cite{T1,Sh,Y}; abelian varieties for $m=1$ \cite{T2}, certain classes of abelian
varieties with arbitrary $m$ ~\cite{Tn,Z2,Z3,LZ}, \cite{Milne2},
``almost all'' K3 surfaces ~\cite{ASD,N,NO}.

Let $Y$ be a {\sl cubic fourfold}, i.e., a smooth projective
hypersurface of degree $3$ in $\PP^5$ defined over $k$. It is
well-known that all odd $\ell$-adic Betti numbers of $Y_a$ do
vanish; it is also known that the second  and sixth Betti numbers
of $Y_a$ are equal to $1$ while its fourth Betti number is $23$
\cite{R}. Let $\Le \in H^2(Y_a,\Ql)(1)$ be the class of a
hyperplane section of $Y$. One may easily check that
$$H^2(Y_a,\Ql)(1)=\Ql\cdot\Le= H^2(Y_a,\Ql)(1)^{G(k)};
 H^6(Y_a,\Ql)(1)=\Ql\cdot\Le^3= H^6(Y_a,\Ql)(1)^{G(k)}.$$
It was proven by Rapoport \cite{R} that the Frobenius automorphism
acts on $H^{4}(X_a,\Ql)(2)$ as a semisimple linear operator.

 Recently N. Levin \cite{L} proved the Tate conjecture when
$X=Y$ is an {\sl ordinary} cubic fourfold. In this case the only
non-trivial case is $m=2$ and the theorem of Levin  asserts that
$H^4(Y_a,\Ql)(2)^{G(k)}$ is generated as $\Ql$-vector subspace by
algebraic classes of codimension $2$ (defined over $k$).

The aim of the present paper is to prove the Tate conjecture for
self-products $X=Y^r$ of an ordinary K3 cubic fourfold $Y$. For
example, if $r=2$ and $X=Y^2$ then the most interesting case is
$m=8$ and we prove, assuming $k$ ``sufficiently large", that
$H^8(X_a,\Ql)(2)^{G(k)}$ is generated as a $\Ql$-vector subspace
by algebraic classes of the following four types: 1) products of
codimension $2$ algebraic cycles on factors $Y$ of $Y^2$; 2) the
classes of graphs of Frobenius and its iterates; 3) the classes of
$Y \times \{\point\}$ and $\{\point \} \times Y$; 4)$\Le \times
\Le^3$ and $\Le^3 \times \Le$.
 The proofs are based on results and ideas of our previous papers ~\cite{Z4,Z5}
 where the Tate conjecture was proven for self-products of ordinary K3 surfaces .

The article is organized as follows. In Section \ref{finite} we
discuss tensor invariants of certain  $\ell$-adic
representations of $G(k)$.
  Section \ref{adic} treats cohomological $\ell$-adic representations.
 The Tate conjecture  is discussed in Section \ref{tate}.
  In Section \ref{prf} we will prove the Tate conjecture for powers of an ordinary
  cubic fourfold.


We write $\Z_+$ for the set $\{0,1,2,\ldots\}$ of non-negative
integers.
 Recall that $G(k)$ is procyclic and has a
canonical generator, namely, the arithmetic Frobenius automorphism
$$ \sigma_k : k(a) \rightarrow k(a), x  \rightarrow x^q$$
where $q$ is the number of elements of $k$. Clearly, $q$ is an integral power of $p$.
Another canonical generator of $G(k)$ is the geometric Frobenius automorphism $\varphi_k = \sigma_k ^{-1}$.




\section{Finite fields and  $\ell$-adic representations}
\label{finite}

Let $k$ be a finite field of characteristic $p$ consisting of $q$
 elements. We keep all notations of the Introduction connected with $k$. If
 $k'\subset k(a)$ is a finite overfield of $k$, then $k'(a)=k(a)$, the Galois group $G(k')$ of $k'$ is
 an open subgroup of finite index $[k':k]$ in $G(k)$ and $\varphi_{k'}=\varphi_k^{[k':k]}$.

 Let $\ell$ be a rational prime different from $p$. We
refer to ~\cite{Mn,T1,Z4} for the definition of the
one-dimensional $\Ql$-vector space $\Ql(1)$ and the corresponding
cyclotomic character
$$\chi_{\ell} : G(k) \rightarrow \Z_{\ell}^\times \subset \Q_{\ell}^{\times} =
\Aut\,\Q_{\ell}(1).$$ This character defines the Galois action on
$\Q_{\ell}(1)$. Notice, that
$$\chi_{\ell}(\sigma_k)=q, \quad \chi_{\ell}(\varphi_k)= q^{-1},.$$
We write $\Q_{\ell}(-1)$ for the  one-dimensional dual vector
space $\Hom_{\Q_{\ell}}(\Q_{\ell}(1), \Q_{\ell})$ with a natural
structure of the dual Galois module defined by the character
$\chi_{\ell}^{-1}$. To each integer $i$ one may attach a certain
one-dimensional $\Q_{\ell}$-vector space $\Q_{\ell}(i)$ with
Galois action defined by $\chi_l^i$. Namely,
$\Q_{\ell}(0)=\Q_{\ell}, \Q_{\ell}(i)=\Q_{\ell}(1)^{\otimes i}$ if
$i$ is positive and $ \Q_{\ell}(i)=\Q_{\ell}(-1)^{\otimes (-i)}$
if $i$ is negative.

For all integers $i,j$ there are natural isomorphisms of Galois modules
$$ \Ql(-i)=\Hom_{\Ql}(\Ql(i),\Ql),\; \Ql(i)\otimes_{\Ql}\Ql(j)=\Ql(i+j).$$

 Let
$$\rho : G(k) \rightarrow \Aut(V)$$
be an $\ell$-adic representation of $G(k)$, i.e., $V$ is a
finite-dimensional $\Q_{\ell}$-vector space and $\rho$ is a
continuous homomorphism \cite{S1}.  Clearly,
$$V^{G(k)}=V^{\rho(\varphi_k)}.$$
To each integer $i$ one may attach the \mbox{\rm   twisted}
$\ell$-adic representation
$$\rho[i] : G(k) \rightarrow \Aut(V(i))$$
where
$$V(i)=V\otimes_{\Ql}\Ql(i),\;
 \rho[i](\sigma)(v\otimes a)=\chi_l^i(\sigma)((\sigma v)\otimes a).$$

For example,
$$\rho[i](\sigma_k)(v\otimes a)=q^i(\rho(\sigma_k)(v)\otimes a),\;
\rho[i](\varphi_k)(v\otimes a)=q^{-i}(\rho(\varphi_k)(v)\otimes
a).$$ We have $V(0)=V, \Ql(i)(j)=\Ql(i+j).$ If
$\rho^*:G(K)\rightarrow \Aut(V^*) $ is the dual $\ell$-adic
representation then there are natural isomorphisms of Galois
modules $(V(i))^*=V^*(-i)$.

\begin{rem}
\label{twist}
 Let $\rho':G(k)\rightarrow \Aut(W)$ be (may be, another)
 $\ell$-adic representation of $G(K)$. Then we obtain natural isomorphisms of
 Galois modules
\begin{gather*}
 V(i)\otimes_{\Ql}W(j)=(V\otimes_{\Ql} W)(i+j),
\rho[i]\otimes \rho'[j]=(\rho\otimes\rho')(i+j),\\
\begin{split}
\Hom_{\Ql}(V,W) &= V^*\otimes_{\Q_{\ell}}W=V^*(-i)\otimes_{\Ql}W(i)\\
&= V(i)^*\otimes_{\Ql}W(i)=\Hom_{\Q_l} (V(i),W(i)).
\end{split}
\end{gather*}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% If $V=W, \rho=\rho'$, then we obtain a canonical Galois-equivariant
% isomorphism of $\Q_l$-algebras
% \begin{gather*}
%\End_{\Q_l}(V) \rightarrow \End_{\Ql}(V(i)), u \rightarrow u_{[i]},\\
% u_{[i]}(x\otimes a) = (ux)\otimes a \quad \mbox{\rm for all } x\in
%V, a \in \Ql(i).
%\end{gather*}
%If $u$ is semisimple then one may easily check that $u_{[i]}$ is
%also semisimple. It is also clear that the characteristic
%polynomials of $u$ and $u_{[i]}$ coincide.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{rem}
\begin{ex}
\label{twist1} If $u=\rho(\sigma)$ for some $\sigma\in G(k)$ then
one may easily check that
$$u_{[i]}=\chi_l^{-i}(\sigma)\rho [i](\sigma) \in \End_{\Q_l}(V(i)).$$
In particular, if $u=\rho(\sigma_k)$ then $u_{[i]}=q^{-i}\rho [i](\sigma_k)$. If
$u=\rho(\varphi_k)$ then $u_{[i]}=q^i\rho [i](\varphi_k)$.
\end{ex}

\begin{thm}[Theorem 3.1.4 of \cite{Z5}]
\label{th314}
  Let us assume that the $\ell$-adic
representation $\rho [i]$ is semisimple and consider the
characteristic
 polynomial
$$P_{\rho [i]}(t)=\Det(\id-t\rho [i](\varphi_k),V(i)).$$
Let  $R$ be the set of reciprocal roots of $P_{\rho [i]}(t)$.
Assume that either $1$ is the unique element of $R$, i.e.,
$R=\{1\}$ is the one-element set or
there exists a non-empty subset $B\subset R$ such that:
\begin{enumerate}
\item $B$ consists of multiplicatively independent elements; in particular,
 it does not contain 1
and does not meet $B^{-1}=\{\alpha^{-1}\mid \alpha \in B\};$
\item  Either
 $R$ coincides with the disjoint union of $B$ and $B^{-1}$ or $R$ coincides
 with the  disjoint union of $B$, $B^{-1}$ and the one-element set $\{1\}$.
\end{enumerate}
Then for each even natural number $2n$ all elements of
$(V(i)^{\otimes 2n})^{G(k)} $ can be presented as a linear
combination of  tensor products of $n$ elements of $(V(i)^{\otimes
2})^{G(k)}.$ Each element of $(V(i)^{\otimes (2n+1)})^{G(k)}$ can
be presented as a linear combination of tensor products of an
element of $V(i)^{G(k)}$ and $n$ elements of $(V(i)^{\otimes
2})^{G(k)}.$
\end{thm}


\begin{defn}
 Recall ~\cite[3.2]{Z4} that $\rho$ is called
{\sl semistable} if it enjoys the following property. If
$u=\rho(\sigma) \in \Aut(V)$ for some $\sigma\in G(k)$ and an
eigenvalue $\alpha$ of $u$ is a root of unity, then $\alpha=1.$ In
fact, in order to make sure that $\rho$ is semistable, it suffices
to inspect the eigenvalues only for $u=\rho(\varphi_k)$
~\cite[3.2.1]{Z4}.
\end{defn}


\begin{rem}
\label{ss}
 If $\rho$ is semistable and $k'$ is a finite overfield
of $k$ then the restriction of $\rho$ to $G(k')$ is also a
semistable $\ell$-adic representation of $G(k')$ and the
invariants of $G(k)$ and $G(k')$ coincide, i.e.,
$V^{G(k)}=V^{G(k')}.$

Conversely, for each $\rho$ there exists a positive integer $r$
such that if an eigenvalue $\alpha$ of $\rho(\varphi_k)$ is a root
of unity then $\alpha^r=1$. Now if $k_r\subset k(a)$ is the degree
$r$ extension of $k$ then every eigenvalue $\beta$ of
$\rho(\varphi_{k_r})=\rho(\varphi_k)^r$ that is a root of unity is
equal to $1$. This means that the restriction of $\rho$ to
$G(k_r)$ is semistable.
\end{rem}









\section{$\ell$-adic cohomology}
\label{adic} Let $Y$ be a smooth projective variety over $k$ and
$Y_a=Y\times k(a)$. Let $\ell$ be a rational prime $\neq p$. The
Galois group $G(k)$ acts on $Y_a=Y\times_k k(a)$ through the
second factor. For each non-negative integer $i$ this action
induces by functoriality the Galois action on the $i$th
$\ell$-adic \'etale cohomology group $H^i(Y_a,\Ql)$. We write
$\rho_{Y,i}$ for the corresponding \mbox{\rm   cohomological}
$l$-adic representation \cite{S2}
$$\rho_{Y,i}: G(k)\rightarrow \Aut(H^i(Y_a,\Q_{\ell})).$$
Let $F_{Y/k}:Y \rightarrow Y$ be the Frobenius endomorphism of the
$k$-scheme $Y$ \cite{T1}, \cite{Mn}. It is defined as the identity
map on points, together with the map $f\rightarrow f^q$ in the
structure sheaf. Let $\Frob_Y=F_{Y/k}\times \id_{k(a)}$ be the
corresponding $k(a)$-endomorphism of $Y_a$.
 We write $(\Frob_Y)_i$ for the endomorphism of $H^i(Y_a,\Ql)$ induced by $\Frob_Y$.

 \begin{rem}
 \label{ext}
If
 $k'\subset k(a)$ is a finite overfield of $k$ and
 $Y'=Y\times_k k'$ is the corresponding smooth projective variety over $k'$, then
 $$Y'_a=Y_a,\quad H^i(Y'_a,\Q_{\ell})=H^i(Y_a,\Q_{\ell})$$
 and $\rho_{Y',i}$ coincides with the restriction of $\rho_{Y,i}$
 to $G(k')$. In particular,
 $$\rho_{Y',i}(\varphi_{k'})=\rho_{Y,i}(\varphi_k)^{[k':k]}.$$
\end{rem}

\begin{rem}
\label{twist2}
 It is well-known ~\cite{T1,Mn} that  $$(\Frob_Y)_i=\rho_{Y,i}(\varphi_k).$$
 \end{rem}

{Let us consider the characteristic polynomial
$$P_{Y,i}(t) = \Det(\id - t \rho_{Y,i}(\varphi_k), H^i(Y_a,\Ql)).$$
A famous theorem of Deligne ~\cite{D2,D3} (conjectured by Weil)
asserts that $P_{Y,i}(t)$ lies in $1+t\Z[t]$, does not depend on
the choice of $l$ and all its (complex) reciprocal roots have
absolute values equal to $q^{i/2}$.

Notice, that in the case of cubic fourfolds this result was proven
earlier than the general case by Rapoport \cite{R} (inspired by
ideas of \cite{D1}). His paper also contain the proof of
semisimplicity of the action of Frobenius in the case of cubic
fourfolds. For Abelian varieties the semisimplicity was proven by
Weil (see \cite{Md}).

 Let $i=2m$ be an even non-negative integer. Let us consider the twisted cohomological $l$-adic representation
$$\rho_{Y,2m}[m]: G(k)\rightarrow \Aut(H^{2m}(Y_a,\Ql)(m)).$$
One may easily check (\cite{Z4}, 4.2) that
$$P_{Y,[m]}(t) = \Det(\id - t \rho_{Y,2m}[m](\varphi_k), H^{2m}(Y_a,\Ql)(m))=
P_{Y,2m}(t/q^m).$$
 Now, the theorem of Deligne implies that
 $P_{Y,[m]}(t)$ lies in $1+t\Z[1/q][t]$, does not depend on the choice of $l$
 and all its (complex) reciprocal roots have absolute values equal to 1.
 In other words, $P_{Y,[m]}(t)$ is a $q$-admissible polynomial in the sense
 of \cite{Z4}. In particular, if $\alpha$ is a (reciprocal) root of $P_{Y,[m]}(t)$
then $\alpha^{-1}$ is also one.

\begin{rem}
\label{extss} Clearly, there exists a positive integer $r$ such
for all roots $\alpha$ of $P_{Y,[m]}(t)$ we have $\alpha^r=1$ if
$\alpha$ is a root of unity. It follows from Remarks \ref{ss} and
\ref{ext} that if $k'$ is the degree $r$ extension $k_r$ of $k$
and $Y'=Y\times_k k'$ then $\rho_{Y',2m}[m]$ is semistable for all
$\ell \ne p$.
\end{rem}

 Let $d=\dim(Y)$. Let us consider the K\"unneth decomposition
$$H^{2d}(Y_a \times Y_a,\Ql)(d)=\oplus_{i=0}^{i=2d}H^{2d-i}(Y_a,\Ql)(d)
\otimes H^i(Y_a,\Ql)$$
of the $d$-twisted middle $l$-adic cohomology group of $Y_a \times Y_a=
(Y \times Y)_a$. Clearly, each element $c \in H^{2d}(Y_a \times Y_a,\Ql)(d)$ can be presented uniquely as a sum
$$c=\oplus_{i=0}^{i=2d} c_i \quad \mbox{\rm with }
 c_i\in H^{2d-i}(Y_a,\Ql)(d) \otimes H^i(Y_a,\Ql).$$
Notice that the K\"unneth decomposition is $G(k)$-equivariant. In particular,
$c \in H^{2d}(Y_a \times Y_a,\Ql)(d)$ is a $G(k)$-invariant if and only if all
$c_i$ are $G(k)$-invariant.

%We write $\Psi_i$ for the intersection pairing
%$$\Psi_i:H^{2d-i}(Y_a,\Ql)(d) \times H^i(Y_a,\Ql) \rightarrow \Ql$$
%which is well-known to be non-degenerate and $G(k)$-invariant
%(Poincar\`e duality \cite{Mn}).

%Let us put
%$$V_i= H^i(Y_a,\Ql) , W_i=H^{2d-i}(Y_a,\Ql)(d).$$
%Then we obtain a non-degenerate $G(k)$-invariant bilinear pairing
%$$\Psi_i : W_i \times V_i \rightarrow \Ql.$$

\begin{rem}
 Let $c \in H^{2d}(Y_a \times Y_a,\Ql)(d)$ be an
 {\sl algebraic cohomology class}, i.e., a linear combination of cohomology classes
 of closed irreducible codimension $d$ subvarieties on $Y_a \times Y_a$ \cite{T1}.
 It follows from results of \cite{KM} that all $c_i$ are also {\sl algebraic cohomology classes}.
\end{rem}

 Let $\mathfrak u$ be a $k(a)$-endomorphism of $Y(a)$. We write
$\Graph_{\mathfrak u}$ for the graph of $\mathfrak u$; it is a
$d$-dimensional irreducible closed subvariety of $Y_a \times Y_a$.
We write $\cl(\Graph_{\mathfrak u})$ for its
 $\ell$-adic cohomology class: it is an element of $H^{2d}(Y_a \times Y_a,\Ql)(d)$.
By functoriality, $\mathfrak u$ induces an endomorphism of
$H^i(Y_a,\Ql)$
 which will be denoted by
$${\mathfrak u}_i \in \End(H^i(Y_a,\Ql) ) =\End(V_i).$$
%It is well-known ~\cite[Ch.6, Lemma 12.2]{Mn}, that
%$${\frak u}_i=\mu_{\Psi_i}( (\cl(\Graph_{\frak u})_i). $$

\begin{ex} If ${\mathfrak u}=\id$ then its graph is the diagonal $\Delta$
and therefore $\id_i$
%$=\mu_{\Psi_i}( (\cl(\Delta))_i)$
is the identity endomorphism of $H^i(Y_a,\Ql)$.
\end{ex}

\begin{ex} If ${\mathfrak u}=\Frob_Y$ then according to \ref{twist2}
$${\mathfrak u}_i=(\Frob_Y)_i = \rho_{Y,i}(\varphi_k).$$
For each positive $j$ the $j$th power $\Frob_Y^j$ of $\Frob_Y$ is defined.
 As usual, if $j=0$, we put $\Frob_Y^j=\id$. Let us put
$${\mathfrak fr}_i =(\cl(\Graph_{\Frob_Y})_i).$$
It is known [7] that ${\mathfrak fr}_i$ can be presented as a
linear combination of $\cl(\Graph_{\Frob_Y^j})$ with rational
coefficients ($j\in\Z_+$). Notice, that it is well-known
~\cite{T1,Mn} that all $\cl((\Graph_{\Frob_Y})_i)$ are
 $G(k)$-invariants.
\end{ex}

\begin{rem} If $d=\dim(Y)=2m$ is even then there is the canonical
 isomorphism
$$H^d(Y_a,\Ql)(d) \otimes H^d(Y_a,\Ql)=
H^{2m}(Y_a,\Ql)(m) \otimes H^{2m}(Y_a,\Ql))(m).$$
%In addition, the corresponding bilinear form
%$$(\Psi_d)_{[m]} : H^{2m}(Y_a,\Ql)(m) \otimes H^{2m}(Y_a,\Ql))(m)
%\rightarrow \Ql $$
%is symmetric, because $d=2m$ is even.
\end{rem}

\begin{rem} If $d=\dim(Y)=2m$ is even then one may easily deduce from
\ref{twist}, \ref{twist1} and \ref{twist2}
 that
$q^m\rho_{Y,2m}[m](\varphi_k) = ((\Frob_Y)_i)_{[m]}$, i.e.,
$$g:=\rho_{Y,2m}[m](\varphi_k) = q^{-m}((\Frob_Y)_i)_{[m]}.$$
%Since $u \rightarrow u_{[m]}$ is an isomorphism of certain algebras (3.1.3),
%$$((\Frob_Y^j)_i)_{[m]}=((\Frob_Y)_i^j)_{[m]}=((\Frob_Y)_i)_{[m]}^j=q^{jm}g^j.$$
\end{rem}

\begin{thm}[Theorem 4.4 of \cite{Z5}]
\label{th44}
  Assume that \ $d=\dim(Y)=2m$ is even,
 $g=\rho_{Y,2m}[m](\varphi_k)$ is a semisimple linear operator and all its
eigenvalues different from 1 are simple.
 Then  the vector subspace of Galois invariants
$${(H^d(Y_a,\Ql)(d)} \otimes {H^d(Y_a,\Ql))^{G(k)}}
 \subset
{H^{2m}(Y_a,\Ql)(m)} \otimes {H^{2m}(Y_a,\Ql))(m)}$$
is generated by
${(H^{2m}(Y_a,\Ql)(m))^{G(k)}} \otimes {(H^{2m}(Y_a,\Ql))(m))^{G(k)}}$
 and all
$(cl(Graph_{\Frob_Y^j}))_d$
with $j\in\Z_+$.
 In particular,
${(H^d(Y_a,\Ql)(d)} \otimes {H^d(Y_a,\Ql))^{G(k)}}$
is contained in the vector subspace
 of
${H^{2d}(Y_a \times Y_a,\Ql)(d)^{G(k)}}$
generated by
 $${(H^{2m}(Y_a,\Ql)(m))^{G(k)}} \otimes {(H^{2m}(Y_a,\Ql))(m))^{G(k)}}$$
 and all
${\cl(\Graph_{\Frob_Y^j})}$ with $j\in\Z_+$.
\end{thm}


\section{The Tate conjecture}
\label{tate}
  We write $\Al^m(Y)$ for the $\Ql$-vector
subspace of $H^{2m}(Y_a,\Ql)(m))$ spanned by the cohomology
classes of all algebraic cycles of codimension $m$ on $Y$. It is
well-known [20] that
$$\Al^m(Y) \subset H^{2m}(Y_a,\Ql)(m))^{G(k)}.$$
Elements of
$H^{2m}(Y_a,\Ql)(m))^{G(k)}$ are called {\sl Tate classes} on  $Y$.

 Tate [19] conjectured that the following assertion holds true.
$$\Al^m(Y) = H^{2m}(Y_a,\Ql)(m))^{G(k)} .\leqno T(Y,m,k,l):$$

\begin{rem} Let $k'$ be a finite algebraic extension of $k$ and $Y':=Y\times_k k'$. It is known
[19] that if the assertion $T(Y',m,k',l)$ holds true then the
assertion  $T(Y,m,k,l)$ also holds true.
\end{rem}



 Recall \cite{Z4}, that $Y$ is called to be of K3 type
in dimension $2m$ if the characteristic polynomial $P_{Y,[m]}(t)$
is of K3 type, i.e. its $p$-adic Newton polygon \cite{K} enjoys
the following properties. There exists a non-zero rational number
$c$ such that the set of slopes is either $\{c,-c\}$ or
$\{c,-c,0\}$. In both cases slopes $c$ and $-c$ must have length
1.

For example, a K3 surface is of K3 type in dimension 2 if and only
if it is ordinary \cite{Z4}. An ordinary Abelian surface is of K3
type in dimension 2.

An ordinary cubic fourfold $Y$ is of K3 type in dimension 4.
Indeed, the Hodge numbers of a cubic fourfold (in dimension $4$)
are as follows \cite{R}.
$$h^{4,0}=h^{0,4}=0, h^{3,1}=h^{1,3}=1, h^{2,2}=21.$$
Since the Hodge polygon of an ordinary cubic fourfold coincides
with the Newton polygon, the $p$-adic Newton polygon of
$P_{Y,4}(t)$ admits the following description. Its set of slopes
is $\{1,2,3\}$; the length of both slopes $1$ and $3$ is $1$ while
the length of slope $2$ is $21$. This implies easily that the
Newton polygon of $P_{Y,[4]}(t)$ is of K3 type with the set of
slopes $\{-1,0,1\}$. (The length of its slopes $1$ and $-1$ is $1$
while the length of slope $0$ is $21$. )


\begin{rem} One may easily define motives of K3 type. The paper \cite{GY}
contains  examples of motives of K3 type arising from Fermat
varieties.
\end{rem}

\begin{thm}
\label{indep}  Let $d=\dim(Y)=2m$ is even. Assume that
 $\rho_{Y,2m}[m]$ is semistable.
Assume also $Y$ is of K3 type in dimension $2m$. We write $a(m,Y)$
for the multiplicity of $1$ viewed as a root of $P_{Y,[m]}(t)$.
Then
$$P_{Y,[m]}(t)=(1-t)^{a(m,Y)} P_{Y,\tr}(t)$$
where the polynomial $P_{Y,\tr}(t) \in \Q[t]$ enjoys the following
properties:
\begin{enumerate}
\item[(i)]
$P_{Y,\tr}(t)$ is irreducible over $\Q$;
\item[(ii)]
%Taking into account, that all the coefficients of $P_{Y,\tr}(t)$
%are real, we obtain that
The the set $R_{Y,\tr}$ of  reciprocal roots of $P_{Y,\tr}(t)$
enjoys the following properties: if $\alpha \in R_{Y,\tr}$ then
$\alpha^{-1}$(= complex conjugate of $\alpha$) also belongs to
$R_{Y,\tr}$. In addition, $R_{Y,tr}$ does not contain roots of
unity.
\item[(iii)]
One may choose a subset $B \subset R_{Y,\tr}$ such that $B$ does
not meet $B^{-1}:=\{\alpha^{-1}\mid \alpha \in B\}$ and
$R_{Y,\tr}$ is the (disjoint) union of $B$ and $B^{-1}$.
\item[(iv)]
In addition, the set $B$ consists of multiplicatively independent
elements.
\end{enumerate}
\end{thm}

\begin{proof}
The semistability means that none of the roots of $P_{Y,\tr}(t)$
is a root of unity. Taking into account, that all the coefficients
of $P_{Y,\tr}(t)$ are real, we obtain (ii). Now the assertion
(iii) follows readily.

Recall that $P_{Y,[m]}(t)$ is of K3 type. It follows easily that
$P_{Y,\tr}(t)$ is also of K3 type. Recall that none of roots of
$P_{Y,\tr}(t)$ is a root of unity. Now the assertions (i) and (iv)
follow easily from general results about polynomials of K3 type
~\cite[th. 2.4.3 and 2.4.4]{Z4}.
\end{proof}

\begin{thm}[Theorem 5.3.1 of \cite{Z5}]
\label{531}
  Let $d=\dim(Y)=2m$ is even. Assume that
 $\rho_{Y,2m}[m]$ is semistable and semisimple.
If $Y$ is of K3 type in dimension $2m$  then ${(H^d(Y_a,\Ql)(d)}
\otimes {H^d(Y_a,\Ql))^{G(k)}}$ is generated as a vector subspace
of ${H^{2m}(Y_a,\Ql)(m)} \otimes {H^{2m}(Y_a,\Ql))(m)}$
 by
${(H^{2m}(Y_a,\Ql)(m))^{G(k)}} \otimes {(H^{2m}(Y_a,\Ql))(m))^{G(k)}}$ and all
$(\cl(\Graph_{\Frob_Y^j}))_d $ with $j\in \Z_+$. In particular,
${(H^d(Y_a,\Ql)(d)} \otimes {H^d(Y_a,\Ql))^{G(k)}}$ is contained in the vector
 subspace of ${H^{2d}(Y_a \times Y_a,\Ql)(n)^{G(k)}}$ generated by
$${(H^{2m}(Y_a,\Ql)(m))^{G(k)}} \otimes {(H^{2m}(Y_a,\Ql))(m))^{G(k)}}$$
 and all
$\cl(Graph_{Frob_Y^j}) $ with $j\in \Z_+$.
\end{thm}

\begin{cor}
\label{square}
 Let $Y$ be an ordinary cubic fourfold over a finite
field $k$ of characteristic $p$. Then there exists a finite
overfield $k'\supset k$  such that the ordinary cubic fourfold
$Y'= Y \times_k k'$ over $k'$  enjoys the following properties for
all primes $\ell \ne p$:

\begin{enumerate}
\item[(i)]
$\rho_{Y',4}[2]$ is semistable;
\item[(ii)]
${(H^4(Y'_a,\Ql)(2)} \otimes_{\Ql} {H^4(Y'_a,\Ql)(2))^{G(k')}}$ is
contained in the $\Ql$-vector
 subspace of ${H^{8}(Y'_a \times Y'_a,\Ql)(4)^{G(k')}}$ generated by
$\Al^2(Y') \otimes_{\Ql} \Al^2(Y')$ and all
$\cl(Graph_{Frob_Y{'}^j}) $ with $j\in \Z_+$.
\item[(iii)]
Let ${\mathfrak l} \in H^2(Y'_a,\Ql)(1)$ be the class of a
hyperplane section. Let ${\mathfrak a}$ be an effective 0-cycle on
$Y'$.
 (For instance, if $Y'(k')=Y(k')$ is non-empty then one may take as ${\mathfrak a}$ any $k'$-rational point on
 $Y$.)
Then $H^8(Y'_a \times Y'_a,\Ql)(4)^{G(k)}$ is generated as a
$\Ql$-vector subspace by  $\Al^2(Y'_a) \otimes_{\Ql} \Al_2(Y'_a)$,
${\mathfrak l} \otimes {\mathfrak l}^3$, ${\mathfrak l}^3 \otimes
{\mathfrak l}$, $\cl(\Graph_{\Frob_Y^j}) \; (j\in \Z_+)$
 and the classes of $Y' \times {\mathfrak a}$ and ${\mathfrak a} \times Y'$.
 In particular, the Tate conjecture holds true for for $Y'\times Y'$ and therefore also for $Y\times Y$.
 \end{enumerate}
\end{cor}

\begin{proof}
Let us choose a finite overfield $k'$ of $k$ such that all
$\rho_{Y',4}[2]$ are semistable: its existence follows from Remark
\ref{extss}.


It was already mentioned that, thanks to the theorem of Rapoport
\cite{R}, $\rho_{Y',4}[2]$ is semisimple. Since, thanks to the
theorem of Levine \cite{L}, $H^4(Y'_a,\Ql)(2))^{G(k')}=\Al^2(Y')$,
the  assertion (ii) follows from Theorem \ref{531}.

The assertion (iii) follows from (ii) combined with the
Galois-invariance of the K\"unneth decomposition for $H^8(Y'_a
\times Y'_a,\Ql)(2)$ and obvious equalities
$$H^8(Y'a,\Ql)(4)=H^8(Y'a,\Ql)(4)^{G(k')}=\Ql\cdot\cl({\mathfrak
a}),$$
$$H^2(Y'a,\Ql)(1)=H^2(Y'a,\Ql)(1)^{G(k')}=\Ql\cdot{\mathfrak
l},$$
$$H^6(Y'a,\Ql)(3)=H^6(Y'a,\Ql)(3)^{G(k')}=\Ql\cdot{\mathfrak
l}^3$$ (where $\cl({\mathfrak a})$ is the cohomology class of
${\mathfrak a}$).
\end{proof}


\begin{rem}
If $Y$ is an ordinary cubic fourfold over a finite field $k$ then
the already mentioned results of Levin  and Rapoport imply that
$$Al^2(Y)=\Num_2(Y)\otimes\Ql$$
where $\Num_2(Y)$ is the group of numerical equivalence classes of
cycles of codimension $2$ on $Y$. In particular, the rank of
$\Num_2(Y)$ coincides with the multiplicity of $1$ viewed as a
root of $P_{Y,[2]}(t)$ ~\cite[Th. 2.9, pp. 74--75]{T2}. Clearly,
if $\rho_{Y_1,4}[2]$ if semistable then the rank of $\Num_2(Y)$ is
$a(2,Y)=23- \deg(P_{Y,tr}(t))$.
\end{rem}

\begin{cor}
\label{product}
 Let $Y$ and $Z$ be  ordinary cubic fourfolds over a finite
field $k$ of characteristic $p$, enjoying the following
properties:

\begin{enumerate}
\item[(i)]
$\rho_{Y,4}[2]$ and $\rho_{Z,4}[2]$ are semistable;
\item[(ii)]
$a(2,Y)\ne a(2,Z)$.
\end{enumerate}
Let $\Le_Y\in H^2(Y_a,\Ql)(1)$ and $\Le_Z\in H^2(Z_a,\Ql)(1)$ are
classes of hyperplane sections of $Y$ and $Z$ respectively. Let
${\mathfrak a}_Y$ and ${\mathfrak a}_Z$ be effective zero cycles
on $Y$ and $Z$ respectively.

 Let us put $W=Y\times Z$. Then $(H^8(W_a,\Ql)(4))^{G(k)}$
is generated as a $\Ql$-vector subspace by  $\Al^2(Y_a)
\otimes_{\Ql} \Al_2(Z_a)$, $\Le_Y \otimes \Le_Z^3$, $\Le_Y^3
\otimes \Le_Z$
 and the classes of $Y \times {\mathfrak a}_Z$ and ${\mathfrak a}_Y \times Z$.
 In particular, the Tate conjecture holds true for for $Y\times
 Z$.
\end{cor}

 \begin{proof}
 It suffices to check that
 $$(H^4(Y_a,\Ql)(2)\otimes
 H^4(Y_a,\Ql)(2)))^{G(k)}=(H^4(Y_a,\Ql)(2))^{G(k)}\otimes
 (H^4(Y_a,\Ql)(2)))^{G(k)}.$$
 In order to do that it suffices to check that if $\alpha$ is a
 root of $P_{Y,[2]}(t)$ and $\beta$ is a
 root of $P_{Z,[2]}(t)$ then $\alpha\beta=1$ if and only if
 $$\alpha=1,\quad \beta=1.$$
 Let us prove it. Suppose $\alpha \ne 1$ and $\beta \ne 1$.
  Then $\alpha$ and $\beta^{-1}$ are roots
 of  $P_{Y,tr}(t)$ and $P_{Z,tr}$ respectively. But $P_{Y,tr}(t)$ and
 $P_{Z,tr}$ are $\Q$-irreducible polynomials with different
 degrees and therefore cannot have common roots. Hence
 $\alpha\ne\beta^{-1}$, i.e., $\alpha\beta\ne 1$.
 \end{proof}

 \begin{rem}
 Similar arguments  prove the Tate conjecture for
 $Y\times S$ where $S$ is an ordinary K3 surface over $k$ with
 semistable $\rho_{Y,2}[1]$ and $a(2,Y)\ne a(1,S)+1$. (The second
 Betti number of a K3 surface is $22$ while the fourth Betti number
 of a cubic fourfold is $23$.) Also by the same token one may
 prove the Tate conjecture for the product of two ordinary K3
 surfaces with different Picard numbers.
 \end{rem}

\section {Powers of fourfolds}
\label{prf}

\begin{thm}
\label{four}
  Let  $Y$ be a smooth geometrically irreducible $4$-dimensional
projective variety over $k$ such that the first and third Betti
numbers of $Y_a$ are zero and the second Betti number of $Y_a$ is
$1$. Let us assume that $\rho_{Y,4}[2]$
 is semisimple and semistable. Assume, in addition,  that
$Y$ is of K3 type in dimension $4$.
 %the assertion $T(Y ,2,k,l)$ holds true.
 Let $r>1$ be an integer and let us put
$X:=Y^r$. Then each cohomology class in
$(H^{2m}(X_a,\Ql)(m))^{G(k)}$ can be presented as a linear
combination of products of pullbacks of Tate
 classes on $Y$ and $Y^2$ with respect to the projection maps
 $X=Y^r \rightarrow Y, X=Y^r \rightarrow Y^2$.
 In particular, if for some prime $\ell$ the Tate conjecture holds true for $Y$ and $Y^2$
 then it is also true for $X$ with the same $\ell$.
\end{thm}

\begin{cor}
\label{cor1}
  Let $Y$ is an ordinary cubic fourfold over a finite field $k$.
  %Assume that $\rho_{Y,4}[2]$ is semistable.
  Let $r>1$ be an integer and let us put
$X:=Y^r$.
% Then each Tate class on $X$ can be
%presented as a linear combination of products of pullbacks of Tate
%classes on $Y^2$ with respect to the projection maps $X=Y^r
%\rightarrow Y^2$. In particular,
Then the Tate conjecture holds true for $X$.
\end{cor}

\begin{proof}[Proof of Corollary \ref{cor1}]
According to Corollary \ref{square} there exists a finite
overfield $k'\supset k$ such that if $Y'=Y\times k'$ then
$\rho_{Y',4}[2]$ is semistable and the Tate conjecture is true for
$Y'^2$. Recall \cite{L} that the Tate conjecture is valid for
$Y'$. Applying Theorem \ref{four} to $Y'/k'$, we conclude that the
Tate conjecture is valid for $Y'^r$. Since $Y'^r=(Y\times
k')^r=Y^r\times k'$, we conclude that the Tate conjecture is valid
for $Y^r=X$.
\end{proof}

\begin{proof}[Proof of Theorem \ref{four}]
Let $\Le \in H^2(Y_a,\Ql)$ be the class of a hyperplane section of
$Y$. Clearly,
$$H^8(Y'_a,\Ql)(4)=(H^8(Y'_a,\Ql)(4))^{G(k')}=\Ql\cdot\cl({\mathfrak
a}),$$
$$H^2(Y'_a,\Ql)(1)=(H^2(Y'_a,\Ql)(1))^{G(k')}=\Ql\cdot{\mathfrak
l},$$
$$H^6(Y'_a,\Ql)(3)=(H^6(Y'_a,\Ql)(3))^{G(k')}=\Ql\cdot{\mathfrak
l}^3$$ (where $\cl({\mathfrak a}$ is the cohomology class of an
effective $0$-cycle ${\mathfrak a}$) on $Y$. In particular, the
cohomology spaces $H^2(Y'_a,\Ql)(1),
H^6(Y'_a,\Ql)(3),H^8(Y'_a,\Ql)(4)$ consist of Tate classes.




We say that $c \in H^{2m}(X_a,\Ql)(m)$ is a {\bf decomposable}
 cohomology class if it can be presented as a linear combination of products
 of pullbacks of Tate classes on $Y$ and $Y^2$ with respect to the projection
 maps
$$X=Y^r \rightarrow Y, \; X=Y^r \rightarrow Y^2.$$
Clearly, linear combinations and $\cup-$products of decomposable cohomology classes are also decomposable one.

Let $r^{\prime} <r$ be a positive integer, $Y^r\rightarrow
Y^{r^{\prime}}$ any projection map. If $c \in
H^{2m}(Y^{r^{\prime}}_a,\Ql)(m)$ is a decomposable cohomology
class on $Y^{r^{\prime}}_a$ then its pullback is a decomposable
cohomology class in $H^{2m}(Y^r_a,\Ql)(m)=H^{2m}(X_a,\Ql)(m)$. If
$r=1$ or $r=2$ then each Tate class on $X_a=Y^r_a$ is decomposable
by obvious reasons.
    Clearly, in order to prove  Theorem \ref{four}, we have to check that  each Galois-invariant
    cohomology class in $H^{2m}(X_a,\Ql)(m)$ is decomposable.

 Let us look more thoroughly at  the cohomology of
$X_a=Y_a^r$. First, notice that the K\"unneth formula combined
with Poincar\'e duality \cite{Mn} implies that under our
assumptions all odd-dimensional cohomology groups of $X_a$ vanish.
In order to describe explicitly the even-dimensional cohomology
groups of $X_a$ let us fix a non-negative integer $m$ and consider
the set ${\mathcal M}(r,m)$ of maps
$$j: \{1,2 \ldots ,r\} \rightarrow \{0,1,2,3,4\} \quad \mbox{\rm with }
\sum_{i=1}^r j(i)=m.$$
Then the K\"unneth formula for $X_a=Y_a^r$ implies easily that
$$H^{2m}(X_a,\Ql)=
\sum_{j \in {\mathcal M}(r,m)}\otimes_{i=1}^r
H^{2j(i)}(Y_a,\Ql).$$

After the proper twist we obtain a canonical isomorphism of Galois modules
$$H^{2m}(X_a,\Ql)(m)=
\sum_{j \in {\mathcal M}(r,m)}\otimes_{i=1}^r
H^{2j(i)}(Y_a,\Ql)(j(i))$$ compatible with $\cup-$products. In
particular,
$$(H^{2m}(X_a,\Ql)(m))^{G(k)}=\sum_{j \in {\mathcal M}(r,m)} H_j^{G(k)} \quad
\mbox{\rm where }
H_j:=\otimes_{i=1}^r H^{2j(i)}(Y_a,\Ql)(j(i)).$$

 The symmetric group ${\mathbf S}_r$ of permutation in
$r$ letters acts on $X=Y^r$ in a natural way. By functoriality, it
acts on $H^{2m}(X_a,\Ql)(m)$ and this action commutes with Galois
action. Clearly, if
 $s \in {\mathbf S}_r$ and $c \in H^{2m}(X_a,\Ql)(m)$ then the cohomology class $c$ is decomposable if and only
 if $s^*c$ is decomposable.
Notice also that
$$s^*H_j=H_{js^{-1}}\; \forall j\in {\mathcal M}(r,m)$$
with the map $js^{-1}:\{1,2 \ldots ,r\} \rightarrow \{0,1,2,3,4\},
\quad
 js^{-1}(i):=j(s^{-1}(i))$.
Of course, the latter formula defines an obvious action of
${\mathbf S}_r $ on ${\mathcal M}(r,m)$. It follows that
$s^*(H_j^{G(k)})=(H_{js^{-1}})^{G(k)}$  for all $j\in {\mathcal
M}(r,m)$; in particular, all Galois-invariant cohomology classes
in $H_j$ are decomposable if and only if  all Galois-invariant
cohomology classes in $H_{js^{-1}}$ are decomposable. This implies
that in order to prove  Theorem \ref{four}, it suffices to check
that all Galois-invariant cohomology classes in $H_j$ are
decomposable for each {\bf non-decreasing} maps $j:\{1,2 \ldots
,r\} \rightarrow \{0,1,2,3,4\}$ from ${\mathcal M}(r,m)$.


So,  Theorem \ref{four} follows from the following assertion.

\begin{lem}
\label{633}
%Lemma 6.3.3
 Let $j$ be a non-decreasing map
$$j:\{1,2 \ldots ,r\} \rightarrow \{0,1,2,3,4\} \quad
\mbox{\rm with } \sum_{i=1}^r j(i)=m.$$ Let $H_j=\otimes_{i=1}^r
H^{2j(i)}(Y_a,\Ql)(j(i))$ be the correspondent K\"unneth chunk of
\linebreak $H^{2m}(X_a,\Ql)(m)=H^{2m}(Y_a^r,\Ql)(m).$ Then each
Galois-invariant cohomology class in $H_j$ is decomposable.
\end{lem}

\begin{proof}[Proof of Lemma \ref{633}]
We use induction by $r$. We already know that the Lemma is true
for  $r=1$ and $r=2$ . So, we may assume that $r>2$.

{\bf Case 1.} Assume that $j(1)<2$, i.e., $j(1)=0$ or $1$. Let us
consider the the projection map $\phi_1:X=Y^r\rightarrow Y$ on the
first  factor, the projection map $\phi:X=Y^r \rightarrow Y^{r-1}$
onto the product of last $(r-1)$ factors and a
 non-decreasing map
$$j^{\prime}:\{1,2 \ldots ,r-1\} \rightarrow \{0,1,2,3,4\}, \quad
j^{\prime}(i):=j(i+1)
%
%
%
\quad \mbox{\rm with } \sum_{i=1}^{r-1} j^{\prime}(i)=m-j(1).$$
Notice
 that $H^{2j(1)}(Y_a,\Ql)(j(1))$
consists of Tate classes and therefore
$\phi_1^*(H^{2j(1)}(Y_a,\Ql)(j(1))$ consists of decomposable
classes.
  Clearly,
\begin{gather*}
\begin{split} H_j &= H^{2j(1)}(Y_a,\Ql)(j(1))\otimes \;\otimes_{i=2}^r
H^{2j(i)}(Y_a,\Ql)(j(i)) \\
&= H^{2j(1)}(Y_a,\Ql)(j(1)) \otimes
H_{j^{\prime}}=\phi^*(H_{j^{\prime}}),
\end{split} \\
H_j^{G(k)}=H^{2j(1)}(Y_a,\Ql)(j(1))\otimes
(H_{j^{\prime}})^{G(k)}= \phi^*((H_{j^{\prime}})^{G(k)})
\end{gather*}
where $H_{j^{\prime}}=\otimes_{i=1}^{r-1}
H^{2j^{\prime}(i)}(Y_a,\Ql)(j^{\prime}(i))$ is a K\"unneth chunk
of $H^{2m-2j(1)}(Y_a^{r-1},\Ql)(m-j(1))$. By induction assumption,
all cohomology classes in $(H_{j^{\prime}})^{G(k)}$ are
decomposable and, therefore, their pullbacks with respect to
$\phi$ are also decomposable. Recall that
$\phi_1^*(H^{2j(1)}(Y_a,\Ql)(j(1))$ consists of decomposable
classes. This ends the proof of the Lemma in the case of $j(1)<2$.

{\bf Case 2.} Assume that $j(r)>2$, i.e. $j(r)=3$ or $4$. Let us
consider the projection map $\phi:X=Y^r \rightarrow Y^{r-1}$ onto
the product of first $(r-1)$ factors,   the projection map
$\phi_r:X=Y^r\rightarrow Y$ on the last factor and a
 non-decreasing map
$$j^{\prime}:\{1,2 \ldots ,r-1\} \rightarrow \{0,1,2,3,4\}, \quad
j^{\prime}(i):=j(i)
%.$$
%Clearly,
%$$
\quad \mbox{\rm with }
 \sum_{i=1}^{r-1} j^{\prime}(i)=m-j(r).$$
Notice that $H^{2j(r)}(Y_a,\Ql)(j(r))$
%$=H^8(Y_a,\Ql)(4)$
consists of Tate classes and therefore
$\phi_r^*(H^{2j(r)}(Y_a,\Ql)(2j(r)))$ consists of decomposable
classes. This implies that
\begin{gather*}
\begin{split} H_j &= \otimes_{i=1}^{r-1} H^{2j(i)}(Y_a,\Ql)(j(i))\:\otimes
H^{2j(i)}(Y_a,\Ql)(j(r)) \\
&= \phi^*(H_{j^{\prime}})\:\otimes
\phi_r^*(H^{2j(r)}(Y_a,\Ql)(2j(r))),
 \end{split} \\
H_j^{G(k)}= (H_{j^{\prime}})^{G(k)}\:\otimes
H^{2j(r)}(Y_a,\Ql)(4)= \phi^*((H_{j^{\prime}})^{G(k)})\:\otimes
\phi_r^*(H^{2j(r)}(Y_a,\Ql)(j(r)))
\end{gather*}
where $H_{j^{\prime}}=\otimes_{i=1}^{r-1}
H^{2j^{\prime}(i)}(Y_a,\Ql)(j^{\prime}(i))$ is a K\"unneth chunk
of $H^{2m-2j(r)}(Y_a^{r-1},\Ql)(m-j(r))$. By induction assumption,
all cohomology classes in $(H_{j^{\prime}})^{G(k)}$ are
decomposable and, therefore, their pullbacks with respect to
$\phi$ are also decomposable. Recall that
$\phi_r^*(H^{2j(r)}(Y_a,\Ql)(2j(r)))$ consists of decomposable
classes. This ends the proof of
 the Lemma in the case of $j(r)>2$.

{\bf Case 3.} Assume that $j(1)>1$ and $j(r)<3$. Since $j$ is
non-decreasing and may take on only values $0,1,2,3,4$, this
implies that $j(i)=2$ for all $i$ and, therefore,
$$H_j=\otimes_{i=1}^r H^{2j(i)}(Y_a,\Ql)(j(i))=
\otimes_{i=1}^r H^4(Y_a,\Ql)(2).$$ So, we have to prove that all
cohomology classes in
$$H_j^{G(k)}=
(\otimes_{i=1}^r H^4(Y_a,\Ql)(2))^{G(k)}$$ are decomposable.
Notice that semistability of $\rho_{Y,2}[1]$ implies that each
(reciprocal) root of $P_{Y,[1]}(t)$ which is a root of unity must
be equal to 1.  Now, the decomposabilty property of elements of
 $(\otimes_{i=1}^r H^2(Y_a,\Ql)(1))^{G(k)}$
 follows from the combination of Theorem \ref{indep} applied to $Y$ and $m=2$ and Theorem \ref{th314} applied to $V=H^4(Y_a,\Ql)$,
$\rho=\rho_{Y,2}$ and $i=2$.
 \end{proof}
  This ends the proof of Theorem \ref{four}.
  \end{proof}




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\bibitem{Z5} Yu. G. Zarhin, {\em  The Tate Conjecture for Powers of Ordinary K3 surfaces Over Finite
Fields}, J. Algebraic Geometry  {\bf 5} (1996), 151--172.


\end{thebibliography}

\end{document}

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\bibitem{N} N. O. Nygaard, {\em The Tate conjecture for ordinary K3 surfaces
 over finite fields}, Invent. Math. {\bf 74} (1983), 213--237.

\bibitem{NO} N. O. Nygaard and A. Ogus, {\em The Tate conjecture for  K3
 surfaces of finite height}, Ann. of Math. {\bf 122} (1985), 461--507.

%\bibitem{PS} I.I. Piatetski-Shapiro and I.R. Shafarevich, {\em The arithmetic
% of K3 surfaces}, Trudy Mat. Inst. Steklov. {\bf 132}(1973), 44--54;
% English transl., Proc. Steklov Inst. Math. {\bf 132}(1975), 45--57.

\bibitem{R} M. Rapoport, {\em Complement \'a l'article de
P.Deligne} ``La conjecture de Weil pour les surfaces K3", Invent.
Math. {\bf 15} (1972), 227--236.

\bibitem{S1} J.-P. Serre, {\em Abelian} $\ell$-{\em adic representations and
 elliptic curves}, Second Edition, Addison-Wesley, 1989.

\bibitem{S2} J.-P. Serre, {\em Repr\'esentations} $\ell$-{\em adiques},
 Algebraic Number Theory (S. Yanaga, ed.),
 Japan Soc. Promotion  Sci., Tokyo, 1977, 177--193.

\bibitem{Sh} T. Shioda, {\em The Hodge conjecture and the Tate conjecture for
 Fermat varieties}, Proc. Japan Academy {\bf 55}(1979), 111--114.

\bibitem{Tn} S. G. Tankeev, {\em On cycles on Abelian varieties of prime
 dimension over finite or number fields}, Math. USSR Izvestija {\bf 22}(1984), 329--337.

\bibitem{T1} J. Tate, {\em Algebraic cycles and poles of zeta functions},
 Arithmetical Algebraic Geometry (O. F. G. Schilling, ed.), Harper and Row, New York, 1965, 93--110.

\bibitem{T2} J. Tate, {\em Endomorphisms of Abelian varieties over finite
 fields}, Invent. Math. {\bf 2}(1966), 134--144.

\bibitem{T3} J. Tate, {\em Conjectures on algebraic cycles in}
$\ell$-{\em adic cohomology}. In: Motives (U. Jannsen, S. Kleiman,
J.-P. Serre, eds.), Proc. Symp. Pure Math. {\bf 55}, Part 1
(1994), 71--83.

%\bibitem{W} A. Weil, {\em Basic Number Theory}, Springer-Verlag, 1967.

\bibitem{Y} N. Yui, {\em Special values of zeta functions of Fermat varieties
 over finite fields}, Number theory New York Seminar 1989-1990, Springer-Verlag, New York, 1991, 251--275.

%\bibitem{Z1} Yu. G. Zarhin, {\em The Brauer group of an Abelian variety over a
% finite field}, Izv. Akad. Nauk SSSR ser. matem. {\bf 46}(1982) ;
% English transl., Math. USSR Izvestija {\bf 20}(1983), 203--234.

\bibitem{Z2} Yu. G. Zarhin, {\em Abelian varieties of K3 type and}
 $\ell$-{\em adic representations},
 Algebraic Geometry and Analytic Geometry Tokyo 1990,
 ICM-90 Satellite Conference Proceedings, Springer-Verlag, Tokyo, 1991, 231--255.

\bibitem{Z3} Yu. G. Zarhin, {\em Abelian varieties of K3 type},
 S\'eminaire de Th\'eorie des Nombres, Paris 1990-91 (S. David, ed.),
 Progress in Math., vol. 108, , Birkh\"auser, 1993, 263-279.


\bibitem{Z4} Yu. G. Zarhin, {\em Transcendental cycles on ordinary K3 surfaces
 over finite fields}, Duke Math. J., {\bf 72}(1993), 65--83.

\bibitem{Z5} Yu. G. Zarhin, {\em  The Tate Conjecture for Powers of Ordinary K3 surfaces Over Finite
Fields}, J. Algebraic Geometry  {\bf 5} (1996), 151--172.


\end{thebibliography}