To prove (<ref>) and (<ref>) we use [1], Theorem 1.1(iii). (See also the discussion in loc. cit. after Remark 1.3.) Let $X$ be the Gushel–Mukai sixfold over $W$ obtained by taking the fibre at $m$ of the universal family over $\tilde{\scrM}$, and let $\gamma \colon X \to \Grass = \Grass(2,V_5)$ be its morphism to the Grassmannian. Writing $X_0$ for the special fibre of $X$, it follows from Corollary <ref> that $H_{\crys}^\bullet(X_0/W)$ is torsion-free. Similarly, it follows from Proposition <ref> that the analogue of Corollary <ref> is true for the Grassmannian; hence also $H_{\crys}^\bullet(\Grass/W)$ is torsion-free. Because $p\geq 5$ ($p\neq 2$ here suffices) we have

\[
H^6(X_{\Kbar},\ZZ_p) = \gamma^* H^6(\Grass_{\Kbar},\ZZ_p) \oplus \tilde{\motP}_{p,m_K}\, .
\]

Applying [1], Theorem 1.1(iii) twice (once to $X$, once to $\Grass$), we obtain (<ref>).

Because $\tilde\motL_{p,s_K}$ is a direct summand of $\ul\End(\tilde\motH_{p,s_K})$ and $\tilde\motL_{\crys,s_0}$ is a direct summand of $\ul\End(\tilde\motH_{\crys,s_0})$, we have

\[
M\bigl(\tilde{\motL}_{p,s_K}\bigr) \otimes_{\frS} W = \Bigl[M\bigl(\tilde{\motL}_{p,s_K}\bigr) \otimes_{\frS} W[1/p]\Bigr] \cap \Bigl[M\bigl(\ul\End(\tilde{\motH}_{p,s_K})\bigr) \otimes_{\frS} W \Bigr]
\]

\[
\tilde\motL_{\crys,s_0} = \tilde\motL_{\crys,s_0}[1/p] \cap \ul\End\bigl(\tilde\motH_{\crys,s_0}\bigr)\, .
\]

Because $M$ is a tensor functor, it therefore suffices to show that $M\bigl(\tilde{\motH}_{p,s_K}\bigr) \otimes_{\frS} W = \tilde{\motH}_{\crys,s_0}$ as submodules of $M\bigl(\tilde{\motH}_{p,s_K}\bigr) \otimes_{\frS} W[1/p] = D_{\crys}(\tilde{\motH}_{p,s_K} \otimes \QQ_p) = \tilde{\motH}_{\crys,s_0}[1/p]$. Because $\tilde{H}$ is obtained as the cohomology in degree $1$ of a Kuga–Satake abelian variety (which has torsion-free cohomology), this again follows from [1], Theorem 1.1(iii).

The isomorphism $\psi_{\dR}$ over $\tilde{\scrM}_E$ extends to an isomorphism of filtered flat vector bundles

\[
\psi_{\dR} \colon (\tilde{\motP}_{\dR} \otimes \scrO_{\tilde{\scrM}_\Lambda},\nabla,\Fil^\bullet) \isomarrow \tilde{\iota}^*(\tilde{\motL}_{\dR}  \otimes \scrO_{\tilde{\scrS}_{N,\Lambda}},\nabla,\Fil^\bullet)
\]

over $\tilde{\scrM}_\Lambda$.

(Recall that $\Lambda = \scrO_{E,\frp}$.)

The argument is similar to the proof of [28], Proposition 5.11. It suffices to show that $\psi_{\dR}$ extends to an isomorphism on the underlying vector bundles; the compatibility with connections and filtrations then follows from the compatibility over $\tilde{\scrM}_E$. Further, we may work Zariski-locally on $\tilde\scrM$, so it suffices to consider an affine open part $U = \Spec(A) \subset \tilde\scrM$ which is faithfully flat over $\ZZ_{(p)}$ such that the vector bundles $\tilde\motP_{\dR}$ and $\tilde\iota^* \tilde\motL_{\dR}$ are free when restricted to $U$. Once we choose bases, the morphism $\psi_{\dR}$ over $U_E$ is given by an invertible matrix $\Psi$ of size $22 \times 22$ with coefficients in $A\otimes_\QQ E$. We have to show that both $\Psi$ and $\Psi^{-1}$ have coefficients in $A_\Lambda = A \otimes_{\ZZ_{(p)}} \Lambda$.

Let $\varpi \in \Lambda$ be a uniformizer. Suppose some matrix coefficient $a = \Psi_{ij}$ does not lie in $A_\Lambda$. Let $\nu >0$ be the minimal integer such that $\varpi^\nu \cdot a \in A_\Lambda$. Now choose a finite field $k$ that contains the residue field $\kappa$ of $\Lambda$, and a $k$-valued point $m_0 \in U_\Lambda(k)$ that does not lie in the zero locus of $\varpi^\nu \cdot a$. Because $\tilde\scrM$ is smooth over $\ZZ_{(p)}$, we can lift $m_0$ to a point $m \in U_\Lambda(\scrO_\scrE)$, where $\scrE \supset \scrO_\scrE$ is as in <ref>. This gives a contradiction with Lemma <ref>. The same argument shows that $\Psi^{-1}$ has coefficients in $A_\Lambda$.

Assume $p \geq 5$. Then the morphism $\tilde{\iota} \colon \tilde{\scrM} \to \tilde{\scrS}_N$ is smooth.

Because $\tilde{\scrM}$ and $\tilde{\scrS}_N$ are smooth over $\ZZ_{(p)}$, it suffices to show that if $m \in \tilde{\scrM}(k)$ for a field $k$, the induced map on tangent spaces

\begin{equation}
\label{eq:TangentMap}
T_{\tilde{\scrM},m} \to T_{\tilde{\scrS}_N,\tilde\iota(m)}
\end{equation}

is surjective. (Use EGA IV, Theorem (17.11.1).) By [29], Proposition 4.16, the tangent sheaf of $\tilde{\scrS}_N$ over $\ZZ_{(p)}$ can be identified with $\ShHom\bigl(\gr^0 \tilde{\motL}_{\dR}, \gr^{-1} \tilde{\motL}_{\dR}\bigr)$. (The graded quotients are of course taken with respect to the Hodge filtration $\Fil^\bullet$). Let $X$ be the fibre of the universal family $\scrX/\tilde{\scrM}$ at $m$. The tangent map (<ref>) factors as

\[
T_{\tilde{\scrM},m} \to H^1(X,\scrT_X) \to \Hom\bigl(\gr^0 \tilde{\motP}_{\dR,\tilde\iota(m)}, \gr^{-1} \tilde{\motP}_{\dR,\tilde\iota(m)}\bigr)\, ;
\]

here we use Corollary <ref>. The first map is surjective by our choice of $\scrM$ (if $X = \CGr(2,V_5) \cap Q$ then every infinitesimal deformation of $X$ is obtained by varying the quadric $Q$), and the second map is surjective by Proposition <ref>.

§ SPECIAL ENDOMORPHISMS AND THEIR LIFTINGS TO CHARACTERISTIC $0$

The goal of this section is to review the notion of a “special endomorphism” that is introduced by Madapusi Pera in [29], Section 5. In what follows, $T$ is a $\ZZ_{(p)}$-scheme and $A/T$ is the abelian scheme corresponding to a $T$-valued point $t \colon T \to \tilde\scrS_N$. Our goal is to define a subspace $L(A) \subset \End(A)$ of “special endomorphisms”. The idea is that an endomorphism $f$ of $A$ is special if the induced endomorphism of $\motH$ is a section of $\tilde\motL \subset \ul{\End}(\motH)$.

We follow [29], Section 5 rather closely, and we only summarise the main results proven there. There are three topics: (1) the definition of the subspace $L(A) \subset \End(A)$, (2) lifting special endomorphisms to characteristic $0$, and (3) realising Tate classes by special endomorphisms. Note that this entire section only deals with the family of motives $\tilde\motL \subset \ul\End(\motH)$ over the Shimura variety $\tilde\scrS_N$. In the next section we will see how these results lead to a proof of the Tate conjecture for Gushel–Mukai sixfolds.

With notation as above, first assume $T$ has characteristic $0$, i.e., $t$ factors through $\tilde\scrS_{N,\QQ}$. Let $A/T$ be the abelian scheme corresponding to $t$, and let $f \in\End(A)$. For $? \in \{\ell,\dR\}$, let $\motH_{?,t}$ be the corresponding realization of $t^*\motH$, and let $f_? \colon \motH_{?,t} \to \motH_{?,t}$ be the endomorphism induced by $f$, which gives a global section of $\ul\End(\motH_{?,t})$. Define $f$ to be ?-special if $f_?$ is a global section of $\tilde\motL_{?,t} \subset \ul\End(\motH_{?,t})$. Further, if $T = \Spec(\CC)$, we say that $f$ is Hodge-special if its Betti realization is a global section of $\tilde\motL_{\infty,t} \subset \ul\End(\motH_{\infty,t})$.

One easily shows ([29], Lemma 5.4) that being $?$-special is independent of which $? \in \{\ell,\dR\}$ we choose and is also equivalent to the condition that the restriction of $f$ to any $\CC$-valued point of $T$ is Hodge-special. We say that $f$ is special if these equivalent conditions are satisfied.

Specialness is invariant under deformations: if $T$ is connected and an endomorphism $f$ is special when restricted to some fibre of $A/T$ then $f$ is special.

Next we consider the case when $p$ is locally nilpotent on $T$. If $k$ is a perfect field of characteristic $p$ and $s \in T(k)$, the restriction $f_s$ of $f$ to the fibre of $A$ at $s$ induces a global section of $\ul\End(\motH_{\crys,s})$, and we say that $f$ is special at $s$ if this section lies in $\tilde\motL_{\crys,s} \subset \ul\End(\motH_{\crys,s})$. One then shows ([29], Lemma 5.9) that, for connected $T$, if $f$ is special at some $s$, it is special at every $s$. Now define $f$ to be special if on every connected component of $T$ it is pointwise special.

With this definition, it is not a priori clear if specialness implies $\ell$-specialness for primes $\ell \neq p$ (which is defined in the same way as above). This is true (see Corollary <ref><ref> below), but it requires some work to prove this.

As a bridge between characteristic $0$ and characteristic $p$, we have the following result.

Let $R$ be a dvr of mixed characteristic $(0,p)$ with perfect residue field, and let $A/R$ be the abelian scheme corresponding to an $R$-valued point of $\tilde\scrS_N$. If $f \in \End(A)$ then the induced endomorphism $f_0$ of the special fibre of $A$ is special if and only if the endomorphism $f_\eta$ of the generic fibre of $A$ is special.

Without loss of generality we may assume $R$ to be complete. The assertion then readily follows by considering the Berthelot–Ogus isomorphism between the de Rham cohomology of the generic fibre and the crystalline cohomology of the special fibre tensored with the fraction field of $R$. Alternatively, it follows from the comparison isomorphism of $p$-adic Hodge theory; cf. the proof of [29], Lemma 5.13.

Crucial for the proof of the Tate conjecture is a result about deformations of special endomorphisms. The situation we consider is that $k$ is a perfect field of characteristic $p$, and $x_0 \in \tilde\scrS_N(k)$. Let $A_0$ be the corresponding abelian variety, and let $\frU$ be the formal completion of $\tilde\scrS_N$ at $x_0$, which is a formal scheme over $W(k)$, non-canonically isomorphic to the formal spectrum of $W(k)[\![t_1,\ldots,t_{20}]\!]$. If $f_0$ is an endomorphism of $A_0$, we denote by $\frU_{f_0} \subset \frU$ the formal subscheme of deformations to which the endomorphism $f_0$ lifts. More precisely, this means that if $R$ is an artinian $W(k)$-algebra with residue field $k$, the $R$-valued points of $\frU_{f_0}$ are the pairs $(x,f)$ consisting of a point $x\in \tilde\scrS_N(R)$ lifting $x_0$ and an endomorphism $f$ of the corresponding abelian variety $A/R$, such that $f$ reduces to $f_0$ modulo the maximal ideal of $R$. Note that, for a given lift $x$, if a lifting $f$ of $f_0$ exists then it is unique. The forgetful morphism $(x,f) \mapsto x$ realizes $\frU_{f_0}$ as a closed formal subscheme of $\frU$.

In the above situation, assume $f_0 \in L(A_0)$, i.e., $f_0$ is a special endomorphism of $A_0$. Then the ideal defining $\frU_{f_0} \subset \frU$ is principal and $\frU_{f_0}$ is flat over $W(k)$.

Let $x_0 \in \tilde\scrS_N(k)$ and $(A_0,f_0)$ be as in <ref>.

*  Assume $f_0$ is special. Then there exists a dvr $R$ which is finite over $W(k)$, an abelian variety $A/R$ and an endomorphism $f$ of $A$, such that $(A,f) \otimes_R R/\frm_R \cong (A_0,f_0) \otimes_k R/\frm_R$.

*  If $f_0$ is a special endomorphism of $A_0$ then it is $\ell$-special for every prime number $\ell \neq p$, i.e., the induced endomorphism $f_\ell \colon \motH_{\ell,x_0} \to \motH_{\ell,x_0}$ corresponds to a global section of $\tilde\motL_{\ell,x_0} \subset \ul\End(\motH_{\ell,x_0})$.

For <ref>, the same as in the proof of [14], Corollary 1.7, gives a finite extension $W(k) \subset R$ and a formal lifting $(\hat{A},\hat{f})$ over $\Spf(R)$. Because the Kuga–Satake abelian variety over $\tilde\scrS_N$ admits a polarization, this formal lifting can be algebraized. Part <ref> follows from <ref> together with Proposition <ref>.

If $k$ is a field and $k \subset \kbar$ is an algebraic closure, we write $k^{\perf} \subset \kbar$ for the perfect closure of $k$. If $V_\ell$ is an $\ell$-adic representation of $\Aut(\kbar/k) = \Gal(\kbar/k^{\perf})$, we write

\[
\Tate(V_\ell) \subset V_\ell
\]

for the subspace of Tate classes; by definition this means that $v \in \Tate(V_\ell)$ if and only if the stabilizer of $v$ in $\Gal(\kbar/k^{\perf})$ is an open subgroup.

Next suppose $k$ is perfect of characteristic $p$ with algebraic closure $\kbar$. We write $K_0(k)$ for the fraction field of $W(k)$ and $\sigma$ for its Frobenius automorphism. Let $(V_p,\phi \colon V_p^{(\sigma)} \isomarrow V_p)$ be an $F$-isocrystal over $k$; here $V_p^{(\sigma)} = V_p \otimes_{K_0(k),\sigma} k$. The space $\Hom_\Isoc\bigl(\mathbf{1},(V_p,\phi)\bigr)$ from the unit isocrystal $\mathbf{1} = (K_0,\sigma)$ to $(V_p,\phi)$ is in bijection with $V_p^{\phi=1} = \bigl\{v\in V_p\bigm| \phi(v\otimes 1) = v\bigr\}$. We could define Tate classes to be the elements in the latter space; but this would not be a natural analogue of the $\ell$-adic Tate classes because $\Hom_\Isoc\bigl(\mathbf{1},(V_p,\phi)\bigr)$ may get bigger after a finite extension of the base field. For this reason we define we define the $\QQ_p$-subspace $\Tate(V_p) \subset V_p \otimes_{K_0(k)} K_0(\kbar)$ by

\[
\Tate(V_p) = \bigcup_{k\subset k^\prime}\; \bigl(V_p \otimes_{K_0(k)} K_0(k^\prime)\bigr)^{(\phi\otimes \sigma) = 1}\, ,
\]

where the union is taken over all finite field extensions $k\subset k^\prime$ inside $\kbar$. (Caution: this is definitely not the same as the space $(V_p \otimes_{K_0(k)} K_0(\kbar))^{(\phi\otimes \sigma)=1}$, which in general is much bigger. The point is that $K_0(\kbar)$ is not the union of all $K_0(k^\prime)$.)

If $(V_p,\phi)$ is an $F$-isocrystal over a finite field $k$ with $p^r$ elements, $\phi^r$ is a $K_0(k)$-linear automorphism of $V_p$ and the natural map

\[
V_p^{\phi=1} \otimes_{\QQ_p} K_0(k) \to \bigl\{v\in V_p\bigm| \phi^r(v) = v\bigr\}
\]

is an isomorphism. It follows from this that the dimension of $\Tate(V_p)$ is equal to the number of eigenvalues of $\phi^r$ (counted with multiplicities) that are roots of unities and that there exists a finite extension $k \subset k^\prime$ such that $\Tate(V_p) = \bigl(V_p \otimes_{K_0(k)} K_0(k^\prime)\bigr)^{(\phi\otimes \sigma) = 1}$.

Let $A_0/k$ be the abelian variety corresponding to a point $x_0 \in \tilde\scrS_N(k)$, where $k$ is a perfect field of characteristic $p$. For $\ell \neq p$ we have the $\ell$-adic realization $\tilde\motL_{\ell,x_0}$; further, we have the $F$-isocrystal $\tilde\motL_{\crys,x_0}[1/p]$. In the next results, the following condition plays a role:

\begin{equation}
\label{eq:ConditionI}
\text{$\dim_{\QQ_\ell}\bigl(\Tate(\tilde\motL_{\ell,x_0})\bigr)$ is independent of~$\ell$ and equals $\dim_{\QQ_p}\bigl(\Tate(\tilde\motL_{\crys,x_0}[1/p])\bigr)$}\, .
\end{equation}

As we shall see in the proof of Theorem <ref>, this condition is satisfied in the situation where we need it.

Let $L(A_{0,\kbar}) \subset \End(A_{0,\kbar})$ be the subspace of special endomorphisms of $A_0 \otimes_k \kbar$. It follows from Corollary <ref><ref> that we have natural maps

\begin{equation}
\label{eq:LtoTatel}
L(A_{0,\kbar}) \otimes \QQ_\ell \to \Tate\bigl(\tilde\motL_{\ell,x_0}\bigr)
\end{equation}

\begin{equation}
\label{eq:LtoTatep}
L(A_{0,\kbar}) \otimes \QQ_p \to \Tate\bigl(\tilde\motL_{\crys,x_0}[1/p]\bigr)\, ,
\end{equation}

which are injective.

A priori, the subspace $L(A_{0,\kbar})$ could be very small. One of the beautiful insights in [28] is that this does not happen, and in fact $L(A_0)$ is as big as one could hope:

Notation and assumptions as above.

*  Assume that $k$ is a finite field and that condition (<ref>) is satsified. Then the maps (<ref>) and (<ref>) are isomorphisms.

*  Assume $k$ is a finitely generated field extension of $\FF_p$ and that there exists a Zariski open $U \subset \tilde\scrS_{N,\FF_p}$ containing $x_0$ such that condition (<ref>) is satsified at all points of $U$ with values in a finite field. Then the maps (<ref>) are isomorphisms for all $\ell \neq p$.

In view of the importance of this result, we outline the proof of part <ref>, following [26], Section 2 and [28], Section 6. Since we assume that condition (<ref>) is satisfied, it suffices to show that (<ref>) is surjective for some choice of $\ell$.

Possibly after replacing $k$ by a finite extension, we may assume that $\End(A_0) = \End(A_{0,\kbar})$ and that the subgroup of $\bar{\QQ}^*$ that is generated by the eigenvalues of Frobenius is torsion-free.

The main ingredients for the argument are the following:

* An algebraic group $I$ over $\QQ$ that acts on the space $L(A_{0,\kbar}) \otimes \QQ$;

* an algebraic group $I_\ell$ over $\QQ_\ell$ that acts on the space $\Tate\bigl(\tilde\motL_{\ell,x_0}\bigr)$;

* a homomorphism $I \otimes \QQ_\ell \hookrightarrow I_\ell$ such that the map $L(A_{0,\kbar}) \otimes \QQ_\ell \to \Tate\bigl(\tilde\motL_{\ell,x_0}\bigr)$ is $I \otimes \QQ_\ell$-equivariant.

We will give further details below. Suppose, then, that the following conditions are satisfied:

*  the representation of $I_\ell$ on $\Tate\bigl(\tilde\motL_{\ell,x_0}\bigr)$ is irreducible;

*  $L(A_{0,\kbar}) \neq 0$;

*  $I \otimes \QQ_\ell \hookrightarrow I_\ell$ is an isomorphism.

Under these assumptions, it is clear that the map (<ref>) is surjective.

Let us now explain how the groups $I$ and $I_\ell$ are defined. We will use notation as in Section <ref>, and as before we denote by $\infty$ the unique complex embedding of $\QQ$. Over the complex Shimura variety $\tilde\scrS_{N,\CC}$ we have a Variation of Hodge Structure $\motH_\infty$ and a local system of algebraic subgroups $\tilde\scrG_\infty \subset \GL(\motH_\infty)$, the fibres of which are isomorphic to the group $\tilde{G}_\QQ = \GSpin(L\otimes \QQ)$. For $\ell \neq p$ we have the $\ell$-adic sheaf $\motH_\ell$ over $\tilde\scrS_N$ (the $\ell$-adic component of $\motH_\et$) and a local system of algebraic groups $\tilde\scrG_\ell \subset \GL(\motH_\ell)$, whose restriction to $\tilde\scrS_{N,\CC}$ is isomorphic, under the comparison isomorphism $i_{\et,\infty}$, to $\tilde\scrG_\infty \otimes \QQ_\ell$. In particular, for $x_0 \in \tilde\scrS_N(k)$ we have $\tilde\scrG_{\ell,x_0} \subset \GL(\motH_{\ell,x_0})$. There is also a $p$-adic analogue for this. A quick way to define it is to pick a lift of $x_0$ to a point $x \in \tilde\scrS_N(W)$ (with $W = W(k)$), and identify $\motH_{\crys,x_0}[1/p]$ with $\motH_{\dR,x} =  \motH_{\infty,x} \otimes_{\QQ} K_0$ (with $K_0 = W[1/p]$); then define $\tilde\scrG_{\crys,x_0} \subset \GL\bigl(\motH_{\crys,x_0}[1/p]\bigr)$ to be $\tilde\scrG_{\infty,x} \otimes K_0$.

In order to define the group $I$, view the group of units in the endomorphism algebra $\End^0(A_{0,\kbar}) = \End(A_{0,\kbar}) \otimes \QQ$ as an algebraic group over $\QQ$. Then define an algebraic subgroup $I \subset \End^0(A_{0,\kbar})^*$ over $\QQ$ by

\[
I = \Biggl\{f \in \End^0(A_{0,\kbar})^* \Biggm|
\vcenter{
\setbox0=\hbox{the induced transformation $f_\ell \in \GL(\motH_{\ell,x_0})$ lies in $\tilde\scrG_{\ell,x_0}$ for all~$\ell \neq p$,}
\copy0
\hbox to\wd0{\hfill and the induced $f_{\crys} \in \GL\bigl(\motH_{\crys,x_0}[1/p]\bigr)$ lies in $\tilde\scrG_{\crys,x_0}$\hfill}
} \Biggr\}\, .
\]

Because the representation of $\tilde\scrG_{\ell,x_0}$ on $\ul\End(\motH_{\ell,x_0})$ maps the subspace $\tilde\motL_{\ell,x_0}$ into itself, and the analogous assertion is true for the crystalline component, the subspace $L(A_{0,\kbar}) \otimes \QQ \subset \End^0(A_{0,\kbar})$ spanned by the special endomorphisms is mapped into itself under left multiplication by elements in $I$.

Next we turn to the group $I_\ell$. Let $q$ be the cardinality of $k$, and let $\gamma_\ell \in \tilde\scrG_{\ell,x_0}(\QQ_\ell)$ be the image of the $q$-power Frobenius element in $\Gal(\kbar/k)$. By Corollary <ref>, $\gamma_\ell$ is semisimple. We define $I_\ell \subset  \tilde\scrG_{\ell,x_0}(\QQ_\ell)$ to be the centralizer of $\gamma_\ell$. Because we have made the base field $k$ large enough, the subspace $\Tate\bigl(\tilde\motL_{\ell,x_0}\bigr) \subset \tilde\motL_{\ell,x_0}$ is simply the eigenspace of the semisimple transformation $\gamma_\ell$ for the eigenvalue $1$. In particular, the action of $I_\ell$ on $\tilde\motL_{\ell,x_0}$ maps the subspace $\Tate\bigl(\tilde\motL_{\ell,x_0}\bigr)$ into itself. As the $q$-power Frobenius endomorphism of $A_0$ is central in $\End(A_0)$, we have a natural inclusion $I \otimes \QQ_\ell \hookrightarrow I_\ell$.

We now consider the conditions <ref>–<ref>. Condition <ref> can be dealt with by a result of Kisin ([26], Corollary 2.1.7), which says that there exists a prime number $\ell \neq p$ (and in fact, infinitely many such) for which $I \otimes \QQ_\ell \hookrightarrow I_\ell$ is an isomorphism. (In loc. cit., this is stated for the identity components, but the argument shows that after making $k$ sufficiently big the groups involved are connected.) Further, an easy lemma ([28], Lemma 6.8) shows that the representation of $I_\ell$ on $\Tate\bigl(\tilde\motL_{\ell,x_0}\bigr)$ is irreducible, provided that $\dim_{\QQ_\ell}\bigl( \Tate(\tilde\motL_{\ell,x_0}) \bigr) \neq 2$. (This last condition is missing in loc. cit., but if the space of Tate classes is $2$-dimensional, the assertion of [28], Lemma 6.8 is wrong.)

At this point we are confronted with two problems: a priori it is not clear why $L(A_{0,\kbar})$ should be nonzero, and the space of Tate classes could be $2$-dimensional. We can bypass these difficulties by redoing the entire construction of the family of motives $\tilde\motL \subset \ul\End(\motH)$ over $\tilde\scrS$ starting with a bigger lattice $L$. (To avoid confusion, note that Theorem <ref> as such is unrelated to Gushel–Mukai varieties.) Thus, one chooses a positive definite integral lattice $L^\prime$ of rank $>2$ with $p \nmid \discr(L^\prime)$, and one redoes the entire construction with $L$ replaced by $L^\sharp = L \perp L^\prime$. Using the notation $?^\sharp$ for objects constructed from $L^\sharp$, we obtain a diagram of (integral canonical models of) Shimura varieties

\[
\begin{tikzcd}[column sep={{{{3em,between origins}}}}]
& \tilde{\scrS}_N^\sharp \ar{rr} && \scrA_{L^\sharp,N}\\
\tilde{\scrS}_N \ar{ur}{i}\ar{rr} && \scrA_{L,N} \ar{ur}
\end{tikzcd}
\]

Over $\tilde{\scrS}_N^\sharp$ we have an abelian scheme $A^\sharp$ which gives rise to a system of realizations $\motH^\sharp$, and we have a submotive $\motL^\sharp \subset \ul\End(\motH^\sharp)$. The pullback of $A^\sharp$ to $\tilde\scrS_N$ (via $i$) is isomorphic to $A \otimes C(L^\prime)$, where of course $A$ is the original (“Kuga–Satake”) abelian scheme over $\tilde\scrS_N$, and where $A \otimes C(L^\prime)$ is Serre's tensor construction (if $r^\prime$ is the rank of the lattice $L^\prime$ this just means that we take the sum of $2^{r^\prime}$ copies of $A$). Correspondingly, $i^*\motH^\sharp \cong \motH \otimes C(L^\prime)$, so that $i^*\bigl(\ul\End(\motH^\sharp)\bigr) \cong \ul\End(\motH) \otimes \End\bigl(C(L^\prime)\bigr)$. In this description, $i^*(\motL^\sharp)$ becomes the submotive

\[
(\motL \otimes \ZZ) \oplus (\unitmot \otimes L^\prime)
\]

where in the second term the unit motive $\unitmot$ should be interpreted as the submotive $\unitmot \cdot \id_{\motH} \subset \ul\End(\motH)$ spanned by the identity endomorphism, and where $L^\prime$ is identified with a subspace of $\End\bigl(C(L^\prime)\bigr)$ via left multiplication in the Clifford algebra. We now take sections at our point $x_0 \in \tilde\scrS_N(k)$ (as in <ref>), and we note that the summand $(\unitmot \otimes L^\prime)$ consists of Tate classes. As explained, grace to Kisin's result that allows us to deal with condition <ref>, this puts us in a situation where conditions <ref>–<ref> are satisfied. The conclusion, then, is that the map $L(A^\sharp_{0,\kbar}) \otimes \QQ_\ell \to \Tate\bigl(\tilde\motL^\sharp_{\ell,x_0}\bigr)$ is surjective, and this implies that (<ref>) is surjective.

§ THE TATE CONJECTURE FOR GUSHEL–MUKAI SIXFOLDS IN CHARACTERISTIC $P \GEQ 5$

We can now put everything together. Let $Y/k$ be a GM sixfold as in the statement of Theorem <ref>. Note that, by our definition of the space of Tate classes, $\Tate\bigl(H^{2i}_{\crys}(Y/K_0(k))(i)\bigr)$ is a subspace of $H^{2i}_{\crys}\bigl(Y_{\kbar}/K_0(\kbar)(i)\bigr)$; see Section <ref>.

The first assertion of <ref> has already been proven in Corollary <ref>. For the proof of the remaining assertions, it suffices to consider the case $i=3$.

We start with a general remark. Let $k$ be a field of characteristic $p$, and suppose we have a $k$-valued point $m_0 \in \tilde\scrM(k)$ which maps to $s_0 = \tilde\iota(m_0) \in \tilde\scrS_N(k)$. Let $X_0/k$ be the Gushel–Mukai sixfold corresponding to $m_0$ and $A_0/k$ the Kuga–Satake abelian variety. For $\ell \neq p$ we have an isomorphism $\tilde\motP_{\ell,m_0} \isomarrow \tilde\motL_{\ell,s_0}$, and if $k$ is perfect we have $\tilde\motP_{\crys,m_0} \isomarrow \tilde\motL_{\crys,s_0}$. Further, $H^6\bigl(X_{0,\kbar},\QQ_\ell(3)\bigr) \cong \QQ_\ell^{\oplus 2} \oplus \tilde\motP_{\ell,m_0}$ as Galois representations and if $k$ is perfect then $H^6_{\crys}(X_0/K_0(k))\bigl(3\bigr) \cong \QQ_p^{\oplus 2} \oplus \tilde\motP_{\crys,m_0}$ as $F$-isocrystals over $k$. If $k$ is a finite field then it follows from the Weil conjectures together with Corollary <ref> that the $\QQ_\ell$-dimension of the space $\Tate\bigl(H^6(X_{0,\kbar},\QQ_\ell(3))\bigr)$ is independent of $\ell \neq p$. (This dimension can be read from the zeta function, which does not depend on $\ell$.) Moreover, by [24], Theorem 1, together with Remark <ref>, this dimension equals the $\QQ_p$-dimension of $\Tate\bigl(H^6_{\crys}\bigl(X_0/K_0(k)\bigr)(3)\bigr)$. Because $\tilde\iota \colon \tilde\scrM \to \tilde\scrS_N$ is smooth (Corollary <ref>), we find that the condition in Theorem <ref><ref> is satisfied, and it follows that whenever $k$ is finitely generated over $\FF_p$, the maps $L(A_{0,\kbar}) \otimes \QQ_\ell \to \Tate(\tilde\motL_{\ell,s_0})$ are isomorphisms.

We now prove the second assertion of <ref>. In doing so, we may replace $k$ by a finitely generated extension. Choose $m_0$ as above such that $Y \cong X_0$, and let $\xi_0$ be a Tate class in $\tilde\motP_{\ell,m_0}$. By the above, it suffices to prove that $\xi_0$ is algebraic if under the isomorphism $\tilde\motP_{\ell,m_0} \isomarrow \tilde\motL_{\ell,s_0}$ it maps to the $\ell$-adic realization of a special endomorphism $f_0$ of $A_0$. Assume this is the case. By Corollary <ref><ref> there exists a finite extension of dvr $W(k^{\perf}) \subset R$ and an $R$-valued point $s \in \tilde\scrS_N(R)$ lifting $s_0$ such that $f_0$ lifts to an endomorphism $f$ of the corresponding abelian variety $A/R$. By Corollary <ref>, there exists $m \in \tilde\scrM(R)$ lifting $m_0$ such that $s = \tilde\iota(m)$. Let $X/R$ be the Gushel–Mukai variety corresponding to $m$. Since we may replace $R$ by a finite extension, we may assume that $R$ contains the ring $\hat\scrO_{E,\frp}$ as in Section <ref>. Let $K$ be the fraction field of $R$, let $X_\eta$ and $A_\eta$ be the generic fibres of $X$ and $A$, and let $f_\eta \in L(A_\eta)$ be the restriction of $f$ to the generic fibre. By Proposition <ref><ref> we have an isomorphism $\tilde\motP_{m_\eta} \isomarrow \tilde\motL_{s_\eta}$ in $\Real(K;\QQ)$. By construction, the endomorphism $f_\eta$ defines an absolute Hodge class in $\tilde\motL_{s_\eta}$. Let $\xi_\eta$ be the corresponding absolute Hodge class in $\tilde\motP_{m_\eta}$. Because the Hodge conjecture is true for Gushel–Mukai sixfolds (see [19], Section 5), $\xi_\eta$ is algebraic, i.e., it is the image of a class in $\CH^3(X_\eta)$. By specialization it follows that $\xi_0$ is algebraic, and we are done.

If $k$ is finite, the same argument (now starting with a Tate class in $\tilde\motP_{\crys,m_0}$) gives part <ref> of the theorem.

§ THE TATE CONJECTURE FOR GUSHEL–MUKAI FOURFOLDS IN CHARACTERISTIC $P\GEQ 5$

In this section we prove the Tate conjecture for Gushel–Mukai fourfolds in characteristic $p\geq 5$, by reducing it to the case of Gushel–Mukai sixfolds. To this end, we show that for a GM fourfold $X$ there exists a “partner” $X^\prime$, which is a GM sixfold, such that $X$ and $X^\prime$ have isomorphic Chow motives in middle degree. The construction uses liftings to characteristic $0$ together with a result from our paper [19] (see Theorem <ref>).

Let $R$ be a ring which is either a field or a dvr (discrete valuation ring). If $V$ is a free $R$-module of finite rank and $V^\vee = \Hom(V,R)$ is its dual, $(\wedge^i V)^\vee$ is canonically isomorphic to $\wedge^i (V^\vee)$ (see for instance [4], Chap. 3,  11, no. 5). The notation $\wedge^i V^\vee$ will therefore not cause any ambiguity.

For symmetric powers one has to be more careful. We shall assume that $2$ is invertible in $R$, and we only need to consider symmetric squares. In this situation, we can use that $\Sym^2(V)^\vee$ is canonically isomorphic to the subspace $\ST^2(V^\vee) \subset (V^\vee)^{\otimes 2}$ of symmetrical tensors (i.e., the subspace of $\frS_2$-invariants in $(V^\vee)^{\otimes}$) and that the composition $\ST^2(V^\vee) \hookrightarrow (V^\vee)^{\otimes 2} \twoheadrightarrow \Sym^2(V^\vee)$ is an isomorphism because $2 \in R^*$. We use this to identify $\Sym^2(V)^\vee$ with $\Sym^2(V^\vee)$.

An important tool in the study of GM varieties is a correspondence between GM data sets and Lagrangian data sets, which first appeared in work of O'Grady [33] and Iliev–Manivel  [21], and which was refined by Debarre–Kuznetsov in [9]. We shall work with a version of this correspondence that is adapted to our needs.

Throughout the discussion, $R$ is a ring which is either a field or a dvr, such that $2 \in R^*$.

* A GM datum of dimension $n\in \{3, 4, 5\}$ over $R$ is a tuple $(V_6,V_5,W,q,\epsilon)$, where

* $V_6$ is a free $R$-module of rank $6$,

* $V_5\subset V_6$ is a direct summand of rank $5$,

* $W\subset \wedge^2 V_5$ is a direct summand of rank $(n+5)$,

* $q \colon V_6 \to \Sym^2(W^\vee)$ is an $R$-linear map,

* $\epsilon\colon \det(V_5) \isomarrow R$ is an isomorphism

such that

\begin{equation}
\label{eq:CompatibilityGMdata}
q(v)\bigl(w \cdot w^\prime\bigr) = \epsilon(v\wedge w\wedge w'), \text{ for all $v\in V_5$ and $w, w^\prime \in W$.}
\end{equation}

(Here we use the isomorphism $\Sym^2(W^\vee) \cong \Sym^2(W)^\vee$.)

* A Lagrangian datum of dimension $n\in \{3, 4, 5\}$ over $R$ is a tuple $(V_6,V_5,A,\epsilon)$, where

* $V_6$ is a free $R$-module of rank $6$,

* $V_5\subset V_6$ is a direct summand of rank $5$,

* $\epsilon\colon \det(V_5) \isomarrow R$ is an isomorphism,

* $A\subset \wedge^3 V_6$ is a Lagrangian submodule with respect to the $\det(V_6)$-valued symplectic form $\wedge\colon \wedge^3 V_6\otimes \wedge^3 V_6 \to \wedge^6 V_6 = \det(V_6)$, such that $A\cap \wedge^3 V_5$ is free of rank $5-n$.

Note that $A$ is assumed to be a direct summand of $\wedge^3 V_6$ (necessarily of rank $10$) and that $A\cap \wedge^3 V_5$ is then automatically a direct summand of both $A$ and $\wedge^3 V_5$.

We trust it is clear what it means for two GM data (resp. Lagrangian data) to be isomorphic.

Let $R$ be a ring which is either a field or a dvr, with $2 \in R^*$, and let $n\in \{3,4,5\}$. There is a natural bijection between the set of isomorphism classes of $n$-dimensional GM data and the set of isomorphism classes of $n$-dimensional Lagrangian data.

When $R$ is a field of characteristic $0$, this is a special case of the more elaborate version in [9], Theorem 3.6. See [6], Theorem 2.2, for a proof in a simplified setting that is closer to the version we consider here. The proofs in loc. cit. also work in characteristic $p \neq 2$. For the reader's convenience we recall how the correspondence works.

Given a GM datum $(V_6,V_5,W,q,\epsilon)$, we want to construct a Lagrangian submodule $A \subset \wedge^3V_6$. Choose an element $v_0 \in V_6$ whose class generates $V_6/V_5$ and define

\begin{equation}
\label{eq:ADef}
A := \Ker\Bigl(\wedge^3 V_5 \oplus (v_0\wedge W) \to W^\vee\Bigr)\, ,
\end{equation}

where $\wedge^3 V_5\oplus (v_0\wedge W)$ is viewed as a submodule of $\wedge^3 V_6$ and the map sends $(\xi, v_0\wedge w)$ to the linear functional $w^\prime \mapsto \epsilon(\xi\wedge w^\prime) + q(v_0)\bigl(w\cdot w^\prime\bigr)$. Using the relation (<ref>) one shows that the submodule $A \subset \wedge^3 V_6$ thus obtained is independent of the choice of $v_0$, and that $A$ is isotropic. As the map $\wedge^3 V_5 \to W^\vee$ that is used in (<ref>) is surjective, $A$ is a direct summand of $\wedge^3 V_6$ of rank equal to $\rk_R(\wedge^3 V_5) + \rk_R(W) - \rk_R(W) = 10 = \frac{1}{2} \rk_R(\wedge^3 V_6)$, so $A$ is  Lagrangian. Further, $\rk_R(A\cap \wedge^3 V_5) = \rk_R(\wedge^3 V_5) - \rk_R(W) = 10 - (n+5) = 5-n$.

In the opposite direction, given a Lagrangian datum $(V_6, V_5, A, \epsilon)$, we want to construct $W$ and $q$. Choose a primitive element $0\neq \lambda \in V_6^\vee$ with $\Ker(\lambda) = V_5$. Let $x\mapsto \lambda \iprod x \colon \wedge^\bullet V_6 \to \wedge^{\bullet-1} V_6$ be the contraction (=interior product) defined by $\lambda$. The kernel of this map is $\wedge^\bullet V_5$ and the image is $\wedge^{\bullet -1} V_5 \subset \wedge^{\bullet-1} V_6$; in particular we have a short exact sequence

\begin{equation}
\label{eq:w3w3w2seq}
0 \tto \wedge^3 V_5 \tto \wedge^3 V_6 \xrightarrow{~ \lambda\, \iprod~} \wedge^2 V_5 \tto 0\, .
\end{equation}

\[
W := \Image\bigl(A \hookrightarrow \wedge^3 V_6  \xrightarrow{~  \lambda\, \iprod~} \wedge^2 V_5\bigr)\, ;
\]

then $W$ is a direct summand of $\wedge^2 V_5$ and because $W \cong A/(A\cap \wedge^3 V_5)$ it is free of rank $10 - \rk_R(A\cap \wedge^3 V_5) = n+5$. For $\xi$, $\xi^\prime \in \wedge^3 V_6$ we have $\xi \wedge \xi^\prime = 0$, and because $\lambda\, \iprod$ is a derivation in the graded sense (see [4], Chap. 3,  11, nos. 6–9) this gives

\[
0 = \lambda \iprod (\xi \wedge \xi^\prime) = (\lambda \iprod \xi) \wedge \xi^\prime - \xi \wedge (\lambda \iprod \xi^\prime)\, ,
\]

so $\xi \wedge (\lambda \iprod \xi^\prime) = (\lambda \iprod \xi) \wedge \xi^\prime = \xi^\prime \wedge (\lambda \iprod \xi)$. Therefore, we obtain a well-defined homomorphism $\tilde{q} \colon V_6 \otimes \Sym^2(A) \to R$ given by

\[
\tilde{q}\bigl(v \otimes (\xi\cdot \xi^\prime)\bigr) = \epsilon\Bigl(v \wedge (\lambda \iprod \xi) \wedge (\lambda \iprod \xi^\prime) - \lambda(v) \cdot \xi \wedge (\lambda \iprod \xi^\prime)\Bigr)\, .
\]

As $\lambda \, \iprod$ is zero on $\wedge^3 V_5$, this map descends to a homomorphism $V_6 \otimes \Sym^2(W) \to R$. Let $q \colon V_6 \to \Sym^2(W^\vee)$ be the induced map. If $v \in V_5$ then $\lambda(v) = 0$, so it follows that $q(v)\bigl(w\cdot w^\prime\bigr) = \epsilon(v\wedge w \wedge w^\prime)$ (write $w = \lambda \iprod \xi$ and $w^\prime = \lambda \iprod \xi^\prime$). This shows that $(V_6, V_5, W, q, \epsilon)$ is a  GM datum. Note that $\lambda$ is unique only up to a scalar in $R^\times$, but a routine calculation shows that $W$ and $q$ do not change under such a rescaling.

One checks without difficulty that the above two constructions are inverse to each other.

Let $(V_6,V_5,W,q,\epsilon)$ be an $n$-dimensional GM datum (with $n \in \{3,4,5\}$) over a ring $R$ as in <ref>. We may then define $X \subset \PP(W)$ over $\Spec(R)$ by

\begin{equation}
\label{eq:GMIntersection}
X := \bigcap_{v\in V_6} Q(v)\, ,
\end{equation}

where $Q(v)$ denotes the quadric in $\PP(W)$ defined by $q(v)$. We refer to $X$ as the GM intersection defined by the GM datum.

It follows from (<ref>) that $\bigl(\PP(W) \cap \Grass(2,V_5)\bigr) \subset \PP(W)$ is cut out by the quadrics $Q(v)$ for $v \in V_5$. Therefore, if $v_0 \in V_6$ is an element whose class generates $V_6/V_5$, we have

\[
X = \PP(W) \cap \Grass(2,V_5) \cap Q(v_0)\, .
\]

If $k$ is a field of characteristic $\neq 2$, every GM variety of Mukai type (as in Definition <ref>) can be realised as a GM intersection (<ref>) over a finite extension of $k$.

A GM intersection as above is not necessarily smooth over $R$, but when it is so, $X$ is an $n$-dimensional GM variety of Mukai type over $R$. Moreover, if $R=k$ is a field, the following proposition, due to Debarre and Kuznetsov, gives a useful criterion for when $X$ is smooth over $k$. In characteristic $0$, this is a special case of [9], Theorem 3.16.

Let $(V_6,V_5,W,q,\epsilon)$ be an $n$-dimensional GM datum over a field $k$ with $\charact(k) \neq 2$. Let $(V_6,V_5,A,\epsilon)$ be the corresponding Lagrangian datum over $k$ and $X/k$ the GM intersection as in (<ref>). Then $X$ is smooth of dimension $n$ over $k$ if and only if $A_{\kbar} \subset \wedge^3 V_{6,\kbar}$ contains no decomposable vectors (i.e., $\PP(A_{\kbar}) \cap \Grass(3,V_{6,\kbar}) = \emptyset$).

Of course, $X/k$ is smooth if and only if $X_{\kbar}/\kbar$ is smooth; so we may assume $k = \kbar$. In this situation the same proof as in [6], Section 2, gives the result. (This proof uses [36], Corollary 1.6, which is valid over an arbitrary field of characteristic $\neq 2$.)

Let $K$ be a field of characteristic $0$. In [9] it is explained that a GM variety $X$ over $K$ gives rise to a “GM data set” $\bigl(V_6(X),V_5(X),W(X),L(X),\mu_X,q_X,\epsilon_X \bigr)$, and a corresponding “Lagrangian data set” $\bigl(V_6(X),V_5(X),A(X)\bigr)$. We shall only use this construction over fields of characteristic $0$. We do use that if $X/K$ (still with $\charact(K) =0$) is obtained as a GM intersection (<ref>) given by a GM datum $(V_6,V_5,W,q,\epsilon)$, and if $(V_6,V_5,A,\epsilon)$ is the corresponding Lagrangian datum obtained as described in (the proof of) Proposition <ref>, we have $\bigl(V_6(X),V_5(X),A(X)\bigr) \cong (V_6,V_5,A)$.

A configuration that arises in the study of GM varieties is that an $n$-dimensional GM variety $X$ of Gushel type can be described as a double cover of another nonsingular variety $M_X^\prime \subset \Grass(2,V_5)$ (a linear section of the Grassmannian), with branch locus a GM variety $Y \subset M_X^\prime$ of Mukai type of dimension $n-1$. In this situation, $X$ and $Y$ are said to be “opposite” GM varieties. It follows from the results in Section 3 of [9] that $\bigl(V_6(X),V_5(X),A(X)\bigr) \cong \bigl(V_6(Y),V_5(Y),A(Y)\bigr)$.

We shall make use of the following result of O'Grady.

Let $k$ be an algebraically closed field of characteristic $\neq 2$. Let $(V_6,V_5,A,\epsilon)$ be a Lagrangian datum over $k$, and assume $A$ contains no decomposable vectors. Then there exists a $5$-dimensional subspace $V_5^\prime \subset V_6$ such that $A \cap \wedge^3 V_5^\prime = 0$.

Define $A^\perp \subset \wedge^3 V_6^\vee$ by $A^\perp = \bigl\{\xi \in \wedge^3 V_6^\vee \bigm| \xi(A) = 0\bigr\}$; this is a Lagrangian subspace of $\wedge^3 V_6^\vee$. First we note that $A^\perp$ contains no decomposable vectors. Indeed if $A^\perp$ contains a decomposable vector, we can choose a basis $e_1,\ldots,e_6$ for $V_6$, with dual basis $\check{e}_1,\ldots,\check{e}_6$ for $V_6^\vee$, such that $\check{e}_{123} = \check{e}_1 \wedge \check{e}_2 \wedge \check{e}_3 \in A^\perp$; because $A^\perp$ is Lagrangian this implies that the $\check{e}_{456}$-coefficient of every vector $\alpha \in A^\perp$ is zero; hence the decomposable vector $e_{456} = e_4 \wedge e_5 \wedge e_6$ lies in $A$, contradicting the assumptions.

Let $U \subset V_6$ be a $5$-dimensional subspace and let $[u] \in \PP(V_6^\vee)$ be the corresponding line in $V_6^\vee$. Then $\wedge^3 U \subset \wedge^3 V_6$ is a Lagrangian subspace and

\[
\bigl(\wedge^3 U \bigr)^\perp = u \wedge \bigl(\wedge^2 V_6^\vee\bigr) = F_u\, ,\qquad \text{where}\quad
F_u := \bigl\{\xi \in \wedge^3 V_6^\vee \bigm| \xi \wedge u = 0\bigr\}\, .
\]

\[
A \cap \wedge^3 U = 0 \iff A^\perp + F_u = \wedge^3 V_6^\vee \iff A^\perp \cap F_u = 0\, ,
\]

the proposition is equivalent to the assertion that there exists a $u \in \PP(V_6^\vee)$ such that $A^\perp \cap F_u = 0$. This is proven by O'Grady in [34], Corollary 2.5; the proof that is given there works over an arbitrary algebraically closed field $k$ with $\charact(k) \neq 2$. (The assumption that $k = \kbar$ is needed because the proof uses a Bertini argument.)

If $k$ is a field, we denote by $\CHM(k)$ the category of Chow motives over $k$ with rational coefficients. In what follows, the notation $\CH$ is used for Chow groups with $\QQ$-coefficients.

If $X/k$ is smooth projective we write $\frh(X)$ for its Chow motive. (Let us specify that we use the cohomological version of the theory, i.e., the functor $\frh$ that we use is the contravariant one.) Further, if $M$ is a Chow motive we have Tate twists $M(n) = M \otimes \unitmot(n)$, for $n \in \ZZ$.

Let now $X$ be a GM variety of dimension $n$ over $k$. Then $X$ admits a natural Chow–Künneth decomposition, that is, a decomposition $[\Delta_X] = \sum_{i=0}^{2n}\, \pi^i_X$ of the diagonal class as a sum of mutually orthogonal projectors $\pi^i_X \in \CH^n(X \times_k X)$, such that for any classical Weil cohomology theory $H^*$, the endomorphism of $H^*(X)$ induced by $\pi^i_X$ is the projector onto the summand $H^i(X)$ in degree $i$. These projectors can be given explictly; we here recall the formulas in the two cases that are most relevant for us, namely GM varieties of dimension $4$ or $6$; see [19], Section 7, for more details. In both cases, all odd projectors are zero: $\pi_X^1 = \pi_X^3 = \cdots = \pi_X^{2n-1} = 0$; so it only remains to describe the even projectors.

* Let $X/k$ be a GM fourfold. Let $H = -\frac{1}{2} \cdot [K_X] \in \CH^1(X)$ be the class of an ample generator. Then $\pt = \frac{1}{10} \cdot H^4 \in \CH^4$ is the class of a point. (The given formulas show that these classes make sense over any field, even if for instance $X$ has no $k$-rational point.) The even Chow–Künneth projectors are given by

\begin{align*}
\pi_X^0 &= \pt \times X  & \pi_X^8 &= X \times \pt\\
\pi_X^2 &= \tfrac{1}{10} \cdot H^3 \times H  & \pi_X^6 &= \tfrac{1}{10} \cdot H \times H^3
\end{align*}

\[
\pi_X^4 = [\Delta_X] - \pi_X^0 - \pi_X^2 - \pi_X^6 - \pi_X^8\, .
\]

* Next consider a GM sixfold $X$. For simplicity of exposition, assume the Gushel morphism $\gamma \colon X \to \Grass(2,V_5)$ is defined over $k$. (The projectors we shall describe can in fact be defined over $k$ without this assumption, but we shall not need this.) Let $\scrU_X$ be the Gushel bundle on $X$, i.e., the pullback under $\gamma$ of the tautological rank $2$ bundle on $\Grass(2,V_5)$. Let $\sigma_{i,j} \in \CH^{i+j}\bigl(\Grass(2,V_5)\bigr)$ (for $3\geq i\geq j\geq 0$) be the Schubert classes. Similar to the previous case we have the class $H = -\frac{1}{4} \cdot [K_X] = \gamma^*(\sigma_{1,0}) \in \CH^1(X)$ of an ample generator and the class $\pt = \frac{1}{10} \cdot H^6 = \frac{1}{2}\cdot \gamma^*(\sigma_{3,3}) \in \CH^6(X)$ of a point. Then

\[
e_1 = H^2 \, ,\qquad e_2 = \gamma^*(\sigma_{1,1}) = c_2(\scrU_X)
\]

are in $\CH^2(X)$. The classes $f_1$, $f_2 \in \CH^4(X)$ given by

\[
f_1 = \tfrac{1}{2}\cdot H^4 - H^2 e_2\, ,\qquad f_2 = -H^4 + \tfrac{5}{2}\cdot H^2 e_2
\]

have the property that $\int_X e_i \cdot f_j = \delta_{i,j}$. The even Chow–Künneth projectors of $X$ are then given by

\begin{align*}
\pi_X^0 &= \pt \times X  & \pi_X^{12} &= X \times \pt\\
\pi_X^2 &= \tfrac{1}{10} \cdot H^5 \times H  & \pi_X^{10} &= \tfrac{1}{10} \cdot H \times H^5\\
\pi_X^4 &= f_1 \times e_1 + f_2 \times e_2 & \pi_X^8 &= e_1 \times f_1 + e_2 \times f_2
\end{align*}

\[
\pi_X^6= [\Delta_X] - \pi_X^0 - \pi_X^2 - \pi_X^4  - \pi_X^8 - \pi_X^{10}-\pi_X^{12}\, .
\]

Theorem <ref> for GM fourfolds follows from the case of sixfolds together with the following result. (Note that it suffices to prove Theorem <ref> after replacing the base field by a finite extension.)

Let $k$ be a field of characteristic $p>2$. Let $X/k$ be a GM fourfold. Then there exists a finite field extension $k \subset k^\prime$ and a GM sixfold $X^\prime/k^\prime$ such that

\begin{equation}
\label{eq:h4h6isom}
\frh^4(X_{k^\prime})\bigl(2\bigr) \cong \frh^6(X^\prime)\bigl(3\bigr)
\end{equation}

in $\CHM(k^\prime)$.

The proof of this result will take up the rest of this section.

In our proof of Theorem <ref>, the GM sixfold $X^\prime$ will be constructed in terms of Lagrangian data. To obtain the desired relation (<ref>), we will construct lifts of $X$ and $X^\prime$ to characteristic $0$ and use a result that was proven in our paper [19], which gives the analogue of (<ref>) in characteristic $0$. The result in characteristic $p$ will then be deduced by a specialisation argument.

Before stating the result from [19] that we need, let us first recall the notion of generalised partners, following [27], Definition 3.5. For this, consider GM varieties $X$ and $X^\prime$ over a field $K$ of characteristic $0$, of dimensions $n$ and $n^\prime$, respectively, such that $n$, $n^\prime \in \{3,4,5,6\}$ have the same parity. Then we say that $X$ and $X^\prime$ are generalised partners if there exists an isomorphism $f \colon V_6(X) \isomarrow V_6(X^\prime)$ such that $\wedge^3 f \colon \wedge^3 V_6(X) \isomarrow \wedge^3 V_6(X^\prime)$ restricts to an isomorphism $A(X) \isomarrow A(X^\prime)$.

Let $X$ and $X^\prime$ be generalised partners over an algebraically closed field $K$ of characteristic $0$, of dimensions $n$ and $n^\prime$, respectively. Then there exists an isomorphism

\[
\frh^n(X) \cong \frh^{n^\prime}(X^\prime)\bigl(\tfrac{n^\prime-n}{2}\bigr)
\]

in $\CHM(K)$.

See [19], Section 8. (There is also a version of this result for so-called generalised dual GM varieties, but we shall not need it here.)

Let $k$ be a field of characteristic $p>2$. Let $(V_6,V_5,A,\epsilon)$ be a Lagrangian datum of dimension $n \in \{3,4,5\}$ over $k$. Then there exists a dvr $R$ with residue field $k$ and fraction field $K$ of characteristic $0$, and a Lagrangian datum $(\scrV_6,\scrV_5,\scrA,\epsilon)$ over $R$ whose reduction modulo the maximal ideal $\frm \subset R$ is isomorphic to $(V_6,V_5,A,\epsilon)$.

Let $R$ be a Cohen ring for $k$. Let $\scrV_6 = R^6$ and fix an isomorphism $\scrV_6 \otimes_R k \isomarrow V_6$, via which we identify $\scrV_6 \otimes_R k$ and $V_6$. Choose any $\epsilon \colon \det(\scrV_6) \isomarrow R$ that lifts the given isomorphism $\det(V_6) \isomarrow k$. Choose a lifting of $V_5$ to a direct summand $\scrV_5 \subset \scrV_6$. By assumption, $L = A\cap \wedge^3 V_5 \subset \wedge^3 V_6$ is $(5-n)$-dimensional. Choose a direct summand $\scrL \subset \wedge^3 \scrV_5$ lifting $L$. Let $\scrL^\perp \subset \wedge^3 \scrV_6$ be the orthogonal complement of $\scrL$ with respect to the symplectic form $\psi \colon \wedge^3 \scrV_6 \times \wedge^3 \scrV_6 \to \det(\scrV_6)$. Then $\scrT = \scrL^\perp/\scrL$ is free of rank $10+2n$ over $R$ and $\psi$ induces a symplectic form $\bar\psi$ on $\scrT$. Writing $T = \scrT \otimes_R k = L^\perp/L$, the subspace $\bar{A} = A/L \subset T$ is Lagrangian with respect to the pairing $\bar\psi$. Because the scheme of Lagrangian subspaces in $\scrT$ is smooth over $R$ we can lift $\bar{A}$ to a Lagrangian submodule $\overline{\scrA} \subset \scrT$. Define $\scrA \subset \scrL^\perp \subset \wedge^3 \scrV_6$ to be the inverse image of $\overline{\scrA}$. By construction, the intersection of $\bar{A}$ and the image of $\wedge^3 V_5$ in $T$ is zero; this implies that $\scrA \cap \wedge^3 \scrV_5 = \scrL$, and therefore $(\scrV_6,\scrV_5,\scrA,\epsilon)$ is a Lagrangian datum of dimension $n$ over $R$.

We are now ready to prove Theorem <ref>, thereby completing the proof of Theorem <ref>. There are several steps in the proof where we may need to replace the base field by a finite extension. In each such step, the notation that was already in place will be retained and will from then on refer to the varieties that are obtained by extension of scalars to the bigger field.

We first consider a GM fourfold $X$ of Mukai type. Possibly after replacing $k$ by a finite extension, there exists a $4$-dimensional GM datum over $k$ such that $X$ is isomorphic to the associated GM intersection as in (<ref>). Let $(V_6,V_5,A,\epsilon)$ be the $4$-dimensional Lagrangian datum over $k$ that corresponds to this GM datum. By Proposition <ref>, there exists a $5$-dimensional subspace $V_5^\prime \subset V_6$ such that $A \cap \wedge^3 V_5^\prime = 0$. As in Lemma <ref>, let $(\scrV_6,\scrV_5,\scrA,\epsilon)$ be a lift of $(V_6,V_5,A,\epsilon)$ over a dvr $R$ of mixed characteristic $(0,p)$. Choose an arbitrary lift $\scrV_5^\prime \subset \scrV_6$ of $V_5^\prime$ to a direct summand of $\scrV_6$, and note that $\scrA \cap \wedge^3 \scrV_5^\prime = 0$. By Proposition <ref>, $\PP(A) \cap \Grass(3,V_6) = \emptyset$. If $K$ is the fraction field of $R$ this implies that $\PP(\scrA_K) \cap \Grass(3,\scrV_{6,K}) = \emptyset$; so neither $\scrA_{\kbar} = A_{\kbar}$ nor $\scrA_{\Kbar}$ contains decomposable vectors.

Let $\scrX/R$ be the GM intersection defined by the GM datum that corresponds, via the bijection of Proposition <ref>, with the Lagrangian datum $(\scrV_6,\scrV_5,\scrA,\epsilon)$. By the fiberwise criterion for smoothness together with Proposition <ref> we see that $\scrX$ is smooth over $R$, and of course its special fibre is $X$. We can do exactly the same with $\scrV_5^\prime$ instead of $\scrV_5$. If $(\scrV_6,\scrV_5^\prime,\scrW^\prime,q^\prime,\epsilon)$ is the GM datum over $R$ corresponding with $(\scrV_6,\scrV_5^\prime,\scrA,\epsilon)$, we obtain a GM fivefold

\begin{equation}
\label{eq:scrYprime}
\scrY^\prime = \bigcap_{v \in \scrV_6}\; \scrQ(v) \quad \subset \PP(\scrW)
\end{equation}

over $R$. Let $\scrM^\prime =  \cap_{v \in \scrV_5^\prime}\; \scrQ(v) = \Grass(2,\scrV_5^\prime) \cap \PP(\scrW)$, and let $\scrX^\prime \to \scrM^\prime$ be a double cover branched along $\scrY^\prime$. (There is, in general, not a unique double cover over $R$, though any two such become isomorphic over an extension of $R$. For the argument we may just choose any double cover of $\scrM^\prime$ branched along $\scrY^\prime$.)

By construction, the generic fibres $\scrX_K$ and $\scrX^\prime_K$ are generalised partners. (To pass from $\scrY^\prime_K$ to $\scrX^\prime_K$, use what was explained in Section <ref>.) It follows from Theorem <ref> that there is a finite extension $K \subset K^\prime$ and an isomorphism $\alpha \colon \frh^4(\scrX_{K^\prime})\bigl(2\bigr) \isomarrow \frh^6(\scrX^\prime_{K^\prime})\bigl(3\bigr)$ in $\CHM(K^\prime)$. Let $R^\prime$ be the normalisation of $R$ in $K^\prime$ (which is again a dvr), and let $k^\prime$ be the residue field of $R^\prime$. By specialisation of the cycles that give the isomorphism $\alpha$ and its inverse we obtain the desired isomorphism $\frh^4(X_{k^\prime})\bigl(2\bigr) \isomarrow \frh^6(\scrX^\prime_{k^\prime})\bigl(3\bigr)$ in $\CHM(k^\prime)$.

The argument in case of a GM fourfold $X/k$ of Gushel type is similar but involves one extra step. In this case there exists, possible after replacing $k$ by a finite extension, a $3$-dimensional GM datum $(V_6,V_5,W,q,\epsilon)$ over $k$ with associated GM intersection

\[
Y = \bigcap_{v \in V_6}\; Q(v) \quad \subset \PP(W)\, ,
\]

such that $X$ can be realised as a double cover of $\cap_{v \in V_5}\; Q(v) = \PP(W) \cap \Grass(2,V_5)$ branched along $Y$. Let $(V_6,V_5,A,\epsilon)$ be the corresponding $3$-dimensional Lagrangian datum over $k$. Now we proceed as in the previous case: by Proposition <ref> there exists a $5$-dimensional subspace $V_5^\prime \subset V_6$ such that $A \cap \wedge^3 V_5^\prime = 0$, using Lemma <ref> we can lift $(\scrV_6,\scrV_5,\scrA,\epsilon)$ to a Lagrangian datum $(V_6,V_5,A,\epsilon)$ over a dvr $R$ of mixed characteristic $(0,p)$, and we choose an arbitrary lift $\scrV_5^\prime \subset \scrV_6$ of $V_5^\prime$ to a direct summand of $\scrV_6$.

Let $(\scrV_6,\scrV_5,\scrW,q,\epsilon)$ be the GM datum over $R$ that corresponds to $(\scrV_6,\scrV_5,\scrA,\epsilon)$. Let $\scrY/R$ be the lift of $Y$ given by this GM datum, and let $\scrX$ be the lift of $X$ to a GM fourfold over $R$ by taking a double cover of $\PP(\scrW) \cap \Grass(2,\scrV_5)$ with branch divisor $\scrY$.

Similarly, let $(\scrV_6,\scrV_5^\prime,\scrW^\prime,q^\prime,\epsilon)$ be the GM datum over $R$ that corresponds to $(\scrV_6,\scrV_5^\prime,\scrA,\epsilon)$, and define the GM fivefold $\scrY^\prime$ over $R$ as in (<ref>). Let $\scrX^\prime$ be a double cover of $\Grass(2,\scrV_5)$ branched along $\scrY^\prime$.

The generic fibres $\scrX_K$ and $\scrX^\prime_K$ are again generalised partners, because $\scrY_K$ is a generalised partner of $\scrY_K^\prime$ and $\scrX_K$ (resp. $\scrX^\prime_K$) is oppositie to $\scrY_K$ (resp. $\scrY^\prime_K$). (See Section <ref> for this notion.) After passing to a finite extension $K \subset K^\prime$, Theorem <ref> gives us an isomorphism $\frh^4(\scrX_{K^\prime})\bigl(2\bigr) \cong \frh^6(\scrX^\prime_{K^\prime})\bigl(3\bigr)$, and by specialisation we then obtain the desired isomorphism $\frh^4(X_{k^\prime})\bigl(2\bigr) \cong \frh^6(\scrX^\prime_{k^\prime})\bigl(3\bigr)$.

This completes the proof of Theorem <ref> and thereby also the proof of Theorem <ref>.

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Radboud University Nijmegen, IMAPP, Nijmegen, The Netherlands

Radboud University Nijmegen, IMAPP, Nijmegen, The Netherlands