Proof. This is [Reference Fulton22, Example 17.4.10].

Proof. This is [Reference Voisin59, Proposition 4.8]. Let us briefly recall the geometric construction of the rational map $\psi$. Let $(l,l') \in F \times F$ be a generic point, so that the two lines $l, l'$ on $Y$ span a three-dimensional linear space in $\mathbb {P}^{5}$. For any point $x \in l$, the plane $\langle x, l'\rangle$ intersects the smooth cubic surface $S = \langle l, l' \rangle \cap Y$ along the union of $l'$ and a residual conic $Q'_x$, which passes through $x$. Then, $\psi (l,l') \in Z$ is the point corresponding to the two-dimensional linear system of twisted cubics on $S$ linearly equivalent to the rational curve $l \cup _x Q'_x$ (this linear system actually contains the $\mathbb {P}^{1}$ of curves $\{ l \cup _x Q'_x \mid x \in l\}$).

Proof. The MCK decomposition is constructed in [Reference Vial54, Theorem 1] (an alternative construction is given in [Reference Negut, Oberdieck and Yin43]). The compatibility with Voisin's filtration $S_\ast$ is stated in [Reference Voisin59, End of Section 4.1]. The formulae for the generators follow from [Reference Voisin59, Theorem 2.5], where it is proven that the filtration $S_\ast$ coincides with the filtration $N_\ast$ defined in [Reference Voisin59, Proposition 2.2].

(The $m=2$ case also follows from [Reference Shen and Vial49].)

Proof. (i) Let us just treat the case $j=2$ (the other cases are only notationally more complicated). Let us write $A^{\ast }(S^{m})^{\mathfrak S_m}$ for the $\mathfrak S_m$-invariant part, where the symmetric group on $m$ elements $\mathfrak S_m$ acts on $S^{m}$ by permutation of the factors. We need to show that the map induced by intersection product

\[ \Bigl(A^{2}_{(2)}(S^{m})^{\mathfrak S_m}\Bigr)^{{\otimes} 2} \to A^{4}_{(4)}(S^{m})^{\mathfrak S_m} \]

is surjective. By definition, an element $a$ on the right-hand side is a symmetric cycle of the form

\[ \bigl( \pi^{2}_S\times \pi^{2}_S\times \pi^{0}_S\times\cdots\times \pi^{0}_S + \hbox{permutations}\bigr){}_\ast (d) \]

(where $d\in A^{4}(S^{m})$). That is,

\[ a = \sum_{\sigma\in\mathfrak S_m} \sigma_\ast (a_1\times a_2\times S\times S\times \cdots\times S)\quad \hbox{in}\ A^{4}(S^{m}), \]

where $a_1, a_2\in A^{2}(S)$ are degree zero $0$-cycles.

Let us define

\begin{align*} b_1& := \sum_{\sigma\in\mathfrak S_m} \sigma_\ast (a_1\times S\times S\times \cdots\times S) \in A^{2}_{(2)}(S^{m})^{\mathfrak S_m},\\ b_2& := \sum_{\sigma\in\mathfrak S_m} \sigma_\ast (S\times a_2\times S\times \cdots\times S) \in A^{2}_{(2)}(S^{m})^{\mathfrak S_m}.\\ \end{align*}

We have that

\[ b_1\cdot b_2= \sum_{\sigma\in\mathfrak S_m} \sum_{\sigma^{\prime}\in\mathfrak S_m} (\sigma\circ\sigma^{\prime})_\ast (a_1\times a_2\times S\times S\times \cdots\times S) \]

is a non-zero multiple of $a$, and so we have proven surjectivity.

(ii) Again, one reduces to the Cartesian product $S^{m}$. That is, one considers

\[ \begin{array}[c]{ccc} & & X\\ & & \downarrow{\scriptstyle f}\\ S^{m} & \xrightarrow{p} & S^{(m)}=S^{m}/\mathfrak S_m\\ \end{array} \]

where $f$ is the Hilbert–Chow morphism and $p$ is the quotient morphism.

There is a commutative diagram

\[ \begin{array}[c]{ccc} A^{j}_{(j)}(X) & \xrightarrow{\cdot h^{2m-j}} & A^{2m}_{(j)}(X)\\ \downarrow {\scriptstyle f_\ast} & & \downarrow {\scriptstyle f_\ast}\\ A^{j}(S^{(m)}) & \xrightarrow{\cdot f_\ast (h)^{2m-j}} & A^{2m}_{}(S^{(m)})\\ \downarrow {\scriptstyle p^{{\ast}}} & & \downarrow {\scriptstyle p^{{\ast}}}\\ A^{j}_{(j)}(S^{m})^{\mathfrak S_m} & \xrightarrow{\cdot p^{{\ast}} f_\ast (h)^{2m-j}} & A^{2m}_{(j)}(S^{m})^{\mathfrak S_m}\\ \end{array} \]

in which composition of two vertical arrows on the left respectively on the right is an isomorphism (NB: the commutativity of the upper square follows from the projection formula).

Thus, it will suffice to prove that for a big $\mathfrak S_m$-invariant divisor $h$ on $S^{m}$, there are isomorphisms

\[{\cdot} h^{2m-j}\colon A^{j}_{(j)}(S^{m})^{\mathfrak S_m} \xrightarrow{\cong} A^{2m}_{(j)}(S^{m})^{\mathfrak S_m}. \]

The case $j=2$ is proven in [Reference Laterveer31, Theorem 3.1]; the general case is only notationally more complicated.

Proof. Item (i) is [Reference Rieß47, Theorem 3.2], while item (ii) is [Reference Rieß47, Lemma 4.4].

In a nutshell, the construction of the correspondence $R_\phi$ is as follows: [Reference Rieß47, Section 2] provides a diagram

where $\mathcal {X}, \mathcal {X}^{\prime }$ are algebraic spaces over a quasi-projective curve $C$ such that the fibres $\mathcal {X}_0, \mathcal {X}^{\prime }_0$ are isomorphic to $X$ respectively $X^{\prime }$, and where $\Psi \colon \mathcal {X} \dashrightarrow \mathcal {X}^{\prime }$ is a birational map inducing an isomorphism $\mathcal {X}_{C\setminus 0}\cong \mathcal {X}^{\prime }_{C\setminus 0}$ and whose restriction $\Psi \vert _{\mathcal {X}_0}$ coincides with $\phi$. The correspondence $R_\phi$ is then defined as the specialization (in the sense of [Reference Fulton22], extended to algebraic spaces) of the graphs of the isomorphisms $\mathcal {X}_c\cong \mathcal {X}^{\prime }_c$, $c\not =0$.

Point (iii) is not stated explicitly in [Reference Rieß47], and can be seen as follows. Let $U\subset X$ be the locus on which $\phi$ induces an isomorphism. The complement $T:=X\setminus U$ has codimension $\ge 2$. Using complete intersections of hypersurfaces, one can find a closed subset $\mathcal {T}\subset \mathcal {X}$ of codimension $2$ such that $\mathcal {T}_0$ contains $T$, i.e. $\Psi$ restricts to an isomorphism on $V:=\mathcal {X}_0\setminus \mathcal {T}_0$. Since specialization commutes with pullback, the restriction of $R_\phi$ to $V\times X^{\prime }$ is the specialization of the graphs of the morphisms $\mathcal {X}_c\setminus \mathcal {T}_c\to \mathcal {X}^{\prime }_c$, $c\not =0$ to $V\times X^{\prime }$, which is exactly the graph of the morphism $\phi \vert _V$, i.e.

\[ R_\phi\vert_{V\times X\prime}= \Gamma_{(\phi\vert_V)}= \bar{\Gamma}_\phi \vert_{V\times X^{\prime}}\quad \hbox{in}\ A^{n}(V\times X^{\prime}). \]

It follows that one has

\[ R_\phi = \bar{\Gamma}_\phi + \gamma\quad \hbox{in}\ A^{n}(X\times X^{\prime}), \]

where $\gamma$ is some cycle supported on $\mathcal {T}_0\times X^{\prime }$. This proves point $(iii)$: $0$-cycles and $1$-cycles on $X$ can be moved to be disjoint of $\mathcal {T}_0$, and so $\gamma _\ast$ acts as zero on $A^{j}(X)$ for $j\ge n-1$. Likewise, $\gamma ^{\ast }$ acts as zero on $A^{j}(X^{\prime })$ for $j\le 1$ for dimension reasons, and so (by inverting the roles of $X$ and $X^{\prime }$) statement $(iii)$ is proven.

Proof. Let $\Gamma \in A^{4}(\mathcal {Y}\times _B \mathcal {Y})$ be a correspondence that is homologically trivial when restricted to the very general fibre. According to [Reference Voisin58, Proposition 1.6] (cf. also [Reference Laterveer, Nagel and Peters34, Proposition 5.1], where the precise form of Voisin's argument applied here is detailed), there exists a cycle $\gamma \in A^{4}(\mathbb {P}^{5}\times \mathbb {P}^{5})$ such that

\[ \Gamma\vert_b = \gamma\vert_b\quad \hbox{in}\ A^{4}(Y_b\times Y_b)\quad \forall\, b\in B. \]

The cycle $\gamma$ is of the form $\sum \nolimits _{j=0}^{4} c_j h^{j}\times h^{4-j}$, where $h\in A^{1}(\mathbb {P}^{5})$ is a hyperplane section and $c_j\in \mathbb {Q}$. The assumptions imply that $\gamma \vert _b$ is homologically trivial for $b\in B$ very general. Since the summands $(h^{j}\times h^{4-j})\vert _b$ live in different summands of the Künneth decomposition of $H^{8}(Y_b\times Y_b)$, each $c_j (h^{j}\times h^{4-j})\vert _b$ must be homologically trivial, which is only possible when $c_j=0$, and so $\gamma =0$.

Proof. The construction, which is a variant of the construction of the transcendental motive of a surface [Reference Kahn, Murre and Pedrini26], is done in [Reference Bolognesi and Pedrini10, Section 2]. Alternatively, one can use the projectors on the algebraic part of cohomology constructed for any variety satisfying the standard conjectures in [Reference Vial53].

Proof. Let us write $F:=F(Y)$ for the Fano variety of lines, and $P\in A^{3}(F\times Y)$ for the universal line. We define a correspondence

(2)\begin{equation} Q:= \bar{\Gamma}_\psi\circ ({}^{t} \Gamma_{p_1}-{}^{t} \Gamma_{p_1})\circ {}^{t} P \in A^{3}(Y\times Z), \end{equation}

where $\psi \colon F\times F\dashrightarrow Z$ is Voisin's rational map, and $p_j\colon F\times F\to F$ is projection to the $j$th factor. The correspondence $Q$ has the following property:

Proof. Since $H^{4}_{tr}(Y)$ is the smallest Hodge structure for which the complexification contains $H^{3,1}(Y)\cong \mathbb {C}$, it suffices to prove there is a non-zero constant ${ c_h}$ such that

\[ \Bigl( c_h {}^{t} Q\circ \Gamma_{h^{6}}\circ Q -\Delta_Y\Bigr){}_\ast=0\colon H^{3,1}_{}(Y) \to H^{3,1}_{}(Y). \]

By definition of $Q$, it suffices to prove there is a non-zero constant $c_h$ making the following diagram commute:

\[ \begin{array}[c]{ccccccc} H^{3,1}(Y) & \xrightarrow{P^{{\ast}}} & H^{2,0}(F) & \xrightarrow{{p_1}^{{\ast}}-{p_2}^{{\ast}}} & H^{2,0}({F\times F}) & \xrightarrow{{\psi}_\ast} & H^{2,0}(Z)\\ & & & & & & \\ \ \ \downarrow{\scriptstyle {1\over c_h}\operatorname{id}} & & \downarrow{\scriptstyle {\cdot (h_F)^{2}}} & & & & \downarrow{\scriptstyle {\cdot h^{6}}}\\ & & & & & & \\ H^{3,1}(Y) & \xleftarrow{P_\ast} & H^{4,2}(F) & \xleftarrow{({p_1})_\ast{-}({p_2})_\ast} & H^{8,6}({F\times F}) & \xleftarrow{{\psi}^{{\ast}}} & H^{8,6}(Z)\\ \end{array} \]

(Here $h_F\in A^{1}(F)$ denotes a polarization,)

The left square of this diagram is well known to be commutative (essentially this is the Abel–Jacobi isomorphism established in [Reference Beauville and Donagi6]). To see that the right-hand square of this diagram commutes, we observe that thanks to Proposition 2.4, we have a relation $\widetilde {\psi }^{\ast }(\omega _Z)=\widetilde {p_1}^{\ast }(\omega _F)-\widetilde {p_2}^{\ast }(\omega _F)$. This implies that the top horizontal line is an isomorphism: more precisely, the composition $\psi _\ast (p_1^{\ast } -p_2^{\ast })$ sends $\omega _F$ to $6\omega _Z\in H^{2,0}(Z)$. Let $\omega _Z^{\vee }\in H^{6,8}(Z)$ be the class dual to $\omega _Z$ under cup product, and let $\bar {\omega }_Z^{\vee }\in H^{8,6}(Z)$ be its complex conjugate. By hard Lefschetz, we know that $\omega _Z\cdot h^{6}\in H^{8,6}(Z)$ is a rational multiple of the class $\bar {\omega }_Z^{\vee }$. By duality, the composition $((p_1)_\ast -(p_2)_\ast ) \psi ^{\ast }$ sends $\omega _Z^{\vee }\in H^{6,8}(Z)$ to ${1\over 6}\omega _F^{\vee }\in H^{0,2}(F)$, and so $\bar {\omega }_Z^{\vee }\in H^{6,8}(Z)$ is sent to ${1\over 6}\bar {\omega }_F^{\vee }\in H^{0,2}(F)$. Using hard Lefschetz on $F$, we can rescale $h_F$ by some rational multiple to make the right-hand square commutative. This proves Lemma 3.3.

Lemma 3.3 implies that one can write

\[ \Bigl( c_h {}^{t} Q\circ \Gamma_{h^{6}}\circ Q -\Delta_Y\Bigr) = \gamma \quad \hbox{in}\ H^{8}(Y\times Y), \]

where $\gamma$ is a completely decomposed cycle, i.e. a cycle supported on a finite union of subvarieties $V_i\times W_i\subset Y\times Y$, with $\dim V_i+\dim W_i=4$. Since the composition of the completely decomposed cycle $\gamma$ with $\pi ^\textrm {prim}_Y$ is zero (for any smooth cubic $Y$), this implies the necessary vanishing (3), and so the theorem is proven.

Proof. (i) Claim 3.2 guarantees that

\[ {}^{t} \pi^{tr}_Z= c_h Q\circ \pi^{tr}_Y\circ {}^{t} Q\circ \Gamma_{h^{6}} \in A^{8}(Z\times Z) \]

is idempotent, and that

\[ Q\colon t(Y):=(Y,\pi^{tr}_Y,0) \to (Z,{}^{t} \pi^{tr}_Z,-1)\quad \hbox{in}\ \mathcal{M}_\textrm{rat} \]

is an isomorphism, with inverse $c_h {}^{t} Q\circ \Gamma _{h^{6}}$. It follows that

\[ ({}^{t} \pi^{tr}_Z)_\ast A^{j}(Z) =0\quad \forall\,j\not=2, \]

and

\[ ({}^{t} \pi^{tr}_Z)_\ast A^{2}(Z) =Q_\ast A^{3}_\textrm{hom}(Y)=:A^{2}_{[2]}(Z). \]

(ii) This is immediate from the proof of Theorem 3.1, where it is shown that

\[ A^{3}_\textrm{hom}(Y)\cong S_3^\textrm{mob} A^{8}(Z)\cap A^{8}_\textrm{hom}(Z)= (\pi^{tr}_Z)_\ast A^{8}(Z)= \operatorname{Im}\Bigl( Q_\ast A^{3}_\textrm{hom}(Y) \xrightarrow{\cdot h^{6}} A^{8}(Z)\Bigr). \]

Proof. Let $F=F(Y)$ be the Fano variety of lines in $Y$. Via Voisin's rational map, one sees as in [Reference Voisin59] that $S_4A^{8}_{}(Z)$ is one-dimensional with generator $\psi _\ast (o_F\times o_F)$, where $o_F\in A^{4}_{(0)}(F)=S_2 A^{4}(F)$ is the distinguished zero-cycle of [Reference Shen and Vial49]. There is a commutative diagram

\[ \begin{array}[c]{ccc} A^{8}(F\times F) & \xrightarrow{\psi_\ast} & A^{8}(Z)\\ & & \\ \downarrow{\scriptstyle( \iota_F)_\ast} & & \downarrow{\scriptstyle \iota_\ast}\\ & & \\ A^{8}(F\times F) & \xrightarrow{\psi_\ast} & A^{8}(Z),\\ \end{array} \]

where $\iota _F$ is the map exchanging the two factors [Reference Camere, Cattaneo and Laterveer13, Remark 2.19]. Since $o_F\times o_F$ is invariant under $\iota _F$, it follows that $\iota$ acts as the identity on $S_4 A^{8}_{}(Z)$.

To prove the statement for $S_3^\textrm {mob} A^{8}(Z)\cap A^{8}_\textrm {hom}(Z)$, let us first ensure that $\iota _\ast$ preserves this subgroup:

Proof. The Hilbert scheme $Z^{\prime }$ has an MCK decomposition [Reference Vial54], and the induced decomposition of $A^{8}(Z^{\prime })$ is compatible with Voisin's orbit filtration $S_\ast$, in the sense that

(5)\begin{equation} \bigoplus_{j=0}^{r} A^{8}_{(2j)}(Z^{\prime}) = S_{4-r} A^{8}(Z^{\prime}) \end{equation}

[Reference Voisin59, End of section 4.1].

Using Rieß’ work (Theorem 2.18), the MCK decomposition of $Z^{\prime }$ induces an MCK decomposition for $Z$, and there exists a correspondence $R_\phi \in A^{8}(Z^{\prime }\times Z)$ inducing an isomorphism of bigraded rings $A^{\ast }_{(\ast )}(Z^{\prime })\cong A^{\ast }_{(\ast )}(Z)$. In order to prove Theorem 3.9, we want to relate the orbit filtration $S_\ast$ on $A^{8}(Z)$ with the ‘motivic’ decomposition $A^{8}_{(\ast )}(Z)$. This is the content of the following lemma.

Proof. This is a translation of Theorem 3.9, using the generators for the various pieces $A^{8}_{(j)}(Z^{\prime })$ given in Theorem 2.15. We know that $\iota ^{\prime }$ preserves these pieces (Theorem 3.9), and so one can check the formula of Theorem 3.11 piece by piece. This is a direct computation. For example, Theorem 2.15 says that $A^{8}_{(8)}(Z^{\prime })$ is generated by expressions of the form

\begin{align*} & [x,y,z,t] - \bigl( [x,y,z,o] + [x,y,t,o] + \cdots\bigr) + \bigl( [x,y,o,o] + [x,z,o,o]+\cdots\bigr)\\ & \quad - \bigl( [x,o,o,o] + [y,o,o,o]+\cdots\bigr) + [o,o,o,o], \end{align*}

and the formula of Theorem 3.11 leaves expressions of this form invariant, which is in agreement with the fact that $\iota ^{\prime }$ is the identity on $A^{8}_{(8)}(Z^{\prime })$ (Theorem 3.9). The verification of the action of $\iota ^{\prime }$ on $A^{8}_{(j)}(Z^{\prime })$, $j<8$ is similar (but easier).

Proof. Given a K3 surface $S$ of Picard number 1 and degree $d>2$, the Hilbert scheme $X=S^{[4]}$ has $\operatorname {Bir}(X)$ a group of order $1$ or $2$ [Reference Debarre16, Section 4.3.1]. On the other hand, the condition $d= (6n^{2}-6n+2)/a^{2}$ for some $n,a\in \mathbb {Z}$ implies (and actually is equivalent to the fact) that $X$ is birational to an LLSS eightfold [Reference Li, Pertusi and Zhao39, Proposition 5.3], and so $\operatorname {Bir}(X)$ is non-trivial. It follows that $X$ as in the theorem must have $\operatorname {Bir}(X)=\mathbb {Z}/2\mathbb {Z}$, and the non-trivial element $\iota \in \operatorname {Bir}(X)$ acts on $A^{8}(X)$ as in Theorem 3.11.

Proof. Recall that $X$ is a quotient variety, and so the Chow groups with $\mathbb {Q}$-coefficients of $X$ have a ring structure (cf. § 2.1).

As noted before, the birationality $Z\to Z^{\prime }$ induces an isomorphism of bigraded rings $A^{\ast }_{(\ast )}(Z)\cong A^{\ast }_{(\ast )}(Z^{\prime })$ (Theorem 2.18). We have seen (equality (12)) that $\iota$ acts as $-\operatorname {id}$ on $A^{2}_{(2)}(Z)$, and so

\[ A^{2}_{(2)}(Z)^{ \iota }=0. \]

Similar reasoning also shows that

\[ A^{2}(Z)^{ \iota } \subset A^{2}_{(0)}(Z). \]

Indeed, let $b\in A^{2}(Z)^{\iota }$, and let $h\in A^{1}(Z)$ be once again a $g$-invariant ample class. Then $b\cdot h^{6}\in A^{8}(Z)^{ \iota }$ and so (in view of Theorem 3.9) we know that $b\cdot h^{6}\in A^{8}_{(0)}(Z)$. Writing the decomposition $b=b_0+b_2$, where $b_j\in A^{2}_{(j)}(Z)$, and invoking the hard Lefschetz type isomorphism (11), we see that $b_2=0$.

The corollary is now readily proven. For instance, the image of the intersection map

\[ \operatorname{Im} \Bigl( A^{3}(Z)^{ \iota }\otimes A^{2}(Z)^{ \iota }\otimes A^{2}(Z)^{ \iota }\otimes A^{1}(Z)^{\iota } \to A^{8}(Z)\Bigr) \]

is contained in

\[ \operatorname{Im} \bigg( \bigg( \bigoplus_{j\le 2} A^{3}_{(j)}(Z)\bigg)\otimes A^{2}_{(0)}(Z)\otimes A^{2}_{(0)}(Z)\otimes A^{1}(Z) \to A^{8}(Z)\bigg) \subset \bigoplus_{j\le 2} A^{8}_{(j)}(Z). \]

On the other hand, the image is obviously $\iota$-invariant, and so

\begin{align*} & \operatorname{Im} \Bigl( A^{3}(Z)^{ \iota }\otimes A^{2}(Z)^{ \iota }\otimes A^{2}(Z)^{ \iota }\otimes A^{1}(Z)^{\iota } \to A^{8}(Z)\Bigr)\\ & \quad \subset A^{8}(Z)^{\iota} \subset A^{8}_{(0)}(Z)\oplus A^{8}_{(4)}(Z) \oplus A^{8}_{(8)}(Z) \end{align*}

(where the second inclusion is equality (13)). It follows that

\[ \operatorname{Im} \Bigl( A^{3}(Z)^{ \iota }\otimes A^{2}(Z)^{ \iota }\otimes A^{2}(Z)^{ \iota }\otimes A^{1}(Z)^{\iota } \to A^{8}(Z)\Bigr) \subset A^{8}_{(0)}(Z)\cong\mathbb{Q}[h^{8}]. \]