Standard conjectures for abelian fourfolds

Abstract

Let A be an abelian fourfold in characteristic p. We prove the standard conjecture of Hodge type for A, namely that the intersection product

$$\begin{aligned} {\mathcal {Z}}^2_{\mathrm {num}}(A)_{{\mathbb {Q}}}\times {\mathcal {Z}}_{\mathrm {num}}^2(A)_{{\mathbb {Q}}} \longrightarrow {\mathbb {Q}}\end{aligned}$$

is of signature $(\rho _2 - \rho _1 +1; \rho _1 - 1)$, with $\rho _n=\dim {\mathcal {Z}}_{\mathrm {num}}^n(A)_{{\mathbb {Q}}}.$ (Equivalently, it is positive definite when restricted to primitive classes for any choice of the polarization.) The approach consists in reformulating this question into a p-adic problem and then using p-adic Hodge theory to solve it. By combining this result with a theorem of Clozel we deduce that numerical equivalence on A coincides with $\ell$-adic homological equivalence on A for infinitely many prime numbers $\ell$. Hence, what is missing among the standard conjectures for abelian fourfolds is $\ell$-independency of $\ell$-adic homological equivalence.

Geometric examples

In this section we discuss several examples to which Theorems 3.18 and 8.1 apply non-trivially. We are particularly interested in exotic classes on abelian fourfolds (Definition 7.1). The main result is Proposition A.1 where we discuss the existence of exotic classes that cannot be lifted to algebraic classes in characteristic zero. The techniques of construction there are inspired by [37] and [56].

Other examples of exotic classes will be found in Remark A.8. We end the section with an example (other than abelian fourfolds) for which the standard conjecture of Hodge type holds true via Theorem 8.1.

Proposition A.1

Let p be a prime number and let K be an imaginary quadratic number field where p does not split. Then there exists an abelian fourfold A over $\overline{{\mathbb {F}}}_p$ verifying the following properties.

• (1) The endomorphism algebra is a number field that can be written as the compositum

  $$\begin{aligned} \mathrm {End}(A)_{\mathbb {Q}}= K \cdot R \end{aligned}$$

  where R is a totally real number field such that $[R:{\mathbb {Q}}]=4$. (In particular A is simple and has a unique CM-structure).

• (2) Consider the motivic decomposition from Proposition 6.7(2). Among the factors ${\mathcal {M}}_I$ of ${\mathfrak {h}}^4(A)$ there exists a (unique) factor M such that M(2) does not contain any Lefschetz class (Definition 5.1) but it is Frobenius invariant (for a model of A defined over a finite field).

• (3) For any CM-lifting of A to characteristic zero (see Theorem 6.10 and Corollary 6.12) the Hodge structure $R_B(M)$ is of type (3, 1), (1, 3).

• (4) For any CM-lifting of A to characteristic zero we have the equality $\mathrm {End}_{\mathrm {NUM}({\mathbb {C}})_{\mathbb {Q}}} (M) = K.$

Proof

The proof is decomposed in a series of lemmas. The final step is Lemma A.7. $\square$

Remark A.2

Let us make some comments on the above proposition.

• (1) If (a model of) the fourfold A verifies the Tate conjecture, then M is an exotic motive in the sense of Sect. 7. In particular, in each characteristic, there should be infinitely many non-isogenous abelian fourfolds having exotic motives.

  If the Tate conjecture was highly false and no such exotic classes existed then Theorem 3.18 would follow directly from the arguments in Sect. 5.

  The Tate conjecture and the standard conjecture of Hodge type should be thought as two independent and different problems. The first is about the construction of algebraic classes, the second is about how they intersect (independently whether there are a lot of algebraic classes or not). It seems likely that a solution of one problem does not imply a solution for the other. For example the proof of the Tate conjecture for divisors on abelian variety [54, Theorem 4] does not imply the standard conjecture of Hodge type for divisor on abelian variety (which is known by a different argument).

• (2) Because of the Hodge types in part (3), the (expected) algebraic classes in positive characteristic cannot be lifted to algebraic classes in characteristic zero.

• (3) Notice that the field $F=K \otimes _{\mathbb {Q}}{\mathbb {Q}}_p$ can be any quadratic extension of ${\mathbb {Q}}_p$. This field F coincides with the one in Proposition 10.8. This shows that the different cases studied in Sects. 11 and 12 were needed.

• (4) The hypothesis that K is totally imaginary is necessary. If K were a real quadratic number field, conditions (3) and (4) in Proposition  A.1 would not be compatible (see the proof of Lemma 7.16).

Lemma A.3

Let p and K be as in Proposition A.1. There exists a totally real number field R such that the following holds:

• (1) The prime p does not split in R.

• (2) The degree $[R:{\mathbb {Q}}]$ is four.

• (3) The field $K \otimes _{\mathbb {Q}}{\mathbb {Q}}_p$ is embeddable in the field $R \otimes _{\mathbb {Q}}{\mathbb {Q}}_p$.

• (4) If ${\tilde{R}}$ is the normal closure of R over ${\mathbb {Q}}$, then $\mathrm {Gal}({\tilde{R}}/{\mathbb {Q}})={\mathfrak {S}}_4$. In particular we have $\mathrm {Gal}({\tilde{R}}/R)={\mathfrak {S}}_3$.

Proof

Same as in [37, §3]. $\square$

Lemma A.4

Let R be as in the above lemma. The following holds:

• (1) The subfields of the compositum $R\cdot K$ are ${\mathbb {Q}},K,R$ and $R\cdot K$.

• (2) The prime p factorises in $R\cdot K$ as

  $$\begin{aligned} p=({\mathfrak {p}}\cdot \bar{{\mathfrak {p}}})^e, \end{aligned}$$

  where ${\mathfrak {p}}$ and $\bar{{\mathfrak {p}}}$ are two prime ideals which are exchanged by complex conjugation and e is the ramification index.

Proof

Let ${\tilde{R}}$ be as in Lemma A.3. The equality $\mathrm {Gal}({\tilde{R}}/{\mathbb {Q}})={\mathfrak {S}}_4$ implies

$$\begin{aligned} \mathrm {Gal}({\tilde{R}}\cdot K/{\mathbb {Q}})={\mathfrak {S}}_4\times {\mathbb {Z}}/2{\mathbb {Z}}\end{aligned}$$

and similarly $\mathrm {Gal}({\tilde{R}}/R)={\mathfrak {S}}_3$ implies $\mathrm {Gal}({\tilde{R}}\cdot K/R \cdot K)={\mathfrak {S}}_3.$ We deduce that the subfields of $R\cdot K$ are in bijection with the subgroups of ${\mathfrak {S}}_4\times {\mathbb {Z}}/2{\mathbb {Z}}$ containing ${\mathfrak {S}}_3$. Those are precisely ${\mathfrak {S}}_3 , {\mathfrak {S}}_4, {\mathfrak {S}}_3\times {\mathbb {Z}}/2{\mathbb {Z}}$ and ${\mathfrak {S}}_4\times {\mathbb {Z}}/2{\mathbb {Z}}$, which implies (1).

Note that the equality $R\cdot K \cong R \otimes _{\mathbb {Q}}K$ holds. Hence we have

$$\begin{aligned} (R\cdot K) \otimes _{\mathbb {Q}}{\mathbb {Q}}_p \cong (K \otimes _{\mathbb {Q}}{\mathbb {Q}}_p) \otimes _{{\mathbb {Q}}_p} (R \otimes _{\mathbb {Q}}{\mathbb {Q}}_p) \cong (R \otimes _{\mathbb {Q}}{\mathbb {Q}}_p)^{\mathrm {Gal}(K \otimes _{\mathbb {Q}}{\mathbb {Q}}_p/{\mathbb {Q}}_p) }, \end{aligned}$$

where the last equality comes from Lemma A.3(3). As ${R \otimes _{\mathbb {Q}}{\mathbb {Q}}_p}$ is a field (by Lemma A.3(1)), the prime p factorizes in $R\cdot K$ as the product of two different primes. Moreover, those two primes are exchanged by $\mathrm {Gal}(K \otimes _{\mathbb {Q}}{\mathbb {Q}}_p/{\mathbb {Q}}_p)$. As complex conjugation generates this Galois group it exchanges these two prime ideals. In particular they must have the same ramification index. This concludes part (2). $\square$

Lemma A.5

Consider the prime ideals ${\mathfrak {p}}$ and $\bar{{\mathfrak {p}}}$ of $R\cdot K$ from the above lemma. Then there exist an integer n, a p-power q and a q-Weil number $\alpha \in R\cdot K$ verifying the following properties:

• (1) The ideal generated by $\alpha$ factorizes as

  $$\begin{aligned} (\alpha )={\mathfrak {p}}^n\cdot \bar{{\mathfrak {p}}}^{3n}. \end{aligned}$$

• (2) For any positive integer s, we have the equality

  $$\begin{aligned} {\mathbb {Q}}(\alpha ^s)= R\cdot K. \end{aligned}$$

• (3) The norm $N_{R\cdot K/K}(\alpha )\in K$ equals $q^2$.

Proof

Consider the ideal $I=({\mathfrak {p}}\cdot \bar{{\mathfrak {p}}}^{3})^e$. We have that $I \cdot {\bar{I}}= p^4$ by Lemma A.4(2). Actually, for any $g\in \mathrm {Gal}({\tilde{R}}\cdot K/{\mathbb {Q}}),$ we have the equality ${g(I) \cdot \overline{g(I)}= p^4}$ of ideals in ${\tilde{R}}\cdot K$. This follows from the case $g=\mathrm {id}$ together with the explicit description of the Galois group in the proof of the lemma above (which implies that complex conjugation is in the center of the Galois group). Hence, we can apply [24, Lemma 1] and deduce that there exists a q-Weil number $\alpha \in R\cdot K$ verifying (1).

Let us show (2). Because of Lemma A.4(1), this amounts to showing that $\alpha ^s$ does not belong to K nor to R. Now, if $\alpha ^s$ belongs to K (or to R) then its factorization in $R\cdot K$ would have the same exponent in ${\mathfrak {p}}$ and in $\bar{{\mathfrak {p}}}$ because there is only one prime above p in K (or in R); see the hypothesis on K in Proposition A.1 (respectively by Lemma A.3(1)).

Let us now show (3). If we compute the norm of the ideal $(\alpha )$ we obtain

$$\begin{aligned} (N_{R\cdot K/K} (\alpha ))= N_{R\cdot K/K} ({\mathfrak {p}}^n\cdot \bar{{\mathfrak {p}}}^{3n}) = N_{R\cdot K/K} ({\mathfrak {p}})^n\cdot N_{R\cdot K/K} ( \bar{{\mathfrak {p}}})^{3n}= {\tilde{p}}^m, \end{aligned}$$

where m is an integer and ${\tilde{p}}$ is the only prime ideal above p in K. Hence, after possibly replacing $\alpha$ by a power, we obtain that the ideal $N_{R\cdot K/K} (\alpha )$ is generated by a power of p. For weight reasons we have

$$\begin{aligned} (N_{R\cdot K/K} (\alpha )) = (q^2). \end{aligned}$$

This is equivalent to the relation $N_{R\cdot K/K} (\alpha ) = \xi \cdot q^2$, where $\xi$ is an invertible element of the ring of integers of K. As the group of invertible elements of the ring of integers of an imaginary quadratic field is finite, after replacing $\alpha$ by a power we get (3). Notice that such a power of $\alpha$ will still have properties (1) and (2). $\square$

Lemma A.6

Let $\alpha$ be a q-Weil number verifying the properties as in the above lemma. Let A be an abelian variety over ${\mathbb {F}}_{q}$ whose isogeny class corresponds to $\alpha$ under the Honda–Tate correspondance [55]. Then the following holds:

• (1) The dimension of A is four.

• (2) A is geometrically simple.

• (3) $\mathrm {End}(A)_{\mathbb {Q}}= \mathrm {End}(A_{\bar{{\mathbb {F}}}_p})_{\mathbb {Q}}= K \cdot R.$

• (4) The slopes of A are (1/4, 3/4)

• (5) There are Frobenius invariant classes in $H_\ell ^4(A)$ which are not of Lefschetz type.

Proof

By [54] the division algebra $\mathrm {End}(A)_{\mathbb {Q}}$ has center equal to ${\mathbb {Q}}(\alpha )$ which is $R\cdot K$ by Lemma A.5. Moreover, by [55, Theorem 1], this division algebra splits at every place except possibly at the places ${\mathfrak {p}}$ and $\bar{{\mathfrak {p}}}$ above p. The local invariants there are computed by the formula in [55, Theorem 1], which gives

$$\begin{aligned} \mathrm {inv}_{\mathfrak {p}}( \mathrm {End}(A)_{\mathbb {Q}})=\frac{v_{\mathfrak {p}}(\alpha )}{v_{\mathfrak {p}}(q)}\cdot [(R\cdot K)_{\mathfrak {p}}:{\mathbb {Q}}_p] \quad \mod {\mathbb {Z}}\end{aligned}$$

and similarly for $\bar{{\mathfrak {p}}}$.

We claim that these local invariants are trivial as well. Indeed, using the factorisation in Lemma A.5(1), we deduce that $\frac{v_{\mathfrak {p}}(\alpha )}{v_{\mathfrak {p}}(q)}=\frac{1}{4}.$ On the other hand, the degree $[(R\cdot K)_{\mathfrak {p}}:{\mathbb {Q}}_p]$ equals four, because $[R\cdot K:{\mathbb {Q}}]=8$ and the two primes above p are exchanged by complex conjugation (Lemma A.4(2)). Altogether we have that $\mathrm {inv}_{\mathfrak {p}}( \mathrm {End}(A)_{\mathbb {Q}})=0$ and similarly one shows $\mathrm {inv}_{\bar{{\mathfrak {p}}}}( \mathrm {End}(A)_{\mathbb {Q}})=0$

Because all the invariants of the $(R\cdot K)$-central algebra $\mathrm {End}(A)_{\mathbb {Q}}$ are trivial, we have $R\cdot K= \mathrm {End}(A)_{\mathbb {Q}}.$ As ${[R\cdot K:{\mathbb {Q}}]=8}$ we deduce (1).

Consider now the abelian variety $A_s$ over ${\mathbb {F}}_{q^s}$ whose isogeny class corresponds to $\alpha ^s$. Following the Honda–Tate correspondance $A_s$ is a simple factor of $A \times _{{\mathbb {F}}_q} {\mathbb {F}}_{q^s}$. On the other hand, all the arguments above work by replacing $\alpha$ by $\alpha ^s$, because of Lemma A.5(2). In particular, $A_s$ has also dimension four and $R\cdot K= \mathrm {End}(A_s)_{\mathbb {Q}}.$ This implies (2) and (3).

One slope has already been computed, namely $\frac{v_P(\alpha )}{v_P(q)}=\frac{1}{4}.$ Duality implies that there is also the slope 3/4. As A has dimension four there are no more slopes.

Let us now show (5). The existence of a class such as the ones claimed is equivalent to the existence of a set I consisting of four Galois conjugates of $\alpha$ whose product equals $q^2$ and such that I is not stable under the action of complex conjugation (see the proof of Lemma 7.7 or [56, § 2]). We claim that the relation

$$\begin{aligned} N_{R\cdot K/K}(\alpha ) = q^2 \end{aligned}$$

(Lemma A.5(3)) gives precisely the existence of those four Galois conjugates. Indeed, let ${\tilde{R}}$ be the normal closure of R over ${\mathbb {Q}}$, by definition we have

$$\begin{aligned} N_{R\cdot K/K}(\alpha ) = \prod _{g \in \mathrm {Hom}_K( R\cdot K , {\tilde{R}}\cdot K)} g(\alpha ). \end{aligned}$$

Hence it is enough to show that complex conjugation does not stabilize the set $J=\{g(\alpha )\}_{g \in \mathrm {Hom}_K( R\cdot K , {\tilde{R}}\cdot K)}.$ As the set J is of size four and the total Galois orbit of $\alpha$ is of size 8 there is an element of $\mathrm {Gal}({\tilde{R}}\cdot K/{\mathbb {Q}})$ which does not stabilize J. On the other hand, thanks to the equality

$$\begin{aligned} \mathrm {Gal}({\tilde{R}}\cdot K/{\mathbb {Q}}) = \mathrm {Gal}({\tilde{R}}\cdot K/K) \times \mathrm {Gal}({\tilde{R}}\cdot K/{\tilde{R}}) \end{aligned}$$

we have that the total Galois group $\mathrm {Gal}({\tilde{R}}\cdot K/{\mathbb {Q}})$ is generated by its subgroup $\mathrm {Gal}({\tilde{R}}\cdot K/K)$ and complex conjugation. As $\mathrm {Gal}({\tilde{R}}\cdot K/K)$ stabilizes J, complex conjugation cannot stabilize it. $\square$

Lemma A.7

Let A be an abelian fourfold which satisfies the properties of the lemma above. Then it also satisfies all the conditions of Proposition A.1.

Proof

Part (1) has already been showed. Part (2) follows from Lemma A.6(5). (Unicity comes from Lemma 7.14.)

Let us now show part (3). Write $\alpha , \beta , \gamma , \delta , q/\alpha , q/\beta , q/\gamma , q/ \delta$ for the eight (distinct) Frobenius eigenvalues and consider the decomposition in eigenlines

$$\begin{aligned} {\mathfrak {h}}^1(A)=V_\alpha \oplus V_\beta \oplus V_\gamma \oplus V_\delta \oplus V_{q/\alpha } \oplus V_{q/\beta } \oplus V_{q/\gamma } \oplus V_{q/ \delta } \end{aligned}$$

as in Proposition 6.6. Among these eight eigenvalues, four have slope 1/4 and four have slope 3/4 and they are exchanged by complex conjugation.

Fix a CM-lifting (Theorem 6.10). The above decomposition in eigenlines will lift as well (Corollary 6.12). Among the eight lines, four will belong to $H^{1,0}$ and four will belong to $H^{0,1}$ and again they are exchanged by complex conjugation. The Shimura–Taniyama formula [55, Lemma 5] implies that there is exactly one eigenvalue, call it $\alpha$, whose slope is 1/4 and such that $V_\alpha \subset H^{1,0}$. Equivalently, there is exactly one eigenvalue, namely $q/\alpha$, whose slope is 3/4 and such that $V_{q/\alpha } \subset H^{0,1}$.

Now decompose $M= M_{\alpha , \beta , \gamma , \delta } \oplus M_{q/\alpha , q/\beta , q/\gamma , q/ \delta }$ via Proposition 6.7. (After possibly renaming the eigenvalues.) With this notation we have the relation

$$\begin{aligned} \alpha \cdot \beta \cdot \gamma \cdot \delta =q^2. \end{aligned}$$

By looking at the p-adic valuation we deduce that, among $\beta , \gamma , \delta$ there is exactly one eigenvalue of slope 1/4, say $\beta$. Hence we have $V_\beta \subset H^{0,1}$ and $V_\gamma , V_\delta \subset H^{1,0}$. Altogether we deduce

$$\begin{aligned} V_\alpha \otimes V_\beta \otimes V_\gamma \otimes V_\delta \subset H^{3,1} \end{aligned}$$

which gives (3).

Let us now show part (4). By construction we can find a quadratic number field $F\subset {\tilde{R}}\cdot K$ such that the motive M decomposes in the category $\mathrm {CHM}(k)_{F}$ into a sum

$$\begin{aligned} {\mathcal {M}}_{I} = M_{I} \oplus M_{{\bar{I}}} \end{aligned}$$

of two motives of rank one (see Proposition 6.7). We first claim that such a field F must be imaginary. If F were contained in ${\mathbb {R}}$ then the Betti realization of the lifting of $M_{I}$ would respect the Hodge symmetry. As it is one dimensional for weight reasons it would be of type (2, 2). This contradicts part (3).

By [26], $D=\mathrm {End}_{\mathrm {NUM}({\mathbb {C}})_{\mathbb {Q}}} (M)$ is a division algebra. By construction, F splits D. We claim that $D=F$. Otherwise we would have an isomorphism $D \otimes F \cong M_{2 \times 2}(F)$ which would imply that $M_{I}$ and $M_{{\bar{I}}}$ are isomorphic as numerical motives. As homological and numerical equivalence is known to coincide for complex abelian varieties [36], this would imply that their Betti realization are isomorphic, which is impossible because of the different Hodge types.

In conclusion, $\mathrm {End}_{\mathrm {NUM}({\mathbb {C}})_{\mathbb {Q}}} (M)$ is an imaginary quadratic field contained in ${\tilde{R}}\cdot K$. On the other hand, there is only one such field (namely K) because of the description of $\mathrm {Gal}({\tilde{R}}\cdot K/{\mathbb {Q}})$ in the proof of Lemma A.4. $\square$

Remark A.8

Let us comment on other examples of exotic motives coming from abelian fourfolds.

• (1) One can construct an abelian fourfold A over a finite field having an exotic motive whose lifting to ${\mathbb {C}}$ has Betti realization of type (2, 2). Such a condition means that the CM-lifting of A over ${\mathbb {C}}$ has Hodge classes which are not Lefschetz. This situation (over ${\mathbb {C}}$) has been classified in [44]. So, any reduction modulo p of their examples will give an abelian fourfold over a finite field of the desired type. (To avoid that the reduction modulo p creates more Lefschetz classes, one can take an ordinary prime).

  As already pointed out, these examples are less interesting for the standard conjecture of Hodge type, see Remark 10.5(2).

• (2) There are no exotic motives over $\overline{{\mathbb {F}}}_p$ (coming from abelian fourfolds) whose lifting to ${\mathbb {C}}$ have Betti realization of type (4, 0), (0, 4). To show this, consider a model of the abelian fourfold over a finite field ${\mathbb {F}}_{q}$. Let I be the set of Frobenius eigenvalues such that the corresponding eigenspaces are lifted into $H^{1,0}$ and ${\bar{I}}$ be the set of Frobenius eigenvalues such that the corresponding eigenspaces are lifted into $H^{0,1}$. If the cohomology group $H^{4,0}\oplus H^{0,4}$ becomes Frobenius invariant over ${\mathbb {F}}_q$, then

  $$\prod _{\alpha \in I} \alpha = \prod _{\beta \in {\bar{I}}} \beta (=q^2).$$

  On the other hand, using the Shimura–Taniyama formula [55, Lemma 5], we have that the p-adic valuation of $\prod _{\alpha \in I} \alpha$ is greater than the one of $\prod _{\beta \in {\bar{I}}} \beta$, except if all Frobenius eigenvalues have the same slopes. In this case the abelian variety would be isogenous to the forth power of a supersingular elliptic curve and hence all algebraic classes would be Lefschetz.

• (3) Having the results of Sect. 7 in mind, the last example that needs to be discussed is that of an abelian fourfold with a four dimensional space of exotic classes. By Lemma 7.14, this reduces to an abelian fourfold over ${\mathbb {F}}_{q^2}$ of the form $X\times E$, where X is an abelian threefold and E is a supersingular elliptic curve on which Frobenius acts as $q \cdot \mathrm {id}$. Now the equations (7.4) and (7.5) imply that the existence of an exotic class on $X\times E$ is equivalent to the existence of an exotic class on $X^2$. There are infinitely many such threefolds X, they have been classified in [56].

Proposition A.9

The standard conjecture of Hodge type holds true for Fermat’s cubic fourfold $X=\{x_0^3+\cdots +x_5^3=0\}\subset {\mathbb {P}}_k^5$ over any field k (of characteristic different from 3).

Proof

Let us first consider X as variety over ${\mathbb {C}}$. By [6, Proposition 11] its Hodge structure decomposes as

$$\begin{aligned} H_B^*(X,{\mathbb {Q}})={\mathbb {Q}}(0) \oplus {\mathbb {Q}}(-1) \oplus {\mathbb {Q}}(-2)^{\oplus 21} \oplus V_B \oplus {\mathbb {Q}}(-3) \oplus {\mathbb {Q}}(-4) \end{aligned}$$

where $V_B$ is a ${\mathbb {Q}}$-Hodge structure of rank 2 and of type (3, 1), (1, 3). As the Hodge conjecture is known for X and its powers [52], this decomposition holds true at the level of homological motives

$$\begin{aligned} M(X)=\mathbb {1}\oplus \mathbb {1}(-1) \oplus \mathbb {1}(-2)^{\oplus 21} \oplus V \oplus \mathbb {1}(-3) \oplus \mathbb {1}(-4). \end{aligned}$$

This implies that the primitive part of the motive is of the form

$$\begin{aligned} {\mathfrak {h}}^{4,\mathrm {prim}}(X)= \mathbb {1}(-2)^{\oplus 20} \oplus ^{\perp } V. \end{aligned}$$

Note that the decomposition is orthogonal with respect to the cup product as the types of the Hodge structures are different. Finally, as the motive of X is finite dimensional [7, Lemma 5.2], this decomposition lifts to the level of Chow motives. (Alternatively, see Remark 8.2(2).)

Let us now work over $\overline{{\mathbb {F}}}_p$. (This is enough for our purpose, thanks to Proposition 3.16). The positivity of the cup product on algebraic classes on the factor $\mathbb {1}(-2)^{\oplus 20}$ is clear as all these classes come from characteristic zero, see Remark 3.13. We are reduced to the study of algebraic classes on the two dimensional motive V. As the characteristic polynomial of Frobenius acting on V is a rational polynomial of degree two, there are either zero or two rational solutions. In the first case the space of algebraic classes on V is reduced to zero hence the standard conjecture of Hodge type holds trivially. In the second case the Fermat variety is supersingular and V is spanned by algebraic cycles [53]. In this case the standard conjecture of Hodge type holds true via Theorem 8.1. (Note that there are infinitely many primes for which the non-trivial case occurs [53, Theorem 2.10]). $\square$

Remark A.10

Let us comment on applications and limits of Theorem 8.1.

• (1) Theorem 8.1 cannot be applied to show the standard conjecture of Hodge type for abelian varieties of dimension at least five. Indeed, let A be a simple abelian variety of dimension g and let $M_I\subset {\mathfrak {h}}^{2i}(A)$ be a factor as constructed in Proposition 6.7. By its very construction, the dimension of $M_I$ is at least g/i as $\mathrm {Gal}({\overline{{\mathbb {Q}}}}/{\mathbb {Q}})$ acts transitively on $\Sigma$, see Notation 6.5. Hence the rank of $M_I$ will never be two (except possibly in middle degree).

• (2) It seems likely that using Theorem 8.1 one can show the standard conjecture of Hodge type for some special varieties as we did in Proposition A.9 for Fermat’s cubic fourfold. On the other hand we do not know examples (other than abelian fourfolds) where this strategy applies for a whole family of varieties and we expect such examples to be rare.⁷ It is rather a miracle, based on the computations of Sect. 7, that for all abelian fourfolds only motives of rank two turn out to be significant.