## 45.11 Weil cohomology theories, II

For us a Weil cohomology theory will be the analogue of a classical Weil cohomology theory (Section 45.7) when the ground field $k$ is not algebraically closed. In Section 45.9 we listed axioms which guarantee our cohomology theory comes from a symmetric monoidal functor on the category of motives over $k$. Missing from our axioms so far are the condition $H^ i(X) = 0$ for $i < 0$ and a condition on $H^{2d}(X)(d)$ for $X$ equidimensional of dimension $d$ corresponding to the classical axioms (A)(c) and (A)(d). Let us first convince the reader that it is necessary to impose such conditions.

Example 45.11.1. Let $k = \mathbf{C}$ and $F = \mathbf{C}$ both be equal to the field of complex numbers. For $X$ smooth projective over $k$ denote $H^{p, q}(X) = H^ q(X, \Omega ^ p_{X/k})$. Let $(H')^*$ be the functor which sends $X$ to $(H')^*(X) = \bigoplus H^{p, q}(X)$ with the usual cup product. This is a classical Weil cohomology theory (insert future reference here). By Proposition 45.7.11 we obtain a $\mathbf{Q}$-linear symmetric monoidal functor $G'$ from $M_ k$ to the category of graded $F$-vector spaces. Of course, in this case for every $M$ in $M_ k$ the value $G'(M)$ is naturally bigraded, i.e., we have

$(G')(M) = \bigoplus (G')^{p, q}(M),\quad (G')^ n = \bigoplus \nolimits _{n = p + q} (G')^{p, q}(M)$

with $(G')^{p, q}$ sitting in total degree $p + q$ as indicated. Now we are going to construct a $\mathbf{Q}$-linear symmetric monoidal functor $G$ to the category of graded $F$-vector spaces by setting

$G^ n(M) = \bigoplus \nolimits _{n = 3p - q} (G')^{p, q}(M)$

We omit the verification that this defines a symmetric monoidal functor (a technical point is that because we chose odd numbers $3$ and $-1$ above the functor $G$ is compatible with the commutativity constraints). Observe that $G(\mathbf{1}(1))$ is still sitting in degree $-2$! Hence by Lemma 45.7.9 we obtain a functor $H^*$, cycle classes $\gamma$, and trace maps satisfying all classical axioms (A), (B), (C), except for possibly the classical axioms (A)(a) and (A)(d). However, if $E$ is an elliptic curve over $k$, then we find $\dim H^{-1}(E) = 1$, i.e., axiom (A)(a) is indeed violated.

Lemma 45.11.2. Assume given (D0), (D1), (D2), and (D3) satisfying (A), (B), and (C). Let $X$ be a smooth projective scheme over $k$. Set $k' = \Gamma (X, \mathcal{O}_ X)$. The following are equivalent

1. there exist finitely many closed points $x_1, \ldots , x_ r \in X$ whose residue fields are separable over $k$ such that $H^0(X) \to H^0(x_1) \oplus \ldots \oplus H^0(x_ r)$ is injective,

2. the map $H^0(\mathop{\mathrm{Spec}}(k')) \to H^0(X)$ is an isomorphism.

If $X$ is equidimensional of dimension $d$, these are also equivalent to

1. the classes of closed points generate $H^{2d}(X)(d)$ as a module over $H^0(X)$.

If this is true, then $H^0(X)$ is a finite separable algebra over $F$.

Proof. We observe that the statement makes sense because $k'$ is a finite separable algebra over $k$ (Varieties, Lemma 33.9.3) and hence $\mathop{\mathrm{Spec}}(k')$ is smooth and projective over $k$. The compatibility of $H^*$ with direct sums (Lemmas 45.9.9 and 45.10.1) shows that it suffices to prove the lemma when $X$ is connected. Hence we may assume $X$ is irreducible and we have to show the equivalence of (1), (2), and (3). Set $d = \dim (X)$. This implies that $k'$ is a field finite separable over $k$ and that $X$ is geometrically irreducible over $k'$, see Varieties, Lemmas 33.9.3 and 33.9.4.

By Lemma 45.8.1 we see that the closed points in (3) may be assumed to have separable residue fields over $k$. By axioms (A)(a) and (A)(b) we see that conditions (1) and (3) are equivalent.

If (2) holds, then pick any closed point $x \in X$ whose residue field is finite separable over $k'$. Then $H^0(\mathop{\mathrm{Spec}}(k')) = H^0(X) \to H^0(x)$ is injective for example by Lemma 45.10.8.

Assume the equivalent conditions (1) and (3) hold. Choose $x_1, \ldots , x_ r \in X$ as in (1). Choose a finite separable extension $k''/k'$. By Lemma 45.10.9 we have

$H^0(X) \otimes _{H^0(\mathop{\mathrm{Spec}}(k'))} H^0(\mathop{\mathrm{Spec}}(k'')) = H^0(X \times _{\mathop{\mathrm{Spec}}(k')} \mathop{\mathrm{Spec}}(k''))$

Thus in order to show that $H^0(\mathop{\mathrm{Spec}}(k')) \to H^0(X)$ is an isomorphism we may replace $k'$ by $k''$. Thus we may assume $x_1, \ldots , x_ r$ are $k'$-rational points (this replaces each $x_ i$ with multiple points, so $r$ is increased in this step). By Lemma 45.10.7 $\gamma (x_1) = \gamma (x_2) = \ldots = \gamma (x_ r)$. By axiom (A)(b) all the maps $H^0(X) \to H^0(x_ i)$ are the same. This means (2) holds.

Finally, Lemma 45.10.2 implies $H^0(X)$ is a separable $F$-algebra if (1) holds. $\square$

Lemma 45.11.3. Assume given (D0), (D1), (D2), and (D3) satisfying (A), (B), and (C). If there exists a smooth projective scheme $Y$ over $k$ such that $H^ i(Y)$ is nonzero for some $i < 0$, then there exists an equidimensional smooth projective scheme $X$ over $k$ such that the equivalent conditions of Lemma 45.11.2 fail for $X$.

Proof. By Lemma 45.9.9 we may assume $Y$ is irreducible and a fortiori equidimensional. If $i$ is odd, then after replacing $Y$ by $Y \times Y$ we find an example where $Y$ is equidimensional and $i = -2l$ for some $l > 0$. Set $X = Y \times (\mathbf{P}^1_ k)^ l$. Using axiom (B)(a) we obtain

$H^0(X) \supset H^0(Y) \oplus H^ i(Y) \otimes _ F H^2(\mathbf{P}^1_ k)^{\otimes _ F l}$

with both summands nonzero. Thus it is clear that $H^0(X)$ cannot be isomorphic to $H^0$ of the spectrum of $\Gamma (X, \mathcal{O}_ X) = \Gamma (Y, \mathcal{O}_ Y)$ as this falls into the first summand. $\square$

Thus it makes sense to finally make the following definition.

Definition 45.11.4. Let $k$ be a field. Let $F$ be a field of characteristic $0$. A Weil cohomology theory over $k$ with coefficients in $F$ is given by data (D0), (D1), (D2), and (D3) satisfying Poincaré duality, the Künneth formula, and compatibility with cycle classes, more precisely, satisfying axioms (A), (B), and (C) of Section 45.9 and in addition such that the equivalent conditions (1) and (2) of Lemma 45.11.2 hold for every smooth projective $X$ over $k$.

By Lemma 45.11.3 this means also that there are no nonzero negative cohomology groups. In particular, if $k$ is algebraically closed, then a Weil cohomology theory as above together with an isomorphism $F \to F(1)$ is the same thing as a classical Weil cohomology theory.

Remark 45.11.5. Let $H^*$ be a Weil cohomology theory (Definition 45.11.4). Let $X$ be a geometrically irreducible smooth projective scheme of dimension $d$ over $k'$ with $k'/k$ a finite separable extension of fields. Suppose that

$H^0(\mathop{\mathrm{Spec}}(k')) = F_1 \times \ldots \times F_ r$

for some fields $F_ i$. Then we accordingly can write

$H^*(X) = \prod \nolimits _{i = 1, \ldots , r} H^*(X) \otimes _{H^0(\mathop{\mathrm{Spec}}(k'))} F_ i$

Now, our final assumption in Definition 45.11.4 tells us that $H^0(X)$ is free of rank $1$ over $\prod F_ i$. In other words, each of the factors $H^0(X) \otimes _{H^0(\mathop{\mathrm{Spec}}(k'))} F_ i$ has dimension $1$ over $F_ i$. Poincaré duality then tells us that the same is true for cohomology in degree $2d$. What isn't clear however is that the same holds in other degrees. Namely, we don't know that given $0 < n < \dim (X)$ the integers

$\dim _{F_ i} H^ n(X) \otimes _{H^0(\mathop{\mathrm{Spec}}(k'))} F_ i$

are independent of $i$! This question is closely related to the following open question: given an algebraically closed base field $\overline{k}$, a field of characteristic zero $F$, a classical Weil cohomology theory $H^*$ over $\overline{k}$ with coefficient field $F$, and a smooth projective variety $X$ over $\overline{k}$ is it true that the betti numbers of $X$

$\beta _ i = \dim _ F H^ i(X)$

are independent of $F$ and the Weil cohomology theory $H^*$?

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