The Stacks project

Lemma 45.9.11. Let $k$ be a field. Let $F$ be a field of characteristic $0$. Given (D0), (D1), (D2), and (D3) satisfying (A), (B), and (C) we can construct a $\mathbf{Q}$-linear functor

\[ G : M_ k \longrightarrow \text{graded }F\text{-vector spaces} \]

of symmetric monoidal categories such that $H^*(X) = G(h(X))$.

Proof. The proof of this lemma is the same as the proof of Lemma 45.7.10; we urge the reader to read the proof of that lemma instead.

By Lemma 45.4.8 it suffices to construct a functor $G$ on the category of smooth projective schemes over $k$ with morphisms given by correspondences of degree $0$ such that the image of $G(c_2)$ on $G(\mathbf{P}^1_ k)$ is an invertible graded $F$-vector space.

Let $X$ be a smooth projective scheme over $k$. There is a canonical decomposition

\[ X = \coprod \nolimits _{0 \leq d \le \dim (X)} X_ d \]

into open and closed subschemes such that $X_ d$ is equidimensional of dimension $d$. By Lemma 45.9.9 we have correspondingly

\[ H^*(X) \longrightarrow \prod \nolimits _{0 \leq d \le \dim (X)} H^*(X_ d) \]

If $Y$ is a second smooth projective scheme over $k$ and we similarly decompose $Y = \coprod Y_ e$, then

\[ \text{Corr}^0(X, Y) = \bigoplus \text{Corr}^0(X_ d, Y_ e) \]

As well we have $X \otimes Y = \coprod X_ d \otimes Y_ e$ in the category of correspondences. From these observations it follows that it suffices to construct $G$ on the category whose objects are equidimensional smooth projective schemes over $k$ and whose morphisms are correspondences of degree $0$. (Some details omitted.)

Given an equdimensional smooth projective scheme $X$ over $k$ we set $G(X) = H^*(X)$. Observe that $G(X) = 0$ if $X = \emptyset $ (Lemma 45.9.5). Thus maps from and to $G(\emptyset )$ are zero and we may and do assume our schemes are nonempty in the discussions below.

Given a correspondence $c \in \text{Corr}^0(X, Y)$ between nonempty equidmensional smooth projective schemes over $k$ we consider the map $G(c) : G(X) = H^*(X) \to G(Y) = H^*(Y)$ given by the rule

\[ a \longmapsto G(c)(a) = \text{pr}_{2, *}(\gamma (c) \cup \text{pr}_1^*a) \]

It is clear that $G(c)$ is additive in $c$ and hence $\mathbf{Q}$-linear. Compatibility of $\gamma $ with pullbacks, pushforwards, and intersection products given by axioms (C)(a), (C)(b), and (C)(c) shows that we have $G(c' \circ c) = G(c') \circ G(c)$ if $c' \in \text{Corr}^0(Y, Z)$. Namely, for $a \in H^*(X)$ we have

\begin{align*} (G(c') \circ G(c))(a) & = \text{pr}^{23}_{3, *}(\gamma (c') \cup \text{pr}^{23, *}_2(\text{pr}^{12}_{2, *}(\gamma (c) \cup \text{pr}^{12, *}_1a))) \\ & = \text{pr}^{23}_{3, *}(\gamma (c') \cup \text{pr}^{123}_{23, *}(\text{pr}^{123, *}_{12}(\gamma (c) \cup \text{pr}^{12, *}_1 a))) \\ & = \text{pr}^{23}_{3, *} \text{pr}^{123}_{23, *}( \text{pr}^{123, *}_{23}\gamma (c') \cup \text{pr}^{123, *}_{12}\gamma (c) \cup \text{pr}^{123, *}_1 a) \\ & = \text{pr}^{23}_{3, *} \text{pr}^{123}_{23, *}( \gamma (\text{pr}^{123, *}_{23}c') \cup \gamma (\text{pr}^{123, *}_{12}c) \cup \text{pr}^{123, *}_1 a) \\ & = \text{pr}^{13}_{3, *} \text{pr}^{123}_{13, *}( \gamma (\text{pr}^{123, *}_{23}c' \cdot \text{pr}^{123, *}_{12}c) \cup \text{pr}^{123, *}_1 a) \\ & = \text{pr}^{13}_{3, *}( \gamma (\text{pr}^{123}_{13, *}(\text{pr}^{123, *}_{23}c' \cdot \text{pr}^{123, *}_{12}c)) \cup \text{pr}^{13, *}_1 a) \\ & = G(c' \circ c)(a) \end{align*}

with obvious notation. The first equality follows from the definitions. The second equality holds because $\text{pr}^{23, *}_2 \circ \text{pr}^{12}_{2, *} = \text{pr}^{123}_{23, *} \circ \text{pr}^{123, *}_{12}$ as follows immediately from the description of pushforward along projections given in Lemma 45.9.2. The third equality holds by Lemma 45.9.1 and the fact that $H^*$ is a functor. The fourth equalith holds by axiom (C)(a) and the fact that the gysin map agrees with flat pullback for flat morphisms (Chow Homology, Lemma 42.59.5). The fifth equality uses axiom (C)(c) as well as Lemma 45.9.1 to see that $\text{pr}^{23}_{3, *} \circ \text{pr}^{123}_{23, *} = \text{pr}^{13}_{3, *} \circ \text{pr}^{123}_{13, *}$. The sixth equality uses the projection formula from Lemma 45.9.1 as well as axiom (C)(b) to see that $ \text{pr}^{123}_{13, *} \gamma (\text{pr}^{123, *}_{23}c' \cdot \text{pr}^{123, *}_{12}c) = \gamma (\text{pr}^{123}_{13, *}( \text{pr}^{123, *}_{23}c' \cdot \text{pr}^{123, *}_{12}c))$. Finally, the last equality is the definition.

To finish the proof that $G$ is a functor, we have to show identities are preserved. In other words, if $1 = [\Delta ] \in \text{Corr}^0(X, X)$ is the identity in the category of correspondences (Lemma 45.3.3), then we have to show that $G([\Delta ]) = \text{id}$. This follows from the determination of $\gamma ([\Delta ])$ in Lemma 45.9.7 and Lemma 45.9.2. This finishes the construction of $G$ as a functor on smooth projective schemes over $k$ and correspondences of degree $0$.

By Lemma 45.9.4 we have that $G(\mathop{\mathrm{Spec}}(k)) = H^*(\mathop{\mathrm{Spec}}(k))$ is canonically isomorphic to $F$ as an $F$-algebra. The K√ľnneth axiom (B)(a) shows our functor is compatible with tensor products. Thus our functor is a functor of symmetric monoidal categories.

We still have to check that the image of $G(c_2)$ on $G(\mathbf{P}^1_ k) = H^*(\mathbf{P}^1_ k)$ is an invertible graded $F$-vector space (in particular we don't know yet that $G$ extends to $M_ k$). By Lemma 45.9.8 we only have nonzero cohomology in degrees $0$ and $2$ both of dimension $1$. We have $1 = c_0 + c_2$ is a decomposition of the identity into a sum of orthogonal idempotents in $\text{Corr}^0(\mathbf{P}^1_ k, \mathbf{P}^1_ k)$, see Example 45.3.7. Further we have $c_0 = a \circ b$ where $a \in \text{Corr}^0(\mathop{\mathrm{Spec}}(k), \mathbf{P}^1_ k)$ and $b \in \text{Corr}^0(\mathbf{P}^1_ k, \mathop{\mathrm{Spec}}(k))$ and where $b \circ a = 1$ in $\text{Corr}^0(\mathop{\mathrm{Spec}}(k), \mathop{\mathrm{Spec}}(k))$, see proof of Lemma 45.4.4. Thus $G(c_0)$ is the projector onto the degree $0$ part. It follows that $G(c_2)$ must be the projector onto the degree $2$ part and the proof is complete. $\square$

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