## 81.26 Bivariant intersection theory

This section is the analogue of Chow Homology, Section 42.33. In order to intelligently talk about higher Chern classes of vector bundles we introduce the following notion, following [FM]. It follows from [Theorem 17.1, F] that our definition agrees with that of [F] modulo the caveat that we are working in different settings.

reference
Definition 81.26.1. In Situation 81.2.1 let $f : X \to Y$ be a morphism of good algebraic spaces over $B$. Let $p \in \mathbf{Z}$. A *bivariant class $c$ of degree $p$ for $f$* is given by a rule which assigns to every morphism $Y' \to Y$ of good algebraic spaces over $B$ and every $k$ a map

\[ c \cap - : \mathop{\mathrm{CH}}\nolimits _ k(Y') \longrightarrow \mathop{\mathrm{CH}}\nolimits _{k - p}(X') \]

where $X' = Y' \times _ Y X$, satisfying the following conditions

if $Y'' \to Y'$ is a proper morphism, then $c \cap (Y'' \to Y')_*\alpha '' = (X'' \to X')_*(c \cap \alpha '')$ for all $\alpha ''$ on $Y''$,

if $Y'' \to Y'$ a morphism of good algebraic spaces over $B$ which is flat of relative dimension $r$, then $c \cap (Y'' \to Y')^*\alpha ' = (X'' \to X')^*(c \cap \alpha ')$ for all $\alpha '$ on $Y'$,

if $(\mathcal{L}', s', i' : D' \to Y')$ is as in Definition 81.22.1 with pullback $(\mathcal{N}', t', j' : E' \to X')$ to $X'$, then we have $c \cap (i')^*\alpha ' = (j')^*(c \cap \alpha ')$ for all $\alpha '$ on $Y'$.

The collection of all bivariant classes of degree $p$ for $f$ is denoted $A^ p(X \to Y)$.

In Situation 81.2.1 let $X \to Y$ and $Y \to Z$ be morphisms of good algebraic spaces over $B$. Let $p \in \mathbf{Z}$. It is clear that $A^ p(X \to Y)$ is an abelian group. Moreover, it is clear that we have a bilinear composition

\[ A^ p(X \to Y) \times A^ q(Y \to Z) \to A^{p + q}(X \to Z) \]

which is associative. We will be most interested in $A^ p(X) = A^ p(X \to X)$, which will always mean the bivariant cohomology classes for $\text{id}_ X$. Namely, that is where Chern classes will live.

Definition 81.26.2. In Situation 81.2.1 let $X/B$ be good. The *Chow cohomology* of $X$ is the graded $\mathbf{Z}$-algebra $A^*(X)$ whose degree $p$ component is $A^ p(X \to X)$.

Warning: It is not clear that the $\mathbf{Z}$-algebra structure on $A^*(X)$ is commutative, but we will see that Chern classes live in its center.

Lemma 81.26.4. In Situation 81.2.1 let $X/B$ be good. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Then the rule that to $f : X' \to X$ assigns $c_1(f^*\mathcal{L}) \cap - : \mathop{\mathrm{CH}}\nolimits _ k(X') \to \mathop{\mathrm{CH}}\nolimits _{k - 1}(X')$ is a bivariant class of degree $1$.

**Proof.**
This follows from Lemmas 81.21.2, 81.19.4, 81.19.2, and 81.23.3.
$\square$

Lemma 81.26.5. In Situation 81.2.1 let $f : X \to Y$ be a morphism of good algebraic spaces over $B$ which is flat of relative dimension $r$. Then the rule that to $Y' \to Y$ assigns $(f')^* : \mathop{\mathrm{CH}}\nolimits _ k(Y') \to \mathop{\mathrm{CH}}\nolimits _{k + r}(X')$ where $X' = X \times _ Y Y'$ is a bivariant class of degree $-r$.

**Proof.**
This follows from Lemmas 81.16.2, 81.10.4, 81.11.1, and 81.22.6.
$\square$

Lemma 81.26.6. In Situation 81.2.1 let $X/B$ be good. Let $(\mathcal{L}, s, i : D \to X)$ be a triple as in Definition 81.22.1. Then the rule that to $f : X' \to X$ assigns $(i')^* : \mathop{\mathrm{CH}}\nolimits _ k(X') \to \mathop{\mathrm{CH}}\nolimits _{k - 1}(D')$ where $D' = D \times _ X X'$ is a bivariant class of degree $1$.

**Proof.**
This follows from Lemmas 81.23.2, 81.22.5, 81.22.6, and 81.23.4.
$\square$

Lemma 81.26.7. In Situation 81.2.1 let $f : X \to Y$ and $g : Y \to Z$ be morphisms of good algebraic spaces over $B$. Let $c \in A^ p(X \to Z)$ and assume $f$ is proper. Then the rule that to $X' \to X$ assigns $\alpha \longmapsto f_*(c \cap \alpha )$ is a bivariant class of degree $p$.

**Proof.**
This follows from Lemmas 81.8.2, 81.11.1, and 81.22.5.
$\square$

Here we see that $c_1(\mathcal{L})$ is in the center of $A^*(X)$.

Lemma 81.26.8. In Situation 81.2.1 let $X/B$ be good. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Then $c_1(\mathcal{L}) \in A^1(X)$ commutes with every element $c \in A^ p(X)$.

**Proof.**
Let $p : L \to X$ be as in Lemma 81.25.2 and let $o : X \to L$ be the zero section. Observe that $p^*\mathcal{L}^{\otimes -1}$ has a canonical section whose vanishing locus is exactly the effective Cartier divisor $o(X)$. Let $\alpha \in \mathop{\mathrm{CH}}\nolimits _ k(X)$. Then we see that

\[ p^*(c_1(\mathcal{L}^{\otimes -1}) \cap \alpha ) = c_1(p^*\mathcal{L}^{\otimes -1}) \cap p^*\alpha = o_* o^* p^*\alpha \]

by Lemmas 81.19.2 and 81.24.1. Since $c$ is a bivariant class we have

\begin{align*} p^*(c \cap c_1(\mathcal{L}^{\otimes -1}) \cap \alpha ) & = c \cap p^*(c_1(\mathcal{L}^{\otimes -1}) \cap \alpha ) \\ & = c \cap o_* o^* p^*\alpha \\ & = o_* o^* p^*(c \cap \alpha ) \\ & = p^*(c_1(\mathcal{L}^{\otimes -1}) \cap c \cap \alpha ) \end{align*}

(last equality by the above applied to $c \cap \alpha $). Since $p^*$ is injective by a lemma cited above we get that $c_1(\mathcal{L}^{\otimes -1})$ is in the center of $A^*(X)$. This proves the lemma.
$\square$

Here a criterion for when a bivariant class is zero.

Lemma 81.26.9. In Situation 81.2.1 let $X/B$ be good. Let $c \in A^ p(X)$. Then $c$ is zero if and only if $c \cap [Y] = 0$ in $\mathop{\mathrm{CH}}\nolimits _*(Y)$ for every integral algebraic space $Y$ locally of finite type over $X$.

**Proof.**
The if direction is clear. For the converse, assume that $c \cap [Y] = 0$ in $\mathop{\mathrm{CH}}\nolimits _*(Y)$ for every integral algebraic space $Y$ locally of finite type over $X$. Let $X' \to X$ be locally of finite type. Let $\alpha \in \mathop{\mathrm{CH}}\nolimits _ k(X')$. Write $\alpha = \sum n_ i [Y_ i]$ with $Y_ i \subset X'$ a locally finite collection of integral closed subschemes of $\delta $-dimension $k$. Then we see that $\alpha $ is pushforward of the cycle $\alpha ' = \sum n_ i[Y_ i]$ on $X'' = \coprod Y_ i$ under the proper morphism $X'' \to X'$. By the properties of bivariant classes it suffices to prove that $c \cap \alpha ' = 0$ in $\mathop{\mathrm{CH}}\nolimits _{k - p}(X'')$. We have $\mathop{\mathrm{CH}}\nolimits _{k - p}(X'') = \prod \mathop{\mathrm{CH}}\nolimits _{k - p}(Y_ i)$ as follows immediately from the definitions. The projection maps $\mathop{\mathrm{CH}}\nolimits _{k - p}(X'') \to \mathop{\mathrm{CH}}\nolimits _{k - p}(Y_ i)$ are given by flat pullback. Since capping with $c$ commutes with flat pullback, we see that it suffices to show that $c \cap [Y_ i]$ is zero in $\mathop{\mathrm{CH}}\nolimits _{k - p}(Y_ i)$ which is true by assumption.
$\square$

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