81.21 Intersecting with an invertible sheaf and rational equivalence

This section is the analogue of Chow Homology, Section 42.28. Applying the key lemma we obtain the fundamental properties of intersecting with invertible sheaves. In particular, we will see that $c_1(\mathcal{L}) \cap -$ factors through rational equivalence and that these operations for different invertible sheaves commute.

Lemma 81.21.1. In Situation 81.2.1 let $X/B$ be good. Assume $X$ integral and $\dim _\delta (X) = n$. Let $\mathcal{L}$, $\mathcal{N}$ be invertible on $X$. Choose a nonzero meromorphic section $s$ of $\mathcal{L}$ and a nonzero meromorphic section $t$ of $\mathcal{N}$. Set $\alpha = \text{div}_\mathcal {L}(s)$ and $\beta = \text{div}_\mathcal {N}(t)$. Then

$c_1(\mathcal{N}) \cap \alpha = c_1(\mathcal{L}) \cap \beta$

in $\mathop{\mathrm{CH}}\nolimits _{n - 2}(X)$.

Proof. Immediate from the key Lemma 81.20.1 and the discussion preceding it. $\square$

Lemma 81.21.2. In Situation 81.2.1 let $X/B$ be good. Let $\mathcal{L}$ be invertible on $X$. The operation $\alpha \mapsto c_1(\mathcal{L}) \cap \alpha$ factors through rational equivalence to give an operation

$c_1(\mathcal{L}) \cap - : \mathop{\mathrm{CH}}\nolimits _{k + 1}(X) \to \mathop{\mathrm{CH}}\nolimits _ k(X)$

Proof. Let $\alpha \in Z_{k + 1}(X)$, and $\alpha \sim _{rat} 0$. We have to show that $c_1(\mathcal{L}) \cap \alpha$ as defined in Definition 81.18.1 is zero. By Definition 81.15.1 there exists a locally finite family $\{ W_ j\}$ of integral closed subspaces with $\dim _\delta (W_ j) = k + 2$ and rational functions $f_ j \in R(W_ j)^*$ such that

$\alpha = \sum (i_ j)_*\text{div}_{W_ j}(f_ j)$

Note that $p : \coprod W_ j \to X$ is a proper morphism, and hence $\alpha = p_*\alpha '$ where $\alpha ' \in Z_{k + 1}(\coprod W_ j)$ is the sum of the principal divisors $\text{div}_{W_ j}(f_ j)$. By Lemma 81.19.4 we have $c_1(\mathcal{L}) \cap \alpha = p_*(c_1(p^*\mathcal{L}) \cap \alpha ')$. Hence it suffices to show that each $c_1(\mathcal{L}|_{W_ j}) \cap \text{div}_{W_ j}(f_ j)$ is zero. In other words we may assume that $X$ is integral and $\alpha = \text{div}_ X(f)$ for some $f \in R(X)^*$.

Assume $X$ is integral and $\alpha = \text{div}_ X(f)$ for some $f \in R(X)^*$. We can think of $f$ as a regular meromorphic section of the invertible sheaf $\mathcal{N} = \mathcal{O}_ X$. Choose a meromorphic section $s$ of $\mathcal{L}$ and denote $\beta = \text{div}_\mathcal {L}(s)$. By Lemma 81.21.1 we conclude that

$c_1(\mathcal{L}) \cap \alpha = c_1(\mathcal{O}_ X) \cap \beta .$

However, by Lemma 81.18.2 we see that the right hand side is zero in $\mathop{\mathrm{CH}}\nolimits _ k(X)$ as desired. $\square$

In Situation 81.2.1 let $X/B$ be good. Let $\mathcal{L}$ be invertible on $X$. We will denote

$c_1(\mathcal{L})^ s \cap - : \mathop{\mathrm{CH}}\nolimits _{k + s}(X) \to \mathop{\mathrm{CH}}\nolimits _ k(X)$

the operation $c_1(\mathcal{L}) \cap -$. This makes sense by Lemma 81.21.2. We will denote $c_1(\mathcal{L}^ s \cap -$ the $s$-fold iterate of this operation for all $s \geq 0$.

Lemma 81.21.3. In Situation 81.2.1 let $X/B$ be good. Let $\mathcal{L}$, $\mathcal{N}$ be invertible on $X$. For any $\alpha \in \mathop{\mathrm{CH}}\nolimits _{k + 2}(X)$ we have

$c_1(\mathcal{L}) \cap c_1(\mathcal{N}) \cap \alpha = c_1(\mathcal{N}) \cap c_1(\mathcal{L}) \cap \alpha$

as elements of $\mathop{\mathrm{CH}}\nolimits _ k(X)$.

Proof. Write $\alpha = \sum m_ j[Z_ j]$ for some locally finite collection of integral closed subspaces $Z_ j \subset X$ with $\dim _\delta (Z_ j) = k + 2$. Consider the proper morphism $p : \coprod Z_ j \to X$. Set $\alpha ' = \sum m_ j[Z_ j]$ as a $(k + 2)$-cycle on $\coprod Z_ j$. By several applications of Lemma 81.19.4 we see that $c_1(\mathcal{L}) \cap c_1(\mathcal{N}) \cap \alpha = p_*(c_1(p^*\mathcal{L}) \cap c_1(p^*\mathcal{N}) \cap \alpha ')$ and $c_1(\mathcal{N}) \cap c_1(\mathcal{L}) \cap \alpha = p_*(c_1(p^*\mathcal{N}) \cap c_1(p^*\mathcal{L}) \cap \alpha ')$. Hence it suffices to prove the formula in case $X$ is integral and $\alpha = [X]$. In this case the result follows from Lemma 81.21.1 and the definitions. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).