## 81.19 Intersecting with an invertible sheaf and push and pull

This section is the analogue of Chow Homology, Section 42.26. In this section we prove that the operation $c_1(\mathcal{L}) \cap -$ commutes with flat pullback and proper pushforward.

Lemma 81.19.1. In Situation 81.2.1 let $X, Y/B$ be good. Let $f : X \to Y$ be a flat morphism of relative dimension $r$. Let $\mathcal{L}$ be an invertible sheaf on $Y$. Assume $Y$ is integral and $n = \dim _\delta (Y)$. Let $s$ be a nonzero meromorphic section of $\mathcal{L}$. Then we have

$f^*\text{div}_\mathcal {L}(s) = \sum n_ i\text{div}_{f^*\mathcal{L}|_{X_ i}}(s_ i)$

in $Z_{n + r - 1}(X)$. Here the sum is over the irreducible components $X_ i \subset X$ of $\delta$-dimension $n + r$, the section $s_ i = f|_{X_ i}^*(s)$ is the pullback of $s$, and $n_ i = m_{X_ i, X}$ is the multiplicity of $X_ i$ in $X$.

Proof. Using sleight of hand we will deduce this from Lemma 81.16.1. (An alternative is to redo the proof of that lemma in the setting of meromorphic sections of invertible modules.) Namely, let $q : T \to Y$ be the morphism of Lemma 81.17.3 constructed using $\mathcal{L}$ on $Y$. We will use all the properties of $T$ stated in this lemma. Consider the fibre product diagram

$\xymatrix{ T' \ar[r]_{q'} \ar[d]_ h & X \ar[d]^ f \\ T \ar[r]^ q & Y }$

Then $q' : T' \to X$ is the morphism constructed using $f^*\mathcal{L}$ on $X$. Then it suffices to prove

$(q')^*f^*\text{div}_\mathcal {L}(s) = \sum n_ i (q')^*\text{div}_{f^*\mathcal{L}|_{X_ i}}(s_ i)$

Observe that $T'_ i = q^{-1}(X_ i)$ are the irreducible components of $T'$ and that $n_ i$ is the multiplicity of $T'_ i$ in $T'$. The left hand side is equal to

$h^*q^*\text{div}_\mathcal {L}(s) = h^*\text{div}_ T(q^*(s))$

by Lemma 81.17.4 (and Lemma 81.10.4). On the other hand, denoting $q'_ i : T'_ i \to X_ i$ the restriction of $q'$ we find that Lemma 81.17.4 also tells us the right hand side is equal to

$\sum n_ i \text{div}_{T_ i}((q'_ i)^*(s_ i))$

In these two formulas the expressions $q^*(s)$ and $(q'_ i)^*(s_ i)$ represent the rational functions corresponding to the pulled back meromorphic sections of $q^*\mathcal{L}$ and $(q'_ i)^*f^*\mathcal{L}|_{X_ i}$ via the isomorphism $\alpha : q^*\mathcal{L} \to \mathcal{O}_ T$ and its pullbacks to spaces over $T$. With this convention it is clear that $(q'_ i)^*(s_ i)$ is the composition of the rational function $q^*(s)$ on $T$ and the morphism $h|_{T'_ i} : T'_ i \to T$. Thus Lemma 81.16.1 exactly says that

$h^*\text{div}_ T(q^*(s)) = \sum n_ i \text{div}_{T_ i}((q'_ i)^*(s_ i))$

as desired. $\square$

Lemma 81.19.2. In Situation 81.2.1 let $X, Y/B$ be good. Let $f : X \to Y$ be a flat morphism of relative dimension $r$. Let $\mathcal{L}$ be an invertible sheaf on $Y$. Let $\alpha$ be a $k$-cycle on $Y$. Then

$f^*(c_1(\mathcal{L}) \cap \alpha ) = c_1(f^*\mathcal{L}) \cap f^*\alpha$

in $\mathop{\mathrm{CH}}\nolimits _{k + r - 1}(X)$.

Proof. Write $\alpha = \sum n_ i[W_ i]$. We will show that

$f^*(c_1(\mathcal{L}) \cap [W_ i]) = c_1(f^*\mathcal{L}) \cap f^*[W_ i]$

in $\mathop{\mathrm{CH}}\nolimits _{k + r - 1}(X)$ by producing a rational equivalence on the closed subspace $f^{-1}(W_ i)$ of $X$. By the discussion in Remark 81.15.3 this will prove the equality of the lemma is true.

Let $W \subset Y$ be an integral closed subspace of $\delta$-dimension $k$. Consider the closed subspace $W' = f^{-1}(W) = W \times _ Y X$ so that we have the fibre product diagram

$\xymatrix{ W' \ar[r] \ar[d]_ h & X \ar[d]^ f \\ W \ar[r] & Y }$

We have to show that $f^*(c_1(\mathcal{L}) \cap [W]) = c_1(f^*\mathcal{L}) \cap f^*[W]$. Choose a nonzero meromorphic section $s$ of $\mathcal{L}|_ W$. Let $W'_ i \subset W'$ be the irreducible components of $\delta$-dimension $k + r$. Write $[W']_{k + r} = \sum n_ i[W'_ i]$ with $n_ i$ the multiplicity of $W'_ i$ in $W'$ as per definition. So $f^*[W] = \sum n_ i[W'_ i]$ in $Z_{k + r}(X)$. Since each $W'_ i \to W$ is dominant we see that $s_ i = s|_{W'_ i}$ is a nonzero meromorphic section for each $i$. By Lemma 81.19.1 we have the following equality of cycles

$h^*\text{div}_{\mathcal{L}|_ W}(s) = \sum n_ i\text{div}_{f^*\mathcal{L}|_{W'_ i}}(s_ i)$

in $Z_{k + r - 1}(W')$. This finishes the proof since the left hand side is a cycle on $W'$ which pushes to $f^*(c_1(\mathcal{L}) \cap [W])$ in $\mathop{\mathrm{CH}}\nolimits _{k + r - 1}(X)$ and the right hand side is a cycle on $W'$ which pushes to $c_1(f^*\mathcal{L}) \cap f^*[W]$ in $\mathop{\mathrm{CH}}\nolimits _{k + r - 1}(X)$. $\square$

Lemma 81.19.3. In Situation 81.2.1 let $X, Y/B$ be good. Let $f : X \to Y$ be a proper morphism. Let $\mathcal{L}$ be an invertible sheaf on $Y$. Assume $X$, $Y$ integral, $f$ dominant, and $\dim _\delta (X) = \dim _\delta (Y)$. Let $s$ be a nonzero meromorphic section $s$ of $\mathcal{L}$ on $Y$. Then

$f_*\left(\text{div}_{f^*\mathcal{L}}(f^*s)\right) = [R(X) : R(Y)]\text{div}_\mathcal {L}(s).$

as cycles on $Y$. In particular

$f_*(c_1(f^*\mathcal{L}) \cap [X]) = c_1(\mathcal{L}) \cap f_*[Y].$

Proof. The last equation follows from the first since $f_*[X] = [R(X) : R(Y)][Y]$ by definition. Proof of the first equaltion. Let $q : T \to Y$ be the morphism of Lemma 81.17.3 constructed using $\mathcal{L}$ on $Y$. We will use all the properties of $T$ stated in this lemma. Consider the fibre product diagram

$\xymatrix{ T' \ar[r]_{q'} \ar[d]_ h & X \ar[d]^ f \\ T \ar[r]^ q & Y }$

Then $q' : T' \to X$ is the morphism constructed using $f^*\mathcal{L}$ on $X$. It suffices to prove the equality after pulling back to $T'$. The left hand side pulls back to

\begin{align*} q^*f_*\left(\text{div}_{f^*\mathcal{L}}(f^*s)\right) & = h_*(q')^*\text{div}_{f^*\mathcal{L}}(f^*s) \\ & = h_*\text{div}_{(q')^*f^*\mathcal{L}}((q')^*f^*s) \\ & = h_*\text{div}_{h^*q^*\mathcal{L}}(h^*q^*s) \end{align*}

The first equality by Lemma 81.11.1. The second by Lemma 81.19.1 using that $T'$ is integral. The third because the displayed diagram commutes. The right hand side pulls back to

$[R(X) : R(Y)]q^*\text{div}_\mathcal {L}(s) = [R(T') : R(T)]\text{div}_{q^*\mathcal{L}}(q^*s)$

This follows from Lemma 81.19.1, the fact that $T$ is integral, and the equality $[R(T') : R(T)] = [R(X) : R(Y)]$ whose proof we omit (it follows from Lemma 81.11.1 but that would be a silly way to prove the equality). Thus it suffices to prove the lemma for $h : T' \to T$, the invertible module $q^\mathcal {L}$ and the section $q^*s$. Since $q^*\mathcal{L} = \mathcal{O}_ T$ we reduce to the case where $\mathcal{L} \cong \mathcal{O}$ discussed in the next paragraph.

Assume that $\mathcal{L} = \mathcal{O}_ Y$. In this case $s$ corresponds to a rational function $g \in R(Y)$. Using the embedding $R(Y) \subset R(X)$ we may think of $g$ as a rational on $X$ and we are simply trying to prove

$f_*\left(\text{div}_ X(g)\right) = [R(X) : R(Y)]\text{div}_ Y(g).$

Comparing with the result of Lemma 81.14.1 we see this true since $\text{Nm}_{R(X)/R(Y)}(g) = g^{[R(X) : R(Y)]}$ as $g \in R(Y)^*$. $\square$

Lemma 81.19.4. In Situation 81.2.1 let $X, Y/B$ be good. Let $p : X \to Y$ be a proper morphism. Let $\alpha \in Z_{k + 1}(X)$. Let $\mathcal{L}$ be an invertible sheaf on $Y$. Then

$p_*(c_1(p^*\mathcal{L}) \cap \alpha ) = c_1(\mathcal{L}) \cap p_*\alpha$

in $\mathop{\mathrm{CH}}\nolimits _ k(Y)$.

Proof. Suppose that $p$ has the property that for every integral closed subspace $W \subset X$ the map $p|_ W : W \to Y$ is a closed immersion. Then, by definition of capping with $c_1(\mathcal{L})$ the lemma holds.

We will use this remark to reduce to a special case. Namely, write $\alpha = \sum n_ i[W_ i]$ with $n_ i \not= 0$ and $W_ i$ pairwise distinct. Let $W'_ i \subset Y$ be the “image” of $W_ i$ as in Lemma 81.7.1. Consider the diagram

$\xymatrix{ X' = \coprod W_ i \ar[r]_-q \ar[d]_{p'} & X \ar[d]^ p \\ Y' = \coprod W'_ i \ar[r]^-{q'} & Y. }$

Since $\{ W_ i\}$ is locally finite on $X$, and $p$ is proper we see that $\{ W'_ i\}$ is locally finite on $Y$ and that $q, q', p'$ are also proper morphisms. We may think of $\sum n_ i[W_ i]$ also as a $k$-cycle $\alpha ' \in Z_ k(X')$. Clearly $q_*\alpha ' = \alpha$. We have $q_*(c_1(q^*p^*\mathcal{L}) \cap \alpha ') = c_1(p^*\mathcal{L}) \cap q_*\alpha '$ and $(q')_*(c_1((q')^*\mathcal{L}) \cap p'_*\alpha ') = c_1(\mathcal{L}) \cap q'_*p'_*\alpha '$ by the initial remark of the proof. Hence it suffices to prove the lemma for the morphism $p'$ and the cycle $\sum n_ i[W_ i]$. Clearly, this means we may assume $X$, $Y$ integral, $f : X \to Y$ dominant and $\alpha = [X]$. In this case the result follows from Lemma 81.19.3. $\square$

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