## 81.23 Gysin homomorphisms

This section is the analogue of Chow Homology, Section 42.30. In this section we use the key formula to show the Gysin homomorphism factor through rational equivalence.

Lemma 81.23.1. In Situation 81.2.1 let $X/B$ be good. Assume $X$ integral and $n = \dim _\delta (X)$. Let $i : D \to X$ be an effective Cartier divisor. Let $\mathcal{N}$ be an invertible $\mathcal{O}_ X$-module and let $t$ be a nonzero meromorphic section of $\mathcal{N}$. Then $i^*\text{div}_\mathcal {N}(t) = c_1(\mathcal{N}) \cap [D]_{n - 1}$ in $\mathop{\mathrm{CH}}\nolimits _{n - 2}(D)$.

Proof. Write $\text{div}_\mathcal {N}(t) = \sum \text{ord}_{Z_ i, \mathcal{N}}(t)[Z_ i]$ for some integral closed subspaces $Z_ i \subset X$ of $\delta$-dimension $n - 1$. We may assume that the family $\{ Z_ i\}$ is locally finite, that $t \in \Gamma (U, \mathcal{N}|_ U)$ is a generator where $U = X \setminus \bigcup Z_ i$, and that every irreducible component of $D$ is one of the $Z_ i$, see Spaces over Fields, Lemmas 71.6.1, 71.6.6, and 71.7.2.

Set $\mathcal{L} = \mathcal{O}_ X(D)$. Denote $s \in \Gamma (X, \mathcal{O}_ X(D)) = \Gamma (X, \mathcal{L})$ the canonical section. We will apply the discussion of Section 81.20 to our current situation. For each $i$ let $\xi _ i \in |Z_ i|$ be its generic point. Let $B_ i = \mathcal{O}_{X, \xi _ i}^ h$. For each $i$ we pick generators $s_ i$ of $\mathcal{L}_{\xi _ i}$ and $t_ i$ of $\mathcal{N}_{\xi _ i}$ over $B_ i$ but we insist that we pick $s_ i = s$ if $Z_ i \not\subset D$. Write $s = f_ i s_ i$ and $t = g_ i t_ i$ with $f_ i, g_ i \in B_ i$. Then $\text{ord}_{Z_ i, \mathcal{N}}(t) = \text{ord}_{B_ i}(g_ i)$. On the other hand, we have $f_ i \in B_ i$ and

$[D]_{n - 1} = \sum \text{ord}_{B_ i}(f_ i)[Z_ i]$

because of our choices of $s_ i$. We claim that

$i^*\text{div}_\mathcal {N}(t) = \sum \text{ord}_{B_ i}(g_ i) \text{div}_{\mathcal{L}|_{Z_ i}}(s_ i|_{Z_ i})$

as cycles. More precisely, the right hand side is a cycle representing the left hand side. Namely, this is clear by our formula for $\text{div}_\mathcal {N}(t)$ and the fact that $\text{div}_{\mathcal{L}|_{Z_ i}}(s_ i|_{Z_ i}) = [Z(s_ i|_{Z_ i})]_{n - 2} = [Z_ i \cap D]_{n - 2}$ when $Z_ i \not\subset D$ because in that case $s_ i|_{Z_ i} = s|_{Z_ i}$ is a regular section, see Lemma 81.17.2. Similarly,

$c_1(\mathcal{N}) \cap [D]_{n - 1} = \sum \text{ord}_{B_ i}(f_ i) \text{div}_{\mathcal{N}|_{Z_ i}}(t_ i|_{Z_ i})$

The key formula (Lemma 81.20.1) gives the equality

$\sum \left( \text{ord}_{B_ i}(f_ i) \text{div}_{\mathcal{N}|_{Z_ i}}(t_ i|_{Z_ i}) - \text{ord}_{B_ i}(g_ i) \text{div}_{\mathcal{L}|_{Z_ i}}(s_ i|_{Z_ i}) \right) = \sum \text{div}_{Z_ i}(\partial _{B_ i}(f_ i, g_ i))$

of cycles. If $Z_ i \not\subset D$, then $f_ i = 1$ and hence $\text{div}_{Z_ i}(\partial _{B_ i}(f_ i, g_ i)) = 0$. Thus we get a rational equivalence between our specific cycles representing $i^*\text{div}_\mathcal {N}(t)$ and $c_1(\mathcal{N}) \cap [D]_{n - 1}$ on $D$. This finishes the proof. $\square$

Lemma 81.23.2. In Situation 81.2.1 let $X/B$ be good. Let $(\mathcal{L}, s, i : D \to X)$ be as in Definition 81.22.1. The Gysin homomorphism factors through rational equivalence to give a map $i^* : \mathop{\mathrm{CH}}\nolimits _{k + 1}(X) \to \mathop{\mathrm{CH}}\nolimits _ k(D)$.

Proof. Let $\alpha \in Z_{k + 1}(X)$ and assume that $\alpha \sim _{rat} 0$. This means there exists a locally finite collection of integral closed subspaces $W_ j \subset X$ of $\delta$-dimension $k + 2$ and $f_ j \in R(W_ j)^*$ such that $\alpha = \sum i_{j, *}\text{div}_{W_ j}(f_ j)$. Set $X' = \coprod W_ i$ and consider the diagram

$\xymatrix{ D' \ar[d]_ q \ar[r]_{i'} & X' \ar[d]^ p \\ D \ar[r]^ i & X }$

of Remark 81.22.3. Since $X' \to X$ is proper we see that $i^*p_* = q_*(i')^*$ by Lemma 81.22.5. As we know that $q_*$ factors through rational equivalence (Lemma 81.16.3), it suffices to prove the result for $\alpha ' = \sum \text{div}_{W_ j}(f_ j)$ on $X'$. Clearly this reduces us to the case where $X$ is integral and $\alpha = \text{div}(f)$ for some $f \in R(X)^*$.

Assume $X$ is integral and $\alpha = \text{div}(f)$ for some $f \in R(X)^*$. If $X = D$, then we see that $i^*\alpha$ is equal to $c_1(\mathcal{L}) \cap \alpha$. This is rationally equivalent to zero by Lemma 81.21.2. If $D \not= X$, then we see that $i^*\text{div}_ X(f)$ is equal to $c_1(\mathcal{O}_ D) \cap [D]_{n - 1}$ in $\mathop{\mathrm{CH}}\nolimits _ k(D)$ by Lemma 81.23.1. Of course capping with $c_1(\mathcal{O}_ D)$ is the zero map. $\square$

Lemma 81.23.3. 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. Let $\mathcal{N}$ be an invertible $\mathcal{O}_ X$-module. Then $i^*(c_1(\mathcal{N}) \cap \alpha ) = c_1(i^*\mathcal{N}) \cap i^*\alpha$ in $\mathop{\mathrm{CH}}\nolimits _{k - 2}(D)$ for all $\alpha \in \mathop{\mathrm{CH}}\nolimits _ k(Z)$.

Lemma 81.23.4. In Situation 81.2.1 let $X/B$ be good. Let $(\mathcal{L}, s, i : D \to X)$ and $(\mathcal{L}', s', i' : D' \to X)$ be two triples as in Definition 81.22.1. Then the diagram

$\xymatrix{ \mathop{\mathrm{CH}}\nolimits _ k(X) \ar[r]_{i^*} \ar[d]_{(i')^*} & \mathop{\mathrm{CH}}\nolimits _{k - 1}(D) \ar[d] \\ \mathop{\mathrm{CH}}\nolimits _{k - 1}(D') \ar[r] & \mathop{\mathrm{CH}}\nolimits _{k - 2}(D \cap D') }$

commutes where each of the maps is a gysin map.

Proof. Denote $j : D \cap D' \to D$ and $j' : D \cap D' \to D'$ the closed immersions corresponding to $(\mathcal{L}|_{D'}, s|_{D'}$ and $(\mathcal{L}'_ D, s|_ D)$. We have to show that $(j')^*i^*\alpha = j^* (i')^*\alpha$ for all $\alpha \in \mathop{\mathrm{CH}}\nolimits _ k(X)$. Let $W \subset X$ be an integral closed subscheme of dimension $k$. We will prove the equality in case $\alpha = [W]$. The general case will then follow from the observation in Remark 81.15.3 (and the specific shape of our rational equivalence produced below). We will deduce the equality for $\alpha = [W]$ from the key formula.

We let $\sigma$ be a nonzero meromorphic section of $\mathcal{L}|_ W$ which we require to be equal to $s|_ W$ if $W \not\subset D$. We let $\sigma '$ be a nonzero meromorphic section of $\mathcal{L}'|_ W$ which we require to be equal to $s'|_ W$ if $W \not\subset D'$. Write

$\text{div}_{\mathcal{L}|_ W}(\sigma ) = \sum \text{ord}_{Z_ i, \mathcal{L}|_ W}(\sigma )[Z_ i] = \sum n_ i[Z_ i]$

and similarly

$\text{div}_{\mathcal{L}'|_ W}(\sigma ') = \sum \text{ord}_{Z_ i, \mathcal{L}'|_ W}(\sigma ')[Z_ i] = \sum n'_ i[Z_ i]$

as in the discussion in Section 81.20. Then we see that $Z_ i \subset D$ if $n_ i \not= 0$ and $Z'_ i \subset D'$ if $n'_ i \not= 0$. For each $i$, let $\xi _ i \in |Z_ i|$ be the generic point. As in Section 81.20 we choose for each $i$ an element $\sigma _ i \in \mathcal{L}_{\xi _ i}$, resp. $\sigma '_ i \in \mathcal{L}'_{\xi _ i}$ which generates over $B_ i = \mathcal{O}_{W, \xi _ i}^ h$ and which is equal to the image of $s$, resp. $s'$ if $Z_ i \not\subset D$, resp. $Z_ i \not\subset D'$. Write $\sigma = f_ i \sigma _ i$ and $\sigma ' = f'_ i\sigma '_ i$ so that $n_ i = \text{ord}_{B_ i}(f_ i)$ and $n'_ i = \text{ord}_{B_ i}(f'_ i)$. From our definitions it follows that

$(j')^*i^*[W] = \sum \text{ord}_{B_ i}(f_ i) \text{div}_{\mathcal{L}'|_{Z_ i}}(\sigma '_ i|_{Z_ i})$

as cycles and

$j^*(i')^*[W] = \sum \text{ord}_{B_ i}(f'_ i) \text{div}_{\mathcal{L}|_{Z_ i}}(\sigma _ i|_{Z_ i})$

The key formula (Lemma 81.20.1) now gives the equality

$\sum \left( \text{ord}_{B_ i}(f_ i) \text{div}_{\mathcal{L}'|_{Z_ i}}(\sigma '_ i|_{Z_ i}) - \text{ord}_{B_ i}(f'_ i) \text{div}_{\mathcal{L}|_{Z_ i}}(\sigma _ i|_{Z_ i}) \right) = \sum \text{div}_{Z_ i}(\partial _{B_ i}(f_ i, f'_ i))$

of cycles. Note that $\text{div}_{Z_ i}(\partial _{B_ i}(f_ i, f'_ i)) = 0$ if $Z_ i \not\subset D \cap D'$ because in this case either $f_ i = 1$ or $f'_ i = 1$. Thus we get a rational equivalence between our specific cycles representing $(j')^*i^*[W]$ and $j^*(i')^*[W]$ on $D \cap D' \cap W$. $\square$

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).