The Stacks project

Lemma 42.32.4. In the situation of Lemma 42.32.2 denote $o : X \to L$ the zero section (see proof of the lemma). Then we have

  1. $o(X)$ is the zero scheme of a regular global section of $p^*\mathcal{L}^{\otimes -1}$,

  2. $o_* : \mathop{\mathrm{CH}}\nolimits _ k(X) \to \mathop{\mathrm{CH}}\nolimits _ k(L)$ as $o$ is a closed immersion,

  3. $o^* : \mathop{\mathrm{CH}}\nolimits _{k + 1}(L) \to \mathop{\mathrm{CH}}\nolimits _ k(X)$ as $o(X)$ is an effective Cartier divisor,

  4. $o^* p^* : \mathop{\mathrm{CH}}\nolimits _ k(X) \to \mathop{\mathrm{CH}}\nolimits _ k(X)$ is the identity map,

  5. $o_*\alpha = - p^*(c_1(\mathcal{L}) \cap \alpha )$ for any $\alpha \in \mathop{\mathrm{CH}}\nolimits _ k(X)$, and

  6. $o^* o_* : \mathop{\mathrm{CH}}\nolimits _ k(X) \to \mathop{\mathrm{CH}}\nolimits _{k - 1}(X)$ is equal to the map $\alpha \mapsto - c_1(\mathcal{L}) \cap \alpha $.

Proof. Since $p_*\mathcal{O}_ L = \text{Sym}^*(\mathcal{L})$ we have $p_*(p^*\mathcal{L}^{\otimes -1}) = \text{Sym}^*(\mathcal{L}) \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes -1}$ by the projection formula (Cohomology, Lemma 20.54.2) and the section mentioned in (1) is the canonical trivialization $\mathcal{O}_ X \to \mathcal{L} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes -1}$. We omit the proof that the vanishing locus of this section is precisely $o(X)$. This proves (1).

Parts (2), (3), and (4) we've seen in the course of the proof of Lemma 42.32.2. Of course (4) is the first formula in Lemma 42.31.1.

Part (5) follows from the second formula in Lemma 42.31.1, additivity of capping with $c_1$ (Lemma 42.25.2), and the fact that capping with $c_1$ commutes with flat pullback (Lemma 42.26.2).

Part (6) follows from Lemma 42.30.3 and the fact that $o^*p^*\mathcal{L} = \mathcal{L}$. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0F96. Beware of the difference between the letter 'O' and the digit '0'.