The Stacks project

Lemma 45.12.5. In the situation above let $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. If $\psi ^2$ is as in Chow Homology, Lemma 42.55.1 and $c^ A$ and $ch^ A$ are as in Propositions 45.12.1 and 45.12.4 then we have $c^ A_ i(\psi ^2(\alpha )) = 2^ i c^ A_ i(\alpha )$ and $ch^ A_ i(\psi ^2(\alpha )) = 2^ i ch^ A_ i(\alpha )$ for all $\alpha \in K_0(\textit{Vect}(X))$.

Proof. Observe that the map $\prod _{i \geq 0} A^ i(X) \to \prod _{i \geq 0} A^ i(X)$ multiplying by $2^ i$ on $A^ i(X)$ is a ring map. Hence, since $\psi ^2$ is also a ring map, it suffices to prove the formulas for additive generators of $K_0(\textit{Vect}(X))$. Thus we may assume $\alpha = [\mathcal{E}]$ for some finite locally free $\mathcal{O}_ X$-module $\mathcal{E}$. By construction of the Chern classes of $\mathcal{E}$ we immediately reduce to the case where $\mathcal{E}$ has constant rank $r$. In this case, we can choose a projective smooth morphism $p : P \to X$ such that restriction $A^*(X) \to A^*(P)$ is injective and such that $p^*\mathcal{E}$ has a finite filtration whose graded parts are invertible $\mathcal{O}_ P$-modules $\mathcal{L}_ j$, see Lemma 45.12.2. Then $[p^*\mathcal{E}] = \sum [\mathcal{L}_ j]$ and hence $\psi ^2([p^*\mathcal{E}]) = \sum [\mathcal{L}_ j^{\otimes 2}]$ by definition of $\psi ^2$. Setting $x_ j = c^ A_1(\mathcal{L}_ j)$ we have

\[ c^ A(\alpha ) = \prod (1 + x_ j) \quad \text{and}\quad c^ A(\psi ^2(\alpha )) = \prod (1 + 2 x_ j) \]

in $\prod A^ i(P)$ and we have

\[ ch^ A(\alpha ) = \sum \exp (x_ j) \quad \text{and}\quad ch^ A(\psi ^2(\alpha )) = \sum \exp (2 x_ j) \]

in $\prod A^ i(P)$. From these formulas the desired result follows. $\square$


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Your email address will not be published.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 0FIA. Beware of the difference between the letter 'O' and the digit '0'.