The Stacks project

Lemma 50.15.4. Let $p : X \to S$ be a morphism of schemes. Let $Y \subset X$ be an effective Cartier divisor. Assume the de Rham complex of log poles is defined for $Y \subset X$ over $S$.

  1. The maps $\wedge : \Omega ^ p_{X/S} \times \Omega ^ q_{X/S} \to \Omega ^{p + q}_{X/S}$ extend uniquely to $\mathcal{O}_ X$-bilinear maps

    \[ \wedge : \Omega ^ p_{X/S}(\log Y) \times \Omega ^ q_{X/S}(\log Y) \to \Omega ^{p + q}_{X/S}(\log Y) \]

    satisfying the Leibniz rule $ \text{d}(\omega \wedge \eta ) = \text{d}(\omega ) \wedge \eta + (-1)^{\deg (\omega )} \omega \wedge \text{d}(\eta )$,

  2. with multiplication as in (1) the map $\Omega ^\bullet _{X/S} \to \Omega ^\bullet _{X/S}(\log (Y)$ is a homomorphism of differential graded $\mathcal{O}_ S$-algebras,

  3. via the maps in (1) we have $\Omega ^ p_{X/S}(\log Y) = \wedge ^ p(\Omega ^1_{X/S}(\log Y))$, and

  4. the map $\text{Res} : \Omega ^\bullet _{X/S}(\log Y) \to \Omega ^\bullet _{Y/S}[-1]$ satisfies

    \[ \text{Res}(\omega \wedge \eta ) = \text{Res}(\omega ) \wedge \eta |_ Y \]

    for $\omega $ a local section of $\Omega ^ p_{X/S}(\log Y)$ and $\eta $ a local section of $\Omega ^ q_{X/S}$.

Proof. This follows by direct calcuation from the local construction of the complex in the proof of Lemma 50.15.2. Details omitted. $\square$


Comments (3)

Comment #6218 by Martin Bright on

In (4), has been defined for ?

Comment #6220 by on

Hmm... yes, that seems bad. I will fix this the next time I go through all the comments,


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Your email address will not be published.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