The Stacks project

Lemma 12.24.3. Let $\mathcal{A}$ be an abelian category. Let $(K^\bullet , F)$ be a filtered complex of $\mathcal{A}$. Assume $\mathcal{A}$ has countable direct sums. Let $(E_ r, d_ r)_{r \geq 0}$ be the spectral sequence associated to $(K^\bullet , F)$.

  1. The map

    \[ d_1^{p, q} : E_1^{p, q} = H^{p + q}(\text{gr}^ p(K^\bullet )) \longrightarrow E_1^{p + 1, q} = H^{p + q + 1}(\text{gr}^{p + 1}(K^\bullet )) \]

    is equal to the boundary map in cohomology associated to the short exact sequence of complexes

    \[ 0 \to \text{gr}^{p + 1}(K^\bullet ) \to F^ pK^\bullet /F^{p + 2}K^\bullet \to \text{gr}^ p(K^\bullet ) \to 0. \]
  2. Assume that $d(F^ pK) \subset F^{p + 1}K$ for all $p \in \mathbf{Z}$. Then $d$ induces the zero differential on $\text{gr}^ p(K^\bullet )$ and hence $E_1^{p, q} = \text{gr}^ p(K^\bullet )^{p + q}$. Furthermore, in this case

    \[ d_1^{p, q} : E_1^{p, q} = \text{gr}^ p(K^\bullet )^{p + q} \longrightarrow E_1^{p + 1, q} = \text{gr}^{p + 1}(K^\bullet )^{p + q + 1} \]

    is the morphism induced by $d$.

Proof. This is clear from the formula given for the differential $d_1^{p, q}$ just above Lemma 12.24.2. $\square$


Comments (3)

Comment #2033 by Yu-Liang Huang on

A typo in (2): the image of should be .

Comment #2034 by Yu-Liang Huang on

The last term of the exact sequence in (1) should be .


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Your email address will not be published.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 012N. Beware of the difference between the letter 'O' and the digit '0'.