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 .

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