Remark 12.21.5 (0AMJ): Variant—The Stacks project

Remark 12.21.5 (Variant). Let $\mathcal{A}$ be an abelian category. Let $S, T : \mathcal{A} \to \mathcal{A}$ be shift functors, i.e., isomorphisms of categories. We will indicate the $n$-fold compositions by $S^ nA$ and $T^ nA$ for $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ and $n \in \mathbf{Z}$. In this situation an exact couple is a datum $(A, E, \alpha , f, g)$ where $A$, $E$ are objects of $\mathcal{A}$ and $\alpha : A \to T^{-1}A$, $f : E \to A$, $g : A \to SE$ are morphisms such that

\[ \xymatrix{ TE \ar[r]^-{Tf} & TA \ar[r]^-{T\alpha } & A \ar[r]^-{g} & SE \ar[r]^{Sf} & SA } \]

is an exact complex. Let's visualize this as follows

\[ \xymatrix{ TA \ar[rrr]_{T\alpha } & & & A \ar[ld]^ g \ar[rrr]_\alpha & & & T^{-1}A \ar[ld]^{T^{-1}g} \\ & TE \ar[lu]^{Tf} \ar@{..}[r] & SE & & E \ar[lu]^ f \ar@{..}[r] & T^{-1}SE } \]

We set $d = g \circ f : E \to SE$. Then $d \circ S^{-1}d = g \circ f \circ S^{-1}g \circ S^{-1}f = 0$ because $f \circ S^{-1}g = 0$. Set $E' = \mathop{\mathrm{Ker}}(d)/\mathop{\mathrm{Im}}(S^{-1}d)$. Set $A' = \mathop{\mathrm{Im}}(T\alpha )$. Let $\alpha ' : A' \to T^{-1}A'$ induced by $\alpha $. Let $f' : E' \to A'$ be induced by $f$ which works because $f(\mathop{\mathrm{Ker}}(d)) \subset \mathop{\mathrm{Ker}}(g) = \mathop{\mathrm{Im}}(T\alpha )$. Finally, let $g' : A' \to TSE'$ induced by “$Tg \circ (T\alpha )^{-1}$”¹.

In exactly the same way as above we find

$\mathop{\mathrm{Ker}}(d) = f^{-1}(\mathop{\mathrm{Ker}}(g)) = f^{-1}(\mathop{\mathrm{Im}}(T\alpha ))$,
$\mathop{\mathrm{Im}}(d) = g(\mathop{\mathrm{Im}}(f)) = g(\mathop{\mathrm{Ker}}(\alpha ))$,
$(A', E', \alpha ', f', g')$ is an exact couple for the shift functors $TS$ and $T$.

We obtain a spectral sequence (as in Remark 12.20.3) with $E_1 = E$, $E_2 = E'$, etc, with $d_ r : E_ r \to T^{r - 1}SE_ r$ for all $r \geq 1$. Lemma 12.21.4 tells us that

\[ SB_{r + 1} = g(\mathop{\mathrm{Ker}}(T^{-r + 1}\alpha \circ \ldots \circ T^{-1}\alpha \circ \alpha )) \]

and

\[ Z_{r + 1} = f^{-1}(\mathop{\mathrm{Im}}(T\alpha \circ T^2\alpha \circ \ldots \circ T^ r\alpha )) \]

in this situation. The description of the map $d_{r + 1}$ is similar to that given in the lemma. (It may be easier to use these explicit descriptions to prove one gets a spectral sequence from such an exact couple.)

[1] This works because $TSE' = \mathop{\mathrm{Ker}}(TSd)/\mathop{\mathrm{Im}}(Td)$ and $Tg(\mathop{\mathrm{Ker}}(T\alpha )) = Tg(\mathop{\mathrm{Im}}(Tf)) = \mathop{\mathrm{Im}}(T(d))$ and $TS(d)(\mathop{\mathrm{Im}}(Tg)) = \mathop{\mathrm{Im}}(TSg \circ TSf \circ Tg) = 0$.

Comments (2)

Comment #9481 by Elías Guisado on June 28, 2024 at 05:21

How one can describe exactly how $g'$ is induced? What I thought is to define $g'$ as the connecting homomorphism arisen from the snake lemma applied to the commutative diagram with exact rows $\require{AMScd} \begin{CD} @.TE@>{Tf}>> TA@>{T\alpha}>>A'@>>> 0\\ @.@V{Td}VV@V{Tg}VV@VV{0}V@.\\ 0@>>>\operatorname{Ker}TS d@>>> TSE@>{\smash{ TSd}}>> TS^2E \end{CD}$ Note that this also gives a description of $g'$ in Lemma 12.21.2 if $T=S=\operatorname{id}_\mathcal{A}$ . I don't know if there is another way, though. The footnote seems to be using something else, instead of the snake lemma.

Comment #9482 by Elías Guisado on June 28, 2024 at 05:28

The diagram should compile like this. I don't know why Mathjax doesn't compile it well here in the Stacks Project (same happened to me in #6829).

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