Section 12.21 (011P): Spectral sequences: exact couples

12.21 Spectral sequences: exact couples

Definition 12.21.1. Let $\mathcal{A}$ be an abelian category.

An exact couple is a datum $(A, E, \alpha , f, g)$ where $A$, $E$ are objects of $\mathcal{A}$ and $\alpha $, $f$, $g$ are morphisms as in the following diagram

\[ \xymatrix{ A \ar[rr]_{\alpha } & & A \ar[ld]^ g \\ & E \ar[lu]^ f & } \]

with the property that the kernel of each arrow is the image of its predecessor. So $\mathop{\mathrm{Ker}}(\alpha ) = \mathop{\mathrm{Im}}(f)$, $\mathop{\mathrm{Ker}}(f) = \mathop{\mathrm{Im}}(g)$, and $\mathop{\mathrm{Ker}}(g) = \mathop{\mathrm{Im}}(\alpha )$.
A morphism of exact couples $t : (A, E, \alpha , f, g) \to (A', E', \alpha ', f', g')$ is given by morphisms $t_ A : A \to A'$ and $t_ E : E \to E'$ such that $\alpha ' \circ t_ A = t_ A \circ \alpha $, $f' \circ t_ E = t_ A \circ f$, and $g' \circ t_ A = t_ E \circ g$.

Lemma 12.21.2. Let $(A, E, \alpha , f, g)$ be an exact couple in an abelian category $\mathcal{A}$. Set

$d = g \circ f : E \to E$ so that $d \circ d = 0$,
$E' = \mathop{\mathrm{Ker}}(d)/\mathop{\mathrm{Im}}(d)$,
$A' = \mathop{\mathrm{Im}}(\alpha )$,
$\alpha ' : A' \to A'$ induced by $\alpha $,
$f' : E' \to A'$ induced by $f$,
$g' : A' \to E'$ induced by “$g \circ \alpha ^{-1}$”.

Then we have

$\mathop{\mathrm{Ker}}(d) = f^{-1}(\mathop{\mathrm{Ker}}(g)) = f^{-1}(\mathop{\mathrm{Im}}(\alpha ))$,
$\mathop{\mathrm{Im}}(d) = g(\mathop{\mathrm{Im}}(f)) = g(\mathop{\mathrm{Ker}}(\alpha ))$,
$(A', E', \alpha ', f', g')$ is an exact couple.

Proof. Omitted. $\square$

Hence it is clear that given an exact couple $(A, E, \alpha , f, g)$ we get a spectral sequence by setting $E_1 = E$, $d_1 = d$, $E_2 = E'$, $d_2 = d' = g' \circ f'$, $E_3 = E''$, $d_3 = d'' = g'' \circ f''$, and so on.

Definition 12.21.3. Let $\mathcal{A}$ be an abelian category. Let $(A, E, \alpha , f, g)$ be an exact couple. The spectral sequence associated to the exact couple is the spectral sequence $(E_ r, d_ r)_{r \geq 1}$ with $E_1 = E$, $d_1 = d$, $E_2 = E'$, $d_2 = d' = g' \circ f'$, $E_3 = E''$, $d_3 = d'' = g'' \circ f''$, and so on.

Lemma 12.21.4. Let $\mathcal{A}$ be an abelian category. Let $(A, E, \alpha , f, g)$ be an exact couple. Let $(E_ r, d_ r)_{r \geq 1}$ be the spectral sequence associated to the exact couple. In this case we have

\[ 0 = B_1 \subset \ldots \subset B_{r + 1} = g(\mathop{\mathrm{Ker}}(\alpha ^ r)) \subset \ldots \subset Z_{r + 1} = f^{-1}(\mathop{\mathrm{Im}}(\alpha ^ r)) \subset \ldots \subset Z_1 = E \]

and the map $d_{r + 1} : E_{r + 1} \to E_{r + 1}$ is described by the following rule: For any (test) object $T$ of $\mathcal{A}$ and any elements $x : T \to Z_{r + 1}$ and $y : T \to A$ such that $f \circ x = \alpha ^ r \circ y$ we have

\[ d_{r + 1} \circ \overline{x} = \overline{g \circ y} \]

where $\overline{x} : T \to E_{r + 1}$ is the induced morphism.

Proof. Omitted. $\square$

Note that in the situation of the lemma we obviously have

\[ B_\infty = g\left(\bigcup \nolimits _ r \mathop{\mathrm{Ker}}(\alpha ^ r)\right) \subset Z_\infty = f^{-1}\left(\bigcap \nolimits _ r \mathop{\mathrm{Im}}(\alpha ^ r)\right) \]

provided $\bigcup \mathop{\mathrm{Ker}}(\alpha ^ r)$ and $\bigcap \mathop{\mathrm{Im}}(\alpha ^ r)$ exist. This produces as limit $E_\infty = Z_\infty / B_\infty $, see Definition 12.20.2.

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 (5)

Comment #1601 by Jean-Michel on August 12, 2015 at 20:23

Hello, I think there is a typo in the statement of lemma 12.18.4. Should be $d_{r+1} \circ \overline{x} = \overline{f \circ y}$ instead of $d_r \circ \bar{x} = \bar{g \circ y}$ . Doesn't make any sense the other way because $g \circ y$ is in $A$ and not in $Z_{r+1}$ .

Comment #1602 by Jean-Michel on August 12, 2015 at 20:24

Hello, I think there is a typo in the statement of lemma 12.18.4. Should be $d_{r+1} \circ \overline{x} = \overline{f \circ y}$ instead of $d_r \circ \bar{x} = \overline{g \circ y}$ . Doesn't make any sense the other way because $g \circ y$ is in $A$ and not in $Z_{r+1}$ .

Comment #1667 by Johan on October 08, 2015 at 13:07

No, I just checked and what it says is correct. Namely, $g : A \to E$ so $g(y) \in E$ . And then $f(g(y)) = 0$ so we see that $g(y) \in Z^{r + 1}$ . To finish the proof, you have to show that if $g(y) \in B^{r + 1}$ , then $x \in Z^{r + 2}$ , etc, etc.

Comment #9478 by Elías Guisado on June 27, 2024 at 10:29

I disagree with the “obviously” before Remark 12.21.5 in “note that in the situation of the lemma we obviously have $B_\infty = g\left(\bigcup\nolimits_r \operatorname{Ker}(\alpha^r)\right) \subset Z_\infty = f^{-1}\left(\bigcap\nolimits_r \operatorname{Im}(\alpha^r)\right)$ .” The non-obvious thing to me is the fact that taking inverse of subobjects commutes with unions (more generally, it will commute with subobject sums) and that taking image of subobjects commutes with intersections.

It is not a difficult matter, though: Given an abelian category $\mathcal{A}$ and an object $X$ of $\mathcal{A}$ , denote $S(X)$ to the class of isomorphism classes of subobjects of $X$ . Then $S(X)$ is partially ordered. Given a morphism $f:X\to Y$ in $\mathcal{A}$ , one can define maps $f:S(X)\rightleftarrows S(Y):f^{-1}$ which are taking image of a subobject and taking inverse image of a subobject along $f$ , see [1, p. 40]. Moreover, the former is left adjoint to the latter [1, Sect. 2.6, Proposition 6.8]. Hence, taking direct images commutes with arbitrary colimits and taking inverse images with arbitrary limits.

References

N. Popescu, Abelian categories with applications to rings and modules

Comment #9479 by Elías Guisado on June 27, 2024 at 11:46

(See also [1, Sect. 2.6, Proposition 6.1].)

The Stacks project

12.21 Spectral sequences: exact couples

Comments (5)

References

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