## 12.21 Spectral sequences: exact couples

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

1. 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 $\mathop{\mathrm{Ker}}(g) = \mathop{\mathrm{Im}}(\alpha )$.

2. 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

1. $d = g \circ f : E \to E$ so that $d \circ d = 0$,

2. $E' = \mathop{\mathrm{Ker}}(d)/\mathop{\mathrm{Im}}(d)$,

3. $A' = \mathop{\mathrm{Im}}(\alpha )$,

4. $\alpha ' : A' \to A'$ induced by $\alpha$,

5. $f' : E' \to A'$ induced by $f$,

6. $g' : A' \to E'$ induced by “$g \circ \alpha ^{-1}$”.

Then we have

1. $\mathop{\mathrm{Ker}}(d) = f^{-1}(\mathop{\mathrm{Ker}}(g)) = f^{-1}(\mathop{\mathrm{Im}}(\alpha ))$,

2. $\mathop{\mathrm{Im}}(d) = g(\mathop{\mathrm{Im}}(f)) = g(\mathop{\mathrm{Ker}}(\alpha ))$,

3. $(A', E', \alpha ', f', g')$ 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

1. $\mathop{\mathrm{Ker}}(d) = f^{-1}(\mathop{\mathrm{Ker}}(g)) = f^{-1}(\mathop{\mathrm{Im}}(T\alpha ))$,

2. $\mathop{\mathrm{Im}}(d) = g(\mathop{\mathrm{Im}}(f)) = g(\mathop{\mathrm{Ker}}(\alpha ))$,

3. $(A', E', \alpha ', f', g')$ 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.)

 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 $TS(d)(\mathop{\mathrm{Im}}(Tg)) = \mathop{\mathrm{Im}}(TSg \circ TSf \circ Tg) = 0$.

Comment #1601 by Jean-Michel on

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

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 on

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.

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