Definition 12.19.1. Let $\mathcal{A}$ be an abelian category. A *differential object* of $\mathcal{A}$ is a pair $(A, d)$ consisting of an object $A$ of $\mathcal{A}$ endowed with a selfmap $d$ such that $d \circ d = 0$. A *morphism of differential objects* $(A, d) \to (B, d)$ is given by a morphism $\alpha : A \to B$ such that $d \circ \alpha = \alpha \circ d$.

## 12.19 Spectral sequences: differential objects

Lemma 12.19.2. Let $\mathcal{A}$ be an abelian category. The category of differential objects of $\mathcal{A}$ is abelian.

**Proof.**
Omitted.
$\square$

Definition 12.19.3. For a differential object $(A, d)$ we denote

its *homology*.

Lemma 12.19.4. Let $\mathcal{A}$ be an abelian category. Let $0 \to (A, d) \to (B, d) \to (C, d) \to 0$ be a short exact sequence of differential objects. Then we get an exact homology sequence

**Proof.**
Apply Lemma 12.12.12 to the short exact sequence of complexes

where the vertical arrows are $d$. $\square$

We come to an important example of a spectral sequence. Let $\mathcal{A}$ be an abelian category. Let $(A, d)$ be a differential object of $\mathcal{A}$. Let $\alpha : (A, d) \to (A, d)$ be an endomorphism of this differential object. If we assume $\alpha $ injective, then we get a short exact sequence

of differential objects. By the Lemma 12.19.4 we get an exact couple

where $g$ is the canonical map and $f$ is the map defined in the snake lemma. Thus we get an associated spectral sequence! Since in this case we have $E_1 = H(A/\alpha A, d)$ we see that it makes sense to define $E_0 = A/\alpha A$ and $d_0 = d$. In other words, we start the spectral sequence with $r = 0$. According to our conventions in Section 12.17 we define a sequence of subobjects

with the property that $E_ r = Z_ r/B_ r$. Namely we have for $r \geq 1$ that

$B_ r$ is the image of $(\alpha ^{r - 1})^{-1}(d A)$ under the natural map $A \to A/\alpha A$,

$Z_ r$ is the image of $d^{-1}(\alpha ^ r A)$ under the natural map $A \to A/\alpha A$, and

$d_ r : E_ r \to E_ r$ is given as follows: given an element $z \in Z_ r$ choose an element $y \in A$ such that $d(z) = \alpha ^ r(y)$. Then $d_ r(z + B_ r + \alpha A) = y + B_ r + \alpha A$.

Warning: It is not necessarily the case that $\alpha A \subset (\alpha ^{r - 1})^{-1}(dA)$, nor $\alpha A \subset d^{-1}(\alpha ^ r A)$. It is true that $(\alpha ^{r - 1})^{-1}(dA) \subset d^{-1}(\alpha ^ r A)$. We have

It is not hard to verify directly that (1) – (3) give a spectral sequence.

Definition 12.19.5. Let $\mathcal{A}$ be an abelian category. Let $(A, d)$ be a differential object of $\mathcal{A}$. Let $\alpha : A \to A$ be an injective selfmap of $A$ which commutes with $d$. The *spectral sequence associated to $(A, d, \alpha )$* is the spectral sequence $(E_ r, d_ r)_{r \geq 0}$ described above.

Remark 12.19.6 (Variant). Let $\mathcal{A}$ be an abelian category and let $S, T : \mathcal{A} \to \mathcal{A}$ be shift functors, i.e., isomorphisms of categories. Assume that $TS = ST$ as functors. Consider pairs $(A, d)$ consisting of an object $A$ of $\mathcal{A}$ and a morphism $d : A \to SA$ such that $d \circ S^{-1}d = 0$. The category of these objects is abelian. We define $H(A, d) = \mathop{\mathrm{Ker}}(d)/\mathop{\mathrm{Im}}(S^{-1}d)$ and we observe that $H(SA, Sd) = SH(A, d)$ (canonical isomorphism). Given a short exact sequence

we obtain a long exact homology sequence

(note the shifts in the boundary maps). Since $ST = TS$ the functor $T$ defines a shift functor on pairs by setting $T(A, d) = (TA, Td)$. Next, let $\alpha : (A, d) \to T^{-1}(A, d)$ be injective with cokernel $(Q, d)$. Then we get an exact couple as in Remark 12.18.5 with shift functors $TS$ and $T$ given by

where $\overline{\alpha } : H(A, d) \to T^{-1}H(A, d)$ is induced by $\alpha $, the map $f : S^{-1}H(Q, d) \to H(A, d)$ is the boundary map and $g : H(A, d) \to TH(Q, d) = TS(S^{-1}H(Q, d))$ is induced by the quotient map $A \to TQ$. Thus we get a spectral sequence as above with $E_1 = S^{-1}H(Q, d)$ and differentials $d_ r : E_ r \to T^ rSE_ r$. As above we set $E_0 = S^{-1}Q$ and $d_0 : E_0 \to SE_0$ given by $S^{-1}d : S^{-1}Q \to Q$. If according to our conventions we define $B_ r \subset Z_ r \subset E_0$, then we have for $r \geq 1$ that

$SB_ r$ is the image of

\[ (T^{-r + 1}\alpha \circ \ldots \circ T^{-1}\alpha )^{-1} \mathop{\mathrm{Im}}(T^{-r}S^{-1}d) \]under the natural map $T^{-1}A \to Q$,

$Z_ r$ is the image of

\[ (S^{-1}T^{-1}d)^{-1} \mathop{\mathrm{Im}}(\alpha \circ \ldots \circ T^{r - 1}\alpha ) \]under the natural map $S^{-1}T^{-1}A \to S^{-1}Q$.

The differentials can be described as follows: if $x \in Z_ r$, then pick $x' \in S^{-1}T^{-1}A$ mapping to $x$. Then $S^{-1}T^{-1}d(x')$ is $(\alpha \circ \ldots \circ T^{r - 1}\alpha )(y)$ for some $y \in T^{r - 1}A$. Then $d_ r(x) \in T^ rSE_ r$ is represented by the class of the image of $y$ in $T^ rSE_0 = T^ rQ$ modulo $T^ rSB_ r$.

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