The Stacks project

12.22 Spectral sequences: differential objects

Definition 12.22.1. 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 $d \circ \alpha = \alpha \circ d$.

slogan

Lemma 12.22.2. Let $\mathcal{A}$ be an abelian category. The category of differential objects of $\mathcal{A}$ is abelian.

Proof. Omitted. $\square$

Definition 12.22.3. For a differential object $(A, d)$ we denote

\[ H(A, d) = \mathop{\mathrm{Ker}}(d)/\mathop{\mathrm{Im}}(d) \]

its homology.

Lemma 12.22.4. 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

\[ \ldots \to H(C, d) \to H(A, d) \to H(B, d) \to H(C, d) \to \ldots \]

Proof. Apply Lemma 12.13.12 to the short exact sequence of complexes

\[ \begin{matrix} 0 & \to & A & \to & B & \to & C & \to & 0 \\ & & \downarrow & & \downarrow & & \downarrow \\ 0 & \to & A & \to & B & \to & C & \to & 0 \\ & & \downarrow & & \downarrow & & \downarrow \\ 0 & \to & A & \to & B & \to & C & \to & 0 \end{matrix} \]

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

\[ 0 \to (A, d) \to (A, d) \to (A/\alpha A, d) \to 0 \]

of differential objects. By the Lemma 12.22.4 we get an exact couple

\[ \xymatrix{ H(A, d) \ar[rr]_{\overline{\alpha }} & & H(A, d) \ar[ld]^ g \\ & H(A/\alpha A, d) \ar[lu]^ f & } \]

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 12.20 we define a sequence of subobjects

\[ 0 = B_0 \subset \ldots \subset B_ r \subset \ldots \subset Z_ r \subset \ldots \subset Z_0 = E_0 \]

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

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

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

  3. $d_ r : E_ r \to E_ r$ 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

\[ E_ r = \frac{d^{-1}(\alpha ^ r A) + \alpha A}{(\alpha ^{r - 1})^{-1}(dA) + \alpha A}. \]

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

Definition 12.22.5. 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 12.22.6 (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

\[ 0 \to (A, d) \to (B, d) \to (C, d) \to 0 \]

we obtain a long exact homology sequence

\[ \ldots \to S^{-1}H(C, d) \to H(A, d) \to H(B, d) \to H(C, d) \to SH(A, d) \to \ldots \]

(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 12.21.5 with shift functors $TS$ and $T$ given by

\[ (H(A, d), S^{-1}H(Q, d), \overline{\alpha }, f, g) \]

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

  1. $SB_ r$ 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 $T^{-1}A \to Q$,

  2. $Z_ r$ 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 $T^ rSB_ r$.


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 011U. Beware of the difference between the letter 'O' and the digit '0'.