## Tag `011M`

## 12.17. Spectral sequences

A nice discussion of spectral sequences may be found in [Eisenbud]. See also [McCleary], [Lang], etc.

Definition 12.17.1. Let $\mathcal{A}$ be an abelian category.

- A
spectral sequence in $\mathcal{A}$is given by a system $(E_r, d_r)_{r \geq 1}$ where each $E_r$ is an object of $\mathcal{A}$, each $d_r : E_r \to E_r$ is a morphism such that $d_r \circ d_r = 0$ and $E_{r + 1} = \mathop{\mathrm{Ker}}(d_r)/\mathop{\mathrm{Im}}(d_r)$ for $r \geq 1$.- A
morphism of spectral sequences$f : (E_r, d_r)_{r \geq 1} \to (E'_r, d'_r)_{r \geq 1}$ is given by a family of morphisms $f_r : E_r \to E'_r$ such that $f_r \circ d_r = d'_r \circ f_r$ and such that $f_{r + 1}$ is the morphism induced by $f_r$ via the identifications $E_{r + 1} = \mathop{\mathrm{Ker}}(d_r)/\mathop{\mathrm{Im}}(d_r)$ and $E'_{r + 1} = \mathop{\mathrm{Ker}}(d'_r)/\mathop{\mathrm{Im}}(d'_r)$.

We will sometimes loosen this definition somewhat and allow $E_{r + 1}$ to be an object with a given isomorphism $E_{r + 1} \to \mathop{\mathrm{Ker}}(d_r)/\mathop{\mathrm{Im}}(d_r)$. In addition we sometimes have a system $(E_r, d_r)_{r \geq r_0}$ for some $r_0 \in \mathbf{Z}$ satisfying the properties of the definition above for indices $\geq r_0$. We will also call this a spectral sequence since by a simple renumbering it falls under the definition anyway. In fact, the cases $r_0 = 0$ and $r_0 = -1$ can be found in the literature.

Given a spectral sequence $(E_r, d_r)_{r \geq 1}$ we define $$ 0 = B_1 \subset B_2 \subset \ldots \subset B_r \subset \ldots \subset Z_r \subset \ldots \subset Z_2 \subset Z_1 = E_1 $$ by the following simple procedure. Set $B_2 = \mathop{\mathrm{Im}}(d_1)$ and $Z_2 = \mathop{\mathrm{Ker}}(d_1)$. Then it is clear that $d_2 : Z_2/B_2 \to Z_2/B_2$. Hence we can define $B_3$ as the unique subobject of $E_1$ containing $B_2$ such that $B_3/B_2$ is the image of $d_2$. Similarly we can define $Z_3$ as the unique subobject of $E_1$ containing $B_2$ such that $Z_3/B_2$ is the kernel of $d_2$. And so on and so forth. In particular we have $$ E_r = Z_r/B_r $$ for all $r \geq 1$. In case the spectral sequence starts at $r = r_0$ then we can similarly construct $B_i$, $Z_i$ as subobjects in $E_{r_0}$. In fact, in the literature one sometimes finds the notation $$ 0 = B_r(E_r) \subset B_{r + 1}(E_r) \subset B_{r + 2}(E_r) \subset \ldots \subset Z_{r + 2}(E_r) \subset Z_{r + 1}(E_r) \subset Z_r(E_r) = E_r $$ to denote the filtration described above but starting with $E_r$.

Definition 12.17.2. Let $\mathcal{A}$ be an abelian category. Let $(E_r, d_r)_{r \geq 1}$ be a spectral sequence.

- If the subobjects $Z_{\infty} = \bigcap Z_r$ and $B_{\infty} = \bigcup B_r$ of $E_1$ exist then we define the
limit^{1}of the spectral sequence to be the object $E_{\infty} = Z_{\infty}/B_{\infty}$.- We say that the spectral sequence
degenerates at $E_r$if the differentials $d_r, d_{r + 1}, \ldots$ are all zero.

Note that if the spectral sequence degenerates at $E_r$, then we have $E_r = E_{r + 1} = \ldots = E_{\infty}$ (and the limit exists of course). Also, almost any abelian category we will encounter has countable sums and intersections.

Remark 12.17.3 (Variant). It is often the case that the terms of a spectral sequence have additional structure, for example a grading or a bigrading. To accomodate this (and to get around certain technical issues) we introduce the following notion. Let $\mathcal{A}$ be an abelian category. Let $(T_r)_{r \geq 1}$ be a sequence of

translationorshiftfunctors, i.e., $T_r : \mathcal{A} \to \mathcal{A}$ is an isomorphism of categories. In this setting aspectral sequenceis given by a system $(E_r, d_r)_{r \geq 1}$ where each $E_r$ is an object of $\mathcal{A}$, each $d_r : E_r \to T_rE_r$ is a morphism such that $T_rd_r \circ d_r = 0$ so that $$ \xymatrix{ \ldots \ar[r] & T_r^{-1}E_r \ar[r]^-{T_r^{-1}d_r} & E_r \ar[r]^-{d_r} & T_rE_r \ar[r]^{T_r d_r} & T_r^2E_r \ar[r] & \ldots } $$ is a complex and $E_{r + 1} = \mathop{\mathrm{Ker}}(d_r)/\mathop{\mathrm{Im}}(T_r^{-1}d_r)$ for $r \geq 1$. It is clear what amorphism of spectral sequencesmeans in this setting. In this setting we can still define $$ 0 = B_1 \subset B_2 \subset \ldots \subset B_r \subset \ldots \subset Z_r \subset \ldots \subset Z_2 \subset Z_1 = E_1 $$ and $Z_\infty$ and $B_\infty$ (if they exist) as above.

- This notation is not universally accepted. In some references an additional pair of subobjects $Z_\infty$ and $B_\infty$ of $E_1$ such that $0 = B_1 \subset B_2 \subset \ldots \subset B_\infty \subset Z_\infty \subset \ldots \subset Z_2 \subset Z_1 = E_1$ is part of the data comprising a spectral sequence! ↑

The code snippet corresponding to this tag is a part of the file `homology.tex` and is located in lines 4045–4173 (see updates for more information).

```
\section{Spectral sequences}
\label{section-spectral-sequence}
\noindent
A nice discussion of spectral sequences may be found in
\cite{Eisenbud}. See also \cite{McCleary}, \cite{Lang}, etc.
\begin{definition}
\label{definition-spectral-sequence}
Let $\mathcal{A}$ be an abelian category.
\begin{enumerate}
\item A {\it spectral sequence in $\mathcal{A}$} is given by a
system $(E_r, d_r)_{r \geq 1}$ where each $E_r$ is an object
of $\mathcal{A}$, each $d_r : E_r \to E_r$ is a morphism such
that $d_r \circ d_r = 0$ and $E_{r + 1} = \Ker(d_r)/\Im(d_r)$
for $r \geq 1$.
\item A {\it morphism of spectral sequences}
$f : (E_r, d_r)_{r \geq 1} \to (E'_r, d'_r)_{r \geq 1}$ is
given by a family of morphisms $f_r : E_r \to E'_r$ such that
$f_r \circ d_r = d'_r \circ f_r$ and such that $f_{r + 1}$
is the morphism induced by $f_r$ via the identifications
$E_{r + 1} = \Ker(d_r)/\Im(d_r)$
and
$E'_{r + 1} = \Ker(d'_r)/\Im(d'_r)$.
\end{enumerate}
\end{definition}
\noindent
We will sometimes loosen this definition somewhat and allow $E_{r + 1}$
to be an object with a given isomorphism
$E_{r + 1} \to \Ker(d_r)/\Im(d_r)$.
In addition we sometimes have a system $(E_r, d_r)_{r \geq r_0}$
for some $r_0 \in \mathbf{Z}$ satisfying the properties of the definition above
for indices $\geq r_0$. We will also call this a spectral sequence since by
a simple renumbering it falls under the definition anyway.
In fact, the cases $r_0 = 0$ and $r_0 = -1$ can be found in the literature.
\medskip\noindent
Given a spectral sequence $(E_r, d_r)_{r \geq 1}$ we define
$$
0 = B_1 \subset B_2 \subset \ldots \subset B_r \subset \ldots
\subset Z_r \subset \ldots \subset Z_2 \subset Z_1 = E_1
$$
by the following simple procedure. Set $B_2 = \Im(d_1)$
and $Z_2 = \Ker(d_1)$. Then it is clear that
$d_2 : Z_2/B_2 \to Z_2/B_2$. Hence we can define $B_3$ as the unique
subobject of $E_1$ containing $B_2$ such that $B_3/B_2$ is the image
of $d_2$. Similarly we can define $Z_3$ as the unique subobject of
$E_1$ containing $B_2$ such that $Z_3/B_2$ is the kernel of $d_2$.
And so on and so forth. In particular we have
$$
E_r = Z_r/B_r
$$
for all $r \geq 1$. In case the spectral sequence starts at $r = r_0$
then we can similarly construct $B_i$, $Z_i$ as subobjects in $E_{r_0}$.
In fact, in the literature one sometimes finds the notation
$$
0 = B_r(E_r) \subset B_{r + 1}(E_r) \subset B_{r + 2}(E_r) \subset \ldots
\subset Z_{r + 2}(E_r) \subset Z_{r + 1}(E_r) \subset Z_r(E_r) = E_r
$$
to denote the filtration described above but starting with $E_r$.
\begin{definition}
\label{definition-limit-spectral-sequence}
Let $\mathcal{A}$ be an abelian category.
Let $(E_r, d_r)_{r \geq 1}$ be a spectral sequence.
\begin{enumerate}
\item If the subobjects $Z_{\infty} = \bigcap Z_r$
and $B_{\infty} = \bigcup B_r$ of $E_1$ exist then we define
the {\it limit}\footnote{This notation is not universally accepted. In some
references an additional pair of subobjects
$Z_\infty$ and $B_\infty$ of $E_1$ such that
$0 = B_1 \subset B_2 \subset \ldots \subset B_\infty \subset Z_\infty
\subset \ldots \subset Z_2 \subset Z_1 = E_1$
is part of the data comprising a spectral sequence!}
of the spectral sequence to be the object
$E_{\infty} = Z_{\infty}/B_{\infty}$.
\item We say that the spectral sequence {\it degenerates at $E_r$}
if the differentials $d_r, d_{r + 1}, \ldots$ are all zero.
\end{enumerate}
\end{definition}
\noindent
Note that if the spectral sequence degenerates at $E_r$, then
we have $E_r = E_{r + 1} = \ldots = E_{\infty}$ (and the limit
exists of course). Also, almost any abelian category we will encounter
has countable sums and intersections.
\begin{remark}[Variant]
\label{remark-allow-translation-functors}
It is often the case that the terms of a spectral sequence have
additional structure, for example a grading or a bigrading.
To accomodate this (and to get around certain technical issues)
we introduce the following notion. Let $\mathcal{A}$ be an
abelian category. Let $(T_r)_{r \geq 1}$ be a
sequence of {\it translation} or {\it shift} functors, i.e.,
$T_r : \mathcal{A} \to \mathcal{A}$ is an isomorphism of categories.
In this setting a {\it spectral sequence} is given by a system
$(E_r, d_r)_{r \geq 1}$ where each $E_r$ is an object of
$\mathcal{A}$, each $d_r : E_r \to T_rE_r$
is a morphism such that $T_rd_r \circ d_r = 0$ so that
$$
\xymatrix{
\ldots \ar[r] &
T_r^{-1}E_r \ar[r]^-{T_r^{-1}d_r} &
E_r \ar[r]^-{d_r} &
T_rE_r \ar[r]^{T_r d_r} &
T_r^2E_r \ar[r] & \ldots
}
$$
is a complex and $E_{r + 1} = \Ker(d_r)/\Im(T_r^{-1}d_r)$ for $r \geq 1$.
It is clear what a {\it morphism of spectral sequences}
means in this setting. In this setting we can still define
$$
0 = B_1 \subset B_2 \subset \ldots \subset B_r \subset \ldots
\subset Z_r \subset \ldots \subset Z_2 \subset Z_1 = E_1
$$
and $Z_\infty$ and $B_\infty$ (if they exist) as above.
\end{remark}
```

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