# Spectral sequence

In homological algebra and algebraic topology, a spectral sequence is a means of computing homology groups by taking successive approximations. Spectral sequences are a generalization of exact sequences, and since their introduction by Jean Leray (1946), they have become important computational tools, particularly in algebraic topology, algebraic geometry and homological algebra.

## Discovery and motivation

Motivated by problems in algebraic topology, Jean Leray introduced the notion of a sheaf and found himself faced with the problem of computing sheaf cohomology. To compute sheaf cohomology, Leray introduced a computational technique now known as the Leray spectral sequence. This gave a relation between cohomology groups of a sheaf and cohomology groups of the pushforward of the sheaf. The relation involved an infinite process. Leray found that the cohomology groups of the pushforward formed a natural chain complex, so that he could take the cohomology of the cohomology. This was still not the cohomology of the original sheaf, but it was one step closer in a sense. The cohomology of the cohomology again formed a chain complex, and its cohomology formed a chain complex, and so on. The limit of this infinite process was essentially the same as the cohomology groups of the original sheaf.

It was soon realized that Leray's computational technique was an example of a more general phenomenon. Spectral sequences were found in diverse situations, and they gave intricate relationships among homology and cohomology groups coming from geometric situations such as fibrations and from algebraic situations involving derived functors. While their theoretical importance has decreased since the introduction of derived categories, they are still the most effective computational tool available. This is true even when many of the terms of the spectral sequence are incalculable.

Unfortunately, because of the large amount of information carried in spectral sequences, they are difficult to grasp. This information is usually contained in a rank three lattice of abelian groups or modules. The easiest cases to deal with are those in which the spectral sequence eventually collapses, meaning that going out further in the sequence produces no new information. Even when this does not happen, it is often possible to get useful information from a spectral sequence by various tricks.

## Formal definition

Fix an abelian category, such as a category of modules over a ring. A spectral sequence is a choice of a nonnegative integer r0 and a collection of three sequences:

1. For all integers rr0, an object Er , called a sheet (as in a sheet of paper), or sometimes a page or a term,
2. Endomorphisms dr : ErEr satisfying dr o dr = 0, called boundary maps or differentials,
3. Isomorphisms of Er+1 with H(Er), the homology of Er with respect to dr.

Usually the isomorphisms between Er+1 and H(Er) are suppressed, and we write equalities instead. Sometimes Er+1 is called the derived object of Er.[citation needed]

The most elementary example is a chain complex C. An object C in an abelian category of chain complexes comes with a differential d. Let r0 = 0, and let E0 be C. This forces E1 to be the complex H(C): At the i'th location this is the i'th homology group of C. The only natural differential on this new complex is the zero map, so we let d1 = 0. This forces E2 to equal E1, and again our only natural differential is the zero map. Putting the zero differential on all the rest of our sheets gives a spectral sequence whose terms are:

• E0 = C
• Er = H(C) for all r ≥ 1.

The terms of this spectral sequence stabilize at the first sheet because its only nontrivial differential was on the zeroth sheet. Consequently, we can get no more information at later steps. Usually, to get useful information from later sheets, we need extra structure on the Er.

In the ungraded situation described above, r0 is irrelevant, but in practice most spectral sequences occur in the category of doubly graded modules over a ring R (or doubly graded sheaves of modules over a sheaf of rings). In this case, each sheet is a doubly graded module, so it decomposes as a direct sum of terms with one term for each possible bidegree. The boundary map is defined as the direct sum of boundary maps on each of the terms of the sheet. Their degree depends on r and is fixed by convention. For a homological spectral sequence, the terms are written ${\displaystyle E_{p,q}^{r}}$ and the differentials have bidegree (− r,r − 1). For a cohomological spectral sequence, the terms are written ${\displaystyle E_{r}^{p,q}}$ and the differentials have bidegree (r, 1 − r). (These choices of bidegree occur naturally in practice; see the example of a double complex below.) Depending upon the spectral sequence, the boundary map on the first sheet can have a degree which corresponds to r = 0, r = 1, or r = 2. For example, for the spectral sequence of a filtered complex, described below, r0 = 0, but for the Grothendieck spectral sequence, r0 = 2. Usually r0 is zero, one, or two.

A morphism of spectral sequences EE' is by definition a collection of maps fr : ErE'r which are compatible with the differentials and with the given isomorphisms between cohomology of the r-th step and the (r + 1)-st sheets of E and E' , respectively.

Let Er be a spectral sequence, starting with say r = 1. Then there is a sequence of subobjects

${\displaystyle 0=B_{0}\subset B_{1}\subset B_{2}\subset \dots \subset B_{r}\subset \dots \subset Z_{r}\subset \dots \subset Z_{2}\subset Z_{1}\subset Z_{0}=E_{1}}$

such that ${\displaystyle E_{r}\simeq Z_{r-1}/B_{r-1}}$; indeed, recursively we let ${\displaystyle Z_{0}=E_{1},B_{0}=0}$ and let ${\displaystyle Z_{r},B_{r}}$ be so that ${\displaystyle Z_{r}/B_{r-1},B_{r}/B_{r-1}}$ are the kernel and the image of ${\displaystyle E_{r}{\overset {d_{r}}{\to }}E_{r}.}$

We then let ${\displaystyle Z_{\infty }=\cap _{r}Z_{r},B_{\infty }=\cup _{r}B_{r}}$ and

${\displaystyle E_{\infty }=Z_{\infty }/B_{\infty }}$;

it is called the limiting term. (Of course, such ${\displaystyle E_{\infty }}$ need not exist in the category, but this is usually a non-issue since for example in the category of modules such limits exist or since in practice a spectral sequence one works with tends to degenerate; there are only finitely many inclusions in the sequence above.)

## Visualization

The E2 sheet of a cohomological spectral sequence

A doubly graded spectral sequence has a tremendous amount of data to keep track of, but there is a common visualization technique which makes the structure of the spectral sequence clearer. We have three indices, r, p, and q. For each r, imagine that we have a sheet of graph paper. On this sheet, we will take p to be the horizontal direction and q to be the vertical direction. At each lattice point we have the object ${\displaystyle E_{r}^{p,q}}$.

It is very common for n = p + q to be another natural index in the spectral sequence. n runs diagonally, northwest to southeast, across each sheet. In the homological case, the differentials have bidegree (−rr − 1), so they decrease n by one. In the cohomological case, n is increased by one. When r is zero, the differential moves objects one space down or up. This is similar to the differential on a chain complex. When r is one, the differential moves objects one space to the left or right. When r is two, the differential moves objects just like a knight's move in chess. For higher r, the differential acts like a generalized knight's move.

## Worked-out examples

When learning spectral sequences for the first time, it is often helpful to work with simple computational examples. For more formal and complete discussions, see the sections below. For the examples in this section, it suffices to use this definition: one says a spectral sequence converges to H with an increasing filtration F if ${\displaystyle E_{p,q}^{\infty }=F_{p}H_{p+q}/F_{p-1}H_{p+q}}$. The examples below illustrate how one relates such filtrations with the E2-term in the forms of exact sequences; many exact sequences in applications (e.g., Gysin sequence) arise in this fashion.

### 2 columns and 2 rows

Let ${\displaystyle E_{p,q}^{r}}$ be a spectral sequence such that ${\displaystyle E_{p,q}^{2}=0}$ for all p other than 0, 1. The differentials on the second page have degree (-2, 1) and therefore they are all zero; i.e., the spectral sequence degenerates: ${\displaystyle E^{\infty }=E^{2}}$. Say, it converges to H with a filtration

${\displaystyle 0=F_{-1}H_{n}\subset F_{0}H_{n}\subset \dots \subset F_{n}H_{n}=H_{n}}$

such that ${\displaystyle E_{p,q}^{\infty }=F_{p}H_{p+q}/F_{p-1}H_{p+q}}$. Then ${\displaystyle F_{0}H_{n}=E_{0,n}^{2}}$, ${\displaystyle F_{1}H_{n}/F_{0}H_{n}=E_{1,n-1}^{2}}$, ${\displaystyle F_{2}H_{n}/F_{1}H_{n}=0}$, ${\displaystyle F_{3}H_{n}/F_{2}H_{n}=0}$, etc. Thus, there is the exact sequence:[1]

${\displaystyle 0\to E_{0,n}^{2}\to H_{n}\to E_{1,n-1}^{2}\to 0}$.

Next, let ${\displaystyle E_{p,q}^{r}}$ be a spectral sequence whose second page consists only of two lines q = 0, 1. This need not degenerate at the second page but it still degenerates at the third page as the differentials there have degree (-3, 2). Note ${\displaystyle E_{p,0}^{3}=\operatorname {ker} (d:E_{p,0}^{2}\to E_{p-2,1}^{2})}$, as the denominator is zero. Similarly, ${\displaystyle E_{p,1}^{3}=\operatorname {coker} (d:E_{p+2,0}^{2}\to E_{p,1}^{2})}$. Thus,

${\displaystyle 0\to E_{p,0}^{\infty }\to E_{p,0}^{2}{\overset {d}{\to }}E_{p-2,1}^{2}\to E_{p-2,1}^{\infty }\to 0}$.

Now, say, the spectral sequence converges to H with a filtration F as in the previous example. Since ${\displaystyle F_{p-2}H_{p}/F_{p-3}H_{p}=E_{p-2,2}^{\infty }=0}$, ${\displaystyle F_{p-3}H_{p}/F_{p-4}H_{p}=0}$, etc., we have: ${\displaystyle 0\to E_{p-1,1}^{\infty }\to H_{p}\to E_{p,0}^{\infty }\to 0}$. Putting everything together, one gets:[2]

${\displaystyle \cdots \to H_{p+1}\to E_{p+1,0}^{2}{\overset {d}{\to }}E_{p-1,1}^{2}\to H_{p}\to E_{p,0}^{2}{\overset {d}{\to }}E_{p-2,1}^{2}\to H_{p-1}\to \dots .}$

### Wang sequence

The computation in the previous section generalizes in a straightforward way. Consider a fibration over a sphere:

${\displaystyle F{\overset {i}{\to }}E{\overset {p}{\to }}S^{n}}$

with n at least 2. There is the Serre spectral sequence:

${\displaystyle E_{p,q}^{2}=H_{p}(S^{n};H_{q}(F))\Rightarrow H_{p+q}(E)}$;

that is to say, ${\displaystyle E_{p,q}^{\infty }=F_{p}H_{p+q}(E)/F_{p-1}H_{p+q}(E)}$ with some filtration F.

Since ${\displaystyle H_{p}(S^{n})}$ is nonzero only when p is zero or n and equal to Z in that case, we see ${\displaystyle E_{p,q}^{2}}$ consists of only two lines p = 0, n and moreover ${\displaystyle E_{p,q}^{2}=H_{q}(F)}$ for p = 0, n, by the universal coefficient theorem. Clearly, ${\displaystyle E^{\infty }=E^{n+1},E^{n}=E^{2}}$ and by computing ${\displaystyle E^{n+1}}$ we see:

${\displaystyle 0\to E_{n,q-n}^{\infty }\to E_{n,q-n}^{n}{\overset {d}{\to }}E_{0,q-1}^{n}\to E_{0,q-1}^{\infty }\to 0.}$

Now, writing ${\displaystyle H=H(E)}$, since ${\displaystyle F_{1}H_{q}/F_{0}H_{q}=E_{1,q-1}^{\infty }=0}$, etc., we have: ${\displaystyle E_{n,q-n}^{\infty }=F_{n}H_{q}/F_{0}H_{q}}$ and thus, since ${\displaystyle F_{n}H_{q}=H_{q}}$,

${\displaystyle 0\to E_{0,q}^{\infty }\to H_{q}\to E_{n,q-n}^{\infty }\to 0.}$

Putting all calculations together, one gets:[3]

${\displaystyle \dots \to H_{q}(F){\overset {i_{*}}{\to }}H_{q}(E)\to H_{q-n}(F){\overset {d}{\to }}H_{q-1}(F){\overset {i_{*}}{\to }}H_{q-1}(E)\to H_{q-n-1}(F)\to \dots }$

(The Gysin sequence is obtained in a similar way.)

### Low-degree terms

With an obvious notational change, the type of the computations in the previous examples can also be carried out for cohomological spectral sequence. Let ${\displaystyle E_{r}^{p,q}}$ be a first-quadrant spectral sequence converging to H with the decreasing filtration

${\displaystyle 0=F^{n+1}H^{n}\subset F^{n}H^{n}\subset \dots \subset F^{0}H^{n}=H^{n}}$

so that ${\displaystyle E_{\infty }^{p,q}=F^{p}H^{p+q}/F^{p+1}H^{p+q}.}$ Since ${\displaystyle E_{2}^{p,q}}$ is zero if p or q is negative, we have:

${\displaystyle 0\to E_{\infty }^{0,1}\to E_{2}^{0,1}{\overset {d}{\to }}E_{2}^{2,0}\to E_{\infty }^{2,0}\to 0.}$

Since ${\displaystyle E_{\infty }^{1,0}=E_{2}^{1,0}}$ for the same reason and since ${\displaystyle F^{2}H^{1}=0,}$

${\displaystyle 0\to E_{2}^{1,0}\to H^{1}\to E_{\infty }^{0,1}\to 0}$.

Since ${\displaystyle F^{3}H^{2}=0}$, ${\displaystyle E_{\infty }^{2,0}\subset H^{2}}$. Stacking the sequences together, we get the so-called five-term exact sequence:

${\displaystyle 0\to E_{2}^{1,0}\to H^{1}\to E_{2}^{0,1}{\overset {d}{\to }}E_{2}^{2,0}\to H^{2}.}$

## Edge maps and transgressions

Let ${\displaystyle E_{p,q}^{r}}$ be a spectral sequence. If ${\displaystyle E_{p,q}^{r}=0}$ for every q < 0, then it must be that: for r ≥ 2,

${\displaystyle E_{p,0}^{r+1}=\operatorname {ker} (d:E_{p,0}^{r}\to E_{p-r,r-1}^{r})}$

as the denominator is zero. Hence, there is a sequence of monomorphisms:

${\displaystyle E_{p,0}^{r}\to E_{p,0}^{r-1}\to \dots \to E_{p,0}^{3}\to E_{p,0}^{2}}$.

They are called the edge maps. Similarly, if ${\displaystyle E_{p,q}^{r}=0}$ for every p < 0, then there is a sequence of epimorphisms (also called the edge maps):

${\displaystyle E_{0,q}^{2}\to E_{0,q}^{3}\to \dots \to E_{0,q}^{r-1}\to E_{0,q}^{r}}$.

The transgression is a partially-defined map (more precisely, a map from a subobject to a quotient)

${\displaystyle \tau :E_{p,0}^{2}\to E_{0,p-1}^{2}}$

given as a composition ${\displaystyle E_{p,0}^{2}\to E_{p,0}^{p}{\overset {d}{\to }}E_{0,p-1}^{p}\to E_{0,p-1}^{2}}$, the first and last maps being the inverses of the edge maps.[4]

For a spectral sequence ${\displaystyle E_{r}^{p,q}}$ of cohomological type, the analogous statements hold. If ${\displaystyle E_{r}^{p,q}=0}$ for every q < 0, then there is a sequence of epimorphisms

${\displaystyle E_{2}^{p,0}\to E_{3}^{p,0}\to \dots \to E_{r-1}^{p,0}\to E_{r}^{p,0}}$.

And if ${\displaystyle E_{r}^{p,q}=0}$ for every p < 0, then there is a sequence of monomorphisms:

${\displaystyle E_{r}^{0,q}\to E_{r-1}^{0,q}\to \dots \to E_{3}^{0,q}\to E_{2}^{0,q}}$.

The transgression is a not necessarily well-defined map:

${\displaystyle \tau :E_{2}^{0,q-1}\to E_{2}^{q,0}}$

induced by ${\displaystyle d:E_{q}^{0,q-1}\to E_{q}^{q,0}}$.

## Multiplicative structure

A cup product gives a ring structure to a cohomology group, turning it into a cohomology ring. Thus, it is natural to consider a spectral sequence with a ring structure as well. Let ${\displaystyle E_{r}^{p,q}}$ be a spectral sequence of cohomological type. We say it has multiplicative structure if (i) ${\displaystyle E_{r}}$ are (doubly graded) differential graded algebras and (ii) the multiplication on ${\displaystyle E_{r+1}}$ is induced by that on ${\displaystyle E_{r}}$ via passage to cohomology.

A typical example is the cohomological Serre spectral sequence for a fibration ${\displaystyle F\to E\to B}$, B simply connected, when the coefficient group is a "ring" R. It has the multiplicative structure such that the limiting term ${\displaystyle E_{\infty }}$ is isomorphic as graded algebra over R to the associated graded algebra of H(E; R), the latter having a ring structure induced by cup product.[5] The multiplicative structure on a spectral sequence can be very useful in calculating the spectral sequence; for some concrete example, see the page on the Serre spectral sequence.

## Constructions of spectral sequences

Spectral sequences can be constructed by various ways. In algebraic topology, an exact couple is perhaps the most common tool for the construction. In algebraic geometry, spectral sequences are usually constructed from filtrations of cochain complexes.

### Exact couples

The most powerful technique for the construction of spectral sequences is William Massey's method of exact couples. Exact couples are particularly common in algebraic topology, where there are many spectral sequences for which no other construction is known. In fact, all known spectral sequences can be constructed using exact couples.[citation needed] Despite this they are unpopular in abstract algebra, where most spectral sequences come from filtered complexes. To define exact couples, we begin again with an abelian category. As before, in practice this is usually the category of doubly graded modules over a ring. An exact couple is a pair of objects A and C, together with three homomorphisms between these objects: f : AA, g : AC and h : CA subject to certain exactness conditions:

• Image f = Kernel g
• Image g = Kernel h
• Image h = Kernel f

We will abbreviate this data by (A, C, f, g, h). Exact couples are usually depicted as triangles. We will see that C corresponds to the E0 term of the spectral sequence and that A is some auxiliary data.

To pass to the next sheet of the spectral sequence, we will form the derived couple. We set:

• d = g o h
• A' = f(A)
• C' = Ker d / Im d
• f' = f|A', the restriction of f to A'
• h' : C'A' is induced by h. It is straightforward to see that h induces such a map.
• g' : A'C' is defined on elements as follows: For each a in A', write a as f(b) for some b in A. g'(a) is defined to be the image of g(b) in C'. In general, g' can be constructed using one of the embedding theorems for abelian categories.

From here it is straightforward to check that (A', C', f', g', h') is an exact couple. C' corresponds to the E1 term of the spectral sequence. We can iterate this procedure to get exact couples (A(n), C(n), f(n), g(n), h(n)). We let En be C(n) and dn be g(n) o h(n). This gives a spectral sequence.

For a simple example, see the Bockstein spectral sequence.

### The spectral sequence of a filtered complex

A very common type of spectral sequence comes from a filtered cochain complex. This is a cochain complex C together with a set of subcomplexes FpC, where p ranges across all integers. (In practice, p is usually bounded on one side.) We require that the boundary map is compatible with the filtration; this means that d(FpCn) ⊆ FpCn+1. We assume that the filtration is descending, i.e., FpCFp+1C. We will number the terms of the cochain complex by n. Later, we will also assume that the filtration is Hausdorff or separated, that is, the intersection of the set of all FpC is zero, and that the filtration is exhaustive, that is, the union of the set of all FpC is the entire chain complex C.

The filtration is useful because it gives a measure of nearness to zero: As p increases, FpC gets closer and closer to zero. We will construct a spectral sequence from this filtration where coboundaries and cocycles in later sheets get closer and closer to coboundaries and cocycles in the original complex. This spectral sequence is doubly graded by the filtration degree p and the complementary degree q = np. (The complementary degree is often a more convenient index than the total degree n. For example, this is true of the spectral sequence of a double complex, explained below.)

We will construct this spectral sequence by hand. C has only a single grading and a filtration, so we first construct a doubly graded object from C. To get the second grading, we will take the associated graded object with respect to the filtration. We will write it in an unusual way which will be justified at the E1 step:

${\displaystyle Z_{-1}^{p,q}=Z_{0}^{p,q}=F^{p}C^{p+q}}$
${\displaystyle B_{0}^{p,q}=0}$
${\displaystyle E_{0}^{p,q}={\frac {Z_{0}^{p,q}}{B_{0}^{p,q}+Z_{-1}^{p+1,q-1}}}={\frac {F^{p}C^{p+q}}{F^{p+1}C^{p+q}}}}$
${\displaystyle E_{0}=\bigoplus _{p,q\in {\mathbf {Z} }}E_{0}^{p,q}}$

Since we assumed that the boundary map was compatible with the filtration, E0 is a doubly graded object and there is a natural doubly graded boundary map d0 on E0. To get E1, we take the homology of E0.

${\displaystyle {\bar {Z}}_{1}^{p,q}=\ker d_{0}^{p,q}:E_{0}^{p,q}\rightarrow E_{0}^{p,q+1}=\ker d_{0}^{p,q}:F^{p}C^{p+q}/F^{p+1}C^{p+q}\rightarrow F^{p}C^{p+q+1}/F^{p+1}C^{p+q+1}}$
${\displaystyle {\bar {B}}_{1}^{p,q}={\mbox{im }}d_{0}^{p,q-1}:E_{0}^{p,q-1}\rightarrow E_{0}^{p,q}={\mbox{im }}d_{0}^{p,q-1}:F^{p}C^{p+q-1}/F^{p+1}C^{p+q-1}\rightarrow F^{p}C^{p+q}/F^{p+1}C^{p+q}}$
${\displaystyle E_{1}^{p,q}={\frac {{\bar {Z}}_{1}^{p,q}}{{\bar {B}}_{1}^{p,q}}}={\frac {\ker d_{0}^{p,q}:E_{0}^{p,q}\rightarrow E_{0}^{p,q+1}}{{\mbox{im }}d_{0}^{p,q-1}:E_{0}^{p,q-1}\rightarrow E_{0}^{p,q}}}}$
${\displaystyle E_{1}=\bigoplus _{p,q\in {\mathbf {Z} }}E_{1}^{p,q}=\bigoplus _{p,q\in {\mathbf {Z} }}{\frac {{\bar {Z}}_{1}^{p,q}}{{\bar {B}}_{1}^{p,q}}}}$

Notice that ${\displaystyle {\bar {Z}}_{1}^{p,q}}$ and ${\displaystyle {\bar {B}}_{1}^{p,q}}$ can be written as the images in ${\displaystyle E_{0}^{p,q}}$ of

${\displaystyle Z_{1}^{p,q}=\ker d_{0}^{p,q}:F^{p}C^{p+q}\rightarrow C^{p+q+1}/F^{p+1}C^{p+q+1}}$
${\displaystyle B_{1}^{p,q}=({\mbox{im }}d_{0}^{p,q-1}:F^{p}C^{p+q-1}\rightarrow C^{p+q})\cap F^{p}C^{p+q}}$

and that we then have

${\displaystyle E_{1}^{p,q}={\frac {Z_{1}^{p,q}}{B_{1}^{p,q}+Z_{0}^{p+1,q-1}}}.}$

${\displaystyle Z_{1}^{p,q}}$ is exactly the stuff which the differential pushes up one level in the filtration, and ${\displaystyle B_{1}^{p,q}}$ is exactly the image of the stuff which the differential pushes up zero levels in the filtration. This suggests that we should choose ${\displaystyle Z_{r}^{p,q}}$ to be the stuff which the differential pushes up r levels in the filtration and ${\displaystyle B_{r}^{p,q}}$ to be image of the stuff which the differential pushes up r-1 levels in the filtration. In other words, the spectral sequence should satisfy

${\displaystyle Z_{r}^{p,q}=\ker d_{0}^{p,q}:F^{p}C^{p+q}\rightarrow C^{p+q+1}/F^{p+r}C^{p+q+1}}$
${\displaystyle B_{r}^{p,q}=({\mbox{im }}d_{0}^{p-r+1,q+r-2}:F^{p-r+1}C^{p+q-1}\rightarrow C^{p+q})\cap F^{p}C^{p+q}}$
${\displaystyle E_{r}^{p,q}={\frac {Z_{r}^{p,q}}{B_{r}^{p,q}+Z_{r-1}^{p+1,q-1}}}}$

and we should have the relationship

${\displaystyle B_{r}^{p,q}=d_{0}^{p,q}(Z_{r-1}^{p-r+1,q+r-2}).}$

For this to make sense, we must find a differential dr on each Er and verify that it leads to homology isomorphic to Er+1. The differential

${\displaystyle d_{r}^{p,q}:E_{r}^{p,q}\rightarrow E_{r}^{p+r,q-r+1}}$

is defined by restricting the original differential d defined on ${\displaystyle C^{p+q}}$ to the subobject ${\displaystyle Z_{r}^{p,q}}$.

It is straightforward to check that the homology of Er with respect to this differential is Er+1, so this gives a spectral sequence. Unfortunately, the differential is not very explicit. Determining differentials or finding ways to work around them is one of the main challenges to successfully applying a spectral sequence.

### The spectral sequence of a double complex

Another common spectral sequence is the spectral sequence of a double complex. A double complex is a collection of objects Ci,j for all integers i and j together with two differentials, d I and d II. d I is assumed to decrease i, and d II is assumed to decrease j. Furthermore, we assume that the differentials anticommute, so that d I d II + d II d I = 0. Our goal is to compare the iterated homologies ${\displaystyle H_{i}^{I}(H_{j}^{II}(C_{\bullet ,\bullet }))}$ and ${\displaystyle H_{j}^{II}(H_{i}^{I}(C_{\bullet ,\bullet }))}$. We will do this by filtering our double complex in two different ways. Here are our filtrations:

${\displaystyle (C_{i,j}^{I})_{p}={\begin{cases}0&{\text{if }}i
${\displaystyle (C_{i,j}^{II})_{p}={\begin{cases}0&{\text{if }}j

To get a spectral sequence, we will reduce to the previous example. We define the total complex T(C•,•) to be the complex whose n'th term is ${\displaystyle \bigoplus _{i+j=n}C_{i,j}}$ and whose differential is d I + d II. This is a complex because d I and d II are anticommuting differentials. The two filtrations on Ci,j give two filtrations on the total complex:

${\displaystyle T_{n}(C_{\bullet ,\bullet })_{p}^{I}=\bigoplus _{i+j=n \atop i>p-1}C_{i,j}}$
${\displaystyle T_{n}(C_{\bullet ,\bullet })_{p}^{II}=\bigoplus _{i+j=n \atop j>p-1}C_{i,j}}$

To show that these spectral sequences give information about the iterated homologies, we will work out the E0, E1, and E2 terms of the I filtration on T(C•,•). The E0 term is clear:

${\displaystyle {}^{I}E_{p,q}^{0}=T_{n}(C_{\bullet ,\bullet })_{p}^{I}/T_{n}(C_{\bullet ,\bullet })_{p+1}^{I}=\bigoplus _{i+j=n \atop i>p-1}C_{i,j}{\Big /}\bigoplus _{i+j=n \atop i>p}C_{i,j}=C_{p,q},}$

where n = p + q.

To find the E1 term, we need to determine d I + d II on E0. Notice that the differential must have degree −1 with respect to n, so we get a map

${\displaystyle d_{p,q}^{I}+d_{p,q}^{II}:T_{n}(C_{\bullet ,\bullet })_{p}^{I}/T_{n}(C_{\bullet ,\bullet })_{p+1}^{I}=C_{p,q}\rightarrow T_{n-1}(C_{\bullet ,\bullet })_{p}^{I}/T_{n-1}(C_{\bullet ,\bullet })_{p+1}^{I}=C_{p,q-1}}$

Consequently, the differential on E0 is the map Cp,qCp,q−1 induced by d I + d II. But d I has the wrong degree to induce such a map, so d I must be zero on E0. That means the differential is exactly d II, so we get

${\displaystyle {}^{I}E_{p,q}^{1}=H_{q}^{II}(C_{p,\bullet }).}$

To find E2, we need to determine

${\displaystyle d_{p,q}^{I}+d_{p,q}^{II}:H_{q}^{II}(C_{p,\bullet })\rightarrow H_{q}^{II}(C_{p+1,\bullet })}$

Because E1 was exactly the homology with respect to d II, d II is zero on E1. Consequently, we get

${\displaystyle {}^{I}E_{p,q}^{2}=H_{p}^{I}(H_{q}^{II}(C_{\bullet ,\bullet })).}$

Using the other filtration gives us a different spectral sequence with a similar E2 term:

${\displaystyle {}^{II}E_{p,q}^{2}=H_{q}^{II}(H_{p}^{I}(C_{\bullet ,\bullet })).}$

What remains is to find a relationship between these two spectral sequences. It will turn out that as r increases, the two sequences will become similar enough to allow useful comparisons.

## Convergence, degeneration, and abutment

In the elementary example that we began with, the sheets of the spectral sequence were constant once r was at least 1. In that setup it makes sense to take the limit of the sequence of sheets: Since nothing happens after the zeroth sheet, the limiting sheet E is the same as E1.

In more general situations, limiting sheets often exist and are always interesting. They are one of the most powerful aspects of spectral sequences. We say that a spectral sequence ${\displaystyle E_{r}^{p,q}}$ converges to or abuts to ${\displaystyle E_{\infty }^{p,q}}$ if there is an r(p, q) such that for all rr(p, q), the differentials ${\displaystyle d_{r}^{p-r,q+r-1}}$ and ${\displaystyle d_{r}^{p,q}}$ are zero. This forces ${\displaystyle E_{r}^{p,q}}$ to be isomorphic to ${\displaystyle E_{\infty }^{p,q}}$ for large r. In symbols, we write:

${\displaystyle E_{r}^{p,q}\Rightarrow _{p}E_{\infty }^{p,q}}$

The p indicates the filtration index. It is very common to write the ${\displaystyle E_{2}^{p,q}}$ term on the left-hand side of the abutment, because this is the most useful term of most spectral sequences.

In most spectral sequences, the ${\displaystyle E_{\infty }}$ term is not naturally a doubly graded object. Instead, there are usually ${\displaystyle E_{\infty }^{n}}$ terms which come with a natural filtration ${\displaystyle F^{\bullet }E_{\infty }^{n}}$. In these cases, we set ${\displaystyle E_{\infty }^{p,q}={\mbox{gr}}_{p}E_{\infty }^{p+q}=F^{p}E_{\infty }^{p+q}/F^{p+1}E_{\infty }^{p+q}}$. We define convergence in the same way as before, but we write

${\displaystyle E_{r}^{p,q}\Rightarrow _{p}E_{\infty }^{n}}$

to mean that whenever p + q = n, ${\displaystyle E_{r}^{p,q}}$ converges to ${\displaystyle E_{\infty }^{p,q}}$.

The simplest situation in which we can determine convergence is when the spectral sequences degenerates. We say that the spectral sequences degenerates at sheet r if, for any sr, the differential ds is zero. This implies that ErEr+1Er+2 ≅ ... In particular, it implies that Er is isomorphic to E. This is what happened in our first, trivial example of an unfiltered chain complex: The spectral sequence degenerated at the first sheet. In general, if a doubly graded spectral sequence is zero outside of a horizontal or vertical strip, the spectral sequence will degenerate, because later differentials will always go to or from an object not in the strip.

The spectral sequence also converges if ${\displaystyle E_{r}^{p,q}}$ vanishes for all p less than some p0 and for all q less than some q0. If p0 and q0 can be chosen to be zero, this is called a first-quadrant spectral sequence. This sequence converges because each object is a fixed distance away from the edge of the non-zero region. Consequently, for a fixed p and q, the differential on later sheets always maps ${\displaystyle E_{r}^{p,q}}$ from or to the zero object; more visually, the differential leaves the quadrant where the terms are nonzero. The spectral sequence need not degenerate, however, because the differential maps might not all be zero at once. Similarly, the spectral sequence also converges if ${\displaystyle E_{r}^{p,q}}$ vanishes for all p greater than some p0 and for all q greater than some q0.

The five-term exact sequence of a spectral sequence relates certain low-degree terms and E terms.

## Examples of degeneration

### The spectral sequence of a filtered complex, continued

Notice that we have a chain of inclusions:

${\displaystyle Z_{0}^{p,q}\supseteq Z_{1}^{p,q}\supseteq Z_{2}^{p,q}\supseteq \cdots \supseteq B_{2}^{p,q}\supseteq B_{1}^{p,q}\supseteq B_{0}^{p,q}}$

We can ask what happens if we define

${\displaystyle Z_{\infty }^{p,q}=\bigcap _{r=0}^{\infty }Z_{r}^{p,q},}$
${\displaystyle B_{\infty }^{p,q}=\bigcup _{r=0}^{\infty }B_{r}^{p,q},}$
${\displaystyle E_{\infty }^{p,q}={\frac {Z_{\infty }^{p,q}}{B_{\infty }^{p,q}+Z_{\infty }^{p+1,q-1}}}.}$

${\displaystyle E_{\infty }^{p,q}}$ is a natural candidate for the abutment of this spectral sequence. Convergence is not automatic, but happens in many cases. In particular, if the filtration is finite and consists of exactly r nontrivial steps, then the spectral sequence degenerates after the r'th sheet. Convergence also occurs if the complex and the filtration are both bounded below or both bounded above.

To describe the abutment of our spectral sequence in more detail, notice that we have the formulas:

${\displaystyle Z_{\infty }^{p,q}=\bigcap _{r=0}^{\infty }Z_{r}^{p,q}=\bigcap _{r=0}^{\infty }\ker(F^{p}C^{p+q}\rightarrow C^{p+q+1}/F^{p+r}C^{p+q+1})}$
${\displaystyle B_{\infty }^{p,q}=\bigcup _{r=0}^{\infty }B_{r}^{p,q}=\bigcup _{r=0}^{\infty }({\mbox{im }}d^{p,q-r}:F^{p-r}C^{p+q-1}\rightarrow C^{p+q})\cap F^{p}C^{p+q}}$

To see what this implies for ${\displaystyle Z_{\infty }^{p,q}}$ recall that we assumed that the filtration was separated. This implies that as r increases, the kernels shrink, until we are left with ${\displaystyle Z_{\infty }^{p,q}=\ker(F^{p}C^{p+q}\rightarrow C^{p+q+1})}$. For ${\displaystyle B_{\infty }^{p,q}}$, recall that we assumed that the filtration was exhaustive. This implies that as r increases, the images grow until we reach ${\displaystyle B_{\infty }^{p,q}={\text{im }}(C^{p+q-1}\rightarrow C^{p+q})\cap F^{p}C^{p+q}}$. We conclude

${\displaystyle E_{\infty }^{p,q}={\mbox{gr}}_{p}H^{p+q}(C^{\bullet })}$,

that is, the abutment of the spectral sequence is the p'th graded part of the p+q'th homology of C. If our spectral sequence converges, then we conclude that:

${\displaystyle E_{r}^{p,q}\Rightarrow _{p}H^{p+q}(C^{\bullet })}$

#### Long exact sequences

Using the spectral sequence of a filtered complex, we can derive the existence of long exact sequences. Choose a short exact sequence of cochain complexes 0 → ABC → 0, and call the first map f : AB. We get natural maps of homology objects Hn(A) → Hn(B) → Hn(C), and we know that this is exact in the middle. We will use the spectral sequence of a filtered complex to find the connecting homomorphism and to prove that the resulting sequence is exact. To start, we filter B:

${\displaystyle F^{0}B^{n}=B^{n}}$
${\displaystyle F^{1}B^{n}=A^{n}}$
${\displaystyle F^{2}B^{n}=0}$

This gives:

${\displaystyle E_{0}^{p,q}={\frac {F^{p}B^{p+q}}{F^{p+1}B^{p+q}}}={\begin{cases}0&{\text{if }}p<0{\text{ or }}p>1\\C^{q}&{\text{if }}p=0\\A^{q+1}&{\text{if }}p=1\end{cases}}}$
${\displaystyle E_{1}^{p,q}={\begin{cases}0&{\text{if }}p<0{\text{ or }}p>1\\H^{q}(C^{\bullet })&{\text{if }}p=0\\H^{q+1}(A^{\bullet })&{\text{if }}p=1\end{cases}}}$

The differential has bidegree (1, 0), so d0,q : Hq(C) → Hq+1(A). These are the connecting homomorphisms from the snake lemma, and together with the maps ABC, they give a sequence:

${\displaystyle \cdots \rightarrow H^{q}(B^{\bullet })\rightarrow H^{q}(C^{\bullet })\rightarrow H^{q+1}(A^{\bullet })\rightarrow H^{q+1}(B^{\bullet })\rightarrow \cdots }$

It remains to show that this sequence is exact at the A and C spots. Notice that this spectral sequence degenerates at the E2 term because the differentials have bidegree (2, −1). Consequently, the E2 term is the same as the E term:

${\displaystyle E_{2}^{p,q}\cong {\text{gr}}_{p}H^{p+q}(B^{\bullet })={\begin{cases}0&{\text{if }}p<0{\text{ or }}p>1\\H^{q}(B^{\bullet })/H^{q}(A^{\bullet })&{\text{if }}p=0\\{\text{im }}H^{q+1}f^{\bullet }:H^{q+1}(A^{\bullet })\rightarrow H^{q+1}(B^{\bullet })&{\text{if }}p=1\end{cases}}}$

But we also have a direct description of the E2 term as the homology of the E1 term. These two descriptions must be isomorphic:

${\displaystyle H^{q}(B^{\bullet })/H^{q}(A^{\bullet })\cong \ker d_{0,q}^{1}:H^{q}(C^{\bullet })\rightarrow H^{q+1}(A^{\bullet })}$
${\displaystyle {\text{im }}H^{q+1}f^{\bullet }:H^{q+1}(A^{\bullet })\rightarrow H^{q+1}(B^{\bullet })\cong H^{q+1}(A^{\bullet })/({\mbox{im }}d_{0,q}^{1}:H^{q}(C^{\bullet })\rightarrow H^{q+1}(A^{\bullet }))}$

The former gives exactness at the C spot, and the latter gives exactness at the A spot.

### The spectral sequence of a double complex, continued

Using the abutment for a filtered complex, we find that:

${\displaystyle H_{p}^{I}(H_{q}^{II}(C_{\bullet ,\bullet }))\Rightarrow _{p}H^{p+q}(T(C_{\bullet ,\bullet }))}$
${\displaystyle H_{q}^{II}(H_{p}^{I}(C_{\bullet ,\bullet }))\Rightarrow _{q}H^{p+q}(T(C_{\bullet ,\bullet }))}$

In general, the two gradings on Hp+q(T(C•,•)) are distinct. Despite this, it is still possible to gain useful information from these two spectral sequences.

#### Commutativity of Tor

Let R be a ring, let M be a right R-module and N a left R-module. Recall that the derived functors of the tensor product are denoted Tor. Tor is defined using a projective resolution of its first argument. However, it turns out that Tori(M, N) = Tori(N, M). While this can be verified without a spectral sequence, it is very easy with spectral sequences.

Choose projective resolutions P and Q of M and N, respectively. Consider these as complexes which vanish in negative degree having differentials d and e, respectively. We can construct a double complex whose terms are Ci,j = PiQj and whose differentials are d ⊗ 1 and (−1)i(1 ⊗ e). (The factor of −1 is so that the differentials anticommute.) Since projective modules are flat, taking the tensor product with a projective module commutes with taking homology, so we get:

${\displaystyle H_{p}^{I}(H_{q}^{II}(P_{\bullet }\otimes Q_{\bullet }))=H_{p}^{I}(P_{\bullet }\otimes H_{q}^{II}(Q_{\bullet }))}$
${\displaystyle H_{q}^{II}(H_{p}^{I}(P_{\bullet }\otimes Q_{\bullet }))=H_{q}^{II}(H_{p}^{I}(P_{\bullet })\otimes Q_{\bullet })}$

Since the two complexes are resolutions, their homology vanishes outside of degree zero. In degree zero, we are left with

${\displaystyle H_{p}^{I}(P_{\bullet }\otimes N)={\mbox{Tor}}_{p}(M,N)}$
${\displaystyle H_{q}^{II}(M\otimes Q_{\bullet })={\mbox{Tor}}_{q}(N,M)}$

In particular, the ${\displaystyle E_{p,q}^{2}}$ terms vanish except along the lines q = 0 (for the I spectral sequence) and p = 0 (for the II spectral sequence). This implies that the spectral sequence degenerates at the second sheet, so the E terms are isomorphic to the E2 terms:

${\displaystyle {\mbox{Tor}}_{p}(M,N)\cong E_{p}^{\infty }=H_{p}(T(C_{\bullet ,\bullet }))}$
${\displaystyle {\mbox{Tor}}_{q}(N,M)\cong E_{q}^{\infty }=H_{q}(T(C_{\bullet ,\bullet }))}$

Finally, when p and q are equal, the two right-hand sides are equal, and the commutativity of Tor follows.

## Further examples

Some notable spectral sequences are:

## Notes

1. ^ Weibel, Exercise 5.2.1.; there are typos in the exact sequence, at least in the 1994 edition.
2. ^ Weibel, Exercise 5.2.2.
3. ^ Weibel, Application 5.3.5.
4. ^ May, § 1
5. ^ Hatcher, Section 1.2.
6. ^