Spectral sequence

Fix an abelian category, such as a category of modules over a ring. A cohomological spectral sequence is a choice of a nonnegative integer ${\displaystyle r_{0}}$ and a collection of three sequences:

1. For all integers ${\displaystyle r\geq r_{0}}$, an object ${\displaystyle E_{r}}$, called a sheet (as in a sheet of paper), or sometimes a page or a term;
2. Endomorphisms ${\displaystyle d_{r}\colon E_{r}\to E_{r}}$ satisfying ${\displaystyle d_{r}\circ d_{r}=0}$, called boundary maps or differentials;
3. Isomorphisms of ${\displaystyle E_{r+1}}$ with ${\displaystyle H(E_{r})}$, the homology of ${\displaystyle E_{r}}$ with respect to ${\displaystyle d_{r}}$.

Usually the isomorphisms between ${\displaystyle E_{r+1}}$ and ${\displaystyle H(E_{r})}$ are suppressed, and we write equalities instead. Sometimes ${\displaystyle E_{r+1}}$ is called the derived object of ${\displaystyle E_{r}}$.

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 ${\displaystyle E_{2}}$ to equal ${\displaystyle E_{1}}$, 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_•
• E_r = 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 ${\displaystyle E_{r}}$.

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 E → E' is by definition a collection of maps f_r : E_r → E'_r which are compatible with the differentials and with the given isomorphisms between cohomology of the rth step and the (r+1)th sheets of E and E' , respectively.

Let E_r 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.)

Let ${\displaystyle E_{p,q}^{r}}$ be a homological spectral sequence such that ${\displaystyle E_{p,q}^{2}=0}$ for all p other than 0, 1. Visually, this is the spectral sequence with ${\displaystyle E^{2}}$-page

${\displaystyle {\begin{matrix}&\vdots &\vdots &\vdots &\vdots &\\\cdots &0&E_{0,2}^{2}&E_{1,2}^{2}&0&\cdots \\\cdots &0&E_{0,1}^{2}&E_{1,1}^{2}&0&\cdots \\\cdots &0&E_{0,0}^{2}&E_{1,0}^{2}&0&\cdots \\\cdots &0&E_{0,-1}^{2}&E_{1,-1}^{2}&0&\cdots \\&\vdots &\vdots &\vdots &\vdots &\end{matrix}}}$

The differentials on the second page have degree (-2, 1), so they are of the form

${\displaystyle d_{p,q}^{2}:E_{p,q}^{2}\to E_{p-2,q+1}^{2}}$

These maps are all zero since they are

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

hence the spectral sequence degenerates: ${\displaystyle E^{\infty }=E^{2}}$. Say, it converges to ${\displaystyle H_{*}}$ 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]

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

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 ${\displaystyle F_{\bullet }}$. 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 ${\displaystyle p=0,n}$, hence the ${\displaystyle E^{2}}$-page is given by

${\displaystyle {\begin{matrix}&\vdots &\vdots &\vdots &&\vdots &\vdots &\vdots &\\\cdots &0&E_{0,2}^{2}&0&\cdots &0&E_{n,2}^{2}&0&\cdots \\\cdots &0&E_{0,1}^{2}&0&\cdots &0&E_{n,1}^{2}&0&\cdots \\\cdots &0&E_{0,0}^{2}&0&\cdots &0&E_{n,0}^{2}&0&\cdots \\\end{matrix}}}$

Moreover, since

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

for ${\displaystyle p=0,n}$ by the universal coefficient theorem, the ${\displaystyle E^{2}}$ page looks like

${\displaystyle {\begin{matrix}&\vdots &\vdots &\vdots &&\vdots &\vdots &\vdots &\\\cdots &0&H_{2}(F)&0&\cdots &0&H_{2}(F)&0&\cdots \\\cdots &0&H_{1}(F)&0&\cdots &0&H_{1}(F)&0&\cdots \\\cdots &0&H_{0}(F)&0&\cdots &0&H_{0}(F)&0&\cdots \\\end{matrix}}}$

Since the only non-zero differentials are on the ${\displaystyle E^{n}}$-page, given by

${\displaystyle d_{n,q}^{n}:E_{n,q}^{n}\to E_{0,q+n-1}^{n}}$

which is

${\displaystyle d_{n,q}^{n}:H_{q}(F)\to H_{q+n-1}(F)}$

the spectral sequence converges on ${\displaystyle E^{n+1}=E^{\infty }}$. By computing ${\displaystyle E^{n+1}}$ we get an exact sequence

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

and written out using the homology groups, this is

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

To establish what the two ${\displaystyle E^{\infty }}$-terms are, write ${\displaystyle H=H(E)}$, and 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 ${\displaystyle F_{n}H_{q}=H_{q}}$,

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

This is the exact sequence

${\displaystyle 0\to H_{q}(F)\to H_{q}(E)\to H_{q-n}(F)\to 0.}$

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.)

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:

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

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

Determining these maps are fundamental for computing many differentials in the Serre spectral sequence. For instance the transgression map determines the differential[5]^{pg 540,564}

${\displaystyle d_{n}:E_{n,0}^{n}\to E_{0,n-1}^{n}}$

for the homological spectral spectral sequence, hence on the Serre spectral sequence for a fibration ${\displaystyle F\to E\to B}$ gives the map

${\displaystyle d_{n}:H_{n}(B)\to H_{n-1}(F)}$

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. 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 : A → A, g : A → C and h : C → A 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 E_n be C⁽ⁿ⁾ and d_n be g⁽ⁿ⁾ o h⁽ⁿ⁾. This gives a spectral sequence.

• Serre spectral sequence[9] - used to compute (co)homology of a fibration
• Atiyah–Hirzebruch spectral sequence - used to compute (co)homology of extraordinary cohomology theories, such as K-theory
• Bockstein spectral sequence.
• Spectral sequences of filtered complexes

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 F^pC^•, 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(F^pCⁿ) ⊆ F^pCⁿ⁺¹. We assume that the filtration is descending, i.e., F^pC^• ⊇ F^p+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 F^pC^• is zero, and that the filtration is exhaustive, that is, the union of the set of all F^pC^• is the entire chain complex C^•.

The filtration is useful because it gives a measure of nearness to zero: As p increases, F^pC^• 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 = n − p. (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 d_r on each E_r and verify that it leads to homology isomorphic to E_r+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 E_r with respect to this differential is E_r+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.

• Can be used to construct Mixed Hodge structures[10]

• Hodge–de Rham spectral sequence
• 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 C_i,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:

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

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

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 C_i,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 C_p,q → C_p,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.

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 rth 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 pth graded part of the (p+q)th homology of C. If our spectral sequence converges, then we conclude that:

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

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 → A^• → B^• → C^• → 0, and call the first map f^• : A^• → B^•. 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 d_0,q : H^q(C^•) → H^q+1(A^•). These are the connecting homomorphisms from the snake lemma, and together with the maps A^• → B^• → C^•, 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.

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 H^p+q(T(C_•,•)) are distinct. Despite this, it is still possible to gain useful information from these two spectral sequences.

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 ${\displaystyle \operatorname {Tor} _{i}(M,N)=\operatorname {Tor} _{i}(N,M)}$. While this can be verified without a spectral sequence, it is very easy with spectral sequences.