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 , they have become important computational tools, particularly in algebraic topology, algebraic geometry and homological algebra.

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 , they have become important computational tools, particularly in algebraic topology, algebraic geometry and homological algebra. d_r \circ d_r = 0 , E_{r+1} \cong H_{*}(E_r, d_r) , the homology of E_r with respect to d_r . Usually the isomorphisms are suppressed and we write E_{r+1} = H_{*}(E_r, d_r) instead. An object E_r is called sheet (as in a sheet of paper), or sometimes a page or a term; an endomorphism d_r is called boundary map or differential. Sometimes E_{r+1} is called the derived object of E_r . In reality spectral sequences mostly occur in the category of doubly graded modules over a ring R (or doubly graded sheaves of modules over a sheaf of rings), i.e. every sheet is a bigraded R-module E_r = \bigoplus_{p,q \in \mathbb{Z}^2} E_r^{p,q}. So in this case a cohomological spectral sequence is a sequence \{E_r, d_r\}_{r\geq r_0} of bigraded R-modules \{E_r^{p,q}\}_{p,q} and for every module the direct sum of endomorphisms d_r = (d_r^{p,q} : E_r^{p,q} \to E_r^{p+r,q-r+1})_{p,q \in \mathbb{Z}^2} of bidegree (r,1-r) , such that for every r\geq r_0 it holds that: d_r^{p+r,q-r+1} \circ d_r^{p,q} = 0 , E_{r+1} \cong H_{*}(E_r, d_r) . The notation used here is called complementary degree. Some authors write E_r^{d,q} instead, where d = p + q is the total degree. 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, r 0 = 0, but for the Grothendieck spectral sequence, r 0 = 2. Usually r 0 is zero, one, or two. In the ungraded situation described above, r 0 is irrelevant. The most elementary example in the ungraded situation is a chain complex C • . An object C • in an abelian category of chain complexes naturally comes with a differential d. Let r 0 = 0, and let E 0 be C • . This forces E 1 to be the complex H(C • ): At the th location this is the th homology group of C • . The only natural differential on this new complex is the zero map, so we let d 1 = 0. This forces E_2 to equal 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: E 0 = 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 E_r . 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. An object E_r can be seen as the checkered page of a book. On these sheets, we will take p to be the horizontal direction and q to be the vertical direction. At each lattice point we have the object E_r^{p,q} . Now turning to the next page means taking homology, that is the (r+1) page is a subquotient of the page. The total degree n = p + q runs diagonally, northwest to southeast, across each sheet. In the homological case, the differentials have bidegree , so they decrease n by one. In the cohomological case, n is increased by one. The differentials change their direction with each turn with respect to r. The red arrows demonstrate the case of a first quadrant sequence (see example below), where only the objects of the first quadrant are non-zero. While turning pages, either the domain or the codomain of all the differentials become zero. The set of cohomological spectral sequences form a category: a morphism of spectral sequences f : E \to E' is by definition a collection of maps f_r : E_r \to E'_r which are compatible with the differentials, i.e. f_r \circ d_r = d'_r \circ f_r , and with the given isomorphisms between the cohomology of the r step and the sheets of E and , respectively: f_{r+1}(E_{r+1}) \,=\, f_{r+1}(H(E_r)) \,=\, H(f_r(E_r)) . In the bigraded case, they should also respect the graduation: f_r(E_r^{p,q}) \subset {E'_r}^{p,q}. 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 E 0 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 E 1 term of the spectral sequence. We can iterate this procedure to get exact couples (A (n) , C (n) , f (n) , g (n) , h (n) ). In order to construct a spectral sequence, let E n be C (n) and d n be g (n) o h (n) . A very common type of spectral sequence comes from a filtered cochain complex, as it naturally induces a bigraded object. Consider a cochain complex (C^{\bullet}, d) together with a descending filtration, ... \supset\, F^{-2}C^{\bullet} \,\supset\, F^{-1}C^{\bullet} \supset F^{0}C^{\bullet} \,\supset\, F^{1}C^{\bullet} \,\supset\, F^{2}C^{\bullet} \,\supset\, F^{3}C^{\bullet} \,\supset... \, . We require that the boundary map is compatible with the filtration, i.e. d(F^pC^n) \subset F^pC^{n+1} , and that the filtration is exhaustive, that is, the union of the set of all F^pC^{\bullet} is the entire chain complex C^{\bullet} . Then there exists a spectral sequence with E_0^{p,q} = F^{p}C^{p+q}/F^{p+1}C^{p+q} and E_1^{p,q} = H^{p+q}(F^{p}C^{\bullet}/F^{p+1}C^{\bullet}) . Later, we will also assume that the filtration is Hausdorff or separated, that is, the intersection of the set of all F^pC^{\bullet} is zero. The filtration is useful because it gives a measure of nearness to zero: As p increases, F^pC^{\bullet} 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 . Z_{-1}^{p,q} = Z_0^{p,q} = F^p C^{p+q} B_0^{p,q} = 0 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}} E_0 = \bigoplus_{p,q\in\mathbf{Z}} E_0^{p,q} Since we assumed that the boundary map was compatible with the filtration, E_0 is a doubly graded object and there is a natural doubly graded boundary map d_0 on E_0 . To get E_1 , we take the homology of E_0 . \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} \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} 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}} 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 \bar{Z}_1^{p,q} and \bar{B}_1^{p,q} can be written as the images in E_0^{p,q} of 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} 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 E_1^{p,q} = \frac{Z_1^{p,q}}{B_1^{p,q} + Z_0^{p+1,q-1}}. Z_1^{p,q} are exactly the elements which the differential pushes up one level in the filtration, and B_1^{p,q} are exactly the image of the elements which the differential pushes up zero levels in the filtration. This suggests that we should choose Z_r^{p,q} to be the elements which the differential pushes up r levels in the filtration and B_r^{p,q} to be image of the elements which the differential pushes up r-1 levels in the filtration. In other words, the spectral sequence should satisfy 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} 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} 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 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 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 C^{p+q} to the subobject 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. 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, and . is assumed to decrease i, and is assumed to decrease j. Furthermore, we assume that the differentials anticommute, so that d d + d d = 0. Our goal is to compare the iterated homologies H^\textrm{I}_i(H^\textrm{II}_j(C_{\bullet,\bullet})) and H^\textrm{II}_j(H^\textrm{I}_i(C_{\bullet,\bullet})) . We will do this by filtering our double complex in two different ways. Here are our filtrations: (C_{i,j}^\textrm{I})_p = \begin{cases} 0 & \text{if } i (C_{i,j}^\textrm{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 th term is \bigoplus_{i+j=n} C_{i,j} and whose differential is + . This is a complex because and are anticommuting differentials. The two filtrations on C i,j give two filtrations on the total complex: T_n(C_{\bullet,\bullet})^\textrm{I}_p = \bigoplus_{i+j=n \atop i > p-1} C_{i,j} T_n(C_{\bullet,\bullet})^\textrm{II}_p = \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 E , E , and E terms of the filtration on T(C •,• ). The E term is clear: {}^\textrm{I}E^0_{p,q} = T_n(C_{\bullet,\bullet})^\textrm{I}_p / T_n(C_{\bullet,\bullet})^\textrm{I}_{p+1} = \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 . To find the E term, we need to determine + on E . Notice that the differential must have degree −1 with respect to n, so we get a map d^\textrm{I}_{p,q} + d^\textrm{II}_{p,q} : T_n(C_{\bullet,\bullet})^\textrm{I}_p / T_n(C_{\bullet,\bullet})^\textrm{I}_{p+1} = C_{p,q} \rightarrow T_{n-1}(C_{\bullet,\bullet})^\textrm{I}_p / T_{n-1}(C_{\bullet,\bullet})^\textrm{I}_{p+1} = C_{p,q-1} Consequently, the differential on E is the map C p,q → C p,q−1 induced by + . But has the wrong degree to induce such a map, so must be zero on E . That means the differential is exactly , so we get {}^\textrm{I}E^1_{p,q} = H^\textrm{II}_q(C_{p,\bullet}). To find E , we need to determine d^\textrm{I}_{p,q} + d^\textrm{II}_{p,q} : H^\textrm{II}_q(C_{p,\bullet}) \rightarrow H^\textrm{II}_q(C_{p+1,\bullet}) Because E was exactly the homology with respect to , is zero on E . Consequently, we get {}^\textrm{I}E^2_{p,q} = H^\textrm{I}_p(H^\textrm{II}_q(C_{\bullet,\bullet})). Using the other filtration gives us a different spectral sequence with a similar E term: {}^\textrm{II}E^2_{p,q} = H^\textrm{II}_q(H^{I}_p(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. Let E r be a spectral sequence, starting with say r = 1. Then there is a sequence of subobjects 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 E_r \simeq Z_{r-1}/B_{r-1} ; indeed, recursively we let Z_0 = E_1, B_0 = 0 and let Z_r, B_r be so that Z_r/B_{r-1}, B_r/B_{r-1} are the kernel and the image of E_r \overset{d_r}\to E_r. We then let \textstyle Z_{\infty} = \bigcap_r Z_r, B_{\infty} = \bigcup_r B_r and E_{\infty} = Z_{\infty}/B_{\infty} ; it is called the limiting term. (Of course, such 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.) We say a spectral sequence converges weakly if there is a graded object H^{\bullet} with a filtration F^{\bullet} H^{n} for every n , and for every p there exists an isomorphism E_{\infty}^{p,q} \cong F^pH^{p+q}/F^{p+1}H^{p+q} . It converges to H^{\bullet} if the filtration F^{\bullet} H^{n} is Hausdorff, i.e. \textstyle\bigcap_{p}F^pH^{\bullet}=0 . We write E_r^{p,q} \Rightarrow_p E_\infty^n to mean that whenever p + q = n, E_r^{p,q} converges to E_\infty^{p,q} . We say that a spectral sequence E_r^{p,q} abuts to E_\infty^{p,q} (the spectral sequence abutment ) if for every p,q there is r(p,q) such that for all r \geq r(p,q) , E_r^{p,q} = E_{r(p,q)}^{p,q} . Then E_{r(p,q)}^{p,q} = E_\infty^{p,q} is the limiting term. The spectral sequence is regular or degenerates at r_0 if the differentials d_r^{p,q} are zero for all r \geq r_0 . If in particular there is r_0 \geq 2 , such that the r_0^{th} sheet is concentrated on a single row or a single column, then we say it collapses. In symbols, we write: E_r^{p,q} \Rightarrow_p E_\infty^{p,q} The p indicates the filtration index. It is very common to write the E_2^{p,q} term on the left-hand side of the abutment, because this is the most useful term of most spectral sequences. The spectral sequence of an unfiltered chain complex degenerates at the first sheet (see first example): since nothing happens after the zeroth sheet, the limiting sheet E_{\infty} is the same as E_1 . The five-term exact sequence of a spectral sequence relates certain low-degree terms and E ∞ terms. Z_\infty^{p,q} = \bigcap_{r=0}^\infty Z_r^{p,q}, B_\infty^{p,q} = \bigcup_{r=0}^\infty B_r^{p,q}, E_\infty^{p,q} = \frac{Z_\infty^{p,q}}{B_\infty^{p,q}+Z_\infty^{p+1,q-1}}. 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: 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}) 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 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 Z_\infty^{p,q} = \ker(F^p C^{p+q} \rightarrow C^{p+q+1}) . For 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 B_\infty^{p,q} = \text{im }(C^{p+q-1} \rightarrow C^{p+q}) \cap F^p C^{p+q} . We conclude E_\infty^{p,q} = \mbox{gr}_p H^{p+q}(C^\bull) , 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: E_r^{p,q} \Rightarrow_p H^{p+q}(C^\bull) F^0 B^n = B^n F^1 B^n = A^n F^2 B^n = 0 This gives: E^{p,q}_0 C^q & \text{if } p = 0 \\ A^{q+1} & \text{if } p = 1 \end{cases} E^{p,q}_1 H^{q+1}(A^\bull) & \text{if } p = 1 \end{cases} The differential has bidegree (1, 0), so d 0,q : H q ( ) → H q+1 ( ). These are the connecting homomorphisms from the snake lemma, and together with the maps → → , they give a sequence: \cdots\rightarrow H^q(B^\bull) \rightarrow H^q(C^\bull) \rightarrow H^{q+1}(A^\bull) \rightarrow H^{q+1}(B^\bull) \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 E 2 term because the differentials have bidegree (2, −1). Consequently, the E 2 term is the same as the E ∞ term: E^{p,q}_2 \cong \text{gr}_p H^{p+q}(B^\bull) \text{im } H^{q+1}f^\bull : H^{q+1}(A^\bull) \rightarrow H^{q+1}(B^\bull) &\text{if } p = 1 \end{cases} But we also have a direct description of the E 2 term as the homology of the E 1 term. These two descriptions must be isomorphic: H^q(B^\bull)/H^q(A^\bull) \cong \ker d^1_{0,q} : H^q(C^\bull) \rightarrow H^{q+1}(A^\bull) \text{im } H^{q+1}f^\bull : H^{q+1}(A^\bull) \rightarrow H^{q+1}(B^\bull) \cong H^{q+1}(A^\bull) / (\mbox{im } d^1_{0,q} : H^q(C^\bull) \rightarrow H^{q+1}(A^\bull)) The former gives exactness at the C spot, and the latter gives exactness at the A spot. 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 \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. Choose projective resolutions P_\bull and Q_\bull 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 C_{i,j} = P_i \otimes Q_j and whose differentials are d \otimes 1 and (-1)^\textrm{I}(1 \otimes 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: H^\textrm{I}_p(H^\textrm{II}_q(P_\bull \otimes Q_\bull)) = H^\textrm{I}_p(P_\bull \otimes H^\textrm{II}_q(Q_\bull)) H^\textrm{II}_q(H^\textrm{I}_p(P_\bull \otimes Q_\bull)) = H^\textrm{II}_q(H^\textrm{I}_p(P_\bull) \otimes Q_\bull) Since the two complexes are resolutions, their homology vanishes outside of degree zero. In degree zero, we are left with H^\textrm{I}_p(P_\bull \otimes N) = \operatorname{Tor}_p(M,N) H^\textrm{II}_q(M \otimes Q_\bull) = \operatorname{Tor}_q(N,M) In particular, the E^2_{p,q} terms vanish except along the lines q = 0 (for the spectral sequence) and p = 0 (for the spectral sequence). This implies that the spectral sequence degenerates at the second sheet, so the E ∞ terms are isomorphic to the E 2 terms: \operatorname{Tor}_p(M,N) \cong E^\infty_p = H_p(T(C_{\bull,\bull})) \operatorname{Tor}_q(N,M) \cong E^\infty_q = H_q(T(C_{\bull,\bull})) Finally, when p and q are equal, the two right-hand sides are equal, and the commutativity of Tor follows. The sequence abuts because E_{r+i}^{p,q} = E_r^{p,q} holds for all i\geq 0 if r>p and r>q+1 . To see this, note that either the domain or the codomain of the differential is zero for the considered cases. In visual terms, the sheets stabilize in a growing rectangle (see picture above). 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 E_r^{p,q} vanishes for all p greater than some p_0 and for all q greater than some q_0 . Let E^r_{p, q} be a homological spectral sequence such that E^2_{p, q} = 0 for all p other than 0, 1. Visually, this is the spectral sequence with E^2 -page \begin{matrix} & \vdots & \vdots & \vdots & \vdots & \\ \cdots & 0 & E^2_{0,2} & E^2_{1,2} & 0 & \cdots \\ \cdots & 0 & E^2_{0,1} & E^2_{1,1} & 0 & \cdots \\ \cdots & 0 & E^2_{0,0} & E^2_{1,0} & 0 & \cdots \\ \cdots & 0 & E^2_{0,-1} & E^2_{1,-1} & 0 & \cdots \\ & \vdots & \vdots & \vdots & \vdots & \end{matrix} The differentials on the second page have degree (-2, 1), so they are of the form d^2_{p,q}:E^2_{p,q} \to E^2_{p-2,q+1} These maps are all zero since they are d^2_{0,q}:E^2_{0,q} \to 0 , d^2_{1,q}:E^2_{1,q} \to 0 hence the spectral sequence degenerates: E^{\infty} = E^2 . Say, it converges to H_* with a filtration 0 = F_{-1} H_n \subset F_0 H_n \subset \dots \subset F_n H_n = H_n such that E^{\infty}_{p, q} = F_p H_{p+q}/F_{p-1} H_{p+q} . Then F_0 H_n = E^2_{0, n} , F_1 H_n / F_0 H_n = E^2_{1, n -1} , F_2 H_n / F_1 H_n = 0 , F_3 H_n / F_2 H_n = 0 , etc. Thus, there is the exact sequence: 0 \to E^2_{0, n} \to H_n \to E^2_{1, n - 1} \to 0 . Next, let E^r_{p, q} 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 E^3_{p, 0} = \operatorname{ker} (d: E^2_{p, 0} \to E^2_{p - 2, 1}) , as the denominator is zero. Similarly, E^3_{p, 1} = \operatorname{coker}(d: E^2_{p+2, 0} \to E^2_{p, 1}) . Thus, 0 \to E^{\infty}_{p, 0} \to E^2_{p, 0} \overset{d}\to E^2_{p-2, 1} \to E^{\infty}_{p-2, 1} \to 0 . Now, say, the spectral sequence converges to H with a filtration F as in the previous example. Since F_{p-2} H_{p} / F_{p-3} H_{p} = E^{\infty}_{p-2, 2} = 0 , F_{p-3} H_p / F_{p-4} H_p = 0 , etc., we have: 0 \to E^{\infty}_{p - 1, 1} \to H_p \to E^{\infty}_{p, 0} \to 0 . Putting everything together, one gets: \cdots \to H_{p+1} \to E^2_{p + 1, 0} \overset{d}\to E^2_{p - 1, 1} \to H_p \to E^2_{p, 0} \overset{d}\to E^2_{p - 2, 1} \to H_{p-1} \to \dots. E^2_{p, q} = H_p(S^n; H_q(F)) \Rightarrow H_{p+q}(E) ; that is to say, E^{\infty}_{p, q} = F_p H_{p+q}(E)/F_{p-1} H_{p+q}(E) with some filtration F_\bullet . Since H_p(S^n) is nonzero only when p is zero or n and equal to Z in that case, we see E^2_{p, q} consists of only two lines p = 0,n , hence the E^2 -page is given by \begin{matrix} & \vdots & \vdots & \vdots & & \vdots & \vdots & \vdots & \\ \cdots & 0 & E^2_{0,2} & 0 & \cdots & 0 & E^2_{n,2} & 0 & \cdots \\ \cdots & 0 & E^2_{0,1} & 0 & \cdots & 0 & E^2_{n,1} & 0 & \cdots \\ \cdots & 0 & E^2_{0,0} & 0 & \cdots & 0 & E^2_{n,0} & 0 & \cdots \\ \end{matrix} Moreover, since E^2_{p, q} = H_p(S^n;H_q(F)) = H_q(F) for p = 0,n by the universal coefficient theorem, the E^2 page looks like \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 E^n -page, given by d^n_{n,q}:E^n_{n,q} \to E^n_{0,q+n-1} which is d^n_{n,q}:H_q(F) \to H_{q+n-1}(F) the spectral sequence converges on E^{n+1} = E^{\infty} . By computing E^{n+1} we get an exact sequence 0 \to E^{\infty}_{n, q-n} \to E^n_{n, q-n} \overset{d}\to E^n_{0, q-1} \to E^{\infty}_{0, q-1} \to 0. and written out using the homology groups, this is 0 \to E^{\infty}_{n, q-n} \to H_{q-n}(F) \overset{d}\to H_{q-1}(F) \to E^{\infty}_{0, q-1} \to 0. To establish what the two E^\infty -terms are, write H = H(E) , and since F_1 H_q/F_0 H_q = E^{\infty}_{1, q - 1} = 0 , etc., we have: E^{\infty}_{n, q-n} = F_n H_q / F_0 H_q and thus, since F_0 H_q = E^{\infty}_{0, q} and F_n H_q = H_q as the first and the last term of the filtration, 0 \to E^{\infty}_{0, q} \to H_q \to E^{\infty}_{n, q - n} \to 0. This is the exact sequence 0 \to H_q(F) \to H_q(E) \to H_{q-n}(F)\to 0. Putting all calculations together, one gets: \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.) 0 = F^{n+1} H^n \subset F^n H^n \subset \dots \subset F^0 H^n = H^n so that E_{\infty}^{p,q} = F^p H^{p+q}/F^{p+1} H^{p+q}. Since E_2^{p, q} is zero if p or q is negative, we have: 0 \to E^{0, 1}_{\infty} \to E^{0, 1}_2 \overset{d}\to E^{2, 0}_2 \to E^{2, 0}_{\infty} \to 0. Since E_{\infty}^{1, 0} = E_2^{1, 0} for the same reason and since F^2 H^1 = 0, 0 \to E_2^{1, 0} \to H^1 \to E^{0, 1}_{\infty} \to 0 . Since F^3 H^2 = 0 , E^{2, 0}_{\infty} \subset H^2 . Stacking the sequences together, we get the so-called five-term exact sequence: 0 \to E^{1, 0}_2 \to H^1 \to E^{0, 1}_2 \overset{d}\to E^{2, 0}_2 \to H^2. Let E^r_{p, q} be a spectral sequence. If E^r_{p, q} = 0 for every q E^{r+1}_{p, 0} = \operatorname{ker}(d: E^r_{p, 0} \to E^r_{p-r, r-1}) as the denominator is zero. Hence, there is a sequence of monomorphisms: E^{r}_{p, 0} \to E^{r-1}_{p, 0} \to \dots \to E^3_{p, 0} \to E^2_{p, 0} . They are called the edge maps. Similarly, if E^r_{p, q} = 0 for every p E^2_{0, q} \to E^3_{0, q} \to \dots \to E^{r-1}_{0, q} \to E^r_{0, q} . The transgression is a partially defined map (more precisely, a map from a subobject to a quotient) \tau: E^2_{p, 0} \to E^2_{0, p - 1} given as a composition E^2_{p, 0} \to E^p_{p, 0} \overset{d}\to E^p_{0, p-1} \to E^2_{0, p - 1} , the first and last maps being the inverses of the edge maps. For a spectral sequence E_r^{p, q} of cohomological type, the analogous statements hold. If E_r^{p, q} = 0 for every q E_{2}^{p, 0} \to E_{3}^{p, 0} \to \dots \to E_{r-1}^{p, 0} \to E_r^{p, 0} . And if E_r^{p, q} = 0 for every p 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: \tau: E_2^{0, q-1} \to E_2^{q, 0} induced by d: E_q^{0, q-1} \to E_q^{q, 0} .

Introductory

References

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