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SPECTRAL SEQUENCES 1. Introduction Definition 1. Let a ≥ 1 SPECTRAL SEQUENCES MATTHEW GREENBERG 1. Introduction Definition 1. Let a ≥ 1. An a-th stage spectral (cohomological) sequence consists of the following data: L p,q • bigraded objects Er = E , r ≥ a p,q∈Z r p,q p+r,q−r+1 • differentials dr : Er → Er such that dr(Er ) ⊆ Er satisfying H(Er) = Er+1, i.e., p,q p+r,q−r+1 p,q ker(Er → Er ) Er+1 = p−r,q+r−1 p,q . im(Er → Er ) We usually draw the r-th stage of a spectral sequence in a tabular format with p increasing horizontally to the left and q increasing vertically to the right: 0,1 1,1 2,1 3,1 E0 E0 E0 E0 ··· O O O O 0,0 1,0 2,0 3,0 E0 E0 E0 E0 ··· 0,1 1,1 2,1 3,1 E1 / E1 / E1 / E1 / ··· 0,0 1,0 2,0 3,0 E1 / E1 / E1 / E1 / ··· 0,1 1,1 2,1 3,1 E2 E2 E2 E2 ··· QQQ QQQ QQQ QQQ QQQ QQQ QQQ QQQ QQQ QQQ QQQ QQ QQQ QQQ QQQ 0,0 1,0 Q( 2,0 Q( 3,0 QQ( E2 E2 E2 E2 ··· 2. The spectral sequence of a filtered complex Let K· = F 0K· ⊇ F 1K· ⊇ · · · be a filtered complex (i.e., each object Kn in the complex K· is filtered and the differentials of the complex K· respect the filtration). We set Grp K· = F pK·/F p+1K·. Note that the filtration on K· induces a filtration H(K·) = F 0H(K·) ⊇ F 1H(K·) ⊇ · · · on cohomology in a natural way. Given this setup, one may prove the following. Theorem 2. Suppose that K· is a nonnegative filtered complex (i.e., Kn = 0 if p,q n < 0). Then there exists a spectral sequence Er satisfying 1 2 MATTHEW GREENBERG p,q p p+q • E0 = Gr K p,q p+q p · • E1 = H (Gr K ) p,q p p+q · • Er = Gr H (K ) for r = r(p, q) 0. p,q p p+q · Sketch of proof. Set E0 = Gr K . As differential of the complex K respects p,q p,q+1 p,q the filtration, it induces a map d0 : E0 → E0 . To realize Er , r ≥ 1, one defines the set p,q p p+q p+r p+q+1 Zr = {x ∈ F K | dx ∈ F K } of ‘cocycles modulo F p+rKp+q+1’ and sets p,q p,q Er = Zr /(appropriate coboundaries). The differential d then induces the map dr : Er → Er (it is at least evident from p,q p+r,q−r+1 the above that we get a map Zr → Zr ). It remains to check the desired properties of these object. We omit this; for details see [3, Ch. XX, Proposition 9.1]. The spectral sequence whose existence is asserted in the above theorem is an example of a first quadrant spectral sequance, by definition a spectral sequence such p,q that Er is zero unless p, q ≥ 0. It is easy to see that in a first quadrant spectral p,q p,q sequence, Er = Er+1 = ··· if r > max(p, q + 1). In the situation of the theorem, this stable value is Grp Hp+q(K·). 3. Convergence of spectral sequences p,q Let Er be a spectral sequence, and suppose that for every pair (p, q), the term p,q Er stabilizes as r → ∞ (a first quadrant spectral sequence, for example). Denote p,q n this stable value by E∞ . Let H be a collection of objects with finite filtrations 0 ⊆ F sHn ⊆ · · · ⊆ F tHn = Hn. p,q · p,q p+q We say that Er converges to H , and write Er ⇒ H , if p,q p p+q p+1 p+q p p+q E∞ = F H /F H = Gr H . Example 3. Suppose K· is a nonnegative filtered complex. Then by Theorem 2, we have a convergent spectral sequence p,q p+q p · p+q · E1 = H (Gr K ) ⇒ H (K ). Given this definition of convergence, one is led immediately to ask to what extent the limit of a spectral sequence is determined by the sequence itself. The next two lemmas give some basic though useful results in this direction. p,q p+q Lemma 4. Suppose Er ⇒ H . p,q n n−q0,q0 (1) If E∞ = 0 unless q = q0, then H = E∞ . p,q n p0,n−p0 (2) If E∞ = 0 unless p = p0, then H = E∞ . Proof. We prove (1), the proof of (2) being similar. Consider the filtration 0 ⊆ · · · ⊆ F n−q0+1Hn ⊆ F n−q0 Hn ⊆ F n−q0−1Hn ⊆ · · · ⊆ Hn. p,q Since E∞ = 0 unless q = q0, the only nonzero quotient of this filtration is n−q0 n n−q0+1 n n−q0,q0 n−q0+1 n n−q0 n n F H /F H = E∞ . Thus, F H = 0 and F H = H , implying n n−q0 n n n−q0+1 n n−q0,q0 H = F H = H /F H = E∞ . SPECTRAL SEQUENCES 3 The next result shows that in certain cases the form of the filtration on the limit object is determined by the spectral sequence. p,q p+q Lemma 5. Suppose Er ⇒ H . p,q n (1) If Er is a first quarter spectral sequence, then H has a filtration of the form 0 = F n+1Hn ⊆ F nHn ⊆ · · · ⊆ F 1Hn ⊆ F 0Hn = Hn. p,q −n (2) If Er is a third quarter spectral sequence, then H has a filtration of the form 0 = F 1H−n ⊆ F 0H−n ⊆ · · · ⊆ F −n+1H−n ⊆ F −nH−n = H−n. p,q p,q A third quarter spectral sequence Er is one in which Er = 0 unless p, q ≤ 0. Proof. Again we only prove (1), the proof of (2) being similar. We must show that F n+kHn = 0 if k > 0 and F kHn = Hn if k ≤ 0. Consider the “left tail” of the (finite!) filtration of Hn, 0 ⊆ · · · ⊆ F n+2Hn ⊆ F n+1Hn ⊆ F nHn. p,q As Er is a first quarter spectral sequence, ( n+k,−k n+k n n+k+1 n E∞ = 0 if k > 0 F H /F H = n,0 E∞ if k = 0. Therefore, F n+kHn = 0 if k > 0. One shows that F kHn = Hn if k ≤ 0 in a similar fashion by considering the “right tail” of the filtration. 4. The spectral sequence of a double complex In this section, we treat one of the most common ways spectral sequences arise – from a double complex. Definition 6. A double complex M consists of a bigraded object M = L M p,q p,q∈Z together with differentials d : M p,q → M p+1,q and δ : M p,q → M p,q+1 satisfying d2 = δ2 = dδ + δd = 0. · · Example 7. Let R be a ring (P , dP ) and (Q , dQ) be complexes of R-modules. · · p,q p q Define a double complex M = P ⊗R Q by M = P ⊗R Q , d = dP , and p p q p q+1 δ = (−1) dQ : P ⊗R Q → P ⊗R Q . To each double complex M, we attach a (single) complex Tot M called its total complex defined by M Totn M = M p,q. p+q=n The differential D on this total complex is given by D = d + δ. Notice that D2 = (d + δ)2 = d2 + δ2 + dδ + δd = 0, i.e., (Tot M, D) is a complex. There are two canonical filtrations on the total complex Tot M of a double com- plex M given by M M 0F p Totn M = M r,s and 00F q Totn M = M r,s. r+s=n r+s=n r≥p s≥q By the theorem of Section 2, the filtrations 0F p Totn M and 00F p Totn M determine 0 p,q 00 p,q 0 p,q spectral sequcences Er and Er , respectively. One observes easily that E0 = 4 MATTHEW GREENBERG p,q 0 p,q 0 p,q+1 M and checks (perhaps not so easily) that the differential E0 → E0 arising from the construction of the spectral sequence of a filtration is simply given by δ. 0 p,q q p,· 0 p,q 0 p+1,q Thus, E1 = Hδ (M ). The differential E1 → E1 is induced by d, viewed as p,· p+1,· 0 p,q a homomorphism of complexes M → M , implying that E2 is equal to the p- q p,· th homology group of the complex (Hδ (M ), d). We will write this simply (though p q 00 p,q slightly ambiguously) as Hd (Hδ (M)). One may obtain similar expressions for Er , r = 0, 1, 2. We summarize these important results in the following theorem. Theorem 8. Let M be a double complex with total complex Tot M. Then there 0 p,q 00 p,q exist two spectral sequences Er and Er (corresponding to the two canonical filtrations on Tot M) such that 0 p,q p,q 0 p,q q p,· 0 p,q p q E0 = M , E1 = Hδ (M ), E2 = Hd (Hδ (M)), 00 p,q q,p 00 p,q q ·,p 00 p,q p q E0 = M , E1 = Hd (M ), E2 = Hδ (Hd (M)). 0 p,q 00 p,q Further, if M is a first or third quadrant double complex, then both Er and Er converge to Hp+q(Tot M).
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