Bourbaki

Exercise 14, § 2

Algebra, X, p. 175

Bourbaki marks this one as harder than the rest.

We call spectral sequence of AA-modules the datum of a sequence of AA-complexes (Er,dr)r≥2(E_r, d_r)_{r \geq 2} and of mappings αr:H(Er)→Er+1\alpha_r : H(E_r) \to E_{r+1}, such that:

(i) The grading of the complex ErE_r is the total grading associated with a bigrading

Er=⨁p,q∈ZErp,q(r≥2); E_r = \bigoplus_{p, q \in \mathbf{Z}} E_r^{p, q} \quad (r \geq 2);
one has dr(Erpq)⊂Erp+r,q−r+1d_r(E_r^{pq}) \subset E_r^{p+r, q-r+1}.

(ii) The mappings αr\alpha_r are isomorphisms of bigraded AA-modules. We shall also use the descending bigrading of ErE_r; we shall denote Erpq=Er−p,−qE_r^{pq} = E_r^{-p, -q}.

a) We shall denote here by Zr+1(Er)Z_{r+1}(E_r) (resp. Br+1(Er)B_{r+1}(E_r)) the bigraded AA-module Ker dr\mathrm{Ker}\ d_r (resp. Im dr\mathrm{Im}\ d_r). Let

pr:Er→Er/Br+1(Er) p_r : E_r \to E_r / B_{r+1}(E_r)
be the mapping of passing to the quotient and ir:Er+1→Er/Br+1(Er)i_r : E_{r+1} \to E_r / B_{r+1}(E_r) the mapping composed of αr−1\alpha_r^{-1} and the natural injection. For all r≥2r \geq 2, define by induction on kk the bigraded submodules Zr+k(Er)Z_{r+k}(E_r) and Br+k(Er)B_{r+k}(E_r) of ErE_r by putting:
Zr+k(Er)=pr−1(ir(Zr+k(Er+1)));Br+k(Er)=pr−1(ir(Br+k(Er+1))). Z_{r+k}(E_r) = p_r^{-1}(i_r(Z_{r+k}(E_{r+1}))) ; \qquad B_{r+k}(E_r) = p_r^{-1}(i_r(B_{r+k}(E_{r+1}))).
Prove that one has the inclusions:
0⊂Br+1(Er)⊂Br+2(Er)⊂…⊂Zr+2(Er)⊂Zr+1(Er)⊂Er 0 \subset B_{r+1}(E_r) \subset B_{r+2}(E_r) \subset \ldots \subset Z_{r+2}(E_r) \subset Z_{r+1}(E_r) \subset E_r
and that EkE_k is isomorphic to Zk(Er)/Bk(Er)Z_k(E_r)/B_k(E_r) for k≥r+1k \geq r+1.

We put Z∞(E2)=⋂r≥3Zr(E2)Z_\infty(E_2) = \bigcap_{r \geq 3} Z_r(E_2), B∞(E2)=⋃r≥3Br(E2)B_\infty(E_2) = \bigcup_{r \geq 3} B_r(E_2) and E∞=Z∞(E2)/B∞(E2)E_\infty = Z_\infty(E_2)/B_\infty(E_2). The spectral sequence is said to be regular if the decreasing sequence (Zr(E2pq))r≥2(Z_r(E_2^{pq}))_{r \geq 2} is stationary for every pair (p,q)(p, q).

b) Let E=(Er,dr,αr)r≥2E = (E_r, d_r, \alpha_r)_{r \geq 2} and E′=(Er′,dr′,αr′)r≥2E' = (E'_r, d'_r, \alpha'_r)_{r \geq 2} be two spectral sequences of AA-modules; we call a morphism from EE into E′E' a sequence of morphisms of AA-complexes ur:Er→Er′u_r : E_r \to E'_r, compatible with the bigradations, such that αr′∘H(ur)=ur+1∘αr\alpha'_r \circ H(u_r) = u_{r+1} \circ \alpha_r for all r≥2r \geq 2. Prove that for all r≥2r \geq 2 and all k≥r+1k \geq r+1, uru_r induces a homomorphism from Zk(Er)Z_k(E_r) (resp. Bk(Er)B_k(E_r)) into Zk(Er′)Z_k(E'_r) (resp. Bk(Er′)B_k(E'_r)). In particular, u2u_2 induces homomorphisms Z∞(u2):Z∞(E2)→Z∞(E2′)Z_\infty(u_2) : Z_\infty(E_2) \to Z_\infty(E'_2), B∞(u2):B∞(E2)→B∞(E2′)B_\infty(u_2) : B_\infty(E_2) \to B_\infty(E'_2) and u∞:E∞→E∞′u_\infty : E_\infty \to E'_\infty; all these homomorphisms are compatible with the bigradations.

c) Suppose that u2u_2 is an isomorphism. Now prove that uru_r is an isomorphism for all r≥2r \geq 2, as well as B∞(u2)B_\infty(u_2). If, moreover, the sequences EE and E′E' are regular, Z∞(u2)Z_\infty(u_2) and u∞u_\infty are isomorphisms.

d) Let G=⨁n∈ZGnG = \bigoplus_{n \in \mathbf{Z}} G^n be a graded AA-module, endowed with a decreasing sequence (FpG)p∈Z(F^p G)_{p \in \mathbf{Z}} of graded sub-AA-modules, such that ⋂pFpG=0\bigcap_p F^p G = 0 and ⋃pFpG=G\bigcup_p F^p G = G. We say that the spectral sequence EE approaches (G,(FpG)p∈Z)(G, (F^p G)_{p \in \mathbf{Z}}) (or simply GG) if there exist isomorphisms βpq:E∞pq→(FpG)p+q/(Fp+1G)p+q\beta^{pq} : E_\infty^{pq} \to (F^p G)^{p+q}/(F^{p+1} G)^{p+q} for all

(p,q)∈Z×Z. (p, q) \in \mathbf{Z} \times \mathbf{Z}.

We say that the spectral sequence E converges to G if it approaches G and if, moreover: (i) The spectral sequence E is regular; (ii) For every integer n, there exists an integer p such that (FpG)n=0(F^p G)^n = 0.

Suppose that E (resp. E') converges to G (resp. G'). Let u:E→E′u : E \to E' be a morphism of spectral sequences, v:G→G′v : G \to G' a homomorphism of degree 0, such that v(FpG)⊂FpG′v(F^p G) \subset F^p G' for all p and that the induced homomorphisms vˉp,n:(FpG)n/(Fp+1G)n→(FpG′)n/(Fp+1G′)n\bar{v}^{p,n} : (F^p G)^n/(F^{p+1} G)^n \to (F^p G')^n/(F^{p+1} G')^n satisfy vˉp,p+q∘βp,q=β′p,q∘u∞p,q\bar{v}^{p,p+q} \circ \beta^{p,q} = {\beta'}^{p,q} \circ u_{\infty}^{p,q} for every pair (p,q)(p, q). Now prove that if u2pqu_2^{pq} is an isomorphism for every pair (p,q)(p, q), then vv is an isomorphism.

Solution

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Machine translation of the English, by gpt-5-6-mini, gpt-5-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.