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Isomorphic associated graded objects need not give isomorphic filtered objects

Statement refuted

The assertion “isomorphic associated graded objects imply isomorphic filtered objects” is false. Let $A = Z / 4$ and $B = (Z / 2)^{2}$ . Set $F_{p} A = F_{p} B = 0$ for $p < 0$ , $F_{0} A = {0, 2}$ , $F_{0} B = (Z / 2) \times 0$ , and $F_{p} A = A$ , $F_{p} B = B$ for $p \geq 1$ . These filtered objects have isomorphic graded families and nonisomorphic underlying groups.

Facts & Assumptions

Given: The displayed filtrations on $A = Z / 4$ and $B = (Z / 2)^{2}$ .

[F1]

The graded pieces are $F_{p} / F_{p - 1}$ (Associated graded object of a filtered object).

[F2]

The indicated groups exist with ordinary subgroup quotients and explicit residues (Abelian-group model for spectral-sequence computations).

Counterexample

technique · direct

1.1F1F2

The sets ${0, 2}$ and $(Z / 2) \times 0$ are subgroups by the additions in [F2], so the specified families are finite increasing filtrations. Their degree-zero pieces are isomorphic via $[k]_{2} \mapsto [2 k]_{4}$ for $A$ and $[k]_{2} \mapsto ([k]_{2}, 0)$ for $B$ . The first map is well defined because $2 (k + 2 l) \equiv 2 k (m o d 4)$ , and its two images are exactly $0, 2$ .

2.1F1F2step 1.1

The degree-one quotient of $A$ maps to $Z / 2$ by $[a]_{4} + {0, 2} \mapsto [a]_{2}$ : the two cosets are ${0, 2}$ and ${1, 3}$ , so this is an additive bijection. For $B$ the map $(a, b) + F_{0} B \mapsto b$ is an additive bijection. All other graded pieces are zero, since consecutive filtration pieces agree. Thus the two graded families are isomorphic in each degree.

3.1F2step 2.1∎

Every element of $B$ is killed by $2$ , whereas $[1]_{4}$ in $A$ is not: $2 [1]_{4} = [2]_{4} \neq 0$ . Any additive isomorphism $f : A \to B$ would satisfy $f ([2]_{4}) = 2 f ([1]_{4}) = 0$ , contrary to injectivity. Thus no underlying or filtered-group isomorphism exists.

Source notes

Weibel, §5.2 bounded abutment and extension interpretation; the two finite-group filtrations and all quotient calculations are supplied explicitly here.

Depends on

Used by

Two filtered abelian groups with the same associated graded Example
Collapse solves all extension problems False statement
E infinity is the abutment object False statement
First quadrant support alone identifies the abutment without a filtration False statement
Collapse does not in general split the abutment Proposition

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Sources

Charles A. Weibel, An Introduction to Homological Algebra, Chapter 5 (standard reference, not scraped)
Romyar Sharifi, Homological Algebra (standard reference, not scraped)