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滤过复形的谱序列

发布于 2026-07-23T11:38:34Z · 公开内容更新于 2026-08-17T13:57:09Z

关于https://www.3blue1brown.com/blog/exact-sequence-picturebook/的笔记

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colorlinks=true, linkcolor=blue!55!black, citecolor=blue!55!black, urlcolor=blue!55!black, pdftitle=Spectral Sequences for Filtered Complexes, pdfauthor= \numberwithinequationsection \allowdisplaybreaks\newaliascntpropositiontheorem \aliascntresettheproposition \newaliascntlemmatheorem \aliascntresetthelemma \newaliascntcorollarytheorem \aliascntresetthecorollary \newaliascntdefinitiontheorem \aliascntresetthedefinition \newaliascntconstructiontheorem \aliascntresettheconstruction \newaliascntremarktheorem \aliascntresettheremark

Spectral Sequences for Filtered Complexes
Notes on Ravi Vakil’s Puzzling through exact sequences

Abstract

These notes were written while reading Ravi Vakil’s picture essay Puzzling through exact sequences (also available as a direct PDF). They focus on the part of the essay that constructs the spectral sequence of a filtered complex. Vakil’s symbols $𝐺 𝑝, 𝑞$ denote the small “puzzle pieces” in his diagrams; the purpose of the notation $P r e 𝑝, 𝑞$ and $I m 𝑝, 𝑞$ below is to describe those pieces as canonical subquotients in traditional homological-algebra language.

The intended mathematical route is the one followed in the original working notes: construct the zeroth, first, and second pages explicitly, observe the pattern suggested by the second page, and then carry out the general induction. The elementary lemmas are collected early only as a reference library, so that every subobject calculation used later is justified.

1 Filtered complexes and the basic pieces

Definition 1 (Filtered cochain complex).

Let $(𝐶 ∙, 𝑑 ∙)$ be a cochain complex equipped with a finite filtration

𝐶 ∙ = 𝐹 𝑁 𝐶 ∙ \supset 𝐹 𝑁 - 1 𝐶 ∙ \supset \dots \supset 𝐹 1 𝐶 ∙ \supset 𝐹 0 𝐶 ∙ = 0

such that

𝑑 ∙ (𝐹 𝑝 𝐶 ∙) \subseteq 𝐹 𝑝 𝐶 ∙ + 1 .

Thus the filtration is increasing with the index: $𝐹 𝑝 - 1 𝐶 𝑛 \subseteq 𝐹 𝑝 𝐶 𝑛$ .

Definition 2 (The cumulative image and preimage pieces).

For every $𝑝, 𝑞, 𝑛$ , define

I m 𝑛 𝑝, 𝑞 : = 𝑑 𝑛 (𝐹 𝑝 𝐶 𝑛) \cap 𝐹 𝑞 𝐶 𝑛 + 1, P r e 𝑛 𝑝, 𝑞 : = 𝐹 𝑝 𝐶 𝑛 \cap (𝑑 𝑛) - 1 (𝐹 𝑞 𝐶 𝑛 + 1) .

The object $P r e 𝑛 𝑝, 𝑞$ consists of elements of $𝐹 𝑝 𝐶 𝑛$ whose differential lies in $𝐹 𝑞 𝐶 𝑛 + 1$ , while $I m 𝑛 𝑝, 𝑞$ is the part of $𝑑 𝑛 (𝐹 𝑝 𝐶 𝑛)$ which lies in $𝐹 𝑞 𝐶 𝑛 + 1$ .

Definition 3 (The $𝐺$ -pieces).

The piece $𝐺 𝑛 𝑝, 𝑞$ is the canonically identified subquotient

𝐺 𝑛 𝑝, 𝑞 : = P r e 𝑛 𝑝, 𝑞 P r e 𝑛 𝑝 - 1, 𝑞 + P r e 𝑛 𝑝, 𝑞 - 1 𝑑 𝑛 \leftarrow\leftarrow\leftarrow\leftarrow\leftarrow\leftarrow \to ≅ I m 𝑛 𝑝, 𝑞 I m 𝑛 𝑝 - 1, 𝑞 + I m 𝑛 𝑝, 𝑞 - 1 .

The following six-step filtration diagram illustrates the two presentations of the $𝐺$ -piece in (LABEL:eq:g-piece) and the overlaps among the cumulative $P r e$ - and $I m$ -pieces.

Figure 1: A six-step filtration picture for the

𝐺

-piece, showing its

P r e

-presentation. Can you draw its

I m

-presentation?

1.1 Elementary subobject lemmas

Remark 4 (Reading order).

This subsection is meant to be used as a reference rather than read as a prerequisite course. On a first reading, after learning the definitions of $P r e$ , $I m$ , and $𝐺$ , it is better to continue directly to the constructions of $𝐸 0$ , $𝐸 1$ , and $𝐸 2$ in Definitions˜2, 4, 3 and 3. When a proof cites one of the lemmas below, return here to check the precise subobject identity being used. This preserves the motivating order of the original notes: the first pages reveal the pattern, while the lemmas certify that the calculations are legitimate. The specialized correction-of-representatives identity needed for the general induction is intentionally proved at the exact point where it enters the calculation.

Lemma 5 (Image–preimage identities).

Let $𝑓 : 𝑀 \to 𝑁$ be a morphism of modules, let $𝐴 \subseteq 𝑀$ , and let $𝐵 \subseteq 𝑁$ . Then

𝑓 (𝐴 \cap 𝑓 - 1 (𝐵)) = 𝑓 (𝐴) \cap 𝐵, 𝑓 - 1 (𝑓 (𝐴) + 𝐵) = 𝐴 + 𝑓 - 1 (𝐵) .

Proof.

For (LABEL:eq:image-of-intersection), the inclusion from left to right is immediate. Conversely, if $𝑦 \in 𝑓 (𝐴) \cap 𝐵$ , write $𝑦 = 𝑓 (𝑥)$ with $𝑥 \in 𝐴$ . Since $𝑦 \in 𝐵$ , we have $𝑥 \in 𝑓 - 1 (𝐵)$ .

For (LABEL:eq:preimage-of-sum), the inclusion from right to left is immediate. Conversely, if $𝑓 (𝑥) = 𝑓 (𝑎) + 𝑏$ with $𝑎 \in 𝐴$ and $𝑏 \in 𝐵$ , then $𝑥 - 𝑎 \in 𝑓 - 1 (𝐵)$ , hence $𝑥 \in 𝐴 + 𝑓 - 1 (𝐵)$ . ∎

Lemma 6 (Modular law).

Let $𝑋, 𝑌, 𝑍$ be subobjects of an object $𝑀$ . If $𝑌 \subseteq 𝑋$ , then

𝑋 \cap (𝑌 + 𝑍) = 𝑌 + (𝑋 \cap 𝑍) .

Proof.

The inclusion from right to left is clear. If $𝑥 \in 𝑋 \cap (𝑌 + 𝑍)$ , write $𝑥 = 𝑦 + 𝑧$ with $𝑦 \in 𝑌$ and $𝑧 \in 𝑍$ . Since $𝑥, 𝑦 \in 𝑋$ , also $𝑧 = 𝑥 - 𝑦 \in 𝑋$ , so $𝑧 \in 𝑋 \cap 𝑍$ . ∎

Lemma 7 (Quotient reduction).

Let $𝐴, 𝐵, 𝐾$ be subobjects of $𝑀$ , with $𝐵 \subseteq 𝐴$ . Then there is a natural isomorphism

𝐾 + 𝐴 𝐾 + 𝐵 ≅ 𝐴 𝐴 \cap (𝐾 + 𝐵) .

If $𝐼 \subseteq 𝐴$ , then

𝐾 + 𝐴 𝐾 + 𝐼 ≅ 𝐴 (𝐴 \cap 𝐾) + 𝐼 .

Proof.

The natural map $𝐴 \to (𝐾 + 𝐴) / (𝐾 + 𝐵)$ is surjective and has kernel $𝐴 \cap (𝐾 + 𝐵)$ , which gives (LABEL:eq:quotient-reduction). Equation (LABEL:eq:quotient-reduction-modular) follows from Lemma˜6, since $𝐼 \subseteq 𝐴$ implies $𝐴 \cap (𝐾 + 𝐼) = (𝐴 \cap 𝐾) + 𝐼$ . ∎

Lemma 8 (Basic calculus of $P r e$ and $I m$ ).

For every $𝑝, 𝑞, 𝑛$ ,

𝑑 𝑛 (P r e 𝑛 𝑝, 𝑞) = I m 𝑛 𝑝, 𝑞 .

Consequently, there is a short exact sequence

0 ⟶ 𝐹 𝑝 𝐶 𝑛 \cap k e r 𝑑 𝑛 ⟶ P r e 𝑛 𝑝, 𝑞 𝑑 𝑛 ⟶ I m 𝑛 𝑝, 𝑞 ⟶ 0 .

Moreover, for every $𝑟$ ,

I m 𝑛 𝑝, 𝑞 \subseteq P r e 𝑛 + 1 𝑞, 𝑟 .

If $𝑎 \leq 𝑞$ , then

I m 𝑛 𝑝, 𝑞 \cap P r e 𝑛 + 1 𝑎, 𝑏 = I m 𝑛 𝑝, 𝑎 .

Finally, $P r e$ and $I m$ are monotone in each filtration index. In particular,

𝑝 \leq 𝑞 ⟹ P r e 𝑛 𝑝, 𝑞 = 𝐹 𝑝 𝐶 𝑛 .

Proof.

Equation (LABEL:eq:d-pre-im) is Lemma˜5 applied to $𝑓 = 𝑑 𝑛$ , $𝐴 = 𝐹 𝑝 𝐶 𝑛$ , and $𝐵 = 𝐹 𝑞 𝐶 𝑛 + 1$ . Its kernel is $𝐹 𝑝 𝐶 𝑛 \cap k e r 𝑑 𝑛$ , giving (LABEL:eq:pre-im-ses).

If $𝑦 \in I m 𝑛 𝑝, 𝑞$ , then $𝑦 \in 𝐹 𝑞 𝐶 𝑛 + 1$ and $𝑑 𝑛 + 1 𝑦 = 0$ , so $𝑦 \in P r e 𝑛 + 1 𝑞, 𝑟$ for every $𝑟$ . This proves (LABEL:eq:im-in-pre).

Every element of $I m 𝑛 𝑝, 𝑞$ is closed. Hence, if $𝑎 \leq 𝑞$ ,

I m 𝑛 𝑝, 𝑞 \cap P r e 𝑛 + 1 𝑎, 𝑏 = I m 𝑛 𝑝, 𝑞 \cap 𝐹 𝑎 𝐶 𝑛 + 1 = I m 𝑛 𝑝, 𝑎,

which is (LABEL:eq:im-intersect-pre). The monotonicity assertions follow from the monotonicity of the filtration. If $𝑝 \leq 𝑞$ , then $𝑑 (𝐹 𝑝) \subseteq 𝐹 𝑝 \subseteq 𝐹 𝑞$ , giving (LABEL:eq:pre-trivial). ∎

Lemma 9 (The two presentations of a $𝐺$ -piece).

The map in (LABEL:eq:g-piece) is a well-defined isomorphism.

Proof.

By (LABEL:eq:d-pre-im), $𝑑 𝑛$ sends the two terms in the source denominator onto the two terms in the target denominator, so the induced map is well-defined and surjective.

Suppose $𝑥 \in P r e 𝑛 𝑝, 𝑞$ and the class of $𝑑 𝑛 𝑥$ vanishes in the target. Write

𝑑 𝑛 𝑥 = 𝑢 + 𝑣, 𝑢 \in I m 𝑛 𝑝 - 1, 𝑞, 𝑣 \in I m 𝑛 𝑝, 𝑞 - 1 .

Choose $𝑎 \in P r e 𝑛 𝑝 - 1, 𝑞$ and $𝑏 \in P r e 𝑛 𝑝, 𝑞 - 1$ with $𝑑 𝑛 𝑎 = 𝑢$ and $𝑑 𝑛 𝑏 = 𝑣$ . Then $𝑑 𝑛 (𝑥 - 𝑎 - 𝑏) = 0$ . A closed element belongs to $P r e 𝑛 𝑝, 𝑞 - 1$ , so $𝑥$ lies in the source denominator. ∎

Lemma 10 (Stabilization under $𝑑 2 = 0$ ).

Suppose $𝑏 \leq 𝑎$ . Then, for every $𝑐$ ,

P r e 𝑛 𝑝, 𝑎 \cap (𝑑 𝑛) - 1 (P r e 𝑛 + 1 𝑏, 𝑐) = P r e 𝑛 𝑝, 𝑏 .

Proof.

An element in the left-hand side lies in $𝐹 𝑝 𝐶 𝑛$ , its differential lies in both $𝐹 𝑎 𝐶 𝑛 + 1$ and $𝐹 𝑏 𝐶 𝑛 + 1$ , and its second differential lies in $𝐹 𝑐 𝐶 𝑛 + 2$ . Since $𝑏 \leq 𝑎$ , the first two differential conditions reduce to $𝑑 𝑛 𝑥 \in 𝐹 𝑏 𝐶 𝑛 + 1$ , and the last condition is automatic because $𝑑 𝑛 + 1 𝑑 𝑛 = 0$ . ∎

2 The zeroth and first pages

Remark 1 (Guiding idea for the first pages).

The original approach to the first pages was to work entirely inside the original complex $𝐶 ∙$ : construct subobjects $𝑍 𝑝, 𝑝 + 𝑞 𝑖$ and $𝐵 𝑝, 𝑝 + 𝑞 𝑖$ , identify the corresponding page as a subquotient $𝑍 𝑝, 𝑝 + 𝑞 𝑖 / 𝐵 𝑝, 𝑝 + 𝑞 𝑖$ , and let the original differential $𝑑 𝑞$ induce the differential on that subquotient. This works directly for $𝐸 0$ and $𝐸 1$ , and it also produces the recursive presentation $𝐸 2 ≅ 𝑍 1 / 𝐵 1$ .

The construction of the second page, however, reveals a subtlety. The recursively defined $𝑍 1$ contains redundant representatives from a lower filtration level, and the original differential need not send all of those representatives into the recursively defined target $𝑍 1$ . Before defining $𝑑 2$ , one must therefore replace $𝑍 1 / 𝐵 1$ by an isomorphic normalized subquotient. The same recursive-then-normalized procedure becomes the model for the general induction.

Definition 2 (The zeroth page).

Define

𝐸 𝑝, 𝑝 + 𝑞 0 : = 𝐹 𝑝 𝐶 𝑞 𝐹 𝑝 - 1 𝐶 𝑞,

with differential

𝑑 𝑝, 𝑝 + 𝑞 0 : 𝐸 𝑝, 𝑝 + 𝑞 0 ⟶ 𝐸 𝑝, 𝑝 + 𝑞 + 1 0

induced by $𝑑 𝑞$ .

Proposition 3 (The zeroth differential removes $𝐺 𝑞 𝑝, 𝑝$ ).

There is a natural identification

i m 𝑑 𝑝, 𝑝 + 𝑞 0 ≅ 𝐺 𝑞 𝑝, 𝑝 .

Proof.

By the first isomorphism theorem and Lemma˜8,

i m 𝑑 𝑝, 𝑝 + 𝑞 0 ≅ 𝑑 𝑞 (𝐹 𝑝 𝐶 𝑞) 𝑑 𝑞 (𝐹 𝑝 𝐶 𝑞) \cap 𝐹 𝑝 - 1 𝐶 𝑞 + 1 = I m 𝑞 𝑝, 𝑝 I m 𝑞 𝑝, 𝑝 - 1 .

Since $𝑑 (𝐹 𝑝 - 1) \subseteq 𝐹 𝑝 - 1$ ,

I m 𝑞 𝑝 - 1, 𝑝 = I m 𝑞 𝑝 - 1, 𝑝 - 1 \subseteq I m 𝑞 𝑝, 𝑝 - 1,

so the last quotient is the target-side presentation of $𝐺 𝑞 𝑝, 𝑝$ . ∎

Definition 4 (The first page).

Set

𝑍 𝑝, 𝑝 + 𝑞 0 : = (𝑑 𝑞) - 1 (𝐹 𝑝 - 1 𝐶 𝑞 + 1) \cap 𝐹 𝑝 𝐶 𝑞 = P r e 𝑞 𝑝, 𝑝 - 1, 𝐵 𝑝, 𝑝 + 𝑞 0 : = 𝑑 𝑞 - 1 (𝐹 𝑝 𝐶 𝑞 - 1) + 𝐹 𝑝 - 1 𝐶 𝑞 = I m 𝑞 - 1 𝑝, 𝑝 + 𝐹 𝑝 - 1 𝐶 𝑞 .

Then

𝐸 𝑝, 𝑝 + 𝑞 1 : = 𝐻 (𝐸 0, 𝑑 0) 𝑝, 𝑝 + 𝑞 = 𝑍 𝑝, 𝑝 + 𝑞 0 𝐵 𝑝, 𝑝 + 𝑞 0 .

The map $𝑑 𝑞$ induces

𝑑 𝑝, 𝑝 + 𝑞 1 : 𝐸 𝑝, 𝑝 + 𝑞 1 ⟶ 𝐸 𝑝 - 1, 𝑝 + 𝑞 1 .

Proposition 5 (Well-definedness and image of $𝑑 1$ ).

The map (LABEL:eq:d1) is well-defined, and

i m 𝑑 𝑝, 𝑝 + 𝑞 1 ≅ 𝐺 𝑞 𝑝, 𝑝 - 1 .

Proof.

Using (LABEL:eq:d-pre-im) and (LABEL:eq:im-in-pre),

𝑑 𝑞 (𝑍 𝑝, 𝑝 + 𝑞 0) = I m 𝑞 𝑝, 𝑝 - 1 \subseteq P r e 𝑞 + 1 𝑝 - 1, 𝑝 - 2 = 𝑍 𝑝 - 1, 𝑝 + 𝑞 0, 𝑑 𝑞 (𝐵 𝑝, 𝑝 + 𝑞 0) = 𝑑 𝑞 (𝐹 𝑝 - 1 𝐶 𝑞) \subseteq 𝐵 𝑝 - 1, 𝑝 + 𝑞 0 .

Thus $𝑑 1$ is well-defined.

Its image is

im⁡𝑑𝑝,𝑝+𝑞1≅Im𝑞𝑝,𝑝−1Im𝑞𝑝,𝑝−1∩𝐵𝑝−1,𝑝+𝑞0=Im𝑞𝑝,𝑝−1Im𝑞𝑝−1,𝑝−1+Im𝑞𝑝,𝑝−2by \cref{lem:modular-law,lem:pre-im-calculus}=𝐺𝑞𝑝,𝑝−1.

∎

3 The second page

Construction \theconstruction (Recursive cycle and boundary representatives)

Define

𝑍 𝑝, 𝑝 + 𝑞 1 : = 𝑍 𝑝, 𝑝 + 𝑞 0 \cap (𝑑 𝑞) - 1 (𝐵 𝑝 - 1, 𝑝 + 𝑞 0), 𝐵 𝑝, 𝑝 + 𝑞 1 : = 𝐵 𝑝, 𝑝 + 𝑞 0 + 𝑑 𝑞 - 1 (𝑍 𝑝 + 1, 𝑝 + 𝑞 0) .

Then, by construction and the third isomorphism theorem,

𝐻 (𝐸 1, 𝑑 1) 𝑝, 𝑝 + 𝑞 ≅ 𝑍 𝑝, 𝑝 + 𝑞 1 𝐵 𝑝, 𝑝 + 𝑞 1 .

Remark 1 (Why normalization is needed).

At this point the original strategy has produced the correct recursive quotient $𝑍 𝑝, 𝑝 + 𝑞 1 / 𝐵 𝑝, 𝑝 + 𝑞 1$ , so it already identifies $𝐸 2$ as the homology of $(𝐸 1, 𝑑 1)$ . It is not yet the right presentation for defining $𝑑 2$ directly from $𝑑 𝑞$ : the numerator contains lower-filtration representatives whose differentials need not lie in the recursive target numerator. The calculation below first makes those redundant representatives explicit, and Theorem˜3 then removes them by passing to an isomorphic normalized quotient.

Proposition 2 (Calculation of $𝑍 1$ and $𝐵 1$ ).

One has

𝑍 𝑝, 𝑝 + 𝑞 1 = 𝐹 𝑝 - 1 𝐶 𝑞 + P r e 𝑞 𝑝, 𝑝 - 2, 𝐵 𝑝, 𝑝 + 𝑞 1 = 𝐹 𝑝 - 1 𝐶 𝑞 + I m 𝑞 - 1 𝑝 + 1, 𝑝 .

Proof.

For $𝑍 1$ , Lemma˜5 gives

(𝑑 𝑞) - 1 (𝐵 𝑝 - 1, 𝑝 + 𝑞 0) = (𝑑 𝑞) - 1 (𝑑 𝑞 (𝐹 𝑝 - 1 𝐶 𝑞) + 𝐹 𝑝 - 2 𝐶 𝑞 + 1) = 𝐹 𝑝 - 1 𝐶 𝑞 + (𝑑 𝑞) - 1 (𝐹 𝑝 - 2 𝐶 𝑞 + 1) .

Intersecting with $𝑍 𝑝, 𝑝 + 𝑞 0 = P r e 𝑞 𝑝, 𝑝 - 1$ and applying Lemma˜6 yields (LABEL:eq:z1-calculated).

For $𝐵 1$ , equation (LABEL:eq:d-pre-im) gives

𝑑 𝑞 - 1 (𝑍 𝑝 + 1, 𝑝 + 𝑞 0) = I m 𝑞 - 1 𝑝 + 1, 𝑝 .

Since $I m 𝑞 - 1 𝑝, 𝑝 \subseteq I m 𝑞 - 1 𝑝 + 1, 𝑝$ , equation (LABEL:eq:b1-calculated) follows. ∎

Theorem 3 (Standard form of the second page).

There is a natural isomorphism

𝐸 𝑝, 𝑝 + 𝑞 2 ≅ P r e 𝑞 𝑝, 𝑝 - 2 P r e 𝑞 𝑝 - 1, 𝑝 - 2 + I m 𝑞 - 1 𝑝 + 1, 𝑝 .

Proof.

By (LABEL:eq:e2-recursive) and Proposition˜2,

𝐸 𝑝, 𝑝 + 𝑞 2 ≅ 𝐹 𝑝 - 1 𝐶 𝑞 + P r e 𝑞 𝑝, 𝑝 - 2 𝐹 𝑝 - 1 𝐶 𝑞 + I m 𝑞 - 1 𝑝 + 1, 𝑝 .

Apply Lemma˜7. Since

𝐹 𝑝 - 1 𝐶 𝑞 \cap P r e 𝑞 𝑝, 𝑝 - 2 = P r e 𝑞 𝑝 - 1, 𝑝 - 2

and $I m 𝑞 - 1 𝑝 + 1, 𝑝 \subseteq P r e 𝑞 𝑝, 𝑝 - 2$ by (LABEL:eq:im-in-pre), this gives (LABEL:eq:e2-standard). ∎

Proposition 4 (The second differential removes $𝐺 𝑞 𝑝, 𝑝 - 2$ ).

The original differential induces

𝑑 𝑝, 𝑝 + 𝑞 2 : 𝐸 𝑝, 𝑝 + 𝑞 2 ⟶ 𝐸 𝑝 - 2, 𝑝 + 𝑞 - 1 2, [𝑥] ⟼ [𝑑 𝑞 𝑥],

and

i m 𝑑 𝑝, 𝑝 + 𝑞 2 ≅ 𝐺 𝑞 𝑝, 𝑝 - 2 .

Proof.

Use the standard form (LABEL:eq:e2-standard). By (LABEL:eq:d-pre-im) and (LABEL:eq:im-in-pre),

𝑑 𝑞 (P r e 𝑞 𝑝, 𝑝 - 2) = I m 𝑞 𝑝, 𝑝 - 2 \subseteq P r e 𝑞 + 1 𝑝 - 2, 𝑝 - 4 .

Furthermore,

𝑑 𝑞 (P r e 𝑞 𝑝 - 1, 𝑝 - 2 + I m 𝑞 - 1 𝑝 + 1, 𝑝) = I m 𝑞 𝑝 - 1, 𝑝 - 2,

which belongs to the target denominator. Hence $𝑑 2$ is well-defined.

Its image is

im⁡𝑑𝑝,𝑝+𝑞2≅Im𝑞𝑝,𝑝−2Im𝑞𝑝,𝑝−2∩(Pre𝑞+1𝑝−3,𝑝−4+Im𝑞𝑝−1,𝑝−2)=Im𝑞𝑝,𝑝−2Im𝑞𝑝,𝑝−3+Im𝑞𝑝−1,𝑝−2by \cref{lem:modular-law,lem:pre-im-calculus}=𝐺𝑞𝑝,𝑝−2.

∎

4 The general induction

Remark 1 (The second page suggests the induction).

The construction of the second page suggests the general induction process. In particular, the standardized representatives

̃ 𝑍 𝑝, 𝑝 + 𝑞 1 = P r e 𝑞 𝑝, 𝑝 - 2, ̃ 𝐵 𝑝, 𝑝 + 𝑞 1 = P r e 𝑞 𝑝 - 1, 𝑝 - 2 + I m 𝑞 - 1 𝑝 + 1, 𝑝

show the two changes that occur when a page is turned: the cycle condition forces the differential one filtration level deeper, while the boundary term admits images coming from one filtration level higher. The definitions below are obtained by continuing exactly this pattern.

Definition 2 (Standard representatives for the pages).

For $𝑟 \geq 0$ , set

̃ 𝑍 𝑝, 𝑝 + 𝑞 𝑟 : = P r e 𝑞 𝑝, 𝑝 - 𝑟 - 1, ̃ 𝐵 𝑝, 𝑝 + 𝑞 𝑟 : = P r e 𝑞 𝑝 - 1, 𝑝 - 𝑟 - 1 + I m 𝑞 - 1 𝑝 + 𝑟, 𝑝 .

Before writing the resulting formula for $𝐸 𝑟$ , the following six-step filtration diagram shows how the normalized denominator is cut out of the normalized numerator, leaving the same $𝐸 𝑟$ -piece.

Figure 2: A six-step filtration picture for the standard form

𝐸 𝑝, 𝑝 + 𝑞 𝑟 ≅ P r e 𝑞 𝑝, 𝑝 - 𝑟 / (P r e 𝑞 𝑝 - 1, 𝑝 - 𝑟 + I m 𝑞 - 1 𝑝 + 𝑟 - 1, 𝑝)

. The colored overlaps show how the normalized denominator is cut out of the normalized numerator.

Thus the expected standard form of the $𝑟$ th page is

𝐸 𝑝, 𝑝 + 𝑞 𝑟 ≅ ̃ 𝑍 𝑝, 𝑝 + 𝑞 𝑟 - 1 ̃ 𝐵 𝑝, 𝑝 + 𝑞 𝑟 - 1 = P r e 𝑞 𝑝, 𝑝 - 𝑟 P r e 𝑞 𝑝 - 1, 𝑝 - 𝑟 + I m 𝑞 - 1 𝑝 + 𝑟 - 1, 𝑝 .

Theorem 3 (Filtered-complex spectral sequence).

For every $𝑟 \geq 1$ ,

𝐸 𝑝, 𝑝 + 𝑞 𝑟 ≅ P r e 𝑞 𝑝, 𝑝 - 𝑟 P r e 𝑞 𝑝 - 1, 𝑝 - 𝑟 + I m 𝑞 - 1 𝑝 + 𝑟 - 1, 𝑝 .

The original differential induces

𝑑 𝑝, 𝑝 + 𝑞 𝑟 : 𝐸 𝑝, 𝑝 + 𝑞 𝑟 ⟶ 𝐸 𝑝 - 𝑟, 𝑝 + 𝑞 - 𝑟 + 1 𝑟, [𝑥] ⟼ [𝑑 𝑞 𝑥] .

Moreover,

𝐸 𝑟 + 1 ≅ 𝐻 (𝐸 𝑟, 𝑑 𝑟),

and

i m 𝑑 𝑝, 𝑝 + 𝑞 𝑟 ≅ 𝐺 𝑞 𝑝, 𝑝 - 𝑟 .

Thus the $𝑟$ th differential removes precisely the diagonal of $𝐺$ -pieces with first index minus second index equal to $𝑟$ .

Proof.

The cases $𝑟 = 1$ and $𝑟 = 2$ were established in Definitions˜4 and 3.

Assume (LABEL:eq:er-standard) holds for a fixed $𝑟 \geq 1$ . Define recursive representatives for the next page by

𝑍 𝑝, 𝑝 + 𝑞 𝑟 : = ̃ 𝑍 𝑝, 𝑝 + 𝑞 𝑟 - 1 \cap (𝑑 𝑞) - 1 (̃ 𝐵 𝑝 - 𝑟, 𝑝 + 𝑞 - 𝑟 + 1 𝑟 - 1), 𝐵 𝑝, 𝑝 + 𝑞 𝑟 : = ̃ 𝐵 𝑝, 𝑝 + 𝑞 𝑟 - 1 + 𝑑 𝑞 - 1 (̃ 𝑍 𝑝 + 𝑟, 𝑝 + 𝑞 + 𝑟 - 1 𝑟 - 1) .

By the definitions of kernel and image in a quotient,

𝐻 (𝐸 𝑟, 𝑑 𝑟) 𝑝, 𝑝 + 𝑞 ≅ 𝑍 𝑝, 𝑝 + 𝑞 𝑟 𝐵 𝑝, 𝑝 + 𝑞 𝑟 .

At this point the recursive numerator is converted into its normalized form. The following specialized correction step is stated and proved here, rather than among the preliminary subobject lemmas, because it is exactly the representative-changing argument that drives the induction.

Corollary 4 (The correction step used in the induction).

For every $𝑟 \geq 1$ ,

P r e 𝑞 𝑝, 𝑝 - 𝑟 \cap (𝑑 𝑞) - 1 (P r e 𝑞 + 1 𝑝 - 𝑟 - 1, 𝑝 - 2 𝑟 + I m 𝑞 𝑝 - 1, 𝑝 - 𝑟) = P r e 𝑞 𝑝 - 1, 𝑝 - 𝑟 + P r e 𝑞 𝑝, 𝑝 - 𝑟 - 1 .

Proof of the correction step.

Take $𝑥$ in the left-hand side. Then

𝑑 𝑞 𝑥 = 𝑎 + 𝑢, 𝑎 \in P r e 𝑞 + 1 𝑝 - 𝑟 - 1, 𝑝 - 2 𝑟, 𝑢 \in I m 𝑞 𝑝 - 1, 𝑝 - 𝑟 .

Choose $𝑦 \in 𝐹 𝑝 - 1 𝐶 𝑞$ with $𝑑 𝑞 𝑦 = 𝑢$ . Since $𝑢 \in 𝐹 𝑝 - 𝑟 𝐶 𝑞 + 1$ , in fact $𝑦 \in P r e 𝑞 𝑝 - 1, 𝑝 - 𝑟$ . Moreover,

𝑑 𝑞 (𝑥 - 𝑦) = 𝑎 \in 𝐹 𝑝 - 𝑟 - 1 𝐶 𝑞 + 1,

so $𝑥 - 𝑦 \in P r e 𝑞 𝑝, 𝑝 - 𝑟 - 1$ . Hence $𝑥 = 𝑦 + (𝑥 - 𝑦)$ lies in the right-hand side. Conversely, the differential of an element of $P r e 𝑞 𝑝 - 1, 𝑝 - 𝑟$ lies in $I m 𝑞 𝑝 - 1, 𝑝 - 𝑟$ , while the differential of an element of $P r e 𝑞 𝑝, 𝑝 - 𝑟 - 1$ lies in $P r e 𝑞 + 1 𝑝 - 𝑟 - 1, 𝑝 - 2 𝑟$ because it is closed after one more application of $𝑑$ . Thus the reverse inclusion also holds. ∎

The numerator is

𝑍𝑝,𝑝+𝑞𝑟=Pre𝑞𝑝,𝑝−𝑟∩(𝑑𝑞)−1(Pre𝑞+1𝑝−𝑟−1,𝑝−2𝑟+Im𝑞𝑝−1,𝑝−𝑟)=Pre𝑞𝑝−1,𝑝−𝑟+Pre𝑞𝑝,𝑝−𝑟−1by \cref{cor:correction-step}.

The denominator is

𝐵 𝑝, 𝑝 + 𝑞 𝑟 = P r e 𝑞 𝑝 - 1, 𝑝 - 𝑟 + I m 𝑞 - 1 𝑝 + 𝑟 - 1, 𝑝 + I m 𝑞 - 1 𝑝 + 𝑟, 𝑝 = P r e 𝑞 𝑝 - 1, 𝑝 - 𝑟 + I m 𝑞 - 1 𝑝 + 𝑟, 𝑝,

because $I m 𝑞 - 1 𝑝 + 𝑟 - 1, 𝑝 \subseteq I m 𝑞 - 1 𝑝 + 𝑟, 𝑝$ .

Substituting into (LABEL:eq:next-page-recursive) and applying Lemma˜7 gives

𝐻(𝐸𝑟,𝑑𝑟)𝑝,𝑝+𝑞≅Pre𝑞𝑝,𝑝−𝑟−1Pre𝑞𝑝,𝑝−𝑟−1∩(Pre𝑞𝑝−1,𝑝−𝑟+Im𝑞−1𝑝+𝑟,𝑝)=Pre𝑞𝑝,𝑝−𝑟−1Pre𝑞𝑝−1,𝑝−𝑟−1+Im𝑞−1𝑝+𝑟,𝑝by \cref{lem:modular-law,lem:pre-im-calculus}=̃𝑍𝑝,𝑝+𝑞𝑟̃𝐵𝑝,𝑝+𝑞𝑟.

This proves (LABEL:eq:next-page) and the standard form for $𝐸 𝑟 + 1$ .

We next verify that the differential on $𝐸 𝑟 + 1$ is induced by $𝑑 𝑞$ . From (LABEL:eq:d-pre-im) and (LABEL:eq:im-in-pre),

𝑑 𝑞 (̃ 𝑍 𝑝, 𝑝 + 𝑞 𝑟) = I m 𝑞 𝑝, 𝑝 - 𝑟 - 1 \subseteq P r e 𝑞 + 1 𝑝 - 𝑟 - 1, 𝑝 - 2 𝑟 - 2 = ̃ 𝑍 𝑝 - 𝑟 - 1, 𝑝 + 𝑞 - 𝑟 𝑟, 𝑑 𝑞 (̃ 𝐵 𝑝, 𝑝 + 𝑞 𝑟) = I m 𝑞 𝑝 - 1, 𝑝 - 𝑟 - 1 \subseteq ̃ 𝐵 𝑝 - 𝑟 - 1, 𝑝 + 𝑞 - 𝑟 𝑟 .

Thus $𝑑 𝑟 + 1$ is well-defined, completing the induction.

Finally, use the target standard form

𝐸 𝑝 - 𝑟, 𝑝 + 𝑞 - 𝑟 + 1 𝑟 ≅ P r e 𝑞 + 1 𝑝 - 𝑟, 𝑝 - 2 𝑟 P r e 𝑞 + 1 𝑝 - 𝑟 - 1, 𝑝 - 2 𝑟 + I m 𝑞 𝑝 - 1, 𝑝 - 𝑟 .

Then

im⁡𝑑𝑝,𝑝+𝑞𝑟≅Im𝑞𝑝,𝑝−𝑟Im𝑞𝑝,𝑝−𝑟∩(Pre𝑞+1𝑝−𝑟−1,𝑝−2𝑟+Im𝑞𝑝−1,𝑝−𝑟)=Im𝑞𝑝,𝑝−𝑟Im𝑞𝑝,𝑝−𝑟−1+Im𝑞𝑝−1,𝑝−𝑟by \cref{lem:modular-law,lem:pre-im-calculus}=𝐺𝑞𝑝,𝑝−𝑟.

This proves (LABEL:eq:dr-image-general). ∎

5 Where the lemmas are used

This section is a map for the intended back-and-forth reading described in Remark˜4: begin with the page constructions, and consult the corresponding row below when a cited subobject manipulation is not yet familiar.

Result	Main uses in the construction
Lemma˜5	Computing inverse images in Proposition˜2; converting $𝑑 (P r e)$ into $I m$ .
Lemma˜6	Intersections with sums in Propositions˜5, 2, 4 and 3.
Lemma˜7	Passing from recursive representatives to the standard quotients in Theorems˜3 and 3.
Lemma˜8	Well-definedness of every page differential, calculation of the images, and the inclusions of boundaries in cycles.
Lemma˜9	Identifying the source-side and target-side descriptions of each piece $𝐺 𝑛 𝑝, 𝑞$ .
Lemma˜10	A general simplification of nested $P r e$ -conditions; the main induction instead gives its specialized correction argument directly at the point of use.
Corollary˜4	Inserted inside the proof of Theorem˜3: replace a representative by one whose differential lies one filtration level deeper.

Remark 1 (Index convention).

The second superscript in $𝐸 𝑝, 𝑝 + 𝑞 𝑟$ is chosen so that the underlying cochain degree is always $𝑞$ . Under this convention,

𝑑 𝑟 : 𝐸 𝑝, 𝑝 + 𝑞 𝑟 ⟶ 𝐸 𝑝 - 𝑟, 𝑝 + 𝑞 - 𝑟 + 1 𝑟

raises the cochain degree from $𝑞$ to $𝑞 + 1$ and lowers the filtration index by $𝑟$ .

Remark 2 (Interpretation).

The condition $𝑥 \in P r e 𝑞 𝑝, 𝑝 - 𝑟$ says that $𝑥 \in 𝐹 𝑝 𝐶 𝑞$ and that $𝑑 𝑞 𝑥$ has already fallen $𝑟$ filtration levels. The quotient by $P r e 𝑞 𝑝 - 1, 𝑝 - 𝑟$ removes representatives already coming from the lower filtration level, and the quotient by $I m 𝑞 - 1 𝑝 + 𝑟 - 1, 𝑝$ removes boundaries already visible by page $𝑟$ . The differential $𝑑 𝑟$ detects the first remaining filtration component of $𝑑 𝑞 𝑥$ .

Spectral Sequences for Filtered Complexes Notes on Ravi Vakil’s Puzzling through exact sequences

Abstract

1 Filtered complexes and the basic pieces

Definition 1 (Filtered cochain complex).

Definition 2 (The cumulative image and preimage pieces).

Definition 3 (The 𝐺-pieces).

1.1 Elementary subobject lemmas

Remark 4 (Reading order).

Lemma 5 (Image–preimage identities).

Proof.

Lemma 6 (Modular law).

Proof.

Lemma 7 (Quotient reduction).

Proof.

Lemma 8 (Basic calculus of Pre and Im).

Proof.

Lemma 9 (The two presentations of a 𝐺-piece).

Proof.

Lemma 10 (Stabilization under 𝑑2=0).

Proof.

2 The zeroth and first pages

Remark 1 (Guiding idea for the first pages).

Definition 2 (The zeroth page).

Proposition 3 (The zeroth differential removes 𝐺𝑞𝑝,𝑝).

Proof.

Definition 4 (The first page).

Proposition 5 (Well-definedness and image of 𝑑1).

Proof.

3 The second page

Construction \theconstruction (Recursive cycle and boundary representatives)

Remark 1 (Why normalization is needed).

Proposition 2 (Calculation of 𝑍1 and 𝐵1).

Proof.

Theorem 3 (Standard form of the second page).

Proof.

Proposition 4 (The second differential removes 𝐺𝑞𝑝,𝑝−2).

Proof.

4 The general induction

Remark 1 (The second page suggests the induction).

Definition 2 (Standard representatives for the pages).

Theorem 3 (Filtered-complex spectral sequence).

Proof.

Corollary 4 (The correction step used in the induction).

Proof of the correction step.

5 Where the lemmas are used

Remark 1 (Index convention).

Remark 2 (Interpretation).

Spectral Sequences for Filtered Complexes
Notes on Ravi Vakil’s Puzzling through exact sequences

Definition 3 (The $𝐺$ -pieces).

Lemma 8 (Basic calculus of $P r e$ and $I m$ ).

Lemma 9 (The two presentations of a $𝐺$ -piece).

Lemma 10 (Stabilization under $𝑑 2 = 0$ ).

Proposition 3 (The zeroth differential removes $𝐺 𝑞 𝑝, 𝑝$ ).

Proposition 5 (Well-definedness and image of $𝑑 1$ ).

Proposition 2 (Calculation of $𝑍 1$ and $𝐵 1$ ).

Proposition 4 (The second differential removes $𝐺 𝑞 𝑝, 𝑝 - 2$ ).