## 12.19 Filtrations

A nice reference for this material is [Section 1, HodgeII]. (Note that our conventions regarding abelian categories are different.)

Definition 12.19.1. Let $\mathcal{A}$ be an abelian category.

1. A decreasing filtration $F$ on an object $A$ is a family $(F^ nA)_{n \in \mathbf{Z}}$ of subobjects of $A$ such that

$A \supset \ldots \supset F^ nA \supset F^{n + 1}A \supset \ldots \supset 0$
2. A filtered object of $\mathcal{A}$ is pair $(A, F)$ consisting of an object $A$ of $\mathcal{A}$ and a decreasing filtration $F$ on $A$.

3. A morphism $(A, F) \to (B, F)$ of filtered objects is given by a morphism $\varphi : A \to B$ of $\mathcal{A}$ such that $\varphi (F^ iA) \subset F^ iB$ for all $i \in \mathbf{Z}$.

4. The category of filtered objects is denoted $\text{Fil}(\mathcal{A})$.

5. Given a filtered object $(A, F)$ and a subobject $X \subset A$ the induced filtration on $X$ is the filtration with $F^ nX = X \cap F^ nA$.

6. Given a filtered object $(A, F)$ and a surjection $\pi : A \to Y$ the quotient filtration is the filtration with $F^ nY = \pi (F^ nA)$.

7. A filtration $F$ on an object $A$ is said to be finite if there exist $n, m$ such that $F^ nA = A$ and $F^ mA = 0$.

8. Given a filtered object $(A, F)$ we say $\bigcap F^ iA$ exists if there exists a biggest subobject of $A$ contained in all $F^ iA$. We say $\bigcup F^ iA$ exists if there exists a smallest subobject of $A$ containing all $F^ iA$.

9. The filtration on a filtered object $(A, F)$ is said to be separated if $\bigcap F^ iA = 0$ and exhaustive if $\bigcup F^ iA = A$.

By abuse of notation we say that a morphism $f : (A, F) \to (B, F)$ of filtered objects is injective if $f : A \to B$ is injective in the abelian category $\mathcal{A}$. Similarly we say $f$ is surjective if $f : A \to B$ is surjective in the category $\mathcal{A}$. Being injective (resp. surjective) is equivalent to being a monomorphism (resp. epimorphism) in $\text{Fil}(\mathcal{A})$. By Lemma 12.19.2 this is also equivalent to having zero kernel (resp. cokernel).

Lemma 12.19.2. Let $\mathcal{A}$ be an abelian category. The category of filtered objects $\text{Fil}(\mathcal{A})$ has the following properties:

1. It is an additive category.

2. It has a zero object.

3. It has kernels and cokernels, images and coimages.

4. In general it is not an abelian category.

Proof. It is clear that $\text{Fil}(\mathcal{A})$ is additive with direct sum given by $(A, F) \oplus (B, F) = (A \oplus B, F)$ where $F^ p(A \oplus B) = F^ pA \oplus F^ pB$. The kernel of a morphism $f : (A, F) \to (B, F)$ of filtered objects is the injection $\mathop{\mathrm{Ker}}(f) \subset A$ where $\mathop{\mathrm{Ker}}(f)$ is endowed with the induced filtration. The cokernel of a morphism $f : A \to B$ of filtered objects is the surjection $B \to \mathop{\mathrm{Coker}}(f)$ where $\mathop{\mathrm{Coker}}(f)$ is endowed with the quotient filtration. Since all kernels and cokernels exist, so do all coimages and images. See Example 12.3.13 for the last statement. $\square$

Definition 12.19.3. Let $\mathcal{A}$ be an abelian category. A morphism $f : A \to B$ of filtered objects of $\mathcal{A}$ is said to be strict if $f(F^ iA) = f(A) \cap F^ iB$ for all $i \in \mathbf{Z}$.

This also equivalent to requiring that $f^{-1}(F^ iB) = F^ iA + \mathop{\mathrm{Ker}}(f)$ for all $i \in \mathbf{Z}$. We characterize strict morphisms as follows.

Lemma 12.19.4. Let $\mathcal{A}$ be an abelian category. Let $f : A \to B$ be a morphism of filtered objects of $\mathcal{A}$. The following are equivalent

1. $f$ is strict,

2. the morphism $\mathop{\mathrm{Coim}}(f) \to \mathop{\mathrm{Im}}(f)$ of Lemma 12.3.12 is an isomorphism.

Proof. Note that $\mathop{\mathrm{Coim}}(f) \to \mathop{\mathrm{Im}}(f)$ is an isomorphism of objects of $\mathcal{A}$, and that part (2) signifies that it is an isomorphism of filtered objects. By the description of kernels and cokernels in the proof of Lemma 12.19.2 we see that the filtration on $\mathop{\mathrm{Coim}}(f)$ is the quotient filtration coming from $A \to \mathop{\mathrm{Coim}}(f)$. Similarly, the filtration on $\mathop{\mathrm{Im}}(f)$ is the induced filtration coming from the injection $\mathop{\mathrm{Im}}(f) \to B$. The definition of strict is exactly that the quotient filtration is the induced filtration. $\square$

Lemma 12.19.5. Let $\mathcal{A}$ be an abelian category. Let $f : A \to B$ be a strict monomorphism of filtered objects. Let $g : A \to C$ be a morphism of filtered objects. Then $f \oplus g : A \to B \oplus C$ is a strict monomorphism.

Proof. Clear from the definitions. $\square$

Lemma 12.19.6. Let $\mathcal{A}$ be an abelian category. Let $f : B \to A$ be a strict epimorphism of filtered objects. Let $g : C \to A$ be a morphism of filtered objects. Then $f \oplus g : B \oplus C \to A$ is a strict epimorphism.

Proof. Clear from the definitions. $\square$

Lemma 12.19.7. Let $\mathcal{A}$ be an abelian category. Let $(A, F)$, $(B, F)$ be filtered objects. Let $u : A \to B$ be a morphism of filtered objects. If $u$ is injective then $u$ is strict if and only if the filtration on $A$ is the induced filtration. If $u$ is surjective then $u$ is strict if and only if the filtration on $B$ is the quotient filtration.

Proof. This is immediate from the definition. $\square$

Lemma 12.19.8. Let $\mathcal{A}$ be an abelian category. Let $f : A \to B$, $g : B \to C$ be strict morphisms of filtered objects.

1. In general the composition $g \circ f$ is not strict.

2. If $g$ is injective, then $g \circ f$ is strict.

3. If $f$ is surjective, then $g \circ f$ is strict.

Proof. Let $B$ a vector space over a field $k$ with basis $e_1, e_2$, with the filtration $F^ nB = B$ for $n < 0$, with $F^0B = ke_1$, and $F^ nB = 0$ for $n > 0$. Now take $A = k(e_1 + e_2)$ and $C = B/ke_2$ with filtrations induced by $B$, i.e., such that $A \to B$ and $B \to C$ are strict (Lemma 12.19.7). Then $F^ n(A) = A$ for $n < 0$ and $F^ n(A) = 0$ for $n \geq 0$. Also $F^ n(C) = C$ for $n \leq 0$ and $F^ n(C) = 0$ for $n > 0$. So the (nonzero) composition $A \to C$ is not strict.

Assume $g$ is injective. Then

\begin{align*} g(f(F^ pA)) & = g(f(A) \cap F^ pB) \\ & = g(f(A)) \cap g(F^ p(B)) \\ & = (g \circ f)(A) \cap (g(B) \cap F^ pC) \\ & = (g \circ f)(A) \cap F^ pC. \end{align*}

The first equality as $f$ is strict, the second because $g$ is injective, the third because $g$ is strict, and the fourth because $(g \circ f)(A) \subset g(B)$.

Assume $f$ is surjective. Then

\begin{align*} (g \circ f)^{-1}(F^ iC) & = f^{-1}(F^ iB + \mathop{\mathrm{Ker}}(g)) \\ & = f^{-1}(F^ iB) + f^{-1}(\mathop{\mathrm{Ker}}(g)) \\ & = F^ iA + \mathop{\mathrm{Ker}}(f) + \mathop{\mathrm{Ker}}(g \circ f) \\ & = F^ iA + \mathop{\mathrm{Ker}}(g \circ f) \end{align*}

The first equality because $g$ is strict, the second because $f$ is surjective, the third because $f$ is strict, and the last because $\mathop{\mathrm{Ker}}(f) \subset \mathop{\mathrm{Ker}}(g \circ f)$. $\square$

The following lemma says that subobjects of a filtered object have a well defined filtration independent of a choice of writing the object as a cokernel.

Lemma 12.19.9. Let $\mathcal{A}$ be an abelian category. Let $(A, F)$ be a filtered object of $\mathcal{A}$. Let $X \subset Y \subset A$ be subobjects of $A$. On the object

$Y/X = \mathop{\mathrm{Ker}}(A/X \to A/Y)$

the quotient filtration coming from the induced filtration on $Y$ and the induced filtration coming from the quotient filtration on $A/X$ agree. Any of the morphisms $X \to Y$, $X \to A$, $Y \to A$, $Y \to A/X$, $Y \to Y/X$, $Y/X \to A/X$ are strict (with induced/quotient filtrations).

Proof. The quotient filtration $Y/X$ is given by $F^ p(Y/X) = F^ pY/(X \cap F^ pY) = F^ pY/F^ pX$ because $F^ pY = Y \cap F^ pA$ and $F^ pX = X \cap F^ pA$. The induced filtration from the injection $Y/X \to A/X$ is given by

\begin{align*} F^ p(Y/X) & = Y/X \cap F^ p(A/X) \\ & = Y/X \cap (F^ pA + X)/X \\ & = (Y \cap F^ pA)/(X \cap F^ pA) \\ & = F^ pY/F^ pX. \end{align*}

Hence the first statement of the lemma. The proof of the other cases is similar. $\square$

Lemma 12.19.10. Let $\mathcal{A}$ be an abelian category. Let $A, B, C \in \text{Fil}(\mathcal{A})$. Let $f : A \to B$ and $g : A \to C$ be morphisms. Then there exists a pushout

$\xymatrix{ A \ar[r]_ f \ar[d]_ g & B \ar[d]^{g'} \\ C \ar[r]^{f'} & C \amalg _ A B }$

in $\text{Fil}(\mathcal{A})$. If $f$ is strict, so is $f'$.

Proof. Set $C \amalg _ A B$ equal to $\mathop{\mathrm{Coker}}((1, -1) : A \to C \oplus B)$ in $\text{Fil}(\mathcal{A})$. This cokernel exists, by Lemma 12.19.2. It is a pushout, see Example 12.5.6. Note that $F^ p(C \amalg _ A B)$ is the image of $F^ pC \oplus F^ pB$. Hence

$(f')^{-1}(F^ p(C \amalg _ A B)) = g(f^{-1}(F^ pB))) + F^ pC$

Whence the last statement. $\square$

Lemma 12.19.11. Let $\mathcal{A}$ be an abelian category. Let $A, B, C \in \text{Fil}(\mathcal{A})$. Let $f : B \to A$ and $g : C \to A$ be morphisms. Then there exists a fibre product

$\xymatrix{ B \times _ A C \ar[r]_{g'} \ar[d]_{f'} & B \ar[d]^ f \\ C \ar[r]^ g & A }$

in $\text{Fil}(\mathcal{A})$. If $f$ is strict, so is $f'$.

Proof. This lemma is dual to Lemma 12.19.10. $\square$

Let $\mathcal{A}$ be an abelian category. Let $(A, F)$ be a filtered object of $\mathcal{A}$. We denote $\text{gr}^ p_ F(A) = \text{gr}^ p(A)$ the object $F^ pA/F^{p + 1}A$ of $\mathcal{A}$. This defines an additive functor

$\text{gr}^ p : \text{Fil}(\mathcal{A}) \longrightarrow \mathcal{A}, \quad (A, F) \longmapsto \text{gr}^ p(A).$

Recall that we have defined the category $\text{Gr}(\mathcal{A})$ of graded objects of $\mathcal{A}$ in Section 12.16. For $(A, F)$ in $\text{Fil}(\mathcal{A})$ we may set

$\text{gr}(A) = \text{the graded object of }\mathcal{A}\text{ whose } p\text{th graded piece is }\text{gr}^ p(A)$

and if $\mathcal{A}$ has countable direct sums, then we simply have

$\text{gr}(A) = \bigoplus \text{gr}^ p(A)$

$\text{gr} : \text{Fil}(\mathcal{A}) \longrightarrow \text{Gr}(\mathcal{A}), \quad (A, F) \longmapsto \text{gr}(A).$

Lemma 12.19.12. Let $\mathcal{A}$ be an abelian category.

1. Let $A$ be a filtered object and $X \subset A$. Then for each $p$ the sequence

$0 \to \text{gr}^ p(X) \to \text{gr}^ p(A) \to \text{gr}^ p(A/X) \to 0$

is exact (with induced filtration on $X$ and quotient filtration on $A/X$).

2. Let $f : A \to B$ be a morphism of filtered objects of $\mathcal{A}$. Then for each $p$ the sequences

$0 \to \text{gr}^ p(\mathop{\mathrm{Ker}}(f)) \to \text{gr}^ p(A) \to \text{gr}^ p(\mathop{\mathrm{Coim}}(f)) \to 0$

and

$0 \to \text{gr}^ p(\mathop{\mathrm{Im}}(f)) \to \text{gr}^ p(B) \to \text{gr}^ p(\mathop{\mathrm{Coker}}(f)) \to 0$

are exact.

Proof. We have $F^{p + 1}X = X \cap F^{p + 1}A$, hence map $\text{gr}^ p(X) \to \text{gr}^ p(A)$ is injective. Dually the map $\text{gr}^ p(A) \to \text{gr}^ p(A/X)$ is surjective. The kernel of $F^ pA/F^{p + 1}A \to A/X + F^{p + 1}A$ is clearly $F^{p + 1}A + X \cap F^ pA/F^{p + 1}A = F^ pX/F^{p + 1}X$ hence exactness in the middle. The two short exact sequence of (2) are special cases of the short exact sequence of (1). $\square$

Lemma 12.19.13. Let $\mathcal{A}$ be an abelian category. Let $f : A \to B$ be a morphism of finite filtered objects of $\mathcal{A}$. The following are equivalent

1. $f$ is strict,

2. the morphism $\mathop{\mathrm{Coim}}(f) \to \mathop{\mathrm{Im}}(f)$ is an isomorphism,

3. $\text{gr}(\mathop{\mathrm{Coim}}(f)) \to \text{gr}(\mathop{\mathrm{Im}}(f))$ is an isomorphism,

4. the sequence $\text{gr}(\mathop{\mathrm{Ker}}(f)) \to \text{gr}(A) \to \text{gr}(B)$ is exact,

5. the sequence $\text{gr}(A) \to \text{gr}(B) \to \text{gr}(\mathop{\mathrm{Coker}}(f))$ is exact, and

6. the sequence

$0 \to \text{gr}(\mathop{\mathrm{Ker}}(f)) \to \text{gr}(A) \to \text{gr}(B) \to \text{gr}(\mathop{\mathrm{Coker}}(f)) \to 0$

is exact.

Proof. The equivalence of (1) and (2) is Lemma 12.19.4. By Lemma 12.19.12 we see that (4), (5), (6) imply (3) and that (3) implies (4), (5), (6). Hence it suffices to show that (3) implies (2). Thus we have to show that if $f : A \to B$ is an injective and surjective map of finite filtered objects which induces and isomorphism $\text{gr}(A) \to \text{gr}(B)$, then $f$ induces an isomorphism of filtered objects. In other words, we have to show that $f(F^ pA) = F^ pB$ for all $p$. As the filtrations are finite we may prove this by descending induction on $p$. Suppose that $f(F^{p + 1}A) = F^{p + 1}B$. Then commutative diagram

$\xymatrix{ 0 \ar[r] & F^{p + 1}A \ar[r] \ar[d]^ f & F^ pA \ar[r] \ar[d]^ f & \text{gr}^ p(A) \ar[r] \ar[d]^{\text{gr}^ p(f)} & 0 \\ 0 \ar[r] & F^{p + 1}B \ar[r] & F^ pB \ar[r] & \text{gr}^ p(B) \ar[r] & 0 }$

and the five lemma imply that $f(F^ pA) = F^ pB$. $\square$

Lemma 12.19.14. Let $\mathcal{A}$ be an abelian category. Let $A \to B \to C$ be a complex of filtered objects of $\mathcal{A}$. Assume $\alpha : A \to B$ and $\beta : B \to C$ are strict morphisms of filtered objects. Then $\text{gr}(\mathop{\mathrm{Ker}}(\beta )/\mathop{\mathrm{Im}}(\alpha )) = \mathop{\mathrm{Ker}}(\text{gr}(\beta ))/\mathop{\mathrm{Im}}(\text{gr}(\alpha )))$.

Proof. This follows formally from Lemma 12.19.12 and the fact that $\mathop{\mathrm{Coim}}(\alpha ) \cong \mathop{\mathrm{Im}}(\alpha )$ and $\mathop{\mathrm{Coim}}(\beta ) \cong \mathop{\mathrm{Im}}(\beta )$ by Lemma 12.19.4. $\square$

Lemma 12.19.15. Let $\mathcal{A}$ be an abelian category. Let $A \to B \to C$ be a complex of filtered objects of $\mathcal{A}$. Assume $A, B, C$ have finite filtrations and that $\text{gr}(A) \to \text{gr}(B) \to \text{gr}(C)$ is exact. Then

1. for each $p \in \mathbf{Z}$ the sequence $\text{gr}^ p(A) \to \text{gr}^ p(B) \to \text{gr}^ p(C)$ is exact,

2. for each $p \in \mathbf{Z}$ the sequence $F^ p(A) \to F^ p(B) \to F^ p(C)$ is exact,

3. for each $p \in \mathbf{Z}$ the sequence $A/F^ p(A) \to B/F^ p(B) \to C/F^ p(C)$ is exact,

4. the maps $A \to B$ and $B \to C$ are strict, and

5. $A \to B \to C$ is exact (as a sequence in $\mathcal{A}$).

Proof. Part (1) is immediate from the definitions. We will prove (3) by induction on the length of the filtrations. If each of $A$, $B$, $C$ has only one nonzero graded part, then (3) holds as $\text{gr}(A) = A$, etc. Let $n$ be the largest integer such that at least one of $F^ nA, F^ nB, F^ nC$ is nonzero. Set $A' = A/F^ nA$, $B' = B/F^ nB$, $C' = C/F^ nC$ with induced filtrations. Note that $\text{gr}(A) = F^ nA \oplus \text{gr}(A')$ and similarly for $B$ and $C$. The induction hypothesis applies to $A' \to B' \to C'$, which implies that $A/F^ p(A) \to B/F^ p(B) \to C/F^ p(C)$ is exact for $p \geq n$. To conclude the same for $p = n + 1$, i.e., to prove that $A \to B \to C$ is exact we use the commutative diagram

$\xymatrix{ 0 \ar[r] & F^ nA \ar[r] \ar[d] & A \ar[r] \ar[d] & A' \ar[r] \ar[d] & 0 \\ 0 \ar[r] & F^ nB \ar[r] \ar[d] & B \ar[r] \ar[d] & B' \ar[r] \ar[d] & 0 \\ 0 \ar[r] & F^ nC \ar[r] & C \ar[r] & C' \ar[r] & 0 }$

whose rows are short exact sequences of objects of $\mathcal{A}$. The proof of (2) is dual. Of course (5) follows from (2).

To prove (4) denote $f : A \to B$ and $g : B \to C$ the given morphisms. We know that $f(F^ p(A)) = \mathop{\mathrm{Ker}}(F^ p(B) \to F^ p(C))$ by (2) and $f(A) = \mathop{\mathrm{Ker}}(g)$ by (5). Hence $f(F^ p(A)) = \mathop{\mathrm{Ker}}(F^ p(B) \to F^ p(C)) = \mathop{\mathrm{Ker}}(g) \cap F^ p(B) = f(A) \cap F^ p(B)$ which proves that $f$ is strict. The proof that $g$ is strict is dual to this. $\square$

Comment #520 by Keenan Kidwell on

Definition 0121 makes reference to the intersection and union of a (countable) family of subobjects of an object in an abelian category, but as far as I can tell intersections and unions of subobjects are not defined anywhere. Perhaps this is intentional, but I thought I should mention it.

Comment #521 by on

Yes, I agree that is bad. Hope somebody will fix this (hint, hint). I guess we do discuss subobjects in the section on abelian categories, but we don't point out that you can take their intersection (of two of them I mean). To intersect infinitely many of them you probably would define that as a limit? In fact, it is always a cofiltered limit since you can throw in the collection of finite intersections? Similarly with sums (and unions in the case of an increasing sequence).

Anyway, I think our omission, in this particular case, is not as bad as all that, because the statement $\bigcap F^i(A) = 0$ just means that any morphism $T \to A$ which factors through all the subobjects $F^i(A)$ is zero, in other words, you don't have to know what the intersection is and in particular you don't need to know a priori that it exists.

Comment #4865 by Amanda on

Why is it clear that the morphism groups of Fil($\mathcal{A}$) are subgroups of the morphism groups of $\mathcal{A}$?

Comment #7851 by Sean on

There is some inconsistency in the choice of symbol for indexing, even in the same definition or proof. Here is a list:

n (or m) Definition 0121 (1) Definition 0121 (5) Definition 0121 (6) Definition 0121 (7) * First paragraph of proof of Lemma 05SJ

i Definition 0121 (3) Definition 0121 (8) Definition 0121 (9) Definition 0123 and just below it * Third paragraph of proof of Lemma 05SJ

p Proof of Lemma 0122 Second paragraph of proof of Lemma 05SJ Proof of Lemma 0129 Proof of Lemma 05SM Just below the proof of Lemma 05SN Lemma 05SP and its proof Proof of Lemma 0127 Lemma 05QH

n and p * Proof of Lemma 05QH

Comment #8073 by on

Please comment on each of the items separately (on the page of the tag itself) and tell me exactly what to fix.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).