1.2 Opposite filtrations

Let A be an object of \mathscr{A} endowed with filtrations F and G. By definition, \operatorname{Gr}_F^n(A) is a quotient of a sub-object of A and, as such, is endowed with a filtration induced by G (1.1.10). Passing to the associated graded defines a bigraded object (\operatorname{Gr}_G^n\operatorname{Gr}_F^m(A))_{n,m\in\mathbb{Z}}. By a lemma of Zassenhaus, \operatorname{Gr}_G^n\operatorname{Gr}_F^m(A) and \operatorname{Gr}_F^m\operatorname{Gr}_G^n(A) are canonically isomorphic: if we define the induced filtrations (1.1.10) as quotient filtrations of the induced filtrations on a sub-object, then we have \begin{aligned} \operatorname{Gr}_G^n\operatorname{Gr}_F^m(A) \cong (F^m(A)\cap G^n(A)) &/ ((F^{m+1}(A)\cap G^n(A)) + (F^m(A)\cap G^{n+1}(A))) \\&= \\\operatorname{Gr}_F^m\operatorname{Gr}_G^n(A) \cong (G^n(A)\cap F^m(A)) &/ ((G^{n+1}(A)\cap F^m(A)) + (G^n(A)\cap F^{m+1}(A))) \end{aligned}

Let H be a third filtration of A. It induces a filtration on \operatorname{Gr}_F(A), and thus on \operatorname{Gr}_G\operatorname{Gr}_F(A). It also induces a filtration on \operatorname{Gr}_F\operatorname{Gr}_G(A). We note that these filtrations do not in general correspond to one another under the isomorphism (1.2.1). In the expression \operatorname{Gr}_H\operatorname{Gr}_G\operatorname{Gr}_F(A), G and H thus play a symmetric role, but not F and G.

Two finite filtrations F and \overline{F} on A are said to be n-opposite if \operatorname{Gr}_F^p\operatorname{Gr}_{\overline{F}}^q(A)=0 for p+q\neq n.

If A^{p,q} is a bigraded object of \mathscr{A} such that

  1. A^{p,q}=0 except for a finite number of pairs (p,q), and
  2. A^{p,q}=0 for p+q\neq n

then we define two n-opposite filtrations of A=\sum_{p,q}A^{p,q} by setting F^p(A) = \sum_{p'\geqslant p} A^{p',q'} \tag{1.2.4.1} \overline{F}^q(A) = \sum_{q'\geqslant q} A^{p',q'}. \tag{1.2.4.2}

We have \operatorname{Gr}_F^p\operatorname{Gr}_{\overline{F}}^q(A) = A^{p,q}. \tag{1.2.4.3}

Conversely:

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  1. Let F and \overline{F} be finite filtrations on A. For F and \overline{F} to be n-opposite, it is necessary and sufficient that, for all p,q, [p+q=n+1] \implies [F^p(A)\oplus\overline{F}^q(A) \xrightarrow{\sim}A].

  2. If F and \overline{F} are n-opposite, and if we set \begin{cases} A^{p,q} = 0 &\text{for }p+q\neq n \\A^{p,q} = F^p(A)\cap\overline{F}^q(A) &\text{for }p+q=n \end{cases} then A is the direct sum of the A^{p,q}, and F and \overline{F} come from the bigrading A^{p,q} of A by the procedure of (1.2.4).

Proof.

  1. The condition \operatorname{Gr}_F^p\operatorname{Gr}_{\overline{F}}^q(A)=0 for p+q>n implies that F^p\cap\overline{F}^q=(F^{p+1}\cap\overline{F}^q)+(F^p\cap\overline{F}^{q+1}) for p+q>n. By hypothesis, F^p\cap\overline{F}^q is zero for large enough p+q; by decreasing induction, we thus deduce that the condition \operatorname{Gr}_F^p\operatorname{Gr}_{\overline{F}}^q(A)=0 for p+q>n is equivalent to the condition F^p(A)\cap\overline{F}^q(A)=0 for p+q>n. Dually ((1.1.6), (1.1.7), (1.1.10)), the condition \operatorname{Gr}_F^p\operatorname{Gr}_{\overline{F}}^q(A)=0 for p+q<n is equivalent to the condition $A=Fp(A)+q(A)} for (1-p)+(1-q)>-n, i.e. for p+q\leqslant n+1, and the claim then follows.

  2. If F and \overline{F} are n-opposite, then we can prove by decreasing induction on p that \bigoplus_{p'\geqslant p} A^{p',q'} \xrightarrow{\sim}F^p(A). \tag{1.2.5.1}

    1. For F^p(A)=0, the claim is evident.
    2. The decomposition A=F^{p+1}(A)\oplus\overline{F}^{n-p}(A) induces on F^p(A)\supset F^{p+1}(A) a decomposition F^p(A) = F^{p+1}(A)\oplus(F^p(A)\cap\overline{F}^{n-p}(A)) and we conclude by induction.

    For p small enough, we have F^p(A)=A. By (1.2.5.1), the A^{p,q} thus form a bigrading of A, and F satisfies (1.2.4.1). The fact that \overline{F} satisfies (1.2.4.2) then follows by symmetry.

The constructions (1.2.4) and (1.2.5) establish equivalences of categories that are quasi-inverse to one another between objects of \mathscr{A} endowed with two finite n-opposite filtrations and bigraded objects of \mathscr{A} of the type considered in (1.2.4).

Three finite filtrations W, F, and \overline{F} on A are said to be opposite if \operatorname{Gr}_F^p\operatorname{Gr}_{\overline{F}}^q\operatorname{Gr}_W^n(A) = 0 for p+q+n\neq0.

This condition is symmetric in F and \overline{F}. It implies that F and \overline{F} induce on W^n(A)/W^{n+1}(A) two (-n)-opposite filtrations. We set A^{p,q} = \operatorname{Gr}_F^p\operatorname{Gr}_{\overline{F}}^q\operatorname{Gr}_F^{-p-q}(A) whence decompositions (1.2.4), (1.2.5) W^n(A)/W^{n+1}(A) = \bigoplus_{p+q=-n} A^{p,q} \tag{1.2.7.1} which make \operatorname{Gr}_W(A) into a bigraded object.

Let W, F, and \overline{F} be three finite opposite filtrations, and \sigma a sequence (p_i,q_i)_{i\geqslant0} pairs of integers satisfying

  1. p_i\leqslant p_j and q_i\leqslant q_j for i\geqslant j, and
  2. p_i+q_i=p_0+q_0-i+1 for i>0.

Set p=p_0, q=q_0, n=-p-q, and A_\sigma = \left( \sum_{0\leqslant i}(W^{n+i}(A)\cap F^{p_i}(A)) \right) \cap \left( \sum_{0\leqslant i}(W^{n+i}(A)\cap\overline{F}^{q_i}(A)) \right). Then the projection from W^n(A) to \operatorname{Gr}_W^n(A) induces an isomorphism A_\sigma \xrightarrow{\sim}A^{p,q} \subset \operatorname{Gr}_W^n(A).

Proof. We will prove by induction on k the following claim:

The projection from W^n(A)/W^{n+k} to \operatorname{Gr}_W^n(A) induces an isomorphism from \begin{aligned} \Bigg[ &\left( \sum_{i<k} (W^{n+i}(A)\cap F^{p_i}(A)) + W^{n+k}(A) \right) \\\cap &\left( \sum_{i<k} (W^{n+i}(A)\cap\overline{F}^{q_i}(A)) + W^{n+k}(A) \right) \Bigg] / W^{n+k}(A) \end{aligned} \tag{$*_k$} to A^{p,q}\subset\operatorname{Gr}_W^n(A).

For k=1, this is exactly the definition of A^{p,q}.

By (i) of (1.2.5) we have F^{p_k}(\operatorname{Gr}_W^{n+k}(A)) \oplus \overline{F}^{q_k}(\operatorname{Gr}_W^{n+k}(A)) \xrightarrow{\sim}\operatorname{Gr}_W^{n+k}(A). \tag{1.2.8.1} Set \begin{aligned} B &= \sum_{i<k} (W^{n+i}(A)\cap F^{p_i}(A)) \\C &= \sum_{i<k} (W^{n+i}(A)\cap\overline{F}^{q_i}(A)) \\B' &= (W^{n+k}(A)\cap F^{p_k}(A)) + W^{n+k+1}(A) \\C' &= (W^{n+k}(A)\cap\overline{F}^{q_k}(A)) + W^{n+k+1}(A) \\D &= W^{n+k}(A) \\E &= W^{n+k+1}(A). \end{aligned}

Equation (1.2.8.1) can then be written as \begin{aligned} B'+C' &= D \\B'\cap C' &= E. \end{aligned} We also have, since p_k\leqslant p_i (for i\leqslant k), B\cap D \subset F^{p_k}(A) \cap W^{n+k}(A) \subset B' and, since q_k\leqslant q_i (for i\leqslant k), C\cap D \subset \overline{F}^{q_k}(A)\cap W^{n+k}(A) \subset C'.

The claim (*_{k+1}) then follows from (*_k) and the following lemma.

Let B, C, B', C', D, and E be sub-objects of A. Suppose that \begin{gathered} B'+C' = D \qquad B'\cap C' = E \\B\cap D \subset B' \qquad C\cap D \subset C'. \end{gathered} Then ((B+B')\cap(C+C'))/E \xrightarrow{\sim}((B+D)\cap(C+D))/D.

Proof. To prove surjectivity, we write \begin{aligned} ((B+B')\cap(C+C'))+D &= (((B+B')\cap(C+C'))+B')+C' \\&= ((B+B')\cap(C+C'+B'))+C' \\&= (B+B'+C')\cap(C+C'+B') \\&= (B+D)\cap(C+D). \end{aligned}

To prove injectivity, we write (B+B')\cap(C+C')\cap D = ((B+B')\cap D)\cap((C+C')\cap D). Since B'\subset D, we have \begin{aligned} (B+B')\cap D &= (B\cap D)+B' \\&= B' \end{aligned} and similarly (C+C')\cap D = C' and \begin{aligned} (B+B')\cap(C+C')\cap D &= B'\cap C' \\&= E. \end{aligned}

This finishes the proof of (1.2.8), noting that (1.2.8) is equivalent to (*_k) for large k.

Let \mathscr{A} be an abelian category, and provisionally denote by \mathscr{A}' the category of objects of \mathscr{A} endowed with three opposite filtrations W, F, and \overline{F}. The morphisms in \mathscr{A}' are the morphisms of \mathscr{A} that are compatible with the three filtrations.

  1. \mathscr{A}' is an abelian category.

  2. The kernel (resp. cokernel) of an arrow f\colon A\to B in \mathscr{A}' is the kernel (resp. cokernel) of f in \mathscr{A}, endowed with the filtrations induced by those of A (resp. the quotients of those of B).

  3. Every morphism f\colon A\to B in \mathscr{A}' is strictly compatible with the filtrations W, F, and \overline{F}; the morphism \operatorname{Gr}_W(f) is compatible with the bigradings of \operatorname{Gr}_W(A) and \operatorname{Gr}_W(B); the morphisms \operatorname{Gr}_F(f) and \operatorname{Gr}_{\overline{F}}(f) are strictly compatible with the filtration induced by W.

  4. The “forget the filtrations” functors, \operatorname{Gr}_W, \operatorname{Gr}_F, and \operatorname{Gr}_{\overline{F}}, and \begin{gathered} \operatorname{Gr}_W\operatorname{Gr}_F \simeq \operatorname{Gr}_F\operatorname{Gr}_W \\\simeq \operatorname{Gr}_{\overline{F}}\operatorname{Gr}_F\operatorname{Gr}_W \\\simeq \operatorname{Gr}_{\overline{F}}\operatorname{Gr}_W \simeq \operatorname{Gr}_W\operatorname{Gr}_{\overline{F}} \end{gathered} from \mathscr{A}' to \mathscr{A} are exact.

Denote by \sigma_0(p,q) and \sigma_1(p,q) the sequences \begin{aligned} \sigma_0(p,q) &= (p,q), (p,q), (p,q-1), (p,q-2), (p,q-3), \ldots \\\sigma_0(p,q) &= (p,q), (p,q), (p-1,q), (p-2,q), (p-3,q), \ldots \end{aligned} and, with the notation of (1.2.8), set A_i^{p,q} = A_{\sigma_i(p,q)} \qquad\text{for } i=0,1.

If f\colon A\to B is compatible with W, F, and \overline{F}, then we have f(A_i^{p,q}) \subset B_i^{p,q} \qquad\text{for }i=0,1. \tag{1.2.10.1} Claim (iii) then follows from the following lemma:

The A_i^{p,q} give a bigrading of A. We have W^n(A) = \sum_{n+p+q\leqslant 0} A_i^{p,q} \qquad\text{for }i=0,1 \tag{1.2.11.1} F^p(A) = \sum_{p'\geqslant p} A_0^{p',q'} \tag{1.2.11.2} \overline{F}^q(A) = \sum_{q'\geqslant q} A_1^{p',q'}. \tag{1.2.11.3}

Proof. By symmetry, it suffices to prove the claims concerning i=0. Set A_0=\bigoplus A_0^{p,q} and define filtrations W and F on A_0 by the equations of (1.2.11). The canonical map i from A_0 to A is compatible with the filtrations W and F. Furthermore, by (1.2.8), \operatorname{Gr}_W(i) is an isomorphism, and induces isomorphisms of graded objects \sum_{p+q=n} A_0^{p,q} \xrightarrow{\sim}\operatorname{Gr}_W^{-n}(A) = \sum_{p+q=n} A^{p,q}. \tag{1.2.11.4} The morphism i is thus an isomorphism, and the A_0^{p,q} give a bigrading of A.

Equation (1.2.11.1) then says that \operatorname{Gr}_W(i) is an isomorphism. By (1.2.11.4), \operatorname{Gr}_F\operatorname{Gr}_W(i) is an isomorphism, and thus so too are \operatorname{Gr}_W\operatorname{Gr}_F(i) and \operatorname{Gr}_F(i). Equation (1.2.11.2) then follows.

We now prove (1.2.10). Let f\colon A\to B in \mathscr{A}' and endow K=\operatorname{Ker}(f) with the filtrations induced by those of A. By (1.2.11), \operatorname{Gr}_W(K)\hookrightarrow\operatorname{Gr}_W(A); furthermore, the filtration F (resp. \overline{F}) on K induces on \operatorname{Gr}_W(K) the inverse image filtration of the filtration F on \operatorname{Gr}_W(A). The sub-object \operatorname{Gr}_W(K) of \operatorname{Gr}_W(A) is then compatible with the bigrading of \operatorname{Gr}_W(A): \operatorname{Gr}_W(K) = \bigoplus_{p,q}(\operatorname{Gr}_W(K)\cap A^{p,q}).

We thus deduce that \operatorname{Gr}_F^p\operatorname{Gr}_{\overline{F}}^q\operatorname{Gr}_W^n(K) \hookrightarrow \operatorname{Gr}_F^p\operatorname{Gr}_{\overline{F}}^q\operatorname{Gr}_W^n(A); the filtrations of W, F, and \overline{F} on K are thus opposite, and K is a kernel of f in \mathscr{A}'. This, combined with the dual result, proves (ii).

If f is an arrow of \mathscr{A}', then the canonical morphism from \operatorname{Coim}(f) to \operatorname{Im}(f) is an isomorphism in \mathscr{A}; by (iii), it is also an isomorphism in \mathscr{A}', which is thus abelian.

The “forget the filtrations” functor is exact by (ii). The exactness of the other functors in (iv) follows immediately from (ii), (iii), and (i) or (ii) of (1.1.11).

Let A be an object of \mathscr{A} endowed with a finite increasing filtration W_\bullet, and two finite decreasing filtrations F and \overline{F}. The construction (1.1.3) associates to W_\bullet a decreasing filtration W^\bullet. We say that the filtrations W_\bullet, F, and \overline{F} are } are n-opposite.

(1.2.10) translates trivially to this variation.