I have to deal with unbounded filtrations and want to use the conditional convergence of spectral sequences and the results from

[1]: J. Michael Boardman, Conditionally Convergent Spectral Sequences, March 1999 (http://hopf.math.purdue.edu/Boardman/ccspseq.pdf)

The article uses cohomological spectral sequences derived from the exact couple coming from a cochain complex $C$ and a decreasing filtration $F$ of $C$. The system of inclusions is $$A^s := H(F_s C) \leftarrow A^{s+1}$$ and the pages are denoted by $E^s_r$ for $s\in \mathbb{Z}$ and $r\in \mathbb{N}$ ($r$ is the page number and $s$ the ``filtration degree''). The symbol $A^\infty$ denotes the limit and the symbol $A^{-\infty}$ the colimit. The symbol $RA^\infty$ denotes the right derived module of the limit. I basically work over $\mathbb{R}$.

The following are the two theorems (or their parts) from [1] which I am interested in:

Theorem 6.1 (p.19): Let $C$ be a filtered cochain complex. Suppose that \begin{equation}\label{Eq:Exit}\tag{C1} E^s = 0\quad\text{for all } s>0.\end{equation} If $A^\infty = 0$, then the spectral sequence converges strongly to $A^{-\infty}$.

Theorem 7.2 (p.21): Let $f: C \rightarrow \bar{C}$ be a morphism of filtered cochain complexes and suppose that $E^s$, resp. $\bar{E}^s$ converge conditionally to $A^{-\infty}$, resp. $\bar{A}^{-\infty}$. Suppose, moreover, that \begin{equation}\tag{C2} E^s = \bar{E}^s = 0\quad\text{for all }s<0. \end{equation} If $f$ induces the isomorphisms $E^\infty\simeq \bar{E}^\infty$ and $RE^\infty\simeq R\bar{E}^\infty$, then it induces the isomorphism $H(C)\simeq H(\bar{C})$.

Let me introduce the standard (degree shifted) bigrading on $E_r$ and visualize $E_r^{s,d}$ as sitting at the coordinate $(s,d)$ in plane. The differentials are then $$ d_r: E_r^{s,d}\rightarrow E_r^{s+r,d-r+1}. $$ My questions are the following:

  1. How does Theorem 6.1 generalize if (C1) is replaced by the following condition of exiting differentials? $$ E_r \text{ sit in a half-plane and if we fix any coordinate }(s,d), \text{ then all but finitely many }d_r\text{ starting at }(s,d)\text{ leave the half-plane.}$$

  2. How does Theorem 7.2 generalize if (C2) is replaced by the following condition of entering differentials? $$ E_r \text{ sit in a half-plane and if we fix any coordinate }(s,d), \text{ then all but finitely many }d_r\text{ ending at }(s,d)\text{ start outside of the half-plane.}$$

The author of [1] addresses the questions as follows:

  1. On p.19, Chapter 6 in brackets right before Theorem 6.1:

    ...The results generalize appropriately, as all arguments can be carried out degreewise; the main difficulty is to find notation that would help rather than hinder the exposition

  2. On p.20, Chapter 7 in brackets a couple of paragraphs before Theorem 7.2:

    ...The results remain valid when appropriately modified, as all arguments can be carried out degreewise; the difficulty is to find notation that helps rather than hinders.

How do these theorems generalize precisely? Has it been done anywhere? Thanks!

P.S. I come from differential geometry and am not familiar with the proof methods for spectral sequences at all. I use it merely as a black box.


After a long life in preprint form, Boardman's paper was published in the conference proceedings celebrating his 60th birthday:

       author={Boardman, J. Michael},
       title={Conditionally convergent spectral sequences},
          title={Homotopy invariant algebraic structures},
          address={Baltimore, MD},
          series={Contemp. Math.},
          publisher={Amer. Math. Soc., Providence, RI},

In the $(s,d)$-bigraded case, you can replace Boardman's (left half-plane) condition that $E_1^{s,d} = 0$ for $s > 0$ (or $s > s_0$ for some fixed integer $s_0$) by the (upper half-plane) condition that $E_1^{s,d} = 0$ for all $d < 0$ (or $d < d_0$ for some fixed integer $d_0$).

Similarly, you can replace his (right half-plane) condition that $E_1^{s,d} = 0$ and $\bar E_1^{s,d} = 0$ for $s < 0$ (or $s < s_0$ for some fixed integer $s_0$) by the (lower half-plane) condition that $E_1^{s,d} = 0$ and $\bar E_1^{s,d} = 0$ for $d > 0$ (or $d > d_0$ for some fixed integer $d_0$).

This adjustment is often enough. Do you need to refer to other half-planes than those bounded by a horizontal or a vertical line?

EDIT: The OP added some questions, including one about the case of a right half-plane cohomological bicomplex $(B^{i,j}, d_h, d_v)$, where $B^{i,j} = 0$ for $i<0$, $d_h : B^{i,j} \to B^{i+1,j}$ and $d_v : B^{i,j} \to B^{i,j+1}$. Suppose that $Z^{i,j} \subset B^{i,j}$ is such that $d_h(Z^{i,j}) = 0$, and filter the total complex $$ (C, d) = (\bigoplus_{i,j} B^{i,j}, d_h + d_v) $$ by $$ F^s = \bigoplus_{j-i>s} B^{i,j} \oplus \bigoplus_{j-i=s} Z^{i,j} $$ for all integers $s$. Here $C$ and $F^s$ are graded, with $B^{i,j}$ and $Z^{i,j}$ in degree $i+j$. Then $(F^s, d)$ is a subcomplex of $(C, d)$, and contains $(F^{s+1}, d)$ as a further subcomplex. We get an exact couple in the usual way, with $A^{s,t}$ and $E_1^{s,t}$ equal to the degree $s+t$ parts of $H(F^s, d)$ and $H(F^s/F^{s+1}, d)$, respectively.

I claim that $\lim_s F^s = 0$ and $\lim^1_s F^s = 0$. This can be checked one degree at a time, since $F^s = 0$ in degrees less than $s$. It follows (see Boardman's Theorem 9.2) that $\lim_s A^s = 0$ and $\lim^1_s A^s = 0$, so the spectral sequence is conditionally convergent to the colimit $G = H(C, d)$.

Furthermore, $F^s/F^{s+1}$ and $E^1_s$ are concentrated in degrees $\ge s$, corresponding to $t\ge0$, so this is an upper half-plane cohomological spectral sequence with exciting differentials. Hence it is strongly convergent to $G$, by the modified form of Boardman's Theorem 6.1 that I mentioned above.

To prove the modified form, one does as Boardman says. Let $F^s G$ be the image of $H(F^s, d)$ in $G$. One must check that the filtration $\{F^s G\}_s$ of $G$ is complete Hausdorff and exhaustive, and that the natural inclusion $F^s G/F^{s+1} G \to E_\infty^s$ is an isomorphism. Both claims can be checked one degree at a time, and for each degree the proof of Theorem 6.1(a) carries over. (I do not have Cartan--Eilenberg at hand: I do not recall if they spelled this out.)

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  • $\begingroup$ Thank you very much for your answer! Switching $s$ and $d$ helps for showing strong convergence of the horizontal filtration of a cohomological right half-plane bicomplex. But I have another example: Consider the direct-sum total complex of a cohomological half-plane bicomplex $B^{ij}$ (I draw it in the right half-plane, the vertical differential points upstairs and the horizontal to the right), and let $Z^{ii} \subset B^{ii}$ be subspaces closed under the horizontal differential. Let $F_k = \bigoplus_{i + j > k} B^{ij} \oplus \bigoplus_i Z^{ii}$ be the diagonal filtration. What now? $\endgroup$ – Pavel Jul 23 '19 at 18:43
  • $\begingroup$ Also, please, could you comment on the proof of your statement? Is it, in fact, important that the groups in a half-plane vanish or is the vanishing of the differentials fundamental? P.S. Thanks for the better reference! $\endgroup$ – Pavel Jul 23 '19 at 18:47
  • $\begingroup$ (1) By "let $Z^{ii} \subset B^{ii}$ be subspaces closed under the horizontal differential", do you mean that the horizontal differential maps $Z^{ii}$ to zero? (2) In the definition of $F_k$, is the direct sum with $Z^{ii}$ for all $i$ intended to mean the internal sum in $\bigoplus_{i,j} B^{i,j}$? (3) It looks as if the vertical differential might not map $F_k$ to itself. Don't you want $F_k$ to be a subcomplex with respect to the total differential? $\endgroup$ – John Rognes Jul 23 '19 at 22:04
  • $\begingroup$ (1) I mean $d_h Z^{ii} = 0$. (2) It is the internal sum (I imagine replacing the diagonal with Z^{ii} and discarding all $B^{ij}$'s below). (3) The vertical differential points upstairs, and hence it preserves $F_k$, as far as I can see. Therefore, it should be invariant under the total differential. $\endgroup$ – Pavel Jul 23 '19 at 22:08
  • $\begingroup$ Your answers to (2) and (3) suggest that $i+j>k$ is not the intended condition in the definition of $F_k$. Do you mean $j > i > k$ or something like that? $\endgroup$ – John Rognes Jul 23 '19 at 22:17

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