# Conditionally convergent spectral sequences with exiting and entering differentials

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 $$$$\label{Eq:Exit}\tag{C1} E^s = 0\quad\text{for all } s>0.$$$$ If $$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 $$$$\tag{C2} E^s = \bar{E}^s = 0\quad\text{for all }s<0.$$$$ If $$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:

  \bib{MR1718076}{article}{
author={Boardman, J. Michael},
title={Conditionally convergent spectral sequences},
conference={
title={Homotopy invariant algebraic structures},
date={1998},
},
book={
series={Contemp. Math.},
volume={239},
publisher={Amer. Math. Soc., Providence, RI},
},
date={1999},
pages={49--84},
review={\MR{1718076}},
doi={10.1090/conm/239/03597},
}


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.)

• 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? – Pavel Jul 23 '19 at 18:43
• 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! – Pavel Jul 23 '19 at 18:47
• (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? – John Rognes Jul 23 '19 at 22:04
• (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. – Pavel Jul 23 '19 at 22:08
• 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? – John Rognes Jul 23 '19 at 22:17