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Maybe this is too technical and elementary, but I cannot make up my mind, nor find a reference.

The situation is the following: let $X$ be a double cochain (right half-plane) complex of abelian groups and let

$$ (\mathbf{Tot}^{\prod} X)^n = \prod_{p+q = n}X^{pq} $$

denote its total-product complex. The first filtration on $X$,

$$ {}_I F^s(X) = \begin{cases} X^{pq} & \ \text{if} \ p \geq s , \\\ 0 & \ \text{otherwise} \end{cases} $$

gives you the filtration on $\mathbf{Tot}^\prod X$:

$$ (F^s \mathbf{Tot}^\prod X)^n = \prod_{p+q=n , p\geq s} X^{pq} $$

and you have, as with any filtered differential complex $(A, F, d)$, an induced filtration in cohomology:

$$ F^pHA = \mathbf{im} (HF^pA \longrightarrow HA) \ . $$

My question is the following: is the filtration induced by ${}_I F$ on $H(\mathbf{Tot}^\prod X)$ Hausdorff? That is,

$$ \bigcap_p F^pH(\mathbf{Tot}^\prod X ) = 0 \ ? $$

I couldn't find an answer in the literature. Weibel's book says that in this situation the spectral sequence arising from the first filtration is "convergent". Unfortunately, for Weibel this only means that you have an isomorphism $E_0 HA = E_\infty A$. Cartan-Eilenberg "Homological Algebra" doesn't work with the total-product complex, but with the total-sum one:

$$ (\mathbf{Tot}^{\bigoplus} X)^n = \bigoplus_{p+q = n}X^{pq} $$

For this one, I think, the answer is "yes": if I had some cohomology class

$$ [x] \in \bigcap_p F^pH^n(\mathbf{Tot}^\bigoplus X ) \quad \Longleftrightarrow \quad [x] \in F^pH^n(\mathbf{Tot}^\bigoplus X ) \ \text{for all} \ p $$

then I could find representatives for $[x]$ like

$$ (0, \stackrel{p-1}{\dots}, 0, x^{p,n-q}, x^{p+1, n-q-1}, \dots ) $$

for all $p \geq 0$. Since there is only a finite number of $x^{pq} \neq 0$ for each element of $(\mathbf{Tot}^\bigoplus X)^n$, in a finite number of steps, I can be sure that I can find a representative for $[x]$ which is zero, so $[x]=0$.

But this reasoning doesn't work with $\mathbf{Tot}^\prod X$: you can only state with certainty that, for every $p$ there is some $x_p \in F^p$ and $b_p$ such that

$$ x - x_p = db_p \quad \Longleftrightarrow \quad x- db_p \in F^p \ . $$

These equations have a nice interpretation: if you topologize $\mathbf{Tot}^\prod X$ taking as basic open sets $x + F^p$ for all $x \in \mathbf{Tot}^\prod X$ and $p$, they read:

$$ (db_p) \longrightarrow x \ . $$

That is, $x$ is a limit of coboundaries. But, unless the filtration is finite, this doesn't imply that $x$ itself is a coboundary, does it?

So I could ask my question this way: in this situation, is the set of coboundaries closed?

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2 Answers 2

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No, this filtration is not necessarily Hausdorff. The problem is connected to nonexactneess of the inverse limit functor. Here is a family of examples to illustrate the basic issue.

Suppose $\cdots \to A_2 \to A_1 \to A_0$ is an inverse system of abelian groups. Define a complex by $$ X^{p,-p} = X^{p+1,-p} = A_p $$ for all natural numbers $p$, with $X^{p,q} = 0$ otherwise. The horizontal differentials $X^{p,-p} \to X^{p+1,-p}$ are the identity maps $A_p \to A_p$, and the vertical differentials $X^{p,-p} \to X^{p,1-p}$ are the maps $A_{p} \to A_{p-1}$ from the inverse system for $p > 0$.

The total complex has only two nonzero groups in degrees 0 and 1: $$ \prod A_p \to \prod A_p $$ and the indicated map is the map Milnor introduced whose kernel is $\lim A_p$ and whose cokernel is $\lim^1 A_p$. (Possibly up to sign, depending on what your conventions for a double complex are.) Therefore $$ H^0 Tot^{\Pi} X = \lim A_p,\ H^1 Tot^\Pi X = {\lim}^1 A_p $$ However, the spectral sequence associated to this complex has the amusing behavior that the "diagonal" entries $E_r^{p,-p}$ busily spawn differentials that erase everything on the "superdiagonal" $E_r^{p+1,-p}$. Therefore, the entirety of $H^1(Tot^\Pi X)$ is in the intersection of the images of the filtrations.

(As a mostly nonmathematical aside, $\lim^1$ is the bane of many an argument.)

If you're interested in convergence properties, I'd suggest Boardman's thesis, mentioned in this answer of Tilman's.

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Nice counter-example. I wanted to comment it, but I needed an aswer to do it an explain a doubt. (You can see it below.) –  a.r. Aug 25 '10 at 16:16
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[This should be a comment, but went too long and somewhat involved to write.]

This seems a perfect counter-example: thank you very much. Still, I have some doubts.

If I understand you correctly, you are saying that, since every term in the "superdiagonal" becomes zero eventually at some page of the spectral sequence, then the corresponding "superdiagonal" in the $E_\infty$ page will be zero. So we will have $H^1(\mathbf{Tot}X) = F^0 = F^1 = F^2 = \dots$

But isn't that already true at the $E_2$ page? I mean: aren't the differentials $d_1 : E_1^{p,-p} \longrightarrow E_1^{p+1,-p}$ simply the projections $A_p \longrightarrow A_p/\mathrm{im}\alpha_{p+1}$, where $\alpha_p : A_p \longrightarrow A_{p-1}$ are the morphisms of the inverse system?

If this is so, we're already killing the "superdiagonal" at the $E_2$ page, aren't we?

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(I assume you mean $d_1$ differentials instead of $d_2$.) Remember that the term at $E_1^{p,-p}$ will actually be the kernel of the $d_1$-differential, i.e. the map $A_p \to A_{p-1}$, and can be identified with the map $ker(\alpha_p) \to A_p/im(\alpha_{p+1})$ which is not necessarilly surjective. –  Tyler Lawson Aug 25 '10 at 16:20
argh - I meant the kernel of the $d_0$-differential, not the $d_1$-differential. A nice example is the case of the inverse system where all terms are $\mathbb{Z}$ and all maps are multiplication by p, and then there is only one term on the diagonal at the $E_1$-page. –  Tyler Lawson Aug 25 '10 at 16:23
Yes: I've corrected those $d_2$ and you are right: those $d_1$ aren't necessarilly surjective. I like that particular example with $\mathbb{Z}$! :-D But, in this case, I'm sorry: I have some problems to see how the differentials coming from the diagonal necesseraly erase all the "superdiagonal". –  a.r. Aug 25 '10 at 16:39
Well, you can either track through explicit methods for computing differentials in a double complex (which are pretty straightforward here), or you can consider the subcomplexes which contain only $X^{pq}$ for $q \geq -n$. These have trivial total cohomology, and so you know that the spectral sequence must degenerate for them; naturality of the differentials will then force the behavior you're looking for. –  Tyler Lawson Aug 25 '10 at 19:22
Well, an E_2-cycle is comes an element in the kernel of the vertical differential whose horizontal image is in the image of the vertical differential; using this you can give an explicit determination of d_2. The further differentials are basically similar; you can show "y is the image of x under d_r" by writing down a chain of elements, starting with x, where the horizontal boundary of one coincides with the vertical boundary of the next, until the last remaining horizontal boundary is y. –  Tyler Lawson Sep 1 '10 at 16:27
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