All references below are from McCleary's book, second edition.

Suppose that we have a filtered complex where the filtration is unbounded. Suppose that the associated spectral sequence is weakly convergent, as per Theorem 3.2. Then the spectral sequence converges to the filtered quotients of the cohomology, but this convergence is not unique as stated in the second last paragraph on page 64.

My question is that suppose the filtered complex is weakly convergent and that the induced filtration on the cohomology is Hausdorff, can we then construct the cohomology uniquely up to extension problems? In the book it seems as if the spectral sequence uniquely determines a target only after we take a completion as per Definition 3.11. Basically I don't understand why the completion condition is necessary.



1 Answer 1


Suppose we have $\bigoplus_\mathbb{N}\mathbb{Z}$ and $\prod_\mathbb{N}\mathbb{Z}$. Both modules hava a canonical decreasing filtration where the $k$-th filtration step consists of all elements whose first k coordinates vanish. Both filtrations have the same subquotients. From this information you cannot distinguish $\bigoplus_\mathbb{N}\mathbb{Z}$ and $\prod_\mathbb{N}\mathbb{Z}$, you can only say that their completions are the same.


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