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I find myself in the following situation:

I have a sequence of first quadrant spectral sequences, let's call them $ E(n)_{p,q}^* $, each convergent to $E(n)_{p,q}^\infty$, with spectral sequence morphisms $E(n)_{*,*}^* \to E(n-1)_{*,*}^*$, so we have an inverse directed system of spectral sequences.

Each module of each page is a locally finite graded vector space, so if you define $E_{p,q}^* = \varprojlim_n E(n)_{p,q}^* $, $E_{p,q}^* $ turns out to be a spectral sequence (differentials are the limit of differentials in the original sequences, etc.) by an argument found in a paper by John Carter. However, in this same paper he states that you can't say that $E_{p,q}^*$ converges to $\varprojlim E(n)_{p,q}^\infty$, and his counterexample rests on the fact that his spectral sequences can be non-bounded... Does anyone know if this "convergence" result is true for bounded (or, say, first quadrant) sequences?

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Adding a precise reference to Carter's paper might help! –  Mariano Suárez-Alvarez Feb 1 '12 at 23:39
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If it's true that $E^*_{pq} = \lim E(n)^*_{pq}$, then the convergence question should be true if you have all of them first quadrant... Since then, for each fixed $p$ and $q$, there is some FIXED $N$ such that $E(n)^{\infty}_{pq} = E(n)^N_{pq}$ (likewise for $n=\infty$). –  Dylan Wilson Feb 2 '12 at 1:51
    
Have you seen Mike Boardman's paper, Conditionally Convergent Spectral Sequences? You might be able to use the convergence result of the terms in the sequence to shoehorn something into the limit... (that is just a first though). –  Sean Tilson Feb 2 '12 at 5:00
    
Thanks both Dylan Wilson and Leonid Positselski for their answers! –  Pablo Zadunaisky Feb 2 '12 at 15:22
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All the comments are correct of course (especially "have you seen Mike Boardman's paper"). I'll just add that it is easy to understand the sort of thing that can go wrong. You can have a coherent sequence of elements $x_n \in E(n)^2_{p,q}$ so that $x_n$ supports a non-zero differential of length $k_n$ where $k_n$ approaches infinity with $n$. This gives an element of the inverse limit $E^2$ term which is a permanent cycle, but which is not an element of the inverse limits of the $E^\infty$ terms. –  Hal Sadofsky Feb 14 '12 at 5:51
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1 Answer 1

up vote 8 down vote accepted

The inverse limit of directed systems of locally finite dimensional graded vector spaces is an exact functor. That is why it takes directed systems of spectral sequences to spectral sequences of the inverse limits.

In the case of the first quadrant spectral sequences, the notion of convergence does not involve taking any limits (or more precisely, all the limits involved stabilize at the terms known in advance from the gradings $p$ and $q$; see Dylan Wilson' comment). So the inverse limit of the limit terms $E^\infty(n)$ will be the limit term $E^\infty$ of the inverse limit of the spectral sequences.

The reason spectral sequence limits may not commute with the inverse limits in general must be because defining the term $E^\infty$ may involve passing to the direct limits in some cases. This does not happen for first quadrant spectral sequences.

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Aside from answering the question, it is a crystal clear explanation. Thank you very much, Leonid! –  Pablo Zadunaisky Feb 2 '12 at 15:19
    
"inverse limit of directed systems of locally finite dimensional graded vector spaces is an exact functor" - really? You don't need to assume a Mittag-Leffler condition? –  Justin Curry Aug 19 '12 at 21:27
    
The locally finite-dimensional graded case reduces to the finite-dimensional ungraded case. As for the latter, any projective system of finite-dimensional vector spaces (indexed by the nonnegative integers) satisfies the Mittag-Leffler condition. Because any decreasing chain of subspaces in a finite-dimensional vector space stabilizes. That's called compactness, or something. –  Leonid Positselski Aug 19 '12 at 22:39
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