TITLE: Inverse limit of spectral sequences
QUESTION [5 upvotes]: 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?

REPLY [11 votes]: 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.