All Questions
7 questions
56
votes
5
answers
9k
views
Why are spectral sequences so ubiquitous?
I sort of understand the definition of a spectral sequence and am aware that it is an indispensable tool in modern algebraic geometry and topology. But why is this the case, and what can one do with ...
7
votes
0
answers
541
views
Convergence of a spectral sequence of a double complex
In Weibel's book, a spectral sequence $E^r_{p,q}$ is said to weakly converge to a graded object $H_{\ast}$ if for every $n$ there exists a filtration $\dots \subset F_{r}H_{n} \subset F_{r-1}H_{\ast} \...
7
votes
0
answers
436
views
spectral sequence for a complex with two filtrations
Suppose $(C,d)$ is a chain complex: an abelian group with a map $d:C \to C$ such that $d^2 = 0$ (people like to assume $C$ is graded; if that helps - feel free to do so). A filtration is an ascending ...
6
votes
1
answer
564
views
a question about Bockstein spectral sequence
I find the following theorem for Bockstein spectral sequence at http://pages.vassar.edu/mccleary/files/2011/04/MC10.fin_.pdf, page 459:
Question. for a fixed $k$, if $\beta$ does not hit $H_k(X;\...
6
votes
1
answer
242
views
For which exact couples do associated spectral sequences degenerate at $E_1$?
It is well known that a bigraded exact couple of objects of an abelian category yields a spectral sequence (cf. https://ncatlab.org/nlab/show/exact+couple#SpectralSequencesFromExactCouples). My ...
5
votes
0
answers
521
views
Spectral sequences coming from filtrations (Postnikov systems) in triangulated categories: references and convergence
Let $c^i: M^{\ge i+1}\to M^{\ge i}$ for $i\in \mathbb{Z}$ be an sequence of morphisms in a triangulated category $C$ and assume that $M^{\ge i}$ are equipped with compatible morphisms into an object $...
2
votes
1
answer
186
views
Can I bound the degree of a contracting homotopy in an exact filtered complex?
Suppose that I have given you a bigraded vector space $V = \bigoplus_{i,j} V_{i,j}$. The first grading is a "homological" $\mathbb Z$-grading, and the second is an independent $\mathbb Z$-grading. ...