Timeline for Does a filtered A_N algebra give rise to a multiplicative spectral sequence?
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
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| Jan 28, 2018 at 21:48 | comment | added | algebrachallenged | In the case of $A_2$ (or $A_3$...) algebra, assuming the filtration is bounded, then I take it that it is still true that we have convergence of the multiplicative spectral sequence (i.e. the possibly non-associative multiplication on the $E_\infty$ page is the associated graded of the induced multplication on homology of $A$)? | |
| Jan 27, 2018 at 23:24 | vote | accept | algebrachallenged | ||
| Jan 27, 2018 at 23:22 | comment | added | algebrachallenged | Thanks for your answer. Just to fill in between lines, the answer to the question is "yes" (at least assuming one omits "unital" from the definition of multiplicative spectral sequence) because it follows from the axiom of filtered $A_3$ algebra structure on the first page will be associative. | |
| Jan 27, 2018 at 22:36 | history | answered | John Rognes | CC BY-SA 3.0 |