Skip to main content
9 events
when toggle format what by license comment
Nov 14, 2023 at 9:57 comment added Johannes Hahn No. I mean the direct sum, i.e. the coproduct in the category of $A$-modules.
Nov 12, 2023 at 2:23 comment added locally trivial Even though you write "map $(X_*, \partial)$ to $\bigoplus_{n\in \mathbb{Z}} X_n$" above, do you consider $(...,x_{-1},x_0,x_1,...)$ to be an element of $X$ if infinitely-many $x_i$ are non-zero? i.e., if $A=\mathbb{Q}$ and $X_n=\mathbb{Q}$ for all $n$, then do you consider $(...,1,1,1,...)$ to be an element of $X$?
Nov 24, 2019 at 16:08 vote accept Spencer Dembner
Nov 24, 2019 at 0:51 history edited Johannes Hahn CC BY-SA 4.0
Added another example without the condition $\mathbb{Q}\subseteq A$
Nov 24, 2019 at 0:46 comment added Johannes Hahn I added another, more general construction that works without this simplifying assumption.
Nov 24, 2019 at 0:45 history edited Johannes Hahn CC BY-SA 4.0
Added another example without the condition $\mathbb{Q}\subseteq A$
Nov 23, 2019 at 21:10 comment added Johannes Hahn You need that the eigenvalues $n$ all act differently on $X$. If, say $X$ where $2$-torsion, then $X_1$ and $X_3$ are both contained in the $1$-eigenspace. You could of course weaken that and ask that $X$ is torsionfree as an abelian group for example, but then you would get the full category of all chain complexes.
Nov 23, 2019 at 18:36 comment added Spencer Dembner Thanks, I really like this. Could you explain how the argument uses that A is a $\mathbb{Q}$-algebra?
Nov 23, 2019 at 16:59 history answered Johannes Hahn CC BY-SA 4.0