Chain Complexes – When Do They Decompose as a Direct Sum?

homological-algebra

I've been looking at some thick subcategories in $K^b(R-proj)$ (the homotopy category of bounded chain complexes of projective modules), and, as expected, I'm running into the question of when chain complexes split quite often. I'm wondering what sorts of useful criteria there are for determining when chain complexes split. By "split" I mean decompose as two nontrivial complexes as $A = A_1 \oplus A_2$.

Feel free to strengthen the hypotheses a bit if you need to- these can be complexes of free modules, if you like. I'm just trying to get a sense of how to look at a chain complex and think, "That probably splits…" or, "That probably doesn't…" Also remember that I'm working in the homotopy category, so if the question becomes easier when homotopy equivalent objects are identified, please feel free to use this hypothesis.

Best Answer

One result that guarantees such a decomposition comes from looking at the homological support of such complexes (assuming that $R$ is commutative so we have a tensor product). The homological support of a complex $A$ is just the union of the supports of the $H^i(A)$ as $R$-modules. Then it is a result of Balmer, in the paper Supports and Filtrations in Algebraic Geometry and Modular Representation Theory which is available on his website, that if the homological support of $A$ can be written as a disjoint union of closed subsets $Y_1\cup Y_2$ of $\mathrm{Spec} \;R$ whose complements are quasi-compact then $A\cong A_1\oplus A_2$ in $K^b(R\text{-}\mathrm{proj})$ where the homological support of $A_i$ is $Y_i$.

Another method which works for any triangulated category is if $f\colon A\to B$ is a morphism then the triangle one gets by completing splits giving $B\cong A\oplus \mathrm{cone}(f)$ if and only if the map $\mathrm{cone}(f) \to \Sigma A$ is zero. A reference for this is Corollary 1.2.5 of Neeman's book on triangulated categories (I think I've also put the proof on MO before but I can't remember in which answer, maybe I can hunt it down later).

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