# When do chain complexes decompose as a direct sum?

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.

• Can you please say what you mean by split? I have some guesses, but I'm not at all confident that I know. – Tom Goodwillie Jul 21 '10 at 22:25
• Uh oh- is there a difference between the notions: split as in has a splitting map and split as in decomposes as a direct sum of chain complexes? Because I mean the latter, but I thought they were the same... – Dylan Wilson Jul 21 '10 at 22:29
• So you are asking about how to detect when a complex $A$ decomposes nontrivially as $A_1 \oplus A_2$? The comment about acyclic complexes is a little confusing if this is what you mean since any bounded acyclic complex of projective $R$-modules is contractible. – Greg Stevenson Jul 21 '10 at 22:52
• No, don't worry, I just wanted to make sure. Reading between the lines, I think that by "does this chain complex split?", you mean "is it the direct sum of two (nonzero? acyclic?) chain complexes?" – Tom Goodwillie Jul 21 '10 at 23:34
• Oh, Greg I see your point! I was quoting Weibel, but that was before he was working in the homotopy category. I'll edit the question to take that out and clarify what I mean by "split". Thanks! – Dylan Wilson Jul 21 '10 at 23:42

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).

• Can you quickly remind me what homological support is? Is it just the collection of prime ideals of $K^b(R-proj)$ not containing an element, or is it something different? And I have Neeman's book, you don't need to hunt down the proof- I think it's rather straightforward. – Dylan Wilson Jul 22 '10 at 15:45
• There are two ways to view the homological support on $K^b(R\text{-}\mathrm{proj})$. As you mention one is given by Balmer's support on the spectrum of prime tensor ideals. For $A$ in $K^b(R\text{-}\mathrm{proj})$ we can also consider $\mathrm{supp}_R A = \cup_i \mathrm{supp}_R H^i(A)$ where we are just taking supports as $R$-modules, i.e., the set of primes $\mathfrak{p}$ such that $H^i(A)_{\mathfrak{p}} \neq 0$. These two notions agree, by the universality property of Balmer's support and the fact that homological support classifies thick subcategories. – Greg Stevenson Jul 22 '10 at 21:16
• Thank you! Excuse my inanity; I'm learning things sort of backwards here- looking at Balmer's paper before understanding all the constructions that inspired it :). Anyway, I don't think I'll be getting criteria more useful than the ones you gave, so I'll accept your answer. Thanks again! – Dylan Wilson Jul 23 '10 at 7:03

This is what comes of textbook authors making up terms. In fact, my first guess as to what you meant by "split" was "quasi-isomorphic to its homology" (i.e. to the chain complex with zero boundaries having the same homology). It seems that this was essentially correct.

I assume you know that for $R=\mathbb Z$ and more generally for any ring such that submodules of projectives are projective (i.e. $Ext^i=0$ when $i>1$) every chain complex of modules is split in Weibel's sense and so is quasi-isomorphic to its homology.

I would guess that a chain complex of $R$-modules $C$ must be split if for $a$ less than $b$ we have $Ext^{b-a+1}(H_a(C),H_b(C))=0$. Or at least I would guess that that is true if $C$ is bounded below.

You mentioned bounded complexes. Do you mean bounded above and below?

• I endorse the opening lament! Tom, isn't "quasi-isomorphic to its homology" known as "formal" and the ring such that submodules of projectives are projective ... "hereditary"? – Victor Protsak Jul 22 '10 at 7:18
• Certainly "formal" is used to mean "weakly equivalent to its homology" in various contexts where the chain complexes and their homology have some ring, co-ring, .... structure. It started with Quillen's rational homotopy and happily spread, and no doubt is used for chain complexes of $R$-modules these days. "Hereditary": now that you say it, I have seen that definition, but if you had said to me out of context "suppose $R$ is a hereditary ring" I would not have known what you meant. – Tom Goodwillie Jul 22 '10 at 11:54
• I meant to say it started with Sullivan's rational homotopy. – Tom Goodwillie Jul 25 '10 at 0:48

I believe the question is: "when are the complexes $A^\bullet$ and $\bigoplus H^i(A^\bullet)[-i]$ (with trivial differential) isomorphic in the derived category"? - which as you say is not very profound for an acyclic complex...

A complex with only two nonzero cohomology groups, $H^0$ and $H^n$ say, can be rewritten after truncation as a Yoneda $(n+1)$-extension of $H^n$ by $H^0$ $$0 \to H^0 \to A^0 \to \dots A^n \to H^n \to 0$$ so the obstruction in this case is just an element of $Ext^{n+1}(H^n,H^0)$. For a more general complex one can proceed inductively: first split the 2-extensions $H^{i-1} \to * \to * \to H^i$, then the 3-extensions... but that's not very pleasant. Cohomological dimension may simplify the considerations.

A useful starting point is Deligne's thesis: "Theoreme de Lefschetz et criteres de degenerescence..." (Publ Math IHES 35 (1968) 107-126) where he proves some important splitting results of this kind in a geometric setting, and a later sequel article by Deligne "Decomposition dans la categorie derivee" (in volume 1 of the Motives proceedings, Proc Symp AMS 55). There is also a study of some splittings in de Rham cohomology in the Deligne-Illusie paper on the degeneration of the Hodge-de Rham spectral sequence and Frobenius (Inventiones vol.89). I am afraid these may be too geometric for what you have in mind, though.

• @Dylan: I read your correction after posting this :-( maybe someone else will find the answer useful – Tony Scholl Jul 21 '10 at 23:52

One weak answer i.e. a special case would be that if your abelian category $A$ has global dimension 0 ($Ext^i(X,Y) = 0$ for $i>0$), eg- category of vector spaces, then any complex involving elements of $A$ splits into complexes of cycles, boundary, and homologies.

In case your abelian category has global dimension 1, then complexes in the bounded Derived Category(not necessarily in homotopy category!) of $A$ splits into homologies (with sufficient translations), i.e.

for every $E \epsilon$ $D^b(A)$, $E=\oplus H^i(E)[-i]$.