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Hello,

I read that if $A\subset D$ is the heart of a bounded t structure, D a triangulated category, then D=D^b(A).

Have not been able to find a proper reference for this. Can anyone confirm this result?

Thanks!

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You need a kind of effacability too, see the Beilinson-Bernstein-Deligne article (Faisceax pervers). –  Torsten Ekedahl May 29 '11 at 16:00

2 Answers 2

up vote 8 down vote accepted

You're never going to find a reference because it's false. Just take $D$ to be the stable homotopy category of Postnikov pieces, i.e. spectra with finitely many non-trivial homotopy groups. Take $D_{\geq 0}$ (resp. $D_{\leq 0}$) the full subcategory of spectra with non-trivial homotopy groups concentrated in non-negative (resp. non-positive) degrees. This is a $t$-structure whose heart $D_{\geq 0}\cap D_{\leq 0}$ is equivalent to the category $Ab$ of abelian groups. There's even a functor $D^b(Ab)\rightarrow D$ preserving the $t$-structure (take the canonical $t$-structure on the left) and inducing an equivalence between the hearts, but $D$ cannot be equivalent to $D^b(Ab)$. Actually $D$ has no algebraic model.

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To add to what Fernando Muro said, there are examples where this is true. Beilinson showed in his paper "On the derived category of perverse sheaves" (Springer LNM 1289 pp. 27-41) that it is true when you are talking about the middle perverse t-structure in the constructible derived category of sheaves on schemes. If you look through Chapter 3 of BBD ("Faisceaux pervers") you will see how subtle it is even to define a functor from $D^b(A)$ to D, and you can also see from Beilinson's solo article how different that task is from the task of showing that such a map is an equivalence. That it works at all depends quite a bit on the fact that the perverse t-structure is used.

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It is also true for the usual t-structure on the constructible derived category, at least over $\mathbb{C}$. This was proved by Nori in his paper "Constructible sheaves". –  ulrich May 30 '11 at 6:34
    
ok, thanks a lot. That helps. Or rather, it destroys the proof I had in mind :) –  Carsten May 30 '11 at 7:53
    
@ulrich: you mean, that the bounded derived category of constructible sheaves is equivalent to the "constructible derived category" in which the sheaves in each complex need not be constructible but the cohomology is (and is bounded)? Thanks for reminding me of that; it's the sort of thing I forget about because it just turns into a definition. –  Ryan Reich May 30 '11 at 13:59
    
Yes, that's what I meant. –  ulrich May 31 '11 at 6:05

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