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Suppose that a triangulated category $C$ contains a full additive subcategory $B$ of (strong) generators (i.e. there does not exist a proper strict triangulated subcategory $C'\subset C$ that contains $B$) such that: there are no non-zero $C$-morphisms between $B_1$ and $B_2[i]$ for any $B_1,B_2\in Obj B$ and $i\neq 0$.

Is it true that $C\cong K^b(B)$? Is anything known about this question (in general)?

Upd. I know how to prove this statement for any 'algebraic' triangulated $C$ (i.e. if $C$ admits a differential graded enhancement); this includes all derived categories of sheaves. So, one can reformulate my question as follows: is such a $C$ necessarily algebraic? I know the proof of this fact when $C$ is an $f$-category (in the sense of Beilinson).

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    $\begingroup$ The condition on B looks a bit like the condition on a tilting complex (in the a priori given context of homotopy categories of complexes of modules, however). Suppose you manage to construct an exact equivalence K^b(B) -> C. This also gives a functor C^b(B) -> C by composition. It maps a complex 0 -> B_1 -d-> B_2 -> 0 to the cone X on d in C. A morphism of such complexes should be mapped to a morphism on the cones, say X -> X'. ... $\endgroup$
    – Matthias Künzer
    Commented Dec 1, 2010 at 6:32
  • $\begingroup$ ... Under the given circumstances, is it uniquely determined by the mere condition that it should fit into a morphism of triangles (B_1,B_2,X) -> (B'_1,B'_2,X')? So suppose this morphism is zero on B_1 -> B_2 and B_2 -> B'_2. Does there exist B_1[1]->X' such that (X->B_1[1]->X') =/= 0, but (X->B_1[1]->X'->B'_1[1]) = 0? Reminds me of the examples in Neeman's "Some new axioms...". Of course, a possible nonuniqueness of X->X' does not imply nonexistence of a functor picking such a X->X'. [Comment split because of length.] $\endgroup$
    – Matthias Künzer
    Commented Dec 1, 2010 at 6:38

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Mikhail, if you assume more generally that $C$ is topological (in an appropriate sense) then your claim is also true. As Matthias suggests above, this is a tilting-like theorem, Theorem 5.1.1 in 'Stable model categories are categories of modules' by Schwede and Shipley. This assumption may comprise all examples of interest to you. So far, the very few known examples of exotic model categories do not satisfy your assumptions.

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It follows that $C \cong D^b(mod-B)$. I don't think the latter is equivalent to $K^b(B)$ (by the way, what does $K^b$ denote?) unless in very special situation, like when the homological dimension of $B$ is zero. For example, if $X = Spec A$ is a smooth affine variety and $C = D^b(coh-X)$, then you can take $B = \{O_X\}$. This gives $C \cong D^b(mod-A)$ which is different from $K^b(A)$.

EDIT: This is an answer to the comment below. If you want $K^b(O_X)$ to be equivalent to $D^b(X)$ for a smooth affine $X$ you definitely should require that all projective $A$-modules have a free resolution (I guess it is equivalent to $Pic X = 0$). If this is not true, to get an equivlence you should take a pseudo-abelian envelope of $B$ by adding to it all direct summands.

As for the equivalence $C \cong D^b(mod-B)$, I implicitly assumed that $C$ is algebraic. In this case it is well known.

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    $\begingroup$ $K^b$ denotes bounded $B$-complexes with morphisms up to homotopy equivalence. In your example $B$ would be the category of finite direct sums of $O_X$, and I think that we have an isomorphism in question. More generally, such an isomorphism exists for any triangulated $C$ that admits a differential graded enhancement (i.e. if $C$ is 'algebraic'); this includes all categories of sheaves. Still, could you tell how do you construct the isomorphism you mentioned? $\endgroup$
    – Mikhail Bondarko
    Commented Dec 1, 2010 at 12:30

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