TITLE: The derived category of the heart of a t-structure
QUESTION [17 upvotes]: Suppose $\mathcal{D}$ is a triangulated category and that we are given a $t$-structure $(\mathcal{D}^{\leq 0},\mathcal{D}^{\geq 0})$ on $\mathcal{D}$.  The heart of the $t$-structure, $\mathcal{A}=\mathcal{D}^{\leq 0} \cap \mathcal{D}^{\geq 0}$, is an abelian category.  Is it true in general that $\mathcal{D}=D(\mathcal{A})$ is the derived category of the heart of the given $t$-structure on $\mathcal{D}$?  If not, is there an easy example that shows why not?

REPLY [15 votes]: The most serious problem is that in general there is no natural functor from $D(A)$ to $D$. To construct one you need an additional structure on $D$. There are several approaches here. One was suggested by Beilinson and gives a notion of a filtered triangulated category. Another important approach uses derivators.  
On the other hand, if you have enough structure to construct a functor, then the criterion for it to be an equivalence is rather simple. If I remember right, the necessary and sufficient condition is that each morphism $A \to A'[n]$ in $D$ (with both $A$ and $A'$ in the heart) should be decomposable into a sequence $A \to A_1[1] \to A_2[2] \to \dots \to A_{n-1}[n-1] \to A'[n]$
with all $A_i$ being objects in the heart (in other words, the graded algebra of $Ext$'s should be 1-generated).