This question came up while studying Bridgeland's Stability conditions on K3 surfaces, in the proof of Lemma 6.3.
Suppose $\mathcal{D}$ is a triangulated category, and $(\mathcal{D}^{\leq 0}, \mathcal{D}^{\geq 0})$ is a bounded $t$-structure with heart
$$ \mathcal{B} = \mathcal{D}^{\leq 0} \cap \mathcal{D}^{\geq 0}.$$
Then $\mathcal{B}$ is a full abelian subcategory of $\mathcal{D}$.
My question is:
Does a short exact sequence
$$0 \to A \to B \to C \to 0$$
in $\mathcal{B}$ correspond to a triangle
$$ A \to B \to C \to A[1]$$
in $\mathcal{D}$?
I know that in general $\mathcal{D}(\mathcal{B}) \neq \mathcal{D}$, but I wonder if this might be true in certain cases, e.g. when $\mathcal{D} = \mathcal{D}(\mathcal{A})$ for some abelian category $\mathcal{A}$, and if $(\mathcal{D}^{\leq 0}, \mathcal{D}^{\geq 0})$ is obtained by tilting a torsion pair $(\mathcal{T}, \mathcal{F})$. In that case, we have at least that $\operatorname{Hom}_{\mathcal{D}(\mathcal A)}(X, Y) = \operatorname{Hom}_{\mathcal D (\mathcal B)}(X, Y)$ for $X, Y \in \mathcal B$, see Tilting in Abelian Categories and Quasitilted Algebras by Happel, Reiten and Smalø.
In the context of Bridgeland's paper, there is a short exact sequence
$$ 0 \to A \to \mathcal{O}_x \to B \to 0 \tag{$*$}$$
in some abelian category obtained by tilting. $\mathcal{O}_x$ is the skyscraper sheaf over a point, considered as a $0$-complex.Bridgeland writes that this gives an exact sequence
$$0 \to H^{-1}(B) \to H^0(A) \to \mathcal{O}_x \to H^0(B) \to 0.$$
I don't know how to follow this longer sequence, if $(*)$ does not correspond to a triangle.
Best Answer
Yes this is indeed such a correspondance. The main reason is that, if you have a $t$-structure with heart $\mathcal{B}$ and if $H$ denote the corresponding cohomology, then a triangle in $\mathcal{D}$ gives rise to a long exact sequence in $\mathcal{B}$.
Now, if $A\to B\to C\to A[1]$ is a distinguished triangle in $\mathcal{D}$, such that $A,B,C\in\mathcal{B}$, then the long exact sequence gives in particular $$H^{-1}(C)\to H^0(A)\to H^0(B)\to H^0(C)\to H^1(A)$$ but since $A,B,C$ belong to the heart, this five terms exact sequence is actually $$0\to A\to B\to C\to 0$$ so you have indeed a short exact sequence in $\mathcal{B}$.
Conversely, let $0\to A\to B\to C\to 0$ a short exact sequence in $\mathcal{B}$. Let $C'$ be the cone of $A\to B$. Because the composition $A\to B\to C$ is zero, the morphism $B\to C$ factors through $C'$ (not necessarily uniquely). So we have a commutative diagram :
$$\require{AMScd} \begin{CD} A@>>>B@>>>C'@>>>A[1]\\ @|@|@VVV\\ A@>>>B@>>>C \end{CD} $$ The top row is a distinguished triangle, we want to show that $C'\to C$ is an isomorphism. Now apply the cohomology functor, by naturality we have a commutative diagram : $$\require{AMScd} \begin{CD} 0@>>>H^{-1}(C')@>>>A@>>>B@>>>H^0(C')@>>>0\\ @.@.@|@|@VVV\\ @.0@>>>A@>>>B@>>>C@>>>0 \end{CD} $$
The top row is exact by the long exact sequence of cohomology, with the outer $0$ because $A$ and $B$ are in the heart. The bottom row is also exact by assumption. By diaram chase, $H^{-1}(C')=0$ and $H^0(C')=C$. The long exact sequence also give $H^i(C')=0$ for all $i\neq -1, 0$. It follows that $C'$ is concentrated in degree $0$ and thus that $C'= H^0(C')= C$.