Usually when one has a short exact sequence of bundles,


\begin{eqnarray}
0\rightarrow A \rightarrow B\rightarrow C\rightarrow 0,
\end{eqnarray}
then there is an associated long exact sequence,


\begin{eqnarray}
0\rightarrow S^2A\rightarrow A\otimes B\rightarrow \Lambda^2B\rightarrow \Lambda^2C\rightarrow 0.
\end{eqnarray}
What is the corresponding identity for the following triangle?
\begin{eqnarray}
F^{\bullet}\rightarrow G^{\bullet}\rightarrow H^{\bullet}\rightarrow F^{\bullet}[+1].
\end{eqnarray}


Thank you.




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Maybe it is better to start with a simpler case. What happened when $A$, $B$ and $C$ are coherent sheaves not simply a locally free sheaf.


Then, in principle, the ordinary tensor products should be replaced by the "derived" version. Is there any similar identity in this case?



Antisymmetric product of complexes
You can interpret your identity as saying: if $cone(A \to B) \simeq C$ then $\wedge^2(cone(A \to B)) \simeq \wedge^2 C$, where $\wedge^2$ of complexes is defined using the Koszul symmetric monoidal structure on complexes. As discussed in the comments, this is only true when $2$ is invertible because to prove it we need the operation of taking the alternating isotypy component to be exact.


This interpretation lets you extend your identity to arbitrary complexes of flat sheaves: take $cone(F^\* \to G^\*)$ and apply the $\wedge^2$ functor to get a complex which is quasi-isomorphic to $\wedge^2 H^\*$


