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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.

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.

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?

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} A^{\bullet}\rightarrow B^{\bullet}\rightarrow C^{\bullet}\rightarrow A^{\bullet}[+1]. \end{eqnarray}

Thank you.

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.

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

Thank you.

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} A^{\bullet}\rightarrow B^{\bullet}\rightarrow C^{\bullet}\rightarrow A^{\bullet}[+1]. \end{eqnarray}

Thank you.