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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.

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?

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    $\begingroup$ The sequence you wrote does not seem to be quite correct, if I am not mistaken: you should replace the first term $S^2A$ by the divided power $\Gamma^2A$ (there is no obvious natural map $S^2A\to A\otimes B$ if $2$ is not invertible). In fact, the Koszul exact sequence [Illusie, Complexe Cotangent et Déformations I, V.4.3.2] gives rise to the long exact sequence \[0\to\Gamma^n A\to\Gamma^{n-1}A\otimes B\to\Gamma^{n-2}A\otimes\bigwedge\nolimits^2B\to\dots\to\bigwedge\nolimits^nB\to\bigwedge\nolimits^nC\to0\]and yours is the special case when $n=2$. $\endgroup$
    – Z. M
    Commented Jun 23, 2021 at 13:31
  • $\begingroup$ That's right. I'm just writing the $n=2$ case, and it can be checked that $S^2A$ is indeed in the kernel of $A\otimes B \rightarrow \Lambda^2 B$. $\endgroup$
    – MKR
    Commented Jun 24, 2021 at 0:11

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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 $\wedge^2 H^*$

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  • $\begingroup$ Thanks a lot! This was really helpful, I used to cook up a formula to give the correct answer. However, it is possible to prove it now. Is this idea well known? I'm a physicist so I don't have a good overview of this topic... $\endgroup$
    – MKR
    Commented Jun 24, 2021 at 8:29
  • $\begingroup$ @MKR I'm glad its useful. I think that the definition of exterior and symmetric powers of complexes qualifies as well-known, though I don't have a great reference to point to $\endgroup$
    – user1092847
    Commented Jun 24, 2021 at 15:31

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