Vanishing natural transformation exact triangle

Question

This question is a follow-up to this question I asked some time ago. Let $X$ be a smooth projective variety of dimension $n$ over $\mathbb{C}$. Let $\omega \in H^{n}(X,K_X)$, $\omega \neq 0$. Let $$A \longrightarrow B \longrightarrow C \longrightarrow A[1]$$

be an exact triangle in $D^b(X)$. Assume that the morphism: $$\mathrm{id}_A \otimes \omega : A \longrightarrow A \otimes K_X[n]$$ vanishes in $\mathrm{Hom}(A,A\otimes K_X[n])$ and that the morphism: $$\mathrm{id}_B \otimes \omega : B \longrightarrow B \otimes K_X [n]$$ vanishes in $\mathrm{Hom}(B,B\otimes K_X[n])$. I'd like to know if the morphism: $$\mathrm{id}_C \otimes \omega : C \longrightarrow C \otimes K_X[n]$$ vanishes in $\mathrm{Hom}(C,C\otimes K_X[n])$?

I know that there is no five lemma for zero-morphisms and that if this statement is true it is not just a formal consequence of the definition of a triangulated category (see the answer to the question I am referring to in the intro).

On the other hand, seen the very special nature of the involved morphisms, I tought it might be true. Or perhaps this is plain wrong and a counter-example would be greatly appreciated.

Thanks in advance!

Answer 1

So I think I have an argument which shows that if $\mathrm{id}_{A} \otimes \omega : A \longrightarrow A \otimes K_X[n]$ is zero and $\mathrm{id}_{B} \otimes \omega : B \longrightarrow B \otimes K_X[n]$ is zero then $\mathrm{id}_{C} \otimes \omega : C \longrightarrow C \otimes K_X[n]$ is not necessarily zero. This argument uses crucially the result of Thomason that was brought to my attention by Jeremy Rickard (who I heartily thank for that).

Let $X$ be a K3 surface and let $\mathcal{T}$ be the cone of the natural map: $$\mathcal{O}_X \longrightarrow \mathcal{O}_X[2]$$ induced by $\omega$. The object $\mathcal{T}$ has rank $0$ in $D^b(X)$ and one easily computes that $\mathrm{End}(\mathcal{T}, \mathcal{T}) = \mathbb{C}$. Assume that the morphism $\mathrm{id}_{\mathcal{T}} \otimes \omega : \mathcal{T} \longrightarrow \mathcal{T}[2]$ is not zero. Then it is a generator of $\mathrm{Ext}^2(\mathcal{T},\mathcal{T}) = \mathrm{End}(\mathcal{T})^* = \mathbb{C}$. This is impossible. Indeed, we have $$\mathrm{Trace}(\mathrm{id}_{\mathcal{T}} \otimes \omega) = \mathrm{rank}(\mathcal{T}).\omega = 0$$ but Serre duality implies that: $$\mathrm{Trace} : \mathrm{Ext}^2(\mathcal{T}, \mathcal{T}) \longrightarrow H^2(X,\mathcal{O}_X)$$ is surjective. This shows that $\mathrm{id}_{\mathcal{T}} \otimes \omega = 0$ in $\mathrm{Ext}^{2}(\mathcal{T}, \mathcal{T})$.

Now, assume that for any exact triangle $A \longrightarrow B \longrightarrow C$ in $D^b(X)$ such that $\mathrm{id}_{A} \otimes \omega : A \longrightarrow A[2]$ is zero and $\mathrm{id}_{B} \otimes \omega : B \longrightarrow B[2]$ is zero, we have $\mathrm{id}_{C} \otimes \omega : C \longrightarrow C[2]$ is also zero. (I denote this hypothesis by $(*)$).

Then, the subcategory of $D^b(X)$ consisting of objects $S$ such $\mathrm{id}_{S} \otimes \omega$ is zero would be a thick $\otimes$-triangulated subcategory of $D^b(X)$ (see the comment of Jeremy Rickard to the question).

It has been checked that $\mathcal{T}$ is in this category and the cohomological support of $\mathcal{T}$ is $X$. Hence, by Thomason classification of thick $\otimes$-triangulated subcategories of $D^b(X)$, we deduce that for any $S \in D^b(X)$ such that the cohomological support of $S$ is $X$, we have the vanishing of $\mathrm{id}_{S} \otimes \omega : S \longrightarrow S \otimes[2]$. This is clearly absurd and hypothesis $(*)$ is incorrect.