For a smooth projective variety $X$, Serre duality gives an exact autoequivalence on the derived category : $$ S_X : D^\flat(X) \to D^\flat(X), \hspace{3em} S_X(-) = - \otimes \omega_X[\dim X] $$ satisfying $$ \hspace{7em} \operatorname{RHom}^\bullet (\mathcal{A}, \mathcal{B}) \cong \operatorname{RHom}^\bullet(\mathcal{B}, S_X(\mathcal{A})) \hspace{7em} (\ast) $$ Applying this to the identity $\operatorname{id} : \mathscr{O}_X \to \mathscr{O}_X$, we should obtain a map $S_X(\textrm{id}) : \mathscr{O}_X \to \omega_X[\dim X]$ representing the cohomology class of the volume form — what I am having trouble doing is trying to figure out what the cone of this map explicitly is. We know that the cohomology of $\operatorname{Cone}(S_X(\textrm{id}))$ is certainly trivial by applying $\operatorname{RHom}^\bullet(\mathscr{O}_X, -)$ to the triangle: $$ \mathscr{O}_X \to \omega_X[\dim X] \to \operatorname{Cone}(S_X(\textrm{id})) $$ and using the fact that $(\ast)$ is an isomorphism, but other than that I'm not sure how to describe it.
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3$\begingroup$ In (*) one of the sides must be dualized. Consequently, there is no canonical morphism $\mathcal{O}_X \to \omega_X[\dim X]$ in general. Moreover, it is not true that the cohomology of $\mathrm{Cone}(S_X(\mathrm{id}))$ is zero. $\endgroup$– SashaCommented Jan 19, 2023 at 4:33
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$\begingroup$ @Sasha cant believe I overlooked that, thanks for correcting. I guess forgetting everything else, $S_X$ is a functor so it takes the identity (morphism) to the identity (morphism), i.e $S_X(id) = id_{S_X} : \omega[n] \to \omega[n]$. But then I suppose $\operatorname{Cone}(S_X(id)) = 0$ by TR1 which is a lot less interesting than what I had in mind $\endgroup$– cdsbCommented Jan 19, 2023 at 5:11
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2$\begingroup$ Sure, the cone of the identity is zero. However, if $h^0(X,\mathcal{O}_X) = 1$, Serre duality shows that $h^n(X,\omega_X) = 1$, hence still there is a unique (up to rescaling) morphism $\mathcal{O}_X \to \omega_X[n]$, so the question "what is its cone?" makes sense. $\endgroup$– SashaCommented Jan 19, 2023 at 5:25
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2$\begingroup$ The cohomology of the cone is not usually zero. However, it is zero for projective space $\mathbb{P}^n$. In that case the cone is the Koszul complex of the tautological surjection $\mathcal{O}(-1)^{\oplus(n+1)} \twoheadrightarrow \mathcal{O}$. For arbitrary $X$, by the Noether normalization theorem (or other arguments), there exists a finite surjective morphism $f:X\to \mathbb{P}^n$. Pullback the Koszul complex and "pushout" by the transpose of the derivative, $f^*\omega_{\mathbb{P}^n} \to \omega_X$. $\endgroup$– Jason StarrCommented Jan 19, 2023 at 12:19
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