A characterization on coherent sheaves inside $D^b(X)$

Question

Consider a smooth projective variety of ample $\omega_X$, how can I quickly see that $$\textbf{Coh}(X)=\{\mathcal{F}^{\bullet}\mid\text{Hom}(\omega_X^{\otimes i},\mathcal{F}^\bullet[n])=0\text{ for } n \neq0\text{ and }i\ll0\}\subset D^b(X)$$ This is used in the proof of Bondal—Orlov reconstruction theorem.

Answer 1

First, one has $$ \mathrm{Hom}(\omega_X^{\otimes i}, F[n]) \cong \mathbb{H}^n(X, F \otimes \omega_X^{\otimes -i}). $$ Next, there is the hypercohomology spectral sequence $$ H^q(X, \mathcal{H}^p(F) \otimes \omega_X^{\otimes -i}) \Rightarrow \mathbb{H}^{p+q}(X, F \otimes \omega_X^{\otimes -i}). $$ Now, since $\omega_X$ is ample and $i \ll 0$, for each (of the finite number) of nonzero sheaves $\mathcal{H}^p(F)$ the twist $\mathcal{H}^p(F) \otimes \omega_X^{\otimes -i}$ has no higher cohomology and is globally generated, therefore the spectral sequence degenerates and gives $$ \mathbb{H}^{n}(X, F \otimes \omega_X^{\otimes -i}) \cong H^0(X, \mathcal{H}^n(F) \otimes \omega_X^{\otimes -i}). $$ Thus, the left hand side vanishes for all $n \ne 0$ if and only if $\mathcal{H}^n(F) = 0$ for all such $n$.