What is the cone of $\mathcal{F}\to i_*i^*\mathcal{F}$ for a divisor $i: D\hookrightarrow X$?

Question

Let $X$ be a smooth projective variety of dimension $n$ and $i: D\hookrightarrow X$ be a smooth, anti-canonical divisor in $X$. In other words,$[D]+[\omega_X]=[0]$ where $\omega_X$ is the canonical bundle of $X$.

Let $i^*: D^b_{coh}(X)\to D^b_{coh}(D)$ and $i_*: D^b_{coh}(D)\to D^b_{coh}(X)$ be the derived pullback and pushforward functors. By adjunction, for any $ \mathcal{F}\in D^b_{coh}(X)$, we have a morphism $\mathcal{F}\to i_*i^*\mathcal{F}$.

Now the question is how to complete the above morphism to an exact triangle in $D^b_{coh}(X)$. In particular, do we have an exact triangle like $$ \mathcal{F}\to i_*i^*\mathcal{F}\to \mathcal{F}\otimes^L_{\mathcal{O}_X}\omega_X[1]\to \mathcal{F}[1]? $$

I also wonder if we can solve the same problem if we do not require that the divisor $D$ is anti-canonical.

Answer 1

For any Cartier divisor $i \colon D \hookrightarrow X$ and any $F \in D^b(X)$ there is an exact triangle $$ F \otimes \mathcal{O}_X(-D) \to F \to i_*i^*F. $$ It can be obtained by tensoring the exact sequence $$ 0 \to \mathcal{O}_X(-D) \to \mathcal{O}_X \to i_*\mathcal{O}_D \to 0 $$ with $F$ and using the projection formula $$ F \otimes i_*\mathcal{O}_D \cong i_*i^*F. $$