How to complete $i^*i_*\mathcal{F}\to \mathcal{F}$ into an exact triangle for a smooth divisor $i: X\hookrightarrow Y$?

Question

Let $Y$ be a smooth algebraic variety and $i: X\hookrightarrow Y$ be a smooth divisor. We consider the derived functors $i^*: D^b_{coh}(Y)\to D^b_{coh}(X)$ and $i_*: D^b_{coh}(X)\to D^b_{coh}(Y)$. By adjunction, for any $\mathcal{F}\in D^b_{coh}(X)$ we have a canonical morphism $i^*i_*\mathcal{F}\to \mathcal{F}$ in $D^b_{coh}(X)$.

My question is: Can we complete $i^*i_*\mathcal{F}\to \mathcal{F}$ to an exact triangle in $D^b_{coh}(X)$? I am particularly interest in the case that $Y$ is Calabi-Yau and I guess that in this case the expected exact triangle should be $$ i^*i_*\mathcal{F}\to \mathcal{F}\to \mathcal{F}\otimes\omega_X^{-1} $$ but I cannot prove or disprove it.

Answer 1

The cone of $i^*i_*\mathcal{F} \to \mathcal{F}$ is isomorphic to $\mathcal{F} \otimes \mathcal{O}_X(-X)[2]$.

EDIT. Let me write an argument for a sheaf $F$. Consider the distinguished triangle $$ i^*i_*F \to F \to F'. $$ We need to identify $F'$. Applying $i_*$ we obtain $$ i_*i^*i_*F \to i_*F \to i_*F'.\tag{*} $$ By projection formula and the Koszul resolution of $i_*\mathcal{O}_X$, we have $$ i_*i^*i_*F \cong i_*F \otimes i_*\mathcal{O}_X \cong \mathrm{Cone}(i_*F \otimes \mathcal{O}_Y(-X) \to i_*F), $$ and since the map in the right-hand side is obviously zero, we conclude that $$ i_*i^*i_*F \cong i_*F \oplus i_*F \otimes \mathcal{O}_Y(-X)[1]. $$ Now the first map in $(*)$ is the projection to the direct summand, hence $$ i_*F' \cong i_*F \otimes \mathcal{O}_Y(-X)[2] \cong i_*(F \otimes \mathcal{O}_X(-X)[2]). $$ Since $i_*$ is fully faithful on the category of sheaves, it follows from this that $$ F' \cong F \otimes \mathcal{O}_X(-X)[2]. $$