All functors are derived and all categories are bounded derived categories of coherent sheaves. Suppose that we have got an inclusion of a smooth divisor $j:D\rightarrow X$ in a smooth projective variety. Is it true that $$j^*j_*F=F\otimes j^*j_*O_D?$$
[Math] Derived push-forward and pull back.
ag.algebraic-geometry
Related Solutions
I don't think this is true in general.
For instance,
Set $f : X \to Y = \mathbb{A}^2$ be the blowup of the origin.
Set $Y' = \text{Spec} k$ and then fix $u : Y' \to Y$ to be the inclusion of the origin (note that $u$ is not flat).
It follows that $X'$ is the reduced exceptional divisor of $f$, in other words, $X'$ is a copy of $\mathbb{P}^1$.
Consider a sheaf $G = O_X(n X')$ for some integer $n > 0$ (here I am viewing $X'$ as a divisor in $X$). We pullback and pushforward in two ways.
Now $f_* G = O_Y$ since we are just allowing a pole of some order over a point. Therefore, $u^* f_* G = u^* O_Y = O_{Y'} = k$.
On the other hand, $v^* G = v^* O_X(nX') = O_{X'}(-n)$ since $X'$ has self-intersection -1 on $X$. Thus $$ g_* v^* O_X(nX') = g_* O_{X'}(-n) = H^0(X', O_{X'}(-n)) = 0. $$
So since $0 \neq k$ it's not true.
For computations you it is better to replace the first argument by a locally free resolution. This allows to compute the local $R{\mathcal H}om$. Then you can use the local-to-global spectral sequence to compute the global $RHom$. In your particular example, as everything happens in a neighborhood of $x$ and since $x$ is a locally complete intersection in $x$, you can find a vector bundle $E$ and it section in a neighborhood of $x$ such that its zero locus is $x$. Then the Koszul complex $$ 0 \to \Lambda^n E^* \to \dots \to \Lambda^2 E^* \to E^* \to O_X \to 0 $$ is a locally free resolution of the skyscraper. This shows that $$ {\mathcal E}xt^i(k(x),k(x)) = \Lambda^i E_x $$ and hence $$ Ext^i(k(x),k(x)) = \Lambda^i E_x $$ as well. Moreover, one has $E_x \cong T_xX$, so the final answer is $$ RHom(k(x),k(x)) = \oplus_{i=0}^n \Lambda^iT_xX[-i]. $$
EDIT. To compute local Ext's you just apply ${\mathcal H}om$ to the Koszul resolution. You will get a complex $$ 0 \to k(x) \to E\otimes k(x) \to \Lambda^2E \otimes k(x) \to \dots \to \Lambda^nE \otimes k(x) \to 0. $$ Since the map in the Koszul complex are given by wedging with the section of $E$ which vanish at $x$, it follows that in the above complex all maps are zero, so its cohomology sheaves are just its terms.
In this particular case you see that all local Ext's are sheaves supported at a point, so they don't have higher cohomology and the local-to-global spectral sequence degenerates in the second term automatically. In general it does not degenerate so quickly, so you may have to calculate the higher differentials.
Best Answer
The answer is no. The simplest example that I know is $X = P^3$, $D = P^1\times P^1$, $F = O(0,1)$. In this case $j^*j_*O_D = O_D \oplus O_D(-2)[1]$ and hence $F\otimes j^*j_*O_D = F \oplus F(-2)[1]$, while $j^*j_*F$ fits into a triangle $$ F(-2)[1] \to j^*j_*F \to F $$ which is not split. To see this note that $j_*F$ has a resolution of the form $$ 0 \to O_X(-1)^2 \to O_X^2 \to j_*F \to 0, $$ which gives a distinguished triangle $$ O_D(-1,-1)^2 \to O_D^2 \to j^*j_*F. $$ It follows easily from this that $Hom(F,j^*j_*F) = 0$, which shows that $F$ is not a direct summand of $j^*j_*F$.