Do we have the following isomorphism for $\mathcal{Ext}$?

Question

Let $X$ be a smooth variety (over $\mathbb{C}$) and $\Delta: X \rightarrow X \times X$ be the diagonal embedding and $p_1: X\times X\rightarrow X, ~p_2: X\times X\rightarrow X$ be the projections to the first and second components. Let $E$ be a finite dimensional vector bundle on $X$. We define $$ E_{\Delta}:=\Delta_*E $$ to be a sheaf on $X\times X$. In particular we have $\mathcal{O}_{\Delta}:=\Delta_*\mathcal{O}$. We have the sheaf Ext functor $\mathcal{Ext}$ on $X\times X$.

$\bf{My~question}$ is: do we have the isomorphism $$ {Rp_1}_*\mathcal{Ext}_{X\times X}^{\bullet}(E_{\Delta},E_{\Delta})\cong {Rp_1}_*\mathcal{Ext}_{X\times X}^{\bullet}(\mathcal{O}_{\Delta},\mathcal{O}_{\Delta})\otimes \mathcal{End}(E) ? $$

I know that when $X$ is affine, they are isomorphic since $E$ is then free. But does it also hold in the general case?

Answer 1

Note that $E \cong \Delta^*p_1^*(E)$ since $p_1\circ\Delta = 1_X$. Therefore by the projection formula $\Delta_*E = \Delta_*\Delta^*p_1^*(E) = p_1^*(E)\otimes\Delta_*O_X$, hence \begin{align*} \mathcal{Ext}(\Delta_*E,\Delta_*E) & = \mathcal{Ext}(p_1^*(E)\otimes\Delta_*O_X,p_1^*(E)\otimes\Delta_*O_X) \\ & = p_1^*(E^*)\otimes p_1^*(E)\otimes\mathcal{Ext}(\Delta_*O_X,\Delta_*O_X) = \\ & = p_1^*\mathcal{End}(E)\otimes\mathcal{Ext}(\Delta_*O_X,\Delta_*O_X), \end{align*} hence by the projection formula \begin{align*} Rp_{1*}\mathcal{Ext}(\Delta_*E,\Delta_*E) & = Rp_{1*}(p_1^*\mathcal{End}(E)\otimes\mathcal{Ext}(\Delta_*O_X,\Delta_*O_X)) \\ & = \mathcal{End}(E)\otimes Rp_{1*}\mathcal{Ext}(\Delta_*O_X,\Delta_*O_X)). \end{align*}