Is the dual of a compact generator also a compact generator of the derived category of a variety?

Question

Let $X$ be a variety (or more generally a quasi-compact, separated scheme) and $D(X)$ be the derived category of complexes of $\mathcal{O}_X$-modules with quasi-coherent cohomologies. Let $\mathcal{E}$ be a compact generator of $D(X)$. It is well-known that $\mathcal{E}$ is a perfect complex (Generators and representability Theorm 3.1.1). Moreover it is also well-known that $\mathcal{E}$ is quasi-isomorphic to a strictly perfect complex $\mathcal{S}$, i.e. each $\mathcal{S}^n$ is a locally free $\mathcal{O}_X$-module with finite rank and $\mathcal{S}^n=0$ for $|n|\gg 0$. So without loss of generality we can assume $\mathcal{E}$ itself is strictly perfect.

Now we consider the dual of $\mathcal{E}$, $\mathcal{E}^{\vee}$, i.e. $\mathcal{E}^{\vee,n}=\mathcal{H}om(\mathcal{E}^{-n},\mathcal{O}_X)$ and the differential is defined naturally. In fact $\mathcal{E}^{\vee}$ could be considered as $\mathcal{RH}om(\mathcal{E},\mathcal{O}_X)$, see Stack Project Lemma 21.35.9.

My question is: is $\mathcal{E}^{\vee}$ also a compact generator of the derived category $D(X)$? Why?

Answer 1

Let me try to sketch an argument (though I am not quite sure in details).

A theorem of Neeman implies that a compact object $M$ in a compactly generated triangulated category $T$ is a generator if and only if the smallest full dense subcategegory $T$ containing $M$ equals with the subcategory of compact objects of $T$. Now, the latter condition remains true when you apply duality (since compactness is preserved by duality; isn't it in your case?).