let $X$ denote a smooth complex algebraic variety. Let $D_{rh}(X)$ denote the category of regular holonomic $D$-modules on $X$ and $D_{rh}^b(D(X))$ denote the bounded derived category of $D$-modules on $X$ with regular holonomic homology. Let $X^{an}$ denote the associated complex manifold and let $Mod_c(X^{an})$ denote the category of (algebraically) constructible sheaves on $X^{an}$. Also let $D^b_c(X^{an}, \mathbb{C})$ denote the bounded derived category of complex sheaves on $X$ with constructible homology. There is a (left exact, contravariant) solution functor:

$$Sol: D_{rh}(X) \rightarrow Mod_c(X^{an})$$ given by $Sol(M):= Hom_{D(X^{an})}(M^{an}, \mathcal{O}_{X^{an}}).$ The Riemann-Hilbert Correspondence asserts that this induces an anti-equivalence of categories:

$$RSol: D_{rh}^b(D(X)) \cong D^b_c(X^{an}, \mathbb{C}).$$

Now because the categories of $D$-modules on $X$ and complex sheaves on $X^{an}$ have enough injectives we can think about these derived categories as the homotopy categories of the DG-categories of complexes whose objects are injective with bounded homology. Thus $D_{rh}^b(D(X))$ is the homotopy category of the DG-category $K_{rh}^b(D(X))$, whose objects are injective chain complexes with bounded, regular holonomic homology. Similarly $D^b_c(X^{an}, \mathbb{C})$ is the homotopy category of the DG-category $K^b(X^{an},\mathbb{C})$, whose objects are injective chain complexes with bounded, constructible homology. The solution function naturally gives a functor:

$$Sol_{DG}: K_{rh}^b(D(X)) \rightarrow K^b(X^{an},\mathbb{C}).$$

Passing to the homotopy categories gives the Riemann-Hilbert Correspondence. My question is the following: Can the Riemann-Hilbert Correspondence be lifted to the DG setting? In other words, is $Sol_{DG}$ an equivalence of DG-categories?

neverbe an equivalence of derived categories that doesn't lift to the DG level, but you should be really shocked if you encounter one. If you can define the functors on the DG level, they can't suddenly stop being equivalences (after all, for free resolutions, quasi-isomorphism is the same as homotopy equivalence). The thing that could go wrong (though it very rarely does) is that the functors don't lift. $\endgroup$ – Ben Webster♦ Mar 22 '12 at 2:52