Complexes of sheaves with locally constant cohomology versus $C_{*}(\Omega M)$-modules

Let $M$ be a nice, connected topological space. Assume it is a manifold, if you like.

There are two rather similar looking differential-graded (dg) categories that one can associate to $M$ that capture something about the homotopy type of $M$. The first is the full dg category of complexes of sheaves of vector spaces (over some field) having locally constant cohomology sheaves. The other is the dg category of dg modules with finite dimensional cohomology over the dg algebra of chains $C_{*}(\Omega M)$ on the based loop space $\Omega M$.

Ways in which these two categories are similar. First, they both have an obvious t-structure for which the heart consists of locally constant sheaves of vector spaces. For the category built from sheaves, this obvious. For the algebraically defined category, just note that $H_{0}(\Omega M)$ is the group algebra of $\pi_{1} M$. Second, I've seen some computations showing that the mapping complexes between objects in the hearts are quasi-isomorphic.

Given these two pieces of evidence, I would guess that the two dg categories should be (quasi) equivalent. So my first question is whether this is in fact the case, and if so, how to write down a pair of (quasi) inverse functors between these categories. Is this written down anywhere?

• Somewhat related: there is a statement that looks something like "$C_{\ast}(\Omega M)$ and $C^{\ast}(M)$ are Koszul dual" but I don't know exactly what hypotheses on $M$ are necessary. I was told that things work out most nicely when $M$ is simply connected. – Qiaochu Yuan Jan 2 '14 at 1:02
• Yes, I think that is related. But it would be nice to see that spelled out somewhere. Note that perfect modules over $C^{*}(M)$ are naturally identified with complexes of sheaves having finite dimensional, constant cohomology sheaves, since $C^{*}(M)=RHom(K_M,K_M)$, where $K_M$ is the constant sheaf for the field $K$. But I'm interested in locally constant cohomology sheaves, so I think $C^{*}(M)$ is not enough. – user44937 Jan 2 '14 at 1:20