After reading Cohen and Voronov's notes on string topology, one can find the following construction: Suppose we have a topological space $X$ with continuous action of $S^1$. This means we have a map $\rho: S^1 \times X \to X$. If we choose a fundamental class $[S^1]$ of the circle, we can form the operator $\Delta: H_\ast(X) \to H_{\ast+1}(X)$ by setting $\Delta(a) = \rho_\ast([S^1] \times a)$. For dimensional reasons, $\Delta$ squares to zero and therefore turns $H_\ast(X)$ into a cochain complex. Let's denote the cohomology of this complex by $H_\ast^\Delta(X)$. Because in this cohomology we have as representatives homology classes in $X$ which are annihilated by the action of $S^1$ "in a homological sense", one would think this is relevant to the $S^1$-equivariant homology of $X$.

However, a few simple examples show that $\Delta$-cohomology and equivariant homology are certainly not equal. For example, take the point with trivial action, then $H^\Delta_\ast(pt)$ is $\mathbb{Z}$ in degree 0 and zero in all other degrees, but $H^{S^1}_\ast(pt) = \mathbb{Z}[a_2]$ where $|a_2| = 2$. Another example is $LS^1$.

So my question is: is there a different (hopefully geometric) description of $\Delta$-cohomology? Is it related to $S^1$-equivariant homology in any way?

If this is not possible in the general case, is it at least possible for $X$ of the form $LM$, with $M$ a manifold? My main motivation for considering $\Delta$-cohomology is string topology, where $\Delta$ is also known as the BV-operator.