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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.

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By $G$-equivariant homology of $X$ you mean the homology of the Borel construction or homotopy orbit space $EG\times_GX$. (I only mention this because the same phrase can refer to other kinds of homology for $G$-spaces, satisfying a weaker kind of homotopy axiom. This "Borel homology" gives you an isomorphism for every $G$-equivariant map $X\to Y$ that is nonequivariantly a homotopy equivalence.)

Yes, the operator that you call $\Delta$ is related to $S^1$-homology. There is a first-quadrant spectral sequence for the $G$-homology of $X$, coming from the fact that $EG\times _GX$ is a bundle over $BG$ with fiber $X$. The $E^2$ is $H_i(BG;H_j(X)$. It is a spectral sequence of modules for the ring $H^{-*}BG$. When $G=S^1$, the groups are $E^2_{2k,j}=H_j(X)$ if $k\ge 0$ and otherwise $0$, and the differential $d^2_{2k,j}\to E^2_{2k-2,j+1}$ is independent of $k$. This first differential $d^2$ is the operator $\Delta$. The $2$-periodicity continues in $E^r$ until the differentials start crossing the vertical axis. So basically your $\Delta$-homology is $E^3=E^4$ and then there is an operator $d^4$ going from $ker(\Delta)/im(\Delta)$ to $ker(\Delta)/im(\Delta)$ and raising $j$ by $3$; and so on.

This is related to cyclic homology.

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Can one get skupers's $\Delta$ composing the two maps (the only ones with the correct domain and codomain) in the homology Gysin sequence for the fibration $S^1\to X\times EG\to X\times_{S^1}EG$? –  Mariano Suárez-Alvarez Jul 28 '10 at 4:19
Yes, that's true, too. –  Tom Goodwillie Jul 28 '10 at 10:34
Yes, I did mean the Borel construction. –  skupers Jul 28 '10 at 10:36
An example of why $\Delta$ really is naive is to consider the case of $S^3$ acted on by $S^1$ the Hopf fibration or acted upon trivially in both cases $\Delta$ is 0. But equivariant cohomology distinguishes these two cases, the difference is reflected in higher differentials in a spectral sequence. From the point of view of cyclic homology, $\Delta$ should be basically the Connes operator, but you need to remember it's action at the chain level to compute cyclic homology, again because of the possibility of higher operations. –  Daniel Pomerleano Jul 29 '10 at 12:45

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