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Cartan proved that for a connected compact Lie group $G$ the left invariant differential forms yield the correct cohomology of $G$. The same argument works for a connected compact $G$-manifold: the idea is to "average" left invariant forms on $G$ using a Haar measure.

Can we extend this result for non compact infinite dimensional manifolds? In particular, consider the free loop space $LM$ of a manifold $M$; this is an infinite dimensional $S^1$-manifold. Is there a way to compute the cohomology of $LM$ using a model of "invariant forms" and the idea of averaging?

By a result of Chen, we know that iterated integrals of differential forms in $M$ yield the correct cohomology of $LM$. Is this model related to Cartan's story of invariant forms?

These questions are a bit vague, but I guess how to make them precise is part of my question.

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    $\begingroup$ You mean $Diff(S^1)$-manifold? Actually, since you're using a based loop space you probably need the group of diffeomorphisms fixing a point? $\endgroup$
    – Vanessa
    Feb 14, 2012 at 22:21
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    $\begingroup$ You might want to look at Pressley-Segal, section 4.11, for the case of the free loop space. Alternatively, you could try section 5 of Abbaspour and Zeimalian's article on the String bracket and flat connections. $\endgroup$
    – skupers
    Feb 15, 2012 at 1:10
  • $\begingroup$ How do you rotate a based loop? Isn't the image of the base point fixed? $\endgroup$
    – S. Carnahan
    Feb 15, 2012 at 2:12
  • $\begingroup$ Oops! Sorry, I wanted to consider the free loop space. I will edit it now. $\endgroup$ Feb 15, 2012 at 3:45
  • $\begingroup$ @Manuel Rivera: Would it help to try canoical invariant differential forms like the Maurer–Cartan form on LM? There should be a way of "translating" these onto LM from M, since you can always pull-back in cohomology. I do not know if such suggestion is too naive. $\endgroup$
    – Kerry
    Feb 15, 2012 at 4:40

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Iterated integrals define a map $$ \sigma: C(\Omega(M)) \to \Omega(LM) $$ where $C(\Omega(M))$ is the cyclic bar complex of $\Omega(M)$. It has various nice properties; for instance, it induces an isomorphism in cohomology, when $M$ is simply-connected.

Unfortunately, the forms in the image of $\sigma$ are not invariant with respect to the $S^1$-action on $LM$, which probably means that there is no relation of the kind you expect.

However, the forms in the image of $\sigma$ are basic (in particular, invariant) with respect to the action of $Diff([0,1])$ by reparameterization, but the image of $\sigma$ is not even dense in the space of all basic forms. (The image is characterized by a stronger invariance that cannot be formulated in terms of a group action: a kind of thin homotopy invariance).

Just in order to add a layer of confusion, I remark that if $G$ is a compact group acting on $M$, there exists a $G$-equivariant extension of the iterated integral map $\sigma$, introduced by Getzler-Jones-Petrack. It uses the Cartan model for $G$-equivariant forms on $LM$, and an equivariant version of the cyclic bar complex of equivariant forms on $M$.

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  • $\begingroup$ Yes, I actually wrote this question when I started to read Getzler's paper and now that I got to the section you cite it makes more sense. Also, I realized that we can also think about iterated integrals as averaging forms with respect to the circle action. For example, the simplest case is the following: Given a 1-form $\omega$ we have a function on $LM$ given by $\gamma \mapsto \int_{\gamma} \omega$ this function is the same as $\gamma \mapsto \int_0^1 \omega(e_{s*}T_{\gamma})ds$ where $T$ is the vector field on $LM$ generated by the circle action and $e_s: LM \to M$ is evaluation at $s$. $\endgroup$ Feb 15, 2012 at 12:06
  • $\begingroup$ Note that the second integral is $\int_0^1 i(T)e_s^*(\omega)ds$, where $i$ denotes interior multiplication, which is an integral in the space of forms: you are averaging the pullback of $\omega$ with respect to the circle action just as we do for compact $G$-manifolds! Moreover, higher iterated integrals can be thought about as averaging over a simplex... $\endgroup$ Feb 15, 2012 at 12:13

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