How to calculate the DeRham cohomology of the free loop space $LM= C^\infty(S^1,M)$ as a Frechet manifold?.
Edit: It will be enough for me to know: When $H^1_{DR}(LM)$ is not $\{0\}$.
Bounty is ending within 5 hours.
How to calculate the DeRham cohomology of the free loop space $LM= C^\infty(S^1,M)$ as a Frechet manifold?. Edit: It will be enough for me to know: When $H^1_{DR}(LM)$ is not $\{0\}$. Bounty is ending within 5 hours. 


An equivariant de Rham theory in precisely this setting seems to be developed by Leandre in "Equivariant Cohomology, Fock Space and Loop Groups" (2006). (http://www.mth.kcl.ac.uk/staff/fa_rogers/ECFSLG.pdf) 


Assuming that de Rham cohomology is ordinary cohomology (which shouldn't be too hard by the standard sheaftheoretic proof?), $H^1 = Hom(\pi_1,\mathbb{R})$. Now the base point fibration gives rise to a short exact sequence $$ 1 \rightarrow \pi_2(M) \rightarrow \pi_1(LM) \rightarrow \pi_1(M) \rightarrow 1 $$ Hence yes, if $H^1(M;\mathbb{R}) \neq 0$, then also $H^1(LM;\mathbb{R}) \neq 0$. Maybe more importantly, in the simplyconnected case $H^1(LM) = H^2(M)$, something that one can also see by hand. 


This is comment rather than answer: Please check it, whether it makes sense... Corollary 2.6 page 11 of Free Loop space and homology by J.L Loday says that For any simply connected space, there is a functorial isomorphism: $$HH_1 (\Omega^1(M)) \cong H^1(LM)$$ And HochschildKostantRosenberg theorem says that: For a kalgebra $R$, its module of Kähler differentials coincides with its first Hochschild homology $$\Omega_1(R/k)\cong HH_1(R)$$ Now we have by this MO post, a surjective map $\Omega_1(C^\infty(M))\to \Omega^1(M)$. So can we say that $H^1(LM)= \Omega_1(C^\infty(M))$ and if $\Omega^1(M)\neq \{0\}$, we have $H^1(LM)\neq \{0\}$ for simply connected finite dimension manifold $M$. 

