Loop space: De Rham cohomology 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.  
 A: Assuming that de Rham cohomology is ordinary cohomology (which shouldn't be too hard by the standard sheaf-theoretic 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 simply-connected case $H^1(LM) = H^2(M)$, something that one can also see by hand.
A: 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)
A: 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  Hochschild-Kostant-Rosenberg theorem  says that: For a k-algebra $R$, 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 $M$.
