show/hide this revision's text 2 edited body; deleted 89 characters in body

This is comment rather than answer: I am not comfortable with many of the following things, but i am writing here. Please check it/Edit 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: $$HH_1 (\Omega^*(M)) \Omega^1(M)) \cong H^1(LM)$$ 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 $M$.
show/hide this revision's text 1

This is comment rather than answer: I am not comfortable with many of the following things, but i am writing here. Please check it/Edit 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: $$HH_1 (\Omega^*(M)) \cong H^1(LM)$$ 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 $M$.