Loop space: De Rham cohomology - MathOverflow

Loop space: De Rham cohomology

2012-08-17T12:02:00Z

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 $\{0\}$ .

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Answer by Igor Rivin for Loop space: De Rham cohomology

2012-08-17T18:15:50Z

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)

Answer by Pradip Mishra for Loop space: De Rham cohomology

2012-08-23T19:53:33Z

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 $M$.

Answer by eigenbunny for Loop space: De Rham cohomology

2012-08-23T20:27:58Z

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.