Loop space: De Rham cohomology - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T17:14:28Z http://mathoverflow.net/feeds/question/104908 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104908/loop-space-de-rham-cohomology Loop space: De Rham cohomology Jonujohn 2012-08-17T12:02:00Z 2012-08-30T10:54:07Z <p>How to calculate the DeRham cohomology of the free loop space $LM= C^\infty(S^1,M)$ as a Frechet manifold?. </p> <p>Edit: It will be enough for me to know: When $H^1_{DR}(LM)$ is not <code>$\{0\}$</code>. </p> <p>Bounty is ending within 5 hours. </p> http://mathoverflow.net/questions/104908/loop-space-de-rham-cohomology/104928#104928 Answer by Igor Rivin for Loop space: De Rham cohomology Igor Rivin 2012-08-17T18:15:50Z 2012-08-17T18:15:50Z <p>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)</p> http://mathoverflow.net/questions/104908/loop-space-de-rham-cohomology/105348#105348 Answer by Pradip Mishra for Loop space: De Rham cohomology Pradip Mishra 2012-08-23T19:53:33Z 2012-08-24T02:23:44Z <p>This is comment rather than answer: Please check it, whether it makes sense...</p> <p>Corollary 2.6 page 11 of <a href="http://www-irma.u-strasbg.fr/~loday/PAPERS/FreeLoop4.pdf" rel="nofollow">Free Loop space and homology by J.L Loday</a> says that For any simply connected space, there is a functorial isomorphism: $$HH_1 (\Omega^1(M)) \cong H^1(LM)$$ And <a href="http://ncatlab.org/nlab/show/Hochschild-Kostant-Rosenberg+theorem" rel="nofollow">Hochschild-Kostant-Rosenberg theorem</a> says that: For a k-algebra $R$, its module of Kähler differentials coincides with its first Hochschild homology $$\Omega_1(R/k)\cong HH_1(R)$$</p> <p>Now we have by this <a href="http://mathoverflow.net/questions/6074/kahler-differentials-and-ordinary-differentials/6138#6138" rel="nofollow">MO post</a>, a surjective map $\Omega_1(C^\infty(M))\to \Omega^1(M)$. </p> <p>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 $M$.</p> http://mathoverflow.net/questions/104908/loop-space-de-rham-cohomology/105352#105352 Answer by eigenbunny for Loop space: De Rham cohomology eigenbunny 2012-08-23T20:27:58Z 2012-08-23T20:27:58Z <p>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.</p>