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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$\begingroup$ This might be of interest math.stackexchange.com/questions/48637/… . Once you computed the homology of the based loop space, it is possible via the free loop space fibration $\Omega M\rightarrow \Lambda M\rightarrow M$ to write down spectral sequence which relates the homologies of $M,\Omega M$, and $\Lambda M$. $\endgroup$– Thomas RotCommented Aug 17, 2012 at 12:28
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1$\begingroup$ I would talk to Andrew Stacey or Patrick Iglesias-Zemmour. $\endgroup$– Steven GubkinCommented Aug 17, 2012 at 12:29
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$\begingroup$ Of course the question I linked is asking fof singular homology. I do not know anything about the De-Rham complex on infinite dimensional spaces. $\endgroup$– Thomas RotCommented Aug 17, 2012 at 12:50
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2$\begingroup$ I would probably start with Chen's papers on iterated integrals, such as ams.org/mathscinet-getitem?mr=380859 $\endgroup$– Andrew StaceyCommented Aug 17, 2012 at 13:30
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1$\begingroup$ What kind of differential forms are you thinking of when you say "de Rham cohomology" in the context Frechet manifold? In case you need this for $LM:=C^0(S^1,M)$ and for singular cohomology then a lot more can be said; the Betti numbers of $LM$ can be computed for a large class of $M$. $\endgroup$– Somnath BasuCommented Aug 23, 2012 at 20:09
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3 Answers
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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)
answered Aug 17, 2012 at 18:15
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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.
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$\begingroup$ It's better to say why the base point fibration (you perhaps mean the loop space fibration $\Omega M\to LM\to M$) gives a short exact sequence. In general, you get a long exact sequence; here it splits to short exact sequence as the fibration has a section. Another point to note is that $LM$ is not connected if $M$ has a fundamental group. $\endgroup$ Commented Aug 23, 2012 at 21:09
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$\begingroup$ @eigenbunny: Note: your short exact sequence comes equipped with a section $\pi_1(M) \to \pi_1(LM)$ induced by the inclusion of constant loops. Hence $\pi_1(LM)$ is a semi-direct product of $\pi_1(M)$ by $\pi_2(M)$. $\endgroup$ Commented Aug 26, 2012 at 4:03
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2$\begingroup$ @John, eigenbunny - If $M$ is not simply connected then $LM$ is not connected; therefore, $\pi_1(LM)$ is meaningless unless you specify which component. The constant section only provides a section for $L_0M \to M$, where $L_0 M$ is the component of constant loops. The components of $LM$ are indexed by the conjugacy classes in $\pi_1(M)$. $\endgroup$ Commented Aug 30, 2012 at 3:27
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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 $M$.
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$\begingroup$ this fails to take into account the internal grading or... in what sense is $\Omega^1$ a ring? see eigenbunny for the correct result. $\endgroup$ Commented Aug 24, 2012 at 4:07