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Notations:

Let $X$ be a manifold, and denote by $LX := C^\infty(S^1,X)$ its loop space. For a loop $\gamma \in LX$ we can think at the tangent space of $LX$ at the point $\gamma$ as the space of loops that are 'arbitrarily closed' to $\gamma$. This motivates the following definition

$$T_{\gamma}LX := \Gamma(S^1, \gamma^*TX),$$

where $\Gamma(S^1, \gamma^*TX)$ denote the space of section of the pullback bundle $\require{AMScd}$ \begin{CD} \gamma^*TX @>>> TX\\ @V V V @VV V\\ S^1 @>>\gamma> X. \end{CD} This is the description of the tangent bundle $TLX \to LX$ of the loop space of $X$. By definition a $k$-differential form on $LX$ is a section of the $k^{th}$ exterior power of $TLX^*$, the cotangent bundle. $$\Omega^k(LX):= \Gamma(LX, \Lambda^kTLX^*).$$

Question:

How is the exterior differential $d$ defined for differential forms on loop space? \begin{CD} \Omega^k(LX) @>d>> \Omega^{k+1}(LX) \end{CD}

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    $\begingroup$ The loop space is an infinite-dimensional manifold. Hence, the exterior differential is defined by the same formula as in the finite-dimensional setting. See for example these notes: math.uni-hamburg.de/home/wockel/data/monastir.pdf $\endgroup$ May 18, 2019 at 19:21
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    $\begingroup$ Minor detail but do you mean $\Omega^k(LX): = \Gamma(LX, \bigwedge^k TLX^\vee)$ (that is, $k$th exterior power of the cotangent bundle) $\endgroup$
    – cgodfrey
    May 19, 2019 at 3:26
  • $\begingroup$ yes, of course. Thank you. $\endgroup$ May 19, 2019 at 7:55
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    $\begingroup$ As Tobias remarks, the real question here is "How do vector fields on $LX$ (according to the given definition) act as derivations on smooth functions on $LX$?" And this is again as in the finite dimensional setting: consider a curve in $LX$ starting at $\gamma$ and with tangent vector at $t=0$ given by $v_\gamma$... $\endgroup$ May 19, 2019 at 17:03
  • $\begingroup$ There's a chapter in Kreigel and Michor's book: the convenient setting of global analysis, where they discuss all the different ways of defining forms. I recommend reading it. $\endgroup$ Nov 5, 2019 at 7:09

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