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I am introducing myself to the topic of iterated integrals using these notes http://www.ihes.fr/%7Ebrown/ColombiaNotes7.pdf

On pag 22 is defined the following operator

$$ D : (X^1)^n \to (\mathcal{A}^*(M))^{\otimes n} $$ $$ [\omega_1|\dots|\omega_n] \mapsto \sum_{i=1}^n [\omega_1|\dots|d\omega_i|\dots|\omega_n] + \sum_{i=1}^{n-1} [\omega_1|\dots|\omega_i \wedge \omega_{i+1}| \dots |\omega_n]$$

and $$B_n(M) = \left\{ \xi = \sum_{l=0}^n \sum_{i_1,\dots,i_l} [\omega_{i_1}|\dots|\omega_{i_l}]\ \mathrm{such\ that}\ D(\xi)=0 \right\},$$

where M is a smooth manifold over the field $k$, $\mathcal{A}^*$ the complex of smooth differential forms and $X$ a suitable sub-complex. (Please refer to the notes for further details).

Then is stated that $$B_1 \sim k \oplus H_{dr}^1(M;k)$$

The notation used to define $D$ is unfortunately somewhat implicit in how it works for $n=1$, so I don't understand how to see this. Using a slavish application of the deinition of D I would instead conclude that $$B_1 = \left\{\xi = f + \omega | df + d\omega = 0 \right\},$$ where f is a function and $\omega$ a form and conclude that $$B_1 \sim k \oplus C^1(M;k),$$ C^1 being the space of closed 1-forms. I do not see where the "up to exact" part comes from.

Can anyone shine some light on me?

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  • $\begingroup$ Since $X^0\cong k$, $d_X^0$ is the $0$ map and $C^1(X)\cong H^1(X)\cong H^1_{dR}(M;k)$, unless I'm misunderstanding something. $\endgroup$ – MTyson Sep 3 '18 at 17:52

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