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How to integrate differential form actually. As far as I know, a differential form is a multilinear function mapping from a vector space to a real number. Let's take $\int_c fdx+gdy$ as an example. It is integrating a differential $1$-form while I don't quite get the meaning of the integral process.

Shouldn't the function act on something, like the tangent vector in the tangent space? For instance $dx^i(e_p^i)=1$ however when we are doing integration, we won't add a tangent vector next to $dx$, why?

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  • $\begingroup$ In your example, isn't $<dx,dy>$ the tangent vector in the tangent space? $\endgroup$ – John Douma Apr 23 '13 at 14:20
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Let $c$ be a one-dimensional submanifold of $\mathbb{R}^d$ with embedding $\iota: c \rightarrow \mathbb{R}^d$, $d \geq 2$. Let $\omega = f\mathrm{d}x + g\mathrm{d}y$ be a differential 1-form on $\mathbb{R}^d$, where $x,y$ denote the first two coordinates. Then $\int_c f\mathrm{d}x + g \mathrm{d}y := \int_c \iota^* \omega$, and the latter is defined via partitions of unity. But suppose $c$ admits a single coordinate chart, say $\gamma: I \rightarrow c$ with $I \subseteq \mathbb{R}$ an open interval. Then $$\int_c \iota^* \omega = \int_I {\gamma^* \iota^* \omega}.$$ $\gamma^* \iota^* \omega$ is a 1-form on $I$, so it is of the form $h \mathrm{d}t$. More specifically, $$h(t) = \omega(\gamma'(t)) = f(\gamma(t)) \gamma_1'(t) + g(\gamma(t)) \gamma_2'(t),$$ (here is where the tangent vector comes in to play!) so $$\int_c f\mathrm{d}x + g \mathrm{d}y = \int_I {(f(\gamma(t)) \gamma_1'(t) + g(\gamma(t)) \gamma_2'(t))} \mathrm{d}t.$$

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