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I am reading Guillemin and Pollack's Differential Topology Page 163:

The coordinate functions $x_1, \dots, x_k$ on $\mathbb{R}^k$ yield $1$-form $dx_1, \dots, dx_k$ on $\mathbb{R}^k$. Check $dx_1, \dots, dx_k$ have the specific action $dx_i(z)(a_1, \dots, a_k) = a_i$.

So I am confused - $dx_i$ is the speed of change on the $x_i$ direction. And then I am not sure how it work with $z \in \mathbb{R}^k$, and $a_i$s?

Thanks

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It is a bit difficult to know how to answer your question without knowing how your text defines the one form $dx_i$. Some might just define it using the equation you've written. Here is some background that might help see why these formulas even make sense.

A differential $k$-form, $\omega$ on a smooth, real manifold $M$ assigns to each point $p\in M$ a multilinear, alternating map:

$$\omega_p:\underbrace{T_pM\times\ldots\times T_pM}_{k \text{ times}}\rightarrow\mathbb{R}$$

where $T_pM$ is the tangent space of $M$ at $p$. Notice that with this definition, $1$-forms are just linear functionals. Now, if $M=\mathbb{R}^n$, $k=1$, and $p=z$, for each $i=1,\ldots,n$, we can define a differential $1$-form, called $dx_i$, as the following linear functional:

$$(dx_i)_z:\mathbb{R}^n\rightarrow\mathbb{R}\\(a_1,\ldots,a_n)\mapsto a_i$$

This is nothing more than the $i$th coordinate function. Here we have used the fact that $T_z\mathbb{R}^n=\mathbb{R}^n$ for all $z$.

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  • $\begingroup$ It feels like it is the definition, but the sentence starts with "Check ...", so I am really confused. Previously, it states "the coordinate functions $x_1, \dots, x_k$ on $\mathbb{R}^k$ yield $1$-form $dx_1, \dots, dx_k$ on $\mathbb{R}^k$". Do I supposed to derive the conclusion from this statement? $\endgroup$ – WishingFish Jul 29 '13 at 20:27
  • $\begingroup$ @WishingFish: It seems then that they just want you to make more explicit the phrase "the coordinate functions $x_1,\ldots,x_k$ yield the $1$-forms $dx_1,\ldots, dx_k$." $\endgroup$ – Jared Jul 29 '13 at 20:38
  • $\begingroup$ But where do I get $dx_i$s send $(a_1,\ldots,a_n)\mapsto a_i$ $\endgroup$ – WishingFish Jul 29 '13 at 20:43
  • $\begingroup$ @WishingFish: That is exactly what a coordinate function does. The $i$th coordinate function picks out the $i$th coordinate of a vector. $\endgroup$ – Jared Jul 29 '13 at 20:45
  • $\begingroup$ Oh!!! So $dx_i$s are exactly coordinate functions acting on the tangent space!! $\endgroup$ – WishingFish Jul 29 '13 at 20:46

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