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I know next to nothing about differential forms, but I saw them being mentioned repeatedly on this site, so I went to Wikipedia to try to understand what a differential form is. Most of it is going pretty smooth, but there's a paragraph I don't understand:

Since any vector $v$ is a linear combination $\sum v^je_j$ of its components, $df$ is uniquely determined by $df_p(e_j)$ for each $j$ and each $p \in U$, which are just the partial derivaties of $f$ on $U$. Thus $df$ provides a way of encoding the partial derivatives of $f$. It can be decoded by noticing that the coordinates $x^1, x^2, \dots , x^n$ are themselves functions on $U$, and so define differential 1-forms $dx^1,dx^2,\dots,dx^n$. Since $\frac{\partial x^i}{\partial x^j} = \delta_{ij}$, the Kronecker delta function, it follows that

$$ df = \sum_{i=1}^n \frac{\partial f}{\partial x^i}dx^i $$

I don't understand the bit about the coordinates being functions themselves. What does that mean?

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up vote 3 down vote accepted

For a vector $y=(y_1,\ldots,y_n) \in \textbf{R}^n$, the i-th coordinate function $x^i$ simply picks out the i-th entry, that is, $x^i(y)=y_i$.

Thus, if $e_1,\ldots,e_n$ is a basis for $\textbf{R}^n$, then any vector $y=(y_1,\ldots,y_n)$ can be written as

$y=\sum_{i=1}^n x^i(y)e_i$.

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So, just to make sure I understand: Suppose we're working in $\mathbb{R}^2$, and instead of $x,y$ I call them $f_1, f_2$ for clarity, so they're functions from $\mathbb{R}^2$ to $\mathbb{R}$. Then we can say that $f_1(a, b) = a$ and $f_2(a,b)=b$? –  Javier Badia May 19 '12 at 3:56
    
Yes, that's correct. –  Jeff May 20 '12 at 22:36
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