# Partial derivative notation: is that a projection function?

Consider the following definition:

Let $(U,\phi)$ be a chart and $f$ a $C^\infty$ function on a manifold $M$ of dimension $n$. As a function into $\mathbb{R}^n$, $\phi$ has $n$ components $x^1,\dots x^n$. This means if $r^1,\dots r^n$ are the standard coordinates on $\mathbb{R}^n$, then $x^i = r^i \circ \phi$. For $p \in U$, we define the partial derivative $\partial f / \partial x^i$ at $p$ to be $$\frac{\partial}{\partial x^i}\bigg|_p f \quad \buildrel {\mathrm{def}}\over{=} \frac{\partial f}{\partial x^i}(p) = \frac{\partial(f \circ \phi^{-1})}{\partial r^i} (\phi(p)) .$$

What does $r^i$ stand for? Is it the projection function $\mathrm{pr}_i$ ?

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You are correct, $r^i$ is the projection function $\operatorname{pr}_i:\mathbb{R}^n\to\mathbb{R}$. – LostInMath Sep 10 '11 at 14:00
@LostInMath: If you post that as an answer, it can get accepted and the question won't remain unanswered. – joriki Sep 10 '11 at 14:52
That definition looks familiar....maybe Tu's Introduction to Smooth Manifolds? – ItsNotObvious Sep 10 '11 at 18:44
@3Sphere: yes, I like very much that book, I believe it is a nice introduction to the subject. It is concise and goes straight to the point. – user14174 Sep 12 '11 at 13:10

You are correct. The function $r^i$ is indeed the projection function $\operatorname{pr}_i:\mathbb{R}^n\to\mathbb{R}$, which maps a $n$-tuple in $\mathbb{R}^n$ to its $i$th coordinate.