I attempt to understand the definition of partial derivatives from An Introduction to Manifolds by Loring Tu (Second Edition, page no. 67). The definition is given below.
My Confusions & Questions
- I am confused about how the following argument works.
The partial derivative $\partial f/\partial x^i$ is $C^{\infty}$ on $U$ because its pullback $(\partial f/\partial x^i) \circ \phi^{-1}$ is $C^{\infty}$ on $\phi(U)$.
My understanding is as follows.
Given that $f: U \to \mathbb{R}$ is $C^{\infty}$ on $U$. According to the definition of a smooth function on a smooth manifold (Definition 6.1. on page no. 59), if $p \in U$, then there exists a chart $(U, \phi)$ about $p$ s.t. $f \circ \phi^{-1}: \phi(U) \to \mathbb{R}$ is $C^{\infty}$ at $\phi(p)$. This conclusion is applicable for all $p \in U$ and it follows that $f \circ \phi^{-1}: \phi(U) \to \mathbb{R}$ is $C^{\infty}$ on $\phi(U)$. (Here, I have used the fact that $U$ being an open set of smooth manifold $M$ of dim $n$ is itself a smooth manifold of dim $n$, so that I can apply Definiton 6.1. .)
Then $f \circ \phi^{-1}$ is $C^{\infty}$ on $\phi(U)$ $\Rightarrow$ $\frac{\partial \left(f \circ \phi^{-1}\right)}{\partial r^i}$ is $C^{\infty}$ on $\phi(U)$ $\Rightarrow$ $\frac{\partial f}{\partial x^i} \circ \phi^{-1}$ is $C^{\infty}$ on $\phi(U)$.
I am not sure how to deduce that $\frac{\partial f}{\partial x^i}$ is $C^{\infty}$ on $U$ from here. Can you please help me to clear up the confusion?
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If it is given that $f: U \to \mathbb{R}$ is $C^{\infty}$ on $U$, then why can't we immediately deduce that $\frac{\partial f}{\partial x^i}$ is $C^{\infty}$ on $U$? Why do we need to use that 'pullback' argument?
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I don't understand why '$:=$' (by definition symbol) is used before $\left.\frac{\partial}{\partial r^i}\right|_{\phi(p)} \left(f \circ \phi^{-1} \right)$. I think it should be an '$=$'sign as the definition of partial derivative of $f$ wrt $x^i$ at $p$ has been used it write it:
$$\left.\frac{\partial}{\partial x^i}\right\vert_p f := \frac{\partial f}{\partial x^i}(p).$$
Best Answer
For your first question:
Note that a real-valued function $F$ on $M$ is smooth if and only if $F \circ \phi^{-1}$ is smooth on $\phi(U)$ for every coordinate chart $(U,\phi)$ on $M$. The reverse implication is easy: If $F\circ\phi^{-1}$ is smooth for every coordinate chart, you can obviously find some chart as in your definition. Conversely, if there is some chart $(V,\psi)$ such that $F\circ\psi^{-1}$ is smooth, then the identity $F\circ\phi^{-1}=F\circ\psi^{-1}\circ(\psi\circ\phi^{-1})$ and the definition of smoothness of charts implies the forward implication.
Now, as you point out, if $f$ is smooth on $M$, then $\frac{\partial(f\circ\phi^{-1})}{\partial r^i}$, and hence $\frac{\partial f}{\partial x^i}\circ\phi^{-1}$, is smooth on $\phi(U)$ for every coordinate chart $(U,\phi)$. Thus, by the first sentence, $\frac{\partial f}{\partial x^i}$ is smooth on $U$.
Note that I didn't really need to worry about arbitrary charts: I only did it this way because of the definition of smoothness you gave. Had you instead started with "$F$ is smooth in a coordinate chart $(U,\phi)$ if and only if $F\circ\phi^{-1}$ is smooth", the result would be immediate from what you wrote.
For your second question:
The issue is that smoothness is defined in terms of pullbacks by coordinate maps. In particular, a function on a manifold is not defined to be smooth if it is infinitely differentiable, since differentiation on manifolds is not yet defined (you are working through the first part of the definition now).