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Consider a smooth map $F: M \rightarrow N$ between smooth manifolds with or without boundary. Choose smooth coordinate charts $(U, \varphi)$ for $M$ containing $p$ and $(V, \psi)$ for $N$ containing $F(p)$ and let $$\hat{F} = \psi \circ F \circ \varphi^{-1} : \varphi(U \cap F^{-1}(V)) \rightarrow \psi(V)$$ and $\hat{p} = \varphi(p)$.

I’m trying to follow the computation of a differential in coordinates in Lee’s smooth manifold book. I understood what the book does when $M$ and $N$ are Euclidean but not for the general case. The computation is $$\mathrm dF_p\left(\frac{\partial}{\partial x^i}\Bigg|_p\right) = \mathrm dF_p \left(\mathrm d(\varphi^{-1})_{\hat{p}} \left(\frac{\partial}{\partial x^i}\Bigg|_{\hat{p}}\right)\right) \\ = \mathrm d(\psi^{-1})_{\hat{F}(\hat{p})} \left(\mathrm d\hat{F}_{\hat{p}}\left(\frac{\partial}{\partial x^i}\Bigg|_{\hat{p}}\right)\right)\\ = \mathrm d(\psi^{-1})_{\hat{F}(\hat{p})} \left(\frac{\partial \hat{F}^j}{\partial x^i}(\hat{p}) \frac{\partial}{\partial y^j}\Bigg|_{\hat{F}(\hat{p}}\right)\\ = \frac{\partial \hat{F}^j}{\partial x^i}(\hat{p}) \frac{\partial}{\partial y^j}\Bigg|_{F(p)}$$ Could anyone explain each of these lines?

I tried reading the same proof in Tu’s smooth manifold book, but it says that the differential is represented by the matrix $[\partial F^i/\partial x^j(p)]$ instead of $\hat{F}$ like in Lee’s proof. Which version is correct: $F$ or $\hat{F}$? Or are they the same?

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  • $\begingroup$ Writing $\partial F^i/\partial x^j$ instead of $\partial \hat {F}^i/\partial x^j$ is just an example of a common abuse of notation: identifying a map with its local coordinate representation. See the comment starting at the bottom of page 33 in my smooth manifolds book. $\endgroup$ Commented Apr 6 at 17:57
  • $\begingroup$ @JackLee That makes a lot of sense thanks! $\endgroup$ Commented Apr 6 at 19:41

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The idea for such computations is simple: $\newcommand{\R}{\mathbb{R}}\newcommand{\pl}{\partial}$

  1. We know how to make them for maps $\hat{F} \colon \R^m \to \R^n$ between Euclidean spaces
  2. We can use maps $\varphi \colon M \to U$, $\psi \colon N \to V$ (I'm oversimplifying a bit) to translate $F \colon M \to N$ into $\hat{F} \colon \R^m \to \R^n$.

Once this is done, one can translate everything else: tangent vectors, differentials etc. Of course, this translation process (going both ways!) leads to complex expressions, making it hard to follow. I think diagrams such as the one below are sometimes useful:

Diagram showing a map M->N and its corresponding map between Euclidean spaces

This is what we call a commutative diagram, which means the compositions $\hat{F} \circ \varphi$ and $\psi \circ F$ are the same thing. Or in other words, $F = \psi^{-1} \circ \hat{F} \circ \varphi$. Looking at the computation you have, we actually consider maps going up, not down, like here:

Diagram showing a map M->N and its corresponding map between Euclidean spaces, but with vertical arrows in the opposite direction

The chain rule (i.e., the rule saying $D(f \circ g) = Df \circ Dg$) lets us translate this commutative diagram into another one about differentials:

Diagram showing differentials of the maps from before

Now we can trace what happens to our favorite tangent vector $\frac{\pl}{\pl x^i}\big|_p$ in the NW corner of the diagram. By definition, it's the image of $\frac{\pl}{\pl x^i}\big|_{\hat{p}}$ (under $d\varphi^{-1}_{\hat{p}}$, of course) in the SW corner. Then we can push it to the SE corner (via the map $d\hat{F}_{\hat{p}}$) and get $d\hat{F}_{\hat{p}}\left(\tfrac{\pl}{\pl x^i}\big|_{\hat{p}}\right)$, which equals $\frac{\pl \hat{F}^j}{\pl x^i}(\hat{p}) \frac{\pl}{\pl y^j}\big|_{\hat{F}(\hat{p})}$ by our Euclidean computation. Finally, we push it to the NE corner (via $d\psi^{-1}_{\hat{F}(\hat{p})}$) and get $\frac{\pl \hat{F}^j}{\pl x^i}(\hat{p}) \frac{\pl}{\pl y^j}\big|_{F(p)}$.

In summary, instead of going "right", we went "down-right-up", but the composition of these three moves gives us $dF$ thanks to the chain rule. Let me point out some advantages of drawing diagrams in such occasions:

  • Our notation for composing maps is quite heavy, whereas a diagram conveys the same meaning visually.
  • In principle, I could omit comments like "via $d\psi^{-1}_{\hat{F}(\hat{p})}$", because our diagram only involves one map (one "arrow") in the direction we are considering, so it's clear anyway.
  • Similarly, we don't necessarily need to mention the points $p$, $\hat{p}$, $\hat{F}(\hat{p})$, $F(p)$, as these are four points "living" in four different corners of the diagram. It's always clear from the context which one we should consider.
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First line is just the definition of $\frac{\partial}{\partial x^i} \big\lvert_p = d\left(\varphi^{-1}\right)_{\hat p} \left( \frac{\partial}{\partial x^i} \big\lvert_{\hat p} \right)$.
Second line is use of the equality $\psi^{-1} \circ F = \hat F \circ \varphi^{-1}$.
Third line is formula for differential in the Euclidean case.
Fourth line is use of linearity of the differential and the definition of $\frac{\partial}{\partial y^i} \big\lvert_p$ analogous to $\frac{\partial}{\partial x^i} \big\lvert_p$.

Regarding your last question, I did not see Tu's proof but I suppose he abuses notation and $\frac{\partial F}{\partial x^i}(p)$ is the same as $\frac{\partial \hat F}{\partial x^i}(\hat p)$ since $M$ and $N$ might not have structure that allows for substraction of points and so it would be hard to define $\frac{\partial F}{\partial x^i}(p)$ in different sense.

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  • $\begingroup$ I don't think I have seen that definition before for the first and fourth lines. Is this the definition of a partial derivative on a manifold? Could you explain how the equality $\psi^{-1} \circ F = \hat F \circ \varphi^{-1}$ is used in the second line? $\endgroup$ Commented Apr 6 at 8:57
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    $\begingroup$ It's equation (3.8) in my copy of the book, first equation in subsection 'Computations in Coordinates'. It is a definition of derivation (tanget vector) related (by the isomorphism $d \left(\varphi^{-1} \right)$) to partial derivative operator in Euclidean space. In first line we have $dF_p \left( d \left( \varphi^{-1} \right)_{\hat p} (\cdot) \right)$. This is the same as $d \left( F \circ \varphi^{-1} \right)_{\hat p} (\cdot)$ which in turn is equal to $d \left( \psi^{-1} \circ \hat F \right)_{\hat p} (\cdot)$. You can see that we started with $\varphi^{-1}$ and ended with $ \psi^{-1}$. $\endgroup$ Commented Apr 6 at 10:57

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