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I am having trouble understanding the chain rule in smooth manifolds (unfortunately that part of the book is an exercise). I see a resemblance to the chain rule in $\mathbb{R}^n$ but do not understand from the definition of the differential $dF_p(v_p)(f) = v_p(f\circ F)$ how to get there?

Ok, made a new try. Is this correct:

$$dF_p\circ v_p\circ f = v_p\circ f\circ F\\ d(G\circ F)_p\circ v_p\circ f = v_p\circ f\circ (G\circ F)=v_p\circ f\circ G\circ F\\ dG_{F(p)}\circ (dF_p\circ v_p)\circ f = (dF_p\circ v_p)\circ f\circ G=\\ =dF_p\circ v_p\circ (f\circ G) = v_p\circ (f\circ G)\circ F=v_p\circ f\circ G\circ F$$

I hope it is, because then pushforward makes sense all of a sudden. I guess it was all the parentheses that confused me.

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1 Answer 1

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Before you try to prove it, you should try to understand why it should be true. In this case, if you have a smooth map $F:M\to N$, its differential $dF_p$ is the linearization of $F$ at $p$. Likewise, if $G:N\to O$, then $dG_q$ is the linearization of $G$ at $q\in N$. So $d(GF)_p$ should take the tangent space $T_pM$ to $T_{G(F(p))}O$ through $T_{F(p)}N$: it should be the linearization of $G$ acting on the image of the linearization of $F$, i.e., $$d(GF)_p = dG_{F(p)}\circ dF_p.$$ Now, as to proof details, your approach is good, but you should probably be a little bit more careful denoting what is acting on what: if $v_p\in T_pM$ and $f:O\to \mathbb{R}$, then \begin{align*} \big(d(G\circ F)_p(v_p)\big)(f) &= v_p( f\circ (G\circ F) )\\ &= v_p\bigg( (f\circ G)\circ F \bigg)\\ &= \big( dF_p(v_p)\big)\bigg( f\circ G\bigg) \\ &= dG_{F(p)}\big(dF_p(v_p)\big)(f)\\ &= \bigg(\big(dG_{F(p)}\circ dF_p\big)(v_p)\bigg)(f) \end{align*} Since $v_p$ and $f$ were arbitrarily chosen, this establishes the result.

In words: if $v_p\in T_pM$ and $f:O\to\mathbb{R}$ are arbitrary, then the action of the vector resulting from differential of $GF$ applied to $v$ on $f$ is the action of $v$ on the composition of $f$ with $GF$. This is the same as the action of $v$ on the composition of $(fG)$ with $F$. That's the same as the action of $dF(v)$ on the function $fG$, which is the same as $dG(dF(v))$ acting on $f$.

You may also find it profitable to take one or two other approaches. One is to take local coordinate charts about $p$, $F(p)$, and $GF(p)$ and apply the fact from Euclidean calculus. Another is to take a curve $\gamma$ through $p$ and function $f$ and compute $\frac{d}{dt}(f\circ G\circ F\circ \gamma)(t)$.

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  • $\begingroup$ The line where you go from $v_p((f\circ G)\circ F)$ to $(dF_p(v_p))(f\circ G)$ seems backwards to me somehow, but thanks for the answer I will read it more carefully when I have more time! $\endgroup$
    – Emil
    Oct 4, 2013 at 12:57
  • $\begingroup$ Since $(dF(v))(g) = v(g\circ F)$, putting $g = f\circ G$ and reading the equality the other direction gives $v( (f\circ G)\circ F) = dF(v)(f\circ G)$. $\endgroup$
    – Neal
    Oct 4, 2013 at 13:01
  • $\begingroup$ @JosuéMolina Fortunately for you, the bar's still open. The third-to-last line describes the vector $dF_p(v_p)$ acting on the function $f\circ G$. Note that a vector acting on a composition is the definition of the differential, so this is equal to: (the image of the vector $dF_p(v_p)$ under the map $dG_{F(p)}$) acting on the function $f$. $\endgroup$
    – Neal
    Feb 22, 2016 at 5:27

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