# Chain rule of differential in smooth manifold

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.

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 $\frac{d}{dt}(f\circ G\circ F\circ \gamma)(t)$.

• 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!
– Emil
Oct 4, 2013 at 12:57
• Since $(dF(v))(g) = v(g\circ F)$, putting $g = f\circ G$ and reading the equality the other direction gives $v( (f\circ G)\circ F) = dF(v)(f\circ G)$.
– Neal
Oct 4, 2013 at 13:01
• @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 $f$.
– Neal
Feb 22, 2016 at 5:27