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I'm struggling to prove the chain rule for differentiable manifolds. I've found a few other similar questions, but they use different formulations that I haven't been able to relate to my own.

What I want to prove is the following claim, with (hopefully) all of the relevant definitions:

Let $X,Y,Z$ be differentiable manifolds, and $f:X\rightarrow Y,g:Y\rightarrow Z$ be differentiable maps. Then at any point $p \in X$, \begin{align} D_p(g\circ f)=D_{f(p)}g \circ D_pf. \end{align} The differential $D_pf$ of the map $f$ at $p\in X$ is defined as \begin{aligned} D_pf:T_pM &\rightarrow T_f(p)N\newline [\gamma]=\left.\frac{d}{dt}\right|_{t=0}\gamma(t)&\mapsto \left.\frac{d}{dt}\right|_{t=0}(f\circ\gamma)(t)=[f \circ \gamma], \end{aligned} and vectors $[\gamma]$ are the (equivalence classes of) derivatives of curves $\gamma:\mathbb{R}\rightarrow X$ such that $\gamma(0)=p$.

I take $(U,\phi)$, $(V,\psi)$, and $(W,\chi)$ to be charts of $X$, $Y$, and $Z$, respectively.

My first try was to insert charts as follows: \begin{aligned} D_p(g\circ f)([\gamma])&= \left.\frac{d}{dt}\right|_{t=0}(g\circ f\circ\gamma)(t)\newline &=\left.\frac{d}{dt}\right|_{t=0}(\chi^{-1} \circ (\chi \circ g\circ \psi^{-1})\circ (\psi \circ f\circ\gamma))(t). \end{aligned} Both the bracketed terms are the maps on $\mathbb{R}^n$ that we can use for the usual chain rule, so if the factor of $\chi^{-1}$ wasn't there, it would be \begin{aligned} &\left.\frac{d}{dt}\right|_{t=0}((\chi \circ g\circ \psi^{-1})\circ (\psi \circ f\circ\gamma))(t)\newline =&D_{\psi(f(p))}(\chi \circ g\circ \psi^{-1})\circ\left.\frac{d}{dt}\right|_{t=0}(\psi \circ f\circ\gamma)(t)\newline =&D_{\psi(f(p))}(\chi \circ g\circ \psi^{-1})\circ D_{f(p)}\psi([f\circ \gamma]), \end{aligned} where \begin{aligned} D_p\phi:T_pX&\rightarrow \mathbb{R}^n\newline [\gamma]&\mapsto [\phi \circ \gamma] \end{aligned} is a vector space isomorphism between the tangent space $T_pX$ and $\mathbb{R}^n$.

So since I needed to 'get rid of' this factor of $\chi^{-1}$, my next try was to use the vector space isomorphism (which I can invert, since it is an isomorphism) to do exactly that. This looks like \begin{aligned} (D_{g(f(p))}\chi)^{-1}\circ(D_{g(f(p))}\chi)\left(D_p(g\circ f)([\gamma])\right)&=(D_{g(f(p))}\chi)^{-1}\circ(D_{g(f(p))}\chi)\left([g \circ f \circ \gamma]\right)\newline &=(D_{g(f(p))}\chi)^{-1}\circ\left([\chi \circ g \circ f \circ \gamma]\right)\newline &=(D_{g(f(p))}\chi)^{-1}\circ\left(\left.\frac{d}{dt}\right|_{t=0}(\chi \circ \chi^{-1} \circ (\chi \circ g\circ \psi^{-1})\circ (\psi \circ f\circ\gamma))(t)\right)\newline &=(D_{g(f(p))}\chi)^{-1}\circ\left(D_{\psi(f(p))}(\chi \circ g\circ \psi^{-1})\circ D_{f(p)}\psi([f\circ \gamma])\right). \end{aligned}

At this point, I don't know what to do. The result I want is \begin{aligned} D_{f(p)}g\circ D_pf([\gamma])&=D_{f(p)}g\circ [f \circ \gamma], \end{aligned} and it looks similar up to the charts I inserted.

How do I get to the end? Or have I gone in a circle and need to take another direction?

Bedge
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    It is a very smart move to turn function $g\circ f\circ \chi$ into composition of functions that are of the form $\mathbb{R}^n\rightarrow \mathbb{R}^m$ so that one can apply chain rule in Euclidean space...But, as you did figure out, you are left with $\chi^{-1}$ and you can not do anything about it... You may have to try another way... Apart from this idea being unsuccessful, I am surprised and very happy to see this kind of approach... :) – Praphulla Koushik Jul 09 '23 at 09:32

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If you’re using the equivalence classes of curves approach this is almost obvious (the only perhaps non-obvious thing is showing that the derivative map $[\gamma]\mapsto [f\circ \gamma]$ is well-defined). There’s no need to go down to charts anymore. The value of both sides of the chain rule expression on $[\gamma]$ is obviously equal to $[g\circ f\circ \gamma]$.

In more detail, \begin{align} D_p(g\circ f)\,([\gamma])&:=[(g\circ f)\circ\gamma]\\ &=[g\circ(f\circ\gamma)]\\ &:=D_{f(p)}g\,([f\circ\gamma])\\ &:=D_{f(p)}g\,\left(D_pf\,([\gamma])\right)\\ &=\left(D_{f(p)}g\circ Df_p\right)([\gamma]), \end{align} and thus $D_p(g\circ f)=D_{f(p)}g\circ D_pf$.

peek-a-boo
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  • Also, while we’re here, let me make the notational remark that I actually don’t like using $D$ (or $d$) to denote the induced map between tangent spaces; I personally reserve that only for the Frechet derivative of maps between open subsets of Banach spaces (eg $\Bbb{R}^n,\Bbb{R}^m$). I’d rather use $Tf_p,Tg_{f(p)}$ etc to denote the induced tangent maps on tangent spaces. Actually, you can even omit the base points if you want because these are fiberwise linear mapping of tangent bundles, so the chain rule simply takes the form $T(g\circ f)=Tg\circ Tf$. – peek-a-boo Jul 09 '23 at 09:58
  • Wow that really is simple, thanks. Out of interest, do you know if it's possible to carry on with my approach (even if there's no need to)? @Praphullah Koushik's comment seems to indicate not. – Bedge Jul 09 '23 at 10:51
  • @Bedge I didn’t read your work initially, but now that I do, you already introduced excessive charts in the equal signs. For example “My first try was to insert charts as follows:” I think you’re being misled by your own notation. For example, notation like $\frac{d}{dt}\bigg|{t=0}(f\circ\gamma)(t)$ makes no since literally speaking, because $f$ takes values in a manifold, so you cannot form difference quotients. Anyway, the reason for your extra $\chi{-1}$ is because you pushed eveyrthing down to the chart, so if you want to ‘get rid of it’, then at the end, just lift to the manifold. – peek-a-boo Jul 09 '23 at 10:55
  • perhaps some of my remarks here and in various linked answers might be helpful (btw what I write as $F_{*p}$ there is $TF_p$ or your $D_pF$, but like I said, I’d reserve $D$ only for Frechet derivatives). See also here and here for how to work with/relate the abstract tangent spaces to concrete $\Bbb{R}^n$ stuff using charts (I have many more answers but too lazy to find them all). – peek-a-boo Jul 09 '23 at 10:58
  • so, in terms of the notation in my links, you actually applied $\Phi_{\chi,(g\circ f)(p)}:T_{(g\circ f)(p)}Z\to\Bbb{R}^{\dim Z}$ to the tangent vector $D_p(g\circ f),([\gamma])$. So, at the end of your calculations, simply apply $\left(\Phi_{\chi,(g\circ f)(p)}\right)^{-1}$. – peek-a-boo Jul 09 '23 at 11:02
  • Ah I think I might be seeing the problem now. So a map $f:X\rightarrow Y$ is defined to be differentiable if the map $x\circ f\circ y^{-1}$ is differentiable (i.e. the Euclidean maps in local charts are differentiable). Furthermore, vectors are defined as the equivalence classes of curves whose derivatives are equal \textit{in coordinates}. So when I used the definition $[\gamma]=\left.\frac{d}{dt}\right|_{t=0}(\gamma)(t)$, this is just wrong, and it makes no sense? Slightly frustrating to be given that in a textbook, perhaps they meant it heuristically... – Bedge Jul 09 '23 at 11:18
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    that’s exactly it. The reason people use (and overload) the notation is because it evokes familiarity. In euclidean spaces, we view $\dot{\gamma}(t)$ as literally being an arrow describing the tangent vector to a curve; here the definition of $\dot{\gamma}(t)$ comes from a difference quotient. In the manifold setting, we still use the notation $\dot{\gamma}(t)$ to mean tangent vector, except now this is no longer (and can no longer be) defined using a difference quotient; it is simply defined as the equivalence class $[s\mapsto \gamma(t+s)]$. – peek-a-boo Jul 09 '23 at 11:22
  • Brilliant, thanks for clarifying! – Bedge Jul 09 '23 at 11:24
  • and notice that $[s\mapsto \gamma(t+s)]$ is itself equal to $T\gamma$ applied to the tangent vector $[s\mapsto s+t]\in T_t\Bbb{R}$. We often denote this simply as $1_t\in T_t\Bbb{R}$ (because if you now use the identity chart on $\Bbb{R}$ to identify via the isomorphism $\Phi_{\text{id},t}:T_t\Bbb{R}\to\Bbb{R}$, then $1_t\in T_t\Bbb{R}$ is mapped to $1\in\Bbb{R}$). So, $\dot{\gamma}$ is the result of “pushing forward along $\gamma$ the unit tangent vector field on $\Bbb{R}$”. – peek-a-boo Jul 09 '23 at 11:24