Suppose that $M$ and $N$ are smooth manifolds of dimension $m$ and $n$ respectively, and that $F: M \to N$ is a smooth map. Fix $p \in M$, and suppose that $(U, \varphi)$ and $(V, \psi)$ are coordinate charts containing $p$ and $F(p)$, respectively.

Let $T_p M$ be the set of all derivations of $C^\infty(M)$ at $p$, $T_{F(p)} N$ to be the set of all derivations of $C^\infty(N)$ at $F(p)$, and let $\widehat{F}$ be the coordinate representation of $F$ with respect to the charts $(U, \varphi)$ and $(V, \psi)$. Then $\varphi$ induces an ordered basis $\left(\frac{\partial}{\partial x^1}\bigg|_p, \frac{\partial}{\partial x^2}\bigg|_p, \dots, \frac{\partial}{\partial x^n}\bigg|_p\right)$ on $T_p M$ where \begin{align} \frac{\partial}{\partial x^i}\bigg|_p : C^\infty(M) \to \mathbb{R}, f \mapsto \frac{\partial}{\partial x^i}\bigg|_p(f \circ \varphi^{-1}), \end{align} for all $1 \leq i \leq n$.

Similarly, $\psi$ induces an ordered basis $\left(\frac{\partial}{\partial x^1}\bigg|_{F(p)}, \frac{\partial}{\partial x^2}\bigg|_{F(p)}, \dots, \frac{\partial}{\partial x^n}\bigg|_{F(p)}\right)$ on $T_{F(p)} N$, where the $\frac{\partial}{\partial x^j}\bigg|_{F(p)}$ are defined in a similar manner to the $\frac{\partial}{\partial x^i}\bigg|_p$, using $\psi$. Let $\alpha: T_p M \to \mathbb{R}^n$ be the coordinate map with respect to the ordered basis $\left(\frac{\partial}{\partial x^i}\bigg|_p\right)_{1 \leq i \leq n}$ on $T_p M$, and $\beta: T_{F(p)} N \to \mathbb{R}^m$ be the coordinate map with respect to the ordered basis $\left(\frac{\partial}{\partial x^i}\bigg|_{F(p)}\right)_{1 \leq i \leq m}$ on $T_{F(p)} N$.

My question is, does the following diagram commute? If so, why?

$\require{AMScd}$ \begin{CD} T_p M @>{dF_p}>> T_{F(p)} N\\ @V{\alpha}VV @VV{\beta}V\\ \mathbb{R}^n @>{J_{\varphi(p)} \widehat{F}}>> \mathbb{R}^m \end{CD} Here $J_{\varphi(p)} \widehat{F}$ denotes the linear map from $\mathbb{R}^n$ to $\mathbb{R}^m$ given by the $m \times n$ Jacobian matrix of $\widehat{F}$ at $\varphi(p)$.

I am trying to state this question in general terms, but I am mainly interested in the case where $N = M$ and $M$ has a global coordinate chart $(M, \varphi)$.

Another reason why I have asked this question is that I would like to see how the concept of the differential of a smooth map relates to the concept of the Jacobian matrix of a smooth map between Euclidean spaces.

  • 2
    $\begingroup$ Observe that $$\beta \circ dF_p \circ \alpha^{-1} (v^1,\dots,v^m)^T = \Big(\sum_i v^i \partial_i \hat{F}^1(\hat{p}), \dots, \sum_i v^i \partial_i \hat{F}^n(\hat{p}) \Big)^T = J_{\hat{p}} \hat{F} (v^1,\dots,v^m)^T.$$ $\endgroup$ Aug 18 '20 at 13:58

It does: in the case $M=\mathbb{R}^m$ and $N=\mathbb{R}^n$, if I write $dF_p:T_p\mathbb{R}^m\to T_{F(p)}\mathbb{R}^n$ the differential of $F$ as a map between derivations and $DF_p:\mathbb{R}^m\to\mathbb{R}^n$ the usual derivative of $F$, then the map $dF_p$ sends $\partial_v|_p$ on $\partial _{DF_p(v)}|_{F(p)}$, where $\partial_w|_q$ is the directional derivative in the direction $v$ at the point $p$, namely $$\partial_w|_qf=\lim\limits_{h\to0}\frac{f(q+hw)-f(q)}{h},$$ which is what the commutativity of your diagram means.

To see it, we use the fact (coming from usual calculus) that for a function $g:\mathbb{R}^p\to\mathbb{R}$, we have $$\partial_v|_pg=Dg_p(v)\,\,(\ast),$$

and the chain rule: thus, for a smooth function $f:\mathbb{R}^n\to\mathbb{R}$, we get

$$dF_p(\partial_v|_p)f:=\partial_v|_p(f\circ F)\overset{(\ast)}{=}D(f\circ F)_p(v)\overset{c.r.}{=}Df_{F(p)}(DF_p(v))\overset{(\ast)}{=}\partial_{DF_p(v)}|_{F(p)}f.$$

Since they are equal on all such functions $f$, the derivations $dF_p(\partial_v|_p)$ and $\partial_{DF_p(v)}|_{F(p)}$ are equal. Thus, $dF_p$ sends $\partial_v|_p$ on $\partial_{DF_p(v)}|_{F(p)}$.


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