partial derivatives and chain rule of functions defined on manifold

Question

For functions on $\mathbb{R}^{n}$, applying chain rule and taking partial derivatives is straight forward. I am a bit confused about how this concept can be extended to functions defined on manifolds. To better explain my question, first I'll give a, rather well known, example of a real valued function on $\mathbb{R}^2$, and then an example of a function on a mainfold, which I need to understand precisely.

Let $f:\mathbb{R}^2\rightarrow\mathbb{R}$, its derivative can be written as, \begin{equation*} \label{eq:1} \begin{aligned} \frac{d}{dt}f(x_{1},x_{2}) &= \frac{d}{dx_{1}}f(x_{1},x_{2}) \frac{d}{dt} x_{1} + \frac{d}{dx_{2}}f(x_{1},x_{2}) \frac{d}{dt} x_{2}\\ &= \left( d_{x_{1}}f \right) \dot{x}_{1} + \left( d_{x_{2}}f \right) \dot{x}_{2} \end{aligned} \end{equation*} One way to interpret $d_{x_{1}}f$ is that it represents how $f$ changes in the direction of the coordinate axis $x_{1}$, also known as Lie derivative.

Now here is the actual question I want to ask. For a function on a manifold, in my case, I have a Lie algebra valued function $f: SO(3)\times SO(3) \rightarrow \mathfrak{so(3)}$, is this precisely correct to write the derivative in the following way? Let $R,R_d\in SO(3)$, \begin{equation*} \label{eq:2} \begin{aligned} \frac{d}{dt}f(R,R_d) &= \frac{d}{d_R}f(R,R_d) \dot R + \frac{d}{dR_{d}}f(R,R_d) \dot R_d\\ &= \left( d_{R}f \right) \dot{R} + \left( d_{R_{d}}f \right) \dot{R}_{d}\\ &= \left( T_{R}f \right) \dot{R} + \left( T_{R_{d}}f \right) \dot{R}_{d}, \end{aligned} \end{equation*} where $T_{R}f$ is the tangent space of $f$ at point $R$. This does not seem correct, or not mathematically precise because of the following two reasons,

1. $d_R f$, strictly speaking, is not the change of $f$ in the direction of the "coordinate" $R$, as this function is geometric, and defined in a coordinate free way. This means the second line of the above equation does not make sense?
2. $T_{R}f$ is the tangent space of $f$ at point $R$, and $T_{R_d}f$ is the tangent space of $f$ at point $R_d$, and we can't simply add vectors belonging to different tangent spaces. This means the third line of the equation does not make sense either?

How can I write this derivative in a mathematically (geometrically) correct way? For some reason I don't want to merge $\dot R$, $\dot R_d$ with the term preceding it. In other words, I want to keep the term appearing because of chain rule separate with the rest of the terms.

The reason I have introduced the notion of tangent spaces is because I think this is the right way to define derivatives of mainfolds. For example let $\mathcal{M}$, and $\mathcal{N}$, be manifolds, and let $f:\mathcal{M}\rightarrow \mathcal{N}$, the derivative, also called the push forward map, is defined as $D_f: T\mathcal{M}\rightarrow T\mathcal{N}$, where $T\mathcal{M}$, and $T\mathcal{N}$ are the tanget bundle of the manifolds $\mathcal{M}$, and $\mathcal{N}$, respectively. Even under this abstract definition of derivative, how can we defined derivative of a function on a manifold, while keeping the terms appearing because of chain rule separate?

Answer 1

$\newcommand{\id}{\text{id}}$ $\newcommand{\vp}{\varphi}$ $\newcommand{\at}{\big|}$ $\newcommand{\wh}{\widehat}$ $\newcommand{\kin}{\text{kin}}$ $\newcommand{\lrat}[1]{\left.#1\right|}$ $\newcommand{\pd}[2]{\partial{#1}/\partial{#2}}$ $\newcommand{\R}{\mathbf R}$ $\newcommand{\C}{\mathbf C}$ $\newcommand{\Q}{\mathbb Q}$ $\newcommand{\Z}{\mathbb Z}$ $\newcommand{\mc}{\mathcal}$ $\newcommand{\set}[1]{\{#1\}}$ $\newcommand{\lrp}[1]{\left(#1\right)}$ $\newcommand{\lrset}[1]{\left\{#1\right\}}$

Notation

For a smooth map $f:\R^n\to \R^m$, we write $Df_a$ to denote the derivative of $f$ at a point $a\in \R^n$.

(Kinematic) Tangent Space

Definition. Let $p$ be a point on a smooth $n$-manifold $M$. Define $\Gamma_p$ as $$ \Gamma_p=\set{\gamma:I\to M~|~I\text{ is open in } \R, 0\in I, \gamma(0)=p, \gamma \text{ is smooth}} $$ We define a relation $\sim_p$ on $\Gamma_p$ as follows: For two elements $\gamma_1,\gamma_2\in \Gamma_p$, write $\gamma_1\sim_p\gamma_2$ if there exists a chart $(U,\vp)$ containing $p$ such that $$ D(\vp\circ \gamma_1)_{0}=D(\vp\circ \gamma_2)_{0} $$ This is an equivalence relation on $\Gamma_p$. The kinematic tangent space $(T_pM)_{\kin}$ at $p$ is the set of all the equivalence classes of $\Gamma_p$ under $\sim_p$. The elements of the kinematic tangent space are called as the kinematic tangent vectors.

Note. There are many ways to define the notion of the tangent space at a point on a smooth manifold. Here we are using one particular way and hence the qualification "kinematic".

Definition. Let $p$ be a point in a smooth $n$-manifold $M$ and let $(U,\vp)$ be a chart on $M$ with $p\in U$. Let $\gamma_1$ and $\gamma_2$ be in $\Gamma_p$. The addition $[\gamma_1]+[\gamma_2]$ in $(T_pM)_{\kin}$ is defined as $[\gamma]$, where $\gamma\in\Gamma_p$ is such that $$ D(\vp\circ\gamma_1)_0+D(\vp\circ\gamma_2)_0=D(\vp\circ\gamma)_0 $$ Given $\lambda\in\R$ and an element $\gamma_0\in\Gamma_p$, the scalar multiplication $\lambda[\gamma']$ in $(T_pM)_{\kin}$ is defined as $[\gamma]$, where $\gamma\in\Gamma_p$ is such that $$ \lambda D(\vp\circ \gamma')_0=D(\vp\circ\gamma)_0 $$

By the following two lemmas, the definition of $[\gamma]$ does not depend on the choice of the chart $(U,\vp)$, and such a $\gamma$ exists. This makes $(T_pM)_{\kin}$ into an $n$-dimensional real vector space.

Lemma 1. Let $p$ be a point in a smooth $n$-manifold $M$ and $(U,\vp)$ and $(V,\psi)$ are two charts on $M$ containing the point $p$. Let $\gamma_1,\gamma_2,\gamma,\gamma'\in \Gamma_p$ and $\lambda$ be any real number. Then $$ D(\vp\circ \gamma_1)_{0}+D(\vp\circ \gamma_2)_{0}=D(\vp\circ \gamma)_{0} \iff D(\psi\circ\gamma_1)_{0}+D(\psi\circ\gamma_2)_{0}=D(\psi\circ\gamma)_{0} $$ and $$ \lambda D(\vp\circ\gamma)_{0}=D(\vp\circ\gamma')_{0} \iff \lambda D(\psi\circ\gamma)_{0}=D(\psi\circ\gamma')_{0} $$

Proof. Trivial by the chain rule. $\blacksquare$

Lemma 2. Let $p$ be a point in a smooth $n$-manifold $M$ and $(U,\vp)$ be a chart on $M$ with $p\in U$. Then $$ \set{D(\vp\circ\gamma)_{0}(1):\gamma\in\Gamma_p}=\R^n $$ Therefore, the map $[\gamma]\mapsto D(\vp\circ \gamma)_0$, where $[\gamma]\in T_pM$ and $(\vp,U)$ is a fixed chart about the point $p$, is a vector space isomorphism,

Proof. Let $ v\in\R^n$. Define $\gamma_{ v}:\R\to M$ as $$ \gamma_{ v}(t)=\vp^{-1}(\vp(p)+t v), \quad \forall t\in\R $$ Note that $\gamma_{ v}$ is a smooth map and thus $\gamma_{ v}\in\Gamma_p$. It is also clear that $D(\vp\circ\gamma_{ v})_{0}= v$, whence the desired result is immediate. $\blacksquare$

Theorem. Let $p$ be a point on a smooth $n$-manifold $M$ and $\gamma_1$ and $\gamma_2$ be two elements of $\Gamma_p$. Then $\gamma_1\sim_p\gamma_2$ if and only if $(f\circ \gamma_1)'(0)=(f\circ\gamma_2)'(0)$ for all smooth functions $f:M\to \R$.

Proof. First assume that $\gamma_1\sim_p \gamma_2$. So there is a chart $(U,\vp)$ with $p\in U$ such that $D(\vp\circ \gamma_1)_{0}=D(\vp\circ \gamma_2)_{0}$. Say $\vp(p)= a\in\R^n$. Let $f:M\to \R$ be any smooth function. We need to show that $(f\circ \gamma_1)'(0)=(f\circ \gamma_2)'(0)$. Note that $$ D(f\circ \gamma_i)_{0} = D((f\circ \vp^{-1})\circ(\vp\circ \gamma_i))_{0} = D(f\circ \vp^{-1})_{ a}D(\vp\circ \gamma_i)_{0}, \quad i=1,2 $$ immediately leading to the desired result. For the other direction let $(U,\vp)$ be any chart about $p$ on $M$. By hypothesis we have $D(\pi_i\circ\vp\circ \gamma_1)_{0}=D(\pi_i\circ\vp\circ\gamma_2)_{0}$ for all $1\leq i\leq n$, whence we have $D(\vp\circ \gamma_1)_{0}=D(\vp\circ \gamma_2)_{0}$ and we are done. Thus $(f\circ \gamma_i)'(0)=(g\circ \gamma_i)'(0)$ for $i=1,2$. $\blacksquare$

Basis of the Tangent Space

Let $M$ be a smooth $n$-manifold and $(U,\vp)$ be a smooth chart on $M$ about a point $p\in M$. Let $( e_1,\ldots, e_n)$ be the standard basis of $\R^n$. For each $1\leq i\leq n$, we define $$ \lrat{\pd{}{\vp_i}}_{p}=[\gamma_i] $$ where $\gamma_i:(-\varepsilon, \varepsilon)\to M$ is a curve with $\gamma_i(0)=p$ such that the composite $\vp\circ \gamma_i:(-\varepsilon, \varepsilon)\to \R^n$ is the curve $t\mapsto te_i$.

Since the usual coordinates $x_1 , \ldots, x_n:\R^n\to \R$ together constitute a smooth chart on $M$, we can now talk about the expression $\lrat{\pd{}{x_i}}_{a}$ for any point $a\in \R^n$.

Definition. Let $M$ be a smooth manifold. Then we can make $TM$ ``act" on $C^\infty(M)$ by defining $$ [\gamma]f=\lrat{\frac{d(f\circ \gamma)}{dt}}_{t=0}$$ It can be easily seen that this is well-defined.

It should be clear from Lemma 2 that $\lrat{\pd{}{\vp_1}}_{p},\ldots, \lrat{\pd{}{\vp_n}}_{p}$ forms a basis for $(T_pM)_\kin$.

The Differential Of a Smooth Map

Definition. Let $F:M\to N$ be a smooth map and $p$ be a point on $M$. The differential of $F$ at $p$ is the map $dF_p:(T_pM)_{\kin}\to (T_{F(p)}N)_{\kin}$ defined as $$dF_p([\gamma])=[F\circ \gamma]$$ for all $[\gamma]\in (T_pM)_{\kin}$.

It should be noted that is $(U, \vp)$ is a chart on $M$ about a point $p\in M$, and $\vp(p)=a$, then. $$ \lrat{\pd{}{\vp_i}}_p= d\vp^{-1}\at_{ a}\lrp{\lrat{\pd{}{x_i}}_{ a}} $$ Further, given any $f\in C^\infty(U)$, we have the following definition $$ \lrat{\pd{}{\vp_i}}_p f= \lrat{\pd{}{x_i}}_{ a}(f\circ\vp^{-1}) = \pd{\wh f}{x_i}(\wh p) $$ where, again, $ a=\vp(p)$.

Exercise. Let $M$ and $N$ be smooth manifolds and $F:M\to N$ be a diffeomorphism. Then $dF_p:T_pM\to T_{F(p)}N$ is an isomorphism.

Exercise. Chain Rule. Let $M$, $N$ and $P$ be smooth manifolds and $F:M\to N$ and $G:N\to P$ be a smooth maps. For each $p\in M$, we have $$d(G\circ F)_p=dG_{F(p)}\circ dF_p$$