Let $M$ be a smooth manifold. We will use the following definitions and results.
Consider a local chart $\,x=\big(x_1,\cdots,x_n\big):\,U\subseteq M\longrightarrow\mathbb R^n\,$ at a point $p\in U$. The partial derivative operator at $p$ is
\begin{align*}
\frac{\partial}{\partial x_i}\bigg|_p:\ C^\infty(p)\,&\longrightarrow\,\mathbb R
\newline f \, &\longmapsto \,\frac{\partial}{\partial x_i}\bigg|_p(f)\,=\,\frac{\partial\big(f\circ x^{-1}\big)}{\partial x_i}\big(x(p)\big)\,=:\,\frac{\partial f}{\partial x_i}(p).
\end{align*}
The tangent space at $p$ of $M$ is
\begin{align*}
\left<\frac{\partial}{\partial x_1}\bigg|_p ,\cdots, \frac{\partial}{\partial x_n}\bigg|_p\right>\,=:\,TM_p.
\end{align*}
For any smooth curve $\,\alpha:\,I\longrightarrow M$, the velocity vector of $\alpha$ at $t_0\in I$ is a map
\begin{align*}
\alpha'(t_0):\ C^\infty(p)\,&\longrightarrow\,\mathbb R
\newline f \, &\longmapsto \, \alpha'(t_0)[f]\,=\,\frac{d\big(f\circ\alpha\big)}{dt}(t_0).
\end{align*}
For any smooth curve $\,\alpha:\,I\longrightarrow M\,$ and local chart $\,x:\,U\longrightarrow\mathbb R^n\,$ at $\alpha(t_0)$, we have
\begin{align}
\alpha'(t_0)\,&=\,\sum_{i=1}^n\big(x_i\circ\alpha\big)'(t_0)\,\frac{\partial}{\partial x_i}\bigg|_p\,\in\,TM_{\alpha(t_0)}
\end{align}
Let $M$ be a smooth $m$-manifold and $N$ be a smooth $n$-manifold. Given a smooth map $f:\,M\longrightarrow N$. The derivative or differential of $f$ at $p$ is given by the linear map
\begin{align*}
df_p:\ TM_p\,&\longrightarrow\,TN_{f(p)}
\newline v\,&\longmapsto\,df_p(v)\,=\,\big(f\circ\alpha \big)'(t_0)
\end{align*}
where $\,\alpha:\,I\longrightarrow M\,$ is some smooth curve satisfies $\alpha(t_0)=p,\ \alpha'(t_0)=v$.
Now, I want to find the explicit formula for $df_p$. Let $\,x=(x_1,\dots,x_m):\ U\subseteq M\longrightarrow\mathbb R^m\,$ be a local chart at $p\in U$ and $\,y=(y_1,\dots,y_n):\ V\subseteq M\longrightarrow\mathbb R^n\,$ be a local chart at $f(p)\in f(U)$. Consider a smooth curve $\,\alpha:\,I\longrightarrow M\,$ such that $\alpha(t_0)=p$ and $\alpha'(t_0)=\displaystyle\frac{\partial}{\partial x_i}\bigg|_p $ Then we have
\begin{align}
df_p\left(\frac{\partial}{\partial x_i}\bigg|_p \right)\,&=\,(f\circ\alpha)'(t_0)
\\ &=\,\sum_{j=1}^n \big(y_j\circ f\circ\alpha \big)'(t_0)\,\frac{\partial}{\partial y_j}\bigg|_{f(\alpha(t_0))}.
\end{align}
Here, I feel like I'm lack of tools to express the composition derivative $\big(y_j\circ f\circ\alpha \big)'(t_0)$. I guess this derivative is something like
\begin{align}
\big(y_j\circ f\circ\alpha \big)'(t_0)\,&=\,\alpha'(t_0)\,(y_j\circ f)\big(\alpha(t_0)\big)
\\ &=\,\frac{\partial}{\partial x_i}\bigg|_{\alpha(t_0)}(y_j\circ f)\big(\alpha(t_0)\big)
\\ &=\,\frac{\partial(y_j\circ f)}{\partial x_i}\big(\alpha(t_0)\big)
\end{align}
but I'm not sure about it. How should I solve this ? Thanks.
Best Answer
I think a more illuminating definition would be the following. Let $M$ and $N$ be smooth manifolds and let $\varphi:M\rightarrow N$ be a map between them. If I want the push-forward of some vector $v\in T_{p}M$ with respect to the map $\varphi$ I would write;
$$d\varphi_{p}(v):=v(-\circ\varphi)|_{p}$$
By inspection, one can see that this is a vector that takes some function $f\in C^{\infty}(N)$ and returns an element of $\mathbb{R}.$ Therefore, $d\varphi_{p}(v)\in T_{\varphi(p)}N$ as expected. Additionally, we can produce the expilict formula for $d\varphi_{p}$ thusly;
$$d\varphi_{p}(v)(f):=v(f\circ\varphi)|_{p}=\sum_{i}v^i\frac{\partial}{\partial x^i}(f\circ\varphi)\big|_{p}$$
$$=\sum_{i}(x^i\circ\alpha)'|_{t_0}\cdot\frac{\partial}{\partial x^i}(f\circ\varphi)\big|_{p}$$
If you now want to express this in terms of the chart components on $N$ we have;
$$=\sum_{i}(x^i\circ\alpha)'|_{t_0}\cdot\frac{\partial}{\partial x^i}(f\circ y^{-1}\circ y\circ\varphi)\big|_{p}$$
$$=\sum_{i}(x^i\circ\alpha)'|_{t_0}\cdot\partial_m(f\circ y^{-1})|_{y(\varphi(p))}\cdot\partial_{i}((y\circ\varphi)^m\circ x^{-1})|_{x(p)}$$
$$=\sum_{i,m}(x^i\circ\alpha)'|_{t_0}\cdot\frac{\partial(y\circ\varphi)^{m}}{\partial x^i}\bigg|_{p}\cdot\frac{\partial}{\partial y^{m}}\bigg|_{\varphi(p)}(f)$$
Or more compactly, we have;
$$d\varphi_{p}(v)=v^i\cdot\frac{\partial(y\circ\varphi)^{m}}{\partial x^i}\bigg|_{p}\cdot\frac{\partial}{\partial y^{m}}\bigg|_{\varphi(p)}$$