Prove that a vector field $X$ is smooth if and only if its coordinates (or components) $X_i$ are smooth for all chart of manifold $M$

differential-geometrysmooth-manifoldstangent-bundle

Question: Prove that a vector field $X$ is smooth if and only if its coordinates (or components) $X_i$ are smooth for all chart of manifold $M$.

Solution: Assume that $X$ is smooth. Take any chart $(U,\varphi)$ with coordinates $(x_1,x_2\cdots,x_m)$. Then $\left.X\right|_U$ is again smooth. Since $X$ is smooth, if we take the charts $(U,\varphi)$ of $M$ and $(\pi^{-1}(U),\Phi)$ of $TM$, then the corresponding local representation

$$\Phi\circ X\circ \varphi^{-1}: \varphi(U)\rightarrow \Phi(\pi^{-1}(U))=\varphi(U)\times\mathbb R $$
is smooth, where

$$\pi:TM\rightarrow M,\quad \pi(v)=x\quad\text{if }v\in T_xM$$
$$\Phi:\pi^{-1}(U)\rightarrow \varphi(U)\times\mathbb R^m$$
defined by $\Phi(v):=(\varphi(\pi(v)),\bar\varphi(v))$
where, $$\bar\varphi:T_xM\rightarrow\mathbb R^m,\quad \bar\varphi(v):=(\varphi\circ\alpha)'(0)$$

Now for every $a \in \varphi(U)$, we have

$$
\begin{align}\left(\Phi \circ X \circ \varphi^{-1}\right)(a)
&=\Phi\left(X\left(\varphi^{-1}(a)\right)\right)\\
& =\Phi(\mathrm{X}(x)) \quad\left[\because \varphi^{-1}(\mathrm{a})=x\right] \\
& =\Phi\left(X_{x}\right)\tag 1 \\
& =\left(\varphi\left(\pi\left(X_{x}\right)\right), \bar{\varphi}\left(X_{x}\right)\right) \text { if } X_{x} \in \mathrm{TM} \\
& =\left(\varphi(x), \bar{\varphi}\left(X_{x}\right)\right) \\
& =(\mathrm{a}, \bar{\varphi}(X(x))) \quad\left[\because \varphi^{-1}(\mathrm{a})=x\Rightarrow \varphi(x)=\mathrm{a}\right] \\
& =\left(\mathrm{a}, \bar{\varphi}\left(\left.X\right|_{U}(x)\right)\right)\\
& = \left(a, \bar\varphi \left( \sum_{i=1}^{m} X_{i}\frac{\partial}{\partial x_{i}}\right)(x)\right)\quad\left[\because \left.X\right|_{U}(x)=\sum_{i=1}^{m} X_{i}\frac{\partial}{\partial x_{i}}\right]\\
& = \left(a, \bar\varphi \left(\left. \sum_{i=1}^{m} X_{i}\frac{\partial}{\partial x_i}\right|_x\right)\right)\\
&=\left(a, \sum_{i=1}^{m} X_{i}(x) \bar{\varphi}\left(\left.\frac{\partial}{\partial x_{i}}\right|_{x}\right)\right) \quad[\because \bar{\varphi} \text{ is linear}]\\
& =\left(a, \sum_{i=1}^{m} X_{i}(x) e_{i}\right)\left[\text { since } \bar{\varphi}\left(\left.\frac{\partial}{\partial x_{i}}\right|_{x}\right)=e_{i}\right] \\
& =\left(a,\left(X_{1}(x), X_{2}(x), \cdots, X_{m}(x)\right)\right) \\
& =\left(a,\left(X_{1}\left(\varphi^{-1}(a)\right), X_{2}\left(\varphi^{-1}(a)\right), \cdots, X_{m}\left(\varphi^{-1}(a)\right)\right)\right) \\
& =\left(a,\left(\left(X_{1} \circ \varphi^{-1}\right)(a), \cdots,\left(X_{m} \circ \varphi^{-1}\right)(a)\right)\right) \\
& =\left(\operatorname{id}_{\varphi(U)}(a),\left(\left(X_{1} \circ \varphi^{-1}\right), \cdots,\left(X_{m} \circ \varphi^{-1}\right)\right)(a)\right) \\
& =\left(\operatorname{id}_{\varphi(U)},\left(X_{1} \circ \varphi^{-1}, \cdots, X_{m} \circ \varphi^{-1}\right)\right)(a)
\end{align}
$$

Therefore, $\Phi_{\circ} \mathrm{X} \circ \varphi^{-1}=\left(\mathrm{id}_{\varphi(U)},\left(\mathrm{X}_{1} \circ \varphi^{-1}, \cdots, \mathrm{X}_{\mathrm{m}} \circ \varphi^{-1}\right)\right)$

Hence the smoothness of $\Phi_{\circ} \mathrm{X} \circ \varphi^{-1}$ implies the smoothness of each $\mathrm{X}_i \circ \varphi^{-1}:\varphi(U)\rightarrow \mathbb R$. Since $\varphi^{-1}$ is diffeomorphism, so $X_i:U\rightarrow\mathbb R$ is smooth.

I copy this from my lecture note where vector field is not described well. I do understand what is tangent bundle, fibre and section. But can't prove theorem from their definition. (As I have no clear idea what vector field is)

  • Some notation is confusing like what $X_x$ is?
  • Why $\bar\varphi$ only act on $\frac{\partial}{\partial x_i}$ and how $\bar\varphi\left( \left.\frac{\partial}{\partial x_i}\right|_x \right)=e_i$?

Best Answer

For $x\in M$ and a section $X$ we have $X(x)=X_x\in T_xM$. Using the chart $(\varphi,U)$, we can write this as

$$X_x=\sum_i X^i(x)\left.\frac{\partial}{\partial x^i}\right\vert_x = \left(\sum_iX^i\frac{\partial}{\partial x^i}\right)(x)$$

with $X^i:M\to\mathbb{R}$.

Since you are familiar with the tangent bundle, recall that the $\bar{\varphi}$ is a linear isomorphism. Usually then one defines the basis of the tangent space $T_xM$ by $\left.\frac{\partial}{\partial x^i}\right\vert_x=\bar{\varphi}^{-1}(e_i)$. Obviously this implies that $$\bar{\varphi}\,(\bar{\varphi}^{-1}(e_i))=\bar{\varphi}\left(\left.\frac{\partial}{\partial x^i}\right\vert_x\right)=e_i$$

Finally, as $\bar{\varphi}$ is linear and $X^i(x)\in \mathbb{R}$, we have

$$\bar{\varphi}(X_x)=\bar{\varphi}\left( \sum_i X^i(x)\left.\frac{\partial}{\partial x^i}\right\vert_x \right) = \sum_i X^i(x)\bar{\varphi}\left(\left.\frac{\partial}{\partial x^i}\right\vert_x\right) = \sum_i X^i(x)e_i $$

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