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I have a question about this answer.

Let $M$ be a a manifold and let $(U, (x^1, \dots, x^n))$ be a chart around a fixed point $m \in M$. The goal is to define a vector field $F$ such that $F(m) = v$ for a fixed $v \in T_mM$.

One can define the map $\Phi(w) = (q, [w])$ from $TM|_U$ to $U\times\mathbb{R}^n$, where $q = \pi(w)$ and $[w]$ is the coordinate vector of $w$ with respect to the basis $\{\frac{\partial}{\partial x^1}|_q, \dots, \frac{\partial}{\partial x^n}|_q\}$ of $T_qM$. Define $F: U \to TM|_U$ as $F(p) = \Phi^{-1}((p, g(p)[v]))$ where $g$ is a certain plateau function around $m$.

My question is: how can I prove that $F$ is differentiable?

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ted
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  • @ted: I hope you don't mind, I edited the question so that the notation matched with the notation I used in the answer you linked to. – Michael Albanese Dec 19 '14 at 04:55

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Let $\alpha = (x^1, \dots, x^n) : U \to \mathbb{R}^n$ and $\beta := (\alpha, \operatorname{id}_{\mathbb{R}^n})\circ\Phi : TM|_U \to \mathbb{R}^n\times\mathbb{R}^n$. Note that $\beta$ is the chart on $TM|_U$ induced by the chart $\alpha$ on $U$.

Let's consider the map $\hat{F} := \beta\circ F\circ\alpha^{-1} : \mathbb{R}^n \to \mathbb{R}^n\times\mathbb{R}^n$; this is the coordinate expression of the map $F$. The map $F$ is smooth if and only if $\hat{F}$ is smooth (this is the definition of smoothness of $F$). Let $\alpha^{-1}(p) = (y^1, \dots, y^n)$ and $[v] = (v^1, \dots, v^n)$, then we have

\begin{align*} \hat{F}(y^1, \dots, y^n) &= (\beta\circ F\circ\alpha^{-1})(y^1, \dots, y^n)\\ &= (\beta\circ F)(\alpha^{-1}(y^1, \dots, y^n))\\ &= (\beta\circ F)(p)\\ &= \beta(F(p))\\ &= \beta(\Phi^{-1}((p, g(p)[v]))\\ &= ((\alpha, \operatorname{id}_{\mathbb{R}^n})\circ\Phi)(\Phi^{-1}((p, g(p)[v])))\\ &= ((\alpha, \operatorname{id}_{\mathbb{R}^n})\circ\Phi\circ\Phi^{-1})((p, g(p)[v]))\\ &= (\alpha, \operatorname{id}_{\mathbb{R}^n})((p, g(p)[v]))\\ &= (\alpha(p), g(p)[v])\\ &= (y^1, \dots, y^n, g(\alpha^{-1}(y^1, \dots, y^n))[v])\\ &= (y^1, \dots, y^n, (g\circ\alpha^{-1})(y^1, \dots, y^n)v^1, \dots, (g\circ\alpha^{-1})(y^1, \dots, y^n)v^n). \end{align*}

Note that $g$ and $\alpha^{-1}$ are smooth by assumption, so $g\circ\alpha^{-1}$ is smooth. As each of the component functions of $\hat{F}$ is smooth, $\hat{F}$ itself is smooth and therefore so is $F$.

Michael Albanese
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