464: Vectors Field Along C∞ Curve Is C∞ iff Operation Result on Any C∞ Function is C∞

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description/proof of that vectors field along $C^{\mathrm{\infty }}$ curve is $C^{\mathrm{\infty }}$ iff operation result on any $C^{\mathrm{\infty }}$ function is $C^{\mathrm{\infty }}$

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Topics

About: $C^{\mathrm{\infty }}$ manifold

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

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Starting Context

• The reader knows a definition of $C^{\mathrm{\infty }}$ vectors field along $C^{\mathrm{\infty }}$ curve.
• The reader admits the proposition that for any $C^{\mathrm{\infty }}$ function on any point open neighborhood of any $C^{\mathrm{\infty }}$ manifold, there exists a $C^{\mathrm{\infty }}$ function on the whole manifold that equals the original function on a possibly smaller neighborhood of the point.
• The reader admits the proposition that for any map between any arbitrary subsets of any $C^{\mathrm{\infty }}$ manifolds with boundary $C^k$ at any point, where $k$ includes $\mathrm{\infty }$, the restriction on any domain that contains the point is $C^k$ at the point.
• The reader admits the proposition that any vectors field is $C^{\mathrm{\infty }}$ if and only if its operation result on any $C^{\mathrm{\infty }}$ function is $C^{\mathrm{\infty }}$.

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Target Context

• The reader will have a description and a proof of the proposition that for any $C^{\mathrm{\infty }}$ manifold with boundary and any $C^{\mathrm{\infty }}$ curve over any interval, any vectors field along the curve is $C^{\mathrm{\infty }}$ if and only if its operation result on any $C^{\mathrm{\infty }}$ function on the manifold with boundary is $C^{\mathrm{\infty }}$.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$M$: $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ manifolds with boundary }\}$
$\mathbb{R}$: $=\text{ the Euclidean }{C}^{\mathrm{\infty }}\text{ manifold }$
$J$: $=(t_1,t_2),[t_1,t_2],(t_1,t_2],\text{ or }[t_1,t_2)\subseteq \mathbb{R}$ such that $t_1<t_2$, as the embedded submanifold with boundary of $\mathbb{R}$
$\gamma$: $:J\to M$, $\in \{\text{ the curves }\}\cap \{\text{ the }{C}^{\mathrm{\infty }}\text{ maps }\}$
$V$: $:J\to \gamma (J)\to TM$, $\in \{\text{ the vectors fields along }\gamma \}$


Statements:
$V\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ maps }\}$
$⟺$
$\mathrm{\forall }f\in C^{\mathrm{\infty }}(M)((Vf)\circ \gamma :J\to \mathbb{R}\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ maps }\})$

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2: Proof

Whole Strategy: Step 1: suppose that $(Vf)\circ \gamma$ is $C^{\mathrm{\infty }}$, and see that $V$ is $C^{\mathrm{\infty }}$ by seeing its components function with respect to some appropriate charts; Step 2: suppose that $V$ is $C^{\mathrm{\infty }}$, and see that $(Vf)\circ \gamma$ is $C^{\mathrm{\infty }}$ by calculating it with respect to some appropriate charts.

Step 1:

Let us suppose that $(Vf)\circ \gamma$ is $C^{\mathrm{\infty }}$.

For any point, $j_0\in J$, there is a chart, $(U_{\gamma (j_0)}^{\prime }\subseteq M,{\varphi }_{\gamma (j_0)}^{\prime }:p\mapsto (x^1,x^2,...,x^n))$, around $\gamma (j_0)$.

$x^k$ is a $C^{\mathrm{\infty }}$ function on $U_{\gamma (j_0)}^{\prime }$, and there is a $C^{\mathrm{\infty }}$ function on $M$, $f_k$, that equals $x^k$ on a possibly smaller open neighborhood, by the proposition that for any $C^{\mathrm{\infty }}$ function on any point neighborhood of any $C^{\mathrm{\infty }}$ manifold, exists a $C^{\mathrm{\infty }}$ function on the whole manifold that equals the original function on a possibly smaller neighborhood of the point. Let us take the possibly smaller chart, $(U_{\gamma (j_0)}\subseteq M,{\varphi }_{\gamma (j_0)}={\varphi }_{\gamma (j_0)}^{\prime }{|}_{U_{\gamma (j_0)}})$ such that $U_{\gamma (j_0)}\subseteq U_{\gamma (j_0)}^{\prime }$.

There is a chart, $(U_{j_0}\subseteq J,i{d}_{j_0})$, such that $\gamma (U_{j_0})\subseteq U_{\gamma (j_0)}$, because $\gamma$ is continuous.

Over $U_{j_0}$, $V(j)=V^l(j)\partial /\partial x^l$.

$V(j)f_k=V^l(j)\partial /\partial x^l{f}_k=V^l(j)\partial /\partial x^l{x}^k=V^l(j){\delta }_l^k=V^k(j)$, which really equals $(V{f}_k)\circ \gamma {|}_{U_{j_0}}$, which is $C^{\mathrm{\infty }}$ by the supposition: the proposition that for any map between any arbitrary subsets of any $C^{\mathrm{\infty }}$ manifolds with boundary $C^k$ at any point, where $k$ includes $\mathrm{\infty }$, the restriction on any domain that contains the point is $C^k$ at the point.

There is the induced chart, $({\pi }^{-1}(U_{\gamma (j_0)})\subseteq TM,\widetilde{{\varphi }_{\gamma (j_0)}})$.

The components function of $V:J\to TM$ with respect $(U_{j_0}\subseteq J,i{d}_{j_0})$ and $({\pi }^{-1}(U_{\gamma (j_0)})\subseteq TM,\widetilde{{\varphi }_{\gamma (j_0)}})$ is $:U_{j_0}\to {\mathbb{R}}^n\times {\varphi }_{\gamma (j_0)}(U_{\gamma (j_0)}),j\mapsto (V^1(j),...,V^n(j),{\gamma }^1(j),...,{\gamma }^n(j))$, which is $C^{\mathrm{\infty }}$ because $V^k$ is $C^{\mathrm{\infty }}$ and $\gamma$ is $C^{\mathrm{\infty }}$.

So, $V$ is $C^{\mathrm{\infty }}$.

Step 2:

Let us suppose that the map, $V:J\to TM$, is $C^{\mathrm{\infty }}$.

For any point, $j_0\in J$, there are a chart, $(U_{\gamma (j_0)}\subseteq M,{\varphi }_{\gamma (j_0)}:p\mapsto (x^1,x^2,...,x^n))$, around $\gamma (j_0)$, and a chart, $(U_{j_0}\subseteq J,i{d}_{j_0})$, around $j_0$ such that $\gamma (U_{j_0})\subseteq U_{\gamma (j_0)}$, because $\gamma$ is continuous.

There is the induced chart, $({\pi }^{-1}(U_{\gamma (j_0)})\subseteq TM,\widetilde{{\varphi }_{\gamma (j_0)}})$.

The components function of $V:J\to TM$ with respect $(U_{j_0}\subseteq J,i{d}_{j_0})$ and $({\pi }^{-1}(U_{\gamma (j_0)})\subseteq TM,\widetilde{{\varphi }_{\gamma (j_0)}})$ is $:U_{j_0}\to {\mathbb{R}}^n\times {\varphi }_{\gamma (j_0)}(U_{\gamma (j_0)}),j\mapsto (V^1(j),...,V^n(j),{\gamma }^1(j),...,{\gamma }^n(j))$, which is $C^{\mathrm{\infty }}$ because $V$ is $C^{\mathrm{\infty }}$, so, $V^l(j)$ is $C^{\mathrm{\infty }}$.

$V(j)=V^l(j)\partial /\partial x^l$.

For any $C^{\mathrm{\infty }}$ function, $f:M\to \mathbb{R}$, over $U_{j_0}$, $V(j)f=V^l(j)\partial /\partial x^{l}f$, where $\partial /\partial x^{l}f$ is a $C^{\mathrm{\infty }}$ function on $U_{\gamma (j_0)}$, by the proposition that any vectors field is $C^{\mathrm{\infty }}$ if and only if its operation result on any $C^{\mathrm{\infty }}$ function is $C^{\mathrm{\infty }}$, so, $(Vf)\circ \gamma {|}_{U_{j_0}}$ is a $C^{\mathrm{\infty }}$ function.

As $(Vf)\circ \gamma$ is $C^{\mathrm{\infty }}$ on an open neighborhood of each point on $J$, $(Vf)\circ \gamma$ is $C^{\mathrm{\infty }}$ on $J$.

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References

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