56: Vectors Field Is C^\infty If and Only If Operation Result on Any C^\infty Function Is C^\infty

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A description/proof of that vectors field is $C^{\mathrm{\infty }}$ if and only if 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: Description
• 2: Proof

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Starting Context

• The reader knows a definition of $C^{\mathrm{\infty }}$ manifold.
• The reader knows a definition of chart.
• The reader knows a definition of $C^{\mathrm{\infty }}$ vectors field.
• 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.

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Target Context

• The reader will have a description and a proof of 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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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: Description

For any $C^{\mathrm{\infty }}$ manifold, $M$, any vectors field, $V$, is $C^{\mathrm{\infty }}$ if and only if its operation result function on any $C^{\mathrm{\infty }}$ function is $C^{\mathrm{\infty }}$.

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2: Proof

Let us suppose that its operation result function on any $C^{\mathrm{\infty }}$ function is $C^{\mathrm{\infty }}$. At any point, $p\in M$, there is a chart, $(U_p,x^1,...,x^n)$. There, $V=V^j\frac{\partial }{\partial x^j}$. Each coordinate function, $x^j:U_p\to \mathbb{R}$, is a $C^{\mathrm{\infty }}$ function on $U_p$, and there is a $C^{\mathrm{\infty }}$ function, $\widetilde{x^j}:M\to \mathbb{R}$, on $M$ that equals $x^j$ on a possibly smaller open neighborhood, $U_p^{\prime }\subseteq U_p$, of $p$, by 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. On $U_p^{\prime }$, $V\widetilde{x^j}=V^i\frac{\partial \widetilde{x^j}}{\partial x^i}=V^j$, $C^{\mathrm{\infty }}$ by the supposition, and $V$ is $C^{\mathrm{\infty }}$ on $U_p^{\prime }$. As $V$ is $C^{\mathrm{\infty }}$ on a neighborhood of any point on $M$, $V$ is $C^{\mathrm{\infty }}$ on $M$.

Let us suppose that $V$ is $C^{\mathrm{\infty }}$. At any point, $p\in M$, there is a chart, $(U_p,x^1,...,x^n)$. There, $V=V^j\frac{\partial }{\partial x^j}$ where $\{V^i\}$ are $C^{\mathrm{\infty }}$. For any $C^{\mathrm{\infty }}$ function, $f$, on $M$, on $U_p$, $Vf=V^j\frac{\partial f}{\partial x^j}$ is $C^{\mathrm{\infty }}$. As $Vf$ is $C^{\mathrm{\infty }}$ on a neighborhood of any point on $M$, it is $C^{\mathrm{\infty }}$ on $M$.

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References

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