365: C^\infty Vectors Field Is Uniquely Defined by Its C^\infty Metric Value Functions with All C^\infty Vectors Fields

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A description/proof of that $C^{\mathrm{\infty }}$ vectors field is uniquely defined by its $C^{\mathrm{\infty }}$ metric value functions with all $C^{\mathrm{\infty }}$ vectors fields

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Topics

About: Riemannian 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 metric.
• The reader admits the proposition that around any point on any Riemannian manifold, there is a chart whose canonical frame is orthonormal.
• The reader admits the proposition that for any $C^{\mathrm{\infty }}$ vectors field on any point neighborhood of any $C^{\mathrm{\infty }}$ manifold, exists a $C^{\mathrm{\infty }}$ vectors field on the whole manifold that equals the original vectors field 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 for any Riemannian manifold, the set of any $C^{\mathrm{\infty }}$ metric value functions with respect to all the vectors fields defines a unique $C^{\mathrm{\infty }}$ vectors field.

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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 Riemannian manifold, $M$, any $C^{\mathrm{\infty }}(M)$ linear map, $f:\mathfrak{X}(M)\to C^{\mathrm{\infty }}(M)$ defines the unique $C^{\mathrm{\infty }}$ vectors field, $V\in \mathfrak{X}(M)$, such that $⟨V,V^{\prime }⟩=f(V^{\prime })$.

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2: Proof

Around any point, $p\in M$, let us take a chart, $(U_p,{\varphi }_p)$, such that $({\partial }_1,{\partial }_2,...,{\partial }_n)$ is an orthonormal frame, which is possible by the proposition that around any point on any Riemannian manifold, there is a chart whose canonical frame is orthonormal. Let us define $V^i:=f({\partial }_i)$ and $V:=V^i{\partial }_i$. $V^{\prime }=V^{\prime i}{\partial }_i$. $⟨V,V^{\prime }⟩=\sum _i{V}^i{V}^{\prime i}=\sum _{i}f({\partial }_i)V^{\prime i}=f(\sum _i{V}^{\prime i}{\partial }_i)=f(V^{\prime })$. So, there is such a $V$ on $U_p$ that satisfies the condition.

For any $p^{\prime }\in U_p$, supposing that there is another $V$ by another chart that contains $p^{\prime }$, taking $V^{\prime }$ such that $V^{\prime i}=1$ and $V^{\prime j}=0$ for $j\ne i$ at $p^{\prime }$ (there is $V^{″}$ such that $V^{″i}=1$ and $V^{″j}=0$ for $j\ne i$ on $U_p$, and there is such a $V^{\prime }$ on $M$, by the proposition that for any $C^{\mathrm{\infty }}$ vectors field on any point neighborhood of any $C^{\mathrm{\infty }}$ manifold, exists a $C^{\mathrm{\infty }}$ vectors field on the whole manifold that equals the original vectors field on a possibly smaller neighborhood of the point), $⟨V,V^{\prime }⟩{|}_{p^{\prime }}=V^i(p^{\prime })=f(V^{\prime })(p^{\prime })=f({\partial }_i)(p^{\prime })$, so, that another $V$ coincides with $V$ in the previous paragraph on the intersection. So, $V$ is well-defined on whole $M$.

As $V$ is locally a $C^{\mathrm{\infty }}$ vectors field around any point on $M$, $V$ is $C^{\mathrm{\infty }}$ on whole $M$.

$V$ is unique, because in whatever way it is constructed, it has to satisfy $V^i=f({\partial }_i)$ on $U_p$, so, that $V$ equals our $V$ on any $U_p$, so, on whole $M$.

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References

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