410: For C∞ Vectors Bundle, Section over Trivializing Open Subset Is C∞ iff Coefficients w.r.t. C∞ Frame over There Are C∞

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description/proof of that for $C^{\mathrm{\infty }}$ vectors bundle, section over trivializing open subset is $C^{\mathrm{\infty }}$ iff coefficients w.r.t. $C^{\mathrm{\infty }}$ frame over there are $C^{\mathrm{\infty }}$

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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
• 3: Note

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Starting Context

• The reader knows a definition of $C^{\mathrm{\infty }}$ vectors bundle of rank $k$.
• The reader knows a definition of restricted $C^{\mathrm{\infty }}$ vectors bundle.
• The reader knows a definition of section of continuous map.
• The reader knows a definition of local $C^{\mathrm{\infty }}$ frame on $C^{\mathrm{\infty }}$ vectors bundle.
• The reader admits the proposition that for any $C^{\mathrm{\infty }}$ vectors bundle, a trivializing open subset is not necessarily a chart open subset, but there is a possibly smaller chart trivializing open subset at each point on any trivializing open subset.
• The reader admits the proposition that for any $C^{\mathrm{\infty }}$ vectors bundle, the trivialization of any chart trivializing open subset induces the canonical chart map.

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Target Context

• The reader will have a description and a proof of the proposition that for any $C^{\mathrm{\infty }}$ vectors bundle, any section over any trivializing open subset is $C^{\mathrm{\infty }}$ if and only if the coefficients of the section with respect to any $C^{\mathrm{\infty }}$ frame over the trivializing open subset are $C^{\mathrm{\infty }}$.

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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:
$(E,M,\pi )$: $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ vectors bundles of rank }k\}$
$U$: $\in \{\text{ the trivializing open subsets of }M\}$
$s$: $:U\to {\pi }^{-1}(U)$, $\in \{\text{ the sections of }\pi {|}_{{\pi }^{-1}(U)}\}$
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Statements:
(
$s\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ maps }\}$
$⟹$
$\mathrm{\forall }\{s_1,s_2,...,s_k\}\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ frames on }{\pi }^{-1}(U)\}(s=s^j{s}_j\text{ where }{s}^j:U\to \mathbb{R}\wedge s^j\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ maps }\})$
)
$\wedge$
(
$\mathrm{\exists }\{s_1,s_2,...,s_k\}\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ frames on }{\pi }^{-1}(U)\}(s=s^j{s}_j\text{ where }{s}^j:U\to \mathbb{R}\wedge s^j\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ maps }\})$
$⟹$
$s\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ maps }\}$
)
//

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2: Proof

Whole Strategy: Step 1: around each point, $p\in U$, take a chart trivializing open subset, $U_p^{\prime }\subseteq M$, such that $U_p^{\prime }\subseteq U$ and the induced chart, $({\pi }^{-1}(U_p^{\prime })\subseteq E,\tilde{\varphi })$; Step 2: suppose that $s^j$ is $C^{\mathrm{\infty }}$, and see that $s$ is $C^{\mathrm{\infty }}$, by seeing that the components of $s$ with respect to the charts are $C^{\mathrm{\infty }}$; Step 3: suppose that $s$ is $C^{\mathrm{\infty }}$, and see that $s^j$ is $C^{\mathrm{\infty }}$, by taking the standard frame and the components of $s$ with respect to the standard frame and seeing $s^j$ s as some $C^{\mathrm{\infty }}$ functions of the components.

Step 1:

$s$ or $s^j$ is $C^{\mathrm{\infty }}$ if and only if it is $C^{\mathrm{\infty }}$ at each point, $p\in U$. So, let us see the $C^{\mathrm{\infty }}$-nesses at $p$.

$C^{\mathrm{\infty }}$-ness of any map at point is always judged with respect to a domain chart around the point and a codomain chart around the image of the point, so, let us take some such charts.

Around $p\in U$, there are a chart trivializing open subset, $U_p^{\prime }\subseteq M$, such that $U_p^{\prime }\subseteq U$, and the induced chart, $({\pi }^{-1}(U_p^{\prime })\subseteq E,\tilde{\varphi })$, by the proposition that for any $C^{\mathrm{\infty }}$ vectors bundle, a trivializing open subset is not necessarily a chart open subset, but there is a possibly smaller chart trivializing open subset at each point on any trivializing open subset and the proposition that for any $C^{\mathrm{\infty }}$ vectors bundle, the trivialization of any chart trivializing open subset induces the canonical chart map. Let $\varphi$ be the chart map on $U_p^{\prime }$, and let $\mathrm{\Phi }$ be the trivialization.

Step 2:

Let us suppose that $s^j$ is $C^{\mathrm{\infty }}$.

Let us see that $s$ is $C^{\mathrm{\infty }}$.

We are going to see that the components of $s$ with respect to the charts are $C^{\mathrm{\infty }}$.

Let $p^{\prime }\in U_p^{\prime }$ be any point.

The coordinates of $s_j(p^{\prime })$ are $\tilde{\varphi }(s_j(p^{\prime }))=({s_j(p^{\prime })}^1,...,{s_j(p^{\prime })}^k,p^{\prime 1},...,p^{\prime d})$, and as $s_j$ is a $C^{\mathrm{\infty }}$ section, ${s_j(p^{\prime })}^k$ is $C^{\mathrm{\infty }}$ with respect to $(p^{\prime 1},...,p^{\prime d})$.

The coordinates of $s(p^{\prime })$ are $\tilde{\varphi }(s(p^{\prime }))=\lambda \circ (\varphi ,id)\circ \mathrm{\Phi }(s(p^{\prime }))$, where $\lambda :{\mathbb{R}}^{d+k}\to {\mathbb{R}}^{d+k},(x^1,...,x^d,x^{d+1},...,x^{d+k})\mapsto (x^{d+1},...,x^{d+k},x^1,...,x^d)$, $=\lambda \circ (\varphi ,id)\circ \mathrm{\Phi }(s^j(p^{\prime })s_j(p^{\prime }))=\lambda \circ (\varphi ,id)(s^j(p^{\prime })\mathrm{\Phi }(s_j(p^{\prime })))$, because any trivialization is a 'vectors spaces - linear morphisms' isomorphism at each fiber, $=(s^j(p^{\prime }){s_j(p^{\prime })}^1,...,s^j(p^{\prime }){s_j(p^{\prime })}^k,p^{\prime 1},p^{\prime 2},...,p^{\prime d})$, and as $s^j(p^{\prime })$ and ${s_j(p^{\prime })}^j$ are $C^{\mathrm{\infty }}$, the components are $C^{\mathrm{\infty }}$ with respect to $p^{\prime 1},p^{\prime 2},...,p^{\prime d}$, which means that $s$ is $C^{\mathrm{\infty }}$.

Step 3:

Let us suppose that $s$ is $C^{\mathrm{\infty }}$.

Let $p^{\prime }\in U_p^{\prime }$ be any point.

There is the canonical frame, $e_1^{\prime },...,e_k^{\prime }$, on $\mathrm{\Phi }({\pi }^{-1}(U_p^{\prime }))=U_p^{\prime }\times {\mathbb{R}}^k$, which means that $e_j^{\prime }:U_p^{\prime }\to U_p^{\prime }\times {\mathbb{R}}^k,p^{\prime }\to (p^{\prime },0,...,0,1,0,...,0)$, where $1$ is the j-th component of ${\mathbb{R}}^k$.

There is the induced $C^{\mathrm{\infty }}$ frame on ${\pi }^{-1}(U_p^{\prime })$, $(e_1,...,e_k)$, where $e_j(p^{\prime }):={\mathrm{\Phi }}^{-1}(e_j^{\prime }(p^{\prime }))$, which is indeed $C^{\mathrm{\infty }}$, because the coordinates of $e_j(p^{\prime })$ are $(0,0,...,1,...,0,p^{\prime 1},p^{\prime 2},...,p^{\prime d})$ where $1$ is the $j$-th component in ${\mathbb{R}}^k$, and is indeed a frame, because ${\mathrm{\Phi }}^{-1}$ is a 'vectors spaces - linear morphisms' isomorphism at each fiber.

Let $s=s^{\prime j}{e}_j$. The coordinates of $s(p^{\prime })$ are $\tilde{\varphi }(s(p^{\prime }))=\tilde{\varphi }(s^{\prime j}(p^{\prime })e_j(p^{\prime }))=\lambda \circ (\varphi ,id)\circ \mathrm{\Phi }(s^{\prime j}(p^{\prime })e_j(p^{\prime }))=\lambda \circ (\varphi ,id)(s^{\prime j}(p^{\prime })\mathrm{\Phi }(e_j(p^{\prime })))=\lambda \circ (\varphi ,id)(s^{\prime j}(p^{\prime })e_j^{\prime }(p^{\prime }))=(s^{\prime 1}(p^{\prime }),...,s^{\prime k}(p^{\prime }),p^{\prime 1},p^{\prime 2},...,p^{\prime d})$. $s$'s being $C^{\mathrm{\infty }}$ is nothing but that the coordinates are $C^{\mathrm{\infty }}$ as functions of $p^{\prime 1},...,p^{\prime d}$, so, $s^{\prime j}$ is $C^{\mathrm{\infty }}$.

Let $s_j=s_j^{\prime l}{e}_l$. $s_j^{\prime l}$ is $C^{\mathrm{\infty }}$, by the previous paragraph. $s=s^j{s}_j=s^j{s}_j^{\prime l}{e}_l$, so, $s^{\prime j}=s^l{s}_l^{\prime j}$. The matrix, $S^{\prime }(p^{\prime })=[s_l^{\prime j}(p^{\prime })]$, is the matrix of the components transformation with respect to the 2 bases, and is invertible. As the components of $S^{\prime }$ are $C^{\mathrm{\infty }}$ with respect to $p^{\prime }$, the components of $S^{\prime -1}$ are $C^{\mathrm{\infty }}$ with respect to $p^{\prime }$: use the Laplace expansion to get the inverse matrix. As $s^j$ is an addition of some multiplications of the components of $S^{\prime -1}$ and $s^{\prime l}$ s, it is $C^{\mathrm{\infty }}$.

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3: Note

This proposition says that if the coefficients of $s$ are $C^{\mathrm{\infty }}$ with respect to any one $C^{\mathrm{\infty }}$ frame (does not need to check for every such frame), the section is $C^{\mathrm{\infty }}$, and if the section is $C^{\mathrm{\infty }}$, the coefficients are $C^{\mathrm{\infty }}$ with respect to any $C^{\mathrm{\infty }}$ frame (so, with respect to every $C^{\mathrm{\infty }}$ frame).

The trivializing open set may seem to be allowed to be any open set if there is a $C^{\mathrm{\infty }}$ frame over there, but in fact, such any open subset has to be a trivializing open set after all in order to have any $C^{\mathrm{\infty }}$ frame, by the proposition that for any $C^{\mathrm{\infty }}$ vectors bundle, any $C^{\mathrm{\infty }}$ frame exists over and only over any trivializing open subset.

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References

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