413: For C∞ Vectors Bundle, C∞ Frame Exists Over and Only Over Trivializing Open Subset

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description/proof of that for $C^{\mathrm{\infty }}$ vectors bundle, $C^{\mathrm{\infty }}$ frame exists over and only over trivializing open subset

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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: Natural Language Description
• 3: Proof

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Starting Context

• The reader knows a definition of $C^{\mathrm{\infty }}$ vectors bundle.
• 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.
• The reader admits the proposition that any bijective linear map is a 'vectors spaces - linear morphisms' isomorphism.
• The reader admits the proposition that for any $C^{\mathrm{\infty }}$ vectors bundle, there is a chart trivializing open cover.
• The reader admits the proposition that for any $C^{\mathrm{\infty }}$ manifold with boundary and its any chart, the restriction of the chart on any open subset domain is a chart.
• The reader admits the proposition that any open subset of any $C^{\mathrm{\infty }}$ trivializing open subset is a $C^{\mathrm{\infty }}$ trivializing open subset.

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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 $C^{\mathrm{\infty }}$ frame exists over and only over any trivializing open subset.

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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:
$(E,M,\pi )$: $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ vectors bundles of rank }k\}$
$U$: $\in \{\text{ the open subsets of }M\}$
//

Statements:
$\mathrm{\exists }\text{ a }{C}^{\mathrm{\infty }}\text{ frame over }U$
$⟺$
$U\in \{\text{ the trivializing open subsets of }M\}$
//

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2: Natural Language Description

For any $C^{\mathrm{\infty }}$ vectors bundle of rank $k$, $(E,M,\pi )$, and any open subset, $U\subseteq M$, any $C^{\mathrm{\infty }}$ frame exists over $U$ if and only if $U$ is a trivializing open subset of $M$.

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3: Proof

Whole Strategy: Step 1: suppose that $U$ is a trivializing open subset, and construct a $C^{\mathrm{\infty }}$ frame over $U$ based on the trivialization: Step 2: suppose that there is a $C^{\mathrm{\infty }}$ frame over $U$, and construct a trivialization over $U$ based on the frame.

Step 1:

Step 1 Strategy: Step 1-1: take a trivialization, $\mathrm{\Phi }$; Step 1-2: take the canonical frame, $e_1^{\prime },...,e_k^{\prime }$, on $\mathrm{\Phi }({\pi }^{-1}(U))=U\times {\mathbb{R}}^k$ and the induced $C^{\mathrm{\infty }}$ frame on ${\pi }^{-1}(U)$, $(e_1,...,e_k)$; Step 1-3: see that it is indeed a $C^{\mathrm{\infty }}$ frame.

Step 1-1:

Let us suppose that $U$ is a trivializing open subset.

There is a trivialization, $\mathrm{\Phi }:{\pi }^{-1}(U)\to U\times {\mathbb{R}}^k$.

Step 1-2:

There is the canonical frame, $e_1^{\prime },...,e_k^{\prime }$, on $\mathrm{\Phi }({\pi }^{-1}(U))=U\times {\mathbb{R}}^k$, which means that $e_j^{\prime }:U\to U\times {\mathbb{R}}^k,p^{\prime }\to (p^{\prime },0,...,0,1,0,...,0)$, where $1$ is the j-th component of ${\mathbb{R}}^k$, which is indeed a frame (not particularly claimed to be $C^{\mathrm{\infty }}$), obviously.

There is the induced $C^{\mathrm{\infty }}$ frame on ${\pi }^{-1}(U)$, $(e_1,...,e_k)$, where $e_j(p^{\prime }):={\mathrm{\Phi }}^{-1}(e_j^{\prime }(p^{\prime }))$.

Step 1-3:

Let us see that it is indeed a $C^{\mathrm{\infty }}$ frame.

$e_j$ is indeed a section, because $\pi (e_j(p^{\prime }))=\pi ({\mathrm{\Phi }}^{-1}(e_j^{\prime }(p^{\prime })))=p^{\prime }$, because $\mathrm{\Phi }{|}_{{\pi }^{-1}(p^{\prime })}:{\pi }^{-1}(p^{\prime })\to \{p^{\prime }\}\times {\mathbb{R}}^k$ is the 'vectors spaces - linear morphisms' isomorphism while $e_j^{\prime }(p^{\prime })\in \{p^{\prime }\}\times {\mathbb{R}}^k$.

$(e_1,...,e_k)$ is indeed linearly independent at each point, because $\mathrm{\Phi }$ is a 'vectors spaces - linear morphisms' isomorphism at each fiber.

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

Around any point, $p\in U$, there is a chart trivializing open subset, $U_p\subseteq M$, such that $U_p\subseteq U$, 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. Let $(U_p\subseteq M,{\varphi }_p)$ be the chart.

There is the induced chart, $({\pi }^{-1}(U_p)\subseteq E,\widetilde{{\varphi }_p})$, by the proposition that for any $C^{\mathrm{\infty }}$ vectors bundle, the trivialization of any chart trivializing open subset induces the canonical chart map.

On the chart, the coordinates of $e_j(p^{\prime })$ are $\widetilde{{\varphi }_p}(e_j(p^{\prime }))=\lambda \circ ({\varphi }_p,id)\circ \mathrm{\Phi }(e_j(p^{\prime }))$, where $\lambda :{\mathbb{R}}^{d+k}\to {\mathbb{R}}^{d+k},(x^1,...,x^d,x^{d+1},...,x^{d+k})\mapsto (x^{d+1},...,x^{d+k},x^1,...,x^d)$, $=\lambda \circ ({\varphi }_p,id)(e_j^{\prime }(p^{\prime }))=\lambda \circ ({\varphi }_p,id)((p^{\prime },0,...,0,1,0,...,0))$, where $1$ is the $j$-th component in ${\mathbb{R}}^k$, $=(0,0,...,1,...,0,p^1,p^2,...,p^d)$. As the components are $C^{\mathrm{\infty }}$ with respect to $p^1,p^2,...,p^d$, $e_j$ is $C^{\mathrm{\infty }}$ at each $p$, which means that $e_j$ is indeed $C^{\mathrm{\infty }}$.

So, $(e_1,...,e_k)$ is indeed a $C^{\mathrm{\infty }}$ frame over $U$.

Step 2:

Step 2 Strategy: Step 2-1: let $s_1,s_2,...,s_k$ be a $C^{\mathrm{\infty }}$ frame over $U$, and express each $b\in {\pi }^{-1}(U)$ as $b=b^j{s}_j(p)$; Step 2-2: define $f:{\pi }^{-1}(U)\to U\times {\mathbb{R}}^k$, as $f(b^j{s}_j(p))=(p,b^1,b^2,...,b^k)$; Step 2-3: see that $f$ is a 'vectors spaces - linear morphisms' isomorphism at each fiber; Step 2-4: see that $f$ is a diffeomorphism.

Step 2-1:

Let us suppose that there is any $C^{\mathrm{\infty }}$ frame, $s_1,s_2,...,s_k$, over $U$.

For each point, $b\in {\pi }^{-1}(U)$, $b=b^j{s}_j(p)$ where $p=\pi (b)$.

Step 2-2:

Let us define the map, $f:{\pi }^{-1}(U)\to U\times {\mathbb{R}}^k$, as $f(b^j{s}_j(p))=(p,b^1,b^2,...,b^k)$, which is well-defined, because $p$ and the components, $b^j$ s, are uniquely determined.

$f$ is obviously bijective and so, there is the inverse, $f^{-1}$.

Step 2-3:

Let us see that $f$ is a 'vectors spaces - linear morphisms' isomorphism at each fiber.

$f{|}_{{\pi }^{-1}(p)}:{\pi }^{-1}(p)\to \{p\}\times {\mathbb{R}}^k$ is linear, because $f(r{b}^j{s}_j(p)+r^{\prime }{b}^{\prime j}{s}_j(p))=f((r{b}^j+r^{\prime }{b}^{\prime j})s_j(p))=(p,r{b}^1+r^{\prime }{b}^{\prime 1},...,r{b}^k+r^{\prime }{b}^{\prime k})=r(p,b^1,...,b^k)+r^{\prime }(p,b^{\prime 1},...,b^{\prime k})=rf(b^j{s}_j(p))+r^{\prime }f(b^{\prime j}{s}_j(p))$.

$f{|}_{{\pi }^{-1}(p)}$ is obviously bijective.

So, $f$ is a 'vectors spaces - linear morphisms' isomorphism at each fiber, by the proposition that any bijective linear map is a 'vectors spaces - linear morphisms' isomorphism.

Step 2-4:

Let us prove that $f$ is a diffeomorphism.

Around any point, $p\in U$, there is a chart trivializing open subset, $U_p\subset M$, such that $U_p\subset U$, because there is a chart trivializing open subset around $p$, by the proposition that for any $C^{\mathrm{\infty }}$ vectors bundle, there is a chart trivializing open cover, and its intersection with $U$ is a one, by the proposition that for any $C^{\mathrm{\infty }}$ manifold with boundary and its any chart, the restriction of the chart on any open subset domain is a chart and the proposition that any open subset of any $C^{\mathrm{\infty }}$ trivializing open subset is a $C^{\mathrm{\infty }}$ trivializing open subset. Let the chart be $(U_p\subseteq M,{\varphi }_p)$. Let the trivialization be $\mathrm{\Phi }:{\pi }^{-1}(U_p)\to U_p\times {\mathbb{R}}^k$.

There is the induced chart, $({\pi }^{-1}(U_p)\subseteq E,\widetilde{{\varphi }_p})$, by the proposition that for any $C^{\mathrm{\infty }}$ vectors bundle, the trivialization of any chart trivializing open subset induces the canonical chart map.

On the chart, $\widetilde{{\varphi }_p}(s_j(p))=({s_j(p)}^1,...,{s_j(p)}^k,p^1,p^2,...,p^d)$, where ${s_j(p)}^l$ is $C^{\mathrm{\infty }}$ because $s_j$ is $C^{\mathrm{\infty }}$.

$\widetilde{{\varphi }_p}(b)=\widetilde{{\varphi }_p}(b^j{s}_j(p^{\prime }))=\lambda \circ ({\varphi }_p,id)\circ \mathrm{\Phi }(b^j{s}_i(p^{\prime }))=\lambda \circ (\varphi ,id)\circ b^j\mathrm{\Phi }(s_j(p^{\prime }))$, because $\mathrm{\Phi }$ is 'vectors spaces - linear morphisms' isomorphic at each fiber, $=(b^j{s_j(p^{\prime })}^1,...,b^j{s_j(p^{\prime })}^k,p^{\prime 1},...,p^{\prime d}):=(c^1,...,c^k,p^{\prime 1},...,p^{\prime d})$.

Let us denote the matrix, $S(p):=[s_j(p{)}^l]$, then, $(c^1,c^2,...,c^k{)}^t=S(b^1,b^2,...,b^k{)}^t$. $S$ is the components function matrix of the 'vectors spaces - linear morphisms' isomorphism of the trivialization on the fiber with respect to the frame basis on ${\pi }^{-1}(p)$ and the canonical basis on $\{p\}\times {\mathbb{R}}^k$, and is invertible. The components of $S$ are $C^{\mathrm{\infty }}$. The components of the inverse matrix, $S^{-1}$, are $C^{\mathrm{\infty }}$: use the Laplace expansion to get the inverse matrix.

On the other hand, there is the chart, $(U_p\times {\mathbb{R}}^k\subseteq U\times {\mathbb{R}}^k,\lambda \circ ({\varphi }_p,id))$.

The components function of $f$, $\lambda \circ ({\varphi }_p,id)\circ f\circ {\widetilde{{\varphi }_p}}^{-1}:\widetilde{{\varphi }_p}({\pi }^{-1}(U_p))\to {\mathbb{R}}^k\times {\varphi }_p(U_p),(c^1,c^2,...,c^k,p^{\prime 1},...,p^{\prime d})\mapsto (b^1,b^2,...,b^k,p^{\prime 1},...,p^{\prime d})$, is $C^{\mathrm{\infty }}$, because $(b^1,b^2,...,b^k{)}^t=S^{-1}(c^1,c^2,...,c^k{)}^t$ while the components of $S^{-1}$ are $C^{\mathrm{\infty }}$ with respect to $(p^{\prime 1},...,p^{\prime d})$. So, $f$ is $C^{\mathrm{\infty }}$ at each $b$.

The components of $f^{-1}$, $\widetilde{{\varphi }_p}\circ f^{-1}\circ ({\varphi }_p,id{)}^{-1}\circ {\lambda }^{-1}:{\mathbb{R}}^k\times {\varphi }_p(U_p)\to \widetilde{{\varphi }_p}({\pi }^{-1}(U_p)),(b^1,b^2,...,b^k,p^{\prime 1},...,p^{\prime d})\mapsto (c^1,c^2,...,c^k,p^{\prime 1},...,p^{\prime d})$, is $C^{\mathrm{\infty }}$, because $(c^1,c^2,...,c^k{)}^t=S(b^1,b^2,...,b^k{)}^t$ while the components of $S$ are $C^{\mathrm{\infty }}$ with respect to $(p^{\prime 1},...,p^{\prime d})$. So, $f^{-1}$ is $C^{\mathrm{\infty }}$ at each $f(b)$.

So, $f$ is diffeomorphic.

So, $f$ is a $C^{\mathrm{\infty }}$ trivialization and $U$ is a trivializing open subset.

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References

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