882: Set of C∞ Sections of C∞ Vectors Bundle Linearly Independent at Point Is Linearly Independent on Open Neighborhood of Point

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description/proof of that set of $C^{\mathrm{\infty }}$ sections of $C^{\mathrm{\infty }}$ vectors bundle linearly independent at point is linearly independent on open neighborhood of point

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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 section of continuous surjection.
• The reader knows a definition of $C^k$ map between arbitrary subsets of $C^{\mathrm{\infty }}$ manifolds with boundary, where $k$ includes $\mathrm{\infty }$.
• The reader knows a definition of linearly independent subset of module.
• The reader knows a definition of neighborhood of point.
• 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 $C^{\mathrm{\infty }}$ 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 the set of any $C^{\mathrm{\infty }}$ sections of any $C^{\mathrm{\infty }}$ vectors bundle that (the set) is linearly independent at a point is linearly independent on an open neighborhood of the point.

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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:
$(E,M,\pi )$: $\in \{\text{ the }d\text{ -dimensional }{C}^{\mathrm{\infty }}\text{ vectors bundles of rank }k\}$
$p$: $\in M$
$\{s_1,...,s_l\}$: $l\le k$, $s_j:M\to E\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ sections }\}$, such that $\{s_1(p),...,s_l(p)\}\in \{\text{ the linearly independent subsets of }{\pi }^{-1}(p)\}$
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Statements:
$\mathrm{\exists }{V}_p\subseteq M\in \{\text{ the open neighborhoods of }p\}(\mathrm{\forall }{p}^{\prime }\in V_p(\{s_1(p^{\prime }),...,s_l(p^{\prime })\}\in \{\text{ the linearly independent subsets of }{\pi }^{-1}(p^{\prime })\}))$
//

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2: Natural Language Description

For any $d$-dimensional $C^{\mathrm{\infty }}$ vectors bundle of rank $k$, $(E,M,\pi )$, any point, $p\in M$, and the set of any $C^{\mathrm{\infty }}$ sections, $\{s_1,...,s_l\}$, such that $l\le k$, $s_j:M\to E$, and $\{s_1(p),...,s_l(p)\}$ is a linearly independent subset of ${\pi }^{-1}(p)$, there is a neighborhood of $p$, $V_p\subseteq M$, such that for each $p^{\prime }\in V_p$, $\{s_1(p^{\prime }),...,s_l(p^{\prime })\}$ is a linearly independent subset of ${\pi }^{-1}(p^{\prime })$.

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3: Proof

Whole Strategy: Step 1: take a chart trivializing open subset of $M$ around $p$, $(U_p\subseteq M,{\varphi }_p)$, and the induced chart, $({\pi }^{-1}(U_p)\subseteq E,\widetilde{{\varphi }_p})$; Step 2: take the components function of $s_j$, $f_j:=\widetilde{{\varphi }_p}\circ s_j\circ {{\varphi }_p}^{-1}$; Step 3: see that $\{{\pi }_k\circ f_1,...,{\pi }_k\circ f_l\}$, where ${\pi }_k$ is the projection into ${\mathbb{R}}^k$, is linearly independent on ${\mathbb{R}}^k$ at ${\varphi }_p(p)$ and that the corresponding matrix is rank $l$, which means that there is a nonzero-determinant $l\times l$ submatrix; Step 4: take an open neighborhood of ${\varphi }_p(p)$, $U_{{\varphi }_p(p)}\subseteq {\varphi }_p(U_p)$, on which the determinant of the submatrix is nonzero; Step 5: take $V_p:={{\varphi }_p}^{-1}(U_{{\varphi }_p(p)})$ and see that $\{s_1,...,s_l\}$ is linearly independent on $V_p$.

Step 1:

Let us take a chart trivializing open subset of $M$ around $p$, $(U_p\subseteq M,{\varphi }_p)$, which is possible by the proposition that for any $C^{\mathrm{\infty }}$ vectors bundle, there is a chart trivializing open cover.

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

Step 2:

Let us take the components function of $s_j$, $f_j:=\widetilde{{\varphi }_p}\circ s_j\circ {{\varphi }_p}^{-1}:{\varphi }_p(U_p)\to \widetilde{{\varphi }_p}({\pi }^{-1}(U_p))$, which is $C^{\mathrm{\infty }}$.

Step 3:

Let ${\pi }_k:{\mathbb{R}}^{d+k}\text{ or }{\mathbb{H}}^{d+k}\to {\mathbb{R}}^k$ be the projection.

Let us think of $\{{\pi }_k\circ f_1,...,{\pi }_k\circ f_l\}$.

Let us see that $\{{\pi }_k\circ f_1({\varphi }_p(p)),...,{\pi }_k\circ f_l({\varphi }_p(p))\}$ is a linearly independent subset of ${\mathbb{R}}^k$.

${\pi }_k\circ f_j({\varphi }_p(p))={\pi }_k\circ \widetilde{{\varphi }_p}\circ s_j\circ {{\varphi }_p}^{-1}({\varphi }_p(p))={\pi }_k\circ \widetilde{{\varphi }_p}\circ s_j(p)={\pi }_k\circ \lambda \circ ({\varphi }_p,id)\circ \mathrm{\Phi }\circ s_j(p)$, where $\mathrm{\Phi }$ is the trivialization by which $\widetilde{{\varphi }_p}$ is induced, so, is 'vectors spaces - linear morphisms' isomorphic at each fiber (so, especially at ${\pi }^{-1}(p)$, where $s_j(p)$ s belong), and so, $\{{\pi }_k\circ f_1({\varphi }_p(p)),...,{\pi }_k\circ f_l({\varphi }_p(p))\}$ is a linearly independent subset of ${\mathbb{R}}^k$ ($\lambda$ is $:{\mathbb{R}}^{d+k}\to {\mathbb{R}}^{d+k},(x^1,...,x^d,x^{d+1},...,x^{d+k})\mapsto (x^{d+1},...,x^{d+k},x^1,...,x^d)$ by the way).

Let us think of the $k\times l$ matrix, $[{\pi }_k\circ f_1({\varphi }_p(p)),...,{\pi }_k\circ f_l({\varphi }_p(p))]$.

$\{{\pi }_k\circ f_1({\varphi }_p(p)),...,{\pi }_k\circ f_l({\varphi }_p(p))\}$'s being linearly independent means that the matrix is rank $l$, which means that there is a nonzero-determinant $l\times l$ submatrix, $C$.

Step 4:

While we took $C$ as the matrix at ${\varphi }_p(p)$, now, let us think $C$ as the map from ${\varphi }_p(U_p)$: it is the $l\times l$ submatrix of $[{\pi }_k\circ f_1,...,{\pi }_k\circ f_l]$.

As each component of $C$ is continuous (in fact, $C^{\mathrm{\infty }}$) with respect to ${\varphi }_p(U_p)$, $detC$ is continuous (in fact, $C^{\mathrm{\infty }}$) with respect to ${\varphi }_p(U_p)$.

So, there is an open neighborhood of ${\varphi }_p(p)$, $U_{{\varphi }_p(p)}\subseteq {\varphi }_p(U_p)$, on which $detC\ne 0$, which means that $\{{\pi }_k\circ f_1,...,{\pi }_k\circ f_l\}$ is linearly independent on $U_{{\varphi }_p(p)}$.

Step 5:

Let us take $V_p:={{\varphi }_p}^{-1}(U_{{\varphi }_p(p)})$.

$V_p\subseteq U_p$ is an open neighborhood of $p$.

On $V_p$, $\{{\pi }_k\circ f_1\circ {\varphi }_p,...,{\pi }_k\circ f_l\circ {\varphi }_p\}=\{{\pi }_k\circ \widetilde{{\varphi }_p}\circ s_1\circ {{\varphi }_p}^{-1}\circ {\varphi }_p,...,{\pi }_k\circ \widetilde{{\varphi }_p}\circ s_l\circ {{\varphi }_p}^{-1}\circ {\varphi }_p\}=\{{\pi }_k\circ \widetilde{{\varphi }_p}\circ s_1,...,{\pi }_k\circ \widetilde{{\varphi }_p}\circ s_l\}$ is linearly independent.

Then, $\{s_1,...,s_l\}=\{{\widetilde{{\varphi }_p}}^{-1}\circ \widetilde{{\varphi }_p}\circ s_1,...,{\widetilde{{\varphi }_p}}^{-1}\circ \widetilde{{\varphi }_p}\circ s_l\}$ is linearly independent on $V_p$, because $\widetilde{{\varphi }_p}=\lambda \circ ({\varphi }_p,id)\circ \mathrm{\Phi }$ and $\mathrm{\Phi }$ is 'vectors spaces - linear morphisms' isomorphic at each fiber (so, especially at each ${\pi }^{-1}(p^{\prime })$ for each $p^{\prime }\in V_p$).

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References

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