description/proof of that set of \(C^\infty\) sections of \(C^\infty\) vectors bundle linearly independent at point is linearly independent on open neighborhood of point
Topics
About: \(C^\infty\) manifold
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
Starting Context
- The reader knows a definition of \(C^\infty\) 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 \(C^\infty\) manifolds with boundary, where \(k\) includes \(\infty\).
- 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 \(C^\infty\) vectors bundle, there is a chart trivializing open cover.
- The reader admits the proposition that for any \(C^\infty\) vectors bundle, the trivialization of any chart trivializing open subset induces the canonical chart map.
Target Context
- The reader will have a description and a proof of the proposition that the set of any \(C^\infty\) sections of any \(C^\infty\) vectors bundle that (the set) is linearly independent at a point is linearly independent on an open neighborhood of the point.
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.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\((E, M, \pi)\): \(\in \{\text{ the } d \text{ -dimensional } C^\infty \text{ vectors bundles of rank } k\}\)
\(p\): \(\in M\)
\(\{s_1, ..., s_l\}\): \(l \le k\), \(s_j: M \to E \in \{\text{ the } C^\infty \text{ sections }\}\), such that \(\{s_1 (p), ..., s_l (p)\} \in \{\text{ the linearly independent subsets of } \pi^{-1} (p)\}\)
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Statements:
\(\exists V_p \subseteq M \in \{\text{ the open neighborhoods of } p\} (\forall p' \in V_p (\{s_1 (p'), ..., s_l (p')\} \in \{\text{ the linearly independent subsets of } \pi^{-1} (p')\}))\)
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2: Natural Language Description
For any \(d\)-dimensional \(C^\infty\) vectors bundle of rank \(k\), \((E, M, \pi)\), any point, \(p \in M\), and the set of any \(C^\infty\) 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 \(p' \in V_p\), \(\{s_1 (p'), ..., s_l (p')\}\) is a linearly independent subset of \(\pi^{-1} (p')\).
3: Proof
Whole Strategy: Step 1: take a chart trivializing open subset of \(M\) around \(p\), \((U_p \subseteq M, \phi_p)\), and the induced chart, \((\pi^{-1} (U_p) \subseteq E, \widetilde{\phi_p})\); Step 2: take the components function of \(s_j\), \(f_j := \widetilde{\phi_p} \circ s_j \circ {\phi_p}^{-1}\); 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 \(\phi_p (p)\) 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 \(\phi_p (p)\), \(U_{\phi_p (p)} \subseteq \phi_p (U_p)\), on which the determinant of the submatrix is nonzero; Step 5: take \(V_p := {\phi_p}^{-1} (U_{\phi_p (p)})\) 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 \(p\), \((U_p \subseteq M, \phi_p)\), which is possible by the proposition that for any \(C^\infty\) vectors bundle, there is a chart trivializing open cover.
Let us take the induced chart, \((\pi^{-1} (U_p) \subseteq E, \widetilde{\phi_p})\), which is possible by the proposition that for any \(C^\infty\) vectors bundle, the trivialization of any chart trivializing open subset induces the canonical chart map.
Step 2:
Let us take the components function of \(s_j\), \(f_j := \widetilde{\phi_p} \circ s_j \circ {\phi_p}^{-1}: \phi_p (U_p) \to \widetilde{\phi_p} (\pi^{-1} (U_p))\), which is \(C^\infty\).
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 \(\{\pi_k \circ f_1 (\phi_p (p)), ..., \pi_k \circ f_l (\phi_p (p))\}\) is a linearly independent subset of \(\mathbb{R}^k\).
\(\pi_k \circ f_j (\phi_p (p)) = \pi_k \circ \widetilde{\phi_p} \circ s_j \circ {\phi_p}^{-1} (\phi_p (p)) = \pi_k \circ \widetilde{\phi_p} \circ s_j (p) = \pi_k \circ \lambda \circ (\phi_p, id) \circ \Phi \circ s_j (p)\), where \(\Phi\) is the trivialization by which \(\widetilde{\phi_p}\) 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, \(\{\pi_k \circ f_1 (\phi_p (p)), ..., \pi_k \circ f_l (\phi_p (p))\}\) 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, \([\pi_k \circ f_1 (\phi_p (p)), ..., \pi_k \circ f_l (\phi_p (p))]\).
\(\{\pi_k \circ f_1 (\phi_p (p)), ..., \pi_k \circ f_l (\phi_p (p))\}\)'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 \(\phi_p (p)\), now, let us think \(C\) as the map from \(\phi_p (U_p)\): 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, \(C^\infty\)) with respect to \(\phi_p (U_p)\), \(det C\) is continuous (in fact, \(C^\infty\)) with respect to \(\phi_p (U_p)\).
So, there is an open neighborhood of \(\phi_p (p)\), \(U_{\phi_p (p)} \subseteq \phi_p (U_p)\), on which \(det C \neq 0\), which means that \(\{\pi_k \circ f_1, ..., \pi_k \circ f_l\}\) is linearly independent on \(U_{\phi_p (p)}\).
Step 5:
Let us take \(V_p := {\phi_p}^{-1} (U_{\phi_p (p)})\).
\(V_p \subseteq U_p\) is an open neighborhood of \(p\).
On \(V_p\), \(\{\pi_k \circ f_1 \circ \phi_p, ..., \pi_k \circ f_l \circ \phi_p\} = \{\pi_k \circ \widetilde{\phi_p} \circ s_1 \circ {\phi_p}^{-1} \circ \phi_p, ..., \pi_k \circ \widetilde{\phi_p} \circ s_l \circ {\phi_p}^{-1} \circ \phi_p\} = \{\pi_k \circ \widetilde{\phi_p} \circ s_1, ..., \pi_k \circ \widetilde{\phi_p} \circ s_l\}\) is linearly independent.
Then, \(\{s_1, ..., s_l\} = \{{\widetilde{\phi_p}}^{-1} \circ \widetilde{\phi_p} \circ s_1, ..., {\widetilde{\phi_p}}^{-1} \circ \widetilde{\phi_p} \circ s_l\}\) is linearly independent on \(V_p\), because \(\widetilde{\phi_p} = \lambda \circ (\phi_p, id) \circ \Phi\) and \(\Phi\) is 'vectors spaces - linear morphisms' isomorphic at each fiber (so, especially at each \(\pi^{-1} (p')\) for each \(p' \in V_p\)).
