description/proof of that for \(C^\infty\) vectors bundle, section over trivializing open subset is \(C^\infty\) iff coefficients w.r.t. \(C^\infty\) frame over there are \(C^\infty\)
Topics
About: \(C^\infty\) manifold
The table of contents of this article
Starting Context
- The reader knows a definition of \(C^\infty\) vectors bundle of rank \(k\).
- The reader knows a definition of restricted \(C^\infty\) vectors bundle.
- The reader knows a definition of section of continuous map.
- The reader knows a definition of local \(C^\infty\) frame on \(C^\infty\) vectors bundle.
- The reader admits the proposition that for any \(C^\infty\) 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 \(C^\infty\) 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 for any \(C^\infty\) vectors bundle, any section over any trivializing open subset is \(C^\infty\) if and only if the coefficients of the section with respect to any \(C^\infty\) frame over the trivializing open subset are \(C^\infty\).
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:
\((E, M, \pi)\): \(\in \{\text{ the } C^\infty \text{ vectors bundles of rank } k\}\)
\(U\): \(\in \{\text{ the trivializing open subsets of } M\}\)
\(s\): \(: U \to \pi^{-1} (U)\), \(\in \{\text{ the sections of } \pi \vert_{\pi^{-1} (U)}\}\)
//
Statements:
(
\(s \in \{\text{ the } C^\infty \text{ maps } \}\)
\(\implies\)
\(\forall \{s_1, s_2, ..., s_k\} \in \{\text{ the } C^\infty \text{ frames on } \pi^{-1} (U)\} (s = s^j s_j \text{ where } s^j: U \to \mathbb{R} \land s^j \in \{\text{ the } C^\infty \text{ maps }\})\)
)
\(\land\)
(
\(\exists \{s_1, s_2, ..., s_k\} \in \{\text{ the } C^\infty \text{ frames on } \pi^{-1} (U)\} (s = s^j s_j \text{ where } s^j: U \to \mathbb{R} \land s^j \in \{\text{ the } C^\infty \text{ maps }\})\)
\(\implies\)
\(s \in \{\text{ the } C^\infty \text{ maps } \}\)
)
//
2: Proof
Whole Strategy: Step 1: around each point, \(p \in U\), take a chart trivializing open subset, \(U'_p \subseteq M\), such that \(U'_p \subseteq U\) and the induced chart, \((\pi^{-1} (U'_p) \subseteq E, \widetilde{\phi})\); Step 2: suppose that \(s^j\) is \(C^\infty\), and see that \(s\) is \(C^\infty\), by seeing that the components of \(s\) with respect to the charts are \(C^\infty\); Step 3: suppose that \(s\) is \(C^\infty\), and see that \(s^j\) is \(C^\infty\), by taking the standard frame and the components of \(s\) with respect to the standard frame and seeing \(s^j\) s as some \(C^\infty\) functions of the components.
Step 1:
\(s\) or \(s^j\) is \(C^\infty\) if and only if it is \(C^\infty\) at each point, \(p \in U\). So, let us see the \(C^\infty\)-nesses at \(p\).
\(C^\infty\)-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, \(U'_p \subseteq M\), such that \(U'_p \subseteq U\), and the induced chart, \((\pi^{-1} (U'_p) \subseteq E, \widetilde{\phi})\), by the proposition that for any \(C^\infty\) 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 \(C^\infty\) vectors bundle, the trivialization of any chart trivializing open subset induces the canonical chart map. Let \(\phi\) be the chart map on \(U'_p\), and let \(\Phi\) be the trivialization.
Step 2:
Let us suppose that \(s^j\) is \(C^\infty\).
Let us see that \(s\) is \(C^\infty\).
We are going to see that the components of \(s\) with respect to the charts are \(C^\infty\).
Let \(p' \in U'_p\) be any point.
The coordinates of \(s_j (p')\) are \(\widetilde{\phi} (s_j (p')) = ({s_j (p')}^1, ..., {s_j (p')}^k, p'^1, ..., p'^d)\), and as \(s_j\) is a \(C^\infty\) section, \({s_j (p')}^k\) is \(C^\infty\) with respect to \((p'^1, ..., p'^d)\).
The coordinates of \(s (p')\) are \(\widetilde{\phi} (s (p')) = \lambda \circ (\phi, id) \circ \Phi (s (p'))\), where \(\lambda: \mathbb{R}^{d + k} \to \mathbb{R}^{d + k}, (x^1, ..., x^d, x^{d + 1}, ..., x^{d + k}) \mapsto (x^{d + 1}, ..., x^{d + k}, x^1, ..., x^d)\), \(= \lambda \circ (\phi, id) \circ \Phi (s^j (p') s_j (p')) = \lambda \circ (\phi, id) (s^j (p') \Phi (s_j (p')))\), because any trivialization is a 'vectors spaces - linear morphisms' isomorphism at each fiber, \(= (s^j (p') {s_j (p')}^1, ..., s^j (p') {s_j (p')}^k, p'^1, p'^2, ..., p'^d)\), and as \(s^j (p')\) and \({s_j (p')}^j\) are \(C^\infty\), the components are \(C^\infty\) with respect to \(p'^1, p'^2, ..., p'^d\), which means that \(s\) is \(C^\infty\).
Step 3:
Let us suppose that \(s\) is \(C^\infty\).
Let \(p' \in U'_p\) be any point.
There is the canonical frame, \(e'_1, ..., e'_k\), on \(\Phi (\pi^{-1} (U'_p)) = U'_p \times \mathbb{R}^k\), which means that \(e'_j: U'_p \to U'_p \times \mathbb{R}^k, p' \to (p', 0, ..., 0, 1, 0, ..., 0)\), where \(1\) is the j-th component of \(\mathbb{R}^k\).
There is the induced \(C^\infty\) frame on \(\pi^{-1} (U'_p)\), \((e_1, ..., e_k)\), where \(e_j (p') := \Phi^{-1} (e'_j (p'))\), which is indeed \(C^\infty\), because the coordinates of \(e_j (p')\) are \((0, 0, ..., 1, ..., 0, p'^1, p'^2, ..., p'^d)\) where \(1\) is the \(j\)-th component in \(\mathbb{R}^k\), and is indeed a frame, because \(\Phi^{-1}\) is a 'vectors spaces - linear morphisms' isomorphism at each fiber.
Let \(s = s'^j e_j\). The coordinates of \(s (p')\) are \(\widetilde{\phi} (s (p')) = \widetilde{\phi} (s'^j (p') e_j (p')) = \lambda \circ (\phi, id) \circ \Phi (s'^j (p') e_j (p')) = \lambda \circ (\phi, id) (s'^j (p') \Phi (e_j (p'))) = \lambda \circ (\phi, id) (s'^j (p') e'_j (p')) = (s'^1 (p'), ..., s'^k (p'), p'^1, p'^2, ..., p'^d)\). \(s\)'s being \(C^\infty\) is nothing but that the coordinates are \(C^\infty\) as functions of \(p'^1, ..., p'^d\), so, \(s'^j\) is \(C^\infty\).
Let \(s_j = s'^l_j e_l\). \(s'^l_j\) is \(C^\infty\), by the previous paragraph. \(s = s^j s_j = s^j s'^l_j e_l\), so, \(s'^j = s^l s'^j_l\). The matrix, \(S' (p') = [s'^j_l (p')]\), is the matrix of the components transformation with respect to the 2 bases, and is invertible. As the components of \(S'\) are \(C^\infty\) with respect to \(p'\), the components of \(S'^{-1}\) are \(C^\infty\) with respect to \(p'\): use the Laplace expansion to get the inverse matrix. As \(s^j\) is an addition of some multiplications of the components of \(S'^{-1}\) and \(s'^l\) s, it is \(C^\infty\).
3: Note
This proposition says that if the coefficients of \(s\) are \(C^\infty\) with respect to any one \(C^\infty\) frame (does not need to check for every such frame), the section is \(C^\infty\), and if the section is \(C^\infty\), the coefficients are \(C^\infty\) with respect to any \(C^\infty\) frame (so, with respect to every \(C^\infty\) frame).
The trivializing open set may seem to be allowed to be any open set if there is a \(C^\infty\) frame over there, but in fact, such any open subset has to be a trivializing open set after all in order to have any \(C^\infty\) frame, by the proposition that for any \(C^\infty\) vectors bundle, any \(C^\infty\) frame exists over and only over any trivializing open subset.
