A description/proof of that for 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.
- The reader knows a definition of restricted \(C^\infty\) vectors bundle.
- The reader knows a definition of \(C^\infty\) section on \(C^\infty\) vectors bundle.
- The reader knows a definition of \(C^\infty\) frame on \(C^\infty\) vectors bundle.
- The reader admits the proposition that for any vectors bundle, a trivializing open subset is not necessarily a chart open subset, but there is a possibly smaller chart trivializing open subset at any point on any trivializing open subset.
- The reader admits the proposition that for any 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 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: Description
For any \(C^\infty\) manifold, \(M\), any vectors bundle, \(\pi: E \rightarrow M\), any trivializing open subset, \(U\), and any section, \(s: U \rightarrow E\vert_U\), \(s\) is \(C^\infty\) if and only if the coefficients of \(s\) with respect to any \(C^\infty\) frame over \(U\), \(s_1, s_2, ..., s_{d'}\), are \(C^\infty\), which is that for \(s = s^i s_i\), \(s^i\) is \(C^\infty\).
2: Proof
Around any point, \(p \in U\), there are a chart trivializing open subset, \(U'_p \subseteq U\), and the induced chart, \((\pi^{-1} (U'_p), \phi'')\), by the proposition that for any vectors bundle, a trivializing open subset is not necessarily a chart open subset, but there is a possibly smaller chart trivializing open subset at any point on any trivializing open subset and the proposition that for any vectors bundle, the trivialization of any chart trivializing open subset induces the canonical chart map. \(\phi\) is the chart map on \(U'_p\), and \(\phi'\) is the trivialization.
Let us suppose that \(s^i\) is \(C^\infty\). Let \(p' \in U'_p\) be any point. The coordinates of \(s_i (p')\) are \(\phi'' (s_i (p')) = (p'^1, p'^2, ..., p'^d, {s_i (p')}^1, {s_i (p')}^2, ..., {s_i (p')}^{d'})\), and as \(s_i\) is a \(C^\infty\) section, \({s_i (p')}^i\) is \(C^\infty\) with respect to \(p'^1, p'^2, ..., p'^d\). The coordinates of \(s (p')\) are \(\phi'' (s (p')) = (\phi, id) \circ \phi' (s (p')) = (\phi, id) \circ \phi' (s^i (p') s_i (p')) = (\phi, id) (s^i (p') \phi' (s_i (p')))\), because any trivialization is 'vectors spaces - linear morphisms' isomorphic at each fiber. And \(= (p'^1, p'^2, ..., p'^d, s^i (p') {s_i (p')}^1, s^i (p') {s_i (p')}^2, ..., s^i (p') {s_i (p')}^{d'})\), and as \(s^i (p')\) and \({s_i (p')}^j\) are \(C^\infty\), the components are \(C^\infty\) with respect to \(p'^1, p'^2, ..., p'^d\), which means that \(s\) is \(C^\infty\).
Let us suppose that \(s\) is \(C^\infty\). Let \(p' \in U'_p\) be any point. For the canonical \(C^\infty\) frame, \(e'_1, e'_2, ..., e'_{d'}\), on \(\phi' (\pi^{-1} (U'_p)) = U'_p \times \mathbb{R}^{d'}\), there is the induced \(C^\infty\) frame, \(e_i (p') := \phi'^{-1} (p', e'_i)\), which is really \(C^\infty\), because the coordinates of \(e_i (p')\) are \((p'^1, p'^2, ..., p'^d, 0, 0, ..., 1, ..., 0)\) where \(1\) is the \(i\)-th component in \(\mathbb{R}^{d'}\). \(s = s'^i e_i\). The coordinates of \(s (p')\) are \(\phi'' (s (p')) = \phi'' (s'^i (p') e_i (p')) = (\phi, id) \circ \phi' (s'^i (p') e_i (p')) = (\phi, id) (s'^i (p') \phi' (e_i (p'))) = (\phi, id) (s'^i (p') e'_i (p')) = (p'^1, p'^2, ..., p'^d, s'^1 (p'), s'^2 (p'), ..., s'^{d'} (p'))\). \(s\)'s being \(C^\infty\) is nothing but the coordinates are \(C^\infty\) as functions of \(p'^1, p'^2, ..., p'^d\), so, \(s'^i\) is \(C^\infty\).
\(s_i = s'^j_i e_j\) where \(s'^j_i\) is \(C^\infty\), by the previous paragraph. \(s = s^i s_i = s^i s'^j_i e_j\), so, \(s'^i = s^j s'^i_j\). The matrix, \(S' (p') = [s'^i_j (p')]\), is the matrix of the components transformation with respect to 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 \(p'\). As \(s^i\) is an addition of some multiplications of the components of \(S'^{-1}\) and \(s'^{i}\), it is \(C^\infty\).
3: Note
"with respect to any \(C^\infty\) frame" in Description means that if the coefficients are so with respect to any one such 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 so with respect to any such frame (so, with respect to every 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 vectors bundle, there is a \(C^\infty\) frame over any open set on the base space if and only if the open subset is a trivializing open set.
