A description/proof of that for vectors bundle, \(C^\infty\) frame exists over and only over trivializing open subset
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 \(C^\infty\) 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 \(C^\infty\) vectors bundle, any \(C^\infty\) frame exists over and only over any trivializing open subset.
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 \(C^\infty\) vectors bundle, \(\pi: E \to M\), and any open subset, \(U \subseteq M\), any \(C^\infty\) frame exists over \(U\) if and only if \(U\) is a trivializing open subset.
2: Proof
Let us suppose that \(U\) is a trivializing open subset. There is the trivialization, \(\phi': \pi^{-1} (U) \to U \times \mathbb{R}^{d'}\). For the canonical \(C^\infty\) frame, \(e'_1, e'_2, ..., e'_{d'}\), on \(U \times \mathbb{R}^{d'}\), there is the induced \(C^\infty\) frame, \(e_i (p) := \phi'^{-1} (p, e'_i)\), which really forms a basis at each fiber because any trivialization is a 'vectors spaces - linear morphisms' isomorphism at each fiber. Let us prove that \(e_i\) is really a \(C^\infty\) section. Around any point, \(p \in U\), there is a chart trivializing open subset, \(U'_p \subseteq U\), 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 any point on any trivializing open subset. There is the chart, \((\pi^{-1} (U'_p), \phi'')\), on \(\pi^{-1} (U)\) induced by the trivialization of the chart trivializing open subset, by the proposition that for any \(C^\infty\) vectors bundle, the trivialization of any chart trivializing open subset induces the canonical chart map. On the chart, the coordinates of \(e_i (p)\) are \(\phi'' (e_i (p)) = (p^1, p^2, ..., p^d, 0, 0, ..., 1, ..., 0)\) where \(1\) is the \(i\)-th component in \(\mathbb{R}^{d'}\). As the components are \(C^\infty\) with respect to \(p^1, p^2, ..., p^d\), \(e_i\) is a \(C^\infty\) section.
Let us suppose that there is any \(C^\infty\) frame, \(s_1, s_2, ..., s_{d'}\), over \(U\). For any point, \(b \in E\vert_U\), \(b = b^i s_i (p)\) where \(p = \pi (b)\). Let us define the map, \(f: E\vert_U \to U \times \mathbb{R}^{d'}\), as \(f (b^i s_i (p)) = (p, b^1, b^2, ..., b^{d'})\). \(f\) is obviously bijective and so, there is the inverse, \(f^{-1}\). \(f\) is fiber preserving and is a 'vectors spaces - linear morphisms' isomorphism at each fiber.
Let us prove that \(f\) is a diffeomorphism. Around any point, \(p \in U\), there is a chart trivializing open subset, \(U'_p \subset U\), because there is a trivializing open subset around \(p\), which contains a chart trivializing open subset around \(p\), 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 any point on any trivializing open subset, whose intersection with \(U\) will be \(U'_p\). Over \(U'_p\), there is the induced chart, \((\pi^{-1} (U'_p), \phi'')\), by the proposition that for any \(C^\infty\) vectors bundle, the trivialization of any chart trivializing open subset induces the canonical chart map. On the chart, \(\phi'' (s_i (p)) = (p^1, p^2, ..., p^{d}, {s_i (p)}^1, {s_i (p)}^2, ..., {s_i (p)}^{d'})\). \(\phi'' (b) = \phi'' (b^i s_i (p)) = (\phi, id) \circ \phi' (b^i s_i (p)) = (\phi, id) \circ b^i \phi' (s_i (p))\) where \(\phi'\) is the trivialization, because \(\phi'\) is 'vectors spaces - linear morphisms' isomorphic at each fiber, \(= (p^1, p^2, ..., p^d, b^i {s_i (p)}^1, b^i {s_i (p)}^2, ..., b^i {s_i (p)}^{d'}) := (p^1, p^2, ..., p^d, c^1, c^2, ..., c^{d'})\). Let us denote the matrix, \(S (p) := [s_i (p)^j]\), then, \((c^1, c^2, ..., c^{d'})^t = S (b^1, b^2, ..., b^{d'})^t\). \(S\) is the components transformation matrix of the 'vectors spaces - linear morphisms' isomorphism of the trivialization at the fiber with respect to the frame basis on \(\pi^{-1} (p)\) and the canonical basis on \(\{p\} \times \mathbb{R}^{d'}\), and is \(C^\infty\) and invertible, and the inverse is \(C^\infty\). On the other hand, there is the chart, \((U'_p \times \mathbb{R}^{d'}, (\phi, id))\), on \(U \times \mathbb{R}^{d'}\), on which the coordinates of \(f (b)\) are \((p^1, p^2, ..., p^d, b^1, b^2, ..., b^{d'})\). So, \(f\) is \((p^1, p^2, ..., p^d, c^1, c^2, ..., c^{d'}) \mapsto (p^1, p^2, ..., p^d, {S^{-1} (p)}^1_i c^i, {S^{-1} (p)}^2_i c^i, ..., {S^{-1} (p)}^{d'}_i c^i)\), components-wise, which is \(C^\infty\). \(f^{-1}\) is \((p^1, p^2, ..., p^d, b^1, b^2, ..., b^{d'}) \mapsto (p^1, p^2, ..., p^d, {S (p)}^1_i b^i, {S (p)}^2_i b^i, ..., {S (p)}^{d'}_i b^i)\), components-wise, which is \(C^\infty\). So, \(f\) is a diffeomorphism.
So, \(U\) is a trivializing open subset.
