description/proof of that for \(C^\infty\) 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
- 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 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.
- The reader admits the proposition that any bijective linear map is a 'vectors spaces - linear morphisms' isomorphism.
- 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 \(C^\infty\) manifold with boundary and its any chart, the restriction of the chart on any open subset domain is a chart.
- The reader admits the proposition that any open subset of any \(C^\infty\) trivializing open subset is a \(C^\infty\) trivializing open subset.
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: Structured Description
Here is the rules of Structured Description.
Entities:
\((E, M, \pi)\): \(\in \{\text{ the } C^\infty \text{ vectors bundles of rank } k\}\)
\(U\): \(\in \{\text{ the open subsets of } M\}\)
//
Statements:
\(\exists \text{ a } C^\infty \text{ frame over } U\)
\(\iff\)
\(U \in \{\text{ the trivializing open subsets of } M\}\)
//
2: Natural Language Description
For any \(C^\infty\) vectors bundle of rank \(k\), \((E, M, \pi)\), and any open subset, \(U \subseteq M\), any \(C^\infty\) frame exists over \(U\) if and only if \(U\) is a trivializing open subset of \(M\).
3: Proof
Whole Strategy: Step 1: suppose that \(U\) is a trivializing open subset, and construct a \(C^\infty\) frame over \(U\) based on the trivialization: Step 2: suppose that there is a \(C^\infty\) frame over \(U\), and construct a trivialization over \(U\) based on the frame.
Step 1:
Step 1 Strategy: Step 1-1: take a trivialization, \(\Phi\); Step 1-2: take the canonical frame, \(e'_1, ..., e'_k\), on \(\Phi (\pi^{-1} (U)) = U \times \mathbb{R}^k\) and the induced \(C^\infty\) frame on \(\pi^{-1} (U)\), \((e_1, ..., e_k)\); Step 1-3: see that it is indeed a \(C^\infty\) frame.
Step 1-1:
Let us suppose that \(U\) is a trivializing open subset.
There is a trivialization, \(\Phi: \pi^{-1} (U) \to U \times \mathbb{R}^k\).
Step 1-2:
There is the canonical frame, \(e'_1, ..., e'_k\), on \(\Phi (\pi^{-1} (U)) = U \times \mathbb{R}^k\), which means that \(e'_j: U \to U \times \mathbb{R}^k, p' \to (p', 0, ..., 0, 1, 0, ..., 0)\), where \(1\) is the j-th component of \(\mathbb{R}^k\), which is indeed a frame (not particularly claimed to be \(C^\infty\)), obviously.
There is the induced \(C^\infty\) frame on \(\pi^{-1} (U)\), \((e_1, ..., e_k)\), where \(e_j (p') := \Phi^{-1} (e'_j (p'))\).
Step 1-3:
Let us see that it is indeed a \(C^\infty\) frame.
\(e_j\) is indeed a section, because \(\pi (e_j (p')) = \pi (\Phi^{-1} (e'_j (p'))) = p'\), because \(\Phi \vert_{\pi^{-1} (p')}: \pi^{-1} (p') \to \{p'\} \times \mathbb{R}^k\) is the 'vectors spaces - linear morphisms' isomorphism while \(e'_j (p') \in \{p'\} \times \mathbb{R}^k\).
\((e_1, ..., e_k)\) is indeed linearly independent at each point, because \(\Phi\) is a 'vectors spaces - linear morphisms' isomorphism at each fiber.
Let us see that \(e_j\) is indeed \(C^\infty\).
Around any point, \(p \in U\), there is a chart trivializing open subset, \(U_p \subseteq M\), such that \(U_p \subseteq U\), 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 each point on any trivializing open subset. Let \((U_p \subseteq M, \phi_p)\) be the chart.
There is the induced chart, \((\pi^{-1} (U_p) \subseteq E, \widetilde{\phi_p})\), 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 \(e_j (p')\) are \(\widetilde{\phi_p} (e_j (p')) = \lambda \circ (\phi_p, id) \circ \Phi (e_j (p'))\), where \(\lambda: \mathbb{R}^{d + k} \to \mathbb{R}^{d + k}, (x^1, ..., x^d, x^{d + 1}, ..., x^{d + k}) \mapsto (x^{d + 1}, ..., x^{d + k}, x^1, ..., x^d)\), \(= \lambda \circ (\phi_p, id) (e'_j (p')) = \lambda \circ (\phi_p, id) ((p', 0, ..., 0, 1, 0, ..., 0))\), where \(1\) is the \(j\)-th component in \(\mathbb{R}^k\), \(= (0, 0, ..., 1, ..., 0, p^1, p^2, ..., p^d)\). As the components are \(C^\infty\) with respect to \(p^1, p^2, ..., p^d\), \(e_j\) is \(C^\infty\) at each \(p\), which means that \(e_j\) is indeed \(C^\infty\).
So, \((e_1, ..., e_k)\) is indeed a \(C^\infty\) frame over \(U\).
Step 2:
Step 2 Strategy: Step 2-1: let \(s_1, s_2, ..., s_k\) be a \(C^\infty\) frame over \(U\), and express each \(b \in \pi^{-1} (U)\) as \(b = b^j s_j (p)\); Step 2-2: define \(f: \pi^{-1} (U) \to U \times \mathbb{R}^k\), as \(f (b^j s_j (p)) = (p, b^1, b^2, ..., b^k)\); Step 2-3: see that \(f\) is a 'vectors spaces - linear morphisms' isomorphism at each fiber; Step 2-4: see that \(f\) is a diffeomorphism.
Step 2-1:
Let us suppose that there is any \(C^\infty\) frame, \(s_1, s_2, ..., s_k\), over \(U\).
For each point, \(b \in \pi^{-1} (U)\), \(b = b^j s_j (p)\) where \(p = \pi (b)\).
Step 2-2:
Let us define the map, \(f: \pi^{-1} (U) \to U \times \mathbb{R}^k\), as \(f (b^j s_j (p)) = (p, b^1, b^2, ..., b^k)\), which is well-defined, because \(p\) and the components, \(b^j\) s, are uniquely determined.
\(f\) is obviously bijective and so, there is the inverse, \(f^{-1}\).
Step 2-3:
Let us see that \(f\) is a 'vectors spaces - linear morphisms' isomorphism at each fiber.
\(f \vert_{\pi^{-1} (p)}: \pi^{-1} (p) \to \{p\} \times \mathbb{R}^k\) is linear, because \(f (r b^j s_j (p) + r' b'^j s_j (p)) = f ((r b^j + r' b'^j) s_j (p)) = (p, r b^1 + r' b'^1, ..., r b^k + r' b'^k) = r (p, b^1, ..., b^k) + r' (p, b'^1, ..., b'^k) = r f (b^j s_j (p)) + r' f (b'^j s_j (p))\).
\(f \vert_{\pi^{-1} (p)}\) is obviously bijective.
So, \(f\) is a 'vectors spaces - linear morphisms' isomorphism at each fiber, by the proposition that any bijective linear map is a 'vectors spaces - linear morphisms' isomorphism.
Step 2-4:
Let us prove that \(f\) is a diffeomorphism.
Around any point, \(p \in U\), there is a chart trivializing open subset, \(U_p \subset M\), such that \(U_p \subset U\), because there is a chart trivializing open subset around \(p\), by the proposition that for any \(C^\infty\) vectors bundle, there is a chart trivializing open cover, and its intersection with \(U\) is a one, by the proposition that for any \(C^\infty\) manifold with boundary and its any chart, the restriction of the chart on any open subset domain is a chart and the proposition that any open subset of any \(C^\infty\) trivializing open subset is a \(C^\infty\) trivializing open subset. Let the chart be \((U_p \subseteq M, \phi_p)\). Let the trivialization be \(\Phi: \pi^{-1} (U_p) \to U_p \times \mathbb{R}^k\).
There is the induced chart, \((\pi^{-1} (U_p) \subseteq E, \widetilde{\phi_p})\), 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, \(\widetilde{\phi_p} (s_j (p)) = ({s_j (p)}^1, ..., {s_j (p)}^k, p^1, p^2, ..., p^d)\), where \({s_j (p)}^l\) is \(C^\infty\) because \(s_j\) is \(C^\infty\).
\(\widetilde{\phi_p} (b) = \widetilde{\phi_p} (b^j s_j (p')) = \lambda \circ (\phi_p, id) \circ \Phi (b^j s_i (p')) = \lambda \circ (\phi, id) \circ b^j \Phi (s_j (p'))\), because \(\Phi\) is 'vectors spaces - linear morphisms' isomorphic at each fiber, \(= (b^j {s_j (p')}^1, ..., b^j {s_j (p')}^k, p'^1, ..., p'^d) := (c^1, ..., c^k, p'^1, ..., p'^d)\).
Let us denote the matrix, \(S (p) := [s_j (p)^l]\), then, \((c^1, c^2, ..., c^k)^t = S (b^1, b^2, ..., b^k)^t\). \(S\) is the components function matrix of the 'vectors spaces - linear morphisms' isomorphism of the trivialization on the fiber with respect to the frame basis on \(\pi^{-1} (p)\) and the canonical basis on \(\{p\} \times \mathbb{R}^k\), and is invertible. The components of \(S\) are \(C^\infty\). The components of the inverse matrix, \(S^{-1}\), are \(C^\infty\): use the Laplace expansion to get the inverse matrix.
On the other hand, there is the chart, \((U_p \times \mathbb{R}^k \subseteq U \times \mathbb{R}^k, \lambda \circ (\phi_p, id))\).
The components function of \(f\), \(\lambda \circ (\phi_p, id) \circ f \circ {\widetilde{\phi_p}}^{-1}: \widetilde{\phi_p} (\pi^{-1} (U_p)) \to \mathbb{R}^k \times \phi_p (U_p), (c^1, c^2, ..., c^k, p'^1, ..., p'^d) \mapsto (b^1, b^2, ..., b^k, p'^1, ..., p'^d)\), is \(C^\infty\), because \((b^1, b^2, ..., b^k)^t = S^{-1} (c^1, c^2, ..., c^k)^t\) while the components of \(S^{-1}\) are \(C^\infty\) with respect to \((p'^1, ..., p'^d)\). So, \(f\) is \(C^\infty\) at each \(b\).
The components of \(f^{-1}\), \(\widetilde{\phi_p} \circ f^{-1} \circ (\phi_p, id)^{-1} \circ \lambda^{-1}: \mathbb{R}^k \times \phi_p (U_p) \to \widetilde{\phi_p} (\pi^{-1} (U_p)), (b^1, b^2, ..., b^k, p'^1, ..., p'^d) \mapsto (c^1, c^2, ..., c^k, p'^1, ..., p'^d)\), is \(C^\infty\), because \((c^1, c^2, ..., c^k)^t = S (b^1, b^2, ..., b^k)^t\) while the components of \(S\) are \(C^\infty\) with respect to \((p'^1, ..., p'^d)\). So, \(f^{-1}\) is \(C^\infty\) at each \(f (b)\).
So, \(f\) is diffeomorphic.
So, \(f\) is a \(C^\infty\) trivialization and \(U\) is a trivializing open subset.
