description/proof of that for \(C^\infty\) manifold with boundary and local \(C^\infty\) frame on tangent vectors bundle, dual local frame is \(C^\infty\)
Topics
About: \(C^infty\) manifold
The table of contents of this article
Starting Context
- The reader knows a definition of tangent vectors bundle over \(C^\infty\) manifold with boundary.
- The reader knows a definition of local \(C^\infty\) frame on \(C^\infty\) vectors bundle.
- The reader knows a definition of dual basis for covectors (dual) space of basis for finite-dimensional vectors space.
- The reader knows a definition of open submanifold with boundary of \(C^\infty\) manifold with boundary.
- The reader admits the proposition that for any \(C^\infty\) vectors bundle, any rough section over any trivializing open subset is \(C^\infty\) if and only if the coefficients of the rough section with respect to any local \(C^\infty\) frame over the trivializing open subset are \(C^\infty\).
- The reader admits the proposition that for any \(C^\infty\) vectors bundle, any \(C^\infty\) frame exists over and only over any trivializing open subset.
- The reader admits the proposition that any rough \(q\)-form over any \(C^\infty\) manifold with boundary is \(C^\infty\) if and only if the operation result on any \(C^\infty\) vectors fields is \(C^\infty\).
Target Context
- The reader will have a description and a proof of the proposition that for any \(C^\infty\) manifold with boundary and any local \(C^\infty\) frame on the tangent vectors bundle, the dual local frame is \(C^\infty\).
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:
\(M\): \(\in \{\text{ the } d \text{ -dimensional } C^\infty \text{ manifolds with boundary }\}\)
\((TM, M, \pi)\): \(= \text{ the tangent vectors bundle }\)
\((TM^*, M, \pi^*)\): \(= \text{ the cotangent vectors bundle }\)
\(U\): \(\in \{\text{ the open subsets of } M\}\)
\(\{v_1, ..., v_d\}\): \(\in \{\text{ the local } C^\infty \text{ frames over } U \text{ on } TM\}\)
\(\{t^1, ..., t^d\}\): \(= \text{ the dual local frame of } \{v_1, ..., v_d\}\)
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Statements:
\(\{t^1, ..., t^d\} \in \{\text{ the local } C^\infty \text{ frames over } U \text{ on } TM^*\}\)
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2: Proof
Whole Strategy: Step 1: see what "the dual local frame of \(\{v_1, ..., v_d\}\)" means; Step 2: see that \(t^j\) is \(C^\infty\).
Step 1:
Let us see what "the dual local frame of \(\{v_1, ..., v_d\}\)" means.
For each \(u \in U\), \(\{v_1 (u), ..., v_d (u)\}\) is a basis for \(\pi^{-1} (u)\).
\(\{t^1 (u), ..., t^d (u)\}\) is the dual basis of \(\{v_1 (u), ..., v_d (u)\}\).
So, we have defined for each \(j \in \{1, ..., d\}\), \(t^j: U \to {\pi^*}^{-1} (U)\).
\(\pi^* (t^j (u)) = u\), so, \(t^j\) is a rough section.
The remaining issue is whether \(t^j\) is \(C^\infty\) (then, inevitably is continuous).
Step 2:
Note that \(U\) is a \(C^\infty\) manifold with boundary as the open submanifold with boundary of \(M\) and \(\pi^{-1} (U)\) and \({\pi^*}^{-1} (U)\) are its tangent vectors bundle and cotangent vectors bundle.
Let \(v: U \to \pi^{-1} (U)\) be any \(C^\infty\) vectors field.
\(v = v^j v_j\), where \(v^j: U \to \mathbb{R}\) is \(C^\infty\), by the proposition that for any \(C^\infty\) vectors bundle, any rough section over any trivializing open subset is \(C^\infty\) if and only if the coefficients of the rough section with respect to any local \(C^\infty\) frame over the trivializing open subset are \(C^\infty\): \(U\) is a trivializing open subset, by the proposition that for any \(C^\infty\) vectors bundle, any \(C^\infty\) frame exists over and only over any trivializing open subset.
For each \(l \in \{1, ..., d\}\), \(t^l (v) = t^l (v^j v_j) = v^j t^l (v_j) = v^j \delta^l_j = v^l\), which is \(C^\infty\).
So, by the proposition that any rough \(q\)-form over any \(C^\infty\) manifold with boundary is \(C^\infty\) if and only if the operation result on any \(C^\infty\) vectors fields is \(C^\infty\), \(t^l\) is \(C^\infty\).
