2025-06-22

1171: \(q\)-Form over \(C^\infty\) Manifold with Boundary Is \(C^\infty\) iff Operation Result on Any \(C^\infty\) Vectors Fields Is \(C^\infty\)

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description/proof of that \(q\)-form over \(C^\infty\) manifold with boundary is \(C^\infty\) iff operation result on any \(C^\infty\) vectors fields is \(C^\infty\)

Topics


About: \(C^\infty\) manifold

The table of contents of this article


Starting Context



Target Context


  • The reader will have a description and a proof of the proposition that any \(q\)-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 \(C^\infty\).

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:
\(M\): \(\in \{\text{ the } C^\infty \text{ manifolds with boundary }\}\)
\(q\): \(\in \mathbb{N} \setminus \{0\}\)
\((T^0_q (TM), M, \pi)\): \(= (0, q) \text{ -tensors bundle over } M\)
\((\Lambda_q (TM), M, \pi)\): \(= q \text{ -covectors bundle over } M\)
\(f\): \(: M \to T^0_q (TM)\) such that \(Ran (f) \subseteq \Lambda_q (TM)\) or \(: M \to \Lambda_q (TM)\), \(\in \{\text{ the sections of } \pi\}\)
//

Statements:
\(f \in \{\text{ the } C^\infty \text{ maps }\}\)
\(\iff\)
\(\forall V_1, ..., V_q \in \{\text{ the } C^\infty \text{ vectors fields over } M\} (f (V_1, ..., V_q): M \to \mathbb{R} \in \{\text{ the } C^\infty \text{ maps }\})\)
//


2: Proof


Whole Strategy: Step 1: when \(: M \to T^0_q (TM)\), see that the proposition holds; Step 2: suppose that \(: M \to \Lambda_q (TM)\); Step 3: suppose that \(f (V_1, ..., V_q) \in \{\text{ the } C^\infty \text{ maps }\}\), and see that \(f\) is \(C^\infty\), by taking an \(r'\)-\(r\)-open-balls charts pair or an \(r'\)-\(r\)-open-half-balls charts pair around each \(m \in M\), \((U'_m \subseteq M, \phi'_m)\) and \((U_m \subseteq M, \phi_m)\), and the induced chart, \((\pi^{-1} (U_m) \subseteq \Lambda_q (TM), \widetilde{\phi_m})\), and seeing that the components function of \(f\) is \(C^\infty\); Step 4: suppose that \(f\) is \(C^\infty\), and see that \(f (V_1, ..., V_q)\) is \(C^\infty\) over a chart, \((U_m \subseteq M, \phi_m)\), around each \(m \in M\).

Step 1:

When \(f\) is \(: M \to T^0_q (TM)\), \(f\) is really just a special type of \((0, q)\)-tensors field, so, the proposition that any \((0, q)\)-tensors field over \(C^\infty\) manifold with boundary is \(C^\infty\) if and only if the operation result on any \(C^\infty\) vectors fields is \(C^\infty\) applies.

Step 2:

Let us suppose that \(f\) is \(: M \to \Lambda_q (TM)\).

Step 3:

Let us suppose that \(f (V_1, ..., V_q) \in \{\text{ the } C^\infty \text{ maps }\}\).

Let \(m \in M\) be any.

Let us take any \(r'\)-\(r\)-open-balls charts pair or any \(r'\)-\(r\)-open-half-balls charts pair around \(m\), \((U'_m \subseteq M, \phi'_m)\) and \((U_m \subseteq M, \phi_m)\), which is possible by the proposition that for any \(C^\infty\) manifold with boundary, each interior point has an \(r'\)-\(r\)-open-balls charts pair and each boundary point has an \(r'\)-\(r\)-open-half-balls charts pair for any positive \(r'\) and \(r\), and take the induced chart, \((\pi^{-1} (U_m) \subseteq \Lambda_q (TM), \widetilde{\phi_m})\).

Let us take \(V_j = V_j^{l_j} \partial / \partial x^{l_j}\) over \(U'_m\) as \(V_j^{l'_j} \equiv 1\) and \(V_j^{l_j} \equiv 0\) for each \({l_j} \neq l'_j\) where \(l'_1 \lt ... \lt l'_q\). \(V_j\) is \(C^\infty\) over \(U'_m\). \(V_j\) is \(C^\infty\) over \(\overline{U_m} \subseteq U'_m\). By the proposition that for any \(C^\infty\) vectors bundle, any \(C^\infty\) section along any closed subset of the base space can be extended to over the whole base space with the support contained in any open neighborhood of the subset, \(V_j\) is extended to over \(M\). The extended \(V_j\) equals the original \(V_j\) over \(\overline{U_m}\) especially over \(U_m\).

Let \(m' \in U_m\) be any.

\(f (m') = \sum_{j_1 \lt ... \lt j_q} f_{j_1, ..., j_q} (m') d x^{j_1} \wedge ... \wedge d x^{j_q}\).

\(f (m') (V_1, ..., V_q) = \sum_{j_1 \lt ... \lt j_q} f_{j_1, ..., j_q} (m') d x^{j_1} \wedge ... \wedge d x^{j_q} (V_1^{l_1} \partial / \partial x^{l_1}, ..., V_q^{l_q} \partial / \partial x^{l_q}) = \sum_{j_1 \lt ... \lt j_q} f_{j_1, ..., j_q} (m') q! Asym d x^{j_1} \otimes ... \otimes d x^{j_q} (V_1^{l_1} \partial / \partial x^{l_1}, ..., V_q^{l_q} \partial / \partial x^{l_q}) = \sum_{j_1 \lt ... \lt j_q} f_{j_1, ..., j_q} (m') q! 1 / q! \sum_{\sigma \in P_{\{1, ..., q\}}} sgn \sigma d x^{j_1} \otimes ... \otimes d x^{j_q} (V_{\sigma_1}^{l_{\sigma_1}} \partial / \partial x^{l_{\sigma_1}}, ..., V_{\sigma_q}^{l_{\sigma_q}} \partial / \partial x^{l_{\sigma_q}}) = \sum_{j_1 \lt ... \lt j_q} f_{j_1, ..., j_q} (m') \sum_{\sigma \in P_{\{1, ..., q\}}} sgn \sigma d x^{j_1} (V_{\sigma_1}^{l_{\sigma_1}} \partial / \partial x^{l_{\sigma_1}}) ... d x^{j_q} (V_{\sigma_q}^{l_{\sigma_q}} \partial / \partial x^{l_{\sigma_q}}) = \sum_{j_1 \lt ... \lt j_q} f_{j_1, ..., j_q} (m') \sum_{\sigma \in P_{\{1, ..., q\}}} sgn \sigma V_{\sigma_1}^{l_{\sigma_1}} \delta^{j_1}_{l_{\sigma_1}} ... V_{\sigma_q}^{l_{\sigma_q}} \delta^{j_q}_{l_{\sigma_q}} = \sum_{j_1 \lt ... \lt j_q} f_{j_1, ..., j_q} (m') \sum_{\sigma \in P_{\{1, ..., q\}}} sgn \sigma V_{\sigma_1}^{j_1} ... V_{\sigma_q}^{j_q} = \sum_{j_1 \lt ... \lt j_q} f_{j_1, ..., j_q} (m') \sum_{\sigma \in P_{\{1, ..., q\}}} sgn \sigma V_1^{j_{\sigma^{-1} (1)}} ... V_q^{j_{\sigma^{-1} (q)}}\): \(V_{\sigma_1}^{j_1} ... V_{\sigma_q}^{j_q}\) is reordered to \(V_1^{m_1} ... V_q^{m_q}\), then, \(V_{\sigma_n}^{j_n} = V_s^{m_s}\), which means that \(\sigma_n = s\), so, \(n = \sigma^{-1} (s)\).

\(V_1^{j_{\sigma^{-1} (1)}} ... V_q^{j_{\sigma^{-1} (q)}}\) can be nonzero only for \(\sigma = id\), because \(l'_1 \lt ... \lt l'_q\).

So, \(= \sum_{j_1 \lt ... \lt j_q} f_{j_1, ..., j_q} (m') V_1^{j_1} ... V_q^{j_q} = f_{l'_1, ..., l'_q} (m')\).

\(f (V_1, ..., V_q): U_m \to \mathbb{R}\) is \(C^\infty\) by the supposition, so, \(f_{l'_1, ..., l'_q}: U_m \to \mathbb{R}\) is \(C^\infty\).

So, each \(f_{j_1, ..., j_q}\) is \(C^\infty\) over \(U_m\), which means that the components function of \(f\) with respect to \((U_m \subseteq M, \phi_m)\) and \((\pi^{-1} (U_m) \subseteq \Lambda_q (TM), \widetilde{\phi_m})\), \(\widetilde{\phi_m} \circ f \circ {\phi_m}^{-1}\) whose components are \(f_{j_1, ..., j_q} \circ {\phi_m}^{-1}\) s, is \(C^\infty\).

So, \(f\) is \(C^\infty\).

Step 4:

Let us suppose that \(f\) is \(C^\infty\).

Let \(m \in M\) be any.

Let us take any chart around \(m\), \((U_m \subseteq M, \phi_m)\), and the induced chart, \((\pi^{-1} (U_m) \subseteq \Lambda_q (TM), \widetilde{\phi_m})\).

Over \(U_m\), \(f = f_{j_1, ..., j_q} d x^{j_1} \wedge... \wedge d x^{j_q}\), where \(f_{j_1, ..., j_q}: U_m \to \mathbb{R}\) is \(C^\infty\), because it is a component of \(\widetilde{\phi_m} \circ f \circ {\phi_m}^{-1} \circ \phi_m\), while the components function of \(f\) with respect to \((U_m \subseteq M, \phi_m)\) and \((\pi^{-1} (U_m) \subseteq \Lambda_q (TM), \widetilde{\phi_m})\), \(\widetilde{\phi_m} \circ f \circ {\phi_m}^{-1}\), is \(C^\infty\) and \(\phi_m\) is \(C^\infty\).

\(V_j = V_j^{l_j} \partial / \partial x^{l_j}\), where \(V_j^{l_j}: U_m \to \mathbb{R}\) is \(C^\infty\).

\(f (V_1, ..., V_q) = \sum_{j_1 \lt ... \lt j_q} f_{j_1, ..., j_q} d x^{j_1} \wedge ... \wedge d x^{j_q} (V_1^{l_1} \partial / \partial x^{l_1}, ..., V_q^{l_q} \partial / \partial x^{l_q}) = \sum_{j_1 \lt ... \lt j_q} f_{j_1, ..., j_q} (m') \sum_{\sigma \in P_{\{1, ..., q\}}} sgn \sigma V_{\sigma_1}^{j_1} ... V_{\sigma_q}^{j_q}\) as before, which is \(C^\infty\).

As \(f (V_1, ..., V_q)\) is \(C^\infty\) over a neighborhood of each \(m \in M\), \(f (V_1, ..., V_q)\) is \(C^\infty\) over \(M\).


References


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