1171: q-Form over C∞ Manifold with Boundary Is C∞ iff Operation Result on Any C∞ Vectors Fields Is C∞

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

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Topics

About: $C^{\mathrm{\infty }}$ manifold

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

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Starting Context

• The reader knows a definition of $q$-form over $C^{\mathrm{\infty }}$ manifold with boundary.
• The reader knows a definition of $r^{\prime }$-$r$-open-balls charts pair around point on $C^{\mathrm{\infty }}$ manifold with boundary.
• The reader knows a definition of $r^{\prime }$-$r$-open-half-balls charts pair around point on $C^{\mathrm{\infty }}$ manifold with boundary.
• The reader admits the proposition that any $(0,q)$-tensors field over $C^{\mathrm{\infty }}$ manifold with boundary is $C^{\mathrm{\infty }}$ if and only if the operation result on any $C^{\mathrm{\infty }}$ vectors fields is $C^{\mathrm{\infty }}$.
• The reader admits the proposition that for any $C^{\mathrm{\infty }}$ manifold with boundary, each interior point has an $r^{\prime }$-$r$-open-balls charts pair and each boundary point has an $r^{\prime }$-$r$-open-half-balls charts pair for any positive $r^{\prime }$ and $r$.
• The reader admits the proposition that for any $C^{\mathrm{\infty }}$ vectors bundle, any $C^{\mathrm{\infty }}$ 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.

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Target Context

• The reader will have a description and a proof of the proposition that any $q$-form over any $C^{\mathrm{\infty }}$ manifold with boundary is $C^{\mathrm{\infty }}$ if and only if the operation result on any $C^{\mathrm{\infty }}$ vectors fields is $C^{\mathrm{\infty }}$.

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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.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$M$: $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ manifolds with boundary }\}$
$q$: $\in \mathbb{N}\setminus \{0\}$
$(T_q^0(TM),M,\pi )$: $=(0,q)\text{ -tensors bundle over }M$
$({\mathrm{\Lambda }}_q(TM),M,\pi )$: $=q\text{ -covectors bundle over }M$
$f$: $:M\to T_q^0(TM)$ such that $Ran(f)\subseteq {\mathrm{\Lambda }}_q(TM)$ or $:M\to {\mathrm{\Lambda }}_q(TM)$, $\in \{\text{ the sections of }\pi \}$
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Statements:
$f\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ maps }\}$
$⟺$
$\mathrm{\forall }{V}_1,...,V_q\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ vectors fields over }M\}(f(V_1,...,V_q):M\to \mathbb{R}\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ maps }\})$
//

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2: Proof

Whole Strategy: Step 1: when $:M\to T_q^0(TM)$, see that the proposition holds; Step 2: suppose that $:M\to {\mathrm{\Lambda }}_q(TM)$; Step 3: suppose that $f(V_1,...,V_q)\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ maps }\}$, and see that $f$ is $C^{\mathrm{\infty }}$, by taking an $r^{\prime }$-$r$-open-balls charts pair or an $r^{\prime }$-$r$-open-half-balls charts pair around each $m\in M$, $(U_m^{\prime }\subseteq M,{\varphi }_m^{\prime })$ and $(U_m\subseteq M,{\varphi }_m)$, and the induced chart, $({\pi }^{-1}(U_m)\subseteq {\mathrm{\Lambda }}_q(TM),\widetilde{{\varphi }_m})$, and seeing that the components function of $f$ is $C^{\mathrm{\infty }}$; Step 4: suppose that $f$ is $C^{\mathrm{\infty }}$, and see that $f(V_1,...,V_q)$ is $C^{\mathrm{\infty }}$ over a chart, $(U_m\subseteq M,{\varphi }_m)$, around each $m\in M$.

Step 1:

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

Step 2:

Let us suppose that $f$ is $:M\to {\mathrm{\Lambda }}_q(TM)$.

Step 3:

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

Let $m\in M$ be any.

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

Let us take $V_j=V_j^{l_j}\partial /\partial x^{l_j}$ over $U_m^{\prime }$ as $V_j^{l_j^{\prime }}\equiv 1$ and $V_j^{l_j}\equiv 0$ for each $l_j\ne l_j^{\prime }$ where $l_1^{\prime }<...<l_q^{\prime }$. $V_j$ is $C^{\mathrm{\infty }}$ over $U_m^{\prime }$. $V_j$ is $C^{\mathrm{\infty }}$ over $\overline{U_m}\subseteq U_m^{\prime }$. By the proposition that for any $C^{\mathrm{\infty }}$ vectors bundle, any $C^{\mathrm{\infty }}$ 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 $m^{\prime }\in U_m$ be any.

$f(m^{\prime })=\sum _{j_1<...<j_q}{f}_{j_1,...,j_q}(m^{\prime })d{x}^{j_1}\wedge ...\wedge d{x}^{j_q}$.

$f(m^{\prime })(V_1,...,V_q)=\sum _{j_1<...<j_q}{f}_{j_1,...,j_q}(m^{\prime })d{x}^{j_1}\wedge ...\wedge d{x}^{j_q}(V_1^{l_1}\partial /\partial x^{l_1},...,V_q^{l_q}\partial /\partial x^{l_q})=\sum _{j_1<...<j_q}{f}_{j_1,...,j_q}(m^{\prime })q!Asymd{x}^{j_1}\otimes ...\otimes d{x}^{j_q}(V_1^{l_1}\partial /\partial x^{l_1},...,V_q^{l_q}\partial /\partial x^{l_q})=\sum _{j_1<...<j_q}{f}_{j_1,...,j_q}(m^{\prime })q!1/q!\sum _{\sigma \in P_{\{1,...,q\}}}sgn\sigma d{x}^{j_1}\otimes ...\otimes d{x}^{j_q}(V_{{\sigma }_1}^{l_{{\sigma }_1}}\partial /\partial x^{l_{{\sigma }_1}},...,V_{{\sigma }_q}^{l_{{\sigma }_q}}\partial /\partial x^{l_{{\sigma }_q}})=\sum _{j_1<...<j_q}{f}_{j_1,...,j_q}(m^{\prime })\sum _{\sigma \in P_{\{1,...,q\}}}sgn\sigma d{x}^{j_1}(V_{{\sigma }_1}^{l_{{\sigma }_1}}\partial /\partial x^{l_{{\sigma }_1}})...d{x}^{j_q}(V_{{\sigma }_q}^{l_{{\sigma }_q}}\partial /\partial x^{l_{{\sigma }_q}})=\sum _{j_1<...<j_q}{f}_{j_1,...,j_q}(m^{\prime })\sum _{\sigma \in P_{\{1,...,q\}}}sgn\sigma V_{{\sigma }_1}^{l_{{\sigma }_1}}{\delta }_{l_{{\sigma }_1}}^{j_1}...V_{{\sigma }_q}^{l_{{\sigma }_q}}{\delta }_{l_{{\sigma }_q}}^{j_q}=\sum _{j_1<...<j_q}{f}_{j_1,...,j_q}(m^{\prime })\sum _{\sigma \in P_{\{1,...,q\}}}sgn\sigma V_{{\sigma }_1}^{j_1}...V_{{\sigma }_q}^{j_q}=\sum _{j_1<...<j_q}{f}_{j_1,...,j_q}(m^{\prime })\sum _{\sigma \in P_{\{1,...,q\}}}sgn\sigma V_1^{j_{{\sigma }^{-1}(1)}}...V_q^{j_{{\sigma }^{-1}(q)}}$: $V_{{\sigma }_1}^{j_1}...V_{{\sigma }_q}^{j_q}$ 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 $l_1^{\prime }<...<l_q^{\prime }$.

So, $=\sum _{j_1<...<j_q}{f}_{j_1,...,j_q}(m^{\prime })V_1^{j_1}...V_q^{j_q}=f_{l_1^{\prime },...,l_q^{\prime }}(m^{\prime })$.

$f(V_1,...,V_q):U_m\to \mathbb{R}$ is $C^{\mathrm{\infty }}$ by the supposition, so, $f_{l_1^{\prime },...,l_q^{\prime }}:U_m\to \mathbb{R}$ is $C^{\mathrm{\infty }}$.

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

So, $f$ is $C^{\mathrm{\infty }}$.

Step 4:

Let us suppose that $f$ is $C^{\mathrm{\infty }}$.

Let $m\in M$ be any.

Let us take any chart around $m$, $(U_m\subseteq M,{\varphi }_m)$, and the induced chart, $({\pi }^{-1}(U_m)\subseteq {\mathrm{\Lambda }}_q(TM),\widetilde{{\varphi }_m})$.

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 $C^{\mathrm{\infty }}$, because it is a component of $\widetilde{{\varphi }_m}\circ f\circ {{\varphi }_m}^{-1}\circ {\varphi }_m$, while the components function of $f$ with respect to $(U_m\subseteq M,{\varphi }_m)$ and $({\pi }^{-1}(U_m)\subseteq {\mathrm{\Lambda }}_q(TM),\widetilde{{\varphi }_m})$, $\widetilde{{\varphi }_m}\circ f\circ {{\varphi }_m}^{-1}$, is $C^{\mathrm{\infty }}$ and ${\varphi }_m$ is $C^{\mathrm{\infty }}$.

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

$f(V_1,...,V_q)=\sum _{j_1<...<j_q}{f}_{j_1,...,j_q}d{x}^{j_1}\wedge ...\wedge d{x}^{j_q}(V_1^{l_1}\partial /\partial x^{l_1},...,V_q^{l_q}\partial /\partial x^{l_q})=\sum _{j_1<...<j_q}{f}_{j_1,...,j_q}(m^{\prime })\sum _{\sigma \in P_{\{1,...,q\}}}sgn\sigma V_{{\sigma }_1}^{j_1}...V_{{\sigma }_q}^{j_q}$ as before, which is $C^{\mathrm{\infty }}$.

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

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References

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