1030: For C∞ Manifold with Boundary and Tangent Vectors Space at Point, Transition of Standard Bases w.r.t. Charts Is This

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description/proof of that for $C^{\mathrm{\infty }}$ manifold with boundary and tangent vectors space at point, transition of standard bases w.r.t. charts is this

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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 tangent vectors space at point on $C^{\mathrm{\infty }}$ manifold with boundary.

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Target Context

• The reader will have a description and a proof of the proposition that for any $C^{\mathrm{\infty }}$ manifold with boundary and the tangent vectors space at any point, the transition of the standard bases with respect to any charts is this.

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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:
$M$: $\in \{\text{ the }d\text{ -dimensional }{C}^{\mathrm{\infty }}\text{ manifolds with boundary }\}$
$m$: $\in M$
$T_{m}M$: $=\text{ the tangent vectors space at }m$
$(U_m\subseteq M,{\varphi }_m)$: $\in \{\text{ the charts for }M\text{ around }m\}$
$(U_m^{\prime }\subseteq M,{\varphi }_m^{\prime })$: $\in \{\text{ the charts for }M\text{ around }m\}$
$B$: $=\text{ the standard basis for }{T}_{m}M$ with respect to $(U_m\subseteq M,{\varphi }_m)$, $=\{\partial /\partial x^1,...,\partial /\partial x^d\}$
$B^{\prime }$: $=\text{ the standard basis for }{T}_{m}M$ with respect to $(U_m^{\prime }\subseteq M,{\varphi }_m^{\prime })$, $=\{\partial /\partial x^{\prime 1},...,\partial /\partial x^{\prime d}\}$
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Statements:
$\partial /\partial x^{\prime j}=\partial x^k/\partial x^{\prime j}\partial /\partial x^k$
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$x$ as a function of $x^{\prime }$ is ${\varphi }_m\circ {{\varphi }_m^{\prime }}^{-1}{|}_{{\varphi }_m^{\prime }(U_m\cap U_m^{\prime })}:{\varphi }_m^{\prime }(U_m\cap U_m^{\prime })\to {\varphi }_m(U_m\cap U_m^{\prime })$.

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2: Proof

Whole Strategy: Step 1: let $\partial /\partial x^{\prime j}=M_j^k\partial /\partial x^k$; Step 2: make the both sides of it operate on $x^l:U_m\cap U_m^{\prime }\to \mathbb{R},p\mapsto {{\varphi }_m(p)}^l$, and see that $M_j^k=\partial x^k/\partial x^{\prime j}$.

Step 1:

Let $\partial /\partial x^{\prime j}=M_j^k\partial /\partial x^k$, which is possible because $B$ is a basis for $T_{m}M$ and $\partial /\partial x^{\prime j}\in T_{m}M$.

Step 2:

$x^l:U_m\cap U_m^{\prime }\to \mathbb{R},p\mapsto {{\varphi }_m(p)}^l$ is a $C^{\mathrm{\infty }}$ function.

$\partial /\partial x^{\prime j}(x^l)=M_j^k\partial /\partial x^k(x^l)=M_j^k{\partial }_k(x^l\circ {{\varphi }_m}^{-1})=M_j^k{\delta }_k^l=M_j^l$. So, $M_j^k=\partial x^k/\partial x^{\prime j}$.

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References

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