1042: For C∞ Map Between C∞ Manifolds with Boundary and Corresponding Charts, Components Function of Differential w.r.t. Standard Bases Is This

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description/proof of that for $C^{\mathrm{\infty }}$ map between $C^{\mathrm{\infty }}$ manifolds with boundary and corresponding charts, components function of differential w.r.t. standard bases 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 differential of $C^{\mathrm{\infty }}$ map between $C^{\mathrm{\infty }}$ manifolds with boundary at point.

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Target Context

• The reader will have a description and a proof of the proposition that for any $C^{\mathrm{\infty }}$ map between any $C^{\mathrm{\infty }}$ manifolds with boundary and any corresponding charts, the components function of the differential of the map with respect to the standard bases 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:
$M_1$: $\in \{\text{ the }{d}_1\text{ -dimensional }{C}^{\mathrm{\infty }}\text{ manifolds with boundary }\}$
$M_2$: $\in \{\text{ the }{d}_2\text{ -dimensional }{C}^{\mathrm{\infty }}\text{ manifolds with boundary }\}$
$f$: $:M_1\to M_2$, $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ maps }\}$
$(U_1\subseteq M_1,{\varphi }_1)$: $\in \{\text{ the charts for }{M}_1\}$
$(U_2\subseteq M_2,{\varphi }_2)$: $\in \{\text{ the charts for }{M}_2\}$, such that $f(U_1)\subseteq U_2$
$m$: $\in U_1$
$d{f}_m$: $:T_m{M}_1\to T_{f(m)}{M}_2$, $=\text{ the differential }$
$\{\partial /\partial x^j|j\in \{1,...,d_1\}\}$: $=\text{ the standard basis for }{T}_m{M}_1$ with respect to $(U_1\subseteq M_1,{\varphi }_1)$
$\{\partial /\partial y^j|j\in \{1,...,d_2\}\}$: $=\text{ the standard basis for }{T}_{f(m)}{M}_2$ with respect to $(U_2\subseteq M_2,{\varphi }_2)$
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Statements:
$\mathrm{\forall }v=v^j\partial /\partial x^j\in T_m{M}_1(d{f}_m(v)=w^j\partial /\partial y^j\text{ where }{w}^j=\partial {\hat{f}}^j/\partial x^l{v}^l\text{ where }\hat{f}={\varphi }_2\circ f\circ {{\varphi }_1}^{-1})$
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In other words, $\hat{f}$ is the components function of $f$ with respect to $(U_1\subseteq M_1,{\varphi }_1)$ and $(U_2\subseteq M_2,{\varphi }_2)$.

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2: Proof

Whole Strategy: Step 1: for any $f^{\prime }\in C^{\mathrm{\infty }}(M_2)$, compute $d{f}_m(v)(f^{\prime })$ as $v(f^{\prime }\circ f)$; Step 2: compare the result of Step 1 with $d{f}_m(v)(f^{\prime })=w^j\partial /\partial y^j(f^{\prime })$.

Step 1:

Let $f^{\prime }\in C^{\mathrm{\infty }}(M_2)$ be any.

$d{f}_m(v)(f^{\prime })=v(f^{\prime }\circ f)=v^j\partial /\partial x^j(f^{\prime }\circ f)=v^j{\partial }_j(f^{\prime }\circ f\circ {{\varphi }_1}^{-1})$, where $f^{\prime }\circ f\circ {{\varphi }_1}^{-1}$ is $:{\varphi }_1(U_1)\subseteq {\mathbb{R}}^{d_1}\to \mathbb{R}$, $=v^j{\partial }_j((f^{\prime }\circ {{\varphi }_2}^{-1})\circ ({\varphi }_2\circ f\circ {{\varphi }_1}^{-1}))$, where $(f^{\prime }\circ {{\varphi }_2}^{-1})\circ ({\varphi }_2\circ f\circ {{\varphi }_1}^{-1})$ is $:{\varphi }_1(U_1)\subseteq {\mathbb{R}}^{d_1}\to {\varphi }_2(U_2)\subseteq {\mathbb{R}}^{d_2}\to \mathbb{R}$, and applying the chain rule, $=v^j{\partial }_l(f^{\prime }\circ {{\varphi }_2}^{-1}){\partial }_j({\varphi }_2\circ f\circ {{\varphi }_1}^{-1}{)}^l=v^j{\partial }_l(f^{\prime }\circ {{\varphi }_2}^{-1}){\partial }_j{\hat{f}}^l={\partial }_j{\hat{f}}^l{v}^j\partial /\partial y^l(f^{\prime })={\partial }_l{\hat{f}}^j{v}^l\partial /\partial y^j(f^{\prime })$, which is just renaming the dummy indexes.

Step 2:

$d{f}_m(v)(f^{\prime })=w^j\partial /\partial y^j(f^{\prime })$.

So, comparing that with the result of Step 1, $w^j\partial /\partial y^j(f^{\prime })={\partial }_l{\hat{f}}^j{v}^l\partial /\partial y^j(f^{\prime })$.

As $f^{\prime }$ is arbitrary and $\{\partial /\partial y^1,...,\partial /\partial y^{d_2}\}$ is a basis, $w^j={\partial }_l{\hat{f}}^j{v}^l$.

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References

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