480: What Chart Induced Basis Vector on C∞ Manifold with Boundary Is

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A description of what chart induced basis vector on $C^{\mathrm{\infty }}$ manifold with boundary is

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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: Note
• 2: Description

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Starting Context

• The reader knows a definition of $C^{\mathrm{\infty }}$ manifold with boundary.
• The reader knows a definition of map between arbitrary subsets of $C^{\mathrm{\infty }}$ manifolds with boundary $C^k$ at point, where $k$ excludes $0$ and includes $\mathrm{\infty }$.

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Target Context

• The reader will have a description of what any chart induced basis vector on any $C^{\mathrm{\infty }}$ manifold with boundary is.

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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: Note

The conclusion is in fact what will be rather prevalently guessed intuitively, but this is about more rigorously confirming that that guess is correct.

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2: Description

For any $C^{\mathrm{\infty }}$ manifold with (possibly empty) boundary, $M$, and any chart, $(U\subseteq M,\varphi )$, what is the chart induced basis vector, $\partial /\partial x^j{|}_p$, at $p\in U$?

This argument is profusely based on a definition of map between arbitrary subsets of $C^{\mathrm{\infty }}$ manifolds with boundary $C^k$ at point, where $k$ excludes $0$ and includes $\mathrm{\infty }$.

$\partial /\partial x^j{|}_p$ is defined to be such that for any $C^{\mathrm{\infty }}$ function, $f:M\to \mathbb{R}$, $\partial /\partial x^j{|}_p(f)=\partial (f\circ {\varphi }^{-1}{)}^{\prime }/\partial x^j{|}_{\varphi (p)}$, where $(f\circ {\varphi }^{-1}{)}^{\prime }:U_{\varphi (p)}^{\prime }\to \mathbb{R}$ is any $C^{\mathrm{\infty }}$ extension of $f\circ {\varphi }^{-1}$, where $U_{\varphi (p)}^{\prime }\subseteq {\mathbb{R}}^d$ is open, such that $(f\circ {\varphi }^{-1}{)}^{\prime }{|}_{U_{\varphi (p)}^{\prime }\cap \varphi (U)}=f\circ {\varphi }^{-1}{|}_{U_{\varphi (p)}^{\prime }\cap \varphi (U)}$.

The result does not really depend on the extension, because while $\varphi (U)\subseteq {\mathbb{H}}^d$ is an open subset of ${\mathbb{H}}^d$, $\partial (f\circ {\varphi }^{-1}{)}^{\prime }/\partial x^j{|}_{\varphi (p)}$ has to equal the one-side partial derivative (when $p$ is on the boundary and $j=d$) or the full partial derivative (otherwise), $\partial (f\circ {\varphi }^{-1})/\partial x^j{|}_{\varphi (p)}$.

Let us confirm that $\partial /\partial x^j{|}_p$ is indeed a derivation. $\partial /\partial x^j{|}_p(fg)=\partial ((fg)\circ {\varphi }^{-1}{)}^{\prime }/\partial x^j{|}_{\varphi (p)}$. But we can take $((fg)\circ {\varphi }^{-1}{)}^{\prime }=(f\circ {\varphi }^{-1}{)}^{\prime }(g\circ {\varphi }^{-1}{)}^{\prime }$, because while there are some $C^{\mathrm{\infty }}$ extensions, $(f\circ {\varphi }^{-1}{)}^{\prime }:U_{f,\varphi (p)}^{\prime }\to \mathbb{R}$ and $(g\circ {\varphi }^{-1}{)}^{\prime }:U_{g,\varphi (p)}^{\prime }\to \mathbb{R}$, $(f\circ {\varphi }^{-1}{)}^{\prime }{|}_{U_{f,\varphi (p)}^{\prime }\cap U_{g,\varphi (p)}^{\prime }}(g\circ {\varphi }^{-1}{)}^{\prime }{|}_{U_{f,\varphi (p)}^{\prime }\cap U_{g,\varphi (p)}^{\prime }}:U_{f,\varphi (p)}^{\prime }\cap U_{g,\varphi (p)}^{\prime }\to \mathbb{R}$ is a $C^{\mathrm{\infty }}$ extension of $fg$. So, $\partial ((fg)\circ {\varphi }^{-1}{)}^{\prime }/\partial x^j{|}_{\varphi (p)}=\partial ((f\circ {\varphi }^{-1}{)}^{\prime }(g\circ {\varphi }^{-1}{)}^{\prime })/\partial x^j{|}_{\varphi (p)}=\partial (f\circ {\varphi }^{-1}{)}^{\prime }/\partial x^j{|}_{\varphi (p)}(g\circ {\varphi }^{-1}{)}^{\prime }(\varphi (p))+(f\circ {\varphi }^{-1}{)}^{\prime }(\varphi (p))\partial (g\circ {\varphi }^{-1}{)}^{\prime }/\partial x^j{|}_{\varphi (p)}=\partial /\partial x^j{|}_p(f)g(p)+f(p)\partial /\partial x^j{|}_p(g)$.

$(\partial /\partial x^1{|}_p,...,\partial /\partial x^d{|}_p)$ is linearly independent, because for $c_1\partial /\partial x^1{|}_p+...+c_d\partial /\partial x^d{|}_p=0$, $(c_1\partial /\partial x^1{|}_p+...+c_d\partial /\partial x^d{|}_p)x^j=c_j=0x^j=0$.

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References

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