771: Tangent Vector at Point on C∞ Manifold with Boundary Is Velocity of C∞ Curve, Especially from Half Closed Interval, Especially as Linear in Coordinates

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description/proof of that tangent vector at point on $C^{\mathrm{\infty }}$ manifold with boundary is velocity of $C^{\mathrm{\infty }}$ curve, especially from half closed interval, especially as linear in coordinates

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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: Natural Language Description
• 3: Note
• 4: Proof

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Starting Context

• The reader knows a definition of tangent vector.
• The reader knows a definition of velocity of $C^{\mathrm{\infty }}$ curve at point on $C^{\mathrm{\infty }}$ manifold with boundary.
• The reader admits the proposition that for any $C^{\mathrm{\infty }}$ manifold with boundary and any chart, the standard basis for the tangent vectors space at each point on the chart domain is indeed a basis.

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Target Context

• The reader will have a description and a proof of the proposition that any tangent vector at any point on any $C^{\mathrm{\infty }}$ manifold with boundary is the velocity of a $C^{\mathrm{\infty }}$ curve, especially from a half closed interval, especially as linear in coordinates.

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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:
$M$: $\in \{\text{ the }d\text{ -dimensional }{C}^{\mathrm{\infty }}\text{ manifolds with boundary }\}$
$m$: $\in M$
$v$: $\in T_{m}M$
$\mathbb{R}$: $=\text{ the Euclidean }{C}^{\mathrm{\infty }}\text{ manifold }$
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Statements:
$\mathrm{\exists }J=[t_0,t_2)\text{ or }(t_1,t_0]\subseteq \mathbb{R},\mathrm{\exists }\gamma :J\to M\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ curves on }M\}(v=d\gamma (d/dt{|}_{t_0}))$
$\wedge$
Especially, $\gamma$ can be take to be such that $t\mapsto {{\varphi }_m}^{-1}(({\varphi }_m(m{)}^1+v^1(t-t_0),...,{\varphi }_m(m{)}^d+v^d(t-t_0)))$, where $(U_m\subseteq M,{\varphi }_m)$ is any chart around $m$ and $v^j$ s are the components of $v$ with respect to the standard basis for the chart
//

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2: Natural Language Description

For any $d$-dimensional $C^{\mathrm{\infty }}$ manifold with boundary, $M$, any point, $m\in M$, and any tangent vector at $m$, $v\in T_{m}M$, there are an interval, $J=[t_0,t_2)\text{ or }(t_1,t_0]\subseteq \mathbb{R}$, and a $C^{\mathrm{\infty }}$ curve, $\gamma :J\to M$, such that $v=d\gamma (d/dt{|}_{t_0})$, and especially, $\gamma$ can be take to be such that $t\mapsto {{\varphi }_m}^{-1}(({\varphi }_m(m{)}^1+v^1(t-t_0),...,{\varphi }_m(m{)}^d+v^d(t-t_0)))$, where $(U_m\subseteq M,{\varphi }_m)$ is any chart around $m$ and $v^j$ s are the components of $v$ with respect to the standard basis for the chart.

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3: Note

A purpose of choosing a non-open interval is to be applicable for all the cases: when $m$ is a boundary point, we may not be able to choose any curve from open interval.

Of course, there are some cases for which $J$ can be an open interval, if you want.

Another purpose is to narrow the area only where the values of $f\in C^{\mathrm{\infty }}(M)$ matter for $vf$: as we can take $\gamma$ as $t\mapsto {{\varphi }_m}^{-1}(({\varphi }_m(m{)}^1+v^1(t-t_0),...,{\varphi }_m(m{)}^d+v^d(t-t_0)))$ while $0\le v^d(t-t_0)$, $vf$ depends on the values of $f$ only on the upper half of the chart domain (even when $m$ is not any boundary point), so to speak.

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4: Proof

Whole Strategy: Step 1: choose any chart around $m$, $(U_m\subseteq M,{\varphi }_m)$; Step 2: express $v$ as $v^j\partial /\partial x^j$; Step 3: choose a $J=[t_0,t_2)\text{ or }(t_1,t_0]$ according to $0\le v^j$ or $v^j<0$ and $\gamma$ as $t\mapsto {{\varphi }_m}^{-1}(({\varphi }_m(m{)}^1+v^1(t-t_0),...,{\varphi }_m(m{)}^d+v^d(t-t_0)))$; Step 4: see that $v=d\gamma (d/dt{|}_{t_0})$; Step 5: make some observations.

Step 1:

Let us choose any chart around $m$, $(U_m\subseteq M,{\varphi }_m)$.

Let $x=(x^1,...,x^d)$ be the coordinates on ${\varphi }_m(U_m)$.

Step 2:

$v=v^j\partial /\partial x^j$, by the proposition that for any $C^{\mathrm{\infty }}$ manifold with boundary and any chart, the standard basis for the tangent vectors space at each point on the chart domain is indeed a basis.

Step 3:

Let us choose $t_0\in \mathbb{R}$ arbitrarily.

When $0\le v^d$, let us choose any $t_2$ such that $t_0<t_2$ and $({\varphi }_m(m{)}^1+v^1(t-t_0),...,{\varphi }_m(m{)}^d+v^d(t-t_0))\in {\varphi }_m(U_m)$ for each $t\in [t_0,t_2)$, which is possible because there is an open ball, $B_{{\varphi }_m(m),ϵ}\subseteq \mathbb{R}\text{ or }\mathbb{H}$, such that $B_{{\varphi }_m(m),ϵ}\subseteq {\varphi }_m(U_m)$ and by taking a small enough $t_2-t_0$, $({\varphi }_m(m{)}^1+v^1(t-t_0),...,{\varphi }_m(m{)}^d+v^d(t-t_0))\in B_{{\varphi }_m(m),ϵ}$: $0\le v^d(t-t_0)$, so, even when $m$ is a boundary point, it holds.

When $v^d<0$, let us choose any $t_1$ such that $t_1<t_0$ and $({\varphi }_m(m{)}^1+v^1(t-t_0),...,{\varphi }_m(m{)}^d+v^d(t-t_0))\in {\varphi }_m(U_m)$ for each $t\in (t_1,t_0]$, which is possible because there is an open ball, $B_{{\varphi }_m(m),ϵ}\subseteq \mathbb{R}\text{ or }\mathbb{H}$, such that $B_{{\varphi }_m(m),ϵ}\subseteq {\varphi }_m(U_m)$ and by taking a small enough $t_0-t_1$, $({\varphi }_m(m{)}^1+v^1(t-t_0),...,{\varphi }_m(m{)}^d+v^d(t-t_0))\in B_{{\varphi }_m(m),ϵ}$: $0\le v^d(t-t_0)$, so, even when $m$ is a boundary point, it holds.

Let us define $\gamma :J\to M,t\mapsto {{\varphi }_m}^{-1}(({\varphi }_m(m{)}^1+v^1(t-t_0),...,{\varphi }_m(m{)}^d+v^d(t-t_0)))$.

$\gamma$ is well-defined because $({\varphi }_m(m{)}^1+v^1(t-t_0),...,{\varphi }_m(m{)}^d+v^d(t-t_0))\subseteq {\varphi }_m(U_m)$.

$\gamma$ is $C^{\mathrm{\infty }}$, because $J\to {\varphi }_m(U_m),t\mapsto ({\varphi }_m(m{)}^1+v^1(t-t_0),...,{\varphi }_m(m{)}^d+v^d(t-t_0))$ is obviously $C^{\mathrm{\infty }}$ and ${{\varphi }_m}^{-1}:{\varphi }_m(U_m)\to M$ is $C^{\mathrm{\infty }}$.

Step 4:

Let us see that $v=d\gamma (d/dt{|}_{t_0})$.

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

$d\gamma (d/dt{|}_{t_0})(f)=d/dt{|}_{t_0}(f\circ \gamma )=d/dt{|}_{t_0}(f\circ {{\varphi }_m}^{-1}(({\varphi }_m(m{)}^1+v^1(t-t_0),...,{\varphi }_m(m{)}^d+v^d(t-t_0))))={\partial }_j(f\circ {{\varphi }_m}^{-1}){|}_{{\varphi }_m(m)}d({\varphi }_m(m{)}^j+v^j(t-t_0))/dt{|}_{t_0}$, by the chain rule (in fact, ${\partial }_d$ and $d/dt{|}_{t_0}$ are one-sided), $=v^j{\partial }_j(f\circ {{\varphi }_m}^{-1}){|}_{{\varphi }_m(m)}$.

$vf=v^j\partial /\partial x^j(f)=v^j{\partial }_j(f\circ {{\varphi }_m}^{-1}){|}_{{\varphi }_m(m)}$: see what any chart induced basis vector on any $C^{\mathrm{\infty }}$ manifold with boundary is.

So, $v=d\gamma (d/dt{|}_{t_0})$.

Step 5:

Let us make some observations.

${\varphi }_m\circ \gamma :J\to {\varphi }_m(U_m)$ is $t\mapsto ({\varphi }_m(m{)}^1+v^1(t-t_0),...,{\varphi }_m(m{)}^d+v^d(t-t_0))$, which is what is meant by $\gamma$'s being "linear in coordinates".

As there can be some multiple curves with the same velocity at $m$, we can take a curve that is not linear in coordinates.

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References

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