339: Curves on Manifold as the C^\infty Right Actions of Curves That Represent Same Vector on Lie Group Represent Same Vector

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A description/proof of that curves on manifold as the $C^{\mathrm{\infty }}$ right actions of curves that represent same vector on Lie group represent same vector

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Topics

About: manifold
About: Lie group
About: right action of Lie group on manifold

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The table of contents of this article

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

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Starting Context

• The reader knows a definition of $C^{\mathrm{\infty }}$ manifold.
• The reader knows a definition of Lie group.
• The reader knows a definition of $C^{\mathrm{\infty }}$ right action of Lie group on $C^{\mathrm{\infty }}$ manifold.

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Target Context

• The reader will have a description and a proof of the proposition that any curves on any $C^{\mathrm{\infty }}$ manifold as the $C^{\mathrm{\infty }}$ right actions of curves that represent the same tangent vector on any Lie group represent the same tangent vector.

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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: Description

For any $C^{\mathrm{\infty }}$ manifold, $M$, with any $C^{\mathrm{\infty }}$ right action, $\mu :M\times G\to M$, of any Lie group, $G$, and any $C^{\mathrm{\infty }}$ curves, $c_1(t)$ and $c_2(t)$, on $G$, that represent the same tangent vector on $G$, the $C^{\mathrm{\infty }}$ curves, $m{c}_1(t)$ and $m{c}_2(t)$ where $m\in M$, on $M$, represent the same tangent vector on $M$.

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2: Proof

$m{c}_i(t)$ is in fact a $C^{\mathrm{\infty }}$ curve, because it is a compound of $C^{\mathrm{\infty }}$ maps as $c_i(t)$ is $C^{\mathrm{\infty }}$ and $\mu$ is $C^{\mathrm{\infty }}$, where m is fixed.

For any chart around $c_1(0)=c_2(0)$ on $G$, ${\frac{d{c}_1^j(t)}{dt}|}_{t=0}={\frac{d{c}_2^j(t)}{dt}|}_{t=0}$ where $c_i^j$ is the $j$ component of $c_i$ by the $G$ chart.

For any chart around $m{c}_1(0)=m{c}_2(0)$ on $M$, ${\frac{d(m{c}_i{)}^j}{dt}|}_{t=0}={\frac{d{\mu }^j(m,c_i)}{dt}|}_{t=0}={\left(\frac{\partial {\mu }^j}{\partial c_i^k}\frac{d{c}_i^k}{dt}\right)|}_{t=0}$ where $(m{c}_i{)}^j={\mu }^j(m,c_i)$ is the $j$ component of $m{c}_i$ by the $M$ chart. As $\frac{\partial {\mu }^j}{\partial c_i^k}$ is a matter of the coordinates function and $c_1(0)=c_2(0)$, ${\frac{\partial {\mu }^j}{\partial c_1^k}|}_{t=0}={\frac{\partial {\mu }^j}{\partial c_2^k}|}_{t=0}$.

As also ${\frac{d{c}_1^j(t)}{dt}|}_{t=0}={\frac{d{c}_2^j(t)}{dt}|}_{t=0}$, ${\frac{d(m{c}_1{)}^j}{dt}|}_{t=0}={\frac{d(m{c}_2{)}^j}{dt}|}_{t=0}$.

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References

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