340: Parameterized Family of Vectors and Curve Induced by C^\infty Right Action of Lie Group Represent Same Vector If . . .

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A description/proof of that parameterized family of vectors and curve induced by $C^{\mathrm{\infty }}$ right action of Lie group represent same vector if . . .

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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 for any $C^{\mathrm{\infty }}$ manifold, M, with any $C^{\mathrm{\infty }}$ right action of any Lie group, G, and any parameterized family of vectors induced from parameterized family of curves and any curve, both on G, that represent the same vector on G, the parameterized family of vectors induced from the right action of the parameterized family of curves on G and the curve as the right action of the curve on G represent the same vector on M.

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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, $G$, any $C^{\mathrm{\infty }}$ parameterized family of $C^{\mathrm{\infty }}$ curves, $c_1(t,t^{\prime })$, where $t$ is the parameter and $c_1(t,0)=g\in G$ without depending on $t$, on $G$ and any $C^{\mathrm{\infty }}$ curve, $c_2(t)$, on $G$ that represent the same tangent vector on $G$ as $\frac{d}{dt}(\frac{\partial c_1(t,t^{\prime })}{\partial t^{\prime }}{|}_{t^{\prime }=0}){|}_{t=0}=\frac{d{c}_2(t)}{dt}{|}_{t=0}$ where $\frac{d}{dt}$ of the left hand side means derivative of parameterized family of vectors while $\frac{\partial }{\partial t}$ and $\frac{d}{dt}$ of the right hand side mean getting tangent vector of curve, the induced parameterized family of vectors and the curve on $M$ represent the same vector on $M$, that is $\frac{d}{dt}(\frac{\partial m{c}_1(t,t^{\prime })}{\partial t^{\prime }}{|}_{t^{\prime }=0}){|}_{t=0}=\frac{dm{c}_2(t)}{dt}{|}_{t=0}$ where $\frac{d}{dt}$s and $\frac{\partial }{\partial t^{\prime }}$ mean likewise.

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2: Proof

$m{c}_1(t,t^{\prime })$ is in fact a $C^{\mathrm{\infty }}$ parameterized family of $C^{\mathrm{\infty }}$ curves, because it is a compound of $C^{\mathrm{\infty }}$ maps as $c_1(t,t^{\prime })$ is $C^{\mathrm{\infty }}$ and $\mu$ is $C^{\mathrm{\infty }}$, where $m$ is fixed.

$m{c}_2(t)$ is in fact a $C^{\mathrm{\infty }}$ curve, because it is a compound of $C^{\mathrm{\infty }}$ maps as $c_2(t)$ is $C^{\mathrm{\infty }}$ and $\mu$ is $C^{\mathrm{\infty }}$, where $m$ is fixed.

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

For any chart around $m{c}_1(t,0)=m{c}_2(0)$ on $M$, the $j$ component, ${(\frac{d}{dt}(\frac{\partial m{c}_1(t,t^{\prime })}{\partial t^{\prime }}{|}_{t^{\prime }=0}){|}_{t=0})}^j=\frac{d}{dt}(\frac{\partial (m{c}_1(t,t^{\prime }){)}^j}{\partial t^{\prime }}{|}_{t^{\prime }=0}){|}_{t=0}=\frac{d}{dt}(\frac{\partial {\mu }^j(m,c_1(t,t^{\prime }))}{\partial t^{\prime }}{|}_{t^{\prime }=0}){|}_{t=0}=\frac{d}{dt}(\frac{{\mu }^j(m,c_1(t,t^{\prime }))}{\partial c_1^i}\frac{c_1^i(t,t^{\prime })}{\partial t^{\prime }}{|}_{t^{\prime }=0}){|}_{t=0}$, but $\frac{\partial {\mu }^j}{\partial c_1^i}$ is a function of $m$ and $c_1$ and $c_1(t,0)$ does not depend on $t$, so, $=\frac{\partial {\mu }^j(m,c_1)}{\partial c_1^i}\frac{d}{dt}(\frac{\partial c_1^i(t,t^{\prime })}{\partial t^{\prime }}{|}_{t^{\prime }=0}){|}_{t=0}=\frac{\partial {\mu }^j(m,c_1)}{\partial c_1^i}{|}_{c_1=g}\frac{d{c}_2^i(t)}{dt}{|}_{t=0}$. On the other hand, ${\left(\frac{dm{c}_2(t)}{dt}{|}_{t=0}\right)}^j=\frac{d(m{c}_2(t){)}^j}{dt}{|}_{t=0}=\frac{d{\mu }^j(m,c_2(t))}{dt}{|}_{t=0}=\frac{\partial {\mu }^j(m,c_2)}{\partial c_2^i}{|}_{c_2=g}\frac{d{c}_2^i(t)}{dt}{|}_{t=0}$. As $\frac{\partial {\mu }^j}{\partial c_k^i}$ is a matter of the coordinates function, ${\frac{\partial {\mu }^j(m,c_1)}{\partial c_1^i}|}_{c_1=g}={\frac{\partial {\mu }^j(m,c_2)}{\partial c_2^i}|}_{c_2=g}$.

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References

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