272: Left-Invariant Vectors Field on Lie Group Is C^infty

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A description/proof of that left-invariant vectors field on Lie group is $C^{\mathrm{\infty }}$

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Topics

About: Lie group
About: vectors field

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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 Lie group.
• The reader knows a definition of $C^{\mathrm{\infty }}$ vectors field.
• The reader admits the proposition that any vectors field is $C^{\mathrm{\infty }}$ if and only if the operation result function on any $C^{\mathrm{\infty }}$ function is $C^{\mathrm{\infty }}$.
• The reader admits the chain rule for differentiation of map.

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Target Context

• The reader will have a description and a proof of the proposition that any left-invariant vectors field on any Lie group is $C^{\mathrm{\infty }}$.

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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 Lie group, G, any left-invariant vectors field, V, is $C^{\mathrm{\infty }}$.

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2: Proof

There is a $C^{\mathrm{\infty }}$ curve, $c:I\to G$, such that $c(0)=e$ and $c^{\prime }(0)=V_e$. For any $g\in G$, gc (t) is a $C^{\mathrm{\infty }}$ curve such that $gc(0)=g$ and $(gc{)}^{\prime }(0)=V_g$, because $(gc{)}^{\prime }(0)=l_{g\ast }{c}^{\prime }(0)=l_{g\ast }{V}_e=V_g$ by the chain rule for differentiation of map. For any $C^{\mathrm{\infty }}$ function, f, $(Vf)(g)=V_{g}f=\frac{df(gc(t))}{dt}{|}_0:=A_1$. $f(gc(t)):I\times G\to \mathbb{R},(t,g)\mapsto f(gc(t))$ is a $C^{\mathrm{\infty }}$ function as a composition of $C^{\mathrm{\infty }}$ maps, $I\times G\to G\times G\to G\to \mathbb{R}$, and $A_1$ is a $C^{\mathrm{\infty }}$ function with respect to g. As the operation result on any $C^{\mathrm{\infty }}$ function is $C^{\mathrm{\infty }}$, V is $C^{\mathrm{\infty }}$.

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References

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