269: Existence of Lie Group Neighborhood Whose Any Point Can Be Connected with Center by Left-Invariant Vectors Field Integral Curve

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A description/proof of existence of Lie group neighborhood whose any point can be connected with center by left-invariant vectors field integral curve

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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
• 3: Note

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Starting Context

• The reader knows a definition of Lie group.
• The reader knows a definition of Lie algebra.
• The reader knows a definition of left-invariant vectors field.
• The reader knows a definition of integral curve of vectors field.
• The reader knows a definition of exponential map on Lie group.
• The reader admits the proposition that on any Lie group there is a diffeomorphic exponential map between a neighborhood of 0 on the Lie algebra and a neighborhood of e on the Lie group.
• The reader admits the proposition that the integral curve of any left-invariant vectors field starting at any point on any Lie group is the integral curve of the vectors field starting at e, left-multiplied by the point.

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Target Context

• The reader will have a description and a proof of the proposition that at any point on any Lie group, there is a neighborhood whose any point can be connected with the center by a left-invariant vectors field integral curve.

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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, and any point, $p_0\in G$, there is a neighborhood, $U_{p_0}$, such that any point, $p_i\in U_{p_0}$, can be connected with $p_0$ by the integral curve, ${\gamma }_i:\mathbb{R}\to G$, of a left-invariant vectors field, $V_i$, starting at $p_0$.

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2: Proof

There are a neighborhood, $U_0$, on the Lie algebra, $\mathfrak{g}$, and a neighborhood, $U_e$, on G such that the exponential map, $exp:U_0\to U_e$, is a diffeomorphism. $U_{p_0}:=\{\mathrm{\forall }p|p_0^{-1}p\in U_e\}$ is an open set because it is the preimage of $U_e$ by the $C^{\mathrm{\infty }}$ map, $f:G\to G,p\mapsto p_0^{-1}p$. For any $p_i\in U_{p_0}$, there is a vector, $l_{p_0\ast ,e}{V}_{i,e}$, at $p_0$ such that the integral curve of the left-invariant vectors field, $V_i$, starting at $p_0$ passes $p_i$, because by choosing $V_{i,e}$ as $p_0^{-1}{p}_i=exp{V}_{i,e}$, $p_i=p_{0}exp{V}_{i,e}={\varphi }_1(p_0)$ by the proposition that the integral curve of any left-invariant vectors field starting at any point on any Lie group is the integral curve of the vectors field starting at e, left-multiplied by the point.

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3: Note

The 1-to-1-ness from the vectors at $p_0$ to $U_{p_0}$ is is not proven here.

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References

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