270: 2 Points on Connected Lie Group Can Be Connected by Finite Left-Invariant Vectors Field Integral Curve Segments

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A description/proof of that 2 points on connected Lie group can be connected by finite left-invariant vectors field integral curve segments

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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 connected topological space.
• The reader knows a definition of path-connected topological space.
• The reader knows a definition of Lie algebra.
• The reader knows a definition of diffeomorphism.
• 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 any connected topological manifold is path-connected.
• 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 any 2 points on any connected Lie group can be connected by a finite number of segments each of which is of an integral curve of a left-invariant vectors field.

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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 connected Lie Group, G, any 2 points, $p_1,p_2\in G$, can be connected by a finite number of segments each of which is of an integral curve, ${\gamma }_i:\mathbb{R}\to G$, of a left-invariant vectors field, $V_i$.

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2: Proof

As G is a $C^{\mathrm{\infty }}$ manifold and is connected, it is path-connected, so take a path, ${\gamma }_1:[r_1,r_2]\to G$, that connects $e$ to $p_1$ as the terminals where e is the unit point of G. 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. At any point, $p_i\in G$, $U_{p_i}:=\{\mathrm{\forall }p|p_i^{-1}p\in U_e\}$ is an open set because it is the preimage of $U_e$ by the $C^{\mathrm{\infty }}$ map, $f:G\to G,p\mapsto p_i^{-1}p$. For any $p\in U_{p_i}$, there is a vector, $l_{p_i\ast ,e}{V}_e$, at $p_i$ such that the integral curve of the left-invariant vectors field, V, starting at $p_i$ passes p, because by choosing $V_e$ as $p_i^{-1}p=exp{V}_e$, $p=p_{i}exp{V}_e={\varphi }_1(p_i)$ 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. Now, ${\gamma }_1$ can be covered by such open sets, and as the path is compact, there is a finite sub cover, $U_{p_{1-0}}=U_e,U_{p_{1-1}},...,U_{p_{1-k}},U_{p_{1-k+1}}=U_{p_1}$. $U_{p_{1-0}}$ shares a point with one of the rest open sets, because otherwise, in the ${\gamma }_1$ subspace topology, $U_{a-0}:=U_{p_{1-0}}\cap {\gamma }_1$ and $U_{r-0}:=({\cup }_{1-i\ne 1-0}{U}_{p_{1-i}})\cap {\gamma }_1$ are open and they would be disjoint and $U_{a-0}\cup U_{r-0}={\gamma }_1$, which means that ${\gamma }_1$ would be disconnected, a contradiction ($\gamma$ is connected as a continuous image of the connected $[r_1,r_2]$). Denoting one of the possibly multiple sharing open sets by $U_{p_{1-c1}}$ and the shared point by $p_{1-s1}$, $p_{1-s1}$ is in $U_{p_{1-0}}$ and so can be connected to e by the integral curve of a left-invariant vectors field and is also in $U_{p_{1-c1}}$ and so can be connected to $p_{1-c1}$ by the integral curves of a left-invariant vectors field; if $p_{1-c1}=p_1$, e is already connected to $p_1$, if not, $U_{p_{1-0}}\cup U_{p_{1-c1}}$ shares a point with one of the rest open sets and denoting one of the possibly multiple sharing open sets by $U_{p_{1-c2}}$ and the shared point by $p_{1-s2}$, $p_{1-s2}$ is in $U_{p_{1-0}}$ or $U_{p_{1-c1}}$ and so can be connected to e or $p_{1-c1}$ and is also in $U_{p_{1-c2}}$ and so can be connected to $p_{1-c2}$; if $p_{1-c2}=p_1$, e is already connected to $p_1$ directly or via $p_{1-c1}$, if not, $U_{p_{1-0}}\cup U_{p_{1-c1}}\cup U_{p_{1-c2}}$ shares a point with one of the rest open sets . . ., and so on; after all, e is connected to $p_1$ by some finite number of segments because $U_{p_{1-ci}}$ becomes $U_{p_1}$ sometime in the finite iteration. Likewise, e is connected to $p_2$ by some finite number of segments each of which is of an integral curve of a left-invariant vectors field. $p_1$ can be connected to $p_2$ via e.

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References

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