A description/proof of existence of Lie group neighborhood whose any point can be connected with center by left-invariant vectors field integral curve
Topics
About: Lie group
About: vectors field
The table of contents of this article
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.
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.
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.
Main Body
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, \(\gamma _i: \mathbb{R} \rightarrow G\), of a left-invariant vectors field, \(V_i\), starting at \(p_0\).
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, \(exp: U_0 \rightarrow U_e\), is a diffeomorphism. \(U_{p_0} := \{\forall p| p_0^{-1} p \in U_e\}\) is an open set because it is the preimage of \(U_e\) by the \(C^\infty\) map, \(f: G \rightarrow G, p \mapsto p_0^{-1} p\). For any \(p_i \in U_{p_0}\), there is a vector, \(l_{p_0*, e} V_{i, e}\), 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 \(p_0^{-1}p_i = exp V_{i, e}\), \(p_i = p_0 exp V_{i, e} = \phi_1 (p_0)\) 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.
3: Note
The 1-to-1-ness from the vectors at \(p_0\) to \(U_{p_0}\) is is not proven here.
