A description/proof of that left-invariant vectors field on Lie group is \(C^\infty\)
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 \(C^\infty\) vectors field.
- The reader admits the proposition that any vectors field is \(C^\infty\) if and only if the operation result function on any \(C^\infty\) function is \(C^\infty\).
- The reader admits the chain rule for differentiation of map.
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 \(C^\infty\).
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, any left-invariant vectors field, V, is \(C^\infty\).
2: Proof
There is a \(C^\infty\) curve, \(c: I \rightarrow G\), such that \(c (0) = e\) and \(c' (0) = V_e\). For any \(g \in G\), gc (t) is a \(C^\infty\) curve such that \(gc (0) = g\) and \((gc)' (0) = V_g\), because \((gc)' (0) = l_{g*} c' (0) = l_{g*} V_e = V_g\) by the chain rule for differentiation of map. For any \(C^\infty\) function, f, \((Vf) (g) = V_g f = \frac{df (gc (t))}{dt}|_{0} := A_1\). \(f (gc (t)): I \times G \rightarrow \mathbb{R}, (t, g) \mapsto f (gc (t))\) is a \(C^\infty\) function as a composition of \(C^\infty\) maps, \(I \times G \rightarrow G \times G \rightarrow G \rightarrow \mathbb{R}\), and \(A_1\) is a \(C^\infty\) function with respect to g. As the operation result on any \(C^\infty\) function is \(C^\infty\), V is \(C^\infty\).
