description/proof of that left-invariant vectors field over Lie group is \(C^\infty\)
Topics
About: group
About: \(C^\infty\) manifold
The table of contents of this article
Starting Context
- The reader knows a definition of left-invariant vectors field over Lie group.
- The reader knows a definition of \(C^\infty\) vectors field over \(C^\infty\) manifold with boundary.
- The reader admits the proposition that any vectors field is \(C^\infty\) if and only if its operation result on any \(C^\infty\) function is \(C^\infty\).
Target Context
- The reader will have a description and a proof of the proposition that any left-invariant vectors field over 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: Structured Description
Here is the rules of Structured Description.
Entities:
\(G\): \(\in \{\text{ the Lie groups }\}\)
\(V\): \(\in \{\text{ the left-invariant vectors fields }\}\)
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Statements:
\(V \in \{\text{ the } C^\infty \text{ maps }\}\)
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2: Proof
Whole Strategy: Step 1: for each \(C^\infty\) function, \(f\), see that \(V (f)\) is a \(C^\infty\) function, by taking a \(C^\infty\) curve, \(c: I \to G\), such that \(c (0) = e\) and \(c' (0) = V_e\), and apply the proposition that any vectors field is \(C^\infty\) if and only if its operation result on any \(C^\infty\) function is \(C^\infty\).
Step 1:
There is a \(C^\infty\) curve, \(c: I \to G\), such that \(c (0) = e\) and \(c' (0) = V_e\).
Let \(g \in G\) be any.
Let \(l_g: G \to G\) be the left-translation by \(g\).
\(l_g \circ c (t)\) is a \(C^\infty\) curve such that \(l_g \circ c (0) = g\) and \((l_g \circ c)' (0) = V_g\), because \((l_g \circ c)' (0) = d l_g c' (0)\), as is well known, \(= d l_g V_1 = V_g\), because \(V\) is left-invariant.
For any \(C^\infty\) function, \(f: G \to \mathbb{R}\), \(V (f) (g) = V_g f = d f (l_g \circ c (t)) / d t \vert_0 := A_1\).
\(f (l_g \circ c (t)): I \times G \to \mathbb{R}, (t, g) \mapsto f (l_g \circ c (t))\) is a \(C^\infty\) map as a composition of \(C^\infty\) maps, \(I \times G \to G \times G \to G \to \mathbb{R}\).
So, \(A_1\) is a \(C^\infty\) map with respect to \(g\).
So, \(V\) is \(C^\infty\), by the proposition that any vectors field is \(C^\infty\) if and only if its operation result on any \(C^\infty\) function is \(C^\infty\).
