Talk:Lie bracket of vector fields

message: The article is going to have to define what J(X) means. Charles Matthews 22:07, 27 March 2007 (UTC)
replies: text: I suspect they mean simply the commutator (Lie bracket) of vector fields, but I'd wait just to be sure. Arcfrk 11:52, 31 March 2007 (UTC)

message: The Lie Bracket, although used in differential geometry, is a construction based in differential topology.  It has no dependance on a metric / norm / or any other type of geometry. Rybu (talk) 05:16, 10 September 2008 (UTC)
thread_title:  Misclassified 

message: The theorem:
"[X,Y]=0\, iff their corresponding flows \phi_X^t, \phi_Y^s\, commute (i.e. \phi_X^t\phi_Y^s = \phi_Y^s\phi_X^t\,)."
is problematic, since there exist counter examples:
Let M= \{(x,y,z) : x^2+y^2 \neq 0\} be the manifold were the vector fields X = (1,0,-y/(x^2+y^2)) and Y = (0,1,x/(x^2+y^2)) are defined. (z sort of measures the change of angle while following the flow of X or Y.) Now we have [X,Y] = 0 but \phi_X^2\phi_Y^2 (-1,-1,0) = (1, 1,  \pi) \neq \phi_Y^2\phi_X^2 (-1,-1,0) = (1, 1, -\pi).
It looks like 3.15 Corollary (page 21) is the corresponding statement of this theorem in the book "Natural operations in differential geometry" (one of the references of the article). I wrote to the author of the book that I have the impression that the statement of the conditions under which this statement holds (3.15 Corollary says "wherever defined") is not restrictive enough (or misleading). He answered: "you are right. wherever defined means here: if one (equiv. both) side is defined on [0,t] \times [0,s],
i.e., on  the rectangle from (0,0) to (t,s).
OR: it holds locally and globally when lifted to simply connected coverings.
This is clear form the proof of 3.15."
It might make sense to mention at least that there are counter examples and provide a reference that is sufficient explicit about this issue. Jakito (talk) 09:52, 3 July 2009 (UTC)
replies: text: That's a good catch, thank you! I have modified the theorem so as to say "for sufficiently small s and t"; that should be safe. I doubt that the author's claim "it holds globally when lifted to simply connected coverings" is true; I believe your above example can be turned into a simply connected one: X and Y generate a foliation of M by Frobenius' theorem, and if we use just one leaf of this foliation (which is simply connected and looks like a spiral staircase winding around the z-axis) and restrict our vector fields to this leaf, we get the same contradiction as above. AxelBoldt (talk) 21:54, 20 April 2013 (UTC)
thread_title:  Commutator theorem 

message: This appears in the Properties section:
*For functions f and g we have
::  [fX, gY] = fg [X,Y] + fX(g) Y - gY(f) X.
To a new reader, this is incomprehensible. What are the domain and range of the functions? What is the action fX? What is the action X(g)Y? I made several naive speculations about possible meanings, but none of them have been plausible so far. Rschwieb (talk) 13:38, 10 April 2013 (UTC)
replies: text: I'll simplify the formula and explain it a bit in the article. The functions f and g are real-valued smooth functions defined on the manifold M. If f is such a function and X is a smooth vector field on M [which assigns to each point p∈M the vector Xp], then fX is the vector field which assigns to p∈M the vector f(p)Xp (where the vector Xp is multiplied with the real number f(p)). Furthermore, X(f) is a new real-valued function defined on M, given by the derivative of f in direction of X, or more explicitely: (X(f))(p) is the derivative of the function f, computed at the point p, in the direction Xp. AxelBoldt (talk) 18:59, 20 April 2013 (UTC)
thread_title:  unexplained notation in Properties section  

message: How does X(Y^i) or Y(X^i) make any sense when X and Y are function from M to TM and Y^i and X^i are function from M to R?  — Preceding unsigned comment added by 92.200.61.85 (talk) 15:21, 2 April 2017 (UTC)
thread_title:  In coordinates