Clarification on Lie Derivatives

  • #1
The standard definition of the lie derivative of X along Y is just

$$(*) \mathcal{L}_YX = \lim_{t\to 0} \frac{X_{\phi(t)} - \phi_{t*}X}{t}$$

where ##\phi_t## is the flow generated by Y. I.e. the limit of the difference between a pushforward of X along Y and X evaluated at a point ##\phi(t)## on the flow of Y. Other than this definition I have seen the expressions

$$(1) \mathcal{L}_Y X = [Y,X]$$

and

$$(2) \mathcal{L}_YX = \nabla_Y X - \nabla_X Y$$

for the Lie Derivative, but I have not seen it stated clearly when (1) and (2) holds. Surely (1) and (2) is equivalent when the connection ##\nabla## is torsion free. However is it generally true that (*) is equivalent with (1) or (2)? What conditions on ##\nabla## must be satisfied in order to identify (*) with (1) (or (2))?
 

Answers and Replies

  • #2
Ben Niehoff
Science Advisor
Gold Member
1,879
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The standard definition of the lie derivative of X along Y is just

$$(*) \mathcal{L}_YX = \lim_{t\to 0} \frac{X_{\phi(t)} - \phi_{t*}X}{t}$$

where ##\phi_t## is the flow generated by Y. I.e. the limit of the difference between a pushforward of X along Y and X evaluated at a point ##\phi(t)## on the flow of Y. Other than this definition I have seen the expressions

$$(1) \mathcal{L}_Y X = [Y,X]$$

and

$$(2) \mathcal{L}_YX = \nabla_Y X - \nabla_X Y$$

for the Lie Derivative, but I have not seen it stated clearly when (1) and (2) holds. Surely (1) and (2) is equivalent when the connection ##\nabla## is torsion free. However is it generally true that (*) is equivalent with (1) or (2)? What conditions on ##\nabla## must be satisfied in order to identify (*) with (1) (or (2))?
(1) and (*) are equivalent, you should be able to show this.

(2) is equivalent to (1) if and only if the torsion vanishes. In fact, the torsion is defined as the difference between (2) and (1):

[tex]T(X,Y) \equiv \nabla_X Y - \nabla_Y X - [X,Y][/tex]
 

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