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Confusion over the definition of Lie Derivative of a Vector Field

  1. Oct 29, 2011 #1
    Hello all, I was hoping someone would be able to clarify this issue I am having with the Lie Derivative of a vector field.

    We define the lie derivative of a vector field [itex]Y[/itex] with respect to a vector field [itex]X[/itex] to be

    [itex]L_X Y :=\operatorname{\frac{d}{dt}} |_{t=0} (\phi_t^*Y)[/itex], where [itex]\phi_t[/itex] is the flow of [itex]X[/itex]. Now how do I know this thing is differentiable? And also, I'm not sure why this thing is a vector field on [itex]M[/itex] because what if our flow is only defined on a smaller open set [itex] V \subseteq M[/itex], and then for [itex] p \in M\setminus V[/itex] surely [itex](L_X Y)_p[/itex] doesn't make sense as a vector in [itex]T_p M[/itex] ?

    Anyway thank you, any help would be appreciated.
     
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  3. Oct 29, 2011 #2

    quasar987

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    First concern: You know that phi varies smoothly with t. Go through the definitions to discover that wrt some fixed basis d/dx^i of T_pM, [itex]((\phi_t)_*Y)(p)=\sum_ia_i(t,p)\frac{ \partial}{ \partial x^i}[/itex] with a_i(t,p) smooth in t.

    Second concern: the flow of any time-independant vector field is defined at all points p of M for a certain time interval [itex](-\epsilon_p,\epsilon_p)[/itex]
     
  4. Oct 31, 2011 #3
    Thanks for your response. I understand the 2nd concern but I still have trouble with the first. I cannot see why that statement is true. I know that [itex]\phi_t ^* Y[/itex] is a vector field, so [itex]\phi_t^* Y |_p \in T_p M[/itex] , i.e.

    [itex]\phi_t^* Y |_p = \displaystyle\sum_i a_i^t (p) \frac{\partial}{\partial x_i }\Big|_p[/itex] where for fixed [itex]t[/itex], [itex]p \mapsto a_i^t (p) [/itex] is smooth. But why for fixed [itex]p[/itex], is [itex] t \mapsto a_i^t(p) [/itex] smooth?

    I have that [itex]\phi_t^* Y |_p (g) = Y(g \circ \phi_t ^{-1})|_{\phi_t(p)} [/itex] incidentally. Thanks for any help
     
  5. Oct 31, 2011 #4

    quasar987

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    You're not "going through the definitions" deep enough. What I was suggesting is for you to find an explicit expression for the a_i(t,p).
     
  6. Oct 31, 2011 #5
    Hello I have tried to do this using the Jacobian but I am not getting anywhere at all
     
  7. Oct 31, 2011 #6
    Hello,

    I have managed to calculate that

    [itex]\phi_{-t}^*\left( \displaystyle\frac{\partial}{\partial x_j} \Bigg|_{\phi_t(p)} \right)= \displaystyle\sum_{i,j} \frac{\partial \phi^i}{\partial x_j}(-t, \phi_t(p)) \frac{\partial}{\partial x_j} \Bigg|_p [/itex],

    although in Loring Tu's book the same calculation gives

    [itex]\phi_{-t}^*\left( \displaystyle\frac{\partial}{\partial x_j} \Bigg|_{\phi_t(p)} \right)= \displaystyle\sum_{i,j} \frac{\partial \phi^i}{\partial x_j}(-t, p) \frac{\partial}{\partial x_j} \Bigg|_p [/itex].

    Which is correct, and should it matter in the argument?

    Thanks for help so far.
     
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