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Lie brackets and derivatives

  1. Jan 29, 2013 #1
    Hi i have two questions:
    1) When asked to prove [itex]\mathcal{L}_{u}\mathcal{L}_{v}W - \mathcal{L}_{v}\mathcal{L}_{u}W = \mathcal{L}_{[u,v]}[/itex].

    I achieved [itex][u,v]w = \mathcal{L}_{[u,v]}[/itex]. This was found by appliying a scalar field <b> to the LHS and simplifying and expanding using + and scalar linearitys to get [itex][u,v]w[/itex] but I am not sure if these are equivalent.

    2) When asked to calculate the Lie bracket [X,Y] where [itex]X=5x^{2}\frac{\partial}{\partial t}-4t\frac{\partial}{\partial x} [/itex] and [itex]Y= y\frac{\partial}{\partial t} + t\frac{\partial}{\partial y} [/itex] is this equivalent to:
    [itex]\left((5x^{2}\frac{\partial}{\partial t} -4t\frac{\partial}{\partial x})\frac{\partial (y\frac{\partial}{\partial t} + t\frac{\partial}{\partial y})}{\partial x^{a}}-(y\frac{\partial}{\partial t} + t\frac{\partial}{\partial y})\frac{\partial (5x^{2}\frac{\partial}{\partial t}-4t\frac{\partial}{\partial x})}{\partial x}\right)\frac{\partial}{\partial x^{b}} [/itex]

    and if so can it be expanded any further I am not so sure but i dont fully understand [itex] \frac{\partial}{\partial x^{a}} [/itex] derivatives.
  2. jcsd
  3. Jan 29, 2013 #2
    never mind figured it out
  4. Jan 29, 2013 #3


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    Homework Helper

    I'm sure you're missing out the vector field W in the RHS of 1).
  5. Jan 29, 2013 #4
    yeah i figured them out and yeah missed it off, thx
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