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Lie derivative of two left invariant vector fields

  1. Nov 18, 2011 #1
    Hi all,
    I was following Nakahara's book and I really got my mind stuck with something. I would appreciate if anybody could help with this.

    The Lie derivative of a vector field [itex]Y[/itex] along the flow [itex]\sigma_t[/itex] of another vector field [itex]X[/itex] is defined as
    [tex] L_X Y=lim_{\epsilon\to0}\frac{1}{\epsilon}\left[Y|_{x}-(\sigma_{\epsilon})_* Y|_{\sigma_{-\epsilon}(x)}\right][/tex]
    wich is equal to the Lie bracket [itex][X,Y][/itex].
    Now when he goes on and defines the left invariant vector field [itex]X[/itex] of a Lie group, this is given by demanding the relation
    [tex]L_{a*}X|_g=X|_{a\,g}[/tex]
    where [itex]L_a[/itex] is the left translation by [itex]a[/itex].

    I guess we can assume that this left translation is the result of a flow of some other field.

    My question is that since every left-invariant vector field defines a one parameter flow, why the Lie bracket of two left invariant vector fields doesn't vanish identically?

    However, we know that the left invariant vector fields define the Lie algebra of the group and this has non-trivial commutation relations.

    Thank you very much for your help.
     
    Last edited: Nov 18, 2011
  2. jcsd
  3. Aug 26, 2015 #2
    This is an old question, but since I stumbled upon it when the same confusion briefly bedeviled me after reading Nakahara, I decided to reply in case others follow a similar path. The key is that $\sigma_{\epsilon} (x)$ does not denote a left action. Instead, $\sigma_{\epsilon}$ denotes the diffeomorphism $\sigma_{t} $ with fixed $t=\epsilon$ from the 1-parameter group of diffeomorphisms $\sigma_{t}$ (see Nakahara section 5.3.1). The diffeomorphism group $\sigma_{t}$ is the flow resulting from a vector field, but under $\sigma_{\epsilon}$ you send each point $x$ of the manifold (the manifold being a Lie group in the case discussed here) not to its image under a left action, but rather to the point the flow has reached after $\epsilon$ time elapses, if you start the flow from the point $x$.
     
    Last edited: Aug 26, 2015
  4. Aug 27, 2015 #3

    WWGD

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    If I may, 1729 , let me latex-format your answer to make it easier to read (we use double pounds before - and after - to Latex-tag.:

    This is an old question, but since I stumbled upon it when the same confusion briefly bedeviled me after reading Nakahara, I decided to reply in case others follow a similar path. The key is that ##\sigma_{\epsilon} (x) ## does not denote a left action. Instead, ##\sigma_{\epsilon} ## denotes the diffeomorphism ## \sigma_{t} ## with fixed ##t=\epsilon ## from the 1-parameter group of diffeomorphisms ##\sigma_{t} ## (see Nakahara section 5.3.1). The diffeomorphism group ##\sigma_{t} ## is the flow resulting from a vector field, but under ##\sigma_{\epsilon} ## you send each point ##x ## of the manifold (the manifold being a Lie group in the case discussed here) not to its image under a left action, but rather to the point the flow has reached after ##\epsilon ## time elapses, if you start the flow from the point ##x ##
     
  5. Aug 27, 2015 #4

    Thank you so much!! I couldn't for the life of me figure it out!!
     
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