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Say, $$L_{V}z^A =0$$ but I don't know much about Lie derivatives except what I saw now through wikipedia http://en.wikipedia.org/wiki/Lie_bracket_of_vector_fields#Definitions that it is (if I am correct) equal to $[V,A]$. Is this right how I compared this to what's on wikipedia? If so then this is dealt with as a commutator mathematically speaking? If not, then how can we expand $L_{V}z^A =0$?

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closed as unclear what you're asking by ACuriousMind, Kyle Kanos, JamalS, Martin, Jim Mar 2 '15 at 15:30

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Please define the notation used within the post - what are $z$,$A$ and $V$? What is the actual question here - if it's a Lie derivative, of course can plug in the definition of a Lie derivative. What do you mean by "is this dealt with as a commutator?"? $\endgroup$ – ACuriousMind Mar 2 '15 at 1:24
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The Lie derivative has a geometrical meaning: it measures the change of a tensor field (including scalar function, vector field and one-form), along the flow of another vector field. For example, the Lie derivative of the metric tensor along a Killing vector is zero (this defines the Killing vector equation). It means that a tensor (for example the metric) is not changing along the Killing vector or mathematically speaking

$$ \mathcal{L}_V g_{\mu\nu}=0.$$

The same can be applied to your case. You equation (3.19) says that the Lie derivative along a Killing vector $V$ of a set of scalars is zero $$ \mathcal{L}_V z^N=0. $$

You can think of it as a condition on your set of scalars. In coordinates it gives $$ \boxed{\mathcal{L}_V z^N=V^\mu \partial_\mu z^N.}$$

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