# Lie derivative in this paper [closed]

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 $L_{V}z^A =0$?

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

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• 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?"? – ACuriousMind Mar 2 '15 at 1:24

## 1 Answer

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 $$\boxed{\mathcal{L}_V z^N=V^\mu \partial_\mu z^N.}$$