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I'm currently working through the math chapters of Norbert Straumann's book on General Relativity. I have trouble understanding the coordinate expression of the Lie derivative of a basis vector. The equation in the book reads:

$$L_X\partial_i = [X,\partial_i] = -X^j_{,i}\partial_j\tag{1}$$

with a vector field $X$, local coordinates $\{x^i\}$ and the base $\{\partial_i:=\partial/\partial x^i\}$.

I don't understand the second step. Writing out the commutator, we get

$$[X,\partial_i] = [X^j\partial_j,\partial_i] = X^j\partial_j\partial_i - \partial_i X^j\partial_j. \tag{2}$$

The second term here is the same as the result given in the book, suggesting that $X^j\partial_j\partial_i$ would be 0. I've thought for quite some time about this, but I still don't understand why this would be the case. Could someone please shed some light on this?

Qmechanic
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Michael Jung
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1 Answers1

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One still has to apply the Leibniz rule to the second term $$-\partial_i X^j\partial_j=-X^j\partial_i\partial_j -X^j_{,i}\partial_j$$ in eq. (2), because the derivative $\partial_i $ also acts further to the right, thereby yielding eq. (1). See also this related Phys.SE post.

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Qmechanic
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  • Thanks for your answer, now everything is clear to me! I was focusing on the first term all the time... – Michael Jung Dec 08 '15 at 06:25
  • Here is another check: If $P$ and $Q$ are differential operators of order $n$ and $m$, respectively, then it is a general fact that the order of the commutator $[P,Q]$ can at most be $n+m-1$. Now in OP's case $n=1=m$. Hence we see that there is no room for the first term $X^j\partial_j\partial_i$ in eq. (2) because the order is too high. – Qmechanic Dec 08 '15 at 22:03