2
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thanks for your answer, now everything is clear to me! I was focusing on the first term all the time... $\endgroup$ – Michael Jung Dec 8 '15 at 6:25
  • $\begingroup$ 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. $\endgroup$ – Qmechanic Dec 8 '15 at 22:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.