The lecturer here wants the viewer to derive the components of the Lie derivative of a (1,1) tensor-field. But even before that I have a little question about the components of the Lie derivative of a vector field:

Careful: 1) and 2) are incorrect!

let $(U,x)$ be a chart and $X,Y$ vector fields on the smooth manifold $(M,\mathcal{O},\mathcal{A})$, I get:

$$ ({L}_X Y)^i = [X,Y]^i = (XY - YX)^i = X^m \left(\frac{\partial}{\partial x^m}\right) Y^i - Y^m \left(\frac{\partial}{\partial x^m}\right) X^i $$ 1) is that correct? I suspect, since we write single components, which are real functions, that I can reorder the terms, as I please (commutativity of multiplication on $C^{\infty}M$)? And the derivatives, which are actually the basis vectors, act on the function to which this thing is applied to anyway, right? So for clarity I could move them to the far right to show this: $$ = X^m Y^i \left(\frac{\partial}{\partial x^m}\right) - Y^m X^i \left(\frac{\partial}{\partial x^m}\right) $$ 2) still correct?

3) Then is there any "rule"/intuition or something why the contraction with the basis is over the "outer" field? If I take $(y \circ x^{-1})^i$ I can write this as $(y^i \circ x^{-1})$, basically the last function is responsible for the component I get out. In the above example it seems to be the first one applied.

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    $\begingroup$ Note that $X^i, Y^i$ are functions. $[X,L]^i$ is the i-th component of the lie derivative and not a vector. In 1) The derivatives you have written act only on the $Y^i$ and $X^i$ and no function as an argument is suppressed. To write it out $[X,Y]^i = X^m \frac{\partial Y^i}{\partial x^m} - Y^m \frac{\partial X^i}{\partial x^m}$, or $[X,Y] = \left(X^m \frac{\partial Y^i}{\partial x^m} - Y^m \frac{\partial X^i}{\partial x^m}\right) \frac{\partial}{\partial x^i}$ $\endgroup$ – s.harp Apr 3 '16 at 10:27
  • $\begingroup$ Thanks for that simple, yet insightful clarification! I totally mixed up the component per se and the component-wise description of the whole vectorfield! The last equation was what I was missing. That basically asnwers 1) .. 3), since the contractions are just applications of the vectors to their argument functions the $X^i$ and $Y^i$, respectively. $\endgroup$ – mike Apr 3 '16 at 11:00
  • $\begingroup$ I've edited the question, so this answer solves the whole thing, and i have another one to ask for covectors. Please submit this as answer, such that I can accept it. $\endgroup$ – mike Apr 3 '16 at 11:30

Note that the components $X^i, Y^i$ of a vector field wrt some coordinates are functions. With $[X,Y]^i$ the i-th component of the Lie derivative is meant, this is not a vector but a function. As such in part 1) the derivatives act only on $X^i$ and $Y^i$, and not on the function in the argument of $[X,Y](f)$. Writing it out:

$$[X,Y]^i=X^m \frac{\partial Y^i}{\partial x^m}- Y^m \frac{\partial X^i}{\partial x^m}$$


$$[X,Y](f)=X(Y(f))-Y(X(f))= \left(X^m \frac{\partial Y^i}{\partial x^m}- Y^m \frac{\partial X^i}{\partial x^m}\right)\frac{\partial f}{\partial x^i} $$

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  • $\begingroup$ Thanks again. If I'm right here, the $X^i$ and $Y^i$ are functions from the manifold to the reals, $X^i : M \to \mathbb{R}$? And if the vector field is smooth, they are elements of $C^{\infty}(M)$. $\endgroup$ – mike Apr 3 '16 at 11:33
  • $\begingroup$ They are functions on the coordinate chart (so functions on an open subset of $\mathbb R^n$), not on the manifold. But they are $C^\infty$ functions that is right. $\endgroup$ – s.harp Apr 3 '16 at 11:34
  • $\begingroup$ Oh, yeah, so $X^i : \mathbb{R}^n \to \mathbb{R}$ such that $x(p) \mapsto X^i |_{x(p)}$ where $p$ is a point in $M$ and $(U,x)$ is the chart? $\endgroup$ – mike Apr 3 '16 at 11:36
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    $\begingroup$ I'm not too sure what you mean right now. If $U \subset \mathbb R^n$ with $x: U \to M$ is the coordinate map, then $X^i: U \to \mathbb R$, and if $p \in x(U) \subset M$ then $X^i \mid_{x^{-1}(p)}$ is the component of $X$ in direction $\partial_{x^i}$ at the point $p$. $\endgroup$ – s.harp Apr 3 '16 at 11:41
  • $\begingroup$ I guess we differ here only in my imprecise use of $U$: I would say $x: U \to A$ where $x(U) = A \subset \mathbb{R}^n$ and $U \subset M$ and $p \in U$. With this definition $X^i : A \to \mathbb{R}$ and $X^i$ now takes the point on the manifold $p$ in coordinates $x(p)$ and maps it to the component you wrote down. In "math" $X^i : x(p) \mapsto X^i|_{x(p)}$. $\endgroup$ – mike Apr 3 '16 at 11:48

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