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I am reading the book Introduction to Smooth Manifold written by M.Lee. I am confusing with the concept of Lie derivative.

We have $\mathcal{L}_XY=[X,Y]$. However we have $D_XY=X(Y^i)\frac{\partial}{\partial x^i}$ in Euclidean space, and $\mathcal{L}_XY=D_XY$. It is obvious that $$\mathcal{L}_XY-\mathcal{L}_YX=[X,Y]$$

Where do I make mistakes?

gaoxinge
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    The statement that $\mathcal L_X Y = D_X Y$ is false. You may be thinking of the case where $Y$ is a function instead of a vector field, where one can still define the notion of Lie derivative $\mathcal L_X$, and you end up having that $\mathcal L_X f = D_X f := Xf$. –  Aug 30 '15 at 02:29
  • @MikeMiller So $\mathcal{L}_XY=D_XY-D_YX$? – gaoxinge Aug 30 '15 at 02:40
  • In the case of vector fields on $\Bbb R^n$, yes. You can also define $\mathcal L_X Y$ this way in terms of charts. The thing is that $\mathcal L_X Y$ is actually a well-defined vector field, that didn't depend on the chart; while $D_X Y$ will depend wildly on what chart you choose. –  Aug 30 '15 at 02:41
  • @MikeMiller Thx. – gaoxinge Aug 30 '15 at 02:42

1 Answers1

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We do not have $\mathcal L_X Y = D_X Y$. The actual statement is $\mathcal L_X Y = D_X Y - D_Y X$ (which, according to some authors, is a definition), where this is defined in charts. This ends up not depending on the choice of chart; but $D_X Y$ does depend on the choice of chart.

The Lie derivative $\mathcal L_X$ makes sense as something that acts on a lot more than just vector fields ($L_X T$ makes sense for $T$ a tensor - so in particular differential forms or even functions). In the case of functions, $$(\mathcal L_X f)(x) = \frac{d}{dt} f(\phi^t(x))\bigg|_{t=0},$$ where $\phi^t$ is the flow of $X$ at time $t$. Because $\gamma(t) = \phi^t(x)$ is the integral curve of $X$ through $x$, by definition this is $X_x f$; so $\mathcal L_X f = Xf$ (which is sometimes denoted $D_X$). This is a context in which "$\mathcal L_X = D_X$" is true; for higher-order tensors the formulae, as above, are more complicated.