What is the definition of Lie derivatives?

[yifli]

Let \varphi be a one-parameter group on a manifold M, and let f be a differentiable function on M, the derivative of f with respect to \varphi is the defined as the limit:

\lim_{t\to 0} \frac{\varphi^*_t[f]-f}{t}(x)=\lim_{t\to 0}\frac{f\circ \varphi_x(t)-f\circ \varphi_x(0)}{t}=D_{\varphi_x}f=X(x)f,
where X(x) is a tangent vector at x and the operator D_\varphi is defined as D_\varphi f=\frac{df\circ \varphi}{dt}\bigg|_{t=0}

I don't understand why D_{\varphi_x}f=X(x)f. According to the chain rule, I would get D_{\varphi_x}f=d_x f \circ d_0 \varphi(x)=X(x)d_x f

 

[quasar987]

Your last expression X(x)d_xf is ill defined, as X(x) is a differential operator on functions on M, whereas d_xf is a 1-form on M.

On the other hand, if you expand d_xf\circ d_0\varphi(x), you get

\sum_i\frac{\partial f}{\partial x^i}\frac{d\varphi^i_x(t)}{dt}(0)=\sum_i \frac{\partial f}{\partial x^i}X^i(x)

which is X(x)f by definition.

 

[yifli]

Your last expression X(x)d_xf is ill defined, as X(x) is a differential operator on functions on M, whereas d_xf is a 1-form on M.

On the other hand, if you expand d_xf\circ d_0\varphi(x), you get

\sum_i\frac{\partial f}{\partial x^i}\frac{d\varphi^i_x(t)}{dt}(0)=\sum_i\frac{\partial f}{\partial x^i}X^i(x)

which is X(x)f by definition.

I read the book again and found out it's just the notation they use:
for any differentiable function f defined about x and any tangent vector \xi they set \xi(f)=D_\varphi(f) where \varphi \in \xi (they define a tangent vector as an equivalence class), so D_{\varphi_x}f=X(x)f

@quasar987: The way you expand d_xf\circ d_0\varphi_x is actually the chain rule in Cartesian space, so it is true only if \varphi:R\rightarrow R^m and f: R^m\rightarrow R.

Moreover, I just realized it's not correct to use the chain rule in this case:
\begin{align*}<br /> D_{\varphi_x}f &amp; = \frac{df\circ \varphi_x}{dt}\bigg|_{t=0} (\mbox{definition of } D_\varphi) \\<br /> &amp; = d_xf \circ d_0 \varphi_x (\mbox{not true because f is defined on a manifold, so the differential of f is not } d_xf. ) <br /> \end{align*}