
#1
Jun2811, 02:23 PM

P: 70

Let [itex]\varphi[/itex] be a oneparameter group on a manifold M, and let [itex]f[/itex] be a differentiable function on M, the derivative of f with respect to [itex]\varphi[/itex] is the defined as the limit:
[tex]\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,[/tex] where [itex]X(x)[/itex] is a tangent vector at x and the operator [itex]D_\varphi[/itex] is defined as [itex]D_\varphi f=\frac{df\circ \varphi}{dt}\bigg_{t=0}[/itex] I don't understand why [itex]D_{\varphi_x}f=X(x)f[/itex]. According to the chain rule, I would get [itex]D_{\varphi_x}f=d_x f \circ d_0 \varphi(x)=X(x)d_x f[/itex] 



#2
Jun2911, 07:51 AM

Sci Advisor
HW Helper
PF Gold
P: 4,768

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 1form on M.
On the other hand, if you expand [itex]d_xf\circ d_0\varphi(x)[/itex], you get [tex]\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)[/tex] which is X(x)f by definition. 



#3
Jun2911, 10:44 AM

P: 70

for any differentiable function f defined about x and any tangent vector [itex]\xi[/itex] they set [itex]\xi(f)=D_\varphi(f)[/itex] where [itex]\varphi \in \xi[/itex] (they define a tangent vector as an equivalence class), so [itex]D_{\varphi_x}f=X(x)f[/itex] @quasar987: The way you expand [itex]d_xf\circ d_0\varphi_x[/itex] is actually the chain rule in Cartesian space, so it is true only if [itex]\varphi:R\rightarrow R^m[/itex] and [itex]f: R^m\rightarrow R[/itex]. Moreover, I just realized it's not correct to use the chain rule in this case: [tex]\begin{align*} D_{\varphi_x}f & = \frac{df\circ \varphi_x}{dt}\bigg_{t=0} (\mbox{definition of } D_\varphi) \\ & = 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. ) \end{align*}[/tex] 


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