- #1
yifli
- 70
- 0
Let [itex]\varphi[/itex] be a one-parameter 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]
[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]
Last edited by a moderator: