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 $D_{\varphi_x}f=d_x f \circ d_0 \varphi(x)=X(x)d_x f$

Last edited by a moderator:
Related Differential Geometry News on Phys.org

quasar987

Homework Helper
Gold Member
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.

Last edited by a moderator:

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*} 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*}

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving