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Let $M\hookrightarrow \mathbb{R}^n$ be a manifold, $\varphi:\mathbb{R}\times M\rightarrow M$ be a smooth function such that $\varphi(0,-)=\text{id}_M$. Then each curve $\varphi(-,x):\mathbb{R}\rightarrow M$ defines a tangent vector $A(x)\in T_x M$.

Let $\phi:I\times M \rightarrow M$ be the one-parameter group whose infinitesimal transformation is $A(x)$.

For any tensor field $T$, its Lie derivative along $A$ at a point $p$ is $(L_A(x)T)_p= \frac{d}{dt}((\phi_t)^* T)_p$. Is this coincide with $\frac{d}{dt}((\varphi_t)^*T)_p$?

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    Unless I'm misunderstanding your question, $\phi = \varphi$. – 2017-02-24
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    How do you prove it? If $¥varphi(s+t)=¥varphi(s)¥circ¥varphi(t)$, I can use the uniqueness of the integral curve of $A(x)$ but this is not assumed. – 2017-02-24
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    Oh, you're right, I assumed $\varphi$ was a flow. – 2017-02-24
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    related question: http://math.stackexchange.com/questions/2154019/divergence-of-a-volume-preserving-function – 2017-03-02

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