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

Is $\text{div}A(x)$ zero?

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    Sounds good to me. If it were positive somewhere, you could choose a little region around that point whose volume increases under the flow. – 2017-02-21
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    How do I prove the increase of volume? – 2017-02-21
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    What's the definition of the divergence? – 2017-02-21
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    If $A=(A_x, A_y, A_z)$, $\text{div} A=\frac{\partial A_x}{\partial x}+\frac{\partial A_y}{\partial y}+\frac{\partial A_z}{\partial z}.$ – 2017-02-21
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    For a vector field on a Riemannian manifold?! – 2017-02-21
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    For the volume element $\Omega$, a vector field $X$ on $M$ and Lie derivative $L_{-}-$, there is a function $\text{div}X$ such that $L_X \Omega = \text{div}X \Omega$. This is the definition of div. – 2017-02-21
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    Right. Now write down the integral of $\Omega$ over the little region and use the definition of $\mathcal L_X$. – 2017-02-21
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    Do I need the assumption $\varphi(s+t,-)=\varphi(s,-)\circ \varphi(t,-)$? – 2017-02-21
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    I can't read your mind, but I don't see why that particular formula is needed. – 2017-02-21
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    I thought the formula $(L_Y T)_p = \frac{d}{dt}|_{t=0} ((\phi_t)^*T)_p$, where $\phi$ is the one-parameter semigroup induced by $Y$, is used. – 2017-02-21
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    Yes, that's right. Now pull the $t$ derivative under (or out of) the integral sign. – 2017-02-21
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    Does $(L_{A(x)}\Omega)_p = \frac{d}{dt}|_{t=0}((\varphi_t)^*\Omega)_p$ hold? My textbook deal with only the case $\phi(s+t)=\phi(s)\circ\phi(t).$ – 2017-02-22
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    I have this question. http://math.stackexchange.com/questions/2157418/definition-of-lie-derivative – 2017-03-02

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