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Suppose $M$ is a manifold and $f:M \rightarrow \bf{R}$ is a $\mathcal{C}^{\infty}$ function. Let $X$ and $Y$ denote vector fields on $M$, and let $\varphi_Y^t$ denote the flow of $Y$. Fix a point $p \in M$, and consider the real valued function $$u(t):=df_{\varphi_Y^t (p)}(X(\varphi_Y^t (p))).$$ When is $u$ identically zero? Is it always zero if the Lie bracket $\mathcal{L}_X Y$ is zero, or does the condition depend on the function $f$?

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I know that it is kind of obvious to say, but we cannot take any function $f$. For instane, constant functions will produce $u(t)=0$, because $\textrm{d}f=0$, independently of the conditions on $X$ and $Y$. I think we may have to ask conditions on $f$ to take some interesting result. – matgaio Apr 22 '12 at 5:41

Let us take a look at an example. Let $M=R^2$, and $Y=x\partial_x+y\partial_y$ and $X=-y\partial_x+x\partial_y$. Then, we have $\phi_t(x,y)=e^t(x,y)$, where $\{\phi_t\}$ is the flow of v.f. $Y$. Then, you can see that $$u(t)=-f_xe^ty+f_ye^tx$$ in which $f_x$ and $f_y$ are partial derivative of an arbitrary smooth function $f$ on $M$ and evaluated at $(e^tx,e^ty)$. Considering $u(0)=0$ implies that $f$ must satisfy the following $$-yf_x+xf_y=0$$ i.e., the Lie derivative of $f$ along v.f. $X$ must be vanished. So, one can guess that if $f$ is constant along the integral curve of $X$, then $u(t)$ is zero?!

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