Flow of time-dependent vector field

Question

Suppose $X_t$ is a time-dependent vector field, with flow $\phi_t$.

Thus,

$$\frac{d}{dt} \phi_t = X_t(\phi_t).$$

Is it true that $$d \phi_t\big(X_t(x)\big) = X_t\big(\phi_t(x)\big)\;?$$

This is true when $X_t$ does not depend on $t$.

Answer 1

No, this equality does not hold. It is not hard to think of counterexamples, but it may be even better to understand the whole picture.

Any time dependent vector field $X_t$ on the manifold $M$ can be thought of as a constant vector field $X$ on $I\times M$. $X$ is simply defined by $$X(t,p)=\frac{\partial}{\partial t}+X_t(p).$$One can verify that the flow of $X$ on $I\times M$ is equivalent to the flow of $X_t$ on $M$. Namely,$$\Phi_t(s,p)=(t+s,\phi_t(p)),$$ where $\Phi$ denotes the flow of $X$. By the last sentence in the posted question, we have $$d\Phi_t\left(\frac{\partial}{\partial t}+X_0(p)\right)=d\Phi_t(X(0,p))=X(\Phi_t(0,p))=X(t,\phi_t(p))=\frac{\partial}{\partial t}+X_t(\phi_t(p)),$$and hence$$d\phi_t(X_0(p))=X_t(\phi_t(p)),$$which is the correct result.

Answer 2

The conclusion of Amitai's answer is false.

For a counterexample, consider a vector field $X_t$ for $t \geq 0$ defined on the real line. Suppose that $X_t$ is the trivial vector field for $t=0$, and is a constant non-trivial vector field for $t = 1$. Then the claimed result at the end of the above answer would be saying that $d \phi_1(X_0(p)) = X_1(\phi_1(p))$ for any real number $p$. But the left hand side is the zero vector while the right hand side is non-zero. The flaw in the reasoning of Amitai's answer has been pointed out in a comment on that answer.

On the other hand, as Amitai suggests, the answer to the original question is indeed no. To give more of a hint at how to construct a counterexample to the original question, let's take the real line again. Then take a vector field $X_t$ on the real line such that $\phi_1(0) = 1$ and such that $X_1(0) = 0$ and $X_1(1) = 1$. The push forward of the zero vector by $d \phi_1$ is certainly not non-zero, thus giving a counterexample.