# Smoothness of vector fields with smooth flow

Suppose you have a Vector field $X$ on a smooth (complete) manifold $M$, whose flow $\phi_X^t$ is, for each time $t\in (-\varepsilon,\varepsilon)$, smooth (of class $C^k$).

Questions: Is $X$ smooth (of class $C^{k-\textrm{something}}$)? Can I control that "something"? What about the case $M=\mathbb{R}^n$?

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I'll just throw this out there and someone can shoot it down if I'm being stupid... but $X(\phi_X^0)$ is the $t$ derivative of $\phi_X^t$ at $t = 0$, so working in local coordinates I think you are asking if $\partial_t f(x,t)|_{t=0}$ is $C^k$ if $f$ is $C^k$ in $x$. In this case the answer is yes. –  Paul Siegel Nov 8 '11 at 12:18
Hi Paul, I'm not sure if I got what you said correctly. Because, as the "$\partial_tf(x,y)|_{t=0}$ is $C^k$ if $f$ is $C^k$ in $x$" is concerned, this is false, at least as I stated it. In fact, just take a function $f(x,t)$ smooth in $x$ and only $C^2$ in $t$, for example $f(x,t)=x+|t|\cdot t\cdot e_1$, or something like that. The point is that such an $f(x,t)$ doesn't come from the flow of a vector field –  Marco Radeschi Nov 8 '11 at 14:00
Ah, I think I understand now: you're asking if the flow $\phi_X(x,t)$ can have strictly more regularity in $x$ than it does in $t$. Certainly the flow is at least as differentiable as $X$ in both variables, but I can't see any reason why $\phi_X$ couldn't accidentally acquire more smoothness in $x$ than it deserves. Nor can I come up with any useful examples. So I'll politely bow out of the conversation. :) –  Paul Siegel Nov 8 '11 at 15:30
A simple example showing that "$f \in C^{1,\infty}(\mathbb{R} \times M,M)$" does not imply "$\partial_t f(0,-) \in C^1(M,M)$" is given by taking $M=\mathbb{R}$ and $f(t,x)=t\sqrt{x^2+t^2}$. –  Benoit Jubin Nov 14 '11 at 23:03

Consider for instance the case of $M:=\mathbb{R}$, and any $C^k$ field $X$ on $M$, that we may identify with a function. Say, such that $0 < a \le X(s)\le b$. The field $X$ has a globally defined flow, conjugated to the translation flow $x\mapsto x+t$: $$\phi^t(x):= h^{-1}(h(x)+t)\,,$$ where the conjugation $h:\mathbb{R}\to\mathbb{R}$ is $$h(x):=\int_0^x\frac{1}{X(s)}ds \, .$$ Thus if $X$ is of class $C^k$, possibly not $C^{k+1}$, $\phi$ is in any case in $C^{k+1}(\mathbb{R}\times \mathbb{R},\mathbb{R})\, .$
By Paul's argument, if phi is of class $C^m$ (in the pair of variables $(t,x)$ ) then X is at least $C^{m−1}$. But if I understand you correctly, you do not want to make assumptions on the regularity of $\partial_t \phi$, right? –  Pietro Majer Nov 8 '11 at 17:41
that is exact. I do not assume any regularity in the $t$ direction. –  Marco Radeschi Nov 8 '11 at 17:49
@Marco: this example doesn't seem to generalize to higher dimensions. Take $X(x) = |x|$: this generates a non-differentiable flow, but if we add another direction $y$ with $X(x,y) = (|x|,1)$, then the vector field is bounded away from zero, but still has non-differentiable flow. –  Jaap Eldering Nov 8 '11 at 18:40
@Jaap: the example refers to the first version of the question, and was meant to show that $\phi\in C^m$ in general does not imply that $X$ is $C^m$ too. –  Pietro Majer Nov 8 '11 at 18:55