9
$\begingroup$

This feels like something I should know, but I have a hard time finding a definite reference.

Let $M$ be a compact (Riemannian) manifold, $k\ge 1$ be an integer and $\alpha\in(0,1)$. When v is a $C^k$ vector field, plenty of references show that its flow $\Phi^t(x)$ is $C^k$ in $(t,x)$. Now, I would like first to consider the $C^{k,\alpha}$ regularity (I need it not to loose too much when I next solve a simple PDE, where this is the "right" regularity to work with), and second I would like to ensure that the derivative of $\Phi^t$ in $t=0$ is $v$ in the $C^{k,\alpha}$ norm; i.e. I want to ensure that there is a family $\varepsilon_t$ of $C^{k,\alpha}$ vector fields going to $0$ in $C^{k,\alpha}$ norm when $t\to0$, such that $$\Phi^t(x) = \exp_x(tv+t\varepsilon_t(x)).$$ I don't doubt any of this, but I would like to have a good reference to cite (and read).

$\endgroup$
3
  • 3
    $\begingroup$ Indeed, references are hard to trace, so that I included some review of $C^{k,\alpha}$ topology in section 2 of my paper arxiv.org/abs/1510.07269. There is no original material there, of course. $\endgroup$ May 9, 2016 at 20:49
  • $\begingroup$ Dear Benoit, have you found a good reference by now? I just bumped into the same issue myself. $\endgroup$ Dec 6, 2019 at 8:24
  • $\begingroup$ @AmitaiYuval: I did not. The paper I was working on ended up written in the smooth setting. $\endgroup$ Dec 7, 2019 at 9:09

1 Answer 1

2
$\begingroup$

Edit: unfortunately I did botch the crucial computation. So, surprisingly, the map $f$ defined below need not be locally Lipschitz in the $C^{k,\alpha}$ space.

Let me give an example in $1$-dimension: let $v(x)=\frac23 |x|^{3/2}\in C^{1,\frac12}$, and consider for small $\varepsilon$ the functions $g(x)=\varepsilon +x$ and $h(x)=x$. Then obviously $\lVert g-h\rVert_{C^{1,\frac12}}=\varepsilon$, but we have for $x>0$ $$(v\circ g-v\circ h)'(x)=\sqrt{\varepsilon+x}-\sqrt{x}=\frac{\varepsilon}{\sqrt{\varepsilon+x}+\sqrt{x}} \to \sqrt{\varepsilon} \quad\mbox{when } x\to 0$$ so $\lVert v\circ g-v\circ h\rVert_{C^{1,\frac12}} \ge \sqrt{\varepsilon}$.

Even if this counter-example does not rule out the possibility that what I asked holds (the flow of $v$ above has the wanted property), it still makes me doubt.


Previous version (hopefully still useful in other regularities)

Since references seem elusive, let me propose a simple proof. The idea is simply to use the Picard–Lindelöf theorem to the right object.

First, since $M$ is compact it has positive injectivity radius, and its (global) exponential map $\exp:TM\to M$ induces a diffeomorphism $(x,u)\mapsto (x,\exp_x(u))$ from a neighborhood of the zero section of $TM$, to a neighborhood of the diagonal in $M\times M$ (of course this is already the Picard–Lindelöf theorem, but in the more classical smooth regularity and without the need for the "global derivative" part of the question). Pulling back by $\exp$, we can identify the diffeomorphisms of $M$ which are uniformly close to the identity with vector fields (which are then uniformly close to zero). We use the letter $\Phi$ to denote a diffeomorphism seen as a point the consequent open set $\Omega$ of $\Gamma^{k,\alpha}$ (the space of $C^{k,\alpha}$ vector fields), and the letter $V$ to denote an element of $\Gamma^{k,\alpha}$, seen as a tangent vector to $\Omega$.

Let $v\in \Gamma^{k,\alpha}$ be our given vector field, and define $f:\Omega\to \Gamma^{k,\alpha}$ by $$f(\Phi) = v\circ \Phi$$ Then $f$ is locally Lipschitz in the $C^{k,\alpha}$ norm, which makes $\Gamma^{k,\alpha}$ a Banach space. This follows from a certainly classical, but somewhat tedious computation I hope I got right (the same needed to show that the $C^{k,\alpha}$ regularity is stable by product and by composition; this last item requires $k\ge1$)

Applying the Picard-Lindelöf theorem to the differential equation $\Phi'(t)=f(\Phi(t))$ in $\Omega\subset \Gamma^{k,\alpha}$ thus yields a unique maximal solution starting at $\Phi(0)=\mathrm{Id}_M$. This solution is obviously the flow of $v$, and now the wanted derivative in $C^{k,\alpha}$ norm follows from the equation itself.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.