Let $X$ be a continuous vector field on a (say compact) manifold $M$, if $X$ has ODE uniqueness then we can define its associated flow $\mathcal F_X:\mathbb R\times M\to M$ uniquely given by $\mathcal F_X(0;x)=x$ and $\frac d{dt}\mathcal F_X(t,x)=X(F_X(t,x))$.

When $X,Y$ are two Lipschitz vector fields on (say compact) manifold $M$, we know if $[X,Y]=0$ then $\mathcal F_X^t\circ \mathcal F_Y^s=\mathcal F_Y^s\circ\mathcal F_X^t$ for all $s,t\in\mathbb R$.

A proof, for example, can be done by using approximation of smooth manifold, and apply formula $\mathcal F_X^{-t}\mathcal F_Y^{-s}\mathcal F_X^t\mathcal F_Y^sx-x=\int_0^t\int_0^s([X,Y](f_{t,\tau,\sigma}))(\mathcal F_X^{t-\sigma}\mathcal F_Y^\tau x)d\tau d\sigma$ where $f_{t,\tau,\sigma}=\mathcal F_X^{-t}\mathcal F_Y^{-\tau}\mathcal F_X^\sigma:\mathbb R^n\to\mathbb R^n$ (Lemma A.8 in this paper). The approximation works because when $X_k\to X,Y_k\to Y$ in Lipschitz space, then $[X_k,Y_k]\xrightarrow{L^\infty}0$ and $\mathcal F_{X_k}\xrightarrow{Lips}\mathcal F_{X}$, $\mathcal F_{Y_k}\xrightarrow{Lips}\mathcal F_Y$, two $L^\infty$-functions are multiplicable, and composition of two Lipschitz function are still Lipschitz.

When vector fields are below Lipschitz, for example, if $X,Y$ are log-Lipschitz, the Lie bracket is still well-defined as distribution. Let $\alpha>\frac12$, if $X,Y\in C^\alpha$, then $[X,Y]$ is a $C^{\alpha-1}$-distribution based on the fact that a $C^\alpha$ function can multiply with a $C^{\alpha-1}$-distribution.

My question is, if $X,Y$ are non-Lipschitz and both has ODE uniqueness, do we still have $\mathcal F_X^t\circ \mathcal F_Y^s=\mathcal F_Y^s\circ\mathcal F_X^t$?

When the thing goes below Lipschitz, the approximation breaks down because we cannot multiply two $C^{-\epsilon}$-functions whenever $\epsilon>0$: When $f\in C^{1-\epsilon}$ and $X\in C^{-\epsilon}$ then $Xf$ is undefined. And from here we know a flow log-Lipschitz vector field is merely $C^{1-\epsilon}$ when the time is small.

I try a lot to use approximation and apply the vanishing Lie bracket condition, but most of the time the bad regularity of composition make things undefined in a function space.

One attempt to make ``the vector field'' defined is to consider the generator of $C_0$-semigroup. Consider $\{\mathcal F_X^t\}_{t\in\mathbb R}$ as a $C_0$-group operator $S_X(t):C^0(M)\to C^0(M)$, given by $S_X(t)f=f\circ\mathcal F_X^t$. So we know $\{S_X(t)\}$ is one parameter subgroup of bounded linear operators on $C^0(M)$ which is strongly continuous. By functional analysis, we know $\frac d{dt}|_{t=0}S(t)$ gives a unbounded operator which is densely defined closed.

Then fix $s\in\mathbb R$, the family $\{S_Y(-s)S_X(t)S_Y(s)\}_{t\in\mathbb R}$ is a $C_0$-group as well. Say its infinitesimal generator be $T_s:D(T_s)\subset C^0(M)\to C^0(M)$. Can we define $\frac d{ds}T_s$ and show that it ``equals'' to $[X,Y]$?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.