Is the solution of a non-stationary differential equation always a diffeomorphism?

Question

Let $\Omega$ be a compact subset of $\mathbb{R}^n$. I read that the Lie algebra of the Lie group $\text{Diff}(\Omega)$, is the algebra of stationary vector fields over $\Omega$, sometimes indicated with $\mathcal{V}(\Omega)$ (Khesin Geometry of infinite dimensional Lie group pag 10).

If we consider the non-stationary case, as the differentiable vector field $\mathbf{v}(t,\mathbf{x})$ that depends on the time parameter as well as on the position in $\Omega$, this gives raise to the non-stationary differential equation: $$ \frac{\text{d} \varphi_{\tau}(\mathbf{x}) }{\text{d}\mathbf{x}}\vert_{t=\tau} = \mathbf{v}(\tau,\varphi_{\tau}(\mathbf{x})). $$ Is the solution of such ODE always a diffeomorphism? Then why the non-stationary vector fields are not included in the Lie algebra of the group of diffeomorphisms?

My guess is that the diffeomorphisms that solves the stationary ODE are in a different space (n+1 dimension) than the diffeomorphisms that solves the non-stationary one. In addition if we consider the diffeomorphism that solves the non-stationary ODE for every value of the time parameter in $\Omega$, this is not anymore a diffeomorphism. How much am I wrong?

Do you have any hints to proving the fact initially stated ($Lie(\text{Diff}(\Omega)) = \mathcal{V}(\Omega)$, where $Lie$ is the Lie algebra of a group).

Thanks!

Answer 1

You have the right intuition: in general nonautonomous differential equations don't give rise to diffeomorphisms on their ambient space. Indeed we need at least one additional dimension for that, although there is no canonical way of adding this or other dimensions.

Some people will tell you that $t'=1$ is canonical, which means that you would look at the autonomous equation $(t',\mathbf x')=(1,\mathbf v(t,\mathbf x))$, but really it is only one of many possibilities. For example this choice makes one loose compactness which depending on our interests may be bad.

On the other hand, it is a somewhat simple exercise in the basic theory of ordinary differential equations to show that in the autonomous case (which you call stationary, really this is quite noncanonical) the two spaces are isomorphic. Basically your identity tells you how to recover $\mathbf v$ from the flow and it is somewhat easy to show that the solutions $\varphi_{\tau}(\mathbf{x})$ of $$ \frac{\text{d} \varphi_{\tau}(\mathbf{x}) }{\text{d}\mathbf{x}}\vert_{t=\tau} = \mathbf{v}(\varphi_{\tau}(\mathbf{x})) $$ are in fact diffeomorphisms. The trickiest part is showing that these are indeed diffeomorphisms and not only homeomorphisms. This is the usual smooth dependence of solutions on the initial conditions (the trickiest is showing this smooth dependence, then it follows immediately that the homeomorphisms are in fact diffeomorphisms).