$k$-jet transitivity of diffeomorphism group

Question

Given a connected smooth manifold $M$ and an invertible jet $\xi \in {\rm inv} J^k_p(M,M)_q$, what are the required conditions for the existence of a diffeomorphism $\phi \in {\rm Diff}(M)$ such that $j^k_p \phi = \xi$? What about if we want $\phi$ to be isotopic to the identity? Given a smooth family $\xi_\epsilon \in {\rm inv} J^k_p(M,M)_q$ can we get a smooth family $\phi_\epsilon \in {\rm Diff}_0 (M)$?

It's clear that we can find such a diffeomorphism from a neighbourhood of $p$ to a neighbourhood of $q$, and also that there must be some smooth map $M \to M$ contacting $\xi$, but I have no idea how to extend the former or make the latter a diffeomorphism. At least in low dimension it feels like the only condition should be that $\xi$ is orientation-preserving, but I have no idea how to establish this.

Answer 1

I think that if you assume $\dim(M)\geq 2$ (?) and that $M$ is connected (or at least that all connected components of $M$ are diffeomorphic), then you can always construct an appropriate diffeomorphism which is isotopic to the identity. I can't give a full proof, but I think I can describe how to get it:

The main input I can give is reducing things to the case that $p=q$. This is done by choosing a smooth injective path $c:[0,1]\to M$ connecting $p$ and $q$, then extending the velocity field $c'$ to a vector field in a tubular neighborhood of $c([0,1])$ the path and then multiplying it by an appropriate bump function. This gives you a vector field with compact support (which thus has a global flow) and agrees with the given velocity field locally around the chosen path. Hence the flow of this vector field at time $t=1$ is a diffeomorphism (with compact support) that maps $p$ to $q$.

Composing with this diffeomorphism you can reduce things to the case $q=p$, in which working with vector fields becomes much easier. Since we are now talking about diffeomorphisms fixing $p$, we can use vector fields vanishing at $p$, and any germ of such a vector field now easily extends to a globally defined, compactly supported vector field, which has a global flow. Now one could either try to study the map from jets of vector fields to jets of diffeomorphism induced by the local flow. The other possibility would be to argue step by step, showing next that any one-jet can be realized by an appropriate vector field and so on.

Answer 2

As pointed out by Andreas we can assume $p=q$. Fix a chart $(U,\psi)$ with $\psi(p)=0$ and let $g : \mathbb R^n \to \mathbb R^n$ be the unique $k^\text{th}$ order polynomial such that $j^k_p (\psi^{-1} g \psi) = \xi$. Note that $g(0) = 0$ because $\xi \in J_p (M,M)_p$. Let $L$ be the linear part of $g$. Restrict the domain of $g$ to some ball $B_R(0)$ small enough that $g$ is an embedding.

Define $g : (0,1] \times B_R \to \mathbb R^n$ by $g_t(x) = g(tx)/t$ and note that $\lim_{t \to 0} g_t = g'(0) = L$, so this $g_t$ defines an isotopy of $g$ to $L$. Any smooth path from $L$ to the identity in $GL(n,\mathbb R)$ (here we need the condition that $\xi$ is orientation-preserving!) is then an isotopy of $L$ to the inclusion map $i :B_R \to \mathbb R^n$; so composing these two isotopies we get a diffeotopy $f : [0,1] \times \mathbb R^n \to \mathbb R^n$ from $g$ to $i$.

Now, let $r_1,r_2 < R$ be small enough that the origin-centred balls $B_{r_1},B_{r_2}$ lie inside $\psi(U)$ and $f(I\times B_{r_1}) \subset B_{r_2}$$^\dagger$; and define the isotopy

$$ F : [0,1] \times \psi^{-1}(B_{r_1}) \to M$$ by $F_t = \psi^{-1} f_t \psi$. Then applying the isotopy extension theorem (see e.g. Hirsch section 8.1) we get a diffeotopy of $M$ that agrees with $F$ on a neighbourhood of $p$, and thus whose endpoint has jet $\xi$ at $p$.

$\dagger$ We can pull this off because 0-fixing polynomials satisfy $g(tx) \le C t g(x)$ for $t\in[0,1]$ and $C$ independent of $t$, so the dilation isotopy can't make the image blow up; and similarly the path in $GL(n,\mathbb R)$ can be chosen so that the operator norm is bounded.