0
$\begingroup$

I wish to know if there is a sort of "constrained Moser Trick". Suppose we have a planar grap $G \subseteq \mathbb{R}^2$, with $(0,0)$ as a vertex. Suppose to have some volume form $\omega = g_\ast(dx \wedge dy)$, where $g$ is a diffeomorphism of the plane that send $0$ into $0$.

By construction, the Moser equation

$\frac{d}{dt}\omega_t + \mathcal{L}_{X_t}\omega_t = 0$ with $\omega_t = \omega(1-t) + tdx\wedge dy$

can be solved, and we can find some $h$ (in this case $= g^{-1}$) such that $h_\ast(\omega) = dx \wedge dy$.

My question is: can we search for a solution $X_t$ that is invariant for $G$? In other word, if $\frac{d}{dt}\psi_t = X_t \circ \psi_t$, then $\psi_t(G) \subseteq G$?

Improve this question
$\endgroup$
9
  • 2
    $\begingroup$ The areas of planar regions cut out by $G$ surely have to be the same for both area forms. $\endgroup$
    – Ben McKay
    Commented Oct 3 at 14:25
  • $\begingroup$ And if we suppose that $G$ is a tree? $\endgroup$
    – Mirko
    Commented Oct 3 at 15:05
  • $\begingroup$ This is a special case of Moser's theorem for manifolds with corners, arxiv.org/abs/1604.07787, under a mild assumption about values of the area forms at the vertex points (they should coincide). To treat the general case, you should read their proof and try to adapt it to your setting. $\endgroup$
    – Moishe Kohan
    Commented Oct 3 at 15:32
  • $\begingroup$ @MoishaKohan I'll look into it. If that does the job, you have just saved my master graduation this month, and hence my phd admission $\endgroup$
    – Mirko
    Commented Oct 3 at 16:26
  • 1
    $\begingroup$ Also, you should be careful about the degree of smoothness of the diffeomorphism, as far as I can tell, you get $C^0$ for free, get $C^1$ relatively easily, but to get $C^\infty$ you need more work. $\endgroup$
    – Moishe Kohan
    Commented Oct 3 at 16:44

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .