3

Recently, I tried to understand how to proof the $n$-transitivity of $\operatorname{Diff}(M)$ acting on a smooth connected manifold $M$. I found a proof here $n$-transitivity of $\operatorname{Diff}(M)$ acting on a smooth manifold $M$ and another article on researchgate which follows the same ideas. However, I still have questions because of some details.

  • How do we prove that we actually can have disjoint embedded curves? What are the hypothesis that we need to prove it? Because in one dimension it is obvious that it doesn't work. Both references take dim $\geq$ 2. But in two dimensions the closed unit disk is also a counterexample if we take points on the boundary (for example $(-1;0) \mapsto (1;0)$ and $(0;-1) \mapsto (0;1)$). So, do we have to add that $M$ is without boundary in our hypothesis? If we suppose those two, what's the proof of that? It doesn't seem elementary... (Or it is?)

  • We took $(x_1, ... x_n)$ to be disctincts points, such as $(y_1, ..., y_n)$. But we need that both sets doesn't share common elements, isn't it? I think it is necessary for the proof, but maybe it is just a misunderstanding on my part of the defintion of $n$-transitivity.

Sov
  • 1,048
  • You can never move a boundary point to the interior via a diffeomorphism. So yes, they want you to consider the interior (a manifold without boundary). 2) Nope, you need not impose any disjointness condition between X and Y (but you do need to demand that each of the $x_i$ are pairwise distinct from one another; you must, independently, demand the same of $Y$.)
    • –  Sep 02 '17 at 04:39
  • Yes, but what if all the points I consider are on the boundary?

    Well, if I consider the $n$-transitivity definition from here, we ask disjointness. Moreover, if those aren't disjoints, then the embedded curves will share common points...

    Thanks for your answer.

     – Sov Sep 02 '17 at 12:26