2
$\begingroup$

Let $M$ be a smooth connected manifold and $\operatorname{Diff}(M)$ the set of diffeomorphisms from $M$ to $M$. I would like to show that this group acts $n$-transitively on $M$.

I started by showing transitivity. I looked at the orbit of one point and showed that is must be both open and closed (relying on the homogeneity of Euclidean space to whom $M$ is locally diffeomorphic and "globalising" via partitions of unity).The result thus follows from connectedness.

Is there some nice way to adapt this argument to obtain $n$-transitivity? Maybe an induction?

Thanks

$\endgroup$
  • $\begingroup$ By $n$-transitively, do you mean that it acts transitively on configurations of $n$ points in $M$? $\endgroup$ – AnonymousCoward Dec 29 '16 at 22:29
  • $\begingroup$ The orbit of the n-uple $(x,...,x)$ is always contained in the diagonal. $\endgroup$ – Tsemo Aristide Dec 29 '16 at 22:38
  • $\begingroup$ Doesn't it just follow from the same argument for $n = 1$ on $M^n$, using the fact that $n$-transitivity holds for $\operatorname{Diff}(\mathbb{R}^N)$? $\endgroup$ – anomaly Sep 2 '17 at 3:04
1
$\begingroup$

Let me outline a construction that works the same way for $n = 1$ and $n > 1$. Assume $M$ is connected.

For motivation, consider first the case $n = 1$ and let $p, q \in M$. Show first that $p, q$ can be connected by an embedded curve $\gamma \colon [0,1] \rightarrow M$ with $\gamma(0) = p, \gamma(1) = q$ (so that the image $\gamma([0,1])$ is a closed one-dimensional embedded manifold with boundary). On $[0,1]$ (with coordinate $t$) with we have vector field $\frac{d}{dt}$ whose time-one flow take $0$ to $1$. If we set $X = \dot{\gamma}(t)$ on $\gamma([0,1])$ and extend $X$ arbitrary to a compactly supported vector field $\tilde{X}$ on $M$ with $\tilde{X}(\gamma(t)) = X(\gamma(t)) = \dot{\gamma}(t)$ then by construction, the curve $\gamma$ is an integral curve of $\tilde{X}$ with $\gamma(0) = p, \gamma(1) = q$ so the time-one flow of $\tilde{X}$ is a global diffeomorphism of $M$ taking $p$ to $q$.

Now assume $n > 1$ and $\dim M > 1$ (otherwise, the result is not true). Show that given distinct $p_1, \dots, p_n$ and distinct $q_1, \dots, q_n$, we can find disjoint embedded curves $\gamma_i \colon [0,1] \rightarrow M$ with $\gamma_i(0) = p_i$ and $\gamma_i(1) = q_i$. Then we have a well-defined vector field on the closed, disconnected, embedded submanifold $\sqcup \gamma_i([0,1])$ (given on each $\gamma_i([0,1])$ like before by $\dot{\gamma_i}(t)$) and so by the extension lemma, we can extend it to a compactly supported global vector field on $M$ whose time-one flow will give us the required result.

$\endgroup$
  • $\begingroup$ How do you get the disjoint embedded curves? How do you prove that's possible? $\endgroup$ – Sov Sep 1 '17 at 20:28
0
$\begingroup$

Hint:for every $x\in M$ Use bump function to construct diffeomorphisms which are equal to the identity on the complement of a neighborhood of $x$ and act transitively on a neighborhood of $x$.

You have a neighborood $U$ of $x$ which is diffeomorphic to a ball $B(0,c)$ of $R^n$. You can find a bump function $f$ defined on $U$ such that the restriction of $f$ to $B(0,c/4)$ is $1$ and $f(y)=0$ if $\|y\|>c/2$. Let $u\in R^n$. You can define the vector field $X_u=fu$ which extend to $M$, and the flow $\phi^u_t$ of $fu$.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.