Suppose $M$ be a connected manifold and $x, y \in M$ are two points. Then I'm trying to show that there is a diffeomeorphism $f$ of $M$ that takes $x$ to $y$.

Since the set of points for which there is a diffeomorphism of $M$ taking $x$ to that point is clopen and $M$ is connected, I think it should be enough to consider the case when both $x$ and $y$ lie in the same chart. But how do I proceed from here? I think I should somehow take a vector field and consider flows, but I'm not really sure.

Can someone please provide a solution? Thanks in advance.


Fix $y \in M$ and define $S_y := \{ x \in M \, | \, \exists \phi \in \mathrm{Diff}_c(M), \phi(x) = y \}$. The set $S_y$ is non-empty (because $y \in S_y$). Let us show that $S_y$ is closed and open. Two properties follow immediately from the definitions:

  1. $x \in S_y$ if and only if $y \in S_x$.
  2. If $x' \in S_x$ and $x \in S_y$ then $x' \in S_y$.

Given $x \in M$, one can find an open subset $U \subseteq M$ and a chart $\varphi \colon U \rightarrow B(0,1)$ around $x$ with $\varphi(x) = \textbf{0}$. Here, $B(\textbf{0},1) \subset \mathbb{R}^n$ and $\varphi = (x^1, \ldots, x^n)$. Given $x' \in U$, denote by $\textbf{c} = \varphi(x')$ the coordinates of $x'$ and choose some $||\textbf{c}|| < r < 1$. The "constant" vector field $X = c^i \frac{\partial}{\partial x^i}$ is well-defined on $U$, and by multiplying $X$ with a bump-function, one can extend $X$ to a globally defined, compactly supported, vector field $\tilde{X}$ with $\tilde{X}|_{\varphi^{-1}(B(0,r))} = X$. Since $\tilde{X}$ has compact support, it generates a globally defined flow $\phi_t$. The curve $\gamma(t) = \varphi^{-1}(t\textbf{c})$ (for $t \in [0,1]$) is an integral curve of $\tilde{X}$ satisfying $\gamma(0) = x$ and $\gamma(1) = x'$ which implies that $\phi_1(x) = x'$. Thus, $x \in S_x'$ and $x' \in S_x$.

Now, let $x \in S_y$. Choosing $U$ as above, we see that for all $x' \in U$ we have $x' \in S_x$ which implies that $x' \in S_y$ and so $S_y$ is open.

Finally, if $x_n \in S_y$ and $x_n \rightarrow x$, then choosing $U$ around $x$ as above, we have $x_n \in S_y$ and $x_n \in U$ for some $n$, which implies that $x \in S_{x_n}$ and so $x \in S_y$.

  • 1
    $\begingroup$ Why bother with showing that $S_y$ is clopen when you explicitly construct the diffeo taking $x$ to $y$? I mean, I'm not complaining, just curious why you did the extra work. $\endgroup$ – Neal Mar 19 '15 at 4:29
  • 3
    $\begingroup$ @Neal He only constructed a diffeomorphism taking $x$ to $x'$, where $x'$ lived in a coordinate neighborhood about $x$. For arbitrary $y$, you need to use the connectedness assumption. Furthermore, there's no need to show $S_y$ is closed, as it was shown to be an equivalence relation, and hence would partition $M$. $\endgroup$ – Matt Mar 19 '15 at 22:49
  • $\begingroup$ @MattRobinson Ah yes, I see your point. To make it precise one needs an embedded path and a tubular neighborhood around that path. $\endgroup$ – Neal Mar 20 '15 at 1:49

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.