Suppose $M$ is a connected smooth manifold and $x,y \in M$ are two points. Is there always a diffeomorphism $\phi: M \rightarrow M$ with $\phi(x)= y$ ?
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$\begingroup$ See topospaces.subwiki.org/wiki/…; in particular, the last argument is nice: it is easy to see that $Aut(M)$-orbits in $M$ are open, so they are also closed, so if the manifold is connected they cannot be many things. $\endgroup$– Mariano Suárez-ÁlvarezCommented Mar 19, 2012 at 4:27
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$\begingroup$ Ok. So on a connected manifold there is at least a homeomorphism. Is it automatic a diffeomorphism, too? I guess but I'm not sure. $\endgroup$– Mark.NeuhausCommented Mar 19, 2012 at 4:36
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$\begingroup$ There is a diffeomorphism of $M$ sending $x$ to $y$ if and only if $x$ and $y$ are in diffeomorphic path-components of $M$. If $M$ is connected, you get the hypothesis for free. This is true for any of the standard structures on manifolds: topological, PL or smooth. It's not true generally when you have more refined structures on manifolds, like Riemann metrics. $\endgroup$– Ryan BudneyCommented Mar 19, 2012 at 4:45
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10$\begingroup$ I like to give this problem as homework in a first year graduate course, so it is unfortunate it now appears at MO. $\endgroup$– Igor BelegradekCommented Mar 19, 2012 at 12:16
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3$\begingroup$ For the openness of the orbits it's enough to exhibit a nontrivial compactly supported diffeomorphism from $\mathbb R^n$ to itself. To get this for $n>1$ it's enough to get it for $n=1$. $\endgroup$– Tom GoodwillieCommented Mar 19, 2012 at 13:56
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2 Answers
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Partially this is a response to Mariano's 2nd comment.
In the smooth manifold case there's actually a really slick proof. Here it is:
Let $\gamma : [0,1] \to M$ be a smooth path in $M$ such that $\gamma(0)=p$ and $\gamma(1)=q$. You can re-consider this map to be an isotopy from the $0$-dimensional submanifold $\{p\}$ to the submanifold $\{q\}$.
The isotopy extension theorem then says, there exists a smooth map
$$G : [0,1] \times M \to M$$
such that the function $G_t : M \to M$ given by $G_t(x) = G(t,x)$ is a diffeomorphism for all $t \in [0,1]$, and $G_0 = Id_M$. Also, it guarantees $G(t,p) = \gamma(t)$. So $G_1 : M \to M$ is a diffeomorphism that sends $p$ to $q$.
The nice thing about this argument is it readily generalizes. For example, take the configuration space of $k$ distinct points in the manifold $M$, $C_k M$.
$$C_k M = \{(p_1,\cdots,p_k) \in M^k : p_i \neq p_j \forall i \neq j\}$$
It's not too hard to argue that if $\dim(M) \geq 2$ and $M$ is connected, then $C_k M$ is connected. So isotopy extension kicks in again and says, if $(p_1, \cdots, p_k)$ are $k$ distinct points in $M$ and $(q_1, \cdots, q_k)$ are also $k$ distinct points, then there exists a diffeomorphism $f : M \to M$ such that $f(p_i) = q_i$ for all $i$. Another way to say this is that $\operatorname{Diff}(M)$ acts $k$-transitively on the manifold.
For $1$-dimensional manifolds, generally $C_k M$ is not connected, and $\operatorname{Diff}(M)$ is not $k$-transitive.
answered Mar 19, 2012 at 5:36
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$\begingroup$ I surely prefer this argument for the generalization to $k$-transitivity! But surely (I don't know the proof) the proof of the isotopy extension theorem involves some explicit construction at least as elaborate as the one needed to prove that diffeos of compact support act transitively on $\mathbb R^n$? $\endgroup$ Commented Mar 19, 2012 at 5:57
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$\begingroup$ Formally it's about equally as difficult, but the proof of the isotopy extension theorem is extremely intuitive. The idea is if $F: [0,1] \times N \to M$ is a smooth 1-parameter family of embeddings, consider the image of $\overline{F} : [0,1] \times M \to [0,1] \times M$ given by $\overline{F}(t,p) = (t,F(t,p))$. $\frac{\partial \overline{F}}{\partial t} \circ \overline{F}^{-1}$ is a vector field on this image so the idea is to extend this vector field to $[0,1] \times M$, and use the flow to define $G$. Constructing the extension requires some bump functions. $\endgroup$ Commented Mar 19, 2012 at 6:02
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$\begingroup$ The domain of $\overline{F}$ was supposed to be $[0,1] \times N$. Er, maybe I shouldn't try to explain the proof here. It's in Hirsch. $\endgroup$ Commented Mar 19, 2012 at 6:03
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1$\begingroup$ The argument for $k$-transitivity shows there is a diffeo $C_k(M)\to C_k(M)$ mapping one $k$-tuple to the other, but not that that diffeo is actually some diffeo $f:M\to M$ acting diagonally, no? $\endgroup$ Commented Apr 24, 2012 at 7:35
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Here's another argument, exploiting the differentiable structure a bit more (in the sense that I don't see a straightforward way to adapt it to the topological/PL case). It might be the standard argument to show that things work locally, but it works globally as well.
Take a smooth, embedded path $\gamma: [0,1]\to M$ so that $\gamma(0)=p$ and $\gamma(1)=q$. Since $\gamma$ is embedded, we can push forward $\partial/\partial t$ and get a vector field $W$ on the image of $\gamma$. Let $V$ be a compactly supported extension of $W$ to all of $M$.
Then it's easy to check that the flow of $V$ at time 1 is a diffeomorphism of $M$ that sends $p$ to $q$ (and is isotopic to the identity).
This argument readily generalises to $k$-transitivity, too: if $\dim M\ge 3$, we can choose generic paths $\gamma_i$ simultaneously, and use the same exact argument. If $M$ is a surface, we need to use the fact that a single embedded path doesn't disconnect $M$.
answered Mar 19, 2012 at 11:13
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1$\begingroup$ I like this one best. :) $\endgroup$ Commented Mar 22, 2012 at 4:33