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A topological space $X$ is called homogeneous, if for every two points $x,y \in X$ there exists a homeomorphism $\phi : X \rightarrow X$ s.t. $\phi(x) = y$.

It is not hard to prove that all connected 2-manifolds are homogeneous. The proof basically comes down to the fact that if $D$ is the open disk in $\mathbb{R}^2$ then for every $x,y \in D$ there exists a homeomorphism $\phi : \bar{D} \rightarrow \bar{D}$ such that $\phi(x) = y$, and $\phi \vert_{\partial D}$ is the identity.

Is it true that a general connected manifold is homogeneous?

PatrickR
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Eric Haengel
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    If the manifold has dimension 2 or larger you can go further and send any collection of $n$ points to any other collection of $n$ points via a homeomorphism. Terminology for this is that $Homeo(M)$ acts on $M$ $n$-transitively for all $n$. – Ryan Budney Dec 08 '11 at 22:22

2 Answers2

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Yes, any connected topological manifold $X$ of arbitrary dimension $n$ is homogeneous .

1) The crucial lemma is that given two points $a,b\in \mathbb B^{\circ}$ in the interior of a closed ball $\mathbb B \subset \mathbb R^n$, there exists a homeomorphism $f: \mathbb B\to \mathbb B$ which is the identity on $\partial \mathbb B$ and such that $f(a)=b$.

2) It then follows that if you fix any point $x_0 \in X$, then the set of points $y\in X$ that can be written $y=F(x_0)$ for some homeomorphism $F:X\to X$ is both open and closed, hence is equal to $X$.
Hence $X$ is homogeneous.
(By the way, an obvious modification of the proof shows that the analogous result is also true for a differential manifold: its diffeomorphisms act transitively on the manifold)

Edit: a fishy image
Let me give a physical model which might help visualize the lemma in 1) ( a totally rigorous and amazingly crisp proof is given in t.b.'s great comment).
Imagine you have a spherical fishbowl completely filled with water and a goldfish sitting somewhere in it.
The lemma says that you can send the goldfish to any preassigned place in the bowl by skilfully (!) shaking the bowl.

Georges Elencwajg
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    Would you happen to know how these maps on $\mathbb{B}$ are constructed? – Eric Haengel Dec 08 '11 at 21:58
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    @Eric: one way to do it is as follows: For $a,b \in \mathbb{B}^\circ$ choose a smooth bump function $0 \leq f \leq 1$ supported in $\mathbb{B}^\circ$ which is constant equal to one in a ball containing both $a$ and $b$. Then look at the vector field $X(x) = f(x) (b-a)$. The time one map $\varphi^1$ of the flow $\varphi^t$ of $X$ will satisfy $\varphi^1(a) = b$ and $\varphi^1$ will fix every point outside the support of $f$. – t.b. Dec 08 '11 at 22:10
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    Dear@t.b. what a wonderful proof! My idea was to replace the interior of the ball by a copy of $\mathbb R^n$, to consider a translation that sends one point to the other and modify it so that the translation dies out at infinity . I could then transport it back to the ball and extend it by the identity of the sphere. However this is messy to write up and I was thinking how to best sum that up to answer Eric's query, when I was saved by your comment. Maybe our ideas are similar but your use of a vector field gives a really miraculous, unimprovably concise proof. Congratulation and many thanks. – Georges Elencwajg Dec 09 '11 at 09:00
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    Do these homeomorphisms generate $Homeo(X)$? – Neal Dec 09 '11 at 11:47
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    @Neal: no. Here are three reasons: 1. they are diffeomorphisms, hence the group they generate is a subgroup of the group of diffeomorphisms. 2. They don't generate all diffeomorphisms either. In fact, by construction they lie in the connected component of the identity of $\mathrm{Diff}{(X)}$, which is a normal subgroup and a proper normal subgroup in general. 3. If $X$ is orientable, those diffeomorphisms are orientation-preserving. – t.b. Dec 09 '11 at 13:02
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    Dear Georges, Thank you for your kind words and enthusiasm. I like that proof, too. I don't remember where I learned it -- probably in my differential geometry course quite a while back. Best wishes, Theo. – t.b. Dec 09 '11 at 13:09
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We prove that any connected $T_2$ locally $n$-Euclidean (each point has a neighborhood homeomorphic to $\mathbb R^n$) space $X$ is homogeneous.

Lemma: Let $B^n$ be an $n$-ball and $p, q \in \operatorname{int}(B^n)$, there exists homeomorphism $\phi$ such that $\phi(p) = q$ and $\phi|_{\partial B^n} = \operatorname{id}$.

Proof: Let $\mu \colon \operatorname{int}(B^n) \to \mathbb R^n$, $\mu(x) = \dfrac x {1 - \left\| x \right\|}$ be a homeomorphism. Define $$ \phi(x) = \begin{cases} \mu^{-1} \bigl( \mu(x) - \mu(p) + \mu(q) \bigr) & x \in \operatorname{int}(B^n) \\ x & x \in \partial B^n, \end{cases} $$ it is easily proved that $\phi$ is continuous on the boundary and so do its inverse, thus be a homeomorphism.

Define a binary relation $\sim$ on $X$, $p \sim q$ if there exists a homeomorphism sending $p$ to $q$, it is easily seen that $\sim$ is a equivalence relation.

Take any $x \in X$, let $U$ be its neighborhood homeomorphic to $\mathbb R^n$, then it contains a copies of $B^n$ such that $x \in \operatorname{int}_U(B^n) = \operatorname{int}(B^n)$.

Let $y \in \operatorname{int}(B^n)$, then it follows from the previous lemma that, there exists homeomorphism $\phi \colon B^n \to B^n$, $\phi(x) = y$, $\phi|_{\partial B^n} = \operatorname{id}$.

Because $B^n$ is compact and $X$ is $T_2$ ($T_2$ used here, otherwise there are counterexamples like Line with two origins), we obtain that $B^n$ is closed in $X$. Also, $X \setminus \operatorname{int}(B^n)$ is closed. Therefore, we can use pasting lemma on $\phi$ and $\operatorname{id}_{X \setminus \operatorname{int}(B^n)}$ to get a continuous function sending $x$ to $y$, which has a continuous inverse (Pasting $\phi^{-1}$ and $\operatorname{id}_{X \setminus \operatorname{int}(B^n)}$).

Hence we have $x \sim y$ for $\forall y \in \operatorname{int}(B^n)$. This implies the equivalence class of $x$ is open in $X$. Finally connectivity gives us the desired result.

Kitsune Kiriha
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  • Could you please explain how connectedness gives the desired result? – Duong Ngo Dec 25 '24 at 04:24
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    Since each equivalence class is open, all equivalence class gives an (disjoint) open partition of $X$. But connected space cannot be partitioned into more than one open subsets. – Kitsune Kiriha Dec 25 '24 at 05:15
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    Thanks for the explanation of how the pasting lemma is applicable. Without Hausdorff, the ball $B^n$ would not necessarily be closed in $X$. – PatrickR Dec 25 '24 at 05:16