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In a question from last week, I was searching for "nice" spaces on which the group of self-homeomorphisms acted $1$-transitively but not $2$-transitively on the space. In that case, I had the condition that the space be connected and Hausdorff.

The example given in answer to that question was related to the "Long Line."

Now I'm asking more concretely:

Can we find a connected infinite sub-space of Euclidean space, $X\subset\mathbb R^n$ for some $n$, on which the entire group of self-homeomorphisms of $X$ acts $1$-transitively on $X$ but not $2$-transitively? Ideally, examples without the axiom of choice, although examples using the axiom of choice would be interesting, too.

[In the last question, someone asked for the definition of $n$-transitivity. Wolfram's MathWorld has a slightly flawed definition, but it works for infinite sets.]

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up vote 4 down vote accepted

I believe I have found an example as the following argument should show. In case someone finds a mistake, or further clarifications are required, comments are of course welcome.

Let $S^1\subseteq\Bbb C$ be the unit circle and let $f:\Bbb R\to S^1\times S^1$ be defined by $f(t)=(e^{it},e^{i\sqrt{2}t})$. Define $X=f(\Bbb R\setminus\{0\})$. (With the usual identification of $\Bbb C=\Bbb R^2$, we have that $X$ is a subspace of $\Bbb R^4$ and since the torus embeds into $\Bbb R^3$ we can also regard $X$ as embedded into $\Bbb R^3$).

Now we shall prove some lemmas to show that $X$ has the properties we seek.

Lemma 1. $X$ is connected.

Proof. Clearly, $X$ has at most two components, since the sets $A=f((-\infty,0))$ and $B=f((0,\infty))$ are connected. To see $X$ is connected, it is enough to show that $A$ and $B$ are not separated. To see this, examine points of the form $f(2k\pi)=(1,e^{2\sqrt{2}k\pi i})$, where $k\in\Bbb N$. These are dense in $\{1\}\times S^1$, because $\sqrt{2}\pi$ is an irrational multiple of $\pi$. The same holds for points of the form $f(-2k\pi)$, where $k\in\Bbb N$. So we have found a subset of $A$ and a subset of $B$ that are both dense in $\{1\}\times S^1$. This implies that $A$ and $B$ are not separated. $\square$

Lemma 2. The action of $\operatorname{Homeo}(X)$ on $X$ is $1$-transitive.

Proof. Write $x\sim y$ whenever there is a homeomorphism $h:X\to X$ such that $h(x)=y$. As we know, this is an equivalence relation. We will show that every two points $x,y\in X$ are equivalent, i.e. in the same orbit of the action. This follows from several simple cases:

Case 1. If there is a $k_0\in\Bbb Z$ such that $a\in f((2k_0\pi,2(k_0+1)\pi))$, then $a\sim f(2k_0\pi+\pi)$.

First, for every $r\in(0,2\pi)$ define a map $g_r:\Bbb R\to\Bbb R$ as follows: $$g_r(t)=\begin{cases}2k\pi+\frac{t-2k\pi}{r}\pi;& \text{if }\exists k\in\Bbb Z:t\in[2k\pi,2k\pi+r]\\2k\pi+\pi+\frac{(t-2k\pi-r)}{2\pi-r}\pi;& \text{if }\exists k\in\Bbb Z:t\in[2k\pi+r,2(k+1)\pi]\end{cases}$$

The relevance of $g_r$ is that it is a homeomorphism of $\Bbb R$ that for $k\in\Bbb Z$ has fixed points $2k\pi$ and $g_r(2k\pi+r)=2k\pi+\pi$ and, most importantly, $g_{r}(t+2k\pi)=g_{r}(t)+2k\pi$ for all $t\in\Bbb R$ and $k\in\Bbb Z$. Now let $a=f(t_0)$, let $r_0=t_0-2k_0\pi$ and define $h:X\to X$ by $$h(f(t))=f(g_{r_0}(t))$$

This is a homeomorphism of $X$. To see this, we define $H:S^1\times S^1\to S^1\times S^1$ by $$H(e^{it},e^{is})=(e^{ig_{r_0}(t)},e^{i(\sqrt{2}g_{r_0}(t)+s-\sqrt{2}t)})$$ This is a well-defined map: suppose $T=t+2k\pi$ and $S=s+2l\pi$ for some $k,l\in\Bbb Z$. We will use the crucial property of $g_{r_0}$ that $g_{r_0}(t+2k\pi)=g_{r_0}(t)+2k\pi$ for all $k\in\Bbb Z$. Let's calculate: $$e^{ig_{r_0}(T)}=e^{ig_{r_0}(t+2k\pi)}=e^{ig_{r_0}(t)+i2k\pi}=e^{ig_{r_0}(t)}$$ and $$\begin{align}e^{i(\sqrt{2}g_{r_0}(T)+S-\sqrt{2}T)}&=e^{i(\sqrt{2}g_{r_0}(t+2k\pi)+s+2l\pi-\sqrt{2}t-2\sqrt{2}k\pi)}\\&=e^{i(\sqrt{2}g_{r_0}(t)+2\sqrt{2}k\pi+s+2l\pi-\sqrt{2}t-2\sqrt{2}k\pi)}\\&=e^{i(\sqrt{2}g_{r_0}(t)+s-\sqrt{2}t)+i2l\pi}\\&=e^{i(\sqrt{2}g_{r_0}(t)+s-\sqrt{2}t)}\end{align}$$ This shows that $H$ is well defined. Its inverse $H^{-1}:S^1\times S^1\to S^1\times S^1$ is defined by $$H^{-1}(e^{iu},e^{iv})=(e^{ig^{-1}_{r_0}(u)},e^{i(\sqrt{2}g^{-1}_{r_0}(u)+v-\sqrt{2}u)})$$ This is again well defined. The proof is basically the same as for $H$, except that this time we use that $g^{-1}_{r_0}(t+2k\pi)=g^{-1}_{r_0}(t)+2k\pi$. This proves that $H$ is a bijection. As both $H$ and $H^{-1}$ are continuous, $H$ is a homeomorphism.

Now, we notice that $h$ is just the restriction of $H$ to $X$. Since the restriction of a homeomorphism is a homeomorphism (onto its image, which in this case is precisely $X$), $h:X\to X$ is itself a homeomorphism.

Case 2. If there is a $k_0\in\Bbb Z$ such that $a\in f((2k_0\frac{\pi}{\sqrt2},2(k_0+1)\frac{\pi}{\sqrt2}))$, then $a\sim f(2k_0\frac{\pi}{\sqrt2}+\frac{\pi}{\sqrt2})$.

The proof of this case is basically the same as in the first case, except that instead of $g_r$ we use a homeomorphism that has fixed points at $2k\frac{\pi}{\sqrt2}$, etc. For example, $t\mapsto \frac1{\sqrt{2}} g_r(\sqrt{2}t)$ should do the job. (The proof is analogous because the $S^1$'s in $S^1\times S^1$ appear in a completely symmetric manner.)

Case 3. If $a=f(t_0)$ for some $t_0>0$ then $a\sim f(-t_0)$.

Here we simply use the map $h:X\to X$ defined by $h(f(t))=f(-t)$.

Now, cases 1 and 2 show that every two points in the $f$-image of an interval of the form $(2k\pi,2(k+1)\pi)$ or $(2k\frac\pi{\sqrt2},2(k+1)\frac\pi{\sqrt2})$ are equivalent. Because of the way how such intervals overlap, we see that every two points in $A$ are equivalent and every two points in $B$ are equivalent. Case 3 then shows that some two points from $A$ and $B$ are equivalent. Conclusion: all points of $X$ are equivalent. $\square$

Lemma 3. The action of $\operatorname{Homeo}(X)$ on $X$ is not $2$-transitive.

Proof. To see this, let $x,y\in A$ and $z\in B$. Then there cannot be any homeomorphism that takes $x$ to $x$ and $y$ to $z$. To see this, it suffices to show that $A$ and $B$ are precisely the path components of $X$. Since they are clearly path-connected, it is enough to show that $X$ is not path connected.

To do this, we shall first define $U_-=X\setminus(S^1\times\{-1\})$ and $U_+=X\setminus(S^1\times\{1\})$. Then the following holds:

  • connected components of $U_+$ are precisely the sets $f((2k\frac\pi{\sqrt2},2(k+1)\frac\pi{\sqrt2}))$ for $k\in\Bbb Z$
  • connected components of $U_-$ are the sets $f((2k\frac\pi{\sqrt2}-\frac\pi{\sqrt2},2k\frac\pi{\sqrt2}+\frac\pi{\sqrt2}))$ for $k\in\Bbb Z\setminus\{0\}$ and because $f(0)\notin X$ two further components: $f((-\frac\pi{\sqrt2},0))$ and $f((0,\frac\pi{\sqrt2}))$

We will now show how to prove the first of these claims. The second one has a similar proof.

Clearly, the sets of the form $f((2k\frac\pi{\sqrt2},2(k+1)\frac\pi{\sqrt2}))$ are connected. They are indeed subsets of $U_+$ because $e^{i\sqrt2t}=1$ if and only if $t=2l\frac\pi{\sqrt2}$ for some $l\in\Bbb Z$ and there is no such $t$ in the interval $(2k\frac\pi{\sqrt2},2(k+1)\frac\pi{\sqrt2})$. Now we just have to show that these are not contained in any larger connected set.

We will show that these are indeed the quasi-components of $U_+$. To show this, define for each $x,y\in\Bbb R$ an open set $U_{x,y}$ as follows: $$U_{x,y}=U_+\cap\{(e^{iw}e^{it},e^{i\sqrt2t})|\;t\in(0,2\frac{\pi}{\sqrt2}),w\in(x,y)\}$$ This is an open set by definition. In fact, it is clopen for most choices of $x$ and $y$, since $(e^{iw}e^{it},e^{i\sqrt2t})\in X$ is only possible if $w=2k\pi+2l\frac{\pi}{\sqrt2}$ for some $k,l\in\Bbb Z$. Indeed, the set of points $(x,y)$ for which $U_{x,y}$ is clopen, is dense in $\Bbb R^2$.

Now, let $C=f((2k\frac\pi{\sqrt2},2(k+1)\frac\pi{\sqrt2}))$ for some $k\in \Bbb Z$. There is a unique $L\in[0,2\pi)$ such that $C=\{(e^{iL}e^{it},e^{i\sqrt2t})|\;t\in(0,2\frac{\pi}{\sqrt2})\}$. By the above, we may thus choose a sequence $(x_n,y_n)_n$ such that for each $n\in\Bbb N$ we have $x_n<L<y_n$, $\lim_{n\to\infty}(y_n-x_n)=0$ and each set $U_{x_n,y_n}$ is clopen. Then $C=\bigcap_{n=1}^{\infty}U_{x_n,y_n}$, which shows that $C$ is a quasi-component. This proves the first bullet point.

Path-disconnectedness now follows easily from the two bullet points above: suppose $\gamma:[0,1]\to X$ is a path from $a\in A$ to $b\in B$. The two bullet points above imply that the (non-empty) preimages $\gamma^{-1}(A)$ and $\gamma^{-1}(B)$ are both open in $[0,1]$. Let $x\in\gamma^{-1}(A)$. Then $x\in U_+$ or $x\in U_-$. Either way, there exists an open interval $V$ containing $x$ such that $\gamma(V)\subseteq f((t_1,t_2))$ for some real numbers $t_1<t_2<0$ and thus $V\subseteq\gamma^{-1}(A)$. The same argument holds for $B$. This is a contradiction, since $[0,1]$ is connected. The proof is thus complete. $\square$

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This looks promising to me. – Olivier Bégassat Jul 25 '12 at 3:56
It's a bit overkill, but as far as showing $A$ and $B$ are each individually 1-transitive, one can note that, up to reparamaterization, they are geodesics of the flat torus and the flat torus is a symmetric space. In a symmetric space, all geodesics are given by 1-parameter families of isometries. – Jason DeVito Jul 25 '12 at 4:08
For connectedness, just use that $X$ is homeomorphic to the set $\mathbb R\setminus\{0\}$ with a sub-topology of the standard topology. A pair $U,V$ of disjoint subsets of this space are also a pair of open subsets under the standard topology, and the only such pair such that $U\cup V$ is the whole space are $\{U,V\}=\{\mathbb R^+,\mathbb R^-\}$. So you need only prove that $\mathbb R^+$ cannot be open in this sub-topology. But that follows by the fact that any neighborhood in $X$ having points from both $A$ and $B$. – Thomas Andrews Jul 25 '12 at 14:39
Since $\mathrm {Homeo}(X)$ has an operation that sends $A$ to $B$, you only need to prove that $\mathrm{Homeo}(X)$ acts transitively on $A$ (or on $B$). I'm still not sure you can get transitivity, however. – Thomas Andrews Jul 25 '12 at 14:57
@ThomasAndrews: I believe the argument is now complete. (Modulo some minor details that should be easy to fill in.) – Dejan Govc Jul 26 '12 at 14:56

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