$\mathbb{R}^{n}$ is a strongly locally homogeneous space.

Question

I am new to the study of CDH topological spaces, I wanted to study basic examples of this type of spaces for example for the fact that $\mathbb{R}^{n}$ is CDH some authors introduce the notion of Strongly Locally Homogeneous spaces.

A space $X$ is SLH if it has a basis of open subsets $\mathcal{U}$ such that if $p$ and $q$ are two point of $U \in \mathcal{U}$ then there is an homeomorphism $h:X\to X$ such that $h(p) = q$ and such that $h(x) = x$ for every $x$ not in $U$.

Unfortunately I have also failed to demonstrate that $\mathbb{R}^{n}$ is strongly locally homogeneous, does anyone have any ideas?

Thanks a lot

Answer 1

$\def\conv{\operatorname{conv}}$ $\def\inte{\operatorname{int}}$

Let $A$ be a finite subset of $\Bbb R^n$ and $$\Lambda_A=\{(\lambda_x)_{x\in A}\in \mathbb R^A: \sum_{x\in A} \lambda_x=1\mbox{ and }\lambda_x\ge 0\mbox{ for each }x\in A\}.$$ Define a map $\pi_A$ from $\Lambda_A$ to the convex hull $\conv A$ of the set $A$ by putting $\pi_A((\lambda_x)_{x\in A})=\sum_{x\in A} \lambda_x x$ for each $(\lambda_x)_{x\in A}\in \Lambda_A $. Since the map $\pi_A$ is continuous and the set $\Lambda_A$ is compact, the set $\conv A=\pi_A(\Lambda_A)$ is compact too. Moreover, if the set $A$ is affinely independent, that is there are no nonzero map $\lambda: A\to \Bbb R$ such that $\sum_{x\in A}\lambda(x)=0$ and $\sum_{x\in A}\lambda(x)x=0$, then the map $\pi_A$ is injective, and so a homeomorphism. Moreover, if $|A|=n+1$, then the set $\conv A$ has nonempty interior $\inte\conv A$ in $\Bbb R^n$, for instance, by the domain invariance theorem. It easily follows that $$\mathcal B=\{\inte\conv A: A\subset\Bbb R^n\mbox{ is an affinely independent set of size }n+1\}$$ is a base of the space $\Bbb R^n$.

We claim that $\mathcal B$ witnesess that $\Bbb R^n$ is a strongly locally homogeneous space. Indeed, let $A\subset\Bbb R^n$ be any affinely independent set of size $n+1$ and $p, q$ be any points of $\inte\conv A$. Then we have $$\conv A=\bigcup_{x\in A} \conv (\{p\} \cup A\setminus\{x\})=\bigcup_{x\in A} \conv(\{q\} \cup A\setminus\{x\}).$$

Let $B$ be any proper subset of $A$. Let $i_B$ be a map from $\Bbb R^{\{p\} \cup B}$ to $\Bbb R^{\{q\} \cup B}$ such that for each point $t=(t_x)_{x\in \{p\} \cup B}$ we have $i_B(t)=(i_B(t)_x)_{x\in \{q\} \cup B}$, where $i_B(t)_x=t_x$ for each $x\in B$ and $i_B(t)_q=t_p$. Clearly, $i_B$ is a homeomorphism. Now define a map $h_B$ from $\conv (\{p\} \cup B)$ to $\conv (\{q\} \cup B)$ as follows. Let $y$ be any point of $\conv (\{p\} \cup B)$. Since the set $\{p\}\cup B$ is affinely independent, the point $y$ has a unique representation $y=\lambda_p p+\sum_{x\in B} \lambda_x x$ with $\lambda_p+\sum_{x\in B} \lambda_x=1$, $\lambda_p\ge 0$, and $\lambda_x\ge 0$ for each $x\in B$. Put $h_B(y)=\lambda_p q+\sum_{x\in B} \lambda_x x$. It is easy to see that $h_B=\pi_{\{q\} \cup B\setminus\{x\}}\circ i_B\circ \pi^{-1}_{\{p\} \cup B\setminus\{x\}}$, so the map $h_B$ is a composition of continuous injections, and therefore a continuous injection too. It follows that $h_B$ is a homeomorphism between compact sets $\conv (\{p\} \cup B)$ and $\conv (\{q\} \cup B)$.

It is easy to see that for each points $x,y\in A$ the maps $h_{A\setminus\{x\}}$ and $h_{A\setminus\{y\}}$ coincide with the map $h_{A\setminus\{x,y\}}$ on the intersection $\conv (\{p\} \cup A\setminus\{x,y\})$ of their domains $\conv (\{p\} \cup A\setminus\{x\})$ and $\conv (\{p\} \cup A\setminus\{y\})$. Therefore there exists a map $h^p_q$ from $\conv A$ to $\conv A$ such that $h^p_q|\conv (\{p\} \cup A\setminus\{x\})= h_{A\setminus\{x\}}$ for each $x\in A$. In particular, the restriction of $h^p_q$ at the boundary of $\conv A$ is the identity map and $h^p_q(p)=q$. Since $\conv A$ is a union of the finite family $\{\conv (\{p\} \cup A\setminus\{x\})\}_{x\in A}$ of its closed subsets, the map $h^p_q$ is continuous. Since $h^p_q$ has a continuous inverse $h^q_p$, it is a homeomorphism of $\conv A$. Then the extension of $h^p_q$ at $\Bbb R^n\setminus\conv A$ by the identity map provides the required homeomorphism $h$.

Answer 2

Denote by $\mathcal{H}(X)$ the group of autohomeomorphism of the topological space $X$.

In particular $\mathbb{R}^{n}$ is SLH.

Answer 3

Sketch of a proof: notice that open $n$-dimensional simplices (things like segments in $\mathbb R$, triangles in $\mathbb R^2$, tetrahedrons in $\mathbb R^3$ etc.) make up a base of $\mathbb R^n$.

Let $S=(x_0x_1\ldots x_n)$ be a closed simplex in $\mathbb R^n$ and $y,z$ two points in the interior of $S$. By taking $n$ out of $n+1$ vertices of $S$ and adding $y$ (or $z$) as a new vertex, we can "triangulate" $S$ and get new simplices $(yx_1x_2\ldots x_n), (x_0yx_2\ldots x_n),\ldots, (x_0x_1\ldots x_{n-1}y)$ (and similarly for $z$).

Now, create a map of $\mathbb R^n\to\mathbb R^n$ as follows:

• Every point outside $S$ maps to itself.
• Every simplex $S_{y,i}:=(x_0x_1\ldots x_{i-1}yx_{i+1}\ldots x_n)$ maps to $S_{z,i}:=(x_0x_1\ldots x_{i-1}zx_{i+1}\ldots x_n)$ by the unique affine transformation that maps $x_j$ to itself (for $j\ne i$) and maps $y$ to $z$.
• Those transformations can be proven to be identities on the $n-1$-dimensional "face" (simplex) $(x_0x_1\ldots x_{i-1}x_{i+1}\ldots x_n)$ of $S_{y,i}$, and they coincide on the intersections of the simplices $S_{y,i}$ and $S_{y,j}$ for $i\ne j$.
• Thus, "gluing" all of those transformations we get a continuous map $\mathbb R^n\to\mathbb R^n$ which is constant outside the open simplex $\text{Int}(S)$ (interior of $S$).

It is also easily shown that this map is a bijection. Its inverse is the analogous map constructed on $S$ with the roles of the points $y$ and $z$ swapped.