A subset $B$ of a group $G$ is called *balanced* if $gBg^{-1}=B$ for all $g\in G$.


An action of a group $G$ on a metric space $X$ is called *ballanced* if for each non-empty balanced subset $B\subset G$ and each $x\in X$ the set $Bx$ coincides with an open or closed ball centered at $x$. 


This means that the topology (and a large piece of the metric structure) of $X$ can be recovered from the ballanced action of the group $G$ on the set $X$.


The term "ballanced" was coined as a mix of two words: "ball" and "balanced".


**Fact.** The standard action of the group $SO(3)$ on the sphere $S^2$ is ballanced.


*Proof.* Take any non-empty balanced subset $B\subset SO(3)$ and observe that any element $g\in B$ is a rotation of the sphere by some angle $\alpha$. Being balanced, the set $B$ contains all possible rotations of the sphere by the angle $\alpha$. Looking at spherical equilateral triangles with base points $x,y$ on the sphere and the angle $\alpha$ at the vertex, we can see that the points $y$ fill a closed spherical disk $D\_g$ centered at $x$. The union $\bigcup\_{g\in B}D\_g$ of such spherical disks is an open or closed spherical disk in $S^2$, which coincides with $Bx$. 


**Problem 1.** For which $n$ the action of the group $SO(n)$ on the sphere $S^{n-1}$ is ballanced? Is it ballanced for $n=5$? Is it ballanced for all odd $n$?


**Remark.** The action of $SO(n)$ on $S^{n-1}$ is ballanced for $n\in\{1,3\}$ but not ballanced for any even $n$ (since the singleton $B=\{-1\}$ consisting of the map $-1:x\mapsto -x$ is balanced in $SO(n)$ but $Bx=\{-x\}$ is not a ball centered at $x$). So, the question actually concerns odd $n$. But for even $n$ we can ask a local version of the ballanced property.


An action of a topological group $G$ on a metric space $X$ will be called *locally ballanced* if there exists a neighborhood $U\subset G$ of the unit such that for any non-empty balanced subset $B\subset U$ and any $x\in X$ the set $Bx$ is an open or closed ball centered at $x$ in the metric space $X$.


**Problem 2.** For which $n$ the action of the group $SO(n)$ on the sphere $S^{n-1}$ is locally ballanced? Is it locally ballanced for $n=4$? Is it locally ballanced for all $n\ge 3$?



Is the action of $SO(n)$ on the sphere $S^{n-1}$ ballanced?
If $n$ is even, then the action is not locally ballanced. You can choose $B$ to be the conjugacy class of an element of $SO(n)$, as close as you like to the identity, that does not have $1$ as an eigenvalue. Now whatever $x$ is, $x$ will not belong to $Bx$.


On the other hand, if $n$ is odd then every element of $SO(n)$ does have $1$ as an eigenvalue, so if $B$ is any conjugacy class then $x\in Bx$. Furthermore, $Bx$ is preserved by all elements of $SO(n)$ that fix $x$, and is connected, and (if $n>1$) does not consist of $x$ alone; and this makes it a closed ball centered at $x$. Therefore for general balanced nonempty $B$ the set $Bx$ is a union of such balls.


