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 $n\ge 3$?

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    $\begingroup$ Is the weird word "ballanced" (with 2 L) invented on purpose? $\endgroup$ – YCor Jul 26 '18 at 20:43
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    $\begingroup$ Your balanced sets are just unions of conjugacy classes, and an action is balanced if any orbit of any conjugacy class is a ball - right? $\endgroup$ – R W Jul 26 '18 at 20:51
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    $\begingroup$ @YCor The (weird) word "ballanced" is invented as a mix of "ball" and "balanced". This is a working terminology. $\endgroup$ – Taras Banakh Jul 26 '18 at 20:59
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    $\begingroup$ @TarasBanakh, you may spend as much time explaining that it's not a typo as talking about the math... Conceivably "ball-anced", with hyphen to emphasize that you realize what you're doing? Otherwise you'll waste time... $\endgroup$ – paul garrett Jul 26 '18 at 21:29
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    $\begingroup$ Actually I started editing the post to fix the typo, then saw it could be a pun with "ball" and asked (possibly somebody else would have corrected). Mentioning explicitly that it's a special spelling would have avoided my question. $\endgroup$ – YCor Jul 26 '18 at 22:23

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.

  • $\begingroup$ Thank you very much for the answer. By the way, it shows that the action of $SO(n)$ of the projective space $\mathbb RP^{n-1}$ is ballanced for every $n\ge 3$. Right? What about other Grassmannian manifolds? $\endgroup$ – Taras Banakh Jul 27 '18 at 5:38

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