# On the minimum distance along an orbit

Let $\Gamma$ be a nontrivial group of isometries of $\mathbb{S}^n$, $n \geq 2$, acting properly discontinuously. For $p \in \mathbb{S}^n$, define

$$r(p) = \min_{g \in \Gamma \setminus\{e\} } d(p, g(p)),$$

that is, $r(p)$ is the minimum distance to $p$ of a point in its orbit. As the sphere is a homogeneous manifold, I expect $r$ to be a constant function, although I couldn't prove it. Any thoughts on it?

• No, it is not a constant function. Hint: Consider $n=1$ and $\Gamma$ generated by a reflection. – Moishe Kohan Dec 8 '17 at 19:51
• You mean the antipodal map? – Eduardo Longa Dec 8 '17 at 19:58
• No, I mean a reflection. – Moishe Kohan Dec 8 '17 at 19:59
• OK, take a group generated by a hyperplane reflection in $R^{n+1}$ where $n$ is your favorite number. – Moishe Kohan Dec 8 '17 at 20:00
• The isometry cannot have fixed points – Eduardo Longa Dec 8 '17 at 20:01

Given a group $G$ acting on a metric space $X$ define the displacement function $\delta_g(x):= d(x, g(x))$, $g\in G$. One way the displacement function of $g$ can be constant is if the centralizer $C(g)$ of $g$ in the isometry group of $X$ acts transitively on $X$. The reason is that $\delta_g= \delta_{fgf^{-1}}$. Hence, for $f\in C(g)$, $$\delta_g(f(x))= d(gf(x), f(x))= d(f^{-1}g f(x), x)= d(g(x), x)= \delta_{g}(x)$$ Therefore, if $C(g)$ acts transtively on $X$, then $\delta_g$ is constant.

Otherwise, $\delta_g$ tends to be nonconstant. For instance, if $g$ is an isometry of the Euclidean space then $\delta_g$ is constant if and only if $g$ is a translation. If $X$ is a hyperbolic space then it's even worse, $\delta_g$ is constant if and only if $g=id$. Now, consider the case when $X=S^{n-1}$ with the standard angular metric $d$.

Lemma. $g\in O(n)$ has constant displacement on $(S^{n-1},d)$ if and only if either $g=\pm id$ (if $n$ is odd) or for any two eigenvalues $\lambda_1, \lambda_2$ of $g$ we have $\lambda_1=\lambda_2^{\pm 1}$ (if $n$ is even). (Note that all the eigenvalues $\lambda$ have absolute value $1$, hence $\lambda^{-1}= \bar\lambda$.)

Proof. I will consider the more interesting case of $n=2k$ and leave you to work out the odd-dimensional case. Suppose that $g\in O(n)$ acts on $S^{n-1}$ with constant displacement function.

Let $\lambda^{\pm 1}_1,...,\lambda^{\pm 1}_k$ denote the complex eigenvalues of $g$ (taken with multiplicity). If some $\lambda_j=1$ then $g$ has a fixed point on $S^n$, hence, $\delta_g$ constant implies that $g=id$. Similarly, if some $\lambda_j=-1$, then for its unit eigenvector $v\in {\mathbb R}^n$, we have $d(v, gv)=\pi$, hence, $d(u, gu)=\pi$ for all $u\in S^{n-1}$, i.e. $g=-id$.

Suppose, therefore, that all eigenvalues of $g$ are non-real.

Let $E_1\oplus ...\oplus E_k={\mathbb R}^{2k}$ be the eigenspace decomposition: Each $E_j$ is 2-dimensional and if we identify it isometrically with the complex plane, then $g$ acts on $E_j$ via multiplication by $\lambda_j$ or $\lambda_j^{-1}$. Next, let's compute the displacements of the restrictions of $g$ to the circles $E_j\cap S^{n-1}$: For a unit vector $u\in E_j$, we have $$d(u, \lambda_j u)= |arg(\lambda_j)|.$$ It follows that the arguments of all the eigenvalues of $g$ have the form $\pm \theta$ for some fixed $\theta\in [-\pi, \pi]$. Hence, $\lambda_j=e^{\pm i\theta}$ for $j=1,...,n$.

The converse direction is equally pleasant: Under suitable identification of ${\mathbb R}^{2k}$ with ${\mathbb C}^k$, $g$ acts as a scalar multiplication $$u\mapsto e^{i\theta}u, u\in {\mathbb C}^k.$$ From this it is easy to see that $g$ acts with the displacemnt $\theta$ on $S^n$. You can also observe that the centralizer of such $g$ in $O(n)$ is $U(k)$ and, hence, acts transitively on $S^{n-1}$. qed

Note that the displacement function of $g\in O(n)$ is clearly real-analytic. Hence, if it is constant on a nonempty open subset of $S^{n-1}$ then $\delta_g$ is constant on the entire sphere.

Now, to your question. You have a finite subgroup $G<O(n)$ acting freely on $S^{n-1}$ and define the function $$r(p)= \min_{g\in G- e} \delta_g(p).$$ This function is piecewise-analytic on $S^{n-1}$: there exists a nowehere dense subset $E\subset S^{-1}$, such that for every component $U$ of $S^{n-1} -E$, there exists $g=g_U\in G$ such that $$r|_U= \delta_g|_U.$$ Now, if the function $r$ were constant on $S^{n-1}$, we would have that for each $g_U$, $\delta_{g_U}$ is constant. So, what's left is to find a finite subgroup $G< O(n)$ acting freely on $S^{n-1}$ such that the only element $g\in G$ with constant $\delta_g$ is the identity. In view of the lemma, it suffices to consider the case of even $n$ and look for $G$ such that every $g\in G-\{e\}$ has at least two eigenvalues with distinct (absolute values of) their arguments.

For instance: Consider a finite group $G$ generated by the orthogonal transformation $$g: (z,w) \mapsto (e^{2\pi i/p}z, (e^{4\pi i/p}w),$$ where $(z,w)\in {\mathbb C}^2$, where $p>3$ is prime. For instance, $p= 7$, which I like more than $p=5$ somehow.

• This is $L(7,2)$, right? – Eduardo Longa Dec 8 '17 at 20:12
• Why it is not constant? – Eduardo Longa Dec 8 '17 at 20:13
• @EduardoLonga: That's an exercise. Compare a point on the $z$-axis with a point slightly off the $z$-axis. – Moishe Kohan Dec 8 '17 at 20:15

I want to add more explanation on Moishe Cohen's example :

i) Consider a curve $c(t):=(e^{2\pi t}z,0)$ in $S^3$. If $U_\varepsilon(c)$ is a tubular neighborhood, then there is a diffeomorphism $$f : S^1\times B_\varepsilon(0) \rightarrow U_\varepsilon(c)$$ s.t. (1) $B_\varepsilon(0)$ is flat $\varepsilon$-ball

(2) $B_\varepsilon (0)$ is identified to $\varepsilon$-ball in $c'(t)^\perp$

(3) $f(t,x)=\exp_{c(t)}\ x$ and

(4) If $U_\varepsilon (c)$ has an free isometric action $g(z,w)=(e^\frac{2\pi i}{p} z,e^\frac{2\pi i l}{p} w)$, then $f$ preserves the free isometric action.

ii) If we fix $x_0\in c'(0)^\perp,\ |x_0|=\varepsilon$, then define $r_t:=\exp_{c(t)} \ x_t$ where $x_t$ is a parallel transport of $x_0$ along $c(t)$. Since $S^3$ is not flat manifold, so $x_0\neq x_\delta$ for small $\delta\neq 0$.

iii) Note that $d(r_\delta,r_0)<2\pi\delta$.

iv) Here all numbers in sequence are natural number. Then there are $p_i,\ u_i$ s.t. $p_i,\ u_i\rightarrow \infty$, $0< u_i<p_i$, $p_i$ is a prime and $\frac{2\pi u_i}{p_i}\rightarrow 2\pi \delta$.

If angle between $x_0$ and $x_\delta$ is $\theta$, where $\theta$ is small, then there is a sequence $p_i,\ r_i$ s.t. $0<r_i<p_i$ and $\frac{2\pi r_i}{p_i}\rightarrow\theta$.

In further there are $l_i,\ m_i$ .t. $l_i,\ m_i\rightarrow \infty$ and $\delta l_i- m_i \rightarrow \frac{\theta }{2\pi}$.

Hence if $g^{u_k}\cdot (z,w)= (\frac{2\pi u_ki}{p_k} z, \frac{2\pi u_kl_k i}{p_k} w)$, then $d(g^{u_k} \cdot r_0, r_\delta)\rightarrow 0$.