## Are packing-homogeneous spaces homogeneous?

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Given a metric space (M,d) define the packing function P(x,R,r) to be the maximum number of non-intersecting balls of radius r with centers in the ball B(x,R). Let’s call M packing-homogeneous if the packing function is independent of the base point x.

Conjecture A complete connected packing-homogeneous space M is homogeneous. That is, the group of isometries acts transitively on M.

Completeness is necessary as Ricky Demer pointed out.

Connectedness is necessary. It is not difficult to construct finite graphs that are packing-homogeneous, but are not homogeneous (vertex-transitive).

Remark 1 It follows from the work of Gleason, Montgomery, Zippin and others on Hilbert’s 5th problem that a complete connected finite-dimensional homogeneous space is a manifold. Thus, a complete connected finite-dimensional packing-homogeneous space that is not a manifold (for example, has fractional Hausdorff dimension) would provide a counter-example to the conjecture.

Remark 2 I hoped to settle this question for length spaces (or inner metric spaces), i.e. when the distance between two points is equal to the length of a shortest path. Every length space is a Gromov-Hausdorff limit of grpahs. I was hoping to establish a connection between some "almost packing-homogeneous" and "almost homogeneous" properties of graphs that in the limit would give me the desired result (in the spirit of what Gromov does in his paper on groups of polynomial growth). Or, to prove the opposite, constructing a convergent sequence of graphs that is "almost packing-homogeneous" but "increasingly inhomogeneous". After some attempts I thought that the problem for graphs may be as difficult as the original conjecture.

Remark 3 As far as I know, the conjecture is open even for Riemannian manifolds of dimension greater than 2.

I was not able to find any papers that would mention this question or try to attack it. How plausible is it that the conjecture is true? Where one would look for counterexamples?

I would also be grateful for any comments that would put this problem in a broader context, or any guesses as to how difficult this problem might be.

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## 1 Answer

Let $d_e$ be the usual metric on $\mathbb{R}^2$ and $S = \{(\frac1n,0) : n \operatorname{is} \operatorname{a} \operatorname{positive} \operatorname{integer}\}$. Let $(X,d)$ be $\mathbb{R}^2-S$ with the metric inherited from $(\mathbb{R}^2,d_e)$.

Clearly, $P_{(X,d)}(x,R,r)\leq P_{(\mathbb{R}^2,d_e)}(x,R,r)$ and $(\mathbb{R}^2,d_e)$ is packing-homogeneous. Consider a ball $B$ in $(\mathbb{R}^2,d_e)$ with center $C$, and disjoint balls of radius $r$ with centers $\{c_i : i\in I\}$ such that $\{c_i : i\in I\} \subseteq B_0$. Bounded subsets of $(\mathbb{R}^2,d_e)$ are totally bounded, so $|\{c_i : i\in I\}| < \infty$. Define $R(\theta)$ as the rotation of $(\mathbb{R},d_e)$ by $\theta$ radians about $C$, clearly these are all isometries which fixe $C$ and $B$ is invariant under. Since the points in $S$ are colinear, $|\{s\in S : d_e(C,c_i) = d_e(C,s)\}| \leq 2$, so if $C$ is not a member of $S$, then $|\{\theta\in (-\pi,\pi] : (r(\theta))(c_i)\in S\}| \leq 2$. This gives us $|\{\theta\in (-\pi,\pi] : (\exists i\in I)((r(\theta))(c_i)\in S)\}| =$ $|\displaystyle\bigcup_{i\in I} \{\theta\in (-\pi,\pi] : (r(\theta))(c_i)\in S\}| \leq |I|\cdot 2 < \infty\cdot 2 = \infty \leq |(-\pi,\pi]|$, so there exists $\theta\in (-\pi,\pi]$ such that $\neg (\exists i\in I)((r(\theta))(c_i)\in S)$. Let $\phi$ be such a $\theta$. Then $(\forall i\in I)((r(\phi))(c_i)\not\in S)$. This shows that if $C$ is not a member of $S$, then there is a packing in $B$ which does not use any points of $S$, so $P_{(\mathbb{R}^2,d_e)}\leq P_{(X,d)}(x,R,r)$. Let $x,y$ be members of $X$, then $P_{(X,d)}(x,R,r)\leq P_{(\mathbb{R}^2,d_e)}(x,R,r) = P_{(\mathbb{R}^2,d_e)}(y,R,r) \leq P_{(X,d)}(y,R,r)$ $\leq P_{(\mathbb{R}^2,d_e)}(y,R,r) = P_{(\mathbb{R}^2,d_e)}(x,R,r)\leq P_{(X,d)}(x,R,r)$, which shows that the packing function on $(X,d)$ is independent of basepoint. Therefore $(X,d)$ is packing-homogeneous.

$(X,d)$ is not locally compact near $(0,0)$, but it is locally compact near $(0,1)$, so $(X,d)$ is not even topologically homogeneous. To get to $(0,0)$ from any point in $(X,d)$, if you start below the x axis travel down to $y=-1$, otherwise travel up to $y=1$, then travel horizontally to $x=0$, then travel vertically to $y=0$. This shows that every point has a path to $(0,0)$, so $(X,d)$ is connected. Therefore $(X,d)$ is a connected packing-homogeneous space which is not homogeneous. QED

Demanding completeness might be enough. It would certainly stop my idea.

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Thank you! This definitely works if we don't require completeness. I will modify my question accordingly. – Yevgeny Liokumovich Oct 4 2010 at 23:58
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