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Is there any sense in which the "epsilon-tail" of an Apollonian disk-packing (by which I mean the union of the disks of radius less than epsilon) exhibits more and more statistical isotropy as epsilon goes to zero?

Here's one example of the kind of thing I mean. Label the disks $D_1,D_2,\dots$ in order of weakly decreasing radius. It seems likely that for all $n$ in a set of density 1, there is a unique largest disk $D'_n$ tangent to disk $D_n$, and the ray from the center of $D_n$ to the center of $D'_n$ determines a unit vector $v_n$. Now we can ask whether the end-points of the unit vectors $v_1,v_2,\dots$ are uniformly distributed on the circle.

I'm interested in finding out what's known about questions like this.

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  • $\begingroup$ Likely you're already familiar with this, but just in case, let me cite the earlier computational question: "Centers of Apollonian Circle Packings." $\endgroup$ – Joseph O'Rourke Oct 15 '14 at 22:07
  • $\begingroup$ @JosephO'Rourke, twwo articles last year in the Bulletin, ams.org/journals/bull/2013-50-02/home.html one by Kontorovich, one by Fuchs. $\endgroup$ – Will Jagy Oct 16 '14 at 4:17
  • $\begingroup$ @JosephO'Rourke, surprised that changing a single multiplier to $$ w^2 + x^2 + y^2 + z^2 = wx+wy+wz+xy+xz+yz $$ gives non-negative solution entries, a single root $(0,1,1,1),$ and a loss of the tree property, although still a partially ordered set... math.stackexchange.com/questions/966156/… $\endgroup$ – Will Jagy Oct 16 '14 at 4:25
  • $\begingroup$ Based on two other questions of yours, I propose another problem of this type: In dimension $d$, let $B$ be a fixed ball, and let $B_1,B_2, \ldots$ be a sequence of balls such that $B_n$ is tangent to $B$ and $d$ balls with smaller index. What can we say about the unit vectors pointing from the center of $B$ to the centers of $B_n$? $\endgroup$ – Hao Chen Nov 10 '14 at 19:29
  • $\begingroup$ I think problems of this type can be attacked (or maybe already studied) in the framework of Kleinian group. Using the Poincar\'e sphere model for hyperbolic space, we take the set of rays from the origin through the points in the orbit of a (cocompact) Kleinian group, then what's the distribution of the direction of the rays? $\endgroup$ – Hao Chen Nov 10 '14 at 19:35
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Hee Oh has replied to my question and given me permission to post her reply here. What follows are her words (so "I" means "Hee") with my reformatting.

What I proved with Shah (Inventiones, 2012) says that for a given Apollonian circle packing $P$, considered as a countable union of circles in the plane, "small circles in $P$ are uniformly distributed with respect to the $\alpha$-dimensional Hausdorff measure of the residual set of $P$, which is the closure of $P$", where $\alpha=1.305...$. For any region $E$ with piecewise smooth boundary the number of circles intersecting $E$ of radius at least $t$ is asymptotic to $c t^{-\alpha} H_{\alpha}(E)$.

Chapter 8 of the following article can be useful: http://gauss.math.yale.edu/~ho2/newMSRI_Oh.pdf

So it seems that your question amounts to asking how the $\alpha$-dimensional Haudorff measure of the residual set of $P$ (${\rm Res}(P$)) behaves. Since $\alpha$ is bigger than 1, a theorem of Marstrand says that a typical point $x$ in ${\rm Res}(P$) -- typical in the sense of the $\alpha$-dimensional Hausdorff measure -- is a condensation point for ${\rm Res}(P$), meaning that for almost all directions $\theta$, $x$ is a limit point of the intersection of $(x, \theta)$ and Res($P$) where $(x, \theta)$ is the ray from $x$ in the direction of $\theta$.

I am referring to Theorem 6.1 of the following paper http://gauss.math.yale.edu/~ho2/BR_JAMS_Final.pdf where we copied a theorem of Marstrand in 1954.

So, you do see circles in almost all directions in the above sense.

Link to Marstrand's paper: http://dx.doi.org/10.1112/plms/s3-4.1.257

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    $\begingroup$ I think it should be "she". $\endgroup$ – Hao Chen Nov 11 '14 at 22:09
  • $\begingroup$ You are completely right! My bad for assuming Hee was a he. $\endgroup$ – James Propp Nov 12 '14 at 2:23

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