OSU Mathematics Seminars and Colloquia
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Thu, Mar 27, 2025
Number Theory Seminar
3:00 PM
MSCS 509
Relations in a Kleinian Circle Packing
David Wright, OSU
[Abstract] [PDF]
Abstract: In this talk, we would like to go through the details of the circle packing which is the limit set of the Kleinian group $G=\langle a,b\rangle$ generated by: \[ a =\begin{bmatrix} \frac{\sqrt{11}-i}{2} & -i\\ -i&0 \end{bmatrix} ,\quad b= \begin{bmatrix} 1&2\\0&1 \end{bmatrix}. \] Previously we have identified this group as the $3/10$ cusp group on the boundary of Maskit's embedding of the Teichmüller space of once-punctured tori. We have also explained that this circle packing has unusual number-theoretic properties. The circle packing consists of the disjoint disk orbits $G(D_0)$ and $G(D_1)$ for the lower half-plane $D_0$ and the disk $D_1$ of radius $\frac{1}{\sqrt{11}}$ which is externally tangent to $D_0$ at $-1$. These correspond to the normalized Hermitian matrices: \[ H_0 =\left[\begin{array}{cc}0 & i \\ -i & 0 \end{array}\right], \quad H_1 =\left[\begin{array}{cc}\sqrt{11} & \sqrt{11}-i \\ \sqrt{11}+i & \sqrt{11} \end{array}\right]. \] Since the image of a disk with Hermitian matrix $H$ under the group element $g\in G$ corresponds to the Hermitian matrix $g(H) = {g^*}^{-1} H g^{-1}$, we have shown by direct calculation that all circles in the packing have Hermitian matrices of the form $ H=\begin{bmatrix} s\sqrt{11} & \frac{m\sqrt{11} +n\,i}2\\ \frac{m\sqrt{11} -n\,i}{2} & t\sqrt{11} \end{bmatrix}$, for nonnegative integers $s,t$ and integers $m,n$ satisfying $m\equiv n\pmod{2}$ and determinant $11st-\frac{11m^2+n^2}{4}=-1$. Thus, all circle curvatures are $s\sqrt{11}$ for nonnegative integers $s$.

In this talk, we wish to describe in more detail the linear and quadratic relations satisfied by finite chains of circles in this packing. We would also like to explain the resemblance of this packing to the Apollonian packing in the following sense. An Apollonian packing begins with three mutually externally tangent disks $D_1$, $D_2$, $D_3$ and the two disks $D_4$, $D_4'$ which are externally tangent to the original three disk chain. In the 3/10 circle packing, there are 5 disjoint disks $D_1$, $D_2$, $D_3$, $D_4$, $D_4'$ which are connected by single disks in between certain pairs. Thus, the packing can be viewed as shrinking the original 5 disks in the Apollonian packing and inserting single disks at places where tangent points were previously located. The picture will make this clearer.

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