Geometric Formula for 2d Ising Zeros: Examples & Numerics

TL;DR

The paper tests a geometric conjecture that zeros of the inhomogeneous 2d Ising partition function on planar graphs can be expressed in terms of the geometry of the dual 2d triangulation embedded in 3d space. It provides analytical checks on simple graphs (Theta, tetrahedron, pyramid, cube) and extensive numerical validations on random sphere triangulations, confirming that the predicted geometric couplings $Y_\ell^{(0)}$ are indeed Ising zeros with high numerical precision, and clarifies the convexity/concavity sign convention via outward normals. It also shows that the proposed formula fails for toroidal topology, indicating topology is crucial and motivating a generalization beyond spherical graphs. The work strengthens the link between 3d quantum gravity amplitudes and 2d Ising zeros, clarifies the role of circle patterns and dihedral angles, and guides future extensions to non-trivial topologies and higher-genus surfaces. Overall, the results provide strong support for the spherical-case geometric zero formula while highlighting limitations and directions for generalization in discrete quantum gravity contexts.

Abstract

A geometric formula for the zeros of the partition function of the inhomogeneous 2d Ising model was recently proposed in terms of the angles of 2d triangulations embedded in the flat 3d space. Here we proceed to an analytical check of this formula on the cubic graph, dual to a double pyramid, and provide a thorough numerical check by generating random 2d planar triangulations. Our method is to generate Delaunay triangulations of the 2-sphere then performing random local rescalings. For every 2d triangulations, we compute the corresponding Ising couplings from the triangle angles and the dihedral angles, and check directly that the Ising partition function vanishes for these couplings (and grows in modulus in their neighborhood). In particular, we lift an ambiguity of the original formula on the sign of the dihedral angles and establish a convention in terms of convexity/concavity. Finally, we extend our numerical analysis to 2d toroidal triangulations and show that the geometric formula does not work and will need to be generalized, as originally expected, in order to accommodate for non-trivial topologies.

Figures (22)

• Figure 1: A 3-valent graph $\Gamma$ (in orange) and its dual triangulation: graph nodes are dual to triangles, and graph links $\ell$ are dual to triangulation edges $e$. The orientation of the links does not matter here and the source and target vertices play the same role in the Ising model. The triangulation is not necessarily in a single 2d plane. It can be folded and have a non-trivial embedding in the 3d space ${\mathbb R}^{3}$, with non-zero dihedral angles between neighbour triangles.
• Figure 2: The two adjacent triangles $\Delta$ and $\widetilde{\Delta}$ sharing the edge $e$, are dual to two graph nodes connected by the graph link $\ell$ (in dotted line). The 2d angles $\varphi_{\ell}$ and $\tilde{\varphi}_{\ell}$ are the triangle angles at the vertices opposite to the edge. The dihedral angle $\theta_{\ell}$ between the two triangles is the angle between their normal vectors, and reflects the non-trivial extrinsic curvature of the embedding of the 2d triangulation in the flat 3d space.
• Figure 3: Unfolding a higher valent graph node into a 3-valent tree by triangulating its dual polygon. One can then apply the geometric formula for Ising zeros on the triangulation and generalize it to graphs of arbitrary valence, as shown in Bonzom:2024zka.
• Figure 4: If a polygon is inscribed in a circle, the opposite triangle angle of a given edge $\varphi_{\ell}$ is always the same whatever the chosen opposite summit, and is automatically equal to half of the center angle $\psi_{\ell}$. Thus working on embedded circle patterns, one gets the generalized formula \ref{['eqn:circlepattern']} for higher valent graphs from the geometric ansatz \ref{['eq_Ising_zeroes']} for 3-valent graphs and their dual 2d triangulations, as shown in Bonzom:2024zka.
• Figure 5: With an anti-clockwise orientation around each triangle, we define the outward normal vectors $\overrightarrow{N}_{ABC}$ and $\overrightarrow{N}_{BAD}$. The scalar product $\overrightarrow{N}_{ABC}\cdot\overrightarrow{N}_{BAD}$ gives the cosine $\cos\theta_{AB}$ of the dihedral angle. Considering the shared edge $\overrightarrow{AB}$, we choose as first triangle the one that has the same orientation as the chosen shared vector - here the triangle (ABC). With this convention, we then consider the cross product $(\overrightarrow{N}_{ABC}\wedge \overrightarrow{N}_{BAD})$. It is automatically collinear to $\overrightarrow{AB}$. If it is oriented in the same direction as $\overrightarrow{AB}$, we have identified a convex configuration (on the left). Otherwise, it is a concave configuration (on the right).