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What will be the sensible extension of the 2D Ising to some curved surface - for instance, for a sphere or even something non-orientable?

For the flat space energy is given by well-known expression: $$ E = \sum_{i, j} J s_i s_j + \sum_ih_i s_i $$ What would make sense for the curved surface, should I imagine the spin as an arrow pointing along the $z$-axis, or the vector normal to the given surface. Also in the term, describing the interaction between the nearest neighbors, the spins $s_i $ now belong to different vector spaces, so it seems that for this expression to make sense, the neighbor should be parallely transported to the $i_{th}$ site. Or this construction doesn't make sense, and one has to work with the full Heisenberg model?

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In fact, the two-dimensional Ising model was first solved by Lars Onsager on a cylinder/torus, which, while lacking intrinsic curvature, has a 'nontrivial' topology and moduli space (you can play with the solution by changing how the cylinder/torus closes on itself.) More generally, given a graph $G = (V, E)$, it is possible to define an Ising model on $G$ by assigning a spin to every vertex and defining the Hamiltonian \begin{align*} H[\{\sigma_i\}_{i\in V};J, h] = -J\sum_{<i,j>\in E}\sigma_i\sigma_j -h\sum_{i\in V}\sigma_i \end{align*} For even greater generality, the coupling $J$ and field $h$ can be allowed to assume a distinct value for each edge and vertex respectively, allowing one to 'simulate' non-orientable topologies by reversing the sign of $J$ along a cut.

There is a vast amount of literature on the behavior of Ising-like models on arbitrary graphs, which are of fundamental interest in theoretical statistics as well as condensed matter physics.

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  • $\begingroup$ Thanks for the comment. Variation of coupling from point to point may be helpful to represent the curvature of the surface. $\endgroup$
    – spiridon_the_sun_rotator
    Aug 10, 2020 at 17:48
  • $\begingroup$ If I am reading this right, we say that the spins are coupled only if they share an edge? So with this definition of the hamiltonian, is there is no notion of distance between vertices? $\endgroup$
    – Tabin
    Aug 10, 2020 at 18:58
  • $\begingroup$ @Tabin It really amounts to how you prefer to conceptualize the model. There are plenty of situations where the notion of a distance could be useful (such as when considering the family of models obtained from a common underlying graph by adding connections between nth-nearest neighbors for n = 1, 2, 3, ... etc.) $\endgroup$
    – TLDR
    Aug 10, 2020 at 19:46
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    $\begingroup$ @Tabin There can a notion of distance, at least in a sense. For example, if two nodes share an edge, then those two nodes are within the range of the nearest-neighbor interaction. $\endgroup$
    – probably_someone
    Aug 11, 2020 at 0:50
  • $\begingroup$ @Tabin : Concerning "distance", it's more that one usually sets $J_{ij}$ as a function of $\|i-j\|$ (or at least as an even function of $i-j$). So the spatial distance between spins still matters in this case. This is particularly important when you consider interactions of infinite-range. $\endgroup$
    – Yvan Velenik
    Aug 11, 2020 at 19:27

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