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Is Ising universallity class on triangular or hexagonal lattice different from universallity class on rectangular lattice?

Is universallty class depends on type of microscopical graph or topology on lattice?

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There are definitely cases where the universality class does depend on the microscopic lattice type. An easy example is an antiferromagnetic Ising model. On a bipartite lattice like the square lattice, it is easy to see that the antiferromagnetic and ferromagnetic Ising models are equivalent, since one can simply redefine $\sigma_i \rightarrow -\sigma_i$ for all sites on one sublattice. On the triangular lattice, however, the antiferromagnetic Ising model exhibits frustration: the three bonds on the edges of a triangle cannot all be simultaneously in their lowest energy state, so the universality class will be different than the ordinary Ising universality class.

Topology is a bit different -- there are cases even in classical mechanics where the topology of the lattice can affect the ground state degeneracy (although this phenomenon is more popular in quantum mechanics), but typically thermodynamic properties are bulk properties and are insensitive to choice of boundary conditions. I'm unaware of an example in which the topology of the lattice affects the universality class.

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