I'm preparing an exam and in a preparation sheet there is an exercise that I just don't know how to deal with. Could someone please explain it to me?
a) Explain why the 3D Ising model on the cubic box $ \{0,\ldots,n\}^3 $ with $ \beta \to +\infty $ with $+1$ boundary conditions on the three faces such that either $x=0,y=0$ or $z=0$ and $-1$ boundary conditions on the three faces such either $x=n$,$y=n$ or $z=n$, corresponds to a dimer model on the hexagonal domain shown in the previous exercise (image below)
b) Harder: if we define a Markov chain dynamics, on the hexagon dimers by randomly performing moves as above (choosing them uniformly among the possible elementary moves available), do we converge to the uniform distribution on dimers? Hint: look at the equilibrium measure.
c) Much harder: What is the difference between this dynamics and the Metropolis dynamics for a corresponding 3D Ising model with $\beta = + \infty$? Why does one give a uniform distribution on dimer configurations an the other one does not?
a) My idea is that I can assign for example $+1$ spin to the left cube here above and the $-1$ spin to the right cube, and then flipping a spin sign if possible is the same as obtaining another configuration of dimer tiling so every possible domino tiling correspond to exactly one possible configuration of the spin in the 3D Ising model. The boundary condition I'm not very sure why... I mean i think define which cube is $-1$ and which one is $+1$ since we can see this little square as an $n\times n \times n$ square and the one on the left "corresponds" to the condition $x=0,y=0,z=0$, similarly to the other one. But maybe the boundary condition are also important for other reason. Why $\beta \to +\infty $ I don't know.
b) I'm not sure I understand the question. If i understood what the question is yes! Because we are choosing one possible configuration of spin $\sigma = (\sigma_x)$ of the 3D Ising model on the cubic box $ \{0,\ldots,n\}^3 $ and since this correspond to a dimer model, i.e. they can be mapped one to one with the hexagonal lattice dimer configurations so $$ \mathbb{P}(\sigma) = \frac{ e^{- \beta H(\sigma)}}{Z} $$ where $ Z = \sum_{ \tilde{\sigma} } e^{- \beta H(\tilde{\sigma})}$, converge as $ \beta \to \infty $ to $ \frac{1}{\text{number of tillings on the hexagonal grid}} $
c) No idea!