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Investigating further questions around this question: Example of sequence of graphs which satisfy the Riemann hypothesis? leads to the partition function $Z$ of the Ising model of the graph defined here: Why is this bipartite graph a partial cube, if it is?

It seems that:

$$Z_{n,\beta} = \sum_{k=0}^N c_{N,k} \cdot \exp(-\beta)^{2(N-2k)}$$

where $N = \sum_{i=1}^{n-1} \omega(i) = |E_{n-1}|$ and the coefficients $c_{N,k}$ are somehow mysterious to me at the moment.

Here are some empirical data, where $t = \exp(-\beta)$:

n N Z_n
1 0 2
2 0 4
3 1 4*t^2 + 4/t^2
4 2 4*t^4 + 4/t^4 + 8
5 3 4*t^6 + 12*t^2 + 12/t^2 + 4/t^6
6 4 4*t^8 + 16*t^4 + 16/t^4 + 4/t^8 + 24
7 6 4*t^12 + 8*t^8 + 28*t^4 + 28/t^4 + 8/t^8 + 4/t^12 + 48
8 7 4*t^14 + 12*t^10 + 36*t^6 + 76*t^2 + 76/t^2 + 36/t^6 + 12/t^10 + 4/t^14
9 8 4*t^16 + 16*t^12 + 48*t^8 + 112*t^4 + 112/t^4 + 48/t^8 + 16/t^12 + 4/t^16 + 152

Question: What are the coefficients $c_{N,k}$?

I am not asking for a proof, just some empirical direction, if it is possible and not so difficult to do.

Here is the python source code for further computations, if needed.

If it is of interest to the question:

The Hamiltonian function is maximal possible:

$$H_n = - \sum_{(u,v) \in E_n} \sigma_u \sigma_v = |E_n| = \sum_{k=1}^n \omega(n) \approx n(\log(\log(n))+B_1)+o(n)$$

where $B_1 \approx 0.26$ is Mertens constant.

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  • $\begingroup$ Also what does maximal possible mean? You mean that $\max_{\sigma} H_n(\sigma)$? You might want to formulate that as $ - \min_{\sigma} -H_n(\sigma)$ that would mean that value we would call $E_n$ and call it the ground state energy of the corresponding antiferromagnetic Ising model. $\endgroup$
    – Frederik Ravn Klausen
    Oct 17, 2022 at 9:52
  • $\begingroup$ @FrederikRavnKlausen: The value of $H_n$ as defined can have maximal value the number of edges of the graph. This is achieved in the cases I am looking at and this is because the shortest paths function $d$ satisfies: $d(i,j)+d(j,k) \equiv d(i,k) \mod (2)$ and this is satisfied because the graphs I am looking at are bipartite with partitions $A_n= \{ v | \sigma_v = (-1)^{d(1,v)} =^! +1 \}$ and $B_n= \{ v | \sigma_v = (-1)^{d(1,v)} =^! -1 \}$ and do not have odd cycles. $\endgroup$
    – mathoverflowUser
    Oct 17, 2022 at 9:58
  • $\begingroup$ @FrederikRavnKlausen: I am interested in cased where the "average magnetization" of the Ising model is computed and equals $0$. This would correspond in number theoretic terms to the prime number theorem. So I am looking for inspiration from physics to maybe apply it to number theory. $\endgroup$
    – mathoverflowUser
    Oct 17, 2022 at 10:03
  • $\begingroup$ @FrederikRavnKlausen: You might be interested in this physics question: physics.stackexchange.com/questions/732418/… $\endgroup$
    – mathoverflowUser
    Oct 17, 2022 at 11:45

1 Answer 1

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In general, the partition function of the Ising model is usually nicer when defining $x = \tanh(\beta)$ instead of $t = \exp(- \beta)$. One explanation for this is that the Ising model partition function is the same as the partition function for the random cluster model and the partition function of the Ising model can be calculated using the loop-O(1)-model with parameter $x = \tanh(\beta)$ and Bernoulli percolation with parameter $x$.

I tried to adopt this point of view here: https://alea.impa.br/articles/v19/19-07.pdf

Let me give just an example with the graph which is just a loop of $n$ vertices and $n$ edges. Then I can find the partition function by taking all even (spanning) subgraphs of the graph (which means subsets of edges such that all vertices have even degree.) In this particular case it is the full subgraph and the empty one. Then the partition function of the Ising model is given as

$$ Z = \sum_{g \text{ even subgraph of } G} x^{o(g)} $$ where $o(g)$ is the number of open edges of $g$ meaning that $o(\emptyset) = 0$ and $o(G) = n$ in the particular case.

Thus, we have that $ Z = 1 + x^n$ in that particular case. I do not really understand the definition of your graphs, but this point of view might be helpful.

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  • $\begingroup$ Thank you very much for your explanation and the linked paper. In case it helps: The graph has vertex set $1\cdots n$ and edges $a \approx b \iff a/b \text{ or } b/a \text{ is prime}$. Can you say something about the line graph on $n$ vertices, without the loop? What is the partition function in this case? $\endgroup$
    – mathoverflowUser
    Oct 14, 2022 at 14:36
  • $\begingroup$ Sorry. In my previous comment, I meant to say "linear graph" of finite size $n$, instead of "line graph". $\endgroup$
    – mathoverflowUser
    Oct 14, 2022 at 16:38
  • $\begingroup$ From your script I find it a bit difficult to see exactly how you define the Hamiltonian. Do you take it as $H(\sigma) = - \sum_{e} \sigma_{e^+} \sigma_{e^-} $ where we sum over all the edges in the graph and their endpoints? Do you have the $-$ sign in front? In any case it is the one dimensional Ising model and you can calculate $Z = \sum_{\sigma} e^{ - \beta H(\sigma)} $ explicitly by summing over the spins from one end to the other. You'll get $2 \cdot (2 \cosh(\beta))^{N-1}$. $\endgroup$
    – Frederik Ravn Klausen
    Oct 17, 2022 at 8:52
  • $\begingroup$ Thanks for the hint with "one dimensional Ising model". The Hamiltonian is defined as you write it. I get for $Z$ in this case: $Z = 4 (t^2+t^{-2})^{n-2}$ where $t = \exp(\beta)$. Is this the same as your result? Thanks for your help! $\endgroup$
    – mathoverflowUser
    Oct 17, 2022 at 9:05
  • $\begingroup$ It is not quite the same as $t^2 + t^{-2} = 2 \cosh(2 \beta)$, but it is more or less the same. $\endgroup$
    – Frederik Ravn Klausen
    Oct 17, 2022 at 9:34

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