I am trying to find a correspondence between 6-vertex model and an Aztec Diamond tiling. Here are the building blocks of the 8-vertex model:

enter image description here

There seems to be more than one correspondence. I found at least two:

And there is likely even more. My goal had been to try to understand if there was equivalence between domino tilings and alternating sign matrix, with 6-vertex as an intermediate object. I wound up just totally confusing myself.

  • ising model
  • alternating sign matrix
  • 6-vertex model
  • domino tiling

enter image description here

Now ASM and Domino tilings in the same space have different numbers of tilings. ELKP shows how to obtain two different Alternating Sign Matrices from the domino tilings.

Every paper seems to give a different answer. My main question is whether the combinatorics models:

  • domino, alternating sign

are the same as the statistical mechanics models? Here I list a few candidates

  • ising, 6-vertex, XXZ etc
  • 1
    Does my reply answer your question? If not let me know what you think is still missing. – Jules Lamers Mar 19 '17 at 20:48
  • no competing answers. i am just a bit surprised – john mangual Mar 19 '17 at 21:19
up vote 7 down vote accepted

In short: unlike the mapping between alternating-sign matrices and the configurations of the six-vertex model with domain-wall boundary conditions, the mapping between the six-vertex model and domino tilings is not one-to-one. This accounts for the fact that the relevant special case of the six-vertex model (with $a=b=c/\sqrt{2}$) corresponds to the 2-enumeration of alternating-sign matrices (ASMs).

On the statistical-mechanical side we're dealing with the special case $a=b<c$, known as the F-model, of the six-vertex model.

(Historical aside: today I finally found out that the name "F-model" was chosen by Rys in honour of his thesis advisor, Fierz, who apparently came up with the model.)

Recall that $c$ is the weight of the two 'saddle-like' arrow configurations on the edges surrounding a vertex: these are vertices 5 and 6 in the OP's figure. The ice rule implies that along any row (and any column) the two vertex configurations of weight $c$ must occur alternatingly, with any number of intermediate vertices of weight $a$ and $b$. The domain-wall boundary conditions further imply that one of the two vertices of weight $c$ occurs less than the other: it must occur precisely one time less in every row and in every column. (Whether this is vertex 5 or 6 in the OP depends on which of the two possible domain-wall boundaries one chooses; the two are related by reversing all arrows.) Of course these facts are precisely what allows one to relate configurations of the six-vertex model with domain walls to ASMs, cf. Kuperberg [arXiv:math/9712207].

On the combinatorial side recall that an $x$-enumeration counts ASMs with weight $x^k$ when the ASM contains $k$ entries equal to $-1$. Now the latter ASM entries precisely correspond to the vertex configuration of weight $c$ that occurs less (cf above). Thus, if a row contains $l$ of these vertices, it must contain precisely $2l+1$ vertices of weight $c$. But this is true for every row of the lattice. The upshot is that if the lattice has size $L\times L$ then the domain-wall partition function of the F-model accounts for the various $x$-enumerations of ASMs:

$$c^L \ Z_L(a=b,c) \quad\longleftrightarrow\quad \text{$c^2$-enumeration of $L\times L$ ASMs} \ . $$

(Since common rescalings of $a,b,c$ only yield a physically unimportant normalization factor for $Z_L$ we can set $a=b=1$ to remove the overall factor in the above correspondence.)

In particular, at the combinatorial or ice point ($a=b=c$) the domain-wall partition function just counts the number of ASMs up to an overall normalization.

By assigning two domino tiles for every $-1$ in the ASM (as in Fig 1 of Zinn-Justin cited in the OP) we thus get a combinatorial interpretation for the 2-enumeration of ASMs (counted by the domain-wall partition function at $a=b=c/\sqrt{2}$) in terms of the number of domino tilings of the so-called Aztec diamond.


PS. Just to mention some more related terminology consider the combination

$$\Delta=\frac{a^2+b^2-c^2}{2\,a\,b}$$

of vertex weights. For the F-model we have $\Delta = 1-(c/a)^2/2$. The ice-model corresponds to $\Delta =1/2$, the 2-enumeration of ASMs to $\Delta = 0$ and their 3-enumeration to $\Delta=-1/2$. The value $\Delta = 0$ is known as the free-fermion point. That this value is quite special is clear from the viewpoint of the XXZ spin chain related to the six-vertex model, whose Hamiltonian is of the form

$$H_{XXZ} = \sum_{j\in\mathbb{Z}/L\mathbb{Z}} (S_j^x \, S_{j+1}^x + S_j^y \, S_{j+1}^y + \Delta \, S_j^z \, S_{j+1}^z) \ \in \ \text{End}((\mathbb{C}^2)^{\otimes L}) \ ,$$

where $\Delta$ now sets the (partial) anisotropy, breaking the $SU(2)$-symmetry of Heisenberg's isotropic XXX spin chain ($\Delta=1$) to the subgroup $U(1)\subseteq SU(2)$ of rotations around the $z$-axis.

  • 3
    Hm, that formula $\Delta = (a^2+b^2-c^2) \, / \, (2ab)$ is familiar to anybody blessed or cursed with the inability to forget high school geometry: by the Law of Cosines, $\Delta = \cos C = $ cosine of the angle opposite side $c$ of a triangle with sides $a,b,c$. In particular the triangle has $C = 120^\circ$, $90^\circ$, $60^\circ$ in the cases $\Delta = -1/2, 0, 1/2$ respectively. Any reason that this triangle should be relevant here? – Noam D. Elkies Mar 13 '17 at 17:12
  • Interesting, I have never thought of it in quite that way. From the point of view of integrability, the combination $\Delta$ arises as the sole way in which the vertex weights $a,b,c$ feature in the (coordinate, say) Bethe eigenvectors---or more algebraically by solving the Yang--Baxter equation. For $-1<\Delta<1$ one often parametrizes the vertex weights as something like $a=\sin(u+\gamma)$, $b=\sin u$, $c=\sin\gamma$ (ignoring the overal scaling), so indeed $\Delta = \cos\gamma$ with $\gamma$ equal to your $C$. (Here $u$ and $\gamma$ are known as the "spectral" and "crossing" parameters.) – Jules Lamers Mar 14 '17 at 10:42
  • Ah, I also know the "overall scaling" factor from high-school contest geometry: it's the diameter of the circumcircle! Law of Sines: $a \, / \sin A = b \, / \sin B = c \, / \sin C = 2R$. – Noam D. Elkies Mar 14 '17 at 15:49
  • Sure, that overall scaling would set the scale of the triangle whose sides are $a,b,c$. NB. To parametrize the F-model in the above way one can take $u=−\eta/2$, $\gamma=\eta+\pi$ and remove the overall sign of $a,b,c$. [I wrote this before but with a typo, so here's the corrected version.] – Jules Lamers Mar 14 '17 at 16:01

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