In the **six-vertex model**, edges in a square lattice are oriented so that the in-degree of each vertex is exactly two. This gives six types of allowable vertices:
$$\begin{array}{cccccc}
 \begin{array}{ccc}
 & \uparrow & \\
 \leftarrow & \bullet & \leftarrow \\
 & \uparrow & \\
 \end{array}
 & \begin{array}{ccc}
 & \downarrow & \\
 \rightarrow & \bullet & \rightarrow \\
 & \downarrow & \\
 \end{array} 
 & \begin{array}{ccc}
 & \uparrow & \\
 \rightarrow & \bullet & \rightarrow \\
 & \uparrow & \\
 \end{array} 
 & \begin{array}{ccc}
 & \downarrow & \\
 \leftarrow & \bullet & \leftarrow \\
 & \downarrow & \\
 \end{array} 
 & \begin{array}{ccc}
 & \downarrow & \\
 \leftarrow & \bullet & \rightarrow \\
 & \uparrow & \\
 \end{array} 
 &\begin{array}{ccc}
 & \uparrow & \\
 \rightarrow & \bullet &\leftarrow \\
 & \downarrow & \\
 \end{array} \\
 [z]&[z]&[az]&[az]&z&z^{-1}
\end{array}$$
If a vertex has label $z$ (along with a global parameter $a$), the vertex is assigned the weight listed above, where $[z]$ means $\frac{z-z^{-1}}{a-a^{-1}}$. As usual, the partition function $Z$ of a lattice is defined to be the sum (over all allowable orientations of the edges) of the product of the individual weights of all the vertices. In a $n\times n$ lattice with column parameters $x\_1,\dots,x\_n$ and row parameters $y\_1,\dots,y\_n$, the label of the vertex in row $i$ and column $j$ is $z\_{ij}=x\_i/y\_j$. Then, the Izergin-Korepin determinant gives an exact formula for the partition function of a lattice in the six-vertex model
$$Z=\frac{\textstyle\prod\_{i=1}^n x\_i/y\_i \prod\_{1\le i,j\le n}[z\_{ij}][az\_{ij}]}{\textstyle \prod\_{1\le i<j\le n}[x\_i/x\_j][y\_j/y\_i]}\det\left([z\_{ij}]^{-1}[az\_{ij}]^{-1}\right).$$ This assumes domain wall boundary conditions, where edges on the left and right are all in and edges are the top and bottom are all out. Kuperberg (1996) famously used the Izergin-Korepin determinant to prove the alternating sign matrix conjecture.


In the **eight-vertex model**, vertices with in-degree 0 or 4 are also allowed.


$$ \begin{array}{cc}
 \begin{array}{ccc}
 & \uparrow & \\
 \leftarrow & \bullet & \rightarrow \\
 & \downarrow & \\
 \end{array} & \begin{array}{ccc}
 & \downarrow& \\
 \rightarrow& \bullet & \leftarrow \\
 & \uparrow & \\
 \end{array} \\
 \text{source} & \text{sink} 
 \end{array}
$$
Question: Is there an assignment for the weights of a sink and source (perhaps with other global parameters) with any type of boundary conditions so that the partition function for a lattice in the eight-vertex model can be similarly represented by a determinant? or, for that matter, any exact formula?



An extension of the Izergin-Korepin determinant to the eight-vertex model
This is essentially a long comment to accompany my answer, which is: not that I know of. 


Before I explain why I think the eight-vertex model has received much less attention in this context (besides that it is technically *much* more involved) I should point out that the closely related 'elliptic solid-on-solid' (SOS) model of Baxter *does* admit determinantal formulae for its partition functions at finite system size. See Rosengren's sum of determinants for the domain-wall case, and Filali's determinant for the case of one (diagonally) reflecting end (also known as 'U-turns') with domain walls on the three other sides. (The latter generalises Tsuchiya's determinant for the six-vertex model with the same boundaries, and Filali--Kitanine's determinant for the trigonometric SOS model.)




---


The additional vertices of eight-vertex model alter the properties of the model quite drastically. In particular, unlike for the six-vertex model, the eight-vertex model's bulk free energy --- characterising the large-lattice asymptotics of the partition function per site --- is independent of the choice of boundary conditions for the eight-vertex model. This is actually quite fun and easy to see, and does not require any parameterization of the vertex weights or quantum integrability.


I claim that any configuration for an $L\times L$ lattice with any choice of boundary conditions, i.e. prescription for the arrows on the 'external' (half)edges sticking out of the lattice, can be embedded in an $(L+2)\times(L+2)$ lattice with any other choice of boundary conditions. (To the best of my knowledge this was first observed by Brascamp, Kunz and Wu, *J Math Phys* **14** '73.)
This claim implies that the differences between the corresponding partition functions are subdominant, yielding only different finite-size corrections (of order $1/L$) to the bulk free energy (which was computed by Baxter).


With the right point of view it is easy to see that this is true. I particularly like the 'line' or 'bond' picture of vertex models, which is just another way to represent all those little arrows along the edges. Let's fix an orientation for the lines of the lattice, say to the right for horizontal lines and upwards for the vertical ones. Then we depict the edges whose arrows go against the orientation with a thick line, and use thin (or dotted) lines for all other edges.


For the six-vertex model, the ice rule says that thick lines may go straight or turn as long as they follow the orientation. This implies that any rectangular (or really: any) portion of the lattice has equally many 'incoming' thick lines, i.e. crossing the left+bottom boundaries, as 'outgoing' thick lines, i.e. leaving the rectangle on the right+top boundaries. This is sometimes called 'line conservation'. (Algebraically it corresponds to the fact that the six-vertex model's R-matrix, and this any operator built from it, preserves the weight-decomposition of the related vector space. It is also the reason why different choices of boundary conditions used at finite length can result in different bulk free energies.)


For the eight-vertex model, however, the sources and sinks from the arrow picture allow the thick lines to go anywhere; the lines still can't just (dis)appear, but they no longer have to follow the orientation. As for rectangular portions of the lattice, the only thing we can now say is that there must be an *even* number of thick lines crossing its boundary. 


It's now easy to verify my claim. Given any configuration on an $L\times L$ lattice, we only need one extra row/column of vertices on each side to connect these pairs of lines, and we can moreover choose any new boundary configuration we like on the external edges of the new $(L+2)\times(L+2)$ lattice.


(The elliptic SOS model that I mention above also obeys the ice rule/line conservation: this is it's key property compared to the eight-vertex model, and comes at the cost of the introduction of a 'dynamical'/height parameter.)


