Two dimensional Ising Model and Dimers

The two-dimensional Ising model is one of the central exactly solvable models in statistical mechanics. At first sight it looks quite different from the dimer model. The variables are spins, not matchings. The partition function is a sum over $\pm1$ configurations, not over pairings of vertices. The interaction is local and energetic: neighboring spins prefer to align or anti-align depending on the coupling. Yet, at zero magnetic field, the model hides a combinatorial structure very close to the one we used for dimers. After a simple algebraic expansion, the Ising partition function becomes a weighted sum over even subgraphs, or closed polygon configurations. These polygons can then be encoded as perfect matchings on a decorated graph, usually called the Fisher graph. Once this translation is made, the same Pfaffian machinery applies: the partition function becomes a Pfaffian, the square of the Pfaffian becomes a determinant, and on the periodic square lattice that determinant is computed by Fourier analysis.

We explain that reduction. We begin with the high-temperature expansion, where the spin sum forces the even-degree condition. Then we explain how the polygon model is converted into a dimer model, and why the same Kasteleyn/Pfaffian sign mechanism appears. Finally, for the square lattice, we diagonalize the resulting periodic operator and derive the Onsager free energy. The guiding point is that the exact solution of the Ising model is not a completely separate miracle from the exact solution of the dimer model. Both are governed by the same analytic pattern: combinatorial configurations become Pfaffians, Pfaffians become determinants, determinants become products of Fourier symbols, and the thermodynamic limit becomes a logarithmic integral.

We consider the nearest-neighbor Ising model on an $M\times N$ square lattice. At each vertex $v$ there is a spin $\sigma_v\in \{+1,-1\}.$ Horizontal edges have coupling $K_1$ and vertical edges have coupling $K_2.$ The zero-field partition function is

$\displaystyle Z_{\text{Ising}}=\sum_{\sigma}\exp\left(K_1\sum_{\langle uv\rangle_h}\sigma_u\sigma_v+K_2\sum_{\langle uv\rangle_v}\sigma_u\sigma_v\right).$

Here the first sum in the exponential runs over horizontal nearest-neighbor pairs and the second over vertical nearest-neighbor pairs. There is no magnetic field term. This absence is essential for the Pfaffian solution in the form discussed here. A magnetic field destroys the simple even-subgraph expansion and the model is no longer solved by this direct Pfaffian method.

The high-temperature expansion

The first step is purely algebraic. For a single edge $e=\langle uv\rangle$ with coupling $K,$ one has

$\displaystyle e^{K\sigma_u\sigma_v}=\cosh K\left(1+\sigma_u\sigma_v\tanh K\right).$

Indeed, if $\sigma_u\sigma_v=1,$ the right side is $\cosh K(1+\tanh K)=e^K.$ If $\sigma_u\sigma_v=-1,$ it is $\cosh K(1-\tanh K)=e^{-K}.$ Define $x=\tanh K_1, y=\tanh K_2.$ Then every horizontal edge contributes a factor $\cosh K_1(1+x\sigma_u\sigma_v),$ and every vertical edge contributes $\cosh K_2(1+y\sigma_u\sigma_v).$ Pulling out all the $\cosh$ factors gives

$\displaystyle Z_{\text{Ising}}=(\cosh K_1)^{E_h}(\cosh K_2)^{E_v}\sum_{\sigma}\prod_{e\in E_h}\left(1+x\sigma_u\sigma_v\right)\prod_{e\in E_v}\left(1+y\sigma_u\sigma_v\right).$

Now expand the products. For each edge we choose either the term $1$ or the term involving $\sigma_u\sigma_v.$ Thus a term in the expansion is specified by a subset $\Gamma$ of edges. If $\Gamma$ contains $h(\Gamma)$ horizontal edges and $v(\Gamma)$ vertical edges, its weight is $x^{h(\Gamma)}y^{v(\Gamma)}.$ Its spin factor is

$\displaystyle \prod_{\langle uv\rangle\in\Gamma}\sigma_u\sigma_v.$

Group this product by vertices. A spin $\sigma_v$ appears once for each chosen edge incident to $v.$ Therefore the exponent of $\sigma_v$ is the degree of $v$ inside the chosen subgraph $\Gamma.$ When we sum over $\sigma_v=\pm1,$ we get zero unless this exponent is even. Indeed, $\sum_{\sigma_v=\pm1}\sigma_v^m=0$ if $m$ is odd, and equals $2$ if $m$ is even.

Thus only those edge subsets survive for which every vertex has even degree. Such subsets are called even subgraphs. On the square lattice they are unions of closed polygonal contours, with the possibility that several contours meet at a vertex of degree $4.$ The spin sum contributes a factor $2^{|V|}$ for every surviving even subgraph. Hence

$\displaystyle Z_{\text{Ising}}=2^{|V|}(\cosh K_1)^{E_h}(\cosh K_2)^{E_v}Z_{\text{even}}(x,y),$

where

$\displaystyle Z_{\text{even}}(x,y)=\sum_{\Gamma\text{ even}}x^{h(\Gamma)}y^{v(\Gamma)}.$

This is the high-temperature expansion. It turns the spin model into a polygon model. The Ising problem is now reduced to computing the generating function of even subgraphs.

The connection with dimers

We now explain the precise correspondence between the Ising high-temperature polygon expansion and a dimer model. Suppose $G$ is a finite planar graph. For the square lattice, every vertex has degree $4$ , but it is useful to describe the construction for a vertex of arbitrary degree $d$ . From the high-temperature expansion, the Ising partition function has been reduced to a sum over even subgraphs $\Gamma\subset E(G)$ , meaning that every vertex of $G$ is incident to an even number of chosen edges. If an edge $e$ is chosen in $\Gamma$ , it receives weight $t_e=\tanh K_e$ . Thus the polygon partition function is

$\displaystyle Z_{\text{even}}(G) =\sum_{\Gamma\text{ even}} \prod_{e\in\Gamma} t_e.$

The Fisher construction builds a new graph, called the Fisher graph and denoted here by $G^F$ , whose perfect matchings encode precisely these even subgraphs. The construction is local at each vertex. Let $v$ be a vertex of $G$ of degree $d$ , and list the incident edges in cyclic order around $v$ as $e_1,\dots,e_d$ . The cyclic order is part of the planar embedding. We replace $v$ by a small decorated graph, often called a Fisher city. This city has $2d$ vertices, denoted

$\displaystyle p_1,\dots,p_d, \quad q_1,\dots,q_d.$

The vertices $q_i$ are the terminals: the original edge $e_i$ will attach to $q_i$ . Inside the city we add, for every $i$ modulo $d$ , the three edges

$\displaystyle p_i p_{i+1}, \quad q_i p_i, \quad q_i p_{i+1}.$

Thus each $q_i$ forms a triangle with the two neighboring $p$ -vertices $p_i$ and $p_{i+1}$ . The $p_i$ themselves form a cycle, and each edge of that cycle is the base of a triangle whose third vertex is $q_i$ . All these internal edges receive weight $1$ .

Now consider an original edge $e=uv$ of $G$ . Suppose $e$ appears as the $i$ -th incident edge at $u$ and as the $j$ -th incident edge at $v$ . In the Fisher graph $G^F$ , we add one external edge joining the terminal $q_{u,i}$ in the city of $u$ to the terminal $q_{v,j}$ in the city of $v$ . This external edge receives weight $t_e$ . Thus original Ising edges become long edges between neighboring Fisher cities, while the small triangular edges inside each city have weight $1$ .

The key local fact is the following parity lemma. Fix one Fisher city of degree $d$ . Suppose a subset $S\subset{1,\dots,d}$ of the terminals $q_i$ is already matched externally, meaning that the corresponding long Fisher edges have been chosen in a dimer configuration. Then those terminal vertices are unavailable for internal matching. The remaining vertices inside the city must be matched using only internal edges. The lemma is: The city can be internally perfectly matched if and only if $\displaystyle |S|$ is even Moreover, for the triangular Fisher city just described, when $|S|$ is even, the number of internal perfect matchings is exactly $2$ , independent of the particular even subset $S$ . When $|S|$ is odd, there is no internal perfect matching.

Let us explain why this is true. The city is a cyclic chain of triangles. A terminal $q_i$ not used by an external dimer must be matched internally to either $p_i$ or $p_{i+1}$ . A terminal $q_i$ used by an external dimer is removed from the local problem, and then the edge $p_i p_{i+1}$ may be used internally. If one starts at any place in the cyclic chain and chooses one of the two possible local matching patterns, the matching is then forced as one moves around the cycle: at each triangle, once one of the adjacent $p$ -vertices has already been matched, the next choice is determined. After going all the way around the city, the final condition is consistent exactly when the number of externally matched terminals is even. If that number is even, the initial binary choice gives two possible internal completions; if it is odd, the propagation returns with the wrong parity and no completion exists.

For the square lattice this local statement is especially concrete. Each original vertex has four incident edges, so the Fisher city has vertices $p_1,p_2,p_3,p_4$ and terminals $q_1,q_2,q_3,q_4$ . The internal edges are the four cycle edges $p_1p_2,p_2p_3,p_3p_4,p_4p_1$ and the eight spoke edges $q_i p_i$ , $q_i p_{i+1}$ . If no external long edge is used at this city, all four $q_i$ must be matched internally, and there are exactly two completions. If exactly two external long edges are used, the two corresponding $q_i$ are already matched outside the city, and the remaining six internal vertices again have exactly two completions. If all four external long edges are used, only the four $p_i$ remain, and the cycle $p_1p_2p_3p_4p_1$ has exactly two perfect matchings. If one or three external long edges are used, an odd number of vertices remains inside the city, or equivalently the forced propagation fails, so no perfect matching is possible. Thus the local perfect-matching condition is exactly the even-degree condition.

This is the whole purpose of the Fisher city. It converts the Ising rule : an even number of polygon edges touches each original vertex, into the dimer rule: every Fisher-graph vertex is matched exactly once.

Now we define the global correspondence. Given a perfect matching $M^F$ of the Fisher graph $G^F$ , look at which long edges are used. Each long edge of $G^F$ corresponds to exactly one original edge of $G$ . Let $\Gamma(M^F)$ be the set of original edges whose corresponding long Fisher edges are present in $M^F$ . At each original vertex $v$ , the number of incident edges of $\Gamma(M^F)$ is exactly the number of terminals in the city of $v$ matched externally. Since the remaining part of the city has a perfect matching, the local lemma implies that this number must be even. Therefore $\Gamma(M^F)$ is an even subgraph of $G$ .

Conversely, start with an even subgraph $\Gamma\subset E(G)$ . For every edge $e\in\Gamma$ , choose the corresponding long Fisher edge in $G^F$ . Then, at every original vertex $v$ , an even number of terminals in the Fisher city of $v$ have already been matched externally. By the local lemma, the remaining vertices in that city can be internally matched, and in fact they can be matched in exactly two ways. Since the cities are disjoint except for the long edges already chosen, these local choices are independent from vertex to vertex. Therefore every even subgraph $\Gamma$ lifts to exactly $2^{|V(G)|}$ perfect matchings of the Fisher graph.

The weights match in the simplest possible way. The only non-unit weights in $G^F$ are the long edges, and the long edge corresponding to $e$ has weight $t_e$ . Therefore every Fisher matching lying above an even subgraph $\Gamma$ has weight

$\displaystyle \prod_{e\in\Gamma} t_e.$

The internal edges have weight $1$ , so they do not change the weight. Since each even subgraph has exactly $2^{|V(G)|}$ internal dimer completions, we obtain the exact partition-function relation

$\displaystyle Z_{\text{dimer}}(G^F) = 2^{|V(G)|} Z_{\text{even}}(G).$

Equivalently,

$\displaystyle Z_{\text{even}}(G) = 2^{-|V(G)|}Z_{\text{dimer}}(G^F).$

Now combine this with the high-temperature expansion of the Ising model. If $t_e=\tanh K_e$ , then

$\displaystyle Z_{\text{Ising}} = 2^{|V(G)|} \prod_{e\in E(G)}\cosh K_e Z_{\text{even}}(G).$

Substituting the Fisher relation gives

$\displaystyle Z_{\text{Ising}} = \prod_{e\in E(G)}\cosh K_e Z_{\text{dimer}}(G^F).$

This formula is very clean: after Fisher’s decoration, the zero-field Ising partition function is the dimer partition function of the Fisher graph, multiplied only by the elementary prefactor $\prod_e\cosh K_e$ . In the anisotropic square lattice, where horizontal edges have coupling $K_1$ and vertical edges have coupling $K_2$ , this reads

$\displaystyle Z_{\text{Ising}} = (\cosh K_1)^{E_h} (\cosh K_2)^{E_v} Z_{\text{dimer}}(G^F),$

with long horizontal Fisher edges weighted by $x=\tanh K_1$ and long vertical Fisher edges weighted by $y=\tanh K_2$ .

This is the Fisher correspondence. It is a precise finite combinatorial transformation between the Ising model and the dimer model. An Ising high-temperature polygon configuration chooses an even set of original edges. The Fisher graph turns each original edge into a possible long dimer. The triangular city at each original vertex enforces the evenness condition because it can be internally matched exactly when an even number of its terminals are already matched externally. The factor $2^{|V(G)|}$ appears because, for this standard Fisher city, every allowed local even choice has exactly two internal completions. Since this factor is independent of the polygon configuration, it is harmless and is absorbed exactly into the high-temperature expansion.

After this correspondence, all the Pfaffian machinery applies. The Fisher graph is planar whenever the original Ising graph is planar. Therefore one may choose a Kasteleyn orientation of $G^F$ and obtain

$\displaystyle Z_{\text{dimer}}(G^F) =|\text{Pf}(K_F)|.$

Consequently,

$\displaystyle Z_{\text{Ising}} = \prod_{e\in E(G)}\cosh K_e |\text{Pf}(K_F)|.$

On a periodic lattice, Fourier transform then diagonalizes the periodic Kasteleyn matrix. Thus the Ising solution follows the same analytic route as the dimer solution:

$\displaystyle \text{spins} \longrightarrow \text{even subgraphs} \longrightarrow \text{Fisher dimers} \longrightarrow \text{Pfaffian} \\ \longrightarrow \text{determinant} \longrightarrow \text{Fourier integral}.$

The Fisher step is the exact bridge between the Ising polygon expansion and the dimer/Pfaffian world.

For the square lattice it is convenient not to write the full Fisher gadget matrix. One can eliminate the internal Fisher degrees of freedom and use an equivalent smaller determinant, the Kac-Ward determinant. It is the same Pfaffian mechanism in compressed form. The Kac-Ward matrix acts on oriented edges rather than on the vertices of the Fisher graph. Its phases keep track of the turning angles of paths, just as the $i$ in the square-lattice dimer matrix kept track of signs around faces.

The Kac-Ward sign mechanism

Let us describe the Kac-Ward matrix .Instead of replacing every Ising vertex by a Fisher gadget and then writing a large dimer Pfaffian, Kac-Ward keeps the original graph but makes the matrix act on oriented edges. This is the main change of viewpoint. The high-temperature expansion gives a sum over even subgraphs, that is, closed polygon configurations. The Kac-Ward matrix is built so that a determinant expansion over closed non-backtracking walks reproduces precisely that even-subgraph generating function, after the correct cancellations.

Let $G$ be a finite planar graph. Every unoriented edge ${u,v}$ is replaced by two oriented edges, written $e=(u\to v)$ and $\bar e=(v\to u)$ . Let $E^{\rightarrow}$ denote the set of all oriented edges. If $e=(u\to v)$ , write $\text{tail}(e)=u$ and $\text{head}(e)=v$ . The Kac-Ward matrix is indexed by oriented edges. Thus its rows and columns are not indexed by vertices, but by directed edges. A transition from an oriented edge $e$ to another oriented edge $e'$ is allowed only when the walk can actually continue: the head of $e$ must equal the tail of $e'$ . In addition, the walk is not allowed to immediately reverse direction, so $e'\neq \bar e$ . Thus a Kac-Ward walk is a non-backtracking walk. It may go straight, turn left, or turn right, but it may not traverse an edge and then immediately traverse the same edge backward.

To define the phase, embed the graph in the plane. Each oriented edge has a direction angle. If $e$ points in direction $\theta(e)$ and $e'$ points in direction $\theta(e')$ , then the signed turning angle from $e$ to $e'$ is

$\displaystyle \alpha(e,e') = \theta(e')-\theta(e) \quad \text{chosen modulo }2\pi\text{ in }(-\pi,\pi).$

For example, on the square lattice, going straight has turning angle $0$ , turning left has angle $\pi/2$ , and turning right has angle $-\pi/2$ . Immediate reversal would have angle $\pi$ or $-\pi$ , but that transition is excluded.

Let $t_e$ be the high-temperature edge weight attached to the unoriented edge underlying $e$ . In the anisotropic square lattice, $t_e=x=\tanh K_1$ for horizontal edges and $t_e=y=\tanh K_2$ for vertical edges. The Kac-Ward transition matrix $T$ is defined by

$\displaystyle T_{e,e'}=\begin{cases} t_{e'}\exp\big(i\alpha(e,e')/2\big), & \text{if }\text{head}(e)=\text{tail}(e')\text{ and }e'\neq\bar e,\\ 0, & \text{otherwise}. \end{cases}$

Notice the placement of $t_{e'}$ . The transition from $e$ to $e'$ pays the weight of the new edge $e'$ . Therefore, if a closed walk uses oriented edges $e_1,e_2,\dots,e_n$ , with $e_{n+1}=e_1$ , then the product of transition weights is

$\displaystyle T_{e_1,e_2}T_{e_2,e_3}\cdots T_{e_n,e_1} = \bigg(\prod_{j=1}^{n}t_{e_j}\bigg) \exp\bigg(\frac{i}{2}\sum_{j=1}^{n}\alpha(e_j,e_{j+1})\bigg).$

Thus every closed non-backtracking walk receives two pieces of data: its ordinary edge weight, and a complex phase determined by its total turning angle. The Kac-Ward theorem says that, for a planar graph with the standard boundary convention,

$\displaystyle Z_{\text{even}}(G)^2 = \det(I-T).$

Here

$\displaystyle Z_{\text{even}}(G) = \sum_{\Gamma\text{ even}}\prod_{e\in\Gamma}t_e$

is the high-temperature polygon generating function. This identity is the compressed determinant form of the same Pfaffian mechanism that appears in the Fisher dimer construction. It is not merely a formal similarity: the Fisher graph Pfaffian and the Kac-Ward determinant are two ways of packaging the same planar sign correction. Let us now explain why a determinant of $I-T$ has anything to do with closed polygon configurations. Start from the identity

$\displaystyle \log\det(I-T) = \text{Tr}\log(I-T) = -\sum_{n\geq1} \frac{1}{n}\text{Tr}(T^n),$

valid first as a formal power series in the edge weights. The trace $\text{Tr}(T^n)$ is a sum over all closed length- $n$ non-backtracking walks in the oriented-edge graph. Indeed, an entry of $T^n$ is a sum over $n$ successive allowed transitions, and taking the trace forces the final oriented edge to return to the initial one. Thus $\log\det(I-T)$ is a weighted sum over closed non-backtracking walks.

Now consider first a simple closed polygon $\gamma$ in the planar graph. Suppose its edge weight is

$\displaystyle w(\gamma) = \prod_{e\in\gamma}t_e.$

There are two orientations of $\gamma$ , clockwise and counterclockwise. Along either orientation, the total turning angle is $2\pi$ or $-2\pi$ , depending on the orientation. Therefore the Kac-Ward phase is

$\displaystyle \exp\bigg(\frac{i}{2}(\pm2\pi)\bigg) = \exp(\pm i\pi) =-1.$

So a simple loop contributes the factor $-w(\gamma)$ to the trace product. This is the first essential point: the half-angle phase converts the geometric fact that a simple closed curve turns by $\pm2\pi$ into the algebraic sign $-1$ . Now look at how this loop contributes to the determinant. If the loop $\gamma$ is primitive and has length $|\gamma|$ , then its $r$ -fold repetition contributes a factor $(-w(\gamma))^r$ . In the logarithmic determinant, repeated loops appear with the usual factor $1/r$ after accounting for cyclic starting points. Thus one obtains a local contribution of the form

$\displaystyle -\sum_{r\geq1} \frac{1}{r} \big(-w(\gamma)\big)^r = \log\big(1+w(\gamma)\big).$

This is the second essential point. The minus sign from the turning phase changes $1-w(\gamma)$ into $1+w(\gamma)$ . Without the half-angle phase, the determinant expansion would have the wrong sign for polygon weights.

If the graph contained only one possible simple loop, this would already prove the idea: exponentiating the logarithmic determinant would give a factor $1+w(\gamma)$ , which means “either do not choose the polygon, or choose it once.” That is exactly how an even-subgraph generating function behaves for a single isolated polygon. For several disjoint simple polygons, the same reasoning gives a product

$\displaystyle \prod_{\gamma} \big(1+w(\gamma)\big),$

and expanding this product gives a sum over collections of disjoint polygons with positive weights. This is already the high-temperature expansion in the special case where even subgraphs are disjoint simple loops. The real graph is more complicated because even subgraphs may touch themselves or meet at vertices, and the trace expansion also contains closed non-backtracking walks with self-intersections. The full Kac-Ward theorem says that the same mechanism still works after all cancellations are taken into account. The half-angle phases are exactly tuned so that the unwanted self-intersecting walk contributions cancel, while the surviving contributions assemble into the square of the even-subgraph generating function.

Why the square? This is an important point. The determinant expansion naturally counts oriented closed-walk structures. The polygon generating function $Z_{\text{even}}$ is unoriented and already sums over all even subgraphs. The Kac-Ward identity is not $Z_{\text{even}}=\det(I-T)$ , but rather

$\displaystyle Z_{\text{even}}^2=\det(I-T).$

Thus the determinant should be thought of as counting two copies of the polygon expansion. This is already visible in the simplest case. Suppose $\gamma$ is a single simple polygon with weight $w(\gamma)=\prod_{e\in\gamma}t_e$ . The even-subgraph partition function for this one loop is $1+w(\gamma)$ . Therefore its square is $(1+w(\gamma))^2=1+2w(\gamma)+w(\gamma)^2$ . The determinant sees exactly these three possibilities because $\gamma$ has two orientations. The clockwise oriented loop contributes one term of weight $w(\gamma)$ , the counterclockwise oriented loop contributes another term of weight $w(\gamma)$ , and choosing both oriented loops contributes $w(\gamma)^2$ . Hence the determinant naturally gives $(1+w(\gamma))^2$ , not $1+w(\gamma)$ . One way to remember this is that the determinant is the square of the Pfaffian object hidden behind the planar sign correction, just as in the dimer model $\text{Pf}(K)^2=\det K$ . The Fisher correspondence gives $Z_{\text{even}}$ as a dimer Pfaffian up to local constants. Squaring that Pfaffian gives a determinant. The Kac-Ward determinant is a smaller determinant that equals this squared Pfaffian expression.

Let us make the closed-walk cancellation idea more concrete. A term in $\text{Tr}(T^n)$ is a closed non-backtracking walk. Such a walk may trace a simple loop, repeat a loop several times, or pass through a vertex more than once. The total phase of a closed walk is

$\displaystyle \exp\bigg(\frac{i}{2}\text{turn}(\omega)\bigg),$

where $\text{turn}(\omega)$ is the total signed turning angle of the walk. For a closed planar curve, this total turn is $2\pi$ times an integer, the winding number of the tangent direction. Hence the phase is always $(-1)^{\text{wind}(\omega)}$ . When a walk decomposes into several loop pieces, the phases multiply. Walks that differ by reversing the order in which self-intersecting loop pieces are traversed have the same ordinary weight but opposite Kac-Ward sign. These pairwise cancellations remove the contributions that do not correspond to genuine even-subgraph choices. What remains is exactly the exponential generating function for even edge sets. Thus the Kac-Ward theorem can be understood as follows. The determinant expansion produces closed non-backtracking walks. The no-backtracking condition prevents immediate cancellations such as walking along an edge and instantly returning. The half-angle factor records the turning of the walk. A simple closed loop gets phase $-1$ , and this changes the logarithmic determinant contribution into $\log(1+w(\gamma))$ , the correct positive polygon factor. More complicated closed walks cancel in pairs or assemble correctly because the same turning-angle sign keeps track of the planar topology of the walk.

For simple cycles this is positive, as shown above. For self-intersecting cycles or for different loop decompositions of the same even edge set, individual terms may have different signs. The important fact is that these signs are not arbitrary. At every self-intersection or multi-visit vertex, the alternative ways of reconnecting the directed strands have the same ordinary edge weight but opposite Kac-Ward phase. Thus the determinant expansion contains a sign-reversing cancellation among the decompositions that do not correspond to the two-copy even-subgraph expansion. The surviving total contribution is exactly the positive contribution required by $Z_{\text{even}}^2$ . One clean way to state the general mechanism is this. Fix a monomial $\prod_e t_e^{m_e}$ , where each $m_e\in{0,1,2}$ . In $Z_{\text{even}}^2$ , its coefficient counts ordered pairs $(\Gamma_1,\Gamma_2)$ of even subgraphs such that each edge $e$ appears in exactly $m_e$ of the two subgraphs. The Kac-Ward determinant gives the same coefficient, but in a different language: it sums over all directed non-backtracking cycle decompositions whose projection to the original graph uses each unoriented edge $e$ exactly $m_e$ times. The half-angle phases make the signed sum of these decompositions equal to the number of such ordered pairs $(\Gamma_1,\Gamma_2)$ . That coefficient identity, for every monomial, is precisely the statement

$\det(I-T)=Z_{\text{even}}^2$ .

This is the exact analogue of the Kasteleyn sign condition for dimers: signs are inserted locally so that global combinatorial objects are counted positively. It is useful to compare the two sign mechanisms directly. In the dimer problem, the Pfaffian is a signed sum over pairings. A Kasteleyn orientation is chosen so that the sign changes around alternating cycles cancel the Pfaffian permutation signs. Then every perfect matching contributes with the same sign. In the Kac-Ward representation of the Ising model, the determinant is expanded through traces, so it is a signed exponential sum over closed non-backtracking walks. The half-angle phases are chosen so that the closed-walk signs collapse to the positive even-subgraph expansion. In both cases, the difficulty is not the weights; the weights are easy. The difficulty is sign. Kasteleyn orientations solve the sign problem for pairings. Kac-Ward turning phases solve the sign problem for closed polygon expansions.

Square Lattice

For the square lattice, the Kac-Ward construction becomes completely explicit because there are only four possible oriented edge directions. We label them $E,\quad N,\quad W,\quad S,$ meaning east, north, west, and south. The Kac-Ward matrix acts not on vertices but on oriented edges. Thus, at each step, the state of the walk remembers the direction of the edge just traversed. From a given oriented edge, the next edge may continue straight or turn left or right, but it may not immediately reverse direction. Thus from $E$ one may go to $E,N,S$ , but not to $W$ ; from $N$ one may go to $N,E,W$ , but not to $S$ ; and similarly for the other directions. This “no immediate backtracking” rule is essential: the closed walks appearing in the determinant expansion are non-backtracking closed walks.

Let the horizontal high-temperature weight be $x=\tanh K_1$ and the vertical high-temperature weight be $y=\tanh K_2$ . Thus a newly traversed horizontal edge contributes weight $x$ , while a newly traversed vertical edge contributes weight $y$ . The Kac-Ward transition also includes a phase depending on the turning angle. If the walk goes straight, the turning angle is $0$ , so the half-angle factor is $e^{0}=1$ . If the walk turns left by $\pi/2$ , the factor is $e^{\pi i/4}$ . If it turns right by $-\pi/2$ , the factor is $e^{-\pi i/4}$ . These are the only three possibilities on the square lattice, since immediate reversal is forbidden.

Now we Fourier transform in space. Let $z=e^{i\theta}$ and $u=e^{i\phi}$ . Translation by one lattice unit east gives a factor $z$ , west gives $z^{-1}$ , north gives $u$ , and south gives $u^{-1}$ . We use the convention that the transition pays the weight and the Fourier translation factor of the new edge. Thus, for example, the transition $E\to E$ means “after moving east, move east again,” so it contributes $xz$ . The transition $E\to N$ turns left and then moves north, so it contributes $yu e^{\pi i/4}$ . The transition $E\to S$ turns right and then moves south, so it contributes $yu^{-1}e^{-\pi i/4}$ . The transition $E\to W$ is forbidden because it is immediate reversal, so its entry is $0$ .

With rows and columns ordered as $E,N,W,S$ , the Fourier block of $I-T$ is therefore

$\displaystyle \mathcal K(z,u)= \begin{pmatrix} 1-xz & -xu e^{-\pi i/4} & 0 & -xu^{-1}e^{\pi i/4}\\ -yz e^{\pi i/4} & 1-yu & -yz^{-1}e^{-\pi i/4} & 0\\ 0 & -xu e^{\pi i/4} & 1-xz^{-1} & -xu^{-1}e^{-\pi i/4}\\ -yz e^{-\pi i/4} & 0 & -yz^{-1}e^{\pi i/4} & 1-yu^{-1} \end{pmatrix}.$

Let us spell out the entries. The diagonal-looking terms $1-xz$ , $1-yu$ , $1-xz^{-1}$ , and $1-yu^{-1}$ come from $I-T$ : the $1$ is the identity matrix, while the negative term is the straight transition in the same direction. The zero entries are exactly the forbidden immediate reversals: $E\to W$ , $N\to S$ , $W\to E$ , and $S\to N$ . The remaining off-diagonal entries are left and right turns. For example, $N\to W$ is a left turn and then a west step, hence it contributes $xz^{-1}e^{\pi i/4}$ , with a minus sign in $I-T$ . Similarly, $N\to E$ is a right turn and then an east step, so it contributes $xz e^{-\pi i/4}$ , again with a minus sign in $I-T$ .

The determinant of this $4\times4$ matrix is the scalar Laurent polynomial that controls the Ising free energy. A direct calculation gives

$\displaystyle \text{det}\mathcal K(z,u) = Q(z,u) =(1+x^2)(1+y^2)-x(1-y^2)\big(z+z^{-1}\big)-y(1-x^2)\big(u+u^{-1}\big).$

It is useful to understand what has happened in this determinant. The block $\mathcal K(z,u)$ remembers oriented-edge directions, so it is a $4\times4$ matrix. But after taking the determinant, all the half-angle phases combine and cancel in a very rigid way. The final answer is not a complicated expression involving $e^{\pi i/4}$ ; it is a real Laurent polynomial in $z,u$ with coefficients depending on $x,y$ . The terms $z+z^{-1}$ record horizontal translation, and the terms $u+u^{-1}$ record vertical translation. The factors $1-y^2$ and $1-x^2$ arise from the interaction between horizontal and vertical turn possibilities in the determinant expansion.

On the unit torus, write $z=e^{i\theta}$ and $u=e^{i\phi}$ . Then $z+z^{-1}=2\cos\theta$ and $u+u^{-1}=2\cos\phi$ . Hence

$\displaystyle Q(e^{i\theta},e^{i\phi})=(1+x^2)(1+y^2)-2x(1-y^2)\cos\theta-2y(1-x^2)\cos\phi.$

This is the square-lattice Ising characteristic polynomial in high-temperature variables. It plays the same role for the Ising model that $P_{\text{dimer}}(z,u)=z+z^{-1}+i(u+u^{-1})$ played for the square-lattice dimer model. In the dimer model with one black and one white vertex per fundamental domain, the Fourier symbol was already a scalar Laurent polynomial. In the Kac-Ward Ising formulation, the Fourier object is first a $4\times4$ matrix because the walk remembers its incoming direction, but its determinant reduces the problem again to a scalar Laurent polynomial.

Torus

Having obtained the scalar Kac-Ward polynomial $Q(z,u)$ , we now use it to compute the torus partition function and then pass to the thermodynamic limit. Take an $M\times N$ square torus, with horizontal coupling $K_1$ and vertical coupling $K_2$ , and write $x=\tanh K_1$ and $y=\tanh K_2$ . The Kac-Ward matrix acts on oriented edges, and on the square lattice there are four oriented directions, $E,N,W,S$ . Thus after Fourier transform, the large periodic matrix $I-T$ decomposes into $4\times4$ Fourier blocks $\mathcal K(z,u)$ , one block for each allowed momentum $(z,u)$ . In a fixed boundary sector, the determinant of the full finite matrix factors as a product of block determinants:

$\displaystyle \text{det}(I-T) = \prod_{(z,u)} \text{det}\mathcal K(z,u)= \prod_{(z,u)} Q(z,u).$

On the unit torus, where $z=e^{i\theta}$ and $u=e^{i\phi}$ , this becomes

$\displaystyle Q(e^{i\theta},e^{i\phi})=(1+x^2)(1+y^2)-2x(1-y^2)\cos\theta-2y(1-x^2)\cos\phi.$

This is the exact analogue of what happened in the dimer calculation. There too, periodicity turned a large determinant into a product over Fourier modes. The difference is that in the simplest bipartite dimer square lattice the Fourier symbol was already a scalar, while here the Kac-Ward operator remembers an oriented direction, so the Fourier object is first a $4\times4$ matrix. Taking its determinant collapses the directional information into the single scalar polynomial $Q(z,u)$ .

On a torus, however, one must not confuse one Fourier sector with the exact finite-torus partition function. As in the dimer problem, there are four boundary sectors. In the horizontal direction the Kac-Ward/Pfaffian variables may be periodic or antiperiodic, and independently in the vertical direction they may also be periodic or antiperiodic. We label the four sectors by $\epsilon,\eta\in{0,1}$ . The allowed momenta in the sector $(\epsilon,\eta)$ are

$\displaystyle \theta_j^{(\epsilon)} = \frac{(2j+\epsilon)\pi}{M}, \quad \phi_k^{(\eta)} = \frac{(2k+\eta)\pi}{N}, \quad 0\leq j<M,\quad 0\leq k<N.$

Thus $\epsilon=0$ means $z^M=1$ , while $\epsilon=1$ means $z^M=-1$ ; similarly, $\eta=0$ means $u^N=1$ , while $\eta=1$ means $u^N=-1$ . In the sector $(\epsilon,\eta)$ , the corresponding Kac-Ward determinant is

$\displaystyle D_{\epsilon,\eta} = \prod_{j=0}^{M-1} \prod_{k=0}^{N-1} Q\big(e^{i\theta_j^{(\epsilon)}},e^{i\phi_k^{(\eta)}}\big).$

The exact finite-torus formula for the even-subgraph partition function is a signed linear combination of the four square roots $D_{\epsilon,\eta}^{1/2}$ . The signs depend on the convention chosen for the Kac-Ward orientation or, equivalently, for the associated Pfaffian orientation. The invariant statement is that the finite-torus expression has the form

$\displaystyle Z_{\text{even}}^{\text{torus}} = \frac12 \sum_{\epsilon,\eta\in{0,1}} s_{\epsilon,\eta}D_{\epsilon,\eta}^{1/2}, \quad s_{\epsilon,\eta}\in{+1,-1}.$

This four-sector formula is important for exact finite-size computations. It is the torus correction to the simpler planar identity. On a simply connected planar graph, the Kac-Ward formula says $Z_{\text{even}}^2=\text{det}(I-T)$ . On a torus, the two noncontractible cycles force four twisted determinants instead of one. But for the infinite-volume free energy density, this finite-sector bookkeeping does not alter the limiting answer. The four momentum grids differ only by half-step shifts. After division by $MN$ , all four logarithmic sums have the same Riemann-sum limit:

$\displaystyle \frac{1}{MN}\log D_{\epsilon,\eta} = \frac{1}{MN} \sum_{j=0}^{M-1} \sum_{k=0}^{N-1} \log Q\big(e^{i\theta_j^{(\epsilon)}},e^{i\phi_k^{(\eta)}}\big) \longrightarrow \frac{1}{(2\pi)^2} \int_0^{2\pi} \int_0^{2\pi} \log Q(e^{i\theta},e^{i\phi}) d\theta d\phi.$

The square root in the finite formula is not an accident. It reflects the basic Kac-Ward square identity. In the planar case, and in each twisted sector on the torus, the determinant is the square of the corresponding Pfaffian-type object. For the even-subgraph partition function this means that the determinant contributes twice the logarithm. Thus, at the level of free energy, the even-subgraph contribution contains one half of the logarithmic determinant. This is why the limiting even-subgraph part is

$\displaystyle \frac12 \frac{1}{(2\pi)^2} \int_0^{2\pi} \int_0^{2\pi} \log Q(e^{i\theta}, e^{i\phi}) d\theta d\phi.$

Now we restore the elementary prefactor from the high-temperature expansion. For one edge $\langle uv\rangle$ , the identity used was

$\displaystyle e^{K\sigma_u\sigma_v} = \cosh K\big(1+\sigma_u\sigma_v\tanh K\big).$

Therefore the full Ising partition function is

$\displaystyle Z_{\text{Ising}} = 2^{|V|} (\cosh K_1)^{E_h} (\cosh K_2)^{E_v} Z_{\text{even}}(x,y), \quad x=\tanh K_1,\quad y=\tanh K_2.$

On the $M\times N$ square torus, there are $|V|=MN$ spins, $E_h=MN$ horizontal edges, and $E_v=MN$ vertical edges. Hence, after dividing $\log Z_{\text{Ising}}$ by $MN$ and letting $M,N\to\infty$ , the free energy per spin is

$\displaystyle f(K_1,K_2) = \log 2 + \log\cosh K_1 + \log\cosh K_2 + \frac12 \frac{1}{(2\pi)^2} \int_0^{2\pi} \int_0^{2\pi} \log Q(e^{i\theta},e^{i\phi}) d\theta d\phi.$

This is the Onsager free energy in high-temperature variables. Each term has a visible origin. The term $\log 2$ comes from the spin sum at each vertex. The terms $\log\cosh K_1$ and $\log\cosh K_2$ come from the edge factors pulled out in the high-temperature expansion. The integral comes from the Kac-Ward determinant. The factor $1/2$ appears because Kac-Ward gives a square, namely a determinant, while the polygon partition function itself is the square root of that determinant.

It is useful to check the formula in two simple regimes. If $K_1=K_2=0$ , then $x=y=0$ and $Q(e^{i\theta},e^{i\phi})=1$ . The integral vanishes, the two $\log\cosh$ terms vanish, and the formula gives $f=\log2$ , as it should, because all $2^{MN}$ spin configurations have equal weight. At very low temperature, $x,y\to1$ . Then $Q(e^{i\theta},e^{i\phi})\to 4$ , so the determinant contribution tends to $\frac12\log4=\log2$ . Since $\log\cosh K_i\sim K_i-\log2$ , the total free energy behaves like $K_1+K_2$ , which is exactly the ground-state energy contribution per spin for the ferromagnetic model.

The critical point

The phase transition occurs when the logarithmic integral develops a singularity. This happens when $Q$ vanishes on the unit torus. Since $Q(e^{i\theta},e^{i\phi})$ is minimized at $\theta=\phi=0$ for positive $x,y,$ the first possible zero occurs at $z=u=1.$ Compute

$\displaystyle Q(1,1)=(1+x^2)(1+y^2)-2x(1-y^2)-2y(1-x^2).$

This expression factors beautifully:

$\displaystyle Q(1,1)=(xy+x+y-1)^2.$

Therefore criticality occurs when

$\displaystyle xy+x+y=1.$

In terms of the original couplings, this is equivalent to

$\displaystyle \sinh(2K_1)\sinh(2K_2)=1.$

Indeed, since $\sinh(2K)=2\tanh K/(1-\tanh^2K),$ the equation $\sinh(2K_1)\sinh(2K_2)=1$ becomes $4xy=(1-x^2)(1-y^2).$ Factoring the difference gives

$\displaystyle (1-x^2)(1-y^2)-4xy=(xy-x-y-1)(xy+x+y-1).$

For $0<x,y<1,$ the first factor is negative and cannot vanish at the physical critical point. Thus the relevant condition is exactly $xy+x+y=1.$

In the isotropic case $K_1=K_2=K,$ so $x=y=t=\tanh K.$ The critical equation becomes

$\displaystyle t^2+2t-1=0.$

The positive solution is

$\displaystyle t_c=\sqrt2-1.$

Therefore

$\displaystyle K_c=\frac12\log(1+\sqrt2).$

This is the famous critical point of the two-dimensional square-lattice Ising model.

The free energy itself remains finite at the critical point, because the logarithmic singularity is integrable in two dimensions. Near $\theta=\phi=0$ at criticality, expand

$\displaystyle \cos\theta=1-\frac{\theta^2}{2}+O(\theta^4),\qquad \cos\phi=1-\frac{\phi^2}{2}+O(\phi^4).$

Since $Q(1,1)=0$ at criticality, the leading behavior is quadratic:

$\displaystyle Q(e^{i\theta},e^{i\phi})\sim c_1\theta^2+c_2\phi^2.$

Thus the integrand behaves like $\log(c_1\theta^2+c_2\phi^2).$ This is integrable, because in polar coordinates it is comparable to $\log r^2,$ and $\int_0^\epsilon r|\log r|,dr<\infty.$ So the free energy is continuous.

However, derivatives of the free energy are more singular. Differentiating with respect to $K$ differentiates $Q$ inside the logarithm. Near the critical point, the second derivative produces the familiar logarithmic divergence of the specific heat. This is one of the signatures of the two-dimensional Ising transition. In this determinant language, the singularity comes directly from a zero of the characteristic polynomial landing on the unit torus.

The parallel with the dimer calculation is now clear. In the square-lattice dimer model, the Fourier multiplier was

$\displaystyle P_{\text{dimer}}(z,u)=z+z^{-1}+i(u+u^{-1}).$

The free energy was the average of $\log|P_{\text{dimer}}|$ over the Fourier torus. The zeros of $P_{\text{dimer}}$ on the unit torus controlled the critical correlations.

For the Ising model, after the high-temperature expansion and the Fisher/Kac-Ward reduction, the corresponding polynomial is

$\displaystyle Q(z,u)=(1+x^2)(1+y^2)-x(1-y^2)(z+z^{-1})-y(1-x^2)(u+u^{-1}).$

The free energy is the average of $\log Q$ over the Fourier torus, with the prefactor $\log2+\log\cosh K_1+\log\cosh K_2.$ The critical point occurs when $Q$ touches zero on the unit torus. Thus the analytic structure is exactly the same kind of structure as before: a combinatorial model becomes a Pfaffian, the Pfaffian squared becomes a determinant, Fourier transform turns the determinant into a product of symbols, and the thermodynamic limit is a logarithmic integral.

Conclusion

The two-dimensional zero-field Ising model is exactly solvable because its high-temperature expansion has an evenness constraint. The spin sum kills every edge subset except those in which every vertex has even degree. Thus the model becomes a polygon model. Fisher’s construction converts that polygon model into a dimer model on a decorated graph. Kasteleyn theory converts the dimer model into a Pfaffian. Equivalently, the Kac-Ward determinant gives a compressed determinant form of the same Pfaffian mechanism.

The calculation follows the same chain as the dimer calculation:

$\displaystyle \text{spins}\Rightarrow\text{even subgraphs}\Rightarrow\text{dimers on a Fisher graph}\Rightarrow\text{Pfaffian}\Rightarrow\text{determinant}\Rightarrow\text{Fourier integral}.$

For the square lattice, the determinant reduces to the polynomial $Q(z,u).$ The infinite-volume free energy is

$\displaystyle f(K_1,K_2)=\log2+\log\cosh K_1+\log\cosh K_2+\frac12\int \int\log Q \frac{d\theta d\phi} {(2\pi)^2}.$

The critical point is where $Q$ first vanishes on the unit torus. In anisotropic form this gives

$\displaystyle \sinh(2K_1)\sinh(2K_2)=1,$

and in the isotropic case it gives

$\displaystyle K_c=\frac12\log(1+\sqrt2).$

Thus the Onsager solution is not a separate miracle from the dimer solution. It is another manifestation of the same analytic mechanism: signs are corrected by a Pfaffian structure, determinants are diagonalized by Fourier transform, and the phase transition is read from the zero set of a Laurent polynomial.

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