Correlations in totally symmetric self-complementary plane partitions

Totally symmetric self-complementary plane partitions (TSSCPPs) are boxed plane partitions with the maximum possible symmetry. We use the well-known representation of TSSCPPs as a dimer model on a honeycomb graph enclosed in one-twelfth of a hexagon with free boundary to express them as perfect matchings of a family of non-bipartite planar graphs. Our main result is that the edges of the TSSCPPs form a Pfaffian point process, for which we give explicit formulas for the inverse Kasteleyn matrix. Preliminary analysis of these correlations are then used to give a precise conjecture for the limit shape of TSSCPPs in the scaling limit.

row and left column (the top-left formula). The latter formula was subsequently improved in [BDFZJ13]. Other refined enumeration formulas include the top two formula [FR09,KR10], the top two and bottom formula [Fis12], the top-left-bottom formula [Fis12,AR13] and the top-left-bottom-right formula [AR13,Beh13]. The problem of computing bulk correlations seems like a difficult and interesting open problem. On the TSSCPP side, no formulas are known for any correlation functions.
On the other hand, correlations for some plane partitions have been established in recent years [OR03,Pet14,BBNV18]. The typical perspective here is to view the plane partition as a rhombus or lozenge tiling. Randomness is introduced by picking each configuration at random from the set of all possible configurations in some prescribed manner, the simplest being picking each configuration uniformly at random which is the case considered here for TSSCPPs. For a specific class of tiling models, interesting probabilistic features are observed when the system size becomes large, such as a macroscopic limit shape, which is a type of law of large numbers result. Around this limit shape, there are still microscopic fluctuations which are believed to be governed by universal probability distributions originating in both statistical mechanics and random matrix theory. This assertion has been proved primarily for domino and lozenge tiling models; see [Gor20] and references therein for details.
To study these fluctuations, one of the more successful approaches has been to study the correlations of an associated particle system to the random tiling model using methods originating from random matrix theory. For many types of tiling models, these correlations are governed by the determinant of some matrix, called the correlation kernel. Probability measures of this form are known as determinantal point processes; see for example [Sos06]. Finding the correlation kernel can be computationally tricky, but there are now some relatively standard approaches such as using the Eynard-Mehta Theorem [Joh03,BR06] which has been particularly useful for those in the Schur process class [Joh05,BF14,Dui13,Pet14,DM15] as well as those that are not Schur processes [DK17, BD19,CDKL20]. Put bluntly, this theorem gives the correlation kernel when the model is expressed in terms of nonintersecting lattice paths with fixed endpoints. There are other approaches for computing correlation kernels, such as vertex operators [OR03,BCC17,BBC + 17] and also the Harish-Chandra/Itzykson-Zuber integral [Nov15] The Eynard-Mehta theorem has a Pfaffian analog where the final positions of the nonintersecting lattice paths are free. In this case, the correlations of the associated particle system to the tiling model are given by a Pfaffian point process; see [BBNV18] for an example where the authors give a formula for the correlation kernel for both symmetric plane partitions and plane overpartitions. The TSSCPP is another example of this and so the Eynard-Mehta theorem immediately shows that the particle system defined through the nonintersecting lattice paths for TSSCPP is a Pfaffian point process with some correlation kernel 1 . Unfortunately, the formula for this correlation kernel is not known, due to computational difficulties in inverting an arbitrary sized matrix that is found in the Eynard-Mehta theorem. In this paper, we use dimer model techniques to settle this problem and find a formula for the inverse Kasteleyn matrix for TSSCPPs, where the inverse Kasteleyn matrix can be heuristically thought of as the dimer model equivalent to the correlation kernel of a particle system. Since lozenge tilings and its associated particle system are in bijection, this implies a formula for the correlation kernel of the associated particle system. Figure 1. The region of the regular hexagon bounded by the blue lines whose tiling is sufficient to determine a totally symmetric selfcomplementary plane partition. The pink and orange lozenges are forced.
The rest of the paper is organized as follows. In Section 2, we convert plane partitions to perfect matchings of a class of non-bipartite graphs bijectively to be able to explain our main results. We then summarize the main results of this article in Section 3, giving a formula for the inverse Kasteleyn matrix and stating a sum rule. Sections 4 and 5 are devoted to computations of this formula at special locations. Section 6 completes the proof of the formula for the inverse Kastelen matrix. The proof of the main result use combinatorial identities whose proof is deferred to Section 7. We present boundary recurrences for the inverse Kasteleyn matrix of independent interest, which we use to prove the sum rule, in Section 8. Finally, we end with heuristics for the limit shape and a precise conjecture in Section 9.

From TSSCPPs to hexagonal graphs
As mentioned earlier, a plane partition inside an a × b × c box can be equivalently viewed as a lozenge (or rhombus) tiling of a hexagonal region of side lengths a, b, c, a, b, c of a triangular lattice [Kup94]. A totally symmetric self-complementary plane partition of order n is then a rhombus tiling of a regular hexagon with side length 2n with the maximum possible symmetry. In this case, all the information about the tiling is contained in (1/12)'th of the hexagon [MRR86,Section 8].
This is illustrated in Figure 1, where the region enclosed by the blue lines is to be tiled with lozenges in a maximal way. This means that among the 2(n − 1) pendant edges, only n − 1 will be matched. This is known as a free boundary condition. Equivalently, we have to find maximum matchings of the dual graph drawn in black. (Recall that a maximum matching of a graph is one which has the largest number of matched edges.) Let T n−1 denote this dual graph for TSSCPPs of size n. Figure 2 shows the dual graphs for TSSCPPs of sizes 3 and 4. Define, for n ≥ 1, Theorem 2.1 (Conjectured in [MRR86], proved in [And94]). The number of maximum matchings of T n is given by A n+1 . Recall that T n has 2n pendant vertices on the right. We now define a related family of graphs G n starting from T n as follows. We add 2n + 1 (resp. 2n + 2) vertices if n is even (resp. odd) in a column on the right of the pendant vertices and connect them in a triangular fashion as illustrated in Figure 3. Notice that if n is odd, the topmost vertex is a leaf, i.e. a pendant vertex. Proposition 2.2. The number of perfect matchings of G n is A n+1 .
Proof. We will construct a bijection between maximum matchings of T n and perfect matchings of G n . Starting with a perfect matching of G n and removing the vertices in the rightmost column, we obtain a maximum matching of T n . To go the other way, we will show that there is a unique way to complete a maximum matching of T n . Consider a maximum matching as shown in Figure 4. We will now match the remaining vertices in G n . We start from the leftmost vertex v 1 and match it to the leftmost available vertex. We then find the leftmost unmatched vertex and match it to the second unmatched vertex on the left. We continue this way until all vertices are matched. To see that we do not run into a contradiction, consider the leftmost available vertex at any stage. Suppose it is v i . If u i is matched, we match v i to v i+1 , and if not, we match it to u i . Suppose it is u i . Then we match it to v i+1 . This is always possible, since v i+1 cannot be matched to v i as u i is unmatched. The rightmost vertex has to be matched since there are an even number of unmatched vertices initially. This completes the proof.

Summary of results
To state our main result, we will need to introduce some notation. We write [x] 2 to denote x mod 2. For a, m ∈ Z with m > 0, we denote by Γ a,a+1,...,a+m a positively oriented contour containing the integers a through to a + m and no other integers. In particular, Γ 0 is a positively oriented circle around the origin with radius less than 1. We denote i = √ −1. We denote Z G to be the number of dimer configurations of the graph G. For a subset of vertices U , we let Z G\U denote the number of dimer configurations on the subgraph of G induced by removing U . When G n is the TSSCPP graph of size n defined above, we write Z n ≡ Z Gn . We will also use the notation Z {v 1 ,...,vm} n ≡ Z Gn\{v 1 ,...,vm} , that is the partition function of the dimer model on the induced graph of G n on the vertex set V n \{v 1 , . . . , v m }. Finally, we will use the convention that m −1 = 0 for all m ∈ N ∪ {0} throughout the paper. For the graph G n = (V n , E n ), the coordinates of the vertices are given by and the edges are given by (3.2) see Figure 5.  The Kasteleyn orientation is chosen so that vertices with odd parity are sinks, those with even parity not on the diagonal are sources, and the edges along the diagonal point towards the origin.
We let b denote the vertex (2n, 2n + 1 − [n] 2 ). The entries of the skew-symmetric Kasteleyn matrix, K n = K n (x, y) x,y∈Vn are given by (3.3) K n ((x 1 , x 2 ), (y 1 , y 2 )) = k n ((x 1 , x 2 ), (y 1 , y 2 )) − k n ((y 1 , y 2 ), ( Here the Kasteleyn orientation is chosen so that the number of counter-clockwise arrows around each face is an odd number. From [Kas61,Kas63], |PfK n | is equal to Z n . Define n is the n'th Catalan number. We now introduce notation helpful for stating the main theorem. Recall that Γ 0 is a positively oriented circle around the origin with radius less than 1. Introduce the following formulas for 0 ≤ i ≤ 2n − 1: We will first state the formula for K −1 n in the special case when the second vertex is b. This will prove useful for the general result. Finally, when n is even, K −1 n ((2n, 2n), b) = −1.
Our proof strategy is as follows. We first establish Theorem 3.1 in Section 4. Once this is established, we use graphical condensation [Kuo06] to establish K −1 n (x, y) for x and y on the top boundary of the TSSCPP. We can then recover the other entries using recursions obtained from the matrix equations K −1 n .K n = K n .K −1 n = I viewed entrywise. The proof is given in Section 6, while postponed proofs of results used in proving Theorem 3.2 are given in Section 7.
This leads to the question whether there is a set of recurrences that establishes K −1 n uniquely, without relying on a priori expressions for K −1 n (·, b). Such recurrences are known for domino tilings of the Aztec diamond [CY14] and are expected to hold for lozenge tilings [Oko09]. In fact, there is an additional recurrence, along with the recurrence from graphical condensation given in Lemma 5.3 which parametrizes K −1 n on the boundary. These two recurrences give a concrete and explicit example of the additional recurrences postulated in [Oko09] for lozenge tilings. These recurrences combined with the recurrences from the matrix equations K n .K −1 n = K −1 n .K n = I give a unique way to determine K −1 n . This additional recurrence is given in Section 8. One consequence of finding K −1 n is a formula to compute local statistics [MPW63,Ken97]. We restate this here in a form that applies to our situation.
The term "sum rule" refers to certain sums of correlation functions that are of importance. In many cases, these sum rules have simple formulas. For the so-called quantum Knizhnik-Zamolodchikov equation, sum rules have been conjectured relating them to the q-enumeration of TSSCPPs by Di Francesco [DF06], and Di Francesco and Zinn-Justin [DFZJ05,ZJDF08] and proved by Zeilberger [Zei07] . Figure 6. A simulation, using Glauber dynamics, of a uniformly random TSSCPP of size 100. Here, we have rotated the hexagonal graph in Figure 3 by π/6. The simulations only show the dimers on this graph, drawn in different three different colors.
We now state a formula for a sum rule for TSSCPPs. For 0 ≤ j ≤ 2n − 1, let Theorem 3.4.
We remark that the entries g b n (j) themselves are not so simple. For example, The formula given in Theorem 3.2 is not in the most convenient form for asymptotic analysis. Nonrigorous computations show that the terms in the formula can be approximated by double contour integral formulas, that share some similarities with lozenge tiling models in the Schur class, e.g. see [GP19] and references therein.
Furthermore, these nonrigorous computations show a limit shape and Airy kernel statistics at the edge, which also appears in simulations; see Figure 6. We will not attempt to pursue these here as the computations appear to be long and involved. However, we will give a conjecture of the limit shape, based on a short computation in Section 9.

Proof of Theorem 3.1
Before giving the proof of Theorem 3.1, we give two combinatorial identities which will be used below. Their proofs are given in Section 7.
C. Krattenthaler has kindly shown us how to derive Theorem 4.1 starting from an identity in the book by Gasper and Rahman [GR04, Equation (3.8.12)]. However, we use Zeilberger's algorithm to prove it in Section 7.
Theorem 4.2. Let n ∈ N and 0 ≤ i ≤ n. Define Then, we have The proofs of Theorem 4.1,Theorem 4.2 are given in Section 7. First, we need the following two results.
Proposition 4.4. From the formulas for K −1 n (·, b) given in (3.8) and (3.9), we have that Proof. We expand out the left side using the definitions of K −1 n (·, b) as given in (3.8) and (3.9) which involves the formulas (3.7) and (3.6). This gives where the first equality follows from simplifying the integrand and the last equality follows from pushing the contour through infinity and computing the residue at r = 1. The claim follows from Theorem 4.1.
We will also repeatedly use the standard integrals, Proof of Theorem 3.1. We will prove the following, assuming (3.8) and (3.9).
(4.9) K n .K −1 The determinant of K n is nonzero because it is the square of the partition function. Therefore the equations from (4.9) are linearly independent and they guarantee the uniqueness of K −1 n (·, b). For the purposes of the proof, we introduce (4.10) We will verify (4.9) in each of the four cases: Expanding out the left side of (4.9) entrywise gives . Write x = (i 1 , i 1 + 2i 2 ) and suppose that x 1 = 0 and x 2 = 2n + 1. Then (4.13) becomes by (3.8) (4.14) where we have used n k + n k−1 = n+1 k in the last step. This verifies (4.9) when x = (x 1 , x 2 ) ∈ E n \D n and x 1 = 0 and x 2 = 2n + 1. We now consider (4.13) when n (0)(−1) i 2 and so by (3.8) we have also verified (4.9) for (0, x 2 ) ∈ E n \D n . Next we consider when x 2 = 2n + 1, which means that (4.13) becomes . Take x + (−1, 0) = (x 1 − 1, 2n + 1) = (i 1 , 2n + 1) so that i 1 = x 1 − 1. This means that we have Using (3.8), we have that (4.15) becomes where the last equation is found from rearranging the relation in (4.14). We need to show that the right side of the above equation is equal to zero. To do so, we expand out the above term using (3.7) which gives (4.17) The inner sum in the integrand equals where the 2 F 1 terms is the Gaussian hypergeometric function. Since the hypergeometric term is non-singular in r, we see that (4.17) is equal to By (4.5), the integral above is equal to i 1 /2−k i 1 +1−2k = 0. We have thus verified (4.9) for x ∈ E n \D n . Case (ii): Next we consider x = (x 1 , x 2 ) ∈ D n provided that x 1 < 2n − 1. For this case, we have that the left side of (4.9) entrywise is equal to (4.20) 1), b). When 0 < x 1 < 2n − 1, the integral contribution in (4.20) from using (3.6) and (3.7) is which verifies (4.9) when x = (x 1 , x 2 ) ∈ D n for 0 < x 1 < 2n. When x 1 = 0, we have that (4.20) becomes 1), b) = 0 by explicit evaluation (both terms have a contribution from k = 0 in the sum in (3.6) and (3.7)). When x = (x 1 , x 2 ) ∈ D n and x 1 = 2n − 1, the left side of (4.9) entrywise is equal to (4.23) 2n), b) if n is even. From Proposition 4.4 and Proposition 4.3, we have that both (4.23) and (4.24) are equal to zero. Case (iii): Next we consider the left side of (4.9) entrywise when x = b when n is odd (the even case is immediate from Proposition 4.3). Expanding out these terms using (3.8), (3.9), (3.7), and (3.6) gives where the third equality follows from the absence of a residue at r = 0 when k = n, the fourth equality follows from (4.5), and we have used Theorem 4.1 for the fifth equality. We also need the left side of (4.9) entrywise when x = (2n, 2n) which is given by where the sum of the first two terms follows from the proceeding computation and we have used the antisymmetry of K −1 n . We have now verified (4.9) for x ∈ E n ∪ {b}. Case (iv): Finally, we verify (4.9) for x ∈ O n . When x = (x 1 , x 2 ) ∈ O n , we have that the left side of (4.9) entrywise equals If x 2 < 2n + 1, using (3.9) and writing x = (x 1 , x 2 ) = (i 1 , i 1 + 2i 2 + 1), we have that (4.27) becomes (4.28) Finally, if x = (x 1 , 2n + 1) then (4.27) becomes . This means that we have Using (3.9), we have that (4.29) becomes where we have used (4.28). We need to show the right side of the above equation equals zero. Expanding the above term out using the formula for (3.6) gives n k=0 p(n, k, 0) 2πi where we have used (4.6) for the second equality. We use Theorem 4.2 on the first term of the last line of the above equation and so the above equation is equal to zero as required.
We have thus verified (4.9) for all x, completing the proof.

Both vertices at the top boundary
We will first establish a formula for K −1 n when both entries are top at the boundary using a series of lemmas in Section 5.
Remark 5.1. We will repeatedly use the following property of perfect matchings. If a leaf is adjacent to a vertex v via an edge with weight 1, then removing both and v from the graph does not change the partition function.
We will make use of the following result below, which follows immediately from Theorem 3.3 by multiplying both sides of the equation in Theorem 3.3 by Z G .
Proposition 5.2 (Graphical Condensation [Kuo06]). Let G be a plane graph with four vertices a, b, c, d that appear in that cyclic order on a face of G. Then Recall that Z {v 1 ,...,vm} n is the partition function of the dimer model on G n with the vertices {v 1 , . . . , v m } removed. Introduce for 0 ≤ i ≤ n − 1, and for 0 ≤ i < j ≤ n − 1, with R n (i, j) = −R n (j, i) if 1 ≤ j < i ≤ n − 1 and R n (i, i) = 0 for 0 ≤ i ≤ n − 1. We use the convention that T n (k) = 0 for k ≥ n, R 0 = 1 and that R n (k, l) = 0 if k ≥ n or l ≥ n.
Lemma 5.3. For 1 ≤ i, j ≤ n − 1, we have Proof. For the purpose of the proof, write i = (2i, 2n + 1) for 1 ≤ i ≤ n − 1 and let a = (0, 2n + 1). For i < j, we apply graphical condensation in Proposition 5.2 2 which gives since a, i, j, b are in cyclic order on the face of TSSCPP. Notice that removing a from V n transforms the graph to V n−1 \{b} since the edges ((0, 0), (1, 1)), ((0, 1+2k), (0, 2+ 2k)), ((1, 2 + 2k), (1, 3 + 2k)) for 0 ≤ k ≤ n − 1 must be matched by Remark 5.1; see Figure 7. Hence, we have Z  . These pendant edges are drawn in dashed. Note that removing a vertex is the equivalent of adding a pendant edge to that vertex. The right figure shows that adding a pendant vertex to a forces the blue dashed dimer as well as the red dimers.
We divide the above equation by Z n−1 Z n and use the definitions of R n and T n to obtain the equation. A similar computation holds for i > j.
To get an explicit expression for R n (i, j) we use generating functions and require some further notation. Introduce Lemma 5.6. For 0 ≤ i, j ≤ n − 1 and n ≥ 1, T n (u)T n−1 (v)vw n − T n−1 (u)T n (v)uw n .

(5.11)
Proof. Multiply the recurrence in Lemma 5.3 by u i v j w n and taking sums gives (5.12) Using R n (0, i) = T n−1 (i − 1) and R n (i, j) = −R n (j, i) we have that vT n−1 (v)w n .
Inserting the above two equations into (5.12) gives uT n−1 (u)T n (u)w n .
We divide both sides by (1 − uwv) and it remains to extract out the coefficient of u i v j w n in the above equation to prove the lemma.
Lemma 5.7. For |u| < 1 where the s-contour S is chosen so that |t 3 zu| pq 2 |s(1+z) 2 | < 1. Proof. Notice that T n (u) = n k=0 T n (k)u k since T n (k) = 0 for k ≥ n and so we use integral expressions to find an explicit expression using geometric sums. We now use the expressions for binomial coefficients and factorials given in (4.6), (4.7), and (4.8), as explained below, in the formula for T n (i) given in Lemma 5.4 by applying these integrals to terms in the expression for (−1) k p(n, k, 0) (3.5). We obtain (5.17) In the above equation, the integrals with respect to p and q are from the two factorial terms in the numerator of (−1) k p(n, k, 0) using (4.7), while the integrals respect to s and t are from the two factorials in the denominator of (−1) k p(n, k, 0) using (4.8). For the integral with respect to z, this comes from rewriting the term (3n − 3k + 2)C n−k in (−1) k p(n, k, 0) as and expressing each of these binomial terms as integrals using (4.5). We choose the s-contour in (5.17) so that (5.19) p|tzu| q|s(1 + z) 2 | < 1, and evaluate the geometric sum t 3(1+n) z 1+n (qs(1 + z) 2 − ptuz) . (5.20) Making the change of variables s → sp 2 q/t 2 gives the result.

Proof of the main result
We will now use the results from the previous sections to prove Theorem 3.2. For the proof, we will need the following result.
Proposition 6.1. For 0 ≤ j ≤ 2n − 1 (6.1) − K −1 n ((0, 0), (j, j + 1)) = K −1 n ((j, j + 1), (0, 0)) = K −1 n ((j, j + 1), b). Proof. The first equality is from antisymmetry of K n . For the second equality, notice that K −1 n ((j, j + 1), (0, 0)) represents the ratio of a signed count on the graph formed from removing the vertices (j, j + 1) and (0, 0) along with their incident edges from the TSSCPP of size n and Z n . The sign arises from having an even number of counterclockwise edges around the face which surrounds the removed vertex (j, j + 1). For each of these dimer configurations (which is part of the signed count) on V n \{(0, 0), (j, j+1)}, remove all dimers incident to the vertices Then, there is a unique way to extend to a dimer configuration on V n \{b, (j, j + 1)} as required. Note that this operation does not change the sign associated to each dimer configuration.
Proof of Theorem 3.2. We will first focus on the formula (3.13) for t 1,1 n (i, j). The formulas for t k,l n (i, j) for (k, l) = (1, 1) are obtained by very similar computations. Once these are found, the rest of entries are obtained from the matrix equations K n .K −1 n = K −1 n .K n = I.
It is not hard to see that (3.14), (3.15) and (3.16) follow from a straightforward applications of K n .K −1 n = K −1 n .K n = I.

Proofs of the combinatorial identities
We now give the proofs of the combinatorial results used in Section 4.
Before proving Theorem 4.2, we need another identity in the sequel, which we prove now.
Proof of Theorem 4.2. First, note that Using (7.2), add the left hand sides of (4.1) and (4.2) to get n k=0 1 2 n−k f n (k) where we have used Theorem 4.1 in the last step. So we have shown that the sum of the left hand sides is what we want.
We now prove (4.1). Our first task is to prove G n,i = 2G n,i+1 . Using Wegschaider's algorithm [Weg97] in the Mathematica package MultiSum, we get the recurrence where ∆ j is the forward difference operator defined above. This is easily verified by a computation. Now, we perform the j-sum on both sides of (7.3). Since the right hand side telescopes, we obtain n+i−2k j=0 (g n,i (k, j) − 2g n,i+1 (k, j)) = g n,i+1 (k, n + i + 1 − 2k) − g n,i (k, n + i + 1 − 2k) We perform the k-sum on both sides of the above equation and compensate for the j = n + i + 1 − 2k term to obtain that G n,i − 2G n,i+1 is equal to where f is defined above. Now, using Theorem 7.1, we obtain that this is equal to 0. All that remains to be done is to prove that G n,n = 1. But this is easily performed since the j-sum becomes and we now use Theorem 4.1. This completes the proof.

Boundary recurrences
Recall that g b n is given in (3.18) and define, for 0 ≤ i < j ≤ 2n − 1, with g n (j, i) = −g n (i, j) and g n (i, i) = 0. The relevant signs given above are evaluated by using a similar argument given in Lemma 5.5; we omit this computation. We obtain some formulas immediately.
Lemma 8.1. We have that for 2 ≤ j ≤ 2n − 1, and g b n (0) = 1. Proof. The first two equations are immediate from writing the definitions of g b n (1) and g n (1, j) as a ratio of partition functions, see also (3.18), and removing the vertex (1, 1) from the graph forces edges to be covered by dimers; see Figure 8.
For the last equation, for each dimer configuration of Z n on V n , remove all dimers incident to the vertices {(i, i), 0 ≤ i ≤ 2n−1}. For each of these dimer configurations, there is a unique way to extend to a dimer configuration on V n and V n \{(0, 0), b}. This shows that Z n = Z {(0,0),b} n as required.
We then have the following system of equations for g n (i, j) and g b n (j): These two recurrences determine K −1 n ((i, i), (j, j)) for 0 ≤ i, j ≤ 2n − 1, and along with the equations K −1 n .K n = K n .K −1 n = I viewed entrywise fully determines the entries of K −1 n . To prove these results, we need two local moves for dimers.
(1) Spider move: Suppose we have a large square with edge weights a, b, c and d (clockwise labelling) on some graph G. This square can be deformed to smaller square with additional edges added between the vertices of the original square and the vertices of the smaller square as shown in Figure 9 to (2) Edge Contraction: If a vertex is incident to two edges each having weight 1, contract the two incident edges. This does not change the partition function; see Figure 10.
Proof of Theorem 8.2. For the purpose of the proof, write i to be the vertex (i, i) for 1 ≤ i ≤ 2n − 1 and let a denote the vertex (0, 2n + 1). We start with the proof of (8.4). The first two conditions are immediate by definition. For 0 ≤ i < j < 2n − 1, we apply graphical condensation, similar to (5.5) in the proof of Lemma 5.3, which gives . Dividing both sides of (8.6) by Z n−1 Z n gives the last condition in (8.4). Next, when i = 0, notice that removing both a and 0 has no impact on the forced edges, which means that Z {a,0,j,b} n = Z {j−2,b} n−1 . Dividing (8.6) by Z n−1 Z n in this case gives (8.7) g n−1 (0, j − 2) = g b n (j) − g b n−1 (j − 2) + g n (0, b) and so (8.8) g n−1 (0, j − 2) + g b n−1 (j − 2) = g b n (j) + g n (0, j). The equality in (8.8) is equal to 1 when n is even which follows immediately from the last equation in Lemma 8.1. We now consider (8.8) when n is odd. To do so, set Notice that due to edges being forced Z {a,0,1,b} n = Z n−1 and Z {a,1} n = 0, where the latter follows since the dimer covering of V n \{a, (1, 1)} is zero as the induced dimers from removing a from the graph are incompatible with removing (1, 1) from the graph; compare Figure 7 and Figure 8. Dividing the above equation by Z n−1 Z n gives (8.10) 1 = g b n (1) + g n (0, 1). We have shown (8.8) is equal to 1 when n is odd, which verifies the third condition in (8.4). Finally, when i = 1 in (8.6), gives   Figure 11. The left graph H 0 is obtained after applying an edge contraction. Now, a square appears where we can apply the spider move. After applying the spider move and contracting edges, we obtain the right graph H 1 .
Proof of Theorem 8.3. The first two conditions follow from the third and first conditions in Lemma 8.1. The last condition in (8.5) is more involved and we illustrate the steps to find a recurrence of the partition function first.
Label this new graph H 1 . This operation gives (8.13) We now proceed iteratively and describe the step from H k−1 to H k from applying the square move for 2 ≤ k ≤ n − 1. On the graph H k−1 , we apply the spider move on the square face whose center is given by (1/2, 2n − 2k + 3/2) and applying edge contraction on the two bottom edges protruding from the new (smaller) square; see Figure 12. Figure 12. The local moves taking H k−1 to H k . In each of the three graphs, the top left vertex is (0, 2n − 2k + 3). All unlabelled edges have weight 1.
We now prove the sum rule.
where we have only kept the term with a residue contribution at r = 0 when evaluating the geometric sum. Notice that the second term on the right side of the above equation does not have a residue contribution at r = 0 when k = n and so we have where the last line follows from pushing the contour through ∞ picking up the residue at r = 1. The result then follows from Theorem 4.1 and from the fact that p(n, n, 0) = (−1) n .

Heuristics for the limit Shape
To obtain the conjectured limit shape formula, we only consider the asymptotics of K −1 n ((x 1 , x 2 ), b) for x 2 ∈ 2Z + 1 for (x 1 , x 2 ) rescaled as given in Conjecture 3.5. Strictly speaking, this term does not contain any probabilistic information but we expect to see a similar structure when analyzing other entries of the inverse Kasteleyn matrix when both terms are close to the limit shape curves. To find these asymptotics, we express the formula for K −1 n ((x 1 , x 2 ), b) as a single contour integral and apply the method of steepest descent. This will give a function whose double roots parameterize the limit shape curves. This is a fairly standard approach in the asymptotics of random tilings, see for instance [Gor20, Lecture 15] for a good exposition, and so we only give a brief outline of the main steps. Even though we can give the computations below in full detail, we cannot analyze the rest of the entries of the inverse Kasteleyn matrix with this method, as they are currently not in the best form for asymptotic analysis.
(9.5) Figure 14. The contours of steepest descent and ascent for s(w). The contour of steepest descent leaves the double critical point with angle +π/3 and passes through the point (X + 2) > 0 and then returns to the double critical point. The crossing points of the real axis can be determined explicitly.
What is important is that it does not influence the saddle point function s(w), nor