The emergence of the electrostatic field as a Feynman sum in random tilings with holes

We consider random lozenge tilings on the triangular lattice with holes $Q_1,...,Q_n$ in some fixed position. For each unit triangle not in a hole, consider the average orientation of the lozenge covering it. We show that the scaling limit of this discrete field is the electrostatic field obtained when regarding each hole $Q_i$ as an electrical charge of magnitude equal to the difference between the number of unit triangles of the two different orientations inside $Q_i$. This is then restated in terms of random surfaces, yielding the result that the average over surfaces with prescribed height at the union of the boundaries of the holes is, in the scaling limit, a sum of helicoids.


Introduction
The study of correlations of holes in random tilings was launched by Fisher and Stephenson [10], who considered in particular the monomer-monomer correlation and obtained exact data suggesting rotational invariance in the scaling limit.Motivated by this, we studied correlations of a finite number of holes of various sizes and shapes in [1], [2] and [3], and found that in the scaling limit they are given by a multiplicative version of the superposition principle for energy in electrostatics.In this paper we consider the discrete field of average tile orientations in a random tiling with holes, and provide the proof of the fact that the electric field comes about as the scaling limit of this discrete field.This result was announced in [4].
Consider the unit equilateral triangular lattice drawn in the plane so that some of the lattice lines are vertical.The union of any two unit triangles that share an edge is called a lozenge.Inspired by Feynman's description of the reflection of light in [9,Ch. 2] and given the connections between lozenge tilings with holes and electrostatics discussed in [2] we construct a discrete vector field as follows.Let ∆ 1 , . . ., ∆ n be holes in the lattice whose boundaries are lattice triangles of even side-lengths.As one ranges over the set of lozenge tilings of the plane with these holes (a portion of such a tiling is illustrated in Figure 5.4(a)), any given left-pointing unit triangle e in the complement of the holes is covered by a lozenge having one of three possible orientations: pointing in the polar direction 0, the polar direction 2π/3, or the polar direction 4π/3.Define F(e) to be the average of these orientations over all tilings (the precise definitions are given in the next section).Now apply a homothety of modulus 1/R to our lattice with holes, and "drag outward" the images of the holes and the image of e so that they shrink to points z 1 , . . ., z n and respectively z 0 in the complex plane, as R → ∞.
Our main result can be phrased as follows.
Theorem.As R → ∞ we have where r i0 is the unit vector pointing from z i in the direction of z 0 and ch(Q) denotes the difference between the number of right-and left-pointing unit triangles in the lattice region Q.
The figure above shows on the left an instance of the field F in the case of two oppositely oriented holes of side two, and on the right the corresponding Coulomb field that it approaches in the scaling limit.
In Section 5 (see Theorem 5.1) we phrase this result in terms of random surfaces.Lozenge tilings of the plane with holes lift to certain discrete multi-sheeted surfaces.The above result implies that the average over these surfaces (an instance of which is illustrated in Figure 5.5) converges in the scaling limit to a sum of helicoids (for lifting surfaces as in Figure 5.5, this limit surface is of the type depicted in Figure 5.7(b); see [8] for an overview on a variety of settings in which helicoids arise).The key to proving this is that we obtain explicit formulas for the placement probabilities of lozenges in the scaling limit.
Phrased this way, our result is similar in spirit to the result of Cohn, Elkies and Propp [5] on the local statistics for random domino tilings of the Aztec diamond, where an explicit surface is found as the scaling limit of the normalized height function.Another related instance of an explicit limiting surface, concerning the placement probability of lozenges in a random tiling of a hexagon, was studied by Cohn, Larsen and Propp in [7].In [6], Cohn, Kenyon and Propp show that the normalized average height function 2 corresponding to domino tilings of regions that scale to an arbitrary simply connected region converges to the unique function that minimizes a certain surface tension integral.These authors also conjectured expressions for the local probability densities of domino configurations in large domino tilings.These conjectures were proved and generalized by Sheffield in [14].Random surfaces corresponding to random dimer coverings of periodic bipartite planar lattices were studied extensively by Kenyon, Okounkov and Sheffield in [13].
One notable difference between our result and the above mentioned ones is the presence of holes.This causes the lifting surfaces to be multi-sheeted.The holes are also responsible for creating the field we are studying -without them the field would be zero.
It may also be worth noting that for instance in [7] the boundary is the boundary of a lattice hexagon, and remains of fixed size throughout the scaling process, while in our case the boundary consists of the union of finitely many boundaries of triangular holes, each of them shrinking to a point in the scaling limit.Related to this is the fact that, unlike the results from the literature mentioned above, we consider the average of un-normalized lifting surfaces -the average of the normalized ones is identically zero.

Definitions and statement of results
Let Q j be a finite union of lattice triangular holes of side two with mutually disjoint interiors, for j = 1, . . ., n (even when Q j is not connected, we still view it as a single generalized hole, often referred to as a multihole).The joint correlation ω(Q 1 , . . ., Q n ) is defined in [3] (and recalled in Section 5 of the present paper) by means of limits of tori; the definition is readily extended to allowing some of the Q j 's to be lozenges.It follows from that definition that if L is any possible lozenge location, the probability that L is occupied by a lozenge in a random lozenge tiling with holes at Q 1 , . . ., Q n is given by the ratio Section 5 for the details).It follows thus that the vector F(e) = F(e; Q 1 , . . ., Q n ) described in the Introduction is given by where L 1 , L 2 and L 3 are the lozenge locations containing the left-pointing unit triangle e and pointing in the 0, 2π/3 and 4π/3 polar directions, respectively, and the e j 's are unit vectors pointing in the directions of the long diagonals of the L j 's.Note that we can specify the location of any left-or right-pointing unit triangle (for short, we will call them left-and right-monomers) by indicating the location of the midpoint of its vertical side.The midpoints of the vertical sides of the unit triangles in our lattice can naturally be coordinatized by pairs of integers using a 60 • coordinate system with axes pointing in the polar directions ±π/3.
Let E a,b be the east-pointing lattice triangle of side 2 whose central monomer has coordinates (a, b).Let W a,b be the similarly defined west-pointing lattice triangle.We regard both of them as holes.
For any q ∈ Q and any strictly increasing list of integers a = (a 1 , . . ., a s ) for which qa i ∈ Z and the E ai,qai 's (equivalently, the W ai,qai 's) are mutually disjoint, i = 1, . . ., s, define the multiholes E q a and W q a by E q a = E a1,qa1 ∪ . . .∪ E as,qas W q a = W a1,qa1 ∪ . . .∪ W as,qas .
For a hole Q in the lattice, let Q(x, y) stand for its translation by the vector (x, y) in our coordinate system.We say that an integer divides a rational number if it divides the numerator of a lowest terms representation of it.
We can now give the precise statement of our main result.
Assume the (x i , y i )'s and (z j , w j )'s are all distinct.
Then for any multiholes E q a1 , . . ., E q am and W q b1 , . . ., W q bn with 3|1 − q, the field defined by (1.1) has orthogonal projections on our coordinate axes with asymptotics and where s i and t j are the lengths of a i and b j , respectively.Furthermore, for any ǫ > 0 and any bounded set B in the plane the implicit constants above are uniform over all choices of the limits for which each distance among the points (x 0 , y 0 ), . . ., (z n , w n ) ∈ B is at least ǫ.
Note that in our oblique coordinate system the Euclidean distance between the points (x, y) and 2 , and the orthogonal projections of (x, y) on the coordinate axes are x + 1 2 y and 1 2 x + y.Note also that the multihole E 1 (0,2,4,...,2s−2) consists of a contiguous horizontal string of right-pointing triangular holes of side two.Due to forced lozenges in its complement, E 1 (0,2,4,...,2s−2) has precisely the same effect as the right-pointing triangular hole of side 2s that contains it; a similar statement holds for W -multiholes.Since ch(E q ai ) = 2s i and ch(W q bj ) = −2t j , one sees that the theorem stated in the Introduction follows as a special case of Theorem 1.1.

Reducing the problem to exact determinant evaluations
One crucial ingredient for proving the results of [3] was an exact determinant formula for the joint correlation of an arbitrary collection of disjoint lattice-triangular holes of size two.The arguments presented there prove in fact a more general statement, which we will need in the current paper.
Our determinant formula involves the coupling function P (x, y), x, y ∈ Z specified by and the symmetries P (x, y) = P (y, x) = P (−x − y − 1, x) (see Kenyon [12]), and the coefficients U s of its asymptotic series where and Since the set of monomers can be partitioned into pairs, m + n is even; hence so is m − n.Thus we can prove the statement by induction on m − n as follows.To relate a given configuration to one for which m − n is 2 units smaller, bring in and extra hole, W (3r, 0), and let r → ∞.The details go through just as in the proof of [3,Proposition 3.2].
In order to obtain the asymptotics of the field F(e) we need to understand the three coefficients in (1.1) when the holes are E a1 , . . ., E am , W b1 , . . ., W bn .The asymptotics of the denominators in the special case x (R) 1 = Rx 1 , . . ., w (R) n = Rw n , x 1 , . . ., w n ∈ 3Z, is worked out in [3,Theorem 8.1].Proposition 2.1 supplies exact determinant expressions for the numerators.Each is the determinant of a block matrix whose blocks consist in turn of (mostly) 2 × 2 blocks.To write them down it will be helpful to define the following five families of matrices.
The second family of matrices is The remaining families consist of one-row or one-column matrices.The third is The fourth family is . . .
and the fifth Let L 1 (x, y), L 2 (x, y), and L 3 (x, y) be the the lozenge locations pointing in the polar directions 0, 2π/3, 4π/3, respectively, and containing the left-monomer (x, y).Let s i and t j be the lengths of the lists a i and b j , respectively, for i = 1, . . ., m, j = 1, . . ., n.Then Proposition 2.1 implies where ) where Denote . (2.12) = δ n (mod 3), we have where and ) (2.16) (there are t pairs of such columns, one for each j = 1, . . ., t), and

.20)
Furthermore, for any ǫ > 0 and any bounded set B, the implicit constant in (2.13) is uniform for all choices of x 0 , . . ., w n for which (x 0 , y 0 ), . . ., (z n , w n ) ∈ B have all mutual distances at least ǫ.
Note that (2.13)-(2.20)(plus 2 analogs) imply already, before evaluating the determinants, that in the scaling limit F(e) does not depend on the spacings between the side-two constituent holes of the multiholes E q ai and W q bj (although interestingly the independence of q is not yet apparent).Proof.By (2.12), (2.8) and (2.10) we have

.21)
It readily follows from the integral expression (2.1) for P that P (0, 0) = 1/3.Regarding the first column of M1 as the sum of two vectors one of which is (1/3, 0, . . ., 0) T and using the linearity of the determinant yields where In the special case shows how to write the main term in the asymptotics of det M as a determinant independent of R.This can be accomplished in the current case of general by essentially the same arguments.Namely, consider the following operation on a square matrix X in which rows i 1 , . . ., i k are of the form f (c 1 ), . . ., f (c k ), respectively, for some vector function f : where D is Newton's divided difference operator, whose powers are defined inductively by This operation has an obvious analog for columns.
As noted in [3, §5] (and as can be seen by looking at (2.11), (2.9), (2.3) and (2.4)), operation (2.24) can be applied a total of 2m + 2n different times to the matrix M : each row of block matrices in the expression for M given by (2.11) and (2.9) provides two opportunities (along the odd-indexed rows and along the even-indexed ones), and each column consisting of A-blocks provides two more.Let M ′ be the matrix obtained from M after applying these 2m + 2n operations.
Since operations (2.24) preserve the determinant (see [ where a k = (a k1 , . . ., a k,s k ) and b l = (b l1 , . . ., b l,t l ) for all k and l.By (2.25) and (2.26) we get (2.27) Propositions 4.1 and 4.5 imply that the implicit constant above is uniform in x 1 , . . ., w n .
Next we turn to the asymptotics of det M 1 .Note that all the 2m + 2n operations of type (2.24) we applied to M are also well-defined as operations on M 1 .Indeed, it is apparent from (2.9) and (2.3)-(2.6)that each of the 2m + 2n times we applied (2.24), the extensions to M 1 of the involved rows or columns of M are also of the form required by this operation.Let M ′ 1 be the matrix obtained from M 1 after applying these 2m + 2n operations.
A calculation similar to the one that gave (2.26) yields (2.28) By the preservation of the determinant we get The same analysis proves also the following result.
= δ n (mod 3), we have where and

.33)
The implicit constant above has the same uniformity property as in Proposition 2.2.

Evaluation of the determinants and proof of Theorem 1.1
The orthogonal projections of the vectors e 1 , e 2 and e 3 of (1.1) on the x-axis of our oblique coordinate system are √ 3/2, − √ 3/2 and 0, respectively.Thus the orthogonal projection of F on the x-axis is (3.2) A quick glance at (2.14) and (2.31) shows that M ′′ 1 and M ′′ 2 differ only in their first rows.Furthermore, by (2.18), (2.20), (2.32) and (2.33), the corresponding first row entries differ just by a factor of ζ −1 .This makes it very tempting to write the numerator on the left hand side of (3.2) as a single determinant, and try to use the method of factor exhaustion (after all, this was the method that proved successful in [ec], where an explicit product expression for a special case of det M ′′ is given; see [3,Theorem 8.1]).However, we were not able to prove (3.2) this way, despite getting frustratingly close (the only missing part was proving divisibility by one last type of linear polynomial divisor).
The proof below derives (3.2) by a certain limiting process from a convenient specialization of [3,Theorem 8.1].This has the advantage of being significantly shorter than a possible proof by factor exhaustion.Still, a solution of the latter kind would be interesting, as it should prove also a conjectured two-parameter generalization of (1.2) that works in particular with ζ being an indeterminate.
Proof.We first prove the statement under the simplifying assumption that α 0 = • • • = δ n = 0.In this case Theorem 8.1 of [3] implies that for any indeterminates x 1 , . . ., x m , y 1 , . . ., y m , z 1 , . . ., z n and w 1 , . . ., w n we have Let ǫ > 0. Note that the matrix in (3.3) depends on four lists of indeterminates -x 1 , . . ., x m , y 1 , . . ., y m , z 1 , . . ., z n , and w 1 , . . ., w n -and two lists of positive integers s 1 , . . ., s m and t 1 , . . ., t n .Replace the indeterminates as follows: the x-list by x 0 , x 1 , . . ., x m , the y-list by y 0 , y 1 , . . ., y m , the z-list by x 0 + ǫ, z 1 , . . ., z n , and the w-list by y 0 , w 1 , . . ., w n .Replace also the s-list by 2, s 1 , s 2 , . . ., s m , and the t-list by 12 2, t 1 , t 2 , . . ., t n .Then (3.3) provides an explicit product expression for the determinant of the matrix ) The resulting product expression for det M ǫ has many common factors with the right hand side of (3.3), which simplify when taking det M ǫ / det M ′′ .After simplification we get Replace the first column in (3.4) by the negative of the sum of the first two columns.Using where Extracting the coefficient of 1/ǫ in the asymptotics of (3.9) as ǫ → 0 and comparing it with the second term in (3.12) gives (3.2).Next we show how the case of general residues α 0 , . . ., δ n modulo 3 reduces to the above case of all zero residues.
Consider the general matrix M ′′ 1 given by (2.14) and (2.16)-(2.20).Except for its first row and column, it consists of 2 × 2 blocks of the type in Lemma 3.3.Furthermore, the corresponding value of a is constant over each block Ȧα k ,β k ,γ l ,δ l x k ,y k ,z l ,w l (s k , t l ), and equals ( , and equals α k − β k .Thus in each 2 × 2 block of type (3.13) in M ′′ 1 the value of the a of Lemma 3.3 is a difference of two quantities, the first being constant along each row of M ′′ 1 , and the second constant along each column of M ′′ 1 .Therefore Lemma 3.3 can be used to transform the general M ′′ 1 matrix by row and column operations into the specialization of M ′′ 1 when all residues are 0 except α 0 and β 0 , and the overall effect on the determinant is that it is multiplied by σ, where σ ∈ {1, −1}.The very same row and column operations transform M ′′ 2 into M ′′ 2 | α1=0,...,δn=0 and M ′′ into M ′′ | α1=0,...,δn=0 , with the effect on their determinants being multiplication by the same σ.Therefore .
(a).Simultaneously replacing Proof.Since ζ 3 = 1, the second row of A(a) is the same as the first row of A(a − 1).The negative of the sum of the entries in the first column of A(a) for all integers k.One similarly checks that the negative of the sum of the entries in the second column of A(a) equals ζ a−2 f (ζ) .This proves (a).Part (b) follows analogously.
By a similar calculation one can easily check the following.

A finer integral asymptotics
Let D denote Newton's divided difference operator, whose powers are defined inductively by We will need the following result on the asymptotics of the coupling function P when acted on in the indicated way by powers of D. Theorem 4.1.Let r n and s n be integers so that lim n→∞ r n /n = u, lim n→∞ s n /n = v, and (u, v) = (0, 0).Then for any integers k, l ≥ 0 and any rational number q with 3|1 − q we have where ζ = e 2πi/3 , D k x acts with respect to a fixed integer sequence a 1 , a 2 , . . ., and D l y acts with respect to an integer sequence b 1 , b 2 , . . .satisfying qb j ∈ Z for all j ≥ 1.Furthermore, for any open set U containing the origin the implicit constant above is uniform for all (u, v) / ∈ U .
We note that Proposition 7.1 of [3] corresponds to the special case when r n = −un and s n = −vn.It will be crucial in our proof of the scaling limit of the average lifting surface to know that the leading term on the right hand side of (4.1) is independent of the way r n /n and s n /n approach their limits.
We deduce Theorem 4.1 from the following auxiliary results.Let P denote the counterclockwise oriented arc of the unit circle connecting ζ = e 2πi/3 to −1; P[ζ, −1) stands for this path minus the point −1.
Lemma 4.2.Let r n and s n be integers, and assume lim n→∞ r n /n = u ≥ ǫ > 0. Then if q ∈ C 0 (P[ζ, −1)) has a pole of finite order at −1, for any integer k ≥ 0 we have where the implicit constant depends only on k, ǫ and q(t) (in particular, it works for all u ≥ ǫ and all integers s n ).

The elementary inequality |e
The above two inequalities combined with (4.3) give However, integration by parts implies Repeated application of this, together with shows that J(r, k) is equal to a sum of k + 1 terms each exponentially small in r, plus for n large enough, it follows that the integral in the second line of (4.4) is majorized for n large enough by 2 2k+3 k!/(nu) k+1 ≤ 2 2k+3 k!/(ǫ) k+1 1/n k+1 , and the proof is complete.
, where q 1 (t) ∈ C 1 (P), k is a non-negative integer, and Furthermore, for any ǫ > 0, each implicit constant above can be chosen to be uniform for u ∈ [ǫ, ∞) and independent of s n .
Proof.For l = 2, 3, . . ., 7, define One readily checks that the Taylor series expansions of ln t and ln(−1 − t) around t = ζ can then be written as Therefore we have Let r, s ∈ Z and denote Using (4.8) and ζ 3 = 1, integration by parts gives Since ζ s−r e (sζ −1 −rζ)(t−ζ) e b(t) = (−1−t) r t s , we see that the upper limit in the first term of the expression in the large curly braces above equals 0 whenever r > k.Thus, for r > k we obtain If |r| ≥ |s| we have A similar argument shows that the above inequality holds in fact also when |s| ≥ |r|.All six fractions in front of the integrals in the expression in curly braces above are thus seen to be majorized in absolute value by 4.
Regard I(r n , s n ) as the sum of the three quantities provided by (4.9).To deduce part (a) of the Lemma, assume k = 0.All six terms of the second quantity are O(1/n 2 ) thanks to (4.10) and an application of Lemma 4.2 with k = 1 (which applies since h ′ 2 (t), . . ., h ′ 7 (t) are all of the form (t − ζ)g(t), where g ∈ C ∞ (P[ζ, −1)) has a simple pole at t = −1), followed by an application of Lemma 3.4 with k = 0 for the resulting integral on the right hand side of (4.6).Finally, the third quantity provided by (4.9) is also O(1/n 2 ), due to the fraction in front of the integral and another application of Lemma 4.2 with k = 0.
For part (b), assume k ≥ 1.Then q(ζ) = 0, and (4.9) provides an expression for I(r n , s n ) as a sum of just two quantities.All six terms in the first quantity are O(1/n k+2 ) due to (4.10) and Lemma 4.2 applied with k replaced by k + 1 (as explained in the previous paragraph, this unit increment comes about by the presence of the h ′ l (t) factors in the integrands).This proves (4.6).The uniformity of the implicit constant follows because both the majorant in (4.10) and the implicit constant in (4.2) are uniform.
Repeated application of part (b) of the above lemma and one final application of part (a) yields the following result.
) has a pole of finite order at t = −1, and q 1 (ζ) = 0. Let r n , s n ∈ Z so that lim n→∞ r n /n = u > 0. Then Furthermore, for any ǫ > 0 the implicit constant is uniform for u ∈ [ǫ, ∞) and independent of s n .
Proof of Theorem 4.1.Suppose first that u < 0. Then (2.1) holds for large enough n, and for 3|1 − q [3, (7.11)] and [3,Lemma 7.4] give Proposition 4.4 applied to the right hand side above yields which is just what (4.1) states.The uniformity of the implicit constant above follows by the uniformity of the implicit constant in (4.11).Since (u, v) = (0, 0), at least one of u < 0, v < 0, and −u − v < 0 is true.The symmetries P (α, β) = P (−α − β − 1, α) and P (α, β) = P (β, α) of the coupling function allow one to use the same arguments that proved the case u < 0 to deduce the other two cases (see the proof of Proposition 7.1 in [3] for details).Proposition 4.5.Let r n , s n ∈ Z so that lim n→∞ r n /n = u and lim n→∞ s n /n = v.Then for any integers k, l ≥ 0 and any rational number q with 3|1 − q we have where D k x acts with respect to some fixed integer sequence a 1 , a 2 , . . . .Given any bounded set B in the plane, the implicit constant can be chosen so that it is uniform for (u, v) ∈ B.
Proof.By [3, (6.8)] one has + monomials in a and b of joint degree < l.
This implies, for 3|1 − q, that where c α,β is independent of r n and s n for all α and β.
On the other hand, the argument that proved [3,Lemma 6.4] implies that for any constants A, B, C ∈ C with implicit constant uniform for (u, v) ranging over any bounded set.Combining (4.14) and (4.15) yields the statement of the proposition.

Interpretation in terms of height functions
Lozenge tilings of regions with no holes are well known to be interpretable as lattice surfaces (see e.g.[15][7]).Regard the unit triangular lattice T on which the tiled region lives as being in a horizontal plane, and let L be a copy of the lattice 3 placed so that one family of its body-diagonals is vertical, intersecting this plane at the vertices of T .Then each segment joining two nearest neighbors of L projects onto a unit lattice segment of T .
Orient the lattice segments of T so that they point in one of the polar directions π/2, −π/6, or −5π/6.Then we can lift a lozenge tiling of a lattice region on T by starting from some lattice point, tracing around its tiles one after another, and at each traversal of a lattice segment s of T , moving either up or down on the corresponding lattice segment of L, according as the traversal respected or violated the orientation of s.Tracing around a lozenge results in going around a lattice square of L whose orthogonal projection on the plane of T is that lozenge.
If the tiling has no holes, the node of L we are finding ourselves at is independent of the way we traced around the tiles to get there, and what results is a lattice surface in L whose lattice square faces are in one to one correspondence with the tiles.
When a hole of non-zero charge is present this ceases to be true.To illustrate this, suppose we have a left-pointing triangular hole of side two in our tiling (see Figure 5.1).Then as we trace its boundary counterclockwise, the traversal of each lattice segment agrees with its orientation.Thus, each complete turn results in six upward steps along lattice segments of L, and leaves us at the node of L which is two cube-body-diagonals higher than where we started (this is illustrated, from three different viewpoints, in Figure 5.2).For a similar right-pointing hole we would end up two such diagonals lower.
In general, if we loop once counterclockwise along a closed walk through the lozenges of a tiling, the ending point of the corresponding walk on L is on the same vertical as its starting point, but a distance of q 3 √ 2 below it, where q is the total charge of the holes we looped around (by a negative distance below we mean the absolute value of that distance above).This makes it impossible to get a single-sheeted lifting surface with no extra boundary in addition to the the lifting of the boundary of the region with holes.
Nevertheless, we can construct a multi-sheeted surface with no additional boundary that lifts any tiling with holes, as follows.Suppose we have a lattice region R on T with a finite number of holes, and let T be a lozenge tiling of it.For each hole, consider a fixed lattice path cut (independent of the tiling T ) from it to the boundary of the region R; let P be the family of lattice paths formed by them.Whenever a step of a lattice path of P crosses a lozenge of T , remove that lozenge from R; let R 0 be the region obtained from R by removing the union T 0 of all such lozenges of T .Now regard T \ T 0 as being a tiling of R 0 in which P ∩ R 0 is part of the boundary.Then T \ T 0 lifts to a single-sheeted lattice surface 5ST in L. The union6 is a multi-sheeted surface having holes above all the lozenges in R \ R 0 .Define S T to be the multisheeted surface obtained from U by filling in these holes with the missing square faces of the lattice L  The detailed definition of the joint correlation ω is the following (see [3]).For j = 1, . . ., k, let Q j be either a lozenge-hole or a lattice triangular hole of side two.
It is enough to define ω(Q 1 , . . ., Q k ) when q = k j=1 ch(Q j ) ≥ 0 (the other case reduces to this by reflection across a vertical lattice line).Our definition is inductive on q: (i).If q = 0, let N be large enough so that the lattice rhombus of side N centered at the origin encloses all Q j 's, and denote by T N the torus obtained from this large lattice rhombus by identifying its opposite sides.Set The above limits exist by Proposition 2.1.Let ∆ 1 , . . ., ∆ k be fixed lattice triangular holes of side two, and let L be a fixed lozenge position.Assume q = k j=1 ch(∆ j ) = 0.In the limit measure of the uniform measures on the tori T N \ ∆ 1 ∪ . . .∪ ∆ k , the probability that L is occupied by a lozenge in a random tiling is This expression for the probability of L being occupied holds in fact for general q.Indeed, suppose this has been established for total charges < q.Then the probability that L is occupied in a random tiling with the extra hole W R,0 in addition to ∆ ) .
However, in the limit R → ∞ this is, by (5.2), the same as the fraction on the right hand side of (5.3), and our statement is proved by induction.Suppose R is a bounded simply connected lattice region on T .Then each tiling of R lifts to a singlesheeted lattice surface on L, and we can define the average lifting surface S av by taking the arithmetic mean of the finitely many heights of these surfaces above each node of T .As pointed out in [CLP], if u and v are nearest neighbors in T so that the lattice segment between them is oriented from u to v (see the second paragraph of this section), then where L uv is the lozenge location whose short diagonal is uv (the factor multiplying the parenthesis on the right hand side arises because the traversal of each unit segment in T results in a change of height on the lifting surface of one third of a body diagonal of a lattice cube of L).Thus, the height of S av at any node of T can be obtained by taking cumulative sums of lozenge occupation probabilities.The regions we are concerned with -complements of finite unions of disjoint lattice triangular holes of side 2 -have an infinite set of lozenge tilings, so we cannot use the arithmetic mean as the definition of the average of the surfaces their tilings lift to.However, since we know the lozenge placement probabilities are given by (5.3), we can turn (5.4) around and use it to define this average surface.
More precisely, consider for each triangular hole a lattice path in T from a point on its boundary to infinity.Let P be the union of these lattice paths; assume they are disjoint.Define Sav to be the lattice surface on L satisfying Sav for any two nearest neighbors u and v of T for which the segment between them is oriented from u to v and is not crossed by any path of P. Define the average lifting surface of the tilings of the complement of 22 the holes by This definition is readily seen to be independent of the family of cuts P. We note that in the case when there are no holes, the average surface can be defined in terms of the translation invariant ergodic measure on the hexagonal lattice (the lattice whose dimer coverings are equivalent to lozenge tilings of the triangular lattice), which follows by a general result of Sheffield (see [14] [13]) to be unique and given by a limit of uniform measures on tori.Due to the presence of holes our setting does not seem to fit that context.
The helicoid H(a, b; c) is the surface whose parametric equations in Cartesian coordinates are  We can define the sum of dotted helicoids analogously, using that the fibers of Ḣs (a, b; sc) above u are of the form f (u) + πcZ.Note that a point on the surface making a complete counterclockwise turn around the "spiral stairwell" on the left ends up two levels higher; a similar turning around the other spiral takes the point one level lower.j /R = w j for 1 ≤ i ≤ m and 1 ≤ j ≤ n.Assume the (x i , y i )'s and (z j , w j )'s are all distinct.
Let E q a1 , . . ., E q am and W q b1 , . . ., W q bn , 3|1−q, be multiholes on T R , and let the lists a i and b j have lengths s i and t j , respectively.Let , . . ., W q bn z (R) n , w (R) n be the average lifting surface of the tilings of the complement of these multiholes on T R .Then, as R → ∞, RS TR av converges to the sum of refined helicoids Proof.In addition to our 60 • oblique coordinate system O in the plane of the unit triangular lattice T , consider also a Cartesian system of coordinates C having the origin at the node of T just below the origin of O , the x-axis in the polar direction 0 and the y-axis in the polar direction π/2.One readily sees that the nodes of T have C-coordinates a We will use Cartesian coordinates to specify points at which S TR av is evaluated, and oblique coordinates for the lozenges whose placement probabilities come up. the average orientation of the lozenge that covers any fixed right-pointing unit triangle!Denote it by F ′ .An argument similar to the one that proved Theorem 1.1 shows that F ′ = −F in the scaling limit.
We conclude by mentioning another, quite different way to define a vector field via random tilings with holes.Let Q 1 , . . ., Q n be a fixed collection of holes on the unit triangular lattice.For any x, y, α, β ∈ Z define T α,β (x, y) := 1 α 2 + αβ + β 2 ω (E x+α,y+β , Q 1 , . . ., Q n ) ω (E x,y , Q 1 , . . ., Q n ) − 1 (i.e., the hole E x,y plays the role of a "test charge," and the effects of its displacement are recorded).Then it can be shown (details will appear in a separate paper) that there exists a vector field T so that in the scaling limit T α,β (x, y) is the orthogonal projection of T(x, y) onto the vector (α, β).Furthermore, up to a constant multiple, the field T turns out to be the same as F. Thus, unlike in physics where the electric field is defined by means of a test charge, in our model there are two different ways to define the corresponding field, and one of them does not use test charges.

. 2 )
Let r(a, b) and l(a, b) denote the right-and left-pointing monomers of coordinates (a, b), respectively.

Lemma 3 . 3 .
Let f be a function defined at ζ and ζ −1 , and let a ∈ Z.Let

Figure 5 . 1 .
Figure 5.1.A tiling with a triangular hole of size two.

Figure 5 . 2 .
Figure 5.2.Three views of ST for the tiling T above.above the lozenges in R \ R 0 .One readily sees that S T is independent of the family of cuts P. The surface corresponding to the tiling in Figure 5.1 is illustrated in Figure 5.3.An instance with three holes is pictured in Figures 5.4 and 5.5.

. 3 )Figure 5 . 3 .
Figure 5.3.(a) One of the two connected components of S T for the tiling in Figure 5.1.(b) The full S T .

Figure 5 . 4 .
Figure 5.4.(a) A tiling T with three holes.(b) A view of ST .

. 6 )
z = cθ The half helicoid H + (a, b; c) is obtained by restricting the range of ρ in the above to (0, ∞).The dotted helicoid Ḣ(a, b; c) is H(a, b; c) minus the vertical axis x = a, y = b.

Figure 5 . 5 . 8 )Figure 5 . 6 .
Figure 5.5.Two views of S T for the tiling T in Figure 5.4(a).For positive integers s define the s-refined half helicoid by
Figure 5.6(b) illustrates the effect of doubling the refinement indices in a sum of refined half helicoids: the new surface is the union of the original surface and a suitable vertical translate of it.A depiction of a sum of three helicoids is given in Figure 5.7.Let T R be the lattice obtained from the triangular lattice T by a homothety around the origin of factor 1 R .integers so that lim R→∞ x (R)i /R = x i , lim R→∞ y (R) i /R = y i , lim R→∞ z (R) j /R = z j and lim R→∞ w (R)