=2 0
−1/(E1√
|x−Q0|)
(2λ)nexp(2λ(x−Q0))
1+O(λ2)+O(|x−Q0|λ3) dλ
= (−1)nn!
(x−Q0)n+1 +O(|x−Q0|−n−3).
Summing up the contributions of the three ranges gives (−1)n
n! ∂nKQ0(x)= 1
(x−Q0)n+1 + τ(x,Q0)
(J(x,Q0)e1e2)n+1 +O(|x−Q0|−n−3).
Acknowledgments The authors thank Richard Kenyon, Christian Mercat, and Mikhail Skopenkov for fruitful discussions and helpful suggestions. In addition, we are grateful to Mikhail Skopenkov for his detailed review of previous versions and his numerous and valuable recommendations.
The first author was partially supported by the DFG Collaborative Research Center TRR 109,
“Discretization in Geometry and Dynamics”. The research of the second author was supported by the Deutsche Telekom Stiftung. Some parts of this paper were written at the IHÉS in Bures-sur-Yvette, the INIMS in Cambridge, and the ESI in Vienna. The second author thanks the EPDI program for the opportunity to stay at these institutes. The stay at the Isaac Newton Institute for Mathematical Sciences was funded through an Engineering and Physical Sciences Research Council Visiting Fellowship, Grant EP/K032208/1.
Proposition 4.1 A planar quad-graph Λ admits a combinatorially equivalent embedding inCwith all rhombic faces if and only if the following two conditions are satisfied:
• No strip crosses itself or is periodic.
• Two distinct strips cross each other at most once.
Remark Let us prove the simpler claim that planar parallelogram-graphs fulfill these two conditions as was already noted by Kenyon in [16]. The underlying reason is that any strip S is monotone with respect to the directioni aS: The coordinates of the endpoints of the edges parallel toaSare strictly increasing or strictly decreasing if they are projected toi aS. Whether the projections are decreasing or increasing depends on the direction in which the faces of S are passed through. Without loss of generality, we assume that the faces of Sare passed through in such a way that the projections of the corresponding coordinates are strictly increasing. ForQ∈S, let SQ denote the semi-infinite part ofS starting in the quadrilateralQthat passes through the faces ofSin the same order.
As a consequence, no strip crosses itself or is periodic. Furthermore, S divides the complex plane into two unbounded regions, to one isaSpointing and to the other
−aS. When a distinct stripScrossesS, it enters a different region determined byS, say it goes to the one to whichaSis pointing. Due to monotonicity, the angle between i aSandaS is less thanπ/2. It follows thatScannot crossSanother time, since it would then go to the region−aSis pointing to, contradicting that the angle between i aSand−aSis greater and not less thanπ/2.
In order to give explicit formulae for the discrete Green’s function and the discrete Cauchy’s kernels in Sects.3.3and3.4, we chose a particular directed path connecting two vertices (or a face and a vertex) by edges of the parallelogram-graphΛ. This path was monotone in one direction. The existence of such a path follows from the following lemma, generalizing a result of [2] to general parallelogram-graphs. The proof bases on the same ideas.
Lemma 4.2 LetΛ be a parallelogram-graph and let v0∈V(Λ) be fixed. For a directed edge e ofΛstarting in v0, consider the subgraph Ue⊂Λthat consists of all vertices and edges of directed paths onΛstarting in v0 whose oriented edges have arguments that can be chosen to lie in[arg(e),arg(e)+π).
Then, the union of all Ue, e a directed edge starting in v0, covers the whole quad-graph. If there is a constantα0>0 such thatαα0 for all interior anglesαof faces ofΛ, then the same statement holds true if[arg(e),arg(e)+π)is replaced by [arg(e),arg(e)+π−α0].
Proof Let us rescale the edges such that all of them have length one. By this, we change neither the combinatorics ofΛnor the size of interior angles.
For a directed edgeestarting inv0, letUe−andUe+denote the (directed) paths onΛstarting inv0, obtained by choosing the directed edge with the least or largest
argument in[arg(e),arg(e)+π)(or[arg(e),arg(e)+π−α0]) at a vertex, respec-tively. We first show that all vertices in betweenUe−andUe+are contained inV(Ue).
Then, it follows thatUeis the conical sector with boundaryUe−andUe+.
Suppose the contrary, i.e., suppose that there is a vertexvbetweenUe−andUe+to whichv0cannot be connected by a directed path of edges whose arguments lie all in the interval[arg(e),arg(e)+π)(or[arg(e),arg(e)+π−α0]). Let the combinato-rial distance betweenv0andvbe minimal among all such vertices.
In the case that interior angles of rhombi are bounded by α0 from below, they are bounded from above byπ−α0. Hence, there is a vertexvadjacent tovsuch that the argument of the directed edgevvlies in[arg(e),arg(e)+π−α0]. Even if interior angles are not uniformly bounded,vcan be chosen in such a way that the argument ofvvlies in[arg(e),arg(e)+π). By construction,vis not inV(Ue), but still betweenUe− andUe+. Let us look at the strip S passing throughvv. Suppose that the common parallelaSpoints fromvtov.
IfSintersectsUe−orUe+, then an edge parallel toaSis contained inUe−orUe+, respectively. By construction,v0andvthen lie on the same side of the stripS.
IfSdoes neither intersectUe−norUe+, then it is completely contained in the left half space determined by the oriented linev0+t e,t ∈R, asUe−andUe+are. Suppose Sintersects the rayv0+taS,t 0. Again, it follows thatv0andvlie on the same side ofS.
It remains the case thatSneither intersectsUe−,Ue+, nor the rayv0+taS,t 0.
Consider the quadrilateral area R in between the parallelsv0+taS,v+taS and v0+t e,v+t e,t∈R. By assumption, the semi-infinite part ofSthat starts with the edgevvand then goes intoRdoes not intersect an edge ofRincident tov0, and by monotonicity, it does not intersectv+taSagain. Now,Λis locally finite, such that only finitely many quadrilaterals of Sare inside P. Thus, S leaves P on the edge v+t e,t ∈R, and it follows thatSseparatesv0andv.
So in any case,Sseparatesv0fromv. Any shortest pathPconnecting both points has to pass throughS. Letwbe the first point ofPthat lies on the same side ofSas vdoes. Any strip passing through an edge on the shortest path connectingwandv onShas to intersectPas well. It follows that replacing all edges ofPon the same side of S asvby the path fromw tovdoes not change its length. But then,v is combinatorially nearer tov0thanv, contradiction.
Finally, we can cyclically order the directed edges starting inv0according to their slopes. Then, the sectorsUeare interlaced, i.e.,Uecontains bothUe+−andUe−+, where e−,e,e+are consecutive according to the cyclic order. As a consequence, the union of all theseUecovers the wholeΛ.
To perform our computations in Sects.3.3and3.4, we needed not only that the interior angles were bounded, but also that the side lengths were bounded. We can relax the latter condition to boundedness of the ratio of side lengths.
Proposition 4.3 LetΛbe a parallelogram-graph and assume that there are con-stantsα0,q0>0such thatαα0and e/eq0for all interior anglesαand two side lengths e,eof any parallelogram ofΛ. Then, E1eE0for all edge lengths
e, where E0:=e0q0N and E1:=e1q0−N with N := 2π/α0/2and e1e0being the side lengths of an arbitrary parallelogram ofΛ.
Proof Let Q∈V(♦) be fixed with edge lengths e1e0 and let Q∈V(♦) be another parallelogram with centerx. In the following, we construct a sequence of n strips such that any two consecutive strips are crossing each other, the first one containsQ, the last one containsQ, andnN. Then, it follows that the side lengths of Qare bounded byE0andE1.
Let S0be a strip containing Q. If Q∈ S0, we are done. Otherwise, we choose the common parallelaS0in such a way thatxlies in the region−aS0 is pointing to.
Letβ0:=arg(aS0). SinceS0is monotone in the directioni aS0and interior angles are uniformly bounded, the rayx+taS0,t >0, intersectsSin exactly one line segment.
Lety0be the first intersection point andQ0a quadrilateral ofScontainingy0. BecauseΛis locally finite, the line segment connectingxandy0intersects only finitely many parallelograms. Through any such parallelogram at most two strips are passing. Thus, only a finite number of strips intersect this line segment. Therefore, we can choose a stripS1intersectingS0Q0in a parallelogramQ0,1such thatS1does not containQand does not intersect the line segment connectingxandy0. Moreover, we require thatQ∈/S0Q0,1. Now, choose the common parallelaS1ofS1in such a way that there is an argumentβ1ofaS1that satisfiesπ+β0> β1> β0. By construction, xlies in the region−aS1is pointing to. Note thatS1Q0,1cannot crossS0a second time.
Suppose we have already constructed the stripSkwith common parallelaSk and argumentβk,k>0, andxlies in the region−aSkis pointing to.Skshall not intersect the line segments connectingxandy0or connectingxandyk−1. Moreover, assume that the semi-infinite partSkQk−1,k starting in the intersectionQk−1,kofSkwithSk−1
does not crossS0.
Letykbe the first intersection of the rayx+taSk,t>0, with a quadrilateralQk
of the stripSk. By the same arguments as above, there exists a stripSk+1
intersect-ing SkQk−1,k∩SkQk that does not containQ and does not intersect the line segments
connectingx andy0 or xandyk. Choose its common parallelaSk+1in such a way that it has an argumentβk+1that satisfiesπ+βk> βk+1> βk. By construction,x lies in the region−aSk+1is pointing to. If the semi-infinite partSkQ+1k,k+1starting in the intersectionQk,k+1withSkdoes not crossS0, then we continue this procedure. For a schematic picture of the proof, see Fig.7.
After at most l:= 2π/α0steps, we end up with a strip Sl such that SlQl−1,l intersects S0. Indeed, let us suppose the contrary, that is, let us suppose that all S2Q1,2, . . . ,SlQl−1,l do not crossS0.
By assumption,βk+π−α0βk+1βk+α0. It follows that the first j such thatβj is greater or equal thanβ0+2πsatisfies j 2π/α0. In addition, we have βj< β0+3π−α0.
By construction,Sj does not intersect the line segment connectingx andyj−1. Moreover, SQj j−1,j cannot cross Sj−1 a second time. It follows that SQj j−1,j cannot intersect the rayx+taSj−1,t >0.
aS0
aS1
aSk
aSk+1
S0
S1
Sk
Sk+1
Q0,1
Qk−1,k
Qk,k+1
x y0
y1
yk
Fig. 7 Schematic picture of the proof of Proposition4.3
Also,SQj j−1,jdoes neither crossS0nor does it intersect the line segment connecting xandy0, so it does not intersect the rayx+taS0,t>0. Thus,SQj j−1,j is contained in the cone with tipxspanned byaSj−1andaS0(with angle less thanπ). This contradicts the monotonicity ofSQj j−1,j into the directioni aSj, because the rayx+taSj,t >0, is not contained in the interior of the cone above.
In summary, we found a cycle ofmstripsS0,S1, . . . ,Sm−1surroundingx, where m2π/α0 +1. Actually,m2π/α0, because theaSk are cyclically ordered.
Since only finitely many strips intersect the stripS0in betweenQandQ0,1, we can assume thatQis contained inS0Qm−1,0.
Thesemstrips determine a bounded regionx is contained in. If Q=Qm−1,0, then we look at the semi-infinite part of the strip S˜0 different from S0 that passes throughQand goes into the interior of the bounded region above. It has to intersect one of the strips S1, . . . ,Sm−1, saySk. Then,S0, . . . ,Sk,S˜0or S˜0,Sk, . . . ,Sm−1,S0
determine a bounded regionxis contained in (Qmay be an element ofS˜0). Clearly, they are at most2π/α0such strips, andQlies on an intersection.
If Q∈ ˜/ S0, then a strip SQcontaining Qhas to cross two different strips of the cycle due to local finiteness. In the same way as above, we can find a cycle of at mostm2π/α0strips S0,S1, . . . ,Sm−1 such that Qlies on one of the strips, saySk, and the intersection ofS0andSm−1isQ. Ifkm/2, then we choose the sequence of stripsS0,S1, . . . ,Sk; otherwise, we takeSm−1,Sm−2, . . . ,Sk. Any two consecutive strips are crossing each other,Qis on the first strip,Qon the last one, and there are at most2π/α0/2of them.
Remark In general, the bound2π/α0/2in the proof is optimal. Indeed, consider n rays emanating from 0 such that the angle between any two neighboring rays is 2π/n. In each of thensegments, choose the quad-graph combinatorially equivalent to the positive octant of the integer lattice that is spanned by two consecutive rays.
For example, if n=4, we obtainZ2. Then, any strip passes through exactly two adjacent segments, andn/2is the optimal bound.
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