Boundary Value Problems on Planar Graphs and Flat Surfaces with integer cone singularities, I: The Dirichlet Problem
BBOUNDARY VALUE PROBLEMS ON PLANAR GRAPHS AND FLATSURFACES WITH INTEGER CONE SINGULARITIES, I: THEDIRICHLET PROBLEM
SA’AR HERSONSKY
Abstract.
Consider a planar, bounded, m -connected region Ω, and let ∂ Ω be its boundary.Let T be a cellular decomposition of Ω ∪ ∂ Ω, where each 2-cell is either a triangle or aquadrilateral. From these data and a conductance function we construct a canonical pair(
S, f ) where S is a genus ( m − singular flat surface tiled by rectangles and f is an energypreserving mapping from T (1) onto S . By a singular flat surface, we will mean a surfacewhich carries a metric structure locally modeled on the Euclidean plane, except at a finitenumber of points. These points have cone singularities, and the cone angle is allowed totake any positive value (see for instance [28] for an excellent survey). Our realization maybe considered as a discrete uniformization of planar bounded regions. Introduction
Consider a planar, bounded, m -connected region Ω, and let ∂ Ω be its boundary. Let T bea cellular decomposition of Ω ∪ ∂ Ω, where each cell is either a triangle or a quadrilateral. Let ∂ Ω = E (cid:116) E , where E is the outermost component of ∂ Ω. Invoke a conductance function on T (1) , making it a finite network , and use it to define a combintorial Laplacian ∆ on T (0) .Let k be a positive constant. Let g be the solution of a Dirichlet boundary value problem (D-BVP) defined on T (0) and determined by requiring that g | E = k, g | E = 0 and ∆ g = 0at every interior vertex of T (0) , i.e. g is combinatorially harmonic . Furthermore, let E ( g )be the Dirichlet energy of g . Following the notation in [13], let ∂g∂n ( x ) denote the normalderivative of g at the vertex x ∈ ∂ Ω. Our first result is:
Theorem 0.1. (Discrete uniformization of a pair of pants)
Let P be a bounded, triply con-nected planar region and let (Ω , ∂ Ω , T ) = ( P , ∂ P = E (cid:116) E , T ) where E = E (cid:116) E . Let S P be a singular flat pair of pants with exactly one singular point Q having π as its coneangle such that (1) Length Euclidean ( S P ) E = (cid:80) x ∈ E ∂g∂n ( x ) , (2) Length Euclidean ( S P ) E = − (cid:80) x ∈ E ∂g∂n ( x ) , (3) Length Euclidean ( S P ) E = − (cid:80) x ∈ E ∂g∂n ( x ) , and (4) the Euclidean length of a shortest geodesic connecting E to either E or E is k ,where ( S P ) E , ( S P ) E and ( S P ) E are the boundary components of S P . Then there exists amapping f which associates to each edge in T (1) a unique Euclidean rectangle in S P , in sucha way that the collection of these rectangles forms a tiling of S P . Furthermore, f is energypreserving in the sense that E ( g ) = Area( S P ) , and f is boundary preserving. Date : November 7, 2018; version 0.0526100. a r X i v : . [ m a t h . G T ] M a y SA’AR HERSONSKY
Throughout this paper a Euclidean rectangle will denote the image under an isometry of aplanar Euclidean rectangle. For instance, most of the image rectangles that we will constructembed in a flat Euclidean cylinder. These will further be glued in a way that will not distortthe Euclidean structure (see § § u, v ] with u ∈ ∂ Ω has one of its edges on acorresponding boundary component of the singular surface.A singular flat, genus zero compact surface with m ≥ Theorem 0.2. (The general case)
Let (Ω , ∂ Ω , T ) be a bounded, m -connected, planar regionwith E = E (cid:116) E . . . (cid:116) E m − . Let S Ω be a ladder of singular pairs of pants such that (1) Length Euclidean ( S Ω ) E = (cid:80) x ∈ E ∂g∂n ( x ) , and (2) Length Euclidean ( S Ω ) E i = − (cid:80) x ∈ E i ∂g∂n ( x ) , for i = 1 , · · · , m − ,where ( S Ω ) E and ( S Ω ) E i , for i = 1 , · · · , m − , are the boundary components of S Ω . Then,there exists a mapping f which associates to each edge in T (1) a unique Euclidean rectanglein S Ω in such a way that the collection of these rectangles forms a tiling of S Ω . Furthermore, f is boundary preserving, and f is energy preserving in the sense that E ( g ) = Area( S Ω ) . The following Corollary is straightforward, thus establishing the statement in the abstractof this paper
Corollary 0.3.
Under the assumptions of Theorem 0.2, there exists a canonical pair ( S, f ) ,where S is a flat surface with conical singularities of genus ( m − tiled by Euclideanrectangles and f is an energy preserving mapping from T (1) into S , in the sense that E ( g ) = Area( S ) . Moreover, S admits a pair of pants decomposition whose dividing curveshave Euclidean lengths given by (1) − (3) of Theorem 0.2.Proof. Given (Ω , ∂ Ω , T ), glue together two copies of S Ω (their existence is guaranteed byTheorem 0.2) along corresponding boundary components. This results in a flat surface S = S Ω (cid:83) ∂ Ω S Ω of genus ( m −
1) and a mapping ¯ f which restricts to f on each copy. (cid:3) In the course of the proofs of Theorem 0.1 and Theorem 0.2, it will become apparentthat the number of singular points and their cone angles, as well as the lengths of shortestgeodesics between boundary curves in the ladder, may be explicitly determined. In partic-ular, the cone angles obtained by our construction are always even integer multiples of π (see Equation (4.12) for the actual computation). Some of these surfaces, those with trivialmonodromy, are called translation surfaces and for excellent accounts see for instance [21],[25] and [30]. Also, the dimensions of each rectangle are determined by the given D-BVPproblem on T (0) . Concretely, for [ u, v ] ∈ T (1) , the associated rectangle will have its heightequal to ( g ( u ) − g ( v )) and its width equal to c ( u, v )( g ( u ) − g ( v )), when g ( u ) > g ( v ). Some ofthe rectangles constructed are not embedded. We will comment on this point (which is alsotransparent in the proofs) in Remark (4.15). In a snapshot, some of the rectangles whicharise from intersection of edges with singular level curves are not embedded.The following theorem is foundational and serves as a building block in the proofs of allof the above theorems. We prove: IRICHLET PROBLEM 3
Theorem 0.4. (Discrete uniformization of an annulus [10])
Let A be an annulus and let (Ω , ∂ Ω , T ) = ( A , ∂ A = E (cid:116) E , T ) . Let S A be a straight Euclidean cylinder with height H = k and circumference C = (cid:88) x ∈ E ∂g∂n ( x ) . Then there exists a mapping f which associates to each edge in T (1) a unique embeddedEuclidean rectangle in S A in such a way that the collection of these rectangles forms a tilingof S A . Furthermore, f is boundary preserving, and f is energy preserving in the sense that E ( g ) = Area( S A ) . Given (Ω , ∂ Ω , T ), we will work with the natural affine structure induced by the cellu-lar decomposition. Let us denote this cell complex endowed with this affine structure by C (Ω , ∂ Ω , T ). There is a common thread in our proofs of the theorems above. Extend g to an affine map ¯ g defined on T . This results in a piecewise linear structure on T . Next,study the level curves of ¯ g on a 2-dimensional complex which is homotopically equivalentto C (Ω , ∂ Ω , T ), embedded in R , obtained by using ¯ g as a height function on C (Ω , ∂ Ω , T ).We will work with the level curves of ¯ g or equivalently, with their projection on C (Ω , ∂ Ω , T ).The topological structure of the level curves associated with the solutions will be carefullystudied. For example, the level curves in the case of an annulus (Theorem 0.4) are simple,piecewise linear closed curves, all of which are in the (free) homotopy class determined by E . One nice consequence of this is that all the rectangles constructed in the proof of The-orem 0.4 are embedded. In the proof of Theorem 0.1, we will show that all the level curvesassociated to values in [0 , k ], for some constant k < k , are simple, piecewise linear closedcurves in either the (free) homotopy class of E or the class determined by E . However,for the value k , it will be proved that the (unique) associated level curve is a figure eight.Furthermore, any level curve of ¯ g for a value which is larger than k and smaller than orequal to k , is a simple closed curve in the (free) homotopy class determined E . The basicidea of Theorem 0.1 is to cut (Ω , ∂ Ω , T ) along the (projection of) a figure eight curve, tileeach cylinder according to Theorem 0.4, and glue back.We will often work with a series of modified boundary value problems. Each is a slightmodification of the initial problem. This important feature of the proofs is due to the factthat the level sets of the original boundary value problem (defined on T (0) ) intersect T (1) ina set which is much larger than T (0) (see § §
4) is that one boundary component of one of the pieces (ormore) in the decomposition will be singular, hence Theorem 0.4 cannot be applied directly.The paper is organized as follows. In § § SA’AR HERSONSKY order to provide a good notion of length for level curves as well as a topological descriptionof them. In §
3, we prove Theorem 0.4, and § Remark . The assertions of Theorem 0.4 may (in principle) be obtained by employingtechniques introduced in the famous paper by Brooks, Smith, Stone and Tutte ([10]), in whichthey study square tilings of rectangles. They define a correspondence between square tilingsof rectangles and planar multigraphs endowed with two poles, a source, and a sink. They viewthe multigraph as a network of resistors in which current is flowing. In their correspondence,a vertex corresponds to a connected component of the union of the horizontal edges of thesquares in the tiling; one edge appears between two such vertices for each square whosehorizontal edges lie in the corresponding connected components. Their approach is based on
Kirckhhoff ’s circuit laws that are widely used in the field of electrical engineering. We foundthe sketch of the proof of Theorem 0.4 given in [10] hard to follow. In fact, another proof of aslight generalization of this theorem was given by Benjamini and Schramm ([7], see also [8] fora related study) using techniques from probability and the dual graph of T . It is interestingto recall that it was Dehn [14, 1903] who was the first to show a relation between squaretiling and electrical networks. In an elegant paper, combining a mixture of geometry andprobability, Kenyon ([23]) used the fact that a resistor network corresponds to a reversibleMarkov chain. He showed a correspondence between planar non-reversible Markov chainsand trapezoid tilings. A completely different method, for the case of tiling a rectangle bysquares, was given using extremal length arguments in [26] by Schramm. One should alsomention that Cannon, Floyd and Parry (see [12]), using extremal length arguments (similarto these in [26]), provide another proof for the existence of tiling by squares. Both [26]and [12] are widely known as “a finite Riemann mapping theorem” and serve as the firststep in Cannon’s combinatorial Riemann mapping theorem ([11]). We include our proof ofTheorem 0.4, which is guided by similar principles to some of the ones mentioned above,yet significantly different in a few points, in order to make this paper self-contained. Inaddition, the important work of Bendito, Carmona and Encinas (see for example [4],[5],[6])on boundary value problems on graphs allows us to use a unified framework to even moregeneral problems. Their work is essential to our applications and we will use parts of it quitefrequently in this paper as well as in its sequels ([19],[20]). Acknowledgement.
Part of this research was conducted while the author visited the Departmentof Mathematics at Princeton University during the Spring of 2005. We express our deepest gratitudefor their generous hospitality and inspiration. We are grateful to G´erard Besson, Francis Bonahon,Gilles Courtois, Dave Gabai, Steven Kerckhoff and Ted Shifrin for enjoyable and helpful discussionson the subject of this paper. The results of this paper and its sequel ([19]) were presented at the G (New Orleans, January 2007) and at the Workshop on Ergodic Theory and Geometry (ManchesterInstitute for Mathematical Sciences, April 2008). We deeply thank the organizers for the invitationsand well-organized conferences. IRICHLET PROBLEM 5 Preliminaries - boundary value problems on graphs
We recall some known facts regarding harmonic functions and boundary value problemson networks. We use the notation of Section 2 in [3]. Let Γ = (
V, E, c ) be a finite network ,that is a simple and finite connected graph with vertex set V and edge set E . Since inthis paper Γ = ( V, E, c ) is induced by (Ω , ∂ Ω , T ), we shall further assume that the graphis planar. Each edge ( x, y ) ∈ E is assigned a conductance c ( x, y ) = c ( y, x ) >
0. Let P ( V ) denote the set of non-negative functions on V . If u ∈ P ( V ), its support is given by S ( u ) = { x ∈ V : u ( x ) (cid:54) = 0 } . Given F ⊂ V we denote by F c its complement in V . Set P ( F ) = { u ∈ P ( V ) : S ( u ) ⊂ F } . The set ∂F = { ( x, y ) ∈ E : x ∈ F, y ∈ F c } is called the edge boundary of F and the set δF = { x ∈ F c : ( x, y ) ∈ E for some y ∈ F } is called the vertex boundary of F . Let ¯ F = F (cid:83) δF and let ¯ E = { ( x, y ) ∈ E : x ∈ F } . Given F ⊂ V , let¯Γ( F ) = ( ¯ F , ¯ E, ¯ c ) be the network such that ¯ c is the restriction of c to ¯ E . We say that x ∼ y if ( x, y ) ∈ ¯ E . For x ∈ ¯ F let k ( x ) denote the degree of x (if x ∈ δ ( F ) the neighbors of x aretaken only from F ).The following are discrete analogues of classical notions in continuous potential theory[16]. Definition 1.1. ([4, Section 3] ) Let u ∈ P ( ¯ F ). Then for x ∈ ¯ F , the function ∆ u ( x ) = (cid:80) y ∼ x c ( x, y ) ( u ( x ) − u ( y )) is called the Laplacian of u at x , (if x ∈ δ ( F ) the neighbors of x are taken only from F ) and the number(1.2) E ( u ) = (cid:88) x ∈ ¯ F ∆ u ( x ) u ( x ) = (cid:88) ( x,y ) ∈ ¯ E c ( x, y )( u ( x ) − u ( y )) , is called the Dirichlet energy of u . A function u ∈ P ( ¯ F ) is called harmonic in F ⊂ V if∆ u ( x ) = 0 , for all x ∈ F .For example when c ( x, y ) ≡ u is harmonic at a vertex x if and only if the value of u at x is the arithmetic average of the value of u on the neighbors of x . A fundamental propertywhich we will often use in the maximum property , asserting that if u is harmonic on V (cid:48) ⊂ V ,where V is a connected subset of vertices having a connected interior, then u attains itsmaximum and minimum on the boundary of V (cid:48) (see [27, Theorem I.35]).For x ∈ δ ( F ), let { y , y , . . . , y m } ∈ F be its neighbors enumerated clockwise. The normalderivative (see [13]) of u at a point x ∈ δF with respect to a set F is(1.3) ∂u∂n ( F )( x ) = (cid:88) y ∼ x, y ∈ F c ( x, y )( u ( x ) − u ( y )) . The following proposition establishes a discrete version of the first classical
Green identity .It plays a crucial role in the proof the main theorem in [18] and is essential in this paper.
Theorem 1.4. ( [3, Prop. 3.1] ) (The first Green identity)
Let F ⊂ V and u, v ∈ P ( ¯ F ) .Then we have that (1.5) (cid:88) ( x,y ) ∈ ¯ E c ( x, y )( u ( x ) − u ( y ))( v ( x ) − v ( y )) = (cid:88) x ∈ F ∆ u ( x ) v ( x ) + (cid:88) x ∈ δ ( F ) ∂u∂n ( F )( x ) v ( x ) . SA’AR HERSONSKY
Remark . In [3], a second Green identity is obtained. In this paper we will use only theone above. In [6] (see in particular Section 2 and Section 3), a systematic study of discretecalculus on n -dimensional (uniform) grids of Euclidean n -space is provided. Their definitionof a tangent space may be adopted to our setting and does not require the notion of directededges.Let T denote a fixed cellular decomposition of Ω ∪ ∂ Ω, and let F ⊂ T (0) be given. Let { c ( x, y ) } ( x,y ) ∈ ¯ E be a fixed conductance function, and let ¯Γ( F ) be the associated network.We are interested in functions that solve a boundary value problem (D-BVP) on ¯Γ( F ). Thefollowing definition is based on [3, Section 3] and [5, Section 4]. Definition 1.7.
Let k > f ∈ P (Ω) such that f is harmonic in F , f | E = k and f | E = 0,for some positive constant k . Remark . The uniqueness and existence of a Dirichlet boundary value solution is guaran-teed by the foundational work in [3, Section 3] and [5, Section 4]. In fact, their work providesa detailed framework for a broader class of boundary value problems on finite networks.A metric on a finite network is a function ρ : V → [0 , ∞ ). In particular, the length ofa path is given by integrating ρ along the path (see [11] and [15] for a different definition).When ρ ≡
1, the familiar distance function on V × V is obtained by setting dist( A, B ) =( (cid:80) x ∈ α − k , where α = ( x, x , . . . x k ) is a path with the smallest possible number ofvertices among all the paths connecting a vertex in A and a vertex in B . We now define a“metric” which will be used throughout this paper. Definition 1.9.
Let F ⊂ V and let f ∈ P ( ¯ F ). The flux-gradient metric is defined by(1.10) ρ ( x ) = ∂f∂n ( F )( x ) , if x ∈ δ ( F ) . This allows us to define a notion of length to any subset of the vertex boundary of F bydeclaring:(1.11) Length( δF ) = (cid:12)(cid:12) (cid:88) x ∈ δF ∂f∂n ( F )( x ) (cid:12)(cid:12) . In the applications of this paper, we will use the second part of the definition in orderto define length of connected components of level curves of the D-BVP solution. In [18,Definition 3.3], we defined a similar metric ( l -gradient metric) proving several length-energyinequalities. IRICHLET PROBLEM 7 topology and geometry of piecewise linear level curve Let G be a polyhedral surface and consider a function f : G (0) → R ∪ { } such that twoadjacent vertices are given different values. Let v ∈ G (0) , and let w , w , . . . , w k be its k neighbors enumerated counterclockwise. Following [24, Section 3], consider the number ofsign changes in the sequence { f ( w ) − f ( v ) , f ( w ) − f ( v ) , . . . , f ( w k ) − f ( v ) , f ( w ) − f ( v ) } ,which we will denote by Sgc f ( v ). The index of v is defined as(2.1) Ind f ( v ) = 1 − Sgc f ( v )2 . Definition 2.2.
A vertex whose index is different from zero will be called singular; otherwisethe vertex is regular. A level set which contains at least one singular vertex will be calledsingular; otherwise the level set is regular.A connection between the combinatorics and the topology is provided by the followingtheorem, which may be considered as a discrete Hopf-Poincar´e Theorem.
Theorem 2.3. ([2, Theorem 1], [24, Theorem 2]) (
An index formula)
We have (2.4) (cid:88) v ∈G Ind f ( v ) = χ ( G ) . Remark . Due to the topological invariance of χ ( G ), note that once the equation above isproved for a triangulated polyhedron, it holds (keeping the same definitions for Sgc f ( · ) andInd f ( · )) for any cellular decomposition of χ ( G ). Also, while the theorem above is stated andproved for a closed polyhedral surface, it is easy to show that it holds in the case of a surfacewith boundary, where there are no singular vertices on the boundary (simply by doublingalong the boundary).Henceforth, we will keep the notation of Section 0 and Section 1. A key ingredient in ourproofs of the theorems stated in the introduction is the ability to define a length for a levelcurve of g . The main difficulty in defining such a quantity (for level curves) is the fact thatthese are not piecewise linear curves of the initial cellular decomposition; hence we cannotdirectly define the weight ρ along them (see Equation (1.10)).Suppose that L is a fixed, simple, closed level curve and let O , O be the two distinctconnected components of L in Ω with L being the boundary of both (this follows by usingthe Jordan curve theorem; see for instance [9, Theorem III.5.G] for an interesting proof).We will call one of them, say O , an interior domain if all the vertices which belong to ithave g -values that are smaller than the g -value of L . The other domain will be called the exterior domain . Note that, by the maximum principle, one of O , O must have all of itsvertices with g -values smaller than L .Let e ∈ T (1) and x = e ∩ L . For x (cid:54)∈ T (0) , we have now created two new edges ( x, v ) and( u, x ). We may assume that v ∈ O and u ∈ O . We now define conductance constants˜ c ( v, x ) and ˜ c ( x, u ) by(2.6) ˜ c ( v, x ) = c ( v, u )( g ( v ) − g ( u )) g ( v ) − g ( x ) and ˜ c ( u, x ) = c ( v, u )( g ( u ) − g ( v )) g ( u ) − g ( x ) . SA’AR HERSONSKY
We repeat the process above for any new vertex formed by the intersection of L with T (1) .By adding all the new vertices and edges, as well as the piecewise arcs of L determined bythe new vertices, we obtain two cellular decompositions, T O of O and T O of O . Also, twoconductance functions are now defined on the one-skeleton of these cellular decompositionsby modifying the conductance function for g according to Equation (2.6) (i.e. changes areoccurring only on new edges). We will denote these by C O and C O respectively (on the arcsof L the conductance is identically zero). Definition 2.7.
For i = 1 ,
2, we let g i denote the solution of the D-BVP determined on theinduced cellular decomposition on O i defined by the following:(1) for any boundary component E of O i and for every vertex x ∈ E , g i ( x ) = g ( x ),(2) the conductance function on O i is C O i , and(3) g i is harmonic in O i .It follows from Equation (2.6), Definition 1.1, and the existence and uniqueness theoremsin [3] that for i = 1 , g i exists and that g i = g | O i . We may now define a flux-gradient metricfor L by using Equations (1.10) and (1.11). However, unlike the situation with the boundarycomponents of Ω, we have two possible choices, i.e. computing the normal derivative along L with respect to O or with respect to O . Since in the applications we will cut Ω along aparticular L and wish to glue the resulting pieces together along L , these lengths computedwith respect to the flux-gradient metric in each domain should be the same. The situationwhen L is not simple is more complicated and will be addressed after a detailed analysisof the topological structure of level curves has been carried out (see the discussion afterRemark 2.19). This analysis is the main core of this section. The next lemma shows thatall level curves are in fact closed. Lemma 2.8.
A level curve for the function g is piecewise linear and closed, and each simplecycle of L contains at least one boundary component of ∂ Ω .Proof. The assertions of the lemma are certainly true for the components of ∂ Ω. Assumenow that L is not one of the these level curves and furthermore that it is not closed. Let q be a boundary point of L . Since g is extended linearly on edges and in an affine fashion ontriangles and quadrilaterals, q may (a priori) be an interior point of an edge of the cellulardecomposition, a vertex, or in the interior of a cell. All of these cases are easily ruled out.It remains to prove that any such level curve contains at least one boundary component of ∂ Ω. If this is not the case, L (being a finite, closed, polygonal planar line) bounds a unionof 2-cells in Ω. This is a violation of the maximum principle for the harmonic function g constructed as in Definition 2.7 ( O being one connected component of the union of the2-cells bounded by L ). (cid:3) Remark . Since g is extended in an affine fashion along edges, it is clear that two disjointlevel curves corresponding to the same g value are at combinatorial distance which is at leastone.The structure of simple curves in the plane can be quite complicated and in fact is notcompletely understood. In our applications, we need only analyze the topological structure ofclosed curves arising as level curves of the affine extension (over triangles and quadrilaterals)of a harmonic function as defined above. We will henceforth work in the piecewise category. IRICHLET PROBLEM 9
Figure 2.10.
Figure 2.11.
Level curvesfor the D-BVP
Figure 2.12.
The surface obtainedby using g as a height function. Definition 2.13.
A generalized piecewise bouquet of circles will denote a union of piecewisebouquets of circles where the intersection of any two bouquets is at most a vertex. Moreover,all such tangencies are exterior, i.e., no circle is contained in the interior of the boundedcomponent of another.
Proposition 2.14.
Two simple cycles which correspond to the same g value either aredisjoint or intersect at a single vertex. Furthermore, under these assumptions, a simplecycle is never contained in the annulus bounded by a different one (i.e. a “tangency” isalways external).Proof. Let L and L be two given simple cycles corresponding to the g -value l . Let A and A be the two (piecewise) annuli which they bound respectively. Assume first that A ∩ A = ∅ and that E = L ∩ L is a piecewise arc. Let I and J be the endpoints of E .As mentioned before (see Lemma 2.8), both I and J are vertices. The link of the vertex I must contain a two cell (a triangle or a quadrilateral) such that an edge between two of itsvertices crosses E . This implies that one of these vertices has g -value which is larger than l and the other has g -value which is smaller than l . This is absurd, since by the maximumprinciple all the vertices in A ∪ A which do not belong to g − ( l ) have g -values that are strictly smaller than l (none of the vertices of these cells belong to g − ( l )). It may alsohappen that L ∩ L = { P , P , . . . , P k } , where k ≥ P i are oriented clockwise. Let t be the arc connecting P to P which belongs to L and whichdoes not pass through any other of the P I ’s. Let t be the arc in L which connects P to P which does not pass through any of the P i ’s. Then t ∪ t is a simple closed level curve whichbounds an annulus (see Lemma 2.8). The link of the vertex P must contain a triangle ora quadrilateral with one of its edges crossing t . As above, this leads to a violation of themaximum principle.Assume now that A ∩ A (cid:54) = ∅ and (without loss of generality) that A ∪ L contains an arc t of L . Let its endpoints be P and Q (they both lie on L ). Then L ∪ (( L \ t ) ∪ E ) consistsof two simple cycles that satisfy all the conditions described in the case above; hence, thiscase may not occur.We now consider the case in which one of the annuli is contained in the other. Withoutloss of generality, assume that A ⊂ A and that E = L ∩ L is a piecewise arc. Let I and J be the endpoints of E . The link of the vertex I must contain a two-cell, included in A ,where one of its edges connects a vertex in A to a vertex in A \ ( A ∪ L ). This edge crosses L ; hence one of these vertices has g -value which is larger than l and the other has g -valuewhich is smaller than l ; this is absurd, since by the maximum principle all the vertices in A which do not belong to g − ( l ) must have their g value strictly smaller than l . It may alsobe the case that L ∩ L = { P , P , . . . , P m } where m ≥ P i are oriented clockwise. Note that we allow all the P i ’s to be a single vertex. Considerationof the link of any of the P i ’s, and an argument similar to the one above, leads again to aviolation of the maximum principle.Finally, assume that L ∩ L = ∅ . Observe that A \ A is an annulus; both of its boundarycurves have g value l . By a construction analogous to the one described before Definition 2.7,we obtain a new BVP problem defined on a cellular decomposition of ( A \ A ) ∪ L ∪ L . Itssolution is harmonic on A \ A and has the same value on the boundary components; henceit is the constant function. The cellular decomposition of the annulus contains edges (partof edges) or vertices (and perhaps both) of the initial cellular decomposition. Moreover, thevalues of the solution and g coincide on ( A \ A ) ∪ L ∪ L . Both cases lead to the conclusionthat connected pieces of edges of the initial cellular decomposition in the annulus have thesame g -value, which is absurd. (cid:3) Theorem 2.15.
Let L be a connected level curve for g . Then each connected component of L is a generalized piecewise bouquet of circles.Proof. First, we only use the fact that L is a closed polygonal line. Let S be the collection ofall the self intersection points of L . By definition, edges of the polygonal (closed) curve L maycross only at points from S . If L is simple, the assertion follows immediately. Assume that L is not simple. It is well known that we may express L as a union of simple close polygonalcurves. The intersection of any two cycles in this decomposition is a union of vertices andedges. Proposition 2.14 forces restrictions on any such decomposition. In particular, anytwo simple cycles in such a decomposition are either disjoint or intersect in a single vertex,none of which is contained in the annuli bounded by the other. The assertion of the theoremfollows immediately. (cid:3) IRICHLET PROBLEM 11
Figure 2.16.
Level curvesin a pair of pants
Figure 2.17.
Level curvesin a 7 connected domain
Figure 2.18.
A non-possible levelcurve in a 6 connected domain
Remark . Our interest here is in the existence of a decomposition of the level curve intosimple pieces. One may also consider the question, or rather a series of questions, regardinga decomposition of the non-simple polygon determined by a general closed polygonal curve.For many deep questions and an excellent survey on this interesting subject (some of thequestions involved are known to be NP ), see for instance [22].The topological structure of a general connected component of a level curve L (Theo-rem 2.15) means that the complement of L in Ω ∪ ∂ Ω is composed of a bounded domain(which is a disjoint union of annuli) in which all vertices have g -values smaller than the g -value of L , and a second domain whose boundary consists of L and possibly other curves in ∂ Ω), where L is in the boundary of both. Following a construction similar to the one preced-ing Definition 2.7, we may now define two modified conductance functions and two solutionsof D-BVP problems. By abuse of notation, we call the first domain O , the second O ,and we let C O , C O , g , g be the analogous quantities. The corresponding two new cellulardecompositions and their union will henceforth be called the induced cellular decomposition. Theorem 2.20.
Let L be a level curve for g . Then the following equality holds (2.21) l O ( L ) = l O ( L ) , where the left-hand side denotes the length of L measured with respect to the flux-gradientmetric induced by g , and the right-hand side denotes the length of L measured with respectto the flux-gradient metric induced by g . Proof.
First, let us assume that L is simple and corresponds to the g -value h . Since L is atcombinatorial distance at least one from any other level curves of g of the same value, the1-combinatorial neighborhood of L in O is comprised of vertices that are of g -values greaterthan h .Let x ∈ T (0) ∩ L . Let y , y be its neighbors in L , { x , . . . , x m } in O and { z , . . . , z k } in O . Since g is harmonic at x we have(2.22) 0 = ∆ g ( x ) = (cid:88) y ∼ x c ( x, y ) ( g ( x ) − g ( y )) . Hence, (since g ( x ) = g ( y ) = g ( y ))(2.23) 0 = m (cid:88) i =1 c ( x, x i ) ( g ( x ) − g ( x i )) + k (cid:88) j =1 c ( x, z j ) ( g ( x ) − g ( z j )) . Let x ∈ T O ∪ T O be a new vertex in L (that is x (cid:54)∈ T (0) ) with v ∈ O and u ∈ O . Wehave(2.24) ∂g ∂n ( O )( x ) = ˜ c ( v, x )( g ( x ) − g ( v )) = c ( u, v )( g ( u ) − g ( v )) , and(2.25) ∂g ∂n ( O )( x ) = ˜ c ( u, x )( g ( x ) − g ( u )) = c ( u, v )( g ( v ) − g ( u )) . By summing the three equations above over all vertices in L , the assertion follows.By combining the arguments above with the topological structure of a general level curveprovided by Theorem 2.15 and a simple modification at singular vertices along L , we willnow treat the second case. Let us assume that L is not simple. Recall that a level curvemay intersect itself only at a vertex. Let S = { v , . . . , v m } be all such intersection pointsand { Ind f ( v ) , . . . , Ind f ( v m ) } be their indices respectively. For a vertex x ∈ L (of the initialtriangulation or the induced one) which is not in S (note that Ind f ( x ) = 0), the argumentsabove leading to Equation 2.22 go through precisely in the same manner and yield (with O , O modified as explained immediately after Remark 2.19) the same conclusion.Let v ∈ S . Then v is in the intersection of C , . . . , C l piecewise simple circles where, l =Sgc g ( v ) / { x i , . . . , x ik ( i ) } be the neighbors of v in O ∩ interior( C i ), { y , . . . , y m } the neighbors of v in L and { z , . . . , z p } in O . Since g is harmonic at v , we may now modifyslightly Equation (2.23). Since g ( v ) = g ( y ) = . . . = g ( y m ), we obtain(2.26) 0 = l (cid:88) j =1 k ( j ) (cid:88) i =1 c ( v, x ji ) (cid:0) g ( v ) − g ( x ji ) (cid:1) + p (cid:88) j =1 c ( v, z j ) ( g ( v ) − g ( z j )) . The assertion of the theorem now follows by summing over all the vertices in L . Finally, if L has several connected components, the assertion of the theorem follows by addition over eachcomponent (which must be at combinatorial distance greater than one from each other). (cid:3) Remark . We now wish to start using Theorem 2.3 and Remark 2.5. One should notethat since boundary components are at the same g -level, a modification is needed. This is IRICHLET PROBLEM 13 done in the following way. Along each boundary component, add a piecewise linear curvehomotopic to it, and at combinatorial distance which is smaller than one. This can be donein such a way that there is exactly one sign change at each new vertex (not taking intoaccount vertices on the boundary). Let ˜Ω ∈ Ω be the region bounded by these curves.Doubling along the boundary of ˜Ω gives a surface without boundary with all of its singularvertices identical to those of g , for which the above mentioned theorem may be applied.We end this section with a useful property of level curves. This property is essential inthe proofs of Theorem 0.1 and Theorem 0.2. In the course of these proofs, we will need toknow that there is a singular level curve which encloses all of the interior components of ∂ Ω. This level curve (which is necessarily singular) is the one along which we will cut thesurface. In the inductive process, we keep cutting along such level curves over domains offewer boundary components until the remaining pieces are annuli.
Proposition 2.28.
With notation as above, there exists a singular level curve which containsin the interior of the domain it bounds, all of the inner boundary components of ∂ Ω . Sucha level curve is unique.Proof. We argue by contradiction. Throughout the proof, we use the topological structure oflevel curves provided by Theorem 2.15. Suppose that the first assertion of this propositionis false. Then every singular level curve omits at least one inner boundary component. Let L be such a curve. Suppose that L omits k ≥ l ≥ k + l = m −
1. Let { v , v , . . . , v n } be the singularvertices that belong to L ; let { w , w , . . . , w m } be the singular vertices in the interior of L .It is easy to see that (for example by induction)(2.29) (cid:88) x ∈{ w ,...,v n } Ind g ( x ) = 1 − l. By Theorem 2.3 we must have additional singular vertices { q , . . . , q s } such that(2.30) (cid:88) y ∈{ q ,...,q s } Ind g ( y ) = (1 − k ) − . Let L be a singular level curve that has q as a singular vertex. Suppose that L does notenclose L in its interior. Then(2.31) Ind g ( q ) ≤ (1 − k ) , with equality if and only if L encloses all k boundary components. Therefore, we must have s = 2, and a singular level curve that passes through q will necessarily enclose all interiorcomponents of ∂ Ω. Hence, we arrive at a contradiction.Assume now that L encloses L and t ≥ k boundary components. Let { q , q , . . . , q s } be the singular vertices on the various cycles on L which do not belong tothe cycle containing L or its interior. It is easy to see that(2.32) (cid:88) z ∈{ q ,q ,...,q s } Ind g ( z ) = (1 − t ) . By Theorem 2.3 we must have additional singular vertices { p , . . . , p n } such that (2.33) (cid:88) t ∈{ p ,...,q n } Ind g ( t ) = − − ( k − t ) . We now repeat the process above with L replacing L , l + t replacing l and k − t replacing k .After finitely many times (at most k − − ∂ Ω. (cid:3) We are now concerned with geometric properties of the set of lengths of level curves. Thisis significant for the applications. For example, even in the case of an annulus (which is thesubject of Theorem 0.4), it is not clear that level curves in the same homotopy class have thesame length. Since we wish to map a given annulus to a straight cylinder, we must verifythis property and its suitable generalizations.
Theorem 2.34.
There exists a finite set of non-negative numbers K such that the followingholds: (1) K contains , and k is its maximal element, (2) K is monotone decreasing, (3) any level curve corresponding to a g -value which does not belong to K is simple, and (4) any component of a level curve, which corresponds to a g -value which is strictlybetween any two values k n > k m in K , such that k n − k m is minimal, and is containedin a unique simple cycle C n determined by k n , has its l -length equal to that of C n . (5) Moreover, the length of C n is equal to the length of the component of g − ( k m ) whichit encloses.Proof. We let K denote the set of critical values of g union k and 0; this gives assertion(1). Since χ (Ω ∪ ∂ Ω) is negative and the index of each singular vertex is less than or equalto −
1, Theorem 2.3 implies that the number of singular vertices, and hence the number ofcorresponding critical g -values, is finite. Assertion (2) follows by ordering. Assertion (3) isthe content of Lemma 2.8. We now turn to proving (4). Let L be a connected componentof a level curve as described in (4). By the structure theory provided in Theorem 2.15 andthe maximum principle, L is contained in a unique simple cycle, part of the bouquet whichcomposes g − ( k m ). Let C n denote this cycle. We first assume that the combinatorial distancebetween L and C n is greater than or equal to 2. We apply the construction that precedesDefinition 2.7 to the level curves C n and L and the annulus A L,C n enclosed by them. Byabuse of notation, we will keep denoting the modified set of vertices by T (0) .Let F = F C n ,L denote the subset of T (0) ∩ A L,C n . Since the combinatorial distance between C n and L is greater than or equal to 2, it follows that(2.35) F (cid:54) = ∅ and that δ ( F ) = { v ∈ T (0) : v ∈ C n ∪ L } . Recall that g , the modified solution of the D-BVP, is defined on F ∪ δ ( F ) by requiringthat g is harmonic (with respect to the modified conductance function) in F , g | L = g | L and g | C n = g | C n . By the uniqueness of the D-BVP solution, it is clear that g = g | F . Let IRICHLET PROBLEM 15 F ◦ denote the set of vertices in F which do not belong to δ ( F ). We apply Proposition 1.4(the first Green Identity) to F with u = g and v ≡
1. Equation (1.5) then gives(2.36) 0 = (cid:88) x ∈ δ ( F ) ∂g ∂n ( F ◦ )( x ) . Hence(2.37) 0 = (cid:88) x ∈ C n ∂g ∂n ( F ◦ )( x ) + (cid:88) x ∈ L ∂g ∂n ( F ◦ )( x ) , which implies that(2.38) (cid:88) x ∈ L ∂g ∂n ( F ◦ )( x ) = − (cid:88) x ∈ C n ∂g ∂n ( F ◦ )( x ) . It follows that, with respect to the flux-gradient metric (see Equation (1.11)), L and C n havethe same length. We now deal with the case that the combinatorial distance between C n and L is less than 2. While we wish to use again the first Green Identity, care is needed since F is empty in this case. We add a vertex on each edge in the modified cellular decompositionthat has a vertex v on L and a vertex u on C n . The value of g ( t ) is defined by the value of g on this edge and we also let(2.39) ˜˜ c ( v, t ) = ˜ c ( v, t )( g ( v ) − g ( u )) g ( v ) − g ( t ) and ˜˜ c ( u, t ) = ˜ c ( u, t )( g ( u ) − g ( v )) g ( u ) − g ( t ) , where ˜ c ( v, t ) and ˜ c ( u, t ) are defined as in equation (2.6) with t replacing x . We keep allconductance functions on other edges unchanged.By applying again the existence and uniqueness theorems in [3], we obtain a solution g of a new D-BVP defined on A L,C n by requiring that g | L = g | L , g | C n = g | C n and that g is harmonic in A L,C n . It follows that g is g on all vertices in T (0) and is modified so as tohave the values of g on all vertices defined above. It is easy to verify (using Equation 2.6and Equation 2.39) that(2.40) ∂g ∂n ( A L,C n )( v ) = ∂g ∂n ( A L,C n )( v ) and ∂g ∂n ( A L,C n )( u ) = ∂g ∂n ( A L,C n )( u ) . Assertion (5) is proved by following the same techniques that were used in proving assertion(4) and Theorem 2.15. (cid:3)
Remark . Observe that once assertion (5) is proved, in order to prove (4), one needs onlyadd at most one vertex on each edge whose vertices lie on C n and g − ( k m ) respectively. Also,the same techniques employed in the proof of assertion (4) allow one to get equality betweenthe sum of the l -lengths of all the components for a non-singular level curve enclosed in asimple cycle and the l -length of this cycle. Finally, the case g − (0) was not stated separately,yet it similarly gives that the sum of the l -lengths of the inner boundary components of ∂ Ω equals the l -length of the outer boundary. We will revisit this last point in the proof ofTheorem 0.1. the case of an annulus In this section we prove Theorem 0.4. The proof consists of two parts. First, we will showthat there is a well-defined mapping from T (1) into a set of (Euclidean) rectangles embeddedin the cylinder S A . The crux of this part is the fact that level curves for g have the sameinduced length (measured with the l metric, see Remark 2.41) and a simple application ofthe maximum principle. Second, we will show that the collection of these rectangles formsa tiling of S A with no gaps and no overlaps. This is a consequence of the first part and anenergy-area computation. The latter follows by our construction, the dimensions of S A , andthe first Green identity (see Theorem 1.4). Figure 3.1.
Level curvesin an annulusGiven a straight Euclidean cylinder of height h , we will endow it with coordinates arisingnaturally from S × [0 , h ]. The boundary component corresponding to h will be called thetop and the other one will be called the bottom. Before providing the proof, we need adefinition which will simplify keeping track of the mapping f . Definition 3.2.
A marker on a straight Euclidean cylinder is a vertical closed interval whichis the isometric image of θ × [ a, b ], for some θ ∈ [0 , π ) and [ a, b ] ⊂ [0 , h ] with a < b . Themarker’s uppermost end-point corresponds to ( θ, b ) and its lowest end-point to ( θ, a ). Proof of Theorem 0.4.
Let S A be a straight Euclidean cylinder with height H = k andcircumference C = (cid:88) x ∈ E ∂g∂n ( x ) . Let L = { L , . . . , L k } be the level sets for g corresponding to the vertices in T (0) arrangedin a descending g -values order. We place a vertex at each intersection of an edge with an L i , i = 1 , . . . , k, , and if necessary more vertices on edges so that any two successive levelcurves in L are at combinatorial distance (at least) two. We call the first group of addedvertices type I and the second type II (recall that conductance along edges are changed aswell, according to the discussion preceding Definition 2.7). In particular, the assertions ofTheorem 2.34 and Remark 2.41 hold. Since χ ( A ) = 0 and the index of a singular vertex isalways negative, the index formula (Theorem 2.3) prevents the existence of singular verticesin A . Therefore, K = { , k } , and furthermore the length of any g -level curve is equal to C . IRICHLET PROBLEM 17
Let the vertices { x , . . . , x p } in E = L be ordered counterclockwise (and on any levelcurve), starting with x . Let { y , y , . . . , y t } be its type I neighbors in the induced cellulardecomposition oriented clockwise (which will always be the ordering for neighbors). Weidentify x with 0 × k in the coordinates above. We associate markers { m x ,y , . . . , m x ,y t } with x in the following way. The length of marker m x ,y s , for s = 1 , . . . , t , is equal to (theconstant) g ( x ) − g ( y s ) and its uppermost end-point is positioned at the top of S A at(3.3) x + (cid:80) s − k =1 c ( x , y k )( g ( x ) − g ( y k )) ∂g∂n ( A )( x ) × π, measured counterclockwise. For each edge e u,v = [ u, v ] with g ( u ) > g ( v ), let Q u,v be aEuclidean rectangle with height equals to g ( u ) − g ( v ) and width equals to c ( u, v )( g ( u ) − g ( v )).We will identify a planar straight Euclidean rectangle and its image, under an isometry, ina straight Euclidean cylinder. For s = 1 , . . . , t , we position Q x ,y s in S A in such a way thatits leftmost edge coincides with m x ,y s . By construction and the position of the markers,(3.4) Q x ,y s ∩ Q x ,y s +1 = m x ,y s +1 . Assume that we have placed markers and rectangles associated to all the vertices up to x k where k < p ; let z be the leftmost neighbor of x k +1 and let Q x k ,v be the rightmost rectangleassociated with x k . We now position the marker m x k +1 ,z , associated with x k +1 and z sothat it is lined with the rightmost edge of Q x k ,v and his upper end-point is at the top of S A .We continue placing markers and rectangles corresponding to the rest of the neighbors of x k +1 . We terminate these steps when k = p . Note that the top of S A is completely coveredby the top of the rectangles constructed above where intersections between any of these (topedges) is either a vertex or empty.For all 1 < n < k , assume that all the markers corresponding to vertices in L n − and theirassociated rectangles have been placed as above in such a way that the following condition,which we call consistent , holds. For [ w, v ] ∈ T (1) with g ( w ) > g ( v ) and s ∈ [ w, v ] a vertex oftype I, the uppermost end-point of the marker m s,v coincides with the lowest end-point ofthe marker m w,s ; moreover, the two rectangles Q w,s , Q s,v tile Q v,w . Informally, this conditionallows us to “continuously extend” rectangles associated with edges that cross level curvesalong these, and therefore will show that that edges in T (1) are mapped (perhaps in severalsteps) onto a unique rectangle.We will now show how to place the markers and rectangles corresponding to the verticesof the level set L n in a consistent way. The uppermost end-point of each marker associatedwith a vertex v ∈ L n in this level set, and any of its neighbors in L n +1 is placed in S A at height g ( v ). Observe that v is a vertex in some [ q i , v ], where q i belongs to L n . Chooseamong all such edges the rightmost (viewed from L n ). Let [ q , v ] be this edge and let m q ,v be its marker. Place the marker of v which corresponds to an edge [ v, w ] with w being theleftmost vertex among the neighbors of v in L n +1 , so that its uppermost end-point coincideswith the lowest end-point of m q ,v . Let s be a vertex of type I on L n . By definition, s is connected to a unique vertex v ∈ L n +1 and to a unique vertex w ∈ L n − . Let s q bethe first vertex to the left of s of type I in L n . By definition, s q is connected to a uniquevertex w p in L n − and to a unique vertex v l ∈ L n +1 . Let { s , . . . , s q − } be the vertices on L n between s and s q and let { w , . . . , w p − } be the vertices on L n − between w and w p . Let Q = Q w ,s ,s q ,w p be the piecewise linear rectangle enclosed by [ w , s ] ∪ [ w p , s q ] ∪ L n − ∪ L n ,and which contains { s , . . . , s q } , and let Q = Q s ,v ,s q ,v l be the piecewise linear rectangleenclosed by [ s , v ] ∪ [ s q , v l ] ∪ L n ∪ L n , and which contains { s , . . . , s q } .In order to prove that the consistent condition holds for all markers and rectangles createdin this step, it suffices to prove it at s q , assuming (with out loss of generality) that thefirst marker we placed in a consistent way is m s ,v . By construction (see in particularEquation (3.3)) we need to prove that p − (cid:88) i =1 ∂g∂n ( Q )( w i ) + ∂g∂n ( Q ) right ( w ) =(3.5) q − (cid:88) i =1 ∂g∂n ( Q )( s i ) + ∂g∂n ( Q )right( s ) , where the subscript “right” indicates that neighbors in the expressions are taken from Q (first line) or Q (second line) only. It is easy to check that since the modification of g isharmonic at each s i , i = 0 , . . . q , and since s and s q are type I vertices, Equation (3.5) holds.To conclude the construction, continue as above, exhausting all all vertices in L n . Bythe maximum principle, our construction, and the fact that all level curves have their flux-gradient length equal to C, it is clear that the union of the rectangles is contained in S A . X X E 1E 2 Qx Q x 1 y Qx ! Figure 3.6.
Several markers and rectangles on the Euclidean cylinder afterthe completion of the construction.We now prove that the collection of rectangles constructed above tiles S A leaving no gaps.Without loss of generality, suppose that the collection of rectangles does not cover a stripof the form [ θ , θ ] × [ h , h ] in S A , where 0 ≤ θ < θ ≤ π and 0 ≤ h < h ≤ k . Byharmonicity, there exists at least one path whose vertices belong to T (0) such that the valuesof g along this path are strictly decreasing from k to 0.In particular, the value h is attained on some edge or vertex of this path. By construction,a gap in a g -level curve (i.e. an arc of level curve which is not covered by the top edges ofrectangles) never occurs when h is the g -value associated to a vertex in the modified cellulardecomposition. Hence, we may assume that h is attained in the interior of an edge. Let L h be the corresponding level curve. Recall that L h is simple and closed (as are all otherlevel curves of g ; see Lemma 2.8). We now follow the construction preceding Definition 2.7 IRICHLET PROBLEM 19 and let { u , u , . . . , u q } be all the new vertices on L h , that is we place a vertex at eachintersection of an edge in ˜ T (1) with L h . As mentioned in the first paragraph of the proof,the flux-gradient length of L h is equal to C. Moreover, this length is equal to(3.7) q (cid:88) i =1 ∂g∂n ( O )( u i ) , where O is the interior of the annulus enclosed by L h and E (see the discussion proceedingDefinition 2.7). In particular, we may now place markers and rectangles associated to thecollection of vertices { u , u , . . . , u q } so that L h is covered by the top edges of these rectan-gles. Recall that up to this moment, markers and rectangles were associated only to verticesof type I in addition to those in T (0) . It is very easy to check that this can be done in aconsistent way (using Equation (2.6)). Since g is extended affinely over edges, every valuebetween h and h is attained by g . Repeating this argument shows that all level curves arecovered by rectangles. Hence the collection of rectangles leaves no gaps in S A .Using an area argument, we now finish the proof by showing that there is no overlapbetween any two of the rectangles. Let R be the union of all the rectangles. By definition,(3.8) Area( R ) = (cid:88) [ x,y ] ∈ ˜ T (1) c ( x, y )( g ( x ) − g ( y ))( g ( x ) − g ( y )) . Note that the sum appearing in the right-hand side of Equation (3.8) is computed overthe induced cellular decomposition. A simple computation (using again Equation (2.6))and the fact that the construction is consistent shows that this sum is equal to the onetaken over [ x, y ] ∈ T (1) . Hence, the right-hand side of this equation is the energy E ( g )(see Definition 1.1). Therefore, by the first Green identity, applied with u = v = g (seeTheorem 1.4), and the dimensions of S A , we have(3.9) E ( g ) = Area( R ) = Area( S A ) . Hence, since the rectangles do not overlap, they must tile S A . It is also evident that themapping f constructed above is energy preserving. (cid:3) the case of an m -connected bounded planar region, m > Pair of pants.
In this subsection we provide a proof of Theorem 0.1. One interestingingredient of the proof is the construction of a good mapping (in the sense of Theorem 0.4)from a planar annulus with one singular boundary component into a Euclidean cylinderhaving one boundary component, a singular curve and the other one which is simple. Theconstruction of the mapping f is then achieved by gluing two Euclidean cylinders to the oneabove along the singular component. Proof of Theorem 0.1.
We extend the solution g to an affine map ¯ g : |P ∪ ∂P | → R + ∪ g as a heightfunction for |P ∪ ∂P | . This gives a two-dimensional polyhedron, denoted by ¯ P g , which ishomotopically equivalent to P . We have(4.1) χ ( ¯ P g ) = χ ( P ) = − . Double ¯ P g along its boundary to obtain a genus 2 closed surface which will be denoted byD P . The Index Theorem (see Theorem 2.3) asserts that(4.2) (cid:88) v ∈ D P Ind g ( v ) = − . It follows from Equation (2.1) and a simple Euler characteristic computation that we musthave a single vertex u ∈ ¯ P g whose index equals −
1, which means that Sgc( u ) = 4. Let L ( u )be the figure eight curve corresponding to the value ¯ g ( u ). Then L ( u ) = L I ∪ L II , where L I and L II are two simple cycles intersecting at u (using here the assertion of Theorem 2.15).Moreover, since E and E are the only inner components of ∂ Ω, by applying Lemma 2.8we may assume that E is contained in L I and E is contained in L II . By cutting along L ( u ),we can decompose Ω ∪ ∂ Ω into three components. The first, denoted by Ω , has boundarycomponents E and L ( u ). The second, denoted by Ω , has boundary components E and L I ,and the third, denoted by Ω , has boundary components E and L II . Note that the interiorsof these components are all homeomorphic to S × (0 , and Ω areeach homeomorphic to an annulus, however χ (Ω ) = −
1. Therefore, Ω may be viewed asan annulus with one singular boundary component, or equivalently as a really short pair ofpants, i.e. one with no cuffs. We now define three D-BVPs on the above domains followingthe discussion preceding Definition 2.7. Note that an essential property of L ( u ), which isgiven by Theorem 2.15, allows us to define a D-BVP on Ω .Let g be the solution on Ω , g on Ω and g on Ω . By Remark 2.41, we have that thelengths of L I and E , with respect to the flux-gradient metric induced by g , are both equalto(4.3) (cid:12)(cid:12) (cid:88) x ∈ E ∂g∂n (Ω )( x ) (cid:12)(cid:12) . Similarly, we have for Ω that the lengths, with respect to the flux-gradient metric inducedby g of L II and E , are equal to(4.4) (cid:12)(cid:12) (cid:88) x ∈ E ∂g∂n (Ω )( x ) (cid:12)(cid:12) . Finally, by applying Theorem 2.34 with C n = E and L = L ( u ), we have that the lengthsof E and L ( u ), with respect to the l gradient metric induced by g , are equal to(4.5) (cid:88) x ∈ E ∂g∂n (Ω )( x ) . We now turn to the construction of ( S P , f ). First, apply Theorem 0.4 to Ω and to Ω .The outputs are two straight Euclidean cylinders S Ω and S Ω , two mappings f Ω : Ω → S and f Ω : Ω → S (where the maps, their domains and the cylinders are described in detail IRICHLET PROBLEM 21
E E1
I II2 E L L f(u)f(u) f(u) f(u)f(u) f(u)f(u) f(u)
Figure 4.6.
The construction of the Euclidean pair of pants with one singular point.in Theorem 0.4). Second, we wish to apply Theorem 0.4 to Ω . However, one boundarycomponent of it ( L ( u )) is singular at u . Note that the construction described in the proof ofTheorem 0.4 will go through for any annulus contained in Ω with one boundary componentbeing E and the other being L g ( L ( u ))+ (cid:15) (with (cid:15) > E ) and their heights converge to k − g ( L ( u )).As (cid:15) →
0, the level curves corresponding to g ( L ( u )) + (cid:15) converge geometrically to L ( u ).Hence, the limiting cylinder S is a Euclidean cylinder with one singular boundary compo-nent (bottom). We call one boundary component of this cylinder top and the other bottom .Concretely, S is obtained by taking a Euclidean cylinder of height k − g ( L ( u )) and circum-ference which is equal to the flux-gradient metric length of E , picking two points on oneboundary component of it at a distance which equals the flux-gradient metric length of L I ,and identifying them. We will abuse notation and denote this point by u . Since identificationoccurs only at the boundary, the tiling persists in the limit. We denote by f Ω the mappinggiven by Theorem 0.4, modified on the bottom as above. Let(4.7) f = f Ω ∪ f Ω ∪ f Ω , where the domain of f is Ω ∪ Ω ∪ Ω and its image is obtained by gluing isometrically S and S to the bottom in such a way that the intersection consists of only one point (seeFigure 4.6), denoted by f ( u ). By Remark 2.41, we have that the length (measured withthe l -induced metric) of the bottom of S is equal to the sum of the lengths of the topsof the two cylinders S and S (one can also obtain this by applying Proposition 1.4 andEquations (4.3)-(4.5) directly). Furthermore, we require the gluing to be consistent; i.e thegluing described above respects rectangles corresponding to edges which cross L ( u ) (see theProof of Theorem 0.4). The fact that this can be done may be justified by basically the same argument appearing before and after Equation (3.5). Note that one consequence of this isthat f ( u ) is uniquely determined (the modified g being harmonic being the key issue).It is easy to check that the only cone angle is at this point and is equal to 4 π . Themappings f Ω , f Ω are energy preserving in the sense explained in Theorem 0.4; the mapping f Ω is also energy preserving (since the identification is done on bottom only). Since thegluing is by isometries, and the cylinders involved meet only at one vertex, the mapping f is energy preserving. (cid:3) The case of an m -connected domain, m > . In this subsection we provide a proofof Theorem 0.2. The proof is a modification of Theorem 0.1 with some bookkeeping andsuccessive changes of the initial D-BVP.
Proof of Theorem 0.4.
Let L be the unique level curve whose existence is guaranteedby Proposition 2.28. Let Ω be the domain bounded by E and L . Observe that Ω ishomeomorphic to S × (0 , L = l ∪ l . . . ∪ l k , where each l i is a simplecycle (for i = 1 , . . . , k ), where any two of these cycles are either disjoint or intersect at asingular vertex in T (0) (see Theorem 2.15). Let U = { u , . . . , u m } be the set of all singularvertices along L with { Ind g ( u ) , . . . , Ind g ( u m ) } being their indices, respectively.The interior of each l i , which will be denoted by l ◦ i , is a q i -connected domain with q i 1, for i = 1 , . . . , k , (unless k = 1). We now modify the initial D-BVP on Ω , l i and l ◦ i as described in the discussion preceding Definition 2.7, and we obtain k + 1 new harmonicfunctions: g which is defined on E , Ω and L , and g , . . . , g k defined on l , . . . , l k and theirinteriors, respectively.Recall that modifications are made by adding vertices to T (0) along the l i ’s and by definingconductance along new edges: those which cross the l i ’s (See Equation (2.6)). By part (5)of Theorem 2.34 and Remark 2.41, we have that the length of L , with respect to the flux-gradient metric induced by g , is equal to the length of E with respect to the flux-gradientmetric induced by g . Hence both are equal to the length of E with respect to the initial g metric. We record this as(4.8) length g,l ( E ) = length g,l ( L ) . By using g i in the interior of each l i , we now choose a singular curve enclosing all of its q i − W = { W , . . . , W n } and the set of singular vertices V = { v , . . . , v p } they contain with theirassociated indices I = { Ind g ( v ) , . . . , Ind g ( u p ) } . Each simple cycle of a singular level curve W j contains s j < q i boundary curves unless W ◦ j is an annulus. As above, we have only addedvertices on the l i ’s, assigning each one of the vertices on a specific l i , the g ( l i )-value. Theconductance constants are changed only along new edges, those which cross l i , according toEquation (2.6). Hence, the set W coincides with the set of singular level curves of g minus L (similar statements hold for V and I ).We repeat this process (at most) finitely many times, modifying (if needed) the cellulardecomposition and defining conductance constants according to Equation (2.6), until theinterior of each simple cycle of each singular level curve is an annulus. At each step, weobtain new harmonic functions defined on domains with fewer boundary components. Anequation analogous to Equation (4.8) holds for each simple cycle which is a component ofa singular level curve and the nearest singular level curve it contains. That is, its length IRICHLET PROBLEM 23 measured by the flux-gradient metric induced by the harmonic functions defined on itsinterior is equal to the length of the singular level curve, both measured by the harmonicmap defined in the interior of the simple cycle (see part (5) of Theorem 2.34). In turn, theyare equal to the length of the simple cycle measured by the harmonic function defined on itsexterior (see Theorem 2.20).It is evident from the proof of Proposition 2.28, the disjointness of level curves whichcorrespond to different g -values, and the maximum principle that there is a well-definedhierarchy on the set of singular level curves. Each component of a singular level curve, say g − ( k m ) , k m ∈ K , is contained in the interior of a domain whose one component is a uniquesimple cycle. This cycle belongs to a unique singular curve, say g − ( k n ) , k n ∈ K , k n > k m (or E ). Simple cycles corresponding to level curves are either contained in a domain boundedby a simple cycle which is part of a singular level curve, or are parts of singular level curvesas is detailed in Theorem 2.15.We now turn to the construction of S Ω and the mapping f . We start with a straightEuclidean cylinder, C E ,L , of height k − g ( L ) and circumference which is equal to(4.9) (cid:88) x ∈ E ∂g∂n ( x ) . Since L is a generalized bouquet of circles, we can select a finite number of points in thebottom and identify subsets of them in such a way that the quotient is topologically andmetrically isomorphic to L . WU V VU UW W U V Figure 4.10. A quotient of a boundary component of a Euclidean cylinderyielding a generalized bouquet of circlesAs in the proof of Theorem 0.1, we wish to apply Theorem 0.4 to Ω ∪ ∂ Ω ( ∂ Ω = E ∪ L ).Since L is singular, we may not apply it directly. Instead, we follow the argument carried outin the proof of Theorem 0.1. All level curves (which are simple) corresponding to g ( L ) + (cid:15) ,with (cid:15) > L (see part (5) of Theorem 2.34); hence,as (cid:15) → 0, the sequence of Euclidean cylinders (guaranteed by Theorem 0.4 applied to thesequence of annuli bounded by these level curves and E ) with their tops being fixed, heightsequal to k − ( g ( L ) + (cid:15) ), and circumferences equal to (cid:80) x ∈ E ∂g∂n ( x ), converges to ˜ C E ,L with the top persisting in the limit and the bottom replaced by its quotient as explained above.We denote the mapping given by Theorem 0.4 and modified as above on the bottom by f Ω .For each l i , i = 1 , . . . k , there are two cases to consider. First, suppose that l ◦ i does notcontain any singular level curve. Without loss of generality, let the unique boundary curve itcontains be E i . By assertion (4) of Theorem 2.34, we conclude that the lengths of E i and l i are equal (measured with the g i -induced flux-gradient metric). Thus, A i = l ◦ i ∪ l i ∪ E i is anannulus with two (non-singular) boundary components. We may therefore apply Theorem 0.4and obtain a mapping f i : A i → C l i ,E i (= S A i ). We now attach the top of the resultingEuclidean cylinder C l i ,E i to l i by an isometry which is consistent. We conclude that thegluing may be taken as such, by first arguing that the length of l i measured with respectto the flux-gradient metric induced by g , is the same as its length with respect to thelength induced by the flux-gradient metric of g i . This is justified directly by employingTheorem 2.20 or by using a series of equalities similar to those in Equations (4.3)-(4.5) andthe reasoning leading to them. The fact that the gluing is consistent follows from the samearguments we have used before.Second, assume that l ◦ i contains a singular curve, say W i ∈ W . Recall that W i is theunique singular curve in l ◦ i which encloses all of the boundary curves in l ◦ i . With l i replacing E , W i replacing L , and g i replacing g , we repeat the construction, yielding a Euclideancylinder with its bottom being a singular curve as described above (in the case of L ), and amapping which we will denote by f l i , W i . The cylinder ˜ C l i ,W i is attached to ˜ C E ,L by gluingits top to the simple cycle corresponding to l i in the quotient of the bottom of ˜ C E ,L , thatis, the singular component of ˜ C E ,L .The above analysis is now carried out, at most finitely many times, for each l i , i = 1 , . . . , k ,until we are left with Euclidean cylinders having both of their boundary components non-singular. In particular, the bottom of each cylinder is the image (under the map given byTheorem 0.4) of E i and its length is equal to(4.11) − (cid:88) x ∈ E i ∂g∂n ( x ) , for i = 1 , · · · , m − . By construction, the images of E and E i , i = 1 , . . . , m , under the maps whose constructionis described above, are the only boundary components of S Ω . The map f is the obviousunion of the collection of mappings constructed above and is analogous to the one appearingin Equation (4.7). It is also evident that f is energy preserving in the sense described before.We now compute the cone angle, φ ( v ), at a singular vertex v with Ind g ( v ) = − n, n ∈ N . Byconstruction, v is the (unique) tangency point of n + 1 Euclidean cylinders, where v belongsto the non-singular component of each such cylinder. Hence, such a cylinder contributes π to the cone angle at v . Recall that each such cylinder is attached (by an isometry) to thequotient of the bottom of another Euclidean cylinder. It is easy to check that the contributionto the cone angle at v from this Euclidean cylinder (with one singular boundary component)is equal to n × π + π . Therefore,(4.12) φ ( v ) = 2( n + 1) π. (cid:3) IRICHLET PROBLEM 25 Remark . One may also prove Theorem 0.2 by an induction on the number of boundarycomponents. However, the assertion of Proposition 2.28 must be used as well as an extensionof g to g over the singular curve L . Theorem 2.20 needs to be used in order to proveequality of the l -length of L according to both the g and the g metric. Overall, we foundthe proof which does not use induction conceptually more gratifying. We could also stop theprocess once a planar pair of pants is encountered, thus using Theorem 0.1 directly. Finally,Theorem 0.1 is of course a special case of the theorem above. Still, we maintain that thisspecial case deserves its own proof. Remark . There is a technical difficulty in our construction if some pair of adjacentvertices of T (0) has the same g -value (the first occurrence is in Equation (2.1)). One maygeneralize the definitions and the index formula to allow rectangles of area zero, as onesolution. For a discussion of this approach and others see [23, Section 5]. Experimentalevidence shows that when the cell decomposition is complicated enough, even when theconductance function is identically equal to 1 and the cells are triangles, such equality rarelyhappens. Remark . The existence of singular curves for g results in the fact that some rectangles arenot embedded in the target. 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