Geodesic Nets: Some Examples and Open Problems
GGEODESIC NETS: SOME EXAMPLES AND OPEN PROBLEMS
ALEXANDER NABUTOVSKY AND FABIAN PARSCH
Abstract.
Geodesic nets on Riemannian manifolds form a natural class ofstationary objects generalizing geodesics. Yet almost nothing is known abouttheir classification or general properties even when the ambient Riemannianmanifold is the Euclidean plane or the round 2-sphere.In the first half of this paper we survey some results and open questions (oldand new) about geodesic nets on Riemannian manifolds. Many of these openquestions are about geodesic nets on the Euclidean plane. The second halfcontains a partial answer for one of these questions, namely, a description of anew infinite family of geodesic nets on the Euclidean plane with 14 boundary(or unbalanced) vertices and arbitrarily many inner (or balanced) vertices ofdegree ≥ Overview
Geodesic nets and multinets: definition.
Let M be a Riemannian mani-fold, S a finite (possibly empty) set of points in M , and G a finite multigraph (or,more formally, a finite 1-dimensional cell complex). A geodesic net modelled on G with vertices S is a smooth embedding f of G into M such that:(1) Every point from S is the image under f of a vertex of G ;(2) For each 1-parametric flow Φ t , t ∈ ( − (cid:15), (cid:15) ) , of diffeomorphisms of M fixingall points of S with Φ = Id , t = 0 is the critical point of the function l ( t )defined as the length of Φ t ( f ( G )).Less formally, geodesic nets on M are critical points (not necessarily local min-ima!) of the length functional on the space of embedded multigraphs into M , wherea certain subset of the set of vertices must be mapped to prescribed points of M .The simplest example of geodesic nets arises when S is a set of two points x, y ,and G = [0 ,
1] the graph with two vertices and one edge. In this case, the geodesicnets modelled on G with vertices x, y are precisely non-self-intersecting geodesicsin M connecting x and y . Self-intersecting geodesics can be modelled on morecomplicated graphs, but if we wish to model them on the same G it makes senseto modify our definition of geodesic nets by allowing f to be only an immersion onthe union of interiors of edges. In other words, one may allow edges to self-intersectand to intersect each other. In particular, we are allowing that two different edgesbetween the same pair of vertices might have the same image. As the result, theimages of edges of G in M acquire multiplicities that can be arbitrary positiveinteger numbers. In this paper we are going to call geodesic nets defined usingimmersions rather than embeddings of multigraphs geodesic multinets . a r X i v : . [ m a t h . M G ] M a r ALEXANDER NABUTOVSKY AND FABIAN PARSCH
Applying the first variation formula for for the length functional we see that theabove definition of a geodesic (multi)net is equivalent to the following:
Definition 1.1.1.
Let S be a (possibly empty) finite set of points in a Riemannianmanifold M . A geodesic net on M consists of a finite set V of points of M (calledvertices) that includes S and a finite set E of non-constant distinct geodesics be-tween vertices (called edges) so that for every vertex v ∈ V \ S the following balanc-ing condition holds: Consider the unit tangent vectors at v to all edges incident to v . Direct each tangent vector from v towards the other endpoint of the edge. Thenthe sum of all these tangent vectors must be equal to 0 ∈ T v M . Further, edgesare not allowed to intersect or self-intersect. Geodesic multinets are defined in thesame way with the two following distinctions: 1) Edges are allowed to intersectand self-intersect; 2) Each edge is endowed with a positive integer multiplicity; thetangent vector to an edge enters the sum in the balancing condition at each itsendpoint with the multiplicity equal to the multiplicity of the corresponding edge.Vertices in S are called boundary or unbalanced , vertices in V \ S are called inner ,or free , or balanced (as the balancing condition must hold only at each vertex in V \ S ). If v is an unbalanced vertex, then the sum of all unit tangent vectors toedges incident to v need not be equal to the zero vector. We call this sum theimbalance vector Imb( v ) at v , and its norm the imbalance , imb( v ), at v . The sumof imbalances (cid:80) v ∈ S imb( v ) over the set of all unbalanced points is called the totalimbalance of the geodesic net. It is convenient to define Imb( v ) also at balancedvertices as zero vectors in T v M .For the rest of the paper we are going to require that no balanced vertex isisolated, that is, has degree zero. As the degree of a balanced vertex clearly cannotbe one, we see that the minimal degree of a balanced vertex becomes two. Thebalancing condition implies that for any balanced vertex of degree 2, its two incidentedges can be merged into a single geodesic. Conversely, we can subdivide each edgeof a geodesic net by inserting as many new balanced vertices of degree 2 as we wish.As now the role of balanced vertices of degree 2 in the classification of geodesic netsis completely clear, we are going to consider below only geodesic nets where allbalanced vertices have degree ≥
3. It is clear that we can add or removegeodesics connecting unbalanced vertices at will without affecting the balancingcondition at a balanced vertex. Therefore, we agree that all considered geodesicnets do not contain edges between unbalanced vertices.
Our final conventionis that we are going to consider here only connected geodesic (multi)nets (asthe classification of disconnected nets obviously reduces to classification of theirconnected components).1.2.
Geodesic nets in Euclidean spaces.
A significant part of this paper will bedevoted to geodesic nets in Euclidean spaces, in particular, in the Euclidean plane.In this case above definition says that a geodesic multinet is a graph G = ( V, E ) in R n such that 1) S is a subset of the set of vertices V , 2) each edge is a straight linesegment between its endpoints, and is endowed with a positive integer multiplicity n ( e ), 3) For each vertex v ∈ V \ S , there is zero imbalance, i.e. (cid:80) e ∈ I ( v ) n ( e ) e (cid:107) e (cid:107) = 0,where I ( v ) denotes the set of edges incident to v , and e denotes an edge regardedas the vector in R n directed from v towards the other endpoint. For geodesic nets,all n ( e ) must be equal to 1, and, in addition, different edges are not allowed to EODESIC NETS: SOME EXAMPLES AND OPEN PROBLEMS 3
Figure 1.2.1.
Examples for balanced vertices of degree 3, 4 and 7.intersect. Figure 1.2.1 depicts examples of balanced points of degrees 3, 4 and 7 in R . It is easy to see that 1) the angles between edges incident to a balanced vertexof degree 3 are always equal to 120 degrees. (This will be true not only for R butfor all ambient Riemannian manifolds M .) 2) A balanced vertex v of degree 4 inthe Euclidean plane is a point of intersection of two straight line segments formedby two pairs of incident edges at v (see Figure 1.2.1).Here are some other easily verified facts about geodesic (multi)nets in Euclideanspaces:(1) Each geodesic multinet is contained in the convex hull of its unbalancedvertices.(2) As a corollary each geodesic (multi)net with two boundary vertices is simplythe straight line segment connecting these points (that can be endowed withany positive integer multiplicity in the case of geodesic multinets).Therefore, the interesting part of the classification of geodesic (multi)nets in theEuclidean plane starts from the case of three boundary points.(3) For each geodesic (multi)net in R n we can consider Imb( v ) as vectors inthe ambient space R n . Therefore, in this case one can also define the totalimbalance vector. Yet this vector is always zero: (cid:88) v ∈ S Imb( v ) = (cid:88) v ∈ V Imb( v ) = 0 , (1 . . N in a Euclidean space its length L ( N ) is givenby the following formula: L ( N ) = − (cid:88) v ∈ S (cid:104) v, Imb( v ) (cid:105) . (1 . . R n . Proof:
In order to prove this formula, first observe that the right hand sidedoes not change when we change the origin of the coordinate system in R n .(This easily follows from the formula (1.2.1).) Therefore, we can assumethat the origin is not on the net. For each positive r let D r denote the ball ALEXANDER NABUTOVSKY AND FABIAN PARSCH
Figure 1.3.1.
A geodesic net with 3 unbalanced vertices and 1balanced vertex (the
Fermat Point ). This is in fact the maximalnumber of balanced vertices when only given 3 unbalanced ver-tices on the plane with a metric of nonpositive curvature. On theother hand, the two Steiner trees for four points are not maximalregarding the number of balanced vertices of a geodesic net withfour unbalanced vertices.of radius r centered at the origin, ∂D r its boundary, E ( r ) the set of edgesof the net intersecting ∂D r . For each e ∈ E ( r ) let e ( r ) denote the point ofintersection of e and ∂D r . Formula (1.2.2) is an immediate corollary of thefollowing formula, when it is applied to very large values of r : L ( N (cid:92) D r ) = (cid:88) e ∈ E ( r ) (cid:104) e ( r ) , e (cid:107) e (cid:107) (cid:105) − (cid:88) v ∈ D r (cid:104) v, Imb( v ) (cid:105) . (1 . . e also as a vector in R n . We choose its directionfrom e ( r ) towards the interior of D r . This formula obviously holds, when r is small, as both sides are equal to zero. Define special values of r asthose, where ∂D r is either tangent to one of the edges or passes throughone of the vertices. There are only finitely many special values of r . Ournext observation which is easy to verify is that the right hand side of (1.2.3)changes continuously, when r passes through its special value. (Obviously,one needs only to check what happens if ∂D r passes through a balancedor an unbalanced vertex.) Now we see that it is sufficient to check thatthe derivatives of the right hand side and the left hand side with respectto r at each non-special point coincide. Each of these derivatives will bea sum over edges in E ( r ). To complete the proof it is sufficient to verifythat the contributions of each edge to both sides are the same. Each suchedge e contributes θ e ( r ) to the derivative of the left hand side, where θ e ( r ) denotes the angle between e and e ( r ). Its contribution to the righthand side is ( r cos θ e ( r )) (cid:48) = cos θ e ( r ) − r sin θ e ( r ) dθ e ( r ) dr = θ e ( r ) , as an easytrigonometric argument implies that dθ e ( r ) dr = − tan θ e ( r ) r . This completes theproof of (1.2.3) and, therefore (1.2.2).1.3.
Steiner trees and locally minimal geodesic nets.
Study of geodesic netswas originally motivated by the following question posed by Gauß: Given a set ofpoints on the plane, connect them by means of a graph of the minimal possiblelength. It is easy to see that this graph is always a geodesic net modelled on a tree(called the Steiner tree). This tree is a geodesic net, where the given points areunbalanced points, but typically it also contains new balanced vertices. It is easyto prove that all balanced vertices of a Steiner tree have degree 3. The first and themost fundamental example is the case of three points A , B , C on the plane forminga triangle with angles < ◦ . In this case there exists the (unique) point O in thetriangle ABC called the Fermat point, such that the angles
AOB , BOC and
COA
EODESIC NETS: SOME EXAMPLES AND OPEN PROBLEMS 5 are all equal to 120 ◦ . The Steiner tree will consist of three edges OA , OB , and OC (see Figure 1.3.1). The Steiner tree on four given (unbalanced) vertices mightinvolve two extra (balanced) vertices (see same figure). A geodesic net (in a Rie-mannian manifold) is called locally minimal if its intersections with all sufficientlysmall balls are Steiner trees (for the set of points formed by all intersection of thegeodesic net with the boundary circle and all unbalanced points in the circle). Forgeodesic nets in the Euclidean plane the local minimality is equivalent to the re-quirement that all balanced points have degree 3. The locally minimal geodesic netsin Euclidean spaces and, more generally, Riemannian manifolds were extensivelyinvestigated by A. Ivanov and V. Tuzhilin (cf. [IT94], [IT16]). (Note that althoughgeneral geodesic net are not locally minimal with respect to this condition, theyare locally minimal in the following less restrictive sense: For each point p on thenet and all sufficiently small r the intersection of the net with the ball of radius r provides the global minimum of the length among all trees of the same shape (i.e.star-shaped with the same number of edges) connecting the boundary points.The idea of minimization of length might seem useful if one wants to constructa geodesic net with the set S of boundary points modelled on a given graph G ,say, in the Euclidean plane as follows: Consider all embeddings of G in the planesuch that all edges are mapped into straight line segments, and a certain set ofvertices is being mapped to S . Yet the positions of other vertices are variable,and we do not insist on balancing condition at any vertices. Now we are going tominimize the total length of all edges of the graph over the set of such embeddings.It is easy to see that the total length will be a convex function and has the uniqueminimum. Moreover, one can start with an arbitrary allowed embedding of G anduse an easy algorithm based on the gradient descent that numerically finds thisminimum which will be always a geodesic net. The problem is that in the processof gradient descent different vertices or edges can merge, and some edges can shrinkto a point. Then thee resulting graph will not be isomorphic to G anymore. Infact, our numerical experiments seem to indicate that if one starts from a randomallowed embedding of G one typically ends at very simple geodesic nets such as, forexample, the geodesic net with just one extra (balanced) vertex in the centre.1.4. Plan of the rest of the paper.
In section 2 we survey closed geodesic netson closed Riemannian manifolds. In section 3 we survey geodesic nets on Euclideanspaces and Riemannian surfaces. The emphasize there will be on (im)possibilityto majorize the number of balanced points in terms of the number of unbalancedpoints ( and possibly, also the total imbalance). Section 4 contains a rather longconstruction of an infinite sequence of geodesic nets on the Euclidean plane with 1boundary vertices and arbitrarily many balanced vertices. This sequence providesa partial answer for one of the questions asked in section 3.2.
Closed geodesic nets.
Geodesic nets with S = ∅ are called closed geodesic nets . The simplest examplesof closed geodesic nets are periodic geodesics (that can be modelled on any cyclicgraph or the multigraph with one vertex and one loop-shaped edge) or, more gen-erally, unions of periodic geodesics. The simplest example of a closed geodesic net ALEXANDER NABUTOVSKY AND FABIAN PARSCH not containing a non-trivial periodic geodesic is modelled on the θ -graph with 2 ver-tices connected by 3 distinct edges. The corresponding closed geodesic net consistsof two vertices connected by 3 distinct geodesics, so that all angles between eachpair of geodesics at each of the vertices are equal to 120 ◦ . J. Hass and F. Morgan([HM96]) proved that for each convex Riemannian S sufficiently close to a roundmetric there exists a closed geodesic net modelled on the θ -graph. It is remarkablethat this is the only known result asserting the existence of closed geodesic netsnot composed of periodic geodesics on an open (in C topology) set of Riemannianmetrics on a closed manifold! Problem 2.0.1.
Is it true that each closed Riemannian manifold contains a closedgeodesic multinet not containing a non-trivial periodic geodesic?The standard Morse-theoretic approach to constructing periodic geodesics failswhen applied to constructing closed geodesic nets, as any gradient-like flow mightmake the underlying multigraph to collapse to a (possibly mutiple) closed curveand, thus, yields only a periodic geodesic.A classification of shapes of closed geodesic nets on specific closed Riemanniansurfaces is aided by the Gauß-Bonnet theorem and the obvious observation that ifa geodesic net on, say, a Riemannian S is modelled on a graph G , then G must beplanar. Using these observations A. Heppes ([Hep99]) classified all closed geodesicnets on the round S , where all vertices have degree 3 (there are just nine possibleshapes). On the other hand, we are not aware of any restrictions on shapes of closedRiemannian manifolds of dimension > ≥ Problem 2.0.2.
Classify all 3-regular graphs G such that the round 3-sphere hasa geodesic net modelled on G .Another reasonable question (which, of course, can also be asked for surfaces) is: Problem 2.0.3.
Is it true that each closed Riemannian manifold of dimension ≥ θ -graph shaped closed geodesic net?To the best of our knowledge, nothing else is known about classification of geo-desic nets on round S . In particular, the answer for the following problem posedby Spencer Becker-Kahn ([BK]) is not known even when M is the round 2-sphere. Problem 2.0.4 (Becker-Kahn) . Let M be a closed Riemannian manifold. Is therea function f M (depending on geometry and topology of M ) such that each closedgeodesic net on M of length L has at most f M ( L ) (balanced) vertices?As we already noticed the set of closed geodesic nets includes periodic geodesicsas well as their unions. Yet the standard “folk” argument involving the compactnessof the set of closed curves of length ≤ x parametrised by the arclength on a closedRiemannian manifold, and a quantitative (Yomdin-style) version of the Sard-Smaletheorem that implies that the set of non-constant periodic geodesics on a genericclosed Riemannian manifold is countable, also implies the set of closed geodesicnets is countable as well. So, closed geodesic nets are also “rare”. This fact might EODESIC NETS: SOME EXAMPLES AND OPEN PROBLEMS 7 be at least partially responsible for the scarcity of examples of closed geodesic netsnot containing periodic geodesics.Surprisingly, many extremely hard open problems about periodic geodesics canbe solved when asked about closed geodesic nets. Here are some results about theexistence of closed geodesic nets with interesting properties:(1) One of the authors (A.N.) and R. Rotman proved the existence of a con-stant c ( n ) such that each closed Riemannian manifold M n contains a closedgeodesic multinet of length ≤ c ( n ) vol ( M n ) n . It also contains a closedgeodesic multinet of length ≤ c ( n ) diameter( M n ) ([NR07]). R. Rotmanlater improved this result and proved that one can choose a closed geodesicmultinet satisfying these estimates that has a shape of a flower, that is con-sist of (possibly) multiple geodesic loops based at the same point (vertex)([Rot11]). (Of course, the balancing (stationarity) condition at this pointmust hold.)(2) Recently L. Guth and Y.Liokumovich ([GL]) proved that for a generic closedRiemannian manifold the union of all closed geodesic multinets must be adense set.Note that these results do not shed any light on the existence of closed geodesicnets that do not include any periodic geodesic on closed manifolds as all closedgeodesic nets in these theorems might be just periodic geodesics. Yet, in dimensions > M n and positive (cid:15) , thereexists a “wide” geodesic loop on M n with the angle greater than π − (cid:15) so that itslength is bounded only in terms of n , (cid:15) and the volume of M n ([Rot]). Alternatively,one can also use the diameter of M n instead of its volume. The proof involvesdemonstrating the existence of closed geodesic multinets with certain properties.(Yet these nets can turn out to be a periodic geodesic in which case the short widegeodesic loop will be a short periodic geodesic as well.)Our last remark about closed geodesic multinets is that in some sense they can beconsidered a better 1-dimensional analog of minimal surfaces in higher dimensionsthan periodic geodesics. Indeed, minimal surfaces tend to develop singularities.Their existence is frequently proven through a version of Morse theory on spacesof cycles, where the resulting minimal surface first arises as a stationary varifold.Similar arguments using the space of 1-cycles lead to proofs of existence of closedgeodesic multinets that can be regarded as a particularly nice class of stationary1-varifolds. We refer the reader to [AA76] for properties of stationary 1-varifoldsincluding a version of formula (1.2.2) (“monotonicity formula”) valid for stationary1-varifolds, and, therefore, closed geodesic multinets on Riemannian manifolds. ALEXANDER NABUTOVSKY AND FABIAN PARSCH α Figure 3.1.1.
An example of a geodesic multi net with three un-balanced vertices and six balanced vertices. Through continuousnesting, the number of balanced vertices can be increased arbi-trarily. However, this is at the expense of additional imbalance atthe three unbalanced vertices. In this example, cos α = 12 /
13 andsin α = 5 / Geodesic nets and multinets in Euclidean spaces and Riemannianmanifolds
Recall, that we agreed to consider only connected geodesic nets with balancedvertices of degree ≥ Geodesic nets and multinets on the Euclidean plane and more gen-eral Riemannian surfaces.
We are going to start from the description of thefollowing example (see Figure 3.1.1):
Example 3.1.1.
Let A A A be a triangle. Denote its angle at A i by α i . Assumethat for each i = 1 , ,
3, cos α i is a rational number. It is easy to produce an infiniteset of such triples of angles using Pythagorean triples of integers. For example, wecan take α = α = 2 arcsin( ), and α = π − α − α . Any choice of angles α i determines the triangle A A A up to a similarity; the exact choice of its sidelengths is not important for us. As cos α i is rational, it can be written as m i n i forpositive integer m i and n i . Let N denote n n n and N i denote (integer) m i Nn i .Further, let 0 < r < r < . . . < r k < < O denote the point of intersection of bisectors of angles α i . Theset of vertices of a geodesic multinet that we are going to describe looks as follows: Ithas three unbalanced vertices A , A and A . To describe its set of balanced verticesconsider k homotheties of A A A with center O using ratios r , . . . , r k . Denotethe corresponding vertices of the homothetic triangles by A j A j A j , j ∈ { , . . . , k } .The set of balanced vertices of the geodesic multinet will include all vertices A ji . EODESIC NETS: SOME EXAMPLES AND OPEN PROBLEMS 9
Observe that for each i = 1 , A ji will subdivide A i A i into k segments.We are going to denote these segments by e ji , j = 1 , . . . , k where the numeration bysuperscripts j goes in the order from A i to A i , so that e i = A i A i and e ki = A ki A i .All these segments e ji will be edges of the geodesic multinet; the weight of e ji willbe equal to 2 jN i . The set of edges of the multinet will also include all sides of thetriangles A ji A j A j ; all these edges will be endowed with the same weight N . (Ofcourse, we can then divide all weights by their g.c.d, if it is greater than 1.) Nowan easy calculation confirms that we, indeed, constructed (an uncountable familyof) geodesic multinets with 3 unbalanced vertices and 3 k balanced vertices, where k can be arbitrarily large.However, we would like to make the following observations:(1) The weights of at least some of the edges (e.g. A ki A i ) become unbounded,as k −→ ∞ .(2) In fact, the total imbalance will increase linearly with k , as k −→ ∞ .(3) The condition of rationality of the trigonometric functions of α i is veryrestrictive. We were able to carry out our construction only for a set oftriples of points A , A , A of measure 0 in the space of all vertices oftriangles in the Euclidean plane.Looking at this example, one might be led into thinking that the constructedgeodesic multinets with 3 unbalanced vertices and arbitrarily many balanced ver-tices can be converted into a geodesic net by some sort of a small perturbation,where the balanced vertices are replaced by “clouds” of nearby points (with someextra edges inside each cloud), and all multiple edges are replaced by close but dis-tinct edges running between chosen nearby “copies” of their former endpoints. It iseasy to believe that such a perturbation plus, maybe, some auxiliary constructionwill be sufficient to construct examples of geodesic nets in the plane with 3 unbal-anced vertices and an arbitrary number of balanced vertices. Yet all such hopes areshattered by the following theorem of one of the authors (F.P.): Theorem 3.1.2 ([Par18]) . A geodesic net with unbalanced vertices A , A , A in the Euclidean plane has exactly one balanced vertex O at the Fermat point of thethree unbalanced vertices and three edges OA i . Moreover, this assertion is true forgeodesic nets with unbalanced vertices on any non-positively curved Riemannian R . Note that [Par18] contains an example demonstrating that this assertion is nolonger true without the sign restriction on the curvature of the Riemannian plane.Yet it is not known if the assertion is still true if the integral of the positive part ofthe curvature is sufficiently small. (The example for positive curvature constructedin [Par18] requires total curvature at least π .)The striking contrast between Example 3.1.1 of geodesic multinets with 3 un-balanced vertices and the extreme rigidity of geodesic nets with three unbalancedvertices on the Euclidean plane leads to some intriguing open questions such as: Problem 3.1.3.
Let Σ be the set of all triples S of points of the Euclidean planesuch that there exist geodesic multinets with 3 unbalanced vertices at S and ar-bitrarily many balanced vertices. Is it true that Σ is a set of measure zero (in( R ) )? Problem 3.1.4.
Is there a function f ( n ) such that for each geodesic multinet withthree boundary vertices in the Euclidean plane such the that multiplicities of alledges do not exceed n the number of balanced vertices does not exceed f ( n )? Problem 3.1.5.
Classify all geodesic multinets in the Euclidean plane with 3unbalanced vertices.3.2.
Geodesic nets in the Euclidean plane and -space with unbalancedvertices. We are going to start from the following remark. Given several points A , . . . , A k in the Euclidean space R n , there is always the unique point O ∈ R n (called Fermat point) such that the sum of distances (cid:80) ki =1 dist( x, A i ) attains itsglobal minimum at O . (This fact is an immediate corollary of the convexity of thefunction (cid:80) i dist( A i , x ) . ) For three points forming a triangle with the angles < ◦ , O is the point such that all angles A i OA j are equal to 120 ◦ . For four points in theplane at the vertices of a convex quadrilateral, O is the point of intersection of thetwo diagonals. For four vertices of a regular tetrahedron, O is its center. If O isnot one of the points A i , then the star-shaped tree formed by all edges OA i is ageodesic net with unbalanced vertices A , . . . , A k and the only balanced vertex at O . If A , A , A , A are, say, vertices of a square or a rectangle close to a square onehas two other well-known and “obvious” geodesic nets with unbalanced vertices at A i ; Both these nets are shaped (see Figure 1.3.1): They have two new balancedvertices O and O connected by an edge. Each balanced vertex is connected byedges with a pair of unbalanced vertices, so that all three angles at either O or O are 120 degrees, and each of the four unbalanced vertices is connected with exactlyone balanced vertex. There are three ways to partition a set of four vertices intotwo unordered pairs, yet only those where the unbalanced vertices in each pair areconnected by a side of the convex quadrilateral can “work”. Of course, the locationsof balanced points O , O will be different for the two ways to partition the set offour sides of the quadrilateral into pairs. (Note that exactly the same idea worksfor the regular tetrahedron: Each of three pairs of opposite edges gives rise to ashaped geodesic net with two balanced vertices.)It had been observed in [Par18] that given vertices A , . . . , A of a convex quadri-lateral close to a square but in general position, one can combine the star-shapednet with one balanced point at the point of intersection of diagonals, the twoshaped nets and four star-shaped geodesic nets with unbalanced points at the ver-tices of each of 4 triangles formed by all triples of four vertices A i one obtains ageodesic net with 28 balanced vertices (see Figure 3.2.1). (One obtains some extrabalanced vertices as points of intersection of edges of geodesic nets that are beingcombined.) This example might seem like a strong indication that no analog ofTheorem 3.1.2 for geodesic nets with 4 unbalanced vertices is possible. Yet one candefine irreducible geodesic nets on a given set S of unbalanced vertices as geodesicnets such that no subgraph formed by a proper subset of the set of edges (with allincident vertices) is a geodesic net with the same set S of unbalanced vertices. Itis clear that classification of geodesic nets boils down to the classification of irre-ducible nets. As so far we have only two “obvious” isomorphism types of geodesicnets with 4 unbalanced vertices (namely, X-shaped and shaped trees), one mightstill suspect that there exists an easy classification of geodesic nets with 4-verticeson the Euclidean plane. EODESIC NETS: SOME EXAMPLES AND OPEN PROBLEMS 11
Yet the situation changed (at least for us) after one of the authors discovered anew example of an irreducible geodesic net with 4 unbalanced vertices at four ver-tices of the square and 16 balanced vertices (see Figure 3.2.2) A detailed descriptionof this example can be found in [Par19]. Now a natural next step in classificationof geodesic nets on 4 vertices in the plane will be the following problem:
Problem 3.2.1.
Find an irreducible geodesic net with 4 unbalanced vertices in theEuclidean plane with more than 16 balanced vertices (or prove that such a geodesicnet does not exist.)In fact, we believe that:
Conjecture 3.2.2.
There exist geodesic nets in the Euclidean plane with unbal-anced vertices and an arbitrarily large number of balanced vertices. (Moreover, wewill not be surprised if this assertion is already true in the case when the set ofunbalanced vertices coincides with the set of vertices of a square). Note that we are not aware of any analogs of this geodesic net with 4 unbalancedand 16 balanced vertices when 4 unbalanced vertices are non co-planar points in theEuclidean 3-space, e.g. the vertices of the regular tetrahedron. Yet in this case thereexists (a more obvious) geodesic net with 4 unbalanced vertices A i and 7 balancedvertices obtained as follows (see figure 3.2.3): Start from the star-shaped geodesicnet with the balanced vertex at the center of the regular tetrahedron. For each of 6triangles A i A j O , where i, j run of the set of all unordered distinct pairs of numbers1 , , , A i , A j and O and a new balanced vertex at the center of the triangle A i A j O . Nevertheless, itseems that it is harder to construct irreducible nets with unbalanced vertices at thevertices of a regular tetrahedron than at the vertices of a square. We would not besurprised if the answer for the following question turns out to be positive: Figure 3.2.1.
An example of a geodesic net in the plane with fourunbalanced vertices which is an “overlay” of trees.
Figure 3.2.2.
The example of an irreducible geodesic net withfour unbalanced vertices that is not a tree, as constructed in [Par19]
Problem 3.2.3.
Is there a number N such that each irreducible geodesic net withunbalanced vertices at all vertices of a regular tetrahedron has at most N balancedvertices?3.3. Geodesic nets in the plane: can one bound the number of balancedvertices in terms of the number of unbalanced vertices?
We cannot solveProblem 3.2.1. Yet in the next section, we are going to describe a construction of acertain family of irreducible geodesic multinets G i ( ϕ ) with 14 unbalanced vertices(7 of which a constant and 7 variable) and arbitrarily many balanced vertices. We believe that these geodesic multinets are, in fact, geodesic nets. Our faith is basedon the following facts:(1) We checked numerically that the first 100 geodesic multinets from our listare, indeed, geodesic nets. (The number of balanced vertices of G i ( ϕ ) isgreater or equal than 7 i .)(2) We constructed an sequence of functions ϕ i of one real variable ϕ . If foreach N , some N functions ϕ i ( ϕ ) are pairwise distinct in a neighbourhoodof 0, then our construction, indeed, produces geodesic nets with at least 7 N balanced vertices. The functions ϕ i ( ϕ ) are presented by a very complicatedset of recurrent relations. Whenever there seem to be no reason for anypair of these functions to coincide, the formulae are so complicated thatthe proof of this fact eludes us.Note, that while the imbalances at the seven constant vertices are unbounded,the imbalance at 7 variable vertices remain bounded. This leads us to a beliefthat some modification of our construction might lead to elimination of several EODESIC NETS: SOME EXAMPLES AND OPEN PROBLEMS 13
Figure 3.2.3.
A geodesic net in Euclidean three-space, with fourunbalanced and seven balanced verticesvariable unbalanced points leaving us only with seven constant unbalanced points.Moreover, we believe that it is possible that our construction will “survive” smallperturbations of the seven constant unbalanced points. As a result we find that thefollowing conjecture is very plausible:
Conjecture 3.3.1.
1. There exist N and an N -tuple S such that for each N thereexists a geodesic net with S being its set of unbalanced vertices and the number ofbalanced vertices greater than N . 2. Furthermore, there exist not merely one such N -tuple S but a subset of ( R ) N of positive measure (or even a non-empty opensubset) of such N -tuples. In fact, it is quite possible that N = 4.3.4. Gromov’s conjecture.
As we saw, even for the simplest geodesic multinetsin the Euclidean plane, there is no upper bound for the number of balanced verticesin terms of the number of unbalanced vertices. The example mentioned in section3.3 and explained in detail in the next section strongly suggests, that such a bounddoes not exist already for geodesic nets. The length of a geodesic net cannot be ofgreat help either, as we can rescale any geodesic net to an arbitrarily small (or large)length without changing its shape. One appealing conjecture due to M. Gromov isthe following:
Conjecture 3.4.1 (M. Gromov) . The number of balanced vertices of a geodesic netin the Euclidean plane can be bounded above in terms of the number of unbalancedvertices and the total imbalance.
Figure 3.4.1.
This construction shows that there is a sequenceof nets with bounded imbalance, but arbitrarily many balancedverticesIn fact, we do not see any reasons why this conjecture cannot be extended togeodesic multinets. Note that the following simple example demonstrates that onecannot majorize the number of balanced points only in terms of the total imbalancewithout using the number of unbalanced vertices (see figure 3.4.1): Take a copy ofa regular N -gon, and obtain a second copy by rotating it by π N about its center.Take a geodesic net obtained as the union of these two copies of the regular N -gon.The set of unbalanced vertices will consist of 2 N vertices of both copies. Yet assides of two copies intersect, we are going to obtain also 2 N balanced vertices thatarise as points of intersections of various pairs of sides. The imbalance at eachunbalanced vertex is 2 sin πN , so the total imbalance is 4 N sin πN < N πN = 4 π . Wesee that when N −→ ∞ , the number of balanced points also tends to ∞ , yet thetotal imbalance remains uniformly bounded.The above conjecture by Gromov was published in the paper by Y. Mermarian[Mem15] for geodesic nets such that all imbalances are equal to one (in our terms.As in this case the total imbalance is equal to the number of unbalanced vertices, theconjecture is that the number of unbalanced vertices does not exceed the value ofsome function of the number of balanced vertices. Note also that [Mem15] containsthe proof of this restricted version of the conjecture in cases, when the degrees of allbalanced vertices are either all equal to 3, or all are equal to 4.) Yet, the followingsimple observation implies that the restricted form (imbalances equal to 1 at eachunbalanced vertex) is, in fact, equivalent to full Conjecture 3.4.1. The observationis that if v is an unbalanced vertex, then it can be extended by adding less thanimb( v ) + 3 new edges starting at v so that v becomes balanced. Applying thistrick to all imbalanced vertices we replace our original geodesic net by a new one,with the new number of unbalanced vertices not exceeding the sum of the totalimbalance and thrice the number of unbalanced vertices in the original net. In thisnew geodesic net the imbalances of all unbalanced vertices are equal to one. Thus,the restricted version of the conjecture implies the general version. We are going toexplain this observation in the case, when imb( v ) ∈ (0 ,
1) leaving the general caseto the reader. In this case we need to find three new edges starting at v such thattheir angles with the imbalance vector Imb( v ) that we denote α , α and α satisfythe balancing condition that can be written in the scalar form as the system of twoequations: (cid:80) i =1 cos α i = 2 imb( v ) and (cid:80) i =1 sin α i = 0 . It is clear that this system
EODESIC NETS: SOME EXAMPLES AND OPEN PROBLEMS 15 has an uncountable set of solutions. This fact enables us to ensure that none of thenew edges coincide with already existing edges incident to v .Note that formula (1.2.2) implies that the length of a geodesic net does not exceedthe product of its total imbalance and the diameter ( which for geodesic nets in theEuclidean space is always equal to the maximal distance between two unbalancedpoints). Further, we can always rescale a geodesic net in the plane so that itsdiameter becomes equal to 1. In this case its length becomes equal to LD , where L and D are the values of the length and the diameter before the rescaling. Therefore,Conjecture 3.4.1 would follow from the validity of the following conjecture: Conjecture 3.4.2.
There exists a function f of such that each geodesic multinetin the Euclidean plane with n unbalanced vertices, diameter D , and total length L has less than f ( LD , n ) balanced vertices. Now we would like to combine this conjecture with the Becker-Kahn problem2.0.4 and extend it to all Riemannian manifolds. Before doing so, consider theexample of a complete non-compact Riemannian manifold which is a disjoint unionof (smooth) capped cylinders that have a fixed length but are getting thinner. Morespecifically, the cylinders have radii n for all positive integers n but fixed diameterof 1. On any of these cylinders, we can now add N closed geodesics around thewaist of the cylinder, connecting all of them with a single closed geodesic thattravels twice along the diameter of the manifold. Such a net will have 2 N balancedvertices and fixed diameter D = 1, but as long as N/n is small enough, the lengthof the net L gets arbitrarily close to 2. So both L/D and L stay bounded whereasthe number of balanced vertices can be chosen to be arbitrarily large. Note that wecould make this manifold connected by connecting consecutive cylinders by thinnerand thinner tubes of length 1.This example shows that for general Riemannian manifolds, we can’t bound thenumber of unbalanced vertices in terms of L/D , L , and the number of balancedvertices. So we must either bound the injectivity radius of our Riemannian manifold M from below, or, more generally, adjust the length as follows: The adjusted totallength ˜ L of a geodesic net is the sum of integrals over all edges e i parametrized bytheir respective arclengths of e i ( s )) , where inj( e i ( s )) denotes the injectivity radiusof the ambient Riemannian manifold at e i ( s ). If M is a Riemannian manifold witha positive injectivity radius inj, then ˜ L ≤ L inj . Now we can state our most generalconjecture. Conjecture 3.4.3 (Boundedness conjecture for geodesic nets on Riemannian man-ifolds) . Let M be a complete Riemannian manifold. There exists a function f M which depends on M but is invariant with respect to rescalings of M with the fol-lowing property: Let G be a geodesic net on M with total length L , adjusted length ˜ L and diameter D that has n unbalanced vertices. Then its number of balanced ver-tices does not exceed f M ( ˜ L, LD , n ) . In particular, if M has injectivity radius inj > ,then the number of balanced vertices does not exceed f M ( L inj , LD , n ) . The Star
Overview.
The “star” G n ( ϕ ) constructed in this section is a possible exampleof how a geodesic net can be constructed that fulfills the following requirements: • The net has 14 unbalanced vertices (of arbitrary degree) • The net has an arbitrarily large (finite) number of balanced vertices • All edges have weight one (as is required by our definition of geodesic nets)In fact, the third condition is what makes the present construction both interestingbut also quite sophisticated. If we allowed integer weights on our edges, there wouldbe much simpler constructions of geodesic nets with just three unbalanced verticesand an arbitrary number of balanced vertices (see 3.1.1).Our construction will work as follows: First, we will construct a highly symmetricgeodesic net G n (0), layer by layer, that provides for an arbitrarily large number ofbalanced vertices. To arrive at a result as depicted in figure 4.3.2, we first need tobuild a toolbox to be used during the construction.This highly symmetric net has edges of integer weights. That is why we willmake sure that our construction works for a small deviation from the symmetriccase as well, arriving at a net G n ( ϕ ). This deviation is intended to remove anyinteger weights.As it turns out, showing that for some nonzero deviation ϕ ∈ ( − (cid:15), (cid:15) ), none ofthe edges of G n ( ϕ ) “overlap” necessitates a close look at a quite complicated finiterecursive sequence. More precisely, we need to ensure that this sequence neverrepeats. We will present explicit formulas for this sequence as well as numericalresults strongly suggesting that this sequence does in fact never repeat.Assuming that this sequence never repeats, the “star” constructed in this sectionwould therefore be an example for a sequence of geodesic nets with a fixed numberof unbalanced vertices but an arbitrarily large number of balanced vertices.4.2. Construction Toolbox.
We will first build our “toolbox” to facilitate theconstruction of the geodesic net below.4.2.1.
Suspending.
Suspending is a process that adds an additional edge to a vertex v to change its imbalance. Method 4.2.1 (Single-hook suspension) . Consider a vertex v and another vertex P , called the hook . We suspend v from P by adding the edge vP . v P Method 4.2.2 (Two-hook suspension) . Consider a vertex v and two othervertices P , Q such that all three interior angles of the triangle ∆ P vQ are lessthan 120 ◦ . EODESIC NETS: SOME EXAMPLES AND OPEN PROBLEMS 17
There is a unique point F – called the Fermatpoint – inside the triangle ∆ P vQ such that theedges
P F , QF and vF form angles of 120 ◦ at F . It can be constructed as follows: • Let X be the third vertex of the uniqueequilateral triangle that has base P Q and that is lying outside the triangle∆
P vQ . • Let c be the unique circle defined bythe points P , Q and X . • Note that c and the segment Xv in-tersect at two points: X itself and oneother point. That other point is F . P QvX F
That this construction does indeed yield the Fermat point (also known as theToricelli point) is a result of classic Euclidean Geometry.We suspend v from P and Q by adding F and the edges P F , QF and vF .Note that now F is a degree three balanced vertex.The orange angle between Xv and the axis of symmetry of the equilateraltriangle will be denoted by ϕ later. Note that if ϕ = 0, then the picture issymmetric under reflection along vX .4.2.2. Winging.
Winging is a process that turns an unbalanced vertex into a bal-anced vertex.
Method 4.2.3 (Winging a degree 2 vertex) . Consideran unbalanced vertex v of degree 2 with α i being thelarger angle between the two incident edges, i.e. with180 ◦ < α i < ◦ . We can balance this vertex by“spreading wings” as follows: Extend the two incidentedges to the other side of the vertex, resulting in adegree 4 balanced vertex.If β i is the smaller of the two angles between the twonew edges (“wings”), then β i = 360 ◦ − α i . α i β i Method 4.2.4 (Winging a degree 3 vertex) . Consideran unbalanced vertex v of degree 3 such that the totalimbalance (i.e. the sum of the unit vectors parallel toan edge) is less than 2.We can balance this vertex by adding two edges in aunique way as follows: Since the imbalance is a vectorof length less than 2, there is one (and only one) way ofwriting its inverse as the sum of two unit vectors. Addthe two corresponding edges that balance the vertexin this way (these edges might coincide with existingedges). We arrive at a balanced vertex of degree 5. α i β i Note that this construction does not require the picture to be symmetric asin the sketch on the right. However, in the case that it is in fact symmetric,it is important to point out a special relationship: After winging, the picturewill remain symmetric and we also get the following angle relation: If wedenote the smaller of the two angles between the newly added wings by β i ,basic trigonometry yields β i = 2 · arccos(1 / − cos( α i / α i (cid:54) = 2 arccos(1 /
4) the two dashed edges will not coincide with already presentedges since then β i (cid:54) = α i .4.2.3. About algebraic angles.
Recall the following theorem based on Lindemann-Weierstraß:
Theorem 4.2.5.
If the angle α is algebraic (in radians), then cos( α ) and sin( α ) are transcendental. We will fix an angle α > ◦ which will be close to 240 ◦ , but so that α is infact algebraic (and therefore its sine and cosine are transcendental, a property thatwe will need below). We will choose α = 88 /
21 (rad) ≈ . ◦ , but of course anyother algebraic angle closer to 240 ◦ would also work.4.2.4. The parameters n and ϕ . The construction of the geodesic net G n ( ϕ ) relieson two parameters ϕ and n .We will start with an outer circle , that is fixed and doesn’t change under anyof the parameters. We the proceed and construct an inner circle whose deviationfrom the symmetric case is measured by the angle ϕ . This inner circle is the “zerothlayer” of the construction. We will then add a total of n layers, producing moreand more balanced vertices while keeping the number of unbalanced vertices fixed.4.2.5. Outer circle.
The outer circle is given by seven equiangularly distributedpoints on a circle. These seven vertices will be one half of the 14 unbalancedvertices of the resulting net.Note that the whole construction will be scaling invariant, so we can choose anarbitrary radius for the outer circle. We will fix the scale of the picture furtherbelow.Whenever we will use the process of suspending a vertex as defined above, thetwo hooks will be two neighbouring vertices on the outer circle.4.2.6.
Inner circle.
The inner circle is defined as follows: First, we fix α as spec-ified above. Note that this angle will not change under deviation later. Fix twoneighbouring points P and Q on the outer circle and let X be the third vertex ofthe equilateral triangle with base P Q that lies outside the outer circle (see figure4.2.1). Recall that we are provided a deviation angle ϕ ∈ ( − (cid:15), (cid:15) ). Consider thesegment OX (where O is the canter of the outer circle) and rotate this segmentaround X by ϕ There is a unique vertex v on this segment such that ∠ P vQ = α .This is one vertex of the inner circle. The other six vertices of the inner circle arethen provided by rotational symmetry (again, see figure 4.2.1). EODESIC NETS: SOME EXAMPLES AND OPEN PROBLEMS 19 α α XP Q
Figure 4.2.1.
Outer circle and inner circle (= zero-th layer), sym-metric case (left) and deviated case (right). Note that for fixed α ,there is a unique way to construct the inner circle from the outercircle once the deviation ϕ ∈ ( − (cid:15), (cid:15) ) is given. The reference pointfor deviation is given by X , which is the third point of the equi-lateral triangle that has two adjacent points P and Q on the outercircle as base. Note that the seven points on the inner circle arealways deviated by the same angle ϕ . In other words: with de-viation, there is no reflectional symmetry anymore, but rotationalsymmetry is maintained.We then connect the outer and inner circles by edges as depicted. To latersimplify calculations, we now scale the picture so that the radius of the inner circleis 1.4.3. The construction.
Overview of the construction.
The initial setup of outer and inner circle as described above is denoted as G ( ϕ ). It is trivially a geodesic net with 14unbalanced and no balanced vertices.We will now add layers to G ( ϕ ) to arrive at geodesic nets G ( ϕ ) , G ( ϕ ) . . . , G n ( ϕ )with the following properties: • Each G i ( ϕ ) has 14 unbalanced vertices. • The number of balanced vertices goes to infinity as i −→ ∞ .So by choosing n large enough, we get a geodesic net G n ( ϕ ) with 14 unbalancedvertices and N balanced vertices. There is, however, one important caveat: We want the geodesic net to only haveedges of weight one. As stated above: If we allowed for weighted edges, muchsimpler examples could be constructed.In light of that requirement, we will observe the following: • If we do not introduce deviation, i.e. if we fix ϕ = 0, we get highly symmet-ric geodesic nets G i (0), many edges of which will intersect non-transversally.This means that some edges would need to be represented using integerweights. • However, for small ϕ ∈ ( − (cid:15), (cid:15) ), we get geodesic nets G i ( ϕ ) with significantlyless symmetry and for which numerical results strongly suggest that alledges intersect transversally (if at all). So they are in fact nets with edgesof weight one.4.3.2. About the feasibility of the process and smooth dependence.
As we will see inthe constructive process below, there are certain requirements on the behaviour ofangles and lengths that are necessary to make this construction possible. We willproceed as follows: • We will first describe the construction, which will be the same for the non-deviated and the deviated case. This construction will use several suchrequirements. • We will then prove that all requirements are fulfilled for the non-deviatedcase ϕ = 0. • We will observe that each iteration of the construction smoothly dependson the previous one. • Since all our requirements turn out to be restrictions of angles and lengthsto open intervals and since the construction smoothly ( ⇒ continuously)depends on the initial setup, the requirements will therefore also be fulfilledfor small ϕ ∈ ( − (cid:15), (cid:15) ).4.3.3. Iterative process.
We denote the set of vertices on the outer circle by V − and on the inner circle by V and proceed to construct V i for i ≥ V i , each of which is a degree 2 vertex that is adjacent totwo vertices of V i − . Using the 14 connecting edges, we get a 14-gon whose verticesalternate between vertices of V i − and vertices of V i . For the interior angle α i at thevertices V i ( not at the vertices of V i − ), one of the following two cases can occur: α i > ◦ , called Case A; or α i < ◦ , called Case B (We justify α i (cid:54) = 180 ◦ for all i later). Case A: α i > ◦ . In this case, the vertices of V i are unbalanced vertices ofdegree 2 such that we can wing a degree vertex getting an angle β i = 360 ◦ − α i < ◦ as described above. Each wing will end as soon as it intersects with anotherwing. At those seven points of intersection, we fix the seven vertices of V i +1 .Proceed to the next iteration. Examples for Case A in figure 4.3.1 are the first two steps, namely the grey andred vertices.
EODESIC NETS: SOME EXAMPLES AND OPEN PROBLEMS 21
Figure 4.3.1.
First steps of the construction for ϕ = 0. Top left:
Outer vertices ( V − , black) and inner vertices ( V , grey) Top right:
The vertices of V (grey) have been winged and thewings meet at the new vertices of V (red) Bottom left:
The vertices of V (red) have been winged and thewings meet at the new vertices of V (green) Bottom right:
The vertices of V (green) first were suspended(note the edges from green to grey and the double weight on theouter edges) and then were winged. The wings meet at the newvertices of V (blue) After these first three steps , the seven vertices on the outercircle as well as the vertices of V are unbalanced. There are 21balanced vertices indicated. During the construction, we also getadditional “accidental” degree four balanced vertices at points ofintersection. For the next step , the dashed edges would be added to suspend the vertices of V (blue) and then each of them would be “winged”again. Case B: α i < ◦ . In this case, we will first add an outwards edge to eachvertex of V i using suspension. We distinguish two cases by the parity of i . Case B1, i is even: . In this case, by construction, each vertex v of V i is closeto a radial line through the origin and at the half-angle between two outer vertices P and Q (for deviation ϕ = 0, v is in fact on that radial line). Consider the triangle∆ P vQ . It is clear that ∠ vP Q, ∠ P Qv ≤ ◦ < ◦ . Furthermore note that v isinside the inner circle (we will prove this in lemma 4.3.4). Even if v were on theinner circle, we would have ∠ QvP < ◦ (by the choice of α > ◦ ). So since v is inside the inner circle, we still have (cid:93) QvP < ◦ . Consequently, we can do a two-hook suspension of v from the hooks P and Q as described above. And example for Case B1 in figure 4.3.1 is the third step, namely the greenvertices.
Case B2, i is odd: . In this case, by construction, each vertex v of V i is close toa radial line through the origin and one of the vertices P on the outer circle (again,for deviation ϕ = 0, v is on that radial line). We will do a one-hook suspension of v from P by adding their connecting edge. And example for Case B2 in figure 4.3.1 is the fourth step, namely the bluevertices.
After applying case B1 or B2 , each vertex of V i now is a degree 3 unbalancedvertex. We will prove below that it has imbalance of less than 2. Therefore, wecan wing this vertex of degree 3. Each wing will end as soon as it intersects withanother wing. At those seven points of intersection, we fix the seven vertices of V i +1 . Proceed to the next iteration.This describes the whole construction. An example of the non-deviated case( ϕ = 0) can be found in figure 4.3.2. We are left to show that the claims that makethis construction possible are actually true.4.3.4. Helpful lemmata.
The above construction implicitly uses several geometricfacts which we will prove in this section. It is interesting to note that parts of thefollowing lemma could be proven similarly for a construction starting with morethan 7 outer vertices, but fail for 6 vertices. More specifically, we prove α i > ◦ below, which would not be true if we started with 3, 4, 5, 6 vertices. This is thereason for the seemingly arbitrary choice of seven as the “magic number” of theconstruction.Note the following: Lemma 4.3.1.
The positions of all vertices depend smoothly on the deviation angle ϕ ∈ ( − (cid:15), (cid:15) ) .Proof. The outer circle never moves. The definition of the inner circle (which isthe layer V ) makes it clear that the position of the vertices of that layer dependsmoothly on ϕ .Since, to find all the other layers, we are using nothing but suspending andwinging as defined previously, we only need to check those processes. Assume theposition of the vertices up to V i depend smoothly on ϕ . EODESIC NETS: SOME EXAMPLES AND OPEN PROBLEMS 23
Figure 4.3.2. G of the non-deviated case. The 14 circled ver-tices are the only unbalanced vertices of this geodesic net. Notethat some edges have integer weights of more than 1. Those are theradial edges, as well as the edges of the outermost 14-gon. Afterintroducing deviation, these edges will split into weight-one-edges. • The angles of the incoming edges to V i from V i − only depend on the posi-tions of V i and V i − which depend smoothly on ϕ by induction hypothesis. • If one-hook suspension is necessary: P is on the outer circle, so it doesn’tchange under ϕ . Since the position of v depends smoothly on ϕ , so doesthe angle of the hooking edge. So the imbalance of v before winging willchange smoothly. • If two-hook suspension is necessary: P , Q and X don’t change under ϕ .Since the position of v depends smoothly on ϕ , so does the angle of thehooking edge. So the imbalance of v before winging will change smoothly.We now have established that the angles of all incoming edges to the vertices of V i and therefore the imbalance at the vertices of V i after possible suspension dependssmoothly on ϕ . Checking the two possibilities for winging, it is apparent that the α i − β i − x i − α i β i x i α i +1 β i +1 x i +1 Figure 4.3.3.
The interdependences of α i , β i and x i angles of the outgoing wings depend smoothly on the imbalance. Since the verticesof the next layer are defined to be the intersection of those wings, the positions ofthe next layer depend smoothly on ϕ . (cid:3) Note that all the following lemmas assert inequalities regarding angles and dis-tances. In light of the previous lemma, it is therefore enough to prove them for ϕ = 0. By smooth dependence (and therefore continuous dependence), they arethen still true for small ϕ ∈ ( − (cid:15), (cid:15) ).We will first prove the following technical lemma, the usefulness of which will beapparent later. Lemma 4.3.2.
Consider the angles α i and β i as the angle between the incom-ing edges and the angle between the outgoing edges during winging (see the figuresdescribing the winging process). We have:(a) α i (cid:54) = 180 ◦ , / for all i ≥ (b) ◦ < α i < ◦ for all i ≥ (c) ◦ < β i < ◦ for all i ≥ Proof.
As established, it is enough to consider the symmetric case ϕ = 0.Recall that α = 88 / ≈ . β ≈ . ◦ . The formulasfor β i depending on α i were derived above when winging was defined: β i = (cid:40) ◦ − α i α i > ◦ (winging of degree 2 vertex)2 · arccos(1 / − cos( α i / α i < ◦ (winging of degree 3 vertex)Furthermore, since the vertices of V i and the vertices of V i +1 form a 14-gon forwhich the interior angles at V i are β i (outgoing edges) and the interior angles at V i +1 are α i +1 (incoming edges), we have7 β i + 7 α i +1 = 12 · ◦ ⇔ α i +1 = 12 · ◦ − β i For a visualization of the interdependence of these sequences see figure 4.3.3.Note that after proving (a), it is indeed clear that we do not need to consider thecase α i = 180 ◦ . We proceed by induction. Note that (a) starts at i = 0 whereas(b) and (c) start at i = 1:(a) Recall that cos( α ) is not algebraic (by our initial choice of α , see above).We will prove the following fact which will imply the required result: cos( α i )is never algebraic for i ≥
0. The base case is given.
EODESIC NETS: SOME EXAMPLES AND OPEN PROBLEMS 25
To proceed with induction, first consider the case where α i > ◦ andtherefore α i +1 = 12 · ◦ / − ◦ + α i ⇒ cos( α i +1 ) = cos(12 · ◦ / − ◦ + α i ) ⇒ cos( α i +1 ) = cos(12 · ◦ / α i ) ⇒ cos( α i +1 ) = cos(12 · ◦ /
7) cos( α i ) − sin(12 · ◦ /
7) sin( α i ) ⇒ cos( α i +1 ) = cos(12 · ◦ /
7) cos( α i ) − sin(12 · ◦ / (cid:112) − cos ( α i )It follows that cos( α i +1 ) is algebraic if and only if cos( α i ) is algebraic.Similarly, if α i < ◦ , then α i +1 = 12 · ◦ / − · arccos(1 / − cos( α i / ⇒ cos( α i +1 / − · ◦ /
7) = 1 / − cos( α i / ⇒ cos( α i +1 /
2) cos(6 · ◦ /
7) + sin( α i +1 /
2) sin(6 · ◦ /
7) = 1 / − cos( α i / ⇒ cos( α i +1 /
2) cos(6 · ◦ /
7) + (cid:112) − cos ( α i +1 /
2) sin(6 · ◦ /
7) = 1 / − cos( α i / α i +1 /
2) is algebraic if and only if cos( α i ) is algebraic.But note that cos( α i +1 ) = 2 cos ( α i +1 / − α i +1 ) is algebraic if and only if cos( α i +1 /
2) is algebraic. Theclaim follows.(b) The base case for α can be verified by calculation. Now assume 120 ◦ <α i < ◦ and consider α i +1 . We have two cases:If α i > ◦ , then α i +1 = 2160 ◦ − ◦ + α i We can see that 120 ◦ < α i +1 < ◦ is true provided 180 ◦ < α i < ◦ .If α i < ◦ , then α i +1 = 2160 ◦ − · arccos(1 / − cos( α i / α i +1 is an increasing function of α i for 120 ◦ < α i < ◦ and that for α i = 120 ◦ we get α i +1 ≈ ◦ whereas for α i = 180 ◦ we get α i +1 ≈ ◦ . So 120 ◦ < α i +1 < ◦ is indeed the case.(c) The bounds for β i are immediate from the bounds on α i and the formulafor β i above. (cid:3) We can now proceed to prove facts that were relevant to our construction.
Lemma 4.3.3.
In the construction as defined above:(a) The incoming edges never meet at an angle of α i = 180 ◦ , i.e. we alwaysend up with case A or case B as described in the construction.(b) The total imbalance before winging, even after possible suspension, is alwaysless than , i.e. winging is always possible.(c) The outgoing edges produced by winging a degree 3 vertex never coincidewith the incoming edges. ◦ β i / x i x i +1 Law of sines: x i +1 sin β i / x i sin(180 ◦ − ◦ / − β i / Figure 4.3.4.
Finding the relation between x i and x i +1 . Proof.
It is again enough to consider the symmetric case ϕ = 0 since each of theasserted properties can be expressed as an inequality, so they remain true for small ϕ ∈ ( − (cid:15), (cid:15) ).(a) This is given explicitly in the previous lemma.(b) α i is the angle between incoming edges. Since 120 ◦ < α i < ◦ (seeprevious lemma), the imbalance produced by the two incoming edges isalways less than 1. Suspension adds an imbalance of at most 1. Thereforethe total imbalance is less than 2.(c) We are considering the symmetric case. As stated in the definition ofwinging, the edges would only coincide if α i = β i which requires α i =2 arccos(1 / (cid:3) Finally, we asserted that adding a Fermat point for two-hook suspension is alwayspossible if necessary. This assertion was based on the following fact:
Lemma 4.3.4.
All layers V i lie strictly inside the inner circle for i ≥ . Further-more, the radius of the layers goes to zero as i −→ ∞ .Proof. We will again only consider the symmetric case. By smooth dependence on ϕ , the claim follows for the deviated case.Recall that we scaled the construction so that the inner circle is at a radius of x = 1. During the construction as defined above, if we denote by x i the distanceof the vertices at the i -th step from the origin, the claim follows if we prove x i < i ≥ x i −→ i −→ ∞ . Note that in the symmetric case: x i +1 = x i f ( β i ) where f ( β i ) = sin β i / ◦ / − β i /
2) (see figure 4.3.4)where the formula for β i , which in itself depends on α i , can be found above.By brute force calculation (for α = 88 / • x i < ≤ i ≤ • x < . • ◦ < α < ◦ EODESIC NETS: SOME EXAMPLES AND OPEN PROBLEMS 27
We will now proceed to prove x i < i >
9, using the fact that the values of x i are going through loops, at the end of which x i will have decreased. This can beformalized by the following claim: Claim:
Let N ≥ ◦ < α N < ◦ . Then the following is true foreither (cid:96) = 8 or (cid:96) = 9: • ◦ < α N + (cid:96) < ◦ • x N + j ≤ . · x N for all j = 1 , , . . . , (cid:96) −
1, and • x N + (cid:96) < . · x N Note that the lemma follows from inductive application of this claim using N = 9as the base case since 1 . · . < α N + (cid:96) < ◦ since this is always true forany α i ( i ≥ • ◦ < α N < ◦ • Therefore, α N +1 = ◦ − ◦ + α N , implying 120 ◦ < α N +1 < ◦ . • Furthermore, the larger α N is, the larger α N +1 will be. • As long as α N + j < ◦ , we can write α N + j +1 = 2160 ◦ − / − cos( α N + j / (cid:124) (cid:123)(cid:122) (cid:125) =: g ( α N + j ) We can therefore rewrite the sequence α N +1 , α N +2 , α N +3 . . . as α N +1 , g ( α N +1 ) , g ( g ( α N +1 )) , . . . This continues until, for some (cid:96) , α N + (cid:96) = g ( (cid:96) − ( α N +1 ) > ◦ . In otherwords, the behaviour of α N , α N +1 , α N +2 , . . . can be summarized as follows:It starts at α N > ◦ , then α N +1 jumps below 180 ◦ and the sequence α N + i climbs back up until α N + (cid:96) > ◦ . • We already established that the larger α N is, the larger α N +1 will be. Also, g ( α ) is an increasing function for 120 ◦ < α < ◦ . Therefore, the larger α N is, the larger each α N + j will be for j = 1 , . . . , (cid:96) .We observe the following two cases: Case 1: ◦ ≤ α N < ◦ . Checking the extremal cases (183 ◦ and 190 ◦ ), wecan observe that either α N +8 > ◦ or α N +9 > ◦ . We pick (cid:96) = 8 , Case 2: ◦ < α N < ◦ . Again checking the extremal cases (183 ◦ and 183 ◦ ),we can observe that α N +9 will be the first angle above 180 ◦ . So we pick (cid:96) = 9.For both cases, note that x N + j +1 = x N + j f ( β N + j ). We can link the factor f ( β )to the angles α as follows • β = 2 · arccos(1 / − cos( α/ α . • f ( β ) = sin β/ ◦ / − β/ is an increasing function of β , since 120 ◦ < β < ◦ as established in a previous lemma. • Therefore, we can write f ( β ( α )) = h ( α ) and as long as we underesti-mate α , we overestimate h ( α ). We write x N + j = x N f ( β N ) f ( β N +1 ) f ( β N +2 ) · · · f ( β N + j − )= x N h ( α N ) h ( α N +1 ) h ( α N +2 ) · · · h ( α N + j − )= x N h ( α N ) h ( α N +1 ) h ( g ( α N +1 )) · · · h ( g ( j − ( α N +1 ))Recall that g is increasing and that h is decreasing, so as long as we underesti-mate α N (and therefore also α N +1 ), we overestimate x N + j .We can return to the two cases: Case 1 ◦ ≤ α N < ◦ . In this case we need 8 or 9 steps and α N is at least183 ◦ . By the above considerations, we can simply study the case α N = 183 ◦ . Forlarger starting values of α , the values of x can only be smaller. In that context,using the above equation: x N + j ≤ x N h (183 ◦ ) h (131 ◦ ) h ( g (131 ◦ )) · · · h ( g ( j − (131 ◦ ))Now calculations yield that h (183 ◦ ) h (131 ◦ ) h ( g (131 ◦ )) · · · h ( g ( j − (131 ◦ )) will beless than 1 . j = 1 , , . . . , , .
96 for j = 8 ,
9. So (cid:96) = 8 or (cid:96) = 9 has the properties stated in the claim.
Case 2 ◦ < α N < ◦ . In this case we need 9 steps and α N is at least180 ◦ . By the above considerations, we can simply study the case α N = 180 ◦ (as alimiting case, of course α = 180 ◦ never happens). For larger starting values of α ,the values of x can only be smaller. In that context, using the above equation: x N + j ≤ x N h (180 ◦ ) h (128 ◦ ) h ( g (128 ◦ )) · · · h ( g ( j − (128 ◦ ))Now calculations yield that h (180 ◦ ) h (128 ◦ ) h ( g (128 ◦ )) · · · h ( g ( j − (128 ◦ )) will beless than 1 . j = 1 , , . . . , , .
96 for j = 9. So (cid:96) = 9 has theproperties stated in the claim.It is noteworthy that a split like the one at 183 ◦ was necessary. In fact would wehave a starting value of α N = 180 ◦ but only 8 steps, the factor would be greaterthan 1. Hence the casework to show that this case doesn’t happen.This finishes the proof of the claim and the lemma. (cid:3) This concludes the proof that the construction works, both in the symmetriccase ϕ = 0 and for small deviation ϕ ∈ ( − (cid:15), (cid:15) ).4.4. Analyzing the non-deviated construction.
Note that our goal was toconstruct a sequence of nets G i (0) such that(a) There are 14 unbalanced vertices(b) The number of balanced vertices goes to infinity as i −→ ∞ .(c) Some edges might intersect non-transversally (i.e. overlap).The first observation is true as explained above: The only unbalanced verticesare the vertices on the outer ring as well as the seven vertices of the last layer thathas been added.Regarding the second observation, note that lemma 4.3.4 demonstrates that theradius of the layers V i approaches zero as i −→ ∞ . Therefore, increasing thenumber of layers increases the number of balanced vertices (otherwise, there would EODESIC NETS: SOME EXAMPLES AND OPEN PROBLEMS 29 have to be a cyclical phenomenon in the construction, contradicting the fact thatthe radius goes to zero).We now turn towards the third observation, which is what makes the introductionof deviation necessary. Note that the edges of the geodesic net can be categorizedas follows: • Whenever a new layer is added through the process of winging , this adds14 edges from V i − to V i to the net. We call those the layer-connectingedges . This includes the very first 14 edges to set up the outer circle andinner circle as seen in figure 4.2.1. • Whenever we suspend a vertex from a single hook, a single edge is beingadded. We call it the suspension edge . • Whenever we suspend a vertex from two hooks, three edges are being added:two edges from two vertices of the outer circle (the hooks ) to the Fermatpoint, as well as one edge from the Fermat point to the vertex that is beingsuspended – we again call these suspension edges .This now raises the question: Which edges can and will intersect non-transversallyon G i (0) (i.e. if ϕ = 0), either partially or in their entirety? We can observe (seealso figure 4.3.2): • Note that the Fermat point used in the process of suspension is the same foreach layer in the symmetric case. Therefore, two of the three suspensionedges involving the same two hooks are the same from the hooks to theFermat point, every single time these hooks are being used. The thirdsuspension edge then always starts at the same Fermat point and continuesradially to the vertex that is being suspended. • Similarly, if we do a one-hook suspension, the suspension edge is alwaysradial, so all the suspension edges going to the same hook have an overlay. • Layer-connecting edges, on the other hand, can only intersect any otheredge transversally: The Fermat points are outside the inner circle andtherefore never meet the layer-connecting edges. Regarding the radial sus-pension edges, note that layer-connecting edges are never radial, so they arealways transversal to radial edges. Besides that, note that layer-connectingedges always start and end on the boundary of a 260 ◦ /
14 disk sector of theconstruction. So they could only be completely identical to another layer-connecting edge or intersect them transversally (if at all). For the sake ofcontradiction, assume that any layer-connecting edge would coincide withanother layer-connecting edge. This would imply that the layer producedby them must be the same, including the incoming edges. Therefore, layerswould repeat. This would contradict the phenomenon described in lemma4.3.4, namely that the radius of the layers must converge to zero.As established previously, everything depends smoothly on the deviation ϕ . There-fore, any small deviation will maintain transversality where it already is given.Deviation will, however, have to make sure that we split up suspension edges.4.5. Edges under deviation.
As just established above, we only need to be con-cerned with the edges produced using the method of suspension . This section servesto support the following claim:
P QX Rα Figure 4.4.1.
Note the dotted equilateral triangle
XP Q as wellas the unique circular arc given by P , R and Q . Denote by ϕ theangle OXR (where O is the origin/center). The symmetric caseis given by ϕ = 0. To introduce deviation , we would slightlyincrease ϕ , and get a new position for R . This does not changethe value of α (since we go along the circular arc). The samedeviation is done at all seven grey points. This means that thestar remains rotationally symmetric under a rotation by 360 ◦ / EODESIC NETS: SOME EXAMPLES AND OPEN PROBLEMS 31
Whereas for ϕ = 0 , the geodesic nets G i (0) are highly symmetrical and manysuspension edges overlap, for nonzero ϕ ∈ ( − (cid:15), (cid:15) ) , all suspension edges of G i ( ϕ ) intersect transversally (if at all). The sequence of suspension angles ϕ i . To demonstrate this, we will studythe sequence of suspension angles, which is defined as follows, based on the twotypes of suspension that we employ:
Definition 4.5.1 (Suspension angle for one-hook sus-pensions, layers V i for i odd) . Whenever we do a one-hook suspension for a layer V i , we are connecting avertex v to the closest vertex on the outer circle.Denote the center of the outer circle by O and the hookby P . Then we define the suspension angle ϕ i to be theangle ∠ OP v . For clarification, consider the figure onthe right giving a positive suspension angle.Note that ϕ i depends only on the deviation ϕ (which isthe only free parameter of our construction) and that ϕ i (0) = 0 for all layers.We can define this suspension angle for all odd layers,even though in some cases we don’t need to suspend avertex (if case A occurs). P Definition 4.5.2 (Suspension angle for two-hook sus-pensions, layers V i for i even) . Whenever we do a two-hook suspension for a layer V i , we are connecting avertex v to the closest two vertices on the outer circle P and Q through the Fermat point of ∆ P vQ , see thefigure when defining two-hook suspension above).As before, we denote by X the third vertex of the equi-lateral triangle ∆ P QX used for the construction of theFermat point. Let O be the center of the outer circle.Then we define the suspension angle ϕ i to be the an-gle ∠ OXv . For clarification, consider the figure on theright showing a positive suspension angle.Note that ϕ i depends only on the deviation ϕ (which isthe only free parameter of our construction) and that ϕ i (0) = 0 for all layers. Most importantly, for i = 0(the initial layer, aka the inner circle) the suspensionangle is ϕ ( ϕ ) = ϕ .We can define this suspension angle for all even layers,even though in some cases we don’t need to suspend avertex. XP Q
With this definition, we can now make the following observations:
Fact 4.5.3.
Consider ϕ ∈ ( − (cid:15), (cid:15) ) a geodesic net G n ( ϕ ) as constructed above withlayers V , . . . , V n . Then • As established before, only suspension edges could overlap/intersect non-transversally. • For all one-hook suspensions (odd layers), only the edges going from vertices v i , v k of two different layers to the same hook P could overlap. But as longas ϕ i (cid:54) = ϕ k , they will not do so (this is apparent from the figure in thedefinition above). • For all two-hook suspensions (even layers), only the edges suspending ver-tices v i , v k of two different layers from the same two hooks P and Q couldoverlap. But as long as ϕ i (cid:54) = ϕ k , they will not do so (see figure 4.5.1).Note the following important observation: Fact 4.5.4. G n ( ϕ ) consists of finitely many layers, therefore ϕ , ϕ , . . . , ϕ n is a finite sequence.Based on definitions 4.5.1, 4.5.2 and fact 4.5.3, we arrive at the following lemma: Lemma 4.5.5.
If for any fixed ϕ ∈ ( − (cid:15), (cid:15) ) , the sequence ϕ i ( ϕ ) never repeats itself,all edges of the resulting geodesic net G n ( ϕ ) intersect transversally (if at all). Inother words, there are no edges with weight other than one. Note that, in fact it would be enough if the ϕ i are different for the same parity(since even and odd layers never have suspension edges in common).Based on symmetry (see also figure 4.3.2), we can observe XP QvF XP QvF
Figure 4.5.1.
Construction of the Fermat point F to suspend v i ∈ V i from P and Q . ϕ i is the angle between the axis of symmetryof P QX and the segment vX . Observe: Whenever ϕ i (the markedsuspension angle) is different, the segment P F is at a differentangle. The same is true for QF and vF . So we only need toestablish that ϕ i is different at every layer and this implies thatnone of the suspension edges overlap. EODESIC NETS: SOME EXAMPLES AND OPEN PROBLEMS 33
Fact 4.5.6. ϕ i (0) = 0 for all i .Also, since we established smooth dependence of the construction on the devia-tion ϕ before and since this is in fact the only free parameter, we can consider thederivative of ϕ i ( ϕ ). We make the following conjecture: Conjecture 4.5.7. ϕ (cid:48) i (0) is a sequence that never repeats itself. Keeping in mind that ϕ i is a finite sequence, the previous fact and conjecture(i.e. same value at 0, but different derivatives) would then imply the following: Conjecture 4.5.8.
For small nonzero ϕ ∈ ( − (cid:15), (cid:15) ) , the sequence ϕ i ( ϕ ) never repeatsitself. This implies that G n ( ϕ ) is a geodesic net for which all edges have weight one. So this G n ( ϕ ) would in fact fulfill all required conditions:(a) There are 14 unbalanced vertices.(b) For n large enough, we can achieve an arbitrarily large number of balancedvertices.(c) All edges have weight one.4.6. Studying the sequence ϕ (cid:48) i (0) . So we are left with conjecture 4.5.7. For theremainder, we will consider statements we can make about the sequence ϕ (cid:48) i (0).We will do the following: • We will provide an explicit recursive formula for ϕ (cid:48) i (0). • We will present numerical results that strongly suggest that conjecture 4.5.7is true.4.7.
An explicit formula for ϕ (cid:48) i (0) . In the following, all derivatives will be withrespect to ϕ . First note, that ϕ = ϕ and therefore obviously ϕ (cid:48) (0) = 1. We willnow find a recursive formula for ϕ (cid:48) i (0). The crucial question is how ϕ (cid:48) i +1 (0) dependson ϕ (cid:48) i (0).Figure 4.7.1 shows the two cases for i even and i odd. Note that if ϕ = 0,then the picture is symmetric along a horizontal reflection. In both cases, we get ahexagon made of two quadrilaterals.In either case, by the angle sum in the lower quadrilateral: τ i + ϕ i + γ i + ψ i +1 + ( µ i − ϕ i +1 ) = 360 ◦ Note that τ i and µ i are constants that don’t change under ϕ . Therefore, differen-tiation by ϕ leads to: ϕ (cid:48) i + γ (cid:48) i + ψ (cid:48) i +1 − ϕ (cid:48) i +1 = 0 ⇒ ϕ (cid:48) i +1 = ϕ (cid:48) i + γ (cid:48) i + ψ (cid:48) i +1 In the appendix, we will establish the following relationships at ϕ = 0. ϕ (cid:48) = 1 ψ (cid:48) = 12 − sin( π/ − α ) ϕ (cid:48) i +1 = ϕ (cid:48) i + γ (cid:48) i + ψ (cid:48) i +1 ψ (cid:48) i +1 = b i γ (cid:48) i + a i ϕ (cid:48) i γ (cid:48) i = c i ψ (cid:48) i The formulas for the coefficients make use of x i and α i , which were defined previ-ously: a i = − tan α i +1 (cid:32) sin( π/ σ i +1 ) sin( π/ σ i +1 + τ i )sin τ i + − sin( τ i +2 π/ σ i +1 )2 sin τ i tan( α i +1 − σ i +1 ) (cid:33) b i = − tan α i +1 / α i +1 / − σ i +1 ) c i = − α i > ◦ α i / − α i / α i < ◦ τ i = (cid:40) π/ i even29 π/ i odd σ i +1 = arctan sin π/ sin α / x i sin( π/ α / − cos π/ i evenarctan sin π/ sin α / x i sin( π/ α /
2) sin( π/ π/ π/ − cos π/ i odd ϕ i ϕ i ϕ i +1 τ i + ϕ i = π + ϕ i τ i − ϕ i = π − ϕ i γ i δ i ψ i +1 α i +1 µ i − ϕ i +1 ϕ i ϕ i ϕ i +1 τ i + ϕ i = π + ϕ i τ i − ϕ i = π − ϕ i γ i δ i ψ i +1 α i +1 µ i − ϕ i +1 Figure 4.7.1.
We want to relate ϕ (cid:48) i +1 to ϕ (cid:48) i . The upper picturesare considering the case i even, the lower pictures are consideringthe case i odd. All green angles are 60 ◦ . µ is a constant angle. Alldashed lines are stationary, all dotted and solid lines vary over achange of the deviation ϕ . The picture on the right extracts theblue hexagon out of the left picture. Note that in this example, ϕ i is positive whereas ϕ i +1 is negative EODESIC NETS: SOME EXAMPLES AND OPEN PROBLEMS 35 i l oga r i t h m i c s c a l e ϕ i (0) Figure 4.8.1.
The first 100 values of ϕ (cid:48) i (0) on a logarithmic scale,calculated with α = 88 /
21. All points for i even are marked inorange. All points for i odd are marked in green.4.8. Numerical consideration of the sequence ϕ (cid:48) i (0) . While the formulas aboveare all explicit, they are arguably not very “handy” which makes understandingtheir behaviour a challenging task. Recall that all we need is that ϕ (cid:48) i (0) never re-peats. This would then imply that a small deviation from G n (0) to G n ( ϕ ) wouldin fact split up all edges as required.For a better understanding, we used MATLAB to compute the first items ofthe sequence. Figure 4.8.1 shows the first 100 elements of the sequence ϕ (cid:48) i (0), ona logarithmic scale. These calculations lead to the following observations, which inturn support conjecture 4.5.7, saying that ϕ (cid:48) i (0) doesn’t repeat: • The magnitude of the sequence grows exponentially. • The sequence seems to be generally increasing (i.e. increasing with a smallnumber of exceptions) • Since it is enough if the sequence differs for all even i and for all odd i , weget additional “leeway”.In fact, numerical evidence suggests that the first 100 elements of the sequencedo not repeat. We computed the first 100 elements of ϕ (cid:48) i (0) with MATLAB usingvariable precision arithmetic, using between 10 and 100 significant digits. Thefollowing result remained stable under variable precision:min i (cid:54) = j | ϕ (cid:48) i (0) − ϕ (cid:48) j (0) | ≈ . ϕ (cid:48) (0) and ϕ (cid:48) (0). ϕ π − α µλ θψ α ba a Figure 5.1.1.
Dependence of ψ on ϕ Appendix: Finding the formulas for the sequence ϕ (cid:48) i (0)Consider figure 4.7.1 and the following formula we derived previously: ϕ (cid:48) i +1 = ϕ (cid:48) i + γ (cid:48) i + ψ (cid:48) i +1 In this appendix, we intend to do the following: • Find the starting values of ϕ (cid:48) (0) and ψ (cid:48) (0). • Establish that α (cid:48) i (0) = 0 for all i . • Derive a formula for γ (cid:48) i (0) in terms of ψ (cid:48) i (0). • Derive a formula for ψ (cid:48) i +1 (0) in terms of ψ (cid:48) i (0) and ϕ (cid:48) i (0).5.1. Starting values.
Since ϕ is defined to be the suspension angle for the innercircle, which is the zeroth level (compare definition 4.5.2 and compare with theinitial definition of the inner circle), ϕ = ϕ and therefore ϕ (cid:48) (0) = 1To find ψ (cid:48) (0), consider figure 5.1.1, more specifically the isosceles triangle withtwo sides of length a . Using angle sums in triangles: ψ = ( π − ( π − λ − θ )) / − λ = θ/ − λ/ ⇒ ψ (cid:48) = θ (cid:48) / − λ (cid:48) / Finding λ (cid:48) . The law of sines, applied to the two triangles that share the commonside of length b but have two different sides of length a gives ussin λ sin ϕ = ba = sin µ sin π/ µ ⇒ sin λ = 2 sin µ sin ϕ ⇒ cos λ λ (cid:48) = 2 sin µ cos ϕ At ϕ = 0 we also have λ = 0 and therefore λ (cid:48) (0) = 2 sin µ = 2 sin( π − π/ − (2 π − α )) = 2 sin( π/ − α ). Finding θ (cid:48) . Using the angle sum in triangles, we get θ + µ + π/ − ϕ = π andtherefore θ (cid:48) = 1. EODESIC NETS: SOME EXAMPLES AND OPEN PROBLEMS 37
Combining these we arrive at ψ (cid:48) (0) = − sin( π/ − α ).5.2. α (cid:48) i (0) and γ (cid:48) i (0) . Recall that α i is the interior angle at the vertices V i of the14-gon formed by V i − and V i . It is one of the two angles between the incomingedges at the vertex of a layer, before winging (see also methods 4.2.3 and 4.2.4 aswell as the formulas in the proof of 4.3.2). Like all other angles, each α i = α i ( ϕ )is a smooth function of the deviation angle ϕ . Figures 4.7.1 and 5.2.1 show how γ is one of the angles between an outgoing edge and the suspension edge, whereas ψ is one of the angles between an incoming edge and the suspension edge (even if wedo not suspend, as is the case for a degree 4 vertex where α > ◦ , we can stillconsider the angle compared to a hypothetical suspension edge. Lemma 5.2.1.
For all i = 0 , , , . . . , we have the following relationships at ϕ = 0 (a) α (cid:48) i = 0 (b) γ (cid:48) i = c i ψ (cid:48) i where c i = − α i > ◦ α i / − α i / α i < ◦ It is important to emphasize that these relationships between derivatives onlyhold at ϕ = 0, which is, however, enough for us. Proof. (a) For the base case, note that the inner circle is chosen precisely toensure that α = 88 /
21 for any choice of ϕ . So α ( ϕ ) is constant and thebase case follows.For the induction step, recall the following formulas (see proof of 4.3.2)for the symmetric case ϕ = 0: α i +1 = 12 · ◦ − β i β i = (cid:40) ◦ − α i α i > ◦ (winging of degree 2 vertex)2 · arccos(1 / − cos( α i / α i < ◦ (winging of degree 3 vertex)It is worth checking which of these formulas still apply in the deviated case ϕ (cid:54) = 0. α i ψ i γ i δ i β i α i ψ i γ i δ i β i Figure 5.2.1.
The angle relationships at a vertex of V i in the caseof α i > ◦ (left, winging a degree 2 vertex) and in the case of α i < ◦ (right, winging a degree 3 vertex) • The relation α i +1 = · ◦ − β i remains unchanged, since the for-mula is based on the angle sum in the 14-gons during construction.Therefore α (cid:48) i +1 (0) = − β (cid:48) i (0). • The formula for β i in the case of α i > ◦ also applies to the non-symmetric case (see method 4 . . β (cid:48) i (0) = − α (cid:48) i (0) = 0by induction hypothesis and we are done. • The formula for β i in the case of α i < ◦ , however, cannot be usedfor the asymmetric situation (as explained in method 4.2.4, it onlyworks for the symmetric case).So we are left to show that β (cid:48) i (0) = 0 assuming that α (cid:48) i (0) = 0 and α i < ◦ , but we can’t use the given formula.Instead, lets consider the right of figure 5.2.1, depicting this case. Weran rotate the vertex as depicted so that one edge is pointing to the right.Note that in the symmetric case, i.e. when ϕ = 0, the picture is symmetricunder reflection along the horizontal axis. Generally, though, this is notthe case.The vertex is balanced, so we have[1 ,
0] + [cos γ i , sin γ i ] + [cos δ i , − sin δ i ] + [cos ψ i , − sin ψ i ] + [cos( ψ i + α i ) , − sin( ψ i + α i )] = [0 , ⇔ (cid:40) γ i + cos δ i + cos ψ i + cos( ψ i + α i ) = 0sin γ i − sin δ i − sin ψ i − sin( ψ i + α i ) = 0We can derive everything with respect to ϕ and arrive at − γ (cid:48) i sin γ i − δ (cid:48) i sin δ i − ψ (cid:48) i sin ψ i − ( ψ (cid:48) i + α (cid:48) i ) sin( ψ i + α i ) = 0 γ (cid:48) i cos γ i − δ (cid:48) i cos δ i − ψ (cid:48) i cos ψ i − ( ψ (cid:48) i + α (cid:48) i ) cos( ψ i + α i ) = 0We are concerned with the derivatives at ϕ = 0. As pointed out, the pictureis symmetric in that case and we get γ i = δ i as well as ψ i + α i = 2 π − ψ i .By induction hypothesis, we also have α (cid:48) i = 0. So we can simplify to − γ (cid:48) i sin γ i − δ (cid:48) i sin γ i − ψ (cid:48) i sin ψ i − ψ (cid:48) i sin(2 π − ψ i ) = 0 γ (cid:48) i cos γ i − δ (cid:48) i cos γ i − ψ (cid:48) i cos ψ i − ψ (cid:48) i cos(2 π − ψ i ) = 0We can simplify further to − γ (cid:48) i sin γ i − δ (cid:48) i sin γ i − ψ (cid:48) i sin ψ i + ψ (cid:48) i sin ψ i = 0 γ (cid:48) i cos γ i − δ (cid:48) i cos γ i − ψ (cid:48) i cos ψ i − ψ (cid:48) i cos ψ i = 0Note that γ i can’t be a multiple of π at ϕ = 0 since that would imply β i = 0or β i = 2 π which is never the case as established previously. So sin γ i (cid:54) = 0and the two equations do in fact simplify to γ (cid:48) i + δ (cid:48) i = 0( γ (cid:48) i − δ (cid:48) i ) cos γ i = 2 ψ (cid:48) i cos ψ i As is clear from figure 5.2.1, β i = 2 π − ( γ i + δ i ). Therefore β (cid:48) i = − ( γ (cid:48) i + δ (cid:48) i ) = 0as required.(b) If α i > ◦ consider the left of figure 5.2.1 from which it is clear that γ i = 180 ◦ − ψ i . It follows that γ (cid:48) i = − ψ (cid:48) i as stated (this relationship, infact, would be true for any ϕ ∈ ( − (cid:15), (cid:15) ), not just for ϕ = 0). EODESIC NETS: SOME EXAMPLES AND OPEN PROBLEMS 39 If α i < ◦ , we have just established the following at ϕ = 0 γ (cid:48) i + δ (cid:48) i = 0 and ( γ (cid:48) i − δ (cid:48) i ) cos γ i = 2 ψ (cid:48) i cos ψ i ⇒ γ (cid:48) i cos γ i = 2 ψ (cid:48) i cos ψ i ⇒ γ (cid:48) i = cos ψ i cos γ i ψ (cid:48) i Furthermore, again due to symmetries at ϕ = 0cos γ i = cos( π − β i /
2) = − cos( β i /
2) = − cos (cid:18) / − cos( α i / (cid:19) = − (1 / − cos α i / ψ i = cos( π − α i /
2) = − cos α i / γ (cid:48) i = c i ψ (cid:48) i with c i = 2 cos α i / − α i / (cid:3) τ i and σ i . τ i is the “reference angle” from which the suspension angle ϕ i ismeasured. Consider the 14-gon formed by the two outer sides of each of the sevenequilateral triangles built on the outer circle. The interior angle at seven corners is π/
3, the interior angle at the other seven corners is 29 π/
21. These are the valuesfor 2 · τ i even and odd respectively.The angle σ i +1 is depicted in figure 5.4.1. It changes under deviation, but at ϕ = 0 it can be directly calculated from the sequence of x i . If i is even, considerfigure 5.4.2. Note that σ i +1 is one of the angles in a triangle with side x i and angle π/
7. Since σ i +1 is the angle at a vertex of the outer circle , we know one more sideof the triangle. Recall that the inner circle is fixed at radius 1 and consider figure4.2.1, giving the other side as sin α / π/ α / . Using the law of sines, we get:sin σ i +1 = x i sin( σ i +1 + π/
7) sin( π/ α / α / i is odd, a similar argument yieldssin σ i +1 = x i sin( σ i +1 + π/
7) sin( π/ α / α / π/ π/ π/ σ i is at radius sin α / π/ α /
2) sin( π/ π/ π/ .Since σ i +1 < ◦ , we can solve both equations for σ i +1 and get σ i +1 = arctan sin π/ sin α / x i sin( π/ α / − cos π/ i evenarctan sin π/ sin α / x i sin( π/ α /
2) sin( π/ π/ π/ − cos π/ i odd5.4. The formula for ψ (cid:48) i (0) . We are left to establish the following
Lemma 5.4.1.
For all i = 0 , , , . . . at ϕ = 0 , we have ψ (cid:48) i +1 = b i γ (cid:48) i + a i ϕ (cid:48) i where b i = − tan α i +1 tan( α i +1 − σ i +1 ) a i = − tan α i +1 (cid:32) sin( π/ σ i +1 ) sin( π/ σ i +1 + τ i )sin τ i + − sin( τ i +2 π/ σ i +1 )2 sin τ i tan( α i +1 − σ i +1 ) (cid:33) τ i + ϕ i = π + ϕ i τ i − ϕ i = π − ϕ i ξ i γ i ζ i δ i ψ i +1 α i +1 σ i +1 ‘ i L i KKL i c i d i Figure 5.4.1.
A detailed overview of a part of figure 4.7.1
Proof.
Consider the two cases in figure 4.7.1. We will cover the first case in detail( i even). The other case, as given the lower of the two figures, can be deducedin the same way (the pictures only differ in the size of the angles, the underlyingsetup of polygons is the same). For this first case, figure 5.4.1 gives a more detailedoverview. ψ i is a particular angle in the given hexagon. The angles κ i and τ i don’t changeunder ϕ and neither does K (it is the length of the sides of the equilateral trianglesover the outer circle). Note the following: If we consider these angles and lengths • The lengths L i • The angles ϕ i , γ i , δ i then there is a one-to-one correspondence (within an open set) between tuples( L i , ϕ i , γ i , δ i ) and hexagons.Also note that the pictures is symmetric at ϕ = 0, implying γ i (0) = δ i (0) and ϕ i (0) = 0. Since these quantities uniquely define the hexagon and since ψ i +1 is anangle defined by the hexagon, we arrive at ψ (cid:48) i +1 = ∂ψ i +1 ∂L i L (cid:48) i + ∂ψ i +1 ∂γ i γ (cid:48) i + ∂ψ i +1 ∂δ i δ (cid:48) i + ∂ψ i +1 ∂ϕ i ϕ (cid:48) i As usual, we consider these derivatives at ϕ = 0. The proof of the previous lemmaestablished that in that case γ (cid:48) i = − δ (cid:48) i . Also note that in the symmetric picture, achange in L i will not affect ψ i . Therefore ∂ψ i +1 ∂L i at ϕ = 0. Combining all these, we EODESIC NETS: SOME EXAMPLES AND OPEN PROBLEMS 41 arrive at ψ (cid:48) i +1 = (cid:18) ∂ψ i +1 ∂γ i − ∂ψ i +1 ∂δ i (cid:19) γ (cid:48) i + ∂ψ i +1 ∂ϕ i ϕ (cid:48) i So we are left to show that(a) b i = (cid:16) ∂ψ i +1 ∂γ i − ∂ψ i +1 ∂δ i (cid:17) = − tan αi +12 tan( αi +12 − σ i +1 ) (b) a i = ∂ψ i +1 ∂ϕ i = − tan αi +12 sin τ i (cid:16) sin( π/ σ i +1 ) sin( π/ σ i +1 + τ i ) + − sin τ i +sin( τ i +2 π/ σ i +1 )2 tan( αi +12 − σ i +1 ) (cid:17) To do so, we will make extensive use of trigonometric identities and laws.The law of sines provides c i sin ξ i sin ψ i +1 = (cid:96) i = d i sin ζ i sin(4 π − γ i − δ i − ( τ i + ϕ i ) − ( τ i − ϕ i ) − (2 π − κ i ) − ψ i +1 ) ⇒ c i sin ξ i sin(4 π − γ − δ − τ i − (2 π − κ i ) − ψ i +1 ) = d i sin ζ i sin ψ i +1 (a) Deriving with respect to γ i yieldsderivatives w.r.t. γ i c i cos ξ i ξ (cid:48) i sin(4 π − γ i − δ i − τ i − (2 π − κ i ) − ψ i +1 )+ c i sin ξ i cos(4 π − γ i − δ i − τ i − (2 π − κ i ) − ψ i +1 ) ( − − ψ (cid:48) i +1 )= d i sin ζ i cos ψ i +1 ψ (cid:48) i +1 We are considering values at ϕ = 0. Due to symmetry ξ i = ζ i c i = d i ψ i +1 = 4 π − γ − δ − τ i − (2 π − κ i ) − ψ i +1 = π − α i +1 ξ i = γ i + θ where θ is an angle that doesn’t change under γ . Therefore(since we are currently considering derivatives with respect to γ i ) we have ξ (cid:48) i = 1. Combining all these, we arrive at:derivatives w.r.t. γ i (cid:26) c i cos ξ i sin α i +1 − c i sin ξ i cos α i +1 ( − − ψ (cid:48) i +1 )= − c i sin ξ i cos α i +1 ψ (cid:48) i +1 This can be simplified toderivatives w.r.t. γ i (cid:8) cot ξ i tan α i +1 − ( − − ψ (cid:48) i +1 ) = − ψ (cid:48) i +1 So we finally arrive at ∂ψ i +1 ∂γ i = − (cid:16) cot ξ i tan α i +1 (cid:17) Note that at ϕ = 0, we have π = ξ i + σ i +1 + ψ i +1 = ξ i + σ i +1 + π − α i +1 and therefore cot ξ i = cot( α i +1 − σ i +1 ), so we can rewrite this as ∂ψ i +1 ∂γ i = − (cid:18) tan α i +1 tan( α i +1 − σ i +1 ) + 1 (cid:19) Very similar deductions based on derivatives with respect to δ i allow us toarrive at ∂ψ i +1 ∂δ i = 12 (cid:18) tan α i +1 tan( α i +1 − σ i +1 ) − (cid:19) The claim for b i follows. (b) We return to the identity based on the law of sines from above, but willnow consider derivatives with respect to ϕ i . Note that this time, c i , d i , ξ i and ζ i vary whereas γ i and δ i are constant.derivatives w.r.t. ϕ i c (cid:48) i sin ξ i sin(4 π − γ i − δ i − τ i − (2 π − κ i ) − ψ i +1 )+ c i cos ξ i ξ (cid:48) i sin(4 π − γ i − δ i − τ i − (2 π − κ i ) − ψ i +1 )+ c i sin ξ i cos(4 π − γ i − δ i − τ i − (2 π − κ i ) − ψ i +1 ) ( − ψ (cid:48) i +1 )= d (cid:48) i sin ζ i sin ψ i +1 + d i cos ζ i ζ (cid:48) i sin ψ i +1 + d i sin ζ i cos ψ i +1 ψ (cid:48) i +1 We are considering values at ϕ = 0. Due to symmetry ξ i = ζ i c i = d i c (cid:48) i = − d (cid:48) i ξ (cid:48) i = − ζ (cid:48) i ψ i +1 = 4 π − γ i − δ i − τ i − (2 π − κ i ) − ψ i +1 = π − α i +1 ϕ i (cid:26) c (cid:48) i sin ξ i sin α i +1 + c i cos ξ i ξ (cid:48) i sin α i +1 + c i sin ξ i cos α i +1 ψ (cid:48) i +1 = − c (cid:48) i sin ξ i sin α i +1 − c i cos ξ i ξ (cid:48) i sin α i +1 − c i sin ξ i cos α i +1 ψ (cid:48) i +1 And thereforederivatives w.r.t. ϕ i (cid:8) c (cid:48) i sin ξ i sin α i +1 + c i cos ξ i ξ (cid:48) i sin α i +1 + c i sin ξ i cos α i +1 ψ (cid:48) i +1 = 0This can be solved forderivatives w.r.t. ϕ i (cid:110) ψ (cid:48) i +1 = − tan α i +1 (cid:16) c (cid:48) i c i + cot ξ i ξ (cid:48) i (cid:17) We now have to find c (cid:48) i /c i and ξ (cid:48) i . We start with c (cid:48) i /c i . Recall that we arestill considering derivatives with respect to ϕ i . The law of cosines gives usderivatives w.r.t. ϕ i c i = K + L i − KL i cos( τ i + ϕ i ) ⇒ c i c (cid:48) i = 2 KL i sin( τ i + ϕ i ) ⇒ c (cid:48) i c i = Kc i L i c i sin( τ i + ϕ i )At ϕ = 0, we have ϕ i = 0 Kc i = sin( γ i − ξ i )sin τ i L i c i = sin( π − ( γ i − ξ i ) − τ i )sin τ i = sin( γ i − ξ i + τ i )sin τ i We finish with finding ξ (cid:48) i (still as a derivative with respect to ϕ i ). We invokethe law of tangents for the triangle with sides c i , K , L i . L i − KL i + K = tan( ( π − ( τ i + ϕ i ) − ( γ i − ξ i ) − ( γ i − ξ i ))tan( ( π − ( τ i + ϕ i ) − ( γ i − ξ i ) + ( γ i − ξ i )) = tan( ( π − τ i − ϕ i − γ i − ξ i )))tan( ( π − τ i − ϕ i ))Note that the left-hand side doesn’t change under a change of ϕ i . So if wederive (and subsequently multiply by the denominator), we getderivatives w.r.t. ϕ i (cid:26) ( ( π − τ i − ϕ i − γ i − ξ i )))( − ξ (cid:48) i ) tan( ( π − τ i − ϕ i ))+ tan( ( π − τ i − ϕ i − γ i − ξ i ))) sec ( ( π − τ i − ϕ i ))Elementary trigonometric identities lead us toderivatives w.r.t. ϕ i (cid:26) ξ (cid:48) i = 12 − sin( π − τ i − ϕ i − γ i − ξ i ))2 sin( π − τ i − ϕ i )At ϕ = 0, we have ϕ i = 0. So we getderivatives w.r.t. ϕ i (cid:26) ξ (cid:48) i = 12 − sin( τ i + 2( γ i − ξ i ))2 sin τ i EODESIC NETS: SOME EXAMPLES AND OPEN PROBLEMS 43 ξ i γ i ψ i +1 α i +1 σ i +1 π/ π/ x i Figure 5.4.2.
A figure showing the relationship between anglesat ϕ = 0 for i even.We now have explicit formulas for c (cid:48) i /c i as well as ξ (cid:48) i . Combining the aboveresults and using the following two identities at ϕ = 0 (see figure 5.4.2): ξ i = π − ψ i +1 − σ i +1 = α i +1 − σ i +1 π − ( γ i − ξ i ) + π/ σ i +1 = π ⇒ γ i − ξ i = π/ σ i +1 we can finally write ∂ψ i +1 ∂ϕ i = − tan α i +1 (cid:32) sin( π/ σ i +1 ) sin( π/ σ i +1 + τ i )sin τ i + − sin( τ i +2 π/ σ i +1 )2 sin τ i tan( α i +1 − σ i +1 ) (cid:33) (cid:3) Acknowledgements
This research has been partially supported by A. Nabutovsky’s NSERC Dis-covery grant and F. Parsch’s Vanier Canada Graduate Scholarship. The authorswould like to thank Adam Stinchcombe for valuable help with variable precisionarithmetic in MATLAB.
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Department of Mathematics, University of Toronto, 40 St. George Street, Toronto,ON M5S 2E4, Canada
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