Dual-cyclic polytopes of convex planar polygons with fixed vertex angles
DDual-cyclic polytopes of convex planarpolygons with fixed vertex angles
Lyle Ramshaw ∗ and James B. SaxeFebruary 20, 2020 Abstract
If we fix the angles at the vertices of a convex n -gon in the plane, thelengths of its n edges must satisfy two linear constraints in order for itto close up. If we also require unit perimeter, we get vectors of n edgelengths that form a convex polytope of dimension n −
3, each facet of whichconsists of those n -gons in which the length of a particular edge has fallento zero. Bavard and Ghys suggest requiring unit area instead, which givesthem a hyperbolic polytope. Those two polytopes are combinatoriallyequivalent, so either is fine for our purposes.Such a fixed-angles polytope is combinatorially richer when the anglesare well balanced. We say that fixed external angles are majority dominant when every consecutive string of more than half of them sums to morethan π . When n is odd, we show that the fixed-angles polytope for anymajority-dominant angles is dual to the cyclic polytope C n − ( n ); so ithas as many vertices as are possible for any polytope of dimension n − n facets. To extend that result to even n , we require that the anglesalso have dipole tie-breaking : None of the n strings of length n/ π , and the n/ π overlap as much aspossible, all containing a particular angle.Fixing the vertex angles is uncommon, however; people more often fixthe edge lengths. That is harder, in part because fixed-lengths n -gonsmay not be convex, but mostly because fixing the lengths constrains theangles nonlinearly — so the resulting moduli spaces, called polygon spaces ,are curved. Consider the polygon space for some edge lengths that sumto 2 π . Using Schwarz–Christoffel maps, Kapovich and Millson show thatthe subset of that polygon space in which the n -gons are convex andtraversed counterclockwise is homeomorphic to the fixed-angles polytopeabove, for those same fixed values. Each such subset is thus a topologicalpolytope; and it is dual cyclic whenever the fixed lengths are majoritydominant and, for even n , have dipole tie-breaking. ∗ //orcid.org/0000-0002-9113-1473MSC-class: 52B11 (Primary) 51M20, 52B05 (Secondary) a r X i v : . [ m a t h . M G ] F e b Introduction
The external angles at the vertices of a convex planar n -gon must sum to 2 π . Forconvenience, we guarantee that sum via rescaling: Given n positive numbers s through s n with sum S ·· = s + · · · + s n , we consider those n -gons in the Euclideanplane whose k th vertices have external angle 2 πs k /S , for all k . We view the s k cyclically, with s n followed by s ; so we group them into a necklace, written insquare brackets: s = [ s , . . . , s n ] = [ s , . . . , s n , s ].In order for such a fixed-angles n -gon to close up in x and in y , its edgelengths must satisfy two linear constraints; let L s be the subspace of R n wherethose constraints hold, which has dimension n −
2. Since edge lengths must benonnegative, the vectors of edge lengths form the cone K s ·· = L s ∩ R n ≥ .The n -gons along any ray in the cone K s differ only by magnification, and wecan select one representative from each ray by requiring either unit perimeter orunit area, as we discuss in Section 2. If we require unit perimeter, our vectorsof edge lengths form a Euclidean convex polytope of dimension n −
3. Wedenote it A nE [ s ], the letter “ A ” reminding us that the vertex Angles are hereheld fixed. Bavard and Ghys [1] suggest requiring unit area instead, which givesthem a hyperbolic polytope A nH [ s ]. Since the polytopes A nE [ s ] and A nH [ s ] arecombinatorially equivalent, we can, for our combinatorial purposes, denote the fixed-angles polytope for the angles in s simply as A n [ s ], meaning either one.The Euclidean is more elementary, but the hyperbolic is more elegant.The total turn from one edge of a fixed-angles n -gon to another is controlledby the sum of a string of consecutive entries of s ; and it is key how thosesums compare to S/
2. When such a sum equals S/
2, the n -gon has a pair ofantiparallel edges — a case that requires special treatment. The fixed-anglespolytope has a vertex for each way of partitioning s into three strings whosesums are less than S/
2. The three edges that constitute the gaps between thosestrings can extend to form a unit-perimeter (or unit-area) triangle in which theother n − n − n −
3. And thenumber of such simple vertices turns out to be maximized when the angles in s are well balanced in an appropriate sense. Recall [13] that a cyclic polytope C d ( m ), for 1 ≤ d < m , is the convex hullof m points along the moment curve in R d , the curve given parametricallyby t (cid:55)→ ( t, t , . . . , t d ). The combinatorial structure of that convex hull doesn’tdepend upon which m points we choose, as long as they are distinct. The cyclicpolytope C d ( m ) is neighborly , meaning that every set of at most d/ k , as many k -faces as are possiblefor any d -polytope with m vertices. That property has no standard name, butlet’s call it being max-faced . 2he cyclic polytope C d ( m ) is particularly straightforward when m − d istiny. The polytope C d ( d + 1), for example, is just the d -simplex ∆ d , which isself-dual. A standard exercise [13, p. 24] shows that the polytope C d ( d + 2) is(∆ (cid:98) d/ (cid:99) × ∆ (cid:100) d/ (cid:101) ) ∆ , the dual of the Cartesian product of two simplices of nearlyequal dimension. We shed some light on the next-more-complicated case byshowing that A d +3 E [ s ] ∆ , the dual of a fixed-lengths polytope, is an instance of C d ( d + 3) for any d + 3 fixed angles s that are appropriately well balanced.While the polytopes C d ( d + 1) and C d ( d + 2) have lots of symmetries, thelevel of symmetry that remains in C d ( m ) when m ≥ d + 3 depends on theparity of the dimension d [8]. When d is even, the polytope C d ( m ) has 2 m combinatorial automorphisms that dihedrally permute its m vertices. When d is odd, on the other hand, it has just four: We can swap the first and lastvertices, reverse the order of all of the vertices, or both, or neither. For d ofeither parity, there are instances of C d ( m ) whose combinatorial automorphismsall come from Euclidean symmetries. In Section 3, we study how the combinatorial structure of the fixed-angles poly-tope A n [ s ] changes as the angles in s vary. The structure changes each timethat the sum of some substring of s passes through a tie with the sum of thecomplementary substring, that tie generating a pair of antiparallel edges. Ofthe structures before and after that change, the richer is the one in which thelonger substring has the larger sum, so that s is better balanced.We say that the angles in s are majority dominant when every consecutivesubstring of length more than n/ S/
2. The complementarystrings, which are minorities, then sum to less than S/
2. Majority dominancesays nothing about the strings of length precisely n/
2, which exist only when n is even. When n is odd and s is majority dominant, we show that the fixed-angles polytope is dual cyclic by constructing, in Section 4, an explicit dualitybetween the face lattices of A n [ s ] and C n − ( n ).When s is majority dominant but n is even, the dual polytope A n [ s ] ∆ maynot be cyclic, but it is neighborly. If there are no substrings of s that sumto S/
2, it is also simplicial, which implies that it is max-faced, by the McMullenupper-bound theorem [13, p. 254]. We show in Section 5 that A n [ s ] ∆ , beyondbeing max-faced, is precisely an instance of C n − ( n ) if we require that s alsohave dipole tie-breaking , meaning that, of its n substrings of length n/
2, the n/ S/
2. The others thencontain the opposite entry and sum to less than S/ n -gons, whose fixed angles areperfectly balanced? Since the sequence (1 , . . . ,
1) is majority dominant, thefixed-angles polytope A n [1 , . . . ,
1] for equiangular n -gons is dual cyclic when n is odd, with A n [1 , . . . , ∆ being an instance of C n − ( n ). When n is even, The authors thank G¨unter Ziegler for pointing out to them, in a 2017 email, that thefixed-angles polytope A [1 , . . . ,
1] for equiangular heptagons is dual to C (7). A n [1 , . . . , ∆ continues to have symmetries that permute its n vertices dihedrally; it thus can’t be dual cyclic, since C n − ( n ) then has onlyfour combinatorial automorphisms. Instead, the sequence (1 , . . . ,
1) acquiressubstrings that sum to S/ n is even; so the polytope A n [1 , . . . , ∆ , whilestill neighborly, is not simplicial and is hence neither max-faced nor cyclic. Fixing the edge lengths of polygons is more popular than fixing their vertexangles. Fixed edge lengths arise, for example, when analyzing a cyclic planarlinkage with rigid links and revolute joints, one of the many kinds of linkagesstudied in robotics [3]. Fixing the edge lengths leads to harder problems thanfixing the vertex angles, both because the resulting n -gons may not be convexand, more importantly, because the constraints that fixed lengths impose on theangles are nonlinear, so the resulting moduli spaces are curved.For now, we impose convexity as a side constraint. Call an n -gon ccw-convex when it is convex and traversed counterclockwise, so its external angles arenonnegative. Given a necklace s ·· = [ s , . . . , s n ] of positive numbers, we considerthose ccw-convex planar n -gons whose k th edge has length s k , for all k ; andwe say that two n -gons have the same shape when they differ by translation orrotation. The resulting moduli space of shapes is a topological polytope, as wediscuss next; so we call it the fixed-lengths-convex top-polytope for s . We denoteit L nc [ s ], where the “ L ” reminds us that the edge Lengths are here held fixed,and the subscript c means ccw-convex.Kapovich and Millson [9] use Schwarz–Christoffel maps to transform eachccw-convex n -gon whose edge lengths are given by s into an n -gon whose externalangles are proportional to s (this n -gon also being ccw-convex). As we review inSection 6, they thereby construct a homeomorphism h s : L nc [ s ] → A n [ s ] from thefixed-lengths-convex top-polytope L nc [ s ] to the fixed-angles polytope A n [ s ]. Thespace L nc [ s ] is thus homeomorphic to a polytope by a preferred homeomorphism,which makes it a topological polytope . By our earlier results, it is dual to thecyclic polytope C n − ( n ) whenever the fixed lengths s are majority dominantand, if n is even, also have dipole tie-breaking.When s has no substrings with the same sums as their complements, both thefixed-lengths-convex top-polytope L nc [ s ] and the fixed-angles polytope A n [ s ] aresmooth manifolds with corners. And it seems likely that they are actually dif-feomorphic, as smooth manifolds with corners. But we close our final Section 6by warning that the Kapovich–Millson homeomorphism h s : L nc [ s ] → A n [ s ] isnot a diffeomorphism, even in those well-behaved cases. In this paper, we study our polytopes and our top-polytopes one at a time. Butthey are often glued together into larger spaces.In the fixed-angles polytope A n [ . . . , p, q, . . . ], consider the facet where theedge joining the vertices with angles p and q has shrunk to a point. Those4ertices then coalesce into a supervertex with angle p + q ; so that facet is a copyof the polytope A n − [ . . . , p + q, . . . ]. The polytope A n [ . . . , q, p, . . . ], in which p and q have swapped places, also has a copy of A n − [ . . . , p + q, . . . ] as a facet;and we can glue A n [ . . . , p, q, . . . ] to A n [ . . . , q, p, . . . ] along those matching facets.Given n positive numbers that sum to 2 π , suppose that we assemble theminto necklaces in all ( n − n -gons whosevertices have our n specified external angles, in any order. The researchers whodo such gluing use the hyperbolic polytopes, since their more elegant geometrieslead to cone-manifolds that may actually be orbifolds or manifolds [5]. The fixed-lengths world typically develops in the opposite order. Rather thanconstructing the fixed-lengths-convex top-polytopes and gluing them together,people instead construct the big space at the outset. The polygon space [4, 6, 9]for the length vector s is the moduli space of shapes of planar n -gons whose k th edge has length s k , for all k , with no requirement that they be ccw-convex.We denote that space simply as L n [ s ], with no subscript of “ c ”. The polygonspace L n [ s ] is a compact, orientable smooth manifold of dimension n −
3, withthe possible exception of some isolated quadratic singular points. Only as anafter-thought do people partition this manifold, based on the cyclic order of thecomplex phases of the n -gon’s edges, into ( n − We now start our work in earnest by fixing the vertex angles of polygons in theEuclidean plane. Given a necklace s ·· = [ s , . . . , s n ] of n positive numbers withsum S ·· = s + · · · + s n , we define an ˆ s -gon to be an n -gon whose k th vertexhas external angle 2 πs k /S . The hat accent indicates that it is the vertex anglesthat are fixed. When we later use the s k as the fixed lengths of the edges of an n -gon, we will call the result an ¯ s -gon , with a bar accent.While the angle at the k th vertex of an ˆ s -gon is the quotient 2 πs k /S , we referto s k itself as the shangle of that vertex, where “shangle” abbreviates “share ofthe angle”. So the external angles are proportional to the shangles.Each edge of an ˆ s -gon constitutes a gap between elements of the shanglenecklace s . We denote by ε k the edge that connects the vertices with shangles s k − and s k . We describe the shape of an ˆ s -gon as the vector (cid:96)(cid:96)(cid:96) = ( (cid:96) , . . . , (cid:96) n ),where (cid:96) k is the length of the edge ε k . Closure in x and in y imposes two linearconstraints, limiting the vector (cid:96)(cid:96)(cid:96) to a linear subspace L s ⊂ R n of dimension Also, they often want to identify each fixed-angles n -gon with its reflection; so theyassemble their n fixed angles, not into into ( n − n − / The choice of how to offset the numbering of the edges from that of the vertices seemsunavoidably arbitrary. When n is odd, it works well to let k th edge be the one opposite the k th vertex. Indeed, this is quite standard for triangles, with the edges a , b , and c being oppositethe vertices A , B , and C . But n is not always odd. ° ° ° ° ° Figure 1: The fixed-angles polytope A E [ q ] for q = [1 , , , ,
5] is a quadrilateral.Rescaling q to sum to 360 degrees gives [24 , , , , q -gons shown, which all have unit perimeter, are at the four vertices ofthe quadrilateral A E [ q ], the midpoints of its four edges, and the centroid of itsfour vertices (which differs from the centroid of its interior). n −
2. The further constraint that (cid:96) k ≥ k limits the shapes of ˆ s -gonsto the cone K s ·· = L s ∩ R n ≥ .We next select one representative from each ray in K s of similar ˆ s -gons. Oneoption selects, from each ray, the ˆ s -gon that has unit perimeter. The resultinglength vectors form a Euclidean convex polytope that we denote A nE [ s ], lyingin the unit-perimeter ( n − (cid:80) k (cid:96) k = 1. As a more sophisticatedalternative, Bavard and Ghys [1] select, as their representative, the ˆ s -gon withunit area. The functional on the ( n − L s that measures area is aquadratic form with signature (1 , n − L s formsa hyperboloid of two sheets, each sheet of which provides a Minkowski modelfor real hyperbolic ( n − R n ≥ cuts out, fromone of those sheets, a hyperbolic convex polytope A nH [ s ]. We call the polytopes A nE [ s ] and A nH [ s ] the fixed-angles polytopes for the shangle necklace s .As an example, Figure 1 shows the Euclidean fixed-angles polytope A E [ q ]whose shangle necklace is q = [1 , , , , n − q -gon between the vertices with shangles 4 and 5 (and hence with angles 96 ◦ and 120 ◦ ) — the brown edges drawn as the bases in Figure 1 — can’t shrink toa point while maintaining unit perimeter.As we move around in the quadrilateral A E [ q ], the lengths of the ˆ q -gon6dges vary affinely. The length of a ˆ q -gon’s red edge varies linearly with itsdistance away from the red side of A E [ q ], the side along which the red edgeshave shrunk to points; and similarly for green, orange, and blue. As for brown,the quadrilateral A E [ q ] is rotated in Figure 1 so that the length of a ˆ q -gon’sbrown base varies affinely with its height on the page; but the horizontal linewhere that length falls to 0 passes well below the entire quadrilateral A E [ q ]. The structure of the fixed-angles polytope A nE [ s ] depends upon the sums ofconsecutive substrings of the shangle necklace s . We refer to the sum of sucha substring as its weight , as distinguished from its length. Given a necklace s = [ s , . . . , s n ] with total weight s + · · · + s n = S , we call a substring of s either light , tied , or heavy according as its weight is less than, equal to, or greaterthan S/
2. If any substring of s is tied, then its complement is also tied, andthey form a substring tie in s . The width of that tie, which we denote w , is theminimum of the lengths of the two complementary substrings; so w ≤ n/ ε k of the ˆ s -gon shrinks to a point, the vertices at its twoends, which have shangles s k − and s k , coalesce into a supervertex of combinedshangle s k − + s k . This can’t happen when s k − + s k > S/
2, that is, whenthe string ( s k − , s k ) is heavy, since external angles of convex polygons can’texceed π . Restricting the edge ε k to have length 0 constrains us to the subset A n − E [ s (cid:48) ] ⊆ A nE [ s ], where s (cid:48) denotes s with its separate shangles s k − and s k replaced by their sum. That subset is empty when s k − + s k > S/
2, is thesingle point that constitutes all of A nE [ s ] when s k − + s k = S/
2, and is otherwisea facet of A nE [ s ]. More generally, the edges ε i through ε j , for i ≤ j , can shrinkto points simultaneously only when the string ( s i − , . . . , s j ) isn’t heavy.A typical vertex of the fixed-angles polytope A nE [ s ] arises from some way topartition the shangle necklace s into three light substrings. The three edges inthe gaps between those substrings can extend to form a triangular ˆ s -gon of unitperimeter, with the other n − A nE [ s ] is thus a simple vertex, which we call a trigon vertex .Each substring tie in the shangle necklace s generates an atypical vertex.The two edges in the gaps between the tied substrings are antiparallel, since thetotal turn involved in going from each to the other is π . Those two edges canthus grow to length while all of the other edges shrink to length 0, producingan ˆ s -gon that is a digon of unit perimeter. We call the resulting vertex of thefixed-angles polytope A nE [ s ] a digon vertex .If a necklace s is free of substring ties, then all of the vertices of A nE [ s ] aretrigon vertices, so A nE [ s ] is a simple polytope. A digon vertex may or may notbe simple. To explore that question, we investigate the vertex figures of thevertices of A nE [ s ]. 7 .2 The vertex figures The vertex figure of any simple vertex is a simplex, and we can see this concretelyfor a trigon vertex of A nE [ s ]. We cut off a trigon vertex by constraining thelengths of its three long edges to sum to 1 − (cid:15) , for some small (cid:15) . The other n − (cid:15) , and they can partition that (cid:15) among themselves however they like, closure then being achieved by suitableadjustments of the lengths of the long edges. So the vertex figure is a ∆ n − .Now consider a digon vertex that results from a substring tie of width w .Suppose that we constrain the perpendicular distance between the two antipar-allel edges to be some small (cid:15) . The remaining n − w − n − w −
1. The edges in each chain must travelan overall perpendicular distance of (cid:15) , and they can partition that travel amongthemselves however they like. The structure of the left chain thus correspondsto a point in the simplex ∆ w − and that of the right chain to a point in ∆ n − w − .So the overall vertex figure is the product ∆ w − × ∆ n − w − .The product ∆ w − × ∆ n − w − is empty when w = 1, since ∆ − = ∅ ; and itis the simplex ∆ n − when w = 2, since ∆ is a single point. Once w ≥ n − w ≥ n − n − A nE [ s ] arising from a tie of width w ≥ What classes arise for fixed-angles polytopes? The polytope A nE [ s ] is empty whenmax( s , . . . , s n ) > S/
2, because no convex n -gon can have an external anglethat exceeds π . It is a single point when max( s , . . . , s n ) = S/
2. The shanglenecklace s then has a substring tie of width w = 1, and the polytope A nE [ s ] isthe isolated digon vertex arising from that tie. When max( s , . . . , s n ) < S/ π . Given any circle, we can constructan ˆ s -gon whose edges are all tangent to that circle; such a polygon, by theway, is called tangential . And we can rescale our tangential ˆ s -gon to have unitperimeter. Since all of its edges have positive length, a small enough ball aroundthat ˆ s -gon in the unit-perimeter ( n − (cid:80) k (cid:96) k = 1 will lie in the interior of A nE [ s ], which is thus a polytope of the full dimension n − A nE [ s ] has dimension n −
3, all of its trigon vertices are simple, andany digon vertices arising from ties of width 2 are also simple. For example,replacing the necklace q = [1 , , , ,
5] in Figure 1 with q (cid:48) = [2 , , , ,
5] wouldgive a tie of width 2 because 2 + 4 + 3 = 4 + 5. The horizontal line where thebrown base falls to length zero would then have risen to just touch the lowest,orange-to-green vertex of the fixed-angles quadrilateral A E [ q (cid:48) ]; but that vertexwould remain simple.A tie of width w ≥
3, however, gives a digon vertex that is nonsimple,lying on n − n −
3. The first examples of this arisewhen n = 6 and w = 3. Consider, for example, the fixed-angles polyhedron8 E [1 , . . . ,
1] for equiangular hexagons. It is combinatorially a bipyramid overa triangle — geometrically, an equilateral triangle with cube-corners on bothsides. Each pair of antiparallel hexagon edges generates a vertex of the triangle,and four faces of A E [1 , . . . ,
1] meet at that vertex as at a vertex of an octahedron.In summary:
Proposition 1.
Let s = [ s , . . . , s n ] be a necklace of positive numbers and let S = s + · · · + s n be their sum. The Euclidean fixed-angles polytope A nE [ s ] • is empty when max( s , . . . , s n ) > S/ ; • is an isolated point when max( s , . . . , s n ) = S/ , that point being the digonvertex arising from the resulting substring tie of width w = 1 ; • and has the full dimension of n − when max( s , . . . , s n ) < S/ .In the third case, its trigon vertices are all simple, as are any digon verticesarising from ties of width . But a digon vertex, if any, arising from a tie ofwidth w ≥ is nonsimple, having ∆ w − × ∆ n − w − as its vertex figure. We focus on the Euclidean fixed-angles polytope A nE [ s ] in this paper, rather thanon the Bavard–Ghys [1] hyperbolic A nH [ s ], since the Euclidean polytopes aremore elementary. The detailed geometry of A nH [ s ] is prettier than that of A nE [ s ],however. Indeed, each polytope A nH [ s ] is a k -truncated ( n − -orthoscheme , forsome k with 0 ≤ k ≤ Also, there is an elegant formula for the dihedralangle between two facet hyperplanes of A nH [ s ] that are not orthogonal, based onthe cross ratio of the slopes of four of the ˆ s -gon’s edges (proven by Bavard andGhys [1] and nicely explained by Fillastre [5]).A warning about digon vertices in the hyperbolic case: We can get a digonwith unit area only as a limit, with the two antiparallel edges growing to beinfinitely long while all of the other edges shrink to points. A digon vertex of A nH [ s ] is thus an ideal point, located out on the Cayley absolute. That vertex iseither isolated, simple, or nonsimple according as w = 1, w = 2, or w ≥
3, justas in the Euclidean case; but all digon vertices in the hyperbolic case are ideal.As one example of the geometric elegance of the hyperbolic option, considerthe fixed-angles polytope A [ q ]. The Euclidean version A E [ q ] in Figure 1 isnothing special; but its hyperbolic analog A H [ q ] is a Lambert quadrilateral,with three right angles (aka a singly truncated -orthoscheme [5]). The oneangle that isn’t right is at the lowest, orange-to-green vertex, the vertex thatwould have been cut off by a brown edge, were there such an edge.As a second example, we mentioned that A E [1 , . . . , A H [1 , . . . , Truncating a vertex of a polytope typically replaces that vertex with a new facet, with anew normal vector. In a k -truncated orthoscheme, however, k of the hyperplanes that wouldotherwise define facets are so far away from the polytope that their facets don’t arise. A H [1 , . . . ,
1] is orthogonal to the three that it meets along an edge and istangent at infinity to the other two, each of which it touches just at a vertex. Consider a shangle necklace s that has a substring tie, say of width w . Wecan perturb s to eliminate that tie in either of two directions. How do thoseperturbations affect the structure of the fixed-angles polytope A nE [ s ]?A substring tie of width w in s gives each ˆ s -gon a pair of antiparallel edgesthat are connected, say, at their left end, by a chain of w − n − w −
1. If all of the edges in either chainshrink to points, then all of the edges in both chains must so shrink, the twopaired edges must have length , and we are at the associated digon vertex.Suppose that we now perturb s by slightly increasing the shangles in thetied substring of length w . The total turn on the left from one of the pairededges to the other will then exceed π , so it will no longer be possible for all ofthe edges in the left chain to shrink to points simultaneously. That will still bepossible, though, for the n − w − F , ofthe perturbed polytope. The face F has dimension ( n − − ( n − w −
1) = w − π . For perturbations that are small enough, however,adding to that external angle the fixed external angle at the other end of eitherof the paired edges will give a sum that exceeds π ; so neither of the paired edgescan shrink to points. It follows that the face F , viewed in its own right as thefixed-angles polytope of this simplified ( w + 1)-gon, has only w − F is combinatorially a ∆ w − .In a similar way, if we perturb s by slightly increasing the shangles in thesubstring of length n − w , the edges in the right chain can no longer shrink topoints simultaneously; and having those in the left chain shrink simultaneouslylands us on a face of the perturbed polytope that is combinatorially a ∆ n − w − .Thus, perturbing s so as to break the tie causes the digon vertex to expand intoeither a ∆ w − or a ∆ n − w − .If w = n/
2, the two directions in which we can perturb are symmetric. If w < n/
2, the tie already implies a certain imbalance in s . Perturbing so as toexacerbate that imbalance expands the digon vertex into a ∆ w − , while perturb-ing to ameliorate gives us a ∆ n − w − . Shangle necklaces that are better balancedthus generate fixed-angles polytopes that are numerically more complicated.For some examples of this, Figure 2 depicts how the combinatorial structureof the polyhedron A E [1 , , , , , σ ] changes as σ increases from 4 to 12, mostlysimplifying as the necklace becomes more imbalanced: Thurston [12, p. 515] discusses this hyperbolic polyhedron in a related hexagonal context.
1, 3, 1, 4, 2, 4 ] [
1, 3, 1, 4, 2, 5 ] [
1, 3, 1, 4, 2, 6 ] [
1, 3, 1, 4, 2, 7 ][
1, 3, 1, 4, 2, 8 ] [
1, 3, 1, 4, 2, 9 ] [
1, 3, 1, 4, 2, 10 ] [
1, 3, 1, 4, 2, 11 ] ( empty ) [
1, 3, 1, 4, 2, 12 ] Figure 2: The combinatorial structures of the fixed-angles polyhedra A E [1 , , , , , σ ] for integral σ from 4 to 12. The four digon vertices are circled.At σ = 5, a red face of dimension 1 becomes a new face of dimension 1; at σ = 7and at σ = 9, green and magenta faces of dimension 2 become dimension 0; andat σ = 11, the cyan face of dimension 3 becomes dimension − A E [1 , , , , ,
4] has the combinatorial structure of a cube.5: When σ = 5, the substring tie of width w = 3 between (3 , ,
4) and(2 , ,
1) produces a digon vertex where four face-planes concur. As σ passes through 5, the red edge of the cube shrinks to that digon vertexand reemerges as a new edge, but connected differently. A 1-face reemergesas a 1-face because n − w − w − A E [1 , , , , ,
6] can be thought of as a tetrahedron withtwo of its four vertices truncated. This polyhedron is also C (6) ∆ , thedual of the convex hull of six points along a twisted cubic.7: When σ = 7, the substring tie of width 2 between (1 , , ,
4) and (2 , σ passes through 7, the greentriangular face shrinks to that vertex and then reemerges as a trigon ver-tex; so a 2-face becomes a 0-face.8: The polyhedron A E [1 , , , , ,
8] is a tetrahedron with one vertex trun-cated.9: When σ = 9, the substring tie of width 2 between (3 , , ,
2) and (9 , σ passes through 9, the magentatriangular face of C (6) ∆ shrinks to that digon vertex and emerges as atrigon vertex.10: The polyhedron A E [1 , , , , ,
10] is a tetrahedron.111: When σ = 11, the substring tie of width 1 between (1 , , , ,
2) and (11)produces a digon vertex that is an isolated point. As σ passes through 11,the entire tetrahedron shrinks to that vertex and disappears, a 3-facebecoming a ( − A E [1 , , , , ,
12] is empty. n is odd The fixed-angles polytopes A nE [ s ] are combinatorially richer when the entries ofthe necklace s are well balanced. When n is odd, the richest of all are the dualsof cyclic polytopes. Given a shangle necklace s of length n , we call a substring of s short , diametral ,or long according as its length is less than, equal to, or greater than n/
2. We calla shangle necklace majority dominant when all of its short substrings are lightor, equivalently, when all of its long substrings are heavy. Majority dominancedoesn’t say anything about the weights of the diametral substrings, which ariseonly when n is even. Any necklace of the form [1 ± (cid:15), , . . . ,
1] for 0 ≤ (cid:15) < , , M, , , , M ] for large M .For n of either parity, majority dominance has a consequence worth noting: Lemma 2.
When the necklace s is majority dominant, any set of at most ( n − / of the facets of the fixed-angles polytope A nE [ s ] always intersect in anonempty face. Stating that more concisely, the ( n − -dimensional polytope A nE [ s ] ∆ is neighborly.Proof. A facet of A nE [ s ] is the subset where some edge has shrunk to a point.The only thing that can prevent some set of edges from shrinking to pointssimultaneously is when that set includes a string of consecutive edges whosesimultaneous shrinking would produce a supervertex of shangle more than S/ n/ n − / n − / n is odd. Let n ≥ tour to be a length-3 necklace of distinct elementsof Z /n Z . Given some tour [ p, q, r ], suppose that we travel around Z /n Z in theincreasing direction by stepping from p to q , on to r , and then around to p again. We classify a tour as once-around or twice-around , according as those12teps take us once or twice around Z /n Z . In a twice-around tour, each stepfrom one element to another passes over the third on the way.There are lots of ways to write down any tour as an ordered triple of integers( i, j, k ). We can replace i by i + kn for any k without changing the correspondingelement of Z /n Z ; and we can choose any one of the three cyclic shifts ( i, j, k ),( j, k, i ), or ( k, i, j ). Among all of those alternatives, each tour has a unique normal form ( i, j, k ), in which i , j , and k all lie in [1 .. n ] and in which i isthe smallest of the three. The normal form ( i, j, k ) of a once-around tour has1 ≤ i < j < k ≤ n , while that of a twice-around tour has 1 ≤ i < k < j ≤ n .Let m ·· = ( n − /
2, so that n = 2 m + 1 with m ≥
1. We define four sets ofintegers, each of cardinality m : S ·· = { , , . . . , m } L ·· = { m + 1 , m + 2 , . . . , m } ,O ·· = { , , , . . . , n − } , and E ·· = { , , , . . . , n − } . Any nonzero element of Z /n Z is congruent, modulo n , either to an element of S or to an element of L , so we call it either small or large . It is also congruenteither to an element of O or of E , so we call it odd or even .It is easy to see that − S = L , − O = E , 2 S = E , and 2 L = O . Since themultiplicative inverse of 2 in the ring Z /n Z is − m = (1 − n ) /
2, we also have( − m ) E = S and ( − m ) O = L , which implies mE = L and mO = S .The three steps of a tour [ p, q, r ] are the three nonzero differences q − p , r − q ,and p − r . We call a tour small when its three steps are all small elements of Z /n Z ; and we call it odd when its three steps are all odd. Since three elementsof S always sum to less than 2 n , every small tour goes around Z /n Z once. Threeelements of O can’t sum to 2 n either, since their sum is odd; so every odd touralso goes around Z /n Z once. Lemma 3.
The map [ p, q, r ] (cid:55)→ [ − p, − q, − r ] is a bijection from the set ofsmall tours to the set of odd tours. Its inverse is the map [ p, q, r ] (cid:55)→ [ mp, mq, mr ] .Proof. Multiplication by − S to O , while its inverse, which is multipli-cation by m , takes O back to S . And both of those operations distribute overthe subtraction that computes a difference.It follows that there just as many small tours as odd tours. You mightenjoy verifying that there are (cid:0) n +13 (cid:1) of each. That expression also counts, in aregular n -gon, the number of inscribed triangles that contain the n -gon’s center,since the two orientations of any inscribed triangle give us two tours, and thattriangle contains the center just when one of those tours is small. Lemma 4.
When n ≥ is odd and the shangle necklace s is majority dominant,three edges ε i , ε j , and ε k of an ˆ s -gon with ≤ i < j < k ≤ n can be extendedto form a trigon just when the triple ( i, j, k ) is the normal form of a small tour. roof. The three substrings into which the shangle necklace s is cut by the edges ε i , ε j , and ε k are( s i , . . . , s j − ) , ( s j , . . . , s k − ) , and ( s k , . . . , s n , s , . . . , s i − ) , of lengths j − i , k − j , and ( n + i ) − k . Those edges can be extended to forma trigon just when those three substrings are all light. When the necklace s ismajority dominant, that happens just when those substrings are all short, thatis, when their lengths are less than n/
2. And that happens just when the triple( i, j, k ) is the normal form of a small tour.
Lemma 5.
When n ≥ is odd, three vertices v i , v j , and v k of the cyclicpolytope C n − ( n ) with ≤ i < j < k ≤ n lie on the same side of the hyperplanedetermined by the other n − vertices just when the triple ( i, j, k ) is the normalform of an odd tour.Proof. This is essentially Gale’s evenness condition [13, pp. 14–15].The moment curve passes through the vertices v through v n of the cyclicpolytope C n − ( n ) in that order. The n − v i , v j , and v k determine a hyperplane, which the moment curve intersects at those n − v i , v j , and v k lie on the same side of that hyperplanejust when the moment curve intersects it an even number of times during eachpassage from one of those three to another.In passing from v i to v j , the moment curve intersects the hyperplane j − i − v i +1 through v j − . In passing from v j to v k , that number is k − j − v i , v j , and v k lie on the same side ofthat hyperplane just when j − i − k − j − j − i and k − j are odd. But whenever those two are both odd, ( n + i ) − k must beodd as well. So v i , v j , and v k lie on the same side of the hyperplane just whenthe triple ( i, j, k ) is the normal form of an odd tour. Proposition 6.
When n ≥ is odd and the necklace s is majority dominant,the face lattice of the fixed-angles polytope A nE [ s ] is dual to that of the cyclicpolytope C n − ( n ) .Proof. It suffices to establish a one-to-one correspondence between the n facetsof A nE [ s ] and the n vertices of C n − ( n ) with the property that n − A nE [ s ]intersect at a vertex precisely when the corresponding n − C n − ( n )lie on a common facet. We establish such a correspondence by associating thefacet of A nE [ s ] on which the edge ε i shrinks to have length 0 with the vertex v i (cid:48) of the cyclic polytope C n − ( n ) just when i (cid:48) ≡ − i (mod n ).By Lemma 4, the n − A nE [ s ] other than those associated with theedges ε i , ε j , and ε k intersect at a trigon vertex just when the triple ( i, j, k ) isthe normal form of a small tour. By Lemma 5, the n − C n − ( n )other than v i (cid:48) , v j (cid:48) , and v k (cid:48) lie on a common facet just when the triple ( i (cid:48) , j (cid:48) , k (cid:48) )is the normal form of an odd tour. Finally, by Lemma 3, the tour [ i, j, k ] issmall just when the tour [ − i, − j, − k ] is odd.14 Dual cyclic when n is even Suppose now that n is even. We continue to require that the necklace s bemajority dominant, so the polytope A nE [ s ] ∆ is neighborly by Lemma 2. Majoritydominance forbids substring ties of width w < n/
2, but allows ties of width w = n/
2. If we further require that s be entirely free of substring ties, then thepolytope A nE [ s ] will be simple, so its dual A nE [ s ] ∆ will be simplicial, as well asneighborly — and hence max-faced, by the upper-bound theorem [13, p. 254].When n is even, what further condition can we impose on s to guaranteethat A nE [ s ] ∆ , beyond being max-faced, is actually the cyclic polytope C n − ( n )? As long as none of the n diametral substrings are tied, which n/ , , , , ,
4] that starts Figure 2, for example, the heavy diametralsubstrings are (3 , , , , , , A E [1 , , , , ,
4] is combinatorially acube. In the later necklace [1 , , , , , A E [1 , , , , ,
6] is C (6) ∆ , the dual of a cyclic polytope. That maximalasymmetry turns out to be the key.We say that s has dipole tie-breaking when it has a pair of antipodal entries,its light pole and heavy pole , with the property that the n/ n/ , , , , ,
4] has dipole tie-breakingwith light pole 2 and heavy pole 3. This example shows that the light pole neednot be a least entry and the heavy pole need not be a greatest. The heavy polemust exceed the light pole, however.Dipole tie-breaking is the extra requirement that we impose on the necklace s when n is even so that the fixed-angles polytope A nE [ s ] will be dual cyclic. Toshow this, we begin with a lemma about the vertex figures of cyclic polytopes. Lemma 7.
For any d and m , the vertex figure of the last vertex of the cyclicpolytope C d ( m ) is itself cyclic, being an instance of C d − ( m − . If d is even,the vertex figures of all of the vertices of C d ( m ) are cyclic.Proof. We can represent a subset of the vertices of a cyclic polytope as a stringof bits, with 1’s for vertices in the subset and 0’s for the ones not in it, in orderalong the moment curve. By Gale’s evenness condition, a set of vertices of theright cardinality to be a facet actually forms a facet just when, for every two 0’sin that bit string, the number of 1’s between them is even. But adding a new 1at the end of the bit string can’t affect that test. Thus, if V is some subset ofthe vertices of C d − ( m −
1) with | V | = d −
1, then V is a facet of C d − ( m − V ∪ { v m } is a facet of C d ( m ) containing v m . So the vertex figureof v m in C d ( m ) is a C d − ( m − d is even, there are combinatorial automorphisms of C d ( m ) that per-mute its m vertices dihedrally; so all vertices look the same. Proposition 8.
If a necklace s has even length n ≥ , is majority dominant,and has dipole tie-breaking, then the dual A nE [ s ] ∆ of its fixed-angles polytope isan instance of the cyclic polytope C n − ( n ) .Proof. It suffices to show this for the particular necklace s = [1 , . . . , , A nE [ s ] depends only upon which substrings of s are light, tied, and heavy.Going up by one dimension, the facial lattice of A n +1 E [1 , . . . ,
1] is dual to thatof the cyclic polytope C n − ( n + 1), by Prop. 6. Under that duality, each of thefacets of A n +1 E [1 , . . . ,
1] is dual to one of the vertex figures of C n − ( n + 1), whichis a C n − ( n ) by Lemma 7. But such a facet corresponds to the ( n + 1)-gonsin which a particular edge has shrunk to length 0. That causes two vertices tomerge into a supervertex of shangle 2, so each facet of A n +1 E [1 , . . . ,
1] is a copyof A nE [1 , . . . , , When n is even, it is intriguing to consider how the fixed-angles polytope P (cid:15) ·· = A nE [1 , . . . , , (cid:15) ] varies with (cid:15) . For 0 < (cid:15) <
2, that shangle neck-lace is majority dominant and has dipole tie-breaking, so the polytope P (cid:15) is a C n − ( n ) ∆ ; and, since n − (cid:15) drops to 0, however, the polytope P (cid:15) must acquire 2 n symmetries that permuteits n facets dihedrally. Those symmetries appear because the n/ (cid:15) reaches 0 cause n/ P (cid:15) to shrink from small instances of∆ n/ − to points. For example, when n = 6, the polyhedron C (6) ∆ has eightvertices and six faces: two triangles, two quadrilaterals, and two pentagons.When (cid:15) reaches 0, three of its edges shrink to points, those points becoming thevertices of the triangular base of the bipyramid A E [1 , . . . , People fix the edge lengths of their polygons more often than they fix the vertexangles. What can we say about that situation?
Given some necklace s = [ s , . . . , s n ] of positive numbers with S ·· = s + · · · + s n ,we define an ¯ s -gon to be a planar n -gon whose k th edge ε k has length | ε k | = s k ,for all k . The bar accent indicates that it is the edge lengths that we here Of the three edges of C (6) ∆ that shrink, no two are adjacent, two bound each of thepentagonal faces, and one bounds each quadrilateral face. There are two ways to choose sucha triple of edges. L n [ s ] is the moduli space of shapes of ¯ s -gons,while L nc [ s ] is the subset in which the ¯ s -gons are ccw-convex.Given an ¯ s -gon that is ccw-convex, Kapovich and Millson [9] associate thenumber 2 πs k /S with the point on the unit circle where the normalized vector ε k /s k ends. They then use a Schwarz–Christoffel map to conformally transformthe unit disk into an n -gon, the k th of those points becoming a vertex with ex-ternal angle 2 πs k /S . That n -gon is defined only up to an orientation-preservingsimilarity; but we can scale it to have unit perimeter (or unit area). We thustransform each ccw-convex ¯ s -gon into an ˆ s -gon, getting a map h s : L nc [ s ] → A n [ s ]that Kapovich and Millson show to be a homeomorphism. The convex-polygonspace L nc [ s ] is thus homeomorphic to a polytope by a preferred homeomorphism,which makes it a topological polytope . And we conclude: Corollary 9.
When n is odd and the fixed lengths in s are majority dominant,the subset L nc [ s ] of the polygon space L n [ s ] in which the n -gons are ccw-convex isa topological polytope that is combinatorially C n − ( n ) ∆ . The same holds when n is even if the fixed lengths in s also have dipole tie-breaking. Given some n -gon, we can produce various n -gons by reassembling its n directededges, tip to tail, in any of ( n − n -gon that is ccw-convex; so the polygon space L n [ s ] is coveredby the fixed-lengths-convex top-polytopes L nc [ s (cid:48) ], for the ( n − s (cid:48) of s . Those top-polytopes overlap just on their faces, in each of whichsome sets of edges share a common phase (and hence function as superedges ).A subset tie arises in the length vector s when some w ≤ n/ n − w . Eachsubset tie in s generates a quadratic singular point of the polygon space L n [ s ].Such a singular point has a neighborhood in L n [ s ] that is analytically isomorphicto a neighborhood of the origin in the zero set of a quadratic form on R n − whosesignature is ( w − , n − w − When the length vector s has such a subset tie, there are w !( n − w )! cyclicreorderings s (cid:48) of s that convert both of the tied subsets into substrings, thusgiving a necklace s (cid:48) with a substring tie of width w . That substring tie causesthe fixed-angles polytope A nE [ s (cid:48) ] to have a digon vertex, and we refer to thecorresponding point of the fixed-lengths-convex top-polytope L nc [ s (cid:48) ] as a digoncorner . The quadratic singular point of a polygon space L n [ s ] that arises froma subset tie of width w in s is thus a digon corner of each of the w !( n − w )!topological polytopes L nc [ s (cid:48) ] that touch it. Kapovich and Millson only sketch the structure of the singular points in [9], but themethods that they use in [10, 11] to show that the singularities in a moduli space of sphericallinkages are Morse, and hence quadratic, apply to any space of constant curvature. .3 Diffeomorphism issues When the necklace s is free of substring ties, the fixed-angles polytope A nE [ s ]has only simple vertices, so it is a smooth ( n − L nc [ s ], near anypoint in it, the n -gon has at least three external angles that are positive, and wecan use the other n − L nc [ s ] is also a smooth ( n − s has substring ties, there is no hope for a diffeomorphism.That is easy to see for a substring tie of width w ≥
3. The resulting digon vertexof the polytope A nE [ s ] is nonsimple; so both that polytope and the homeomorphictop-polytope L nc [ s ] are not smooth manifolds with corners. For a substring tieof width w = 2, the digon vertex of A nE [ s ] is simple, so that polytope might stillbe a smooth manifold with corners. Once n ≥
5, however, the top-polytope L nc [ s ] is bent, near its digon corner, in a way that prevents it from being asmooth manifold with corners; so the situation remains hopeless.We close with a warning: Even when s is free subset ties, so that somediffeomorphism likely exists, the Kapovich–Millson map h s : L nc [ s ] → A nE [ s ] isnot a diffeomorphism. Consider an ¯ s -gon that approaches the boundary of thetop-polytope L nc [ s ], say because its external angle ρ at the vertex joining itsedges of lengths s k and s k +1 approaches zero. Let r be the length of the edgethat joins the vertices with angles s k and s k +1 in the corresponding ˆ s -gon. Asthe angle ρ goes to zero, the length r does also, but at a slower rate. By theSchwarz–Christoffel integral, r goes to zero like ρ q where q = 1 − s k + s k +1 ) /S ,and q < h s from being a diffeomorphism. References [1] Christophe Bavard and ´Etienne Ghys, Polygones du plan et poly`edreshyperboliques,
Geometriae Dedicata (1992) 207–224, //perso.ens-lyon.fr/ghys/articles/polygonesplan.pdf[2] Michael W. Davis, When are two Coxeter orbifolds diffeomor-phic? Michigan Mathematical Journal (2014) 401–421, //projecteuclid.org/euclid.mmj/1401973057 and arXiv:1306.6046 [math.GT] [3] Michael F. Farber, Invitation to Topogical Robotics . European Mathemat-ical Society, Zurich, 2008.[4] M. Farber and D. Sch¨utz, Homology of planar polygon spaces,
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1] for ccw-convex equilateral hexagons isn’t one of those either. l’Enseignement Math´ematique (2011), 23–56, //doi.org/10.4171/LEM/57-1-2 and arXiv:math/0308187 [math.MG] [6] Jean-Claude Hausmann and Eugenio Rodriguez, The space of cloudsin Euclidean space, Experimental Mathematics , (2004) 31–47,//dx.doi.org/10.1080/10586458.2004.10504521 and arXiv:math/0207107[math.DG] [7] Dominic Joyce, A generalization of manifolds with corners, Advancesin Mathematics (2016) 760–862, //doi.org/10.1016/j.aim.2016.06.004and arXiv:1501.00401 [math.DG] [8] Volker Kaibel and Arnold Wassmer, Automorphism groupsof cyclic polytopes, Chapter 8 of Frank Lutz,
TriangulatedManifolds