aa r X i v : . [ m a t h . M G ] M a y DOMES OVER CURVES
ALEXEY GLAZYRIN ⋆ AND IGOR PAK ⋄ Abstract.
A closed PL-curve is called integral if it is comprised of unit intervals. Kenyon’s problemasks whether for every integral curve γ in R , there is a dome over γ , i.e. whether γ is a boundaryof a polyhedral surface whose faces are equilateral triangles with unit edge lengths. First, we give analgebraic necessary condition when γ is a quadrilateral, thus giving a negative solution to Kenyon’sproblem in full generality. We then prove that domes exist over a dense set of integral curves. Finally,we give an explicit construction of domes over all regular n -gons. Introduction
The study of polyhedra with regular polygonal faces is a classical subject going back to ancient times.It was revived periodically when new tools and ideas have developed, most recently in connection toalgebraic tools in rigidity theory. In this paper we study one of most basic problems in the subject –polyhedral surfaces in R whose faces are congruent equilateral triangles. We prove both positive andnegative results on the types of boundaries these surfaces can have, suggesting a rich theory extendingfar beyond the current state of the art.Formally, let γ ⊂ R be a closed piecewise linear (PL-) curve. We say that γ is integral if it iscomprised of intervals of integer length. Now, let S ⊂ R be a PL-surface realized in R with theboundary ∂S = γ , and with all facets comprised of unit equilateral triangles. In this case we say that S is a unit triangulation or dome over γ , that γ is spanned by S , and that γ can be domed . Question 1.1 (Kenyon, see § . Is every integral closed curve γ ⊂ R spanned by a unit triangulation?In other words, can every such γ be domed? For example, the unit square and the (unit sided) regular pentagon can be domed by a regular pyramidwith triangular faces. Of course, there is no such simple construction for a regular heptagon. Perhaps,surprisingly, the answer to Kenyon’s question is negative in general.A 3-dimensional unit rhombus is a closed curve ρ ⊂ R with four edges of unit length. This is a2-parameter family of space quadrilaterals ρ ( a, b ) parameterized by the diagonals a and b , defined asdistances between pairs of opposite vertices. Theorem 1.2.
Let ρ ( a, b ) ⊂ R be a unit rhombus with diagonals a, b > . Suppose ρ ( a, b ) can bedomed. Then there is a nonzero polynomial P ∈ Q [ x, y ] , such that P ( a , b ) = 0 . In other words, for a, b > Q , the corresponding unit rhombus cannotbe domed, giving a negative answer to Kenyon’s question. In fact, our tools give further examples of aunit rhombi which cannot be domed, such as ρ (cid:0) π , π (cid:1) , see Corollary 4.9.The following result is a positive counterpart to the theorem. We show that the set of integral curvesspanned by a unit triangulation is everywhere dense within the set of all integral curves.Let γ, γ ′ ⊂ R be two integral closed curves of equal length. We assume the vertices of γ, γ ′ aresimilarly labeled (cid:2) v , . . . , v n (cid:3) and (cid:2) v ′ , . . . , v ′ n (cid:3) , giving a parameterizations of the curves. The Fr´echetdistance | γ, γ ′ | F in this case is given by | γ, γ ′ | F = max ≤ i ≤ n | v i , v ′ i | . ⋆ School of Mathematical & Statistical Sciences, University of Texas Rio Grande Valley, Brownsville, TX 78520. Email: [email protected] . ⋄ Department of Mathematics, UCLA, Los Angeles, CA, 90095. Email: [email protected] . Theorem 1.3.
For every integral curve γ ⊂ R and ε > , there is an integral curve γ ′ ⊂ R of equallength, such that | γ, γ ′ | F < ε and γ ′ can be domed. The theorem above does not give a concrete characterization of domed integral curves, and such acharacterization seems difficult (see § Theorem 1.4.
Every regular integral n -gon in the plane can be domed. This gives a new infinite class of regular polygon surfaces , comprised of one regular n -gon and manyunit triangles. See Section 3 for the proof and some previously known special cases. Outline of the paper.
We begin with a technical proof of Theorem 1.3 in Section 2. Our proofis constructive and almost completely self-contained except for the Steinitz Lemma with Bergstr¨omconstant, see § § theory of places , see e.g. [20, Ch. 1] and [25, § bellows conjecture , see [7] (see also [25, § rigidity theory and the EuclideanRamsey theory . Final remarks are given in Section 6.In the Appendix A, we include a negative solution of the question in [12] on the dimension of theflexes of doubly periodic surfaces. This counterexample arose upon careful inspection of our proof ofTheorem 1.2, and is of independent interest (cf. [28]).
Notation.
Let | vw | denote the length between v, w ∈ R . We use [ v . . . v n ], v i ∈ R d , to denote aclosed polygonal curve γ ⊂ R d . We use parentheses notation ( a , . . . , a n ), a i >
0, to denote the edgelengths of γ , i.e. a i = | v i v i +1 | , and a n = | v n v | . Denote by | γ | = a + . . . + a n ∈ N the length of theintegral curve γ .Throughout the paper we consider PL-curves modulo rigid motions. All curves will be in R , integraland closed, unless stated otherwise. Similarly, all PL-surfaces S will have unit triangles, unless statedotherwise. They are realized in R by the vertex coordinates and such realizations have no additionalextrinsic constraints (such as being embedding or immersion, cf. § Integral curves which can be domed are dense
Denote by M n the space of all integral curves of length n in R , modulo rigid motions, which iscompact in the Fr´echet topology. Let D n ⊂ M n denote the subset of integral curves which can bedomed. The goal of this section is to prove Theorem 1.3, which states that D n is dense in M n .The proof goes through several stages of simplification of integral curves, along the following route: integral curves −→ generic curves −→ near planar curves −→ compact near planar curves .Compact curves are curves which fit inside a ball or radius 3 / flips and are obtained by attachingunit rhombi which can be domed. Making these reductions rigorous is somewhat technical and willoccupy much of this section. The rhombi ρ ∈ D will play a special role, so we consider them first.2.1. Dense rhombi.
Throughout the paper, a unit closed curve of length 4 is called a unit rhombus ,or just a rhombus. Each unit rhombus is determined by the diagonals a and b ; we denote such unitrhombus by ρ ( a, b ). Observe that a + b ≤
4, with the equality achieved on plane rhombi.
Lemma 2.1.
Fix the diagonal a , and suppose < a < , a / ∈ Q . Then the set of values of b ≥ forwhich ρ ( a, b ) ∈ D is dense in (cid:2) , √ − a (cid:3) . In particular, for every ε > , there is a unit rhombus ρ = ρ ( a, b ) ∈ D , such that | ρ, ρ ⋄ | F < ε , where ρ ⋄ = ρ (cid:0) a, √ − a (cid:1) is a convex plane unit rhombus. OMES OVER CURVES 3
Proof.
Let c = 1. Consider a unit rhombus ρ = ρ ( a, c ), spanned by two unit triangles. Define ρ = ρ ( a, c ) to be a rhombus obtained by attaching two copies of ρ together, with a common diagonalof length a . Similarly, defined ρ = ρ ( a, c ), etc. Clearly, every rhombus ρ m , m ≥
1, can be domed bysurface with 2 m unit triangles. We have: c n = p − a · | sin nα | , where α := arcsin(4 − a ) − . Thus, set { c m , m ≥ } is dense in (cid:2) , √ − a (cid:3) , for all α / ∈ π Q . Finally, we have α / ∈ π Q , since otherwisesin α = (4 − a ) − ∈ Q , a contradiction with the assumption that a / ∈ Q . (cid:3) The integer m in the proof will be called a multiplier throughout this section. We can now proveTheorem 1.3 for | γ | = 4. Let X := { x > − x ) − ∈ π Q } . Lemma 2.2.
Let ρ = ρ ( a, b ) ⊂ R be a unit rhombus, and let ε > . Then there is a unit rhombus ρ ′ = ρ ( a ′ , b ′ ) ∈ D , such that | ρ, ρ ′ | F < ε . Moreover, if a / ∈ X , one can take a ′ = a .Proof. The second part follows from the above proof of Lemma 2.1. For the first part, choose a / ∈ X , sothat | a − a ′ | < ε . Then apply the construction as above. (cid:3) Reachable curves.
Let us introduce some definitions and notation. Consider two integral curves γ = [ v . . . v k . . . v n ] and γ ′ = [ v . . . v ′ k . . . v n ], such that [ v k − v k v k +1 v ′ k ] ∈ D . In this case we say that γ and γ ′ are k -flip connected , or just flip connected ; write γ → k γ ′ . Two integral curves γ = [ v . . . v n ]and γ ′ = [ v ′ . . . v ′ n ] are called flip equivalent , write γ ∼ γ ′ , if(2.1) γ = γ → k γ → k γ → k . . . → k m γ m = γ ′ , for some integer sequence k := ( k , . . . , k N ), where 1 ≤ k i ≤ N for all i = 1 , . . . , N . See an example inFigure 1. Clearly, if γ ∼ γ ′ and γ ′ ∼ γ ′′ , then γ ∼ γ ′′ . vw v ' w ' wv ' γ γ γ Figure 1.
Sequence of two flips γ → γ → γ , at v and then at w . Here γ =[ . . . vw . . . ], γ = [ . . . v ′ w . . . ], and γ = [ . . . v ′ w ′ . . . ].We say that an integral curve γ ⊂ R of length n is reachable , if for all ε >
0, there is an integralcurve γ ′ ⊂ R of length n , such that | γ, γ ′ | F < ε , and γ ′ ∈ D n . In this notation, Theorem 1.3 claimsthat all integral curves are reachable, while Lemma 2.2 proves this for curves of length 4. Lemma 2.3.
Let γ ∼ γ ′ are flip equivalent integral curves in R . Suppose γ is reachable. Then so is γ ′ . In other words, the lemma says that if γ ∈ M n is a limit point of D n , then so are all flip equivalentcurves γ ′ ∼ γ . Proof.
Since γ ∼ γ ′ , there is a flip sequence k as in (2.1), and a sequence of multipliers m = ( m , . . . , m N ) ∈ Z m counting how many pairs of unit triangles added at each flip. Use positive and negative integers m i todenote clockwise of counterclockwise direction of the flip γ i − → γ i . Thus, pair ( k , m ) uniquely encodesthe combinatorial structure of the flip equivalence. Let(2.2) Φ k , m : M n → M n be the map defining flip sequence as above. By construction, Φ k , m : D n → D n , and Φ k , m ( γ ) = γ ′ .Clearly, the map Φ k , m is a composition of | m | + . . . + | m N | continuous maps, and thus also continuouson M n . Since γ is reachable, there is a sequence (cid:8) γ h t i → γ, t ∈ N (cid:9) of converging curves γ h t i ∈ D n .Thus, we have another sequence of converging curves in D n : (cid:8) Φ k , m (cid:0) γ h t i (cid:1) → γ ′ , t ∈ N (cid:9) , which shows that γ ′ is reachable. (cid:3) ALEXEY GLAZYRIN AND IGOR PAK
Generic curves.
We say that an integral curve γ = [ v . . . v n ] ⊂ R is generic , if all diagonals | v i v j | are algebraically independent over Q . Denote by G n ⊂ M n the set of generic integral curves oflength n . Lemma 2.4.
Let γ ∼ γ ′ , where γ ∈ G n and γ ′ ∈ M n . Then γ ′ ∈ G n . In other words, an integral curve that is flip equivalent to a generic integral curve, is also generic. Theproof is straightforward and follows immediately from the algebraic formulas in the proof of Lemma 2.1.
Lemma 2.5.
Set G n is dense in M n .Proof. Let γ = [ v . . . v n ] ∈ M n be a integral curve in R . Assume for now that no three adjacentvertices are collinear: | v i v ( i +2)mod n | <
2, for all 1 ≤ i ≤ n . Consider a sequence of curves γ = [ v v . . . v n ] [ v ′ v . . . v n ] [ v ′ v ′ . . . v n ] . . . [ v ′ v ′ . . . v ′ n ] = γ ′ , where at each step we perturb one vertex v k to a vertex v ′ k , | v k v ′ k | < ε in such a way, that new coordinatesof v ′ k are algebraically independent on all coordinates of v ′ i , i < k (but not of each other, of course).Observe that the resulting curve γ ′ has now algebraically independent diagonals.Assume now that | v i − v i +1 | = 2. Take a non-degenerate triangle [ v j v i − v i +1 ]. Denote by ζ =[ v i +1 . . . v j ] a segment of γ . Rotate ζ → ζ ′ = [ v ′ i +1 . . . v ′ j − v j ] around v j , so that | v i − v ′ i +1 | = 2 − δ , | v i +1 v ′ i +1 | < δ , and place v ′ i at unit distance from v i − and v ′ i +1 . This results in a new integral curve γ ′ ∈ M n , s.t. | γ, γ ′ | F < δ . Proceed to do this for all i as above. Taking δ > γ ∗ without collinearities, and s.t. | γ, γ ∗ | F < ε . The remaining cases when all vertices of γ lie on a line are straightforward. (cid:3) Planar curves.
An integral curve γ ∈ M n is called planar if it lies in a plane H ⊂ R . Denoteby P n ⊂ M n the set of planar integral curves of length n . Lemma 2.6.
The flip equivalence class of every γ ∈ G n contains a planar curve ξ ∈ P n as its limitpoint. In other words, for every ε >
0, and every generic integral curve γ ∈ G n , there is a generic integralcurve γ ′ ∈ G n and a planar integral curve ξ ∈ P n , such that γ ∼ γ ′ and | γ ′ , ξ | F < ε . Note that thecurve ξ does not have to be generic itself, or be flip equivalent to γ . Proof.
The proof is based on the same idea of using flips to obtain a near-planar curve γ ′ . Let γ =[ v . . . v n ] ∈ G n , and let ϕ : R → R be a generic linear function, i.e. defined by linear equations withcoefficients that are algebraically independent to coordinates of v i . Denote by h k := ϕ ( v k ) the value to ϕ on vertices of γ , 1 ≤ k ≤ n . Cyclically, for all k from 1 to n , make k -flips:(2.3) γ = γ → γ → γ → . . . → n γ n → γ n +1 → γ n +2 → . . . Choose integers m (see the proof of Lemma 2.3), as follows. Consider a flip γ jn + k − = [ . . . w k − w k w k +1 . . . ] → k γ jn + k = [ . . . w k − w ′ k w k +1 . . . ] . By the proof of Lemma 2.1, we can always choose a multiplier m jn + k so that(2.4) 23 α + 13 β < ϕ ( w ′ k ) < α + 23 β , where α := min (cid:8) ϕ ( w k − ) , ϕ ( w k +1 ) (cid:9) , β := max (cid:8) ϕ ( w k − ) , ϕ ( w k +1 ) (cid:9) . Note that we have α = β , since ϕ is generic, so there is always room to make such flip possible.Using (2.4), it is easy to see that there is a limit (cid:0) ϕ ( w ) , . . . , ϕ ( w n ) (cid:1) → ( h, . . . , h ), for some h ∈ R ,Here the limit is when N → ∞ , where N is the number of flips in (2.3). The limit curve ξ is integraland lies in the plane H := { x ∈ R : ϕ ( x ) = h } . Therefore, for N = N ( ε ) large enough, we obtain acurve γ ′ := γ N , such that γ ′ ∼ γ , and | γ ′ , ξ | < ε . By Lemma 2.4, we have γ ′ ∈ G n , which completes theproof. (cid:3) OMES OVER CURVES 5
Packing curves.
Let u , . . . , u n ∈ R d be unit vectors which satisfy u + . . . + u n = 0. The SteinitzLemma famously states, see e.g. [3], that there is always a permutation σ ∈ S n , s.t.(2.5) (cid:12)(cid:12) u σ (1) + . . . + u σ ( k ) (cid:12)(cid:12) ≤ B d for all 1 ≤ k ≤ n, where B d ≤ d is a universal constant which depends only on the dimension d . Bergstr¨om [4] found theoptimal value B = p /
4, see also § γ = [ v . . . v n ] ∈M n be an integral curve in R . We say that γ is B -packing , if | v v i | ≤ B for all 1 ≤ i ≤ n . Lemma 2.7.
Every generic integral curve γ ∈ G n is flip equivalent to a generic integral curve γ ′ ∈ G n that is / -packing. Here the constant B = 3 / p / < B < √ Proof.
Fix ε > γ ∈ G n . Let γ ′ ∼ γ and ξ = [ w , . . . , w n ] ∈ P n be as in the proof of Lemma 2.6,so | γ ′ , ξ | < ε . Define u i = −−−−→ w i w i +1 , 1 ≤ i < n , and u n = −−−→ w n w . Clearly, u i are unit vectors which satisfy u + . . . + u n = 0. By the Steinitz Lemma, there is a permutation σ ∈ S n , s.t. (2.5) holds. Consider areduced factorization of σ into adjacent transpositions ( i, i + 1) ∈ S n : σ = ( k , k + 1) · · · ( k ℓ , k ℓ + 1) , where 1 ≤ k , . . . , k ℓ ≤ n −
1, and ℓ = inv( σ ) is the number of inversions in σ , see e.g. [31]. This reducedfactorization may not be unique, of course.Define a sequence of flips as in (2.1), according to this factorization:(2.6) γ ′ = γ → k γ → k γ → k . . . → k ℓ γ ℓ = γ ′′ . Recall that γ j remain generic by Lemma 2.4. Thus, by the second part of Lemma 2.1 and induction, wecan always choose the multipliers m j so that | γ j , ξ | < ε , 1 ≤ j ≤ ℓ .Now let ε →
0. As in the proof of Lemma 2.3, by continuity of Φ k , m in (2.2), we have the limitplanar curve γ ′′ → ̺ := [ y , . . . , y n ] ∈ P n . By induction on the length ℓ of the factorization, wehave: −−−→ y i y i +1 = u σ ( i ) , 1 ≤ i < n , and −−→ y n y = u σ ( n ) . By the Steinitz Lemma with Bergstr¨om constant B = p /
4, we conclude that for sufficiently small ε >
0, the integral curve γ ′′ is ( B + δ )-packing, forall δ >
0. Taking δ < (3 / − B ), we obtain the result. (cid:3) Remark 2.8.
In notation of the proof above, in a special case of a convex centrally symmetric n -gon ξ ∈ P n , n = 2 k , the sequence of unit vectors is u , . . . , u k , − u , . . . , − u k . Take a permutation whichgives the order u , . . . , u k , − u k , . . . , − u ; the corresponding limit curve ̺ ∈ P n is then degenerate. Forevery reduced factorization as in the proof, the pattern of rhombi used in the flip sequence then definesa zonotopal tilings , see e.g. [25, Exc. 14.25].2.6. Proof of Theorem 1.3.
We prove the result by induction. First, closed integral curve of length 3 isa unit triangle, so Theorem 1.3 is trivially true in this case. The case of length 4 is resolved in Lemma 2.2.Note that by Lemma 2.5 it suffices to prove the theorem only for generic curves γ = [ v . . . v n ] ∈ G n .Formally, we will show for all n ≥
5, the set D n ∩ G n is dense in G n . In fact, to make the inductiveargument work we will need a stronger assumption.Let n = 5, and let γ ∈ G be a generic integral pentagon. By Lemma 2.7, there is γ ′ = [ w , . . . , w ] ∈G , such that γ ′ ∼ γ , and | w w i | ≤ / ≤ i ≤
5. Since | w w | , | w w | , | w w | ≤ √
3, it is easyto see that the triangle Q = [ w w w ] can be covered with a unit circle in the plane H spanned by Q (see e.g. [25, Cor. 1.8]). Then there is a point z ∈ R , s.t. | w z | = | w z | = | w z | = 1, see Figure 2 (left).Apply now Lemma 2.2 to rhombi ρ = [ w w w z ] and ρ = [ w w w z ], to obtain rhombi ρ ′ =[ w w ′ w z ] ∈ D and ρ ′ = [ w w ′ w z ] ∈ D , which satisfy | w w ′ | , | w w ′ | < ε . Attach unit triangle[ w w z ] to rhombi ρ ′ and ρ ′ . This gives the desired pentagon η = [ w w ′ w w w ′ ] ∈ D , s.t. | γ ′ , η | F < ε .Thus, γ ′ is reachable. By Lemma 2.3, then so is γ , as desired.The argument above gives a continuous deformation { η h t i , t ∈ [0 , } , where η h i = γ ′ , and η h t i ∈D ∩ G for all but countably many t >
0. Observe that the construction is flexible enough to allowconvergence of angles on both sides: ∠ w w ′ w → ( ∠ w w w )+, and ∠ w w ′ w → ( ∠ w w w ) − . Thisconclusion is used in the inductive step given below. ALEXEY GLAZYRIN AND IGOR PAK zw γ ' zw w w w w η w w w w w w ρ ρ γ ' Figure 2.
Base of induction n = 5, and step of induction for n = 7.For n ≥
6, we employ a similar argument. Let γ ∈ G n be a generic integral curve as above. ByLemma 2.7, there is γ ′ = [ w , . . . , w n ] ∈ G n , such that γ ′ ∼ γ , and | w w i | ≤ / ≤ i ≤ n . Since | w w | <
2, there is a generic point z ∈ R , such that | w z | = | w z | = 1, see Figure 2 (right). Considerintegral curves η = [ w w w w z ] ∈ G and φ = [ w zw w . . . w n ] ∈ G n − . By induction, there is acontinuous deformation { φ h t i , t ∈ [0 , } , where φ h i = φ and φ h t i ∈ D n − ∩ G n − for all but countablymany t . Without loss of generality, assume that ∠ w h t i z h t i w h t i → ( ∠ w zw )+ as t → n = 5 case, there is a continuous deformation { η h s i , s ∈ [0 , } , where η h i = η ,and η h s i ∈ D ∩ G for all but countably many s >
0. From the observation above, we can assume that ∠ w h s i z h s i w h s i → ( ∠ w zw )+ as s → φ h t i to η h s i with the same angle as above, we obtain a continuous deformation { γ h t i , t ∈ [0 , } , where γ h i = γ ′ and φ h t i ∈ D n − ∩ G n − for all but countably many t . In particular, thecurve γ ′ ∈ G n is reachable. By Lemma 2.3, we conclude that γ is also reachable, as desired. Thiscompletes the proof of the induction step and finishes the proof of the theorem. (cid:3) Regular polygons
Classical domes.
Denote by Q n ⊂ R the regular n -gon with unit sides in the xy -plane with thecenter at the origin O . From the introduction, there is a trivial dome over Q and Q , and domes over Q , Q are given by regular pyramids. Less obviously, a tiling of Q given in Figure 3 (left), gives anatural dome over Q , when square pyramids are added. Similarly, recall that the regular octagon Q and decagon Q are spanned by the surfaces of Johnson solids square cupola and pentagonal cupola ,respectively, see Figure 3 (right) and [16] for details. In fact, both are cuts of the Archimedean solids,see e.g. [9, p. 88]. The faces of both surfaces are regular triangles, squares or pentagons. Adding apyramid to each face we obtain domes over Q and Q . Figure 3.
Left: Tiling giving a dome over Q . Right: Pentagonal cupola giving a dome over Q . The image on the right is available from the
Wikimedia Commons , and is free to use with attribution.
OMES OVER CURVES 7
Proof of Theorem 1.4.
We follow notation in the proof of Lemma 2.1, and employ the symmetryof Q n at every step.First, attach a unit triangle to each side of Q n at angle θ > θ verysmall, to be chosen at a later point. Denote by a the distances between vertices of adjacent unit trianglesand assume that a / ∈ Q . Note that a > n ≥ Q n , attach to adjacent unit edges n rhombi R = ρ n (cid:0) a , ∗ (cid:1) . To simplifythe notation, we use ( ∗ ) for the second diagonal, since it is completely determined by the multiplier m and a , see the proof of Lemma 2.1. Take m large enough and chosen so that R is nearly planar, at anangle θ > θ with the plane. Such m exists by Lemma 2.1 if we assume further that a / ∈ Q .Next, moving along the boundary, denote by a the distances between vertices of adjacent rhombi R ,and observe that a / ∈ Q . Now attach to the adjacent unit edges rhombi R = ρ m (cid:0) a , ∗ (cid:1) , where themultiplier m is large enough and chosen so that R is nearly planar, at an angle θ > θ , see Figure 4.Again, such m exists by Lemma 2.1, if we assume further that a / ∈ Q .Repeat this procedure for k iterations, until the distance β to the vertical z -axis from new rhombivertices satisfies β < p − α /
4. Here α := a k +1 denotes the distance between vertices of adjacentrhombi R k = ρ m k (cid:0) a k , ∗ (cid:1) , and we assume that that α / ∈ Q . The above bound on β corresponds to havingthe projection of the nearly planar rhombus R k +1 cover the origin O , see Figure 4 (center).At this stage, attach to the adjacent unit edges new unit rhombi R = ρ M ( α, ∗ ) in such a way thatthe new vertices are at distance δ > z -axis, see Figure 4 (center). By Lemma 2.1, distance δ > θ > m := ( m , m , . . . , m k , M ) . Since the number of vectors m is countable, the assumptions a i , α / ∈ Q over all m represent countablymany inequalities on θ , so for some θ > S is continuously deformed with θ for every fixed m (cf. the proof ofLemma 2.3). Continuously decreasing θ > z axis. This completes the construction of a dome over Q n for all n ≥ ≤ n ≤ n -gon rQ n , replace unit triangles Q with their scaled version rQ and proceedas above. Now triangulate every copy rQ with r unit triangles Q , completing the construction of adome over rQ n . (cid:3) OQ n O z δβ Figure 4.
Nearly planar tiling of a portion of Q n with rhombi and its vertical slice. Remark 3.1.
Also, one can ask if a version of the arm lemma (see e.g. [25, § m , on a sufficiently small interval θ ∈ ( θ − ǫ, θ + ǫ ), but thisresult is not necessary for the continuity argument in the proof. ALEXEY GLAZYRIN AND IGOR PAK The algebra of squared diagonals
Contractible domes.
As a warm-up to the proof of Theorem 1.2, we first present a short argumentfor the case when the spanning surface S is homeomorphic to a disc. Proposition 4.1.
Let γ ⊂ R be a unit rhombus γ = ρ ( s, t ) , with diagonal lengths s and t . Suppose γ can be domed by a surface homeomorphic to a disc. Then there exists a polynomial P ∈ Q [ x, y ] , suchthat P ( s , t ) = 0 . For the proof of the proposition, we need to consider doubly periodic surfaces homeomorphic to theplane. Let K be a simplicial connected pure 2-dimensional complex with a free action of the groupG = Z ⊕ Z with generators a and b . Assume that G acts as a linear bijection on each simplex of K , andthat the number of orbits of triangles under the action of G is finite. Consider a mapping θ : K → R ,linear on each simplex of K , and equivariant with respect to the action of Z ⊕ Z , such that a and b act by translations with vectors α and β , respectively. Then the pair ( K, θ ) is called a doubly periodictriangular surface . Sometimes, with a slight abuse of notation, we call the surface K as well.Now, let us construct a doubly periodic surface comprised of unit triangles for every unit triangulationof a unit rhombus. Lemma 4.2.
For a unit rhombus γ = ρ ( a, b ) with diagonals s and t , there is a doubly periodic sur-face of unit triangles with two orthogonal periodicity vectors of length s and t , respectively. Moreover,there is such a surface homeomorphic to the plane if the triangulation of the rhombus spans a surfacehomeomorphic to a disc.Proof. First we construct a doubly periodic surface whose cells are either parallel translates of γ or − γ (see Figure 6). This surface is combinatorially equivalent to a tiling of the plane with unit squares. Achessboard coloring of such a tiling makes white squares correspond to parallel translates of γ and blacksquares correspond to parallel translates of − γ . The periodicity vectors of this surface are the vectors ofthe diagonals of γ .Attaching a spanning unit triangulation to each translate of γ and − γ we obtain a doubly periodicpolyhedral surface comprised of unit triangles with required periodicity vectors. Clearly, if a spanningtriangulation of γ is homeomorphic to a disk, the resulting doubly periodic surface is homeomorphic tothe plane. (cid:3) γ γ - Figure 5.
Doubly periodic surface from translates of γ and − γ .At this point we consider only doubly periodic triangular surfaces ( K, θ ) with unit triangular faces.For a fixed complex K , let G ( K ) be the set of all possible Gram matrices formed by vectors α and β forall doubly periodic triangular surfaces ( K, θ ). For a Gram matrix G ∈ G ( K ), we denote its entries by g , g = g , and g . Theorem 4.3 (Gaifullin–Gaifullin [12]) . Let K be a simplicial pure 2-dimensional complex homeomor-phic to R with a free action of the group Z ⊕ Z . Then there is a one-dimensional real affine algebraicsubvariety of R containing G ( K ) .In particular, the entries of each Gram matrix G from G ( K ) satisfy a one-dimensional system of twonon-trivial polynomial equations with integer coefficients: ( p ( g , g , g ) = 0 q ( g , g , g ) = 0 . OMES OVER CURVES 9
Remark 4.4.
In fact, the result in [12] is more general, as the authors consider all polygonal doublyperiodic surfaces homeomorphic to the plane with arbitrary sets of side lengths. In this setting, thecoefficients of polynomials p and q are obtained from the ideal generated by squares of all side lengthsof polygons in the polygonal surface Proof of Proposition 4.1.
If a unit rhombus with diagonals s and t can be spanned by unit trianglesthen, by Lemma 4.2, there is a doubly periodic triangular surface with orthogonal periodicity vectors oflength s and t . The entries of the Gram matrix of periodicity vectors are g = s , g = 0, g = t .By Theorem 4.3, there are two polynomials p , q with integer coefficients vanishing on the entries of theGram matrix. Thus, at least one of the equations p ( s , , t ) = 0 and q ( s , , t ) = 0 is non-trivial, andcan be used for the polynomial P . (cid:3) Theory of places.
There is no result generalizing Theorem 4.3 for doubly periodic surfaces ofnon-trivial topology and, moreover, we will show in Theorem A.2 that such a generalization is not true.However, for our purposes, we do not need two polynomials p and q as in Theorem 4.3. It is sufficientto find at least one polynomial that is non-trivial whenever g = 0. The machinery developed in [12]is based on the proof of the bellows conjecture for orientable 2-dimensional surfaces [7], and is also thebasis for our approach. We use places of fields as the main algebraic instrument of the proof.Let F be a field and b F be F extended by ∞ , i.e. b F = F ∪ {∞} with arithmetic operations extendedto b F by a ± ∞ = ∞ and a ∞ = 0 , for all a ∈ F,a · ∞ = a ∞ for all a ∈ b F \ { } . The expressions 00 , ∞∞ , · ∞ and ∞ ± ∞ are not defined . Let L be a field. A map φ : L → b F is called a place if φ (1) = 1 and φ ( a ± b ) = φ ( a ) ± φ ( b ) , φ ( a · b ) = φ ( a ) · φ ( b ) for all a, b ∈ L, whenever the right-hand side expressions are defined.As a direct consequence of the definition, we have φ (0) = 0 for all places, and φ ( x ) = ∞ for x = 0if and only if φ ( − x ) = ∞ and if and only if φ (1 /x ) = 0. It is also clear that whenever char F = 0, wemust also have char L = 0. Similarly, we have φ ( kx ) = ∞ for a non-zero k ∈ Z , if and only if φ ( x ) = ∞ .We will use the following basic fact on extensions of places. Lemma 4.5 (see e.g. [20, Ch. 1, Thm 1]) . Let L be a field containing a ring R . Let φ be a homomorphismof R in an algebraically closed field Ω , and suppose φ (1) = 1 . Then φ can be extended to a place L → Ω ∪ {∞} . General domes.
For a doubly periodic triangular surface (
K, θ ), let α and β be the periodicityvectors of the surface. The number of orbits under the action of G = Z ⊕ Z is finite, so we choosea representative of each orbit. Let ( x , y , z ) , . . . , ( x N , y N , z N ) be their coordinates in R , and let( x α , y α , z α ), ( x β , y β , z β ) be the coordinates of the periodicity vectors. Define field L as follows: L := Q (cid:0) x , y , z , . . . , x N , y N , z N , x α , y α , z α , x β , y β , z β (cid:1) . Note that L does not depend on the choice of representatives of orbits and the choice of the basis for thelattice Λ.When vertices a, b on the surface ( K, θ ) form an edge, denote by ℓ ab the squared distance betweenthem: ℓ ab := ( x a − x b ) + ( y a − y b ) + ( z a − z b ) . Clearly, ℓ ab ∈ L . For each surface the set of all possible ℓ ab is finite. Let R be the Q -subalgebra of L generated by all ℓ ab of the surface.Let Λ be the lattice generated by α and β . All vectors in Λ can be written as integer linear combinationsof the periodicity vectors, λ = kα + mβ . In case k and m are relatively prime, a vector λ is called primitive . Denote by Λ ∗ the set of primitive vectors λ ∈ Λ. As the first step in the proof of Theorem 1.2, we prove the lemma on finite elements of places.
Lemma 4.6 (Main lemma) . For a doubly periodic triangular surface ( K, θ ) obtained by the constructionin Lemma 4.2, let φ : L → F ∪ {∞} be a place that is finite on all ℓ ab defined by the surface and letchar F = 0 . Then there is a vector λ ∈ Λ , λ = 0 , such that φ is finite on ( λ, λ ) . For the proof, we use the following technical result of Connelly, Sabitov, and Walz [7, Lemma 4], seealso [25, § Theorem 4.7 (Connelly–Sabitov–Walz) . Let u be a vertex of a triangular surface in R and v , . . . , v d , d ≥ , be adjacent to it in this cyclic order, denote also v d +1 = v and v d +2 = v . Let φ be a place thatis defined on Q ( x u , y u , z u , x v , y v , z v , . . . , x v d , y v d , z v d ) and is finite on all ℓ uv i , ℓ v i v i +1 , ≤ i ≤ d . Then φ is finite on at least one of the squared diagonal lengths ℓ v i v i +2 , ≤ i ≤ d . Proof of the Main Lemma 4.6.
The statement of the lemma is true if one of the edges ofthe surface forms a vector from Λ. We define the complexity as a partial ordering of doubly periodictriangular surfaces with the same periodicity lattice Λ. Surfaces with edges from Λ are called the leastcomplex (an example is given in Figure 6). For surfaces without edges from Λ, the ordering is defined asfollows.A surface K is said to be less complex than K , if the Euler characteristic of K / Λ is greater thanthe Euler characteristic of K / Λ. The surface K is less complex than K if χ ( K / Λ) = χ ( K / Λ), and K / Λ has fewer vertices than K / Λ. The surface K is less complex than K if K / Λ and K / Λ havethe same Euler characteristic and the same number of vertices, but the smallest vertex degree of K isless than the smallest vertex degree of K . The proof will proceed by induction on complexity. First case.
Suppose the surface contains the edges ab, bc, ca , but does not contain a triangle [ abc ]. Theclosed curve formed by the edges ab , bc and ca , divides its neighborhood into two components. Thenwe define the surgery along [ abc ] by removing vertices a , b , c , edges ab , bc , ca , and adding two copiesof [ abc ], which we call [ a ′ b ′ c ′ ] and [ a ′′ b ′′ c ′′ ]. We do this in such a way that [ a ′ b ′ c ′ ] and [ a ′′ b ′′ c ′′ ] retainthe incidences of a , b , c in the first and the second component of the neighborhood, respectively. If thissurgery keeps the surface connected, then it increases the Euler characteristic of K/ Λ. If the surgerysplits the surface into two new surfaces then the Euler characteristic for each of them is not smaller thanthe initial Euler characteristic, for both of them there are fewer vertices than for the initial surface, andat least one of them is a connected doubly periodic triangular surface with the periodicity lattice Λ. Wecall the latter the connectivity property , see Figure 6. a' b'c'a bc
Figure 6.
Connecticity property in the
First Case of a doubly periodic surface.The connectivity property of one of the two new surfaces requires some explanation. Let T be thetorus generated by the lattice Λ, i.e. T = R / Λ. Then, initially, K/ Λ ∼ = T D , where D corresponds tothe surface of the dome. After the transformations to the surface described in the proof of the lemmafor the resulting surface K ′ , we obtain that K ′ / Λ can be always represented as T D ′ for some closedsurface D ′ . The surgery described above preserves T in one of the two disconnected components, thus OMES OVER CURVES 11 making its corresponding surface connected (cf. Remark 4.8 below). Since the set of ℓ ab for either surfaceis a subset of the initial set, we can use the inductive step. Second case.
Suppose there are no triples of vertices a, b, c as in the first case. Consider a vertex u of the surface with the smallest degree d adjacent to vertices v , . . . , v d . The smallest degree must beat least 4, because the first case holds otherwise. We use Theorem 4.7 but we have to be careful withapplying it as the field in Theorem 4.7 is not a subfield of the field L defined earlier. The issue is thatsome vertices v i and v j may belong to the same orbit under the action of the lattice Λ.Let R be a Q -subalgebra of K = Q ( x u , y u , z u , x v , y v , z v , . . . , x v d , y v d , z v d ) generated by all ℓ uv i , ℓ v i v i +1 , 1 ≤ i ≤ d . There is a natural homomorphism ψ from R to L mapping all elements of R to theircorresponding expressions in L . Note that this homomorphism is not necessarily defined on all elementsof L . For example, when v = v + α and v = v + α , the image of 1 / ( x v + x v − x v − x v ) is notdefined.At this point, we use Lemma 4.5 and extend ψ to ψ : K → L ∪ {∞} . The place φ can be also extendedto a place φ : L → F ∪ {∞} . In order to construct this extension, we apply Lemma 4.5 to a subring ofall elements of L whose images under φ are finite. For the constructed mapping, φ ( x ) = 0 if φ ( x ) = 0.Subsequently, if φ ( x ) is ∞ , φ ( x ) must be ∞ as well and φ extends the whole place φ .Using φ ( ∞ ) = ∞ , we can define the composition φ ◦ ψ . This composition is the place from K to F ∪ {∞} . Applying Theorem 4.7 to φ ◦ ψ we conclude that there is i such that the composition and,subsequently, φ is finite on ℓ v i v i +2 .For the next step, we substitute two triangles of the surface, uv i v i +1 and uv i +1 v i +2 with uv i v i +2 and v i v i +1 v i +2 simultaneously deleting the edge uv i +1 and adding the edge v i v i +2 . There was no edge v i v i +2 prior to this operation because otherwise the triangle uv i v i +2 would satisfy the case considered above. Atthe same time we make the same operations for all triangles that are the images of uv i v i +1 and uv i v i +2 under the action of Λ. As the result we obtain another surface K ′ such that K ′ / Λ is topologically thesame as K/ Λ and has the same number of vertices but the minimum vertex degree of K ′ is smaller. Theplace φ is still finite on all ℓ ab for edges ab , so all conditions of the lemma still hold.Observe that the operations in both cases decrease the complexity of the surface. Note that thiscannot continue indefinitely since the Euler characteristic is at most 2, and the number of edges andvertex degrees are positive. Thus, at some point we reach the least complex surface for which thestatement of the lemma is true. (cid:3) Remark 4.8.
In the proof of the First Case, the connectivity property fails for general doubly periodicsurfaces. In particular, if the elements of the fundamental group of K/ Λ corresponding to periodicityvectors do not commute, two new surfaces may be both disconnected unions of one-periodic pieces. Anexample is given in Figure 7. Here we show only the bottom half of the surface, which has connectedcomponents periodic along α . The top half is attached to the bottom along red triangles and has similarstructure, but with connected component periodic along β . This observation will prove crucial in theproof of Theorem A.2 in the Appendix. β α Figure 7.
Non-example to the connectivity property in the
First Case for generaldoubly periodic surfaces.
Proof of Theorem 1.2.
Let R ′ be the Q -subalgebra of L obtained by adding all ( λ, λ ) − , λ ∈ Λ ∗ ,to the subalgebra R : R ′ = R (cid:2) ( λ, λ ) − (cid:12)(cid:12) λ ∈ Λ ∗ (cid:3) . Let I ′ be the following ideal in R ′ : I ′ = (cid:0) ( λ, λ ) − (cid:12)(cid:12) λ ∈ Λ ∗ (cid:1) ⊳ R ′ . Assume that I ′ = R ′ . Then, by Krull’s theorem (see e.g. [2]), there exits a maximal ideal I , such that I ′ ⊂ I . Let F = R ′ /I . Since R ′ contains Q , field F must contain Q as well and char F = 0. Let F bean algebraic closure of F . The quotient homomorphism R ′ → F satisfies the conditions of Lemma 4.5for Ω = F so it can be extended to the place φ : R ′ → F ∪ {∞} . The quotient homomorphism is equalto 0 on all ( λ, λ ) − , λ ∈ Λ ∗ . Therefore, the place φ is infinite on ( λ, λ ) for all λ ∈ Λ ∗ . This implies thatthe same holds for all non-zero λ ∈ Λ. On the other hand, the quotient homomorphism is finite on R ′ .Therefore, we get a contradiction with Lemma 4.6. We conclude that the assumption that I ′ = R ′ isfalse.From above, we have that I ′ = R ′ . In particular, this implies that 1 ∈ I ′ :(4.1) 1 = M X i =1 r i ( λ i , λ i )( λ i , λ i ) . . . ( λ ip i , λ ip i ) , where all λ ij ∈ Λ ∗ , and all r i ∈ R . After multiplying by the least common multiple of all denominators,the left hand side of (4.1) becomes Z := N Y j =1 ( λ j , λ j ) = N Y j =1 (cid:0) k j α + m j β, k j α + m j β (cid:1) = N Y j =1 (cid:0) k j ( α, α ) + 2 k j m j ( α, β ) + m j ( β, β ) (cid:1) , where λ j = k j α + m j β .In the same manner we can write down the products in the right hand side of (4.1) times Z . We rewritethe equation via the entries of the Gram matrix of the lattice Λ, which are equal to g = ( α, α ) = s , g = ( β, β ) = t , and g = 0. We also use the fact that polynomial functions r i ∈ R take only rationalvalues on doubly periodic unit triangular surfaces, and denote by q i ∈ Q the value of r i on the surface( K, θ ). We then have:(4.2) N Y j =1 (cid:0) k j s + m j t (cid:1) − M X i =1 q i N i Y j =1 (cid:0) k ij s + m ij t (cid:1) = 0 , where all N i < N . Note that this is the only time in the proof we use the fact that we have unit triangles,and that the periodicity vectors are orthogonal.We conclude that the polynomial P ( s , t ) formed by the equation (4.2) has rational coefficients. Let x ← s and y ← t . From above, the polynomial P ( x, y ) is nonzero sincedeg N Y j =1 (cid:0) k j x + m j y (cid:1) = N, and the degree of all other terms in (4.2) have degrees N i < N . This completes the proof. (cid:3) Further applications.
Note that polynomials P found in the proof of Theorem 1.2 are quitespecial. In some cases, with a more careful analysis, one can conclude non-existence of domes for somerhombi whose diagonals are algebraically dependent over Q . For example, consider the rhombi whoseratio of diagonal lengths is algebraic: Corollary 4.9.
Let s / ∈ Q and t/s ∈ Q . Then the unit rhombus ρ ( s, t ) cannot be domed. For example, the corollary implies that ρ (cid:0) π , π (cid:1) and ρ (cid:0) e √ , e √ (cid:1) cannot be domed. Proof.
Suppose ρ ( s, t ) ∈ D , and let c := t/s ∈ Q . Consider a polynomial P ( s , t ), as in the proof ofTheorem 1.2. Viewed as a polynomial in x = s over Q the leading degree term of P becomes N Y j =1 (cid:0) k j x + m j c x (cid:1) , OMES OVER CURVES 13 so it still has a higher degree than all other terms. Therefore, s ∈ Q , a contradiction. (cid:3) The following is a generalization of the previous corollary:
Corollary 4.10.
Let s / ∈ Q , and let s and t be algebraically dependent with the minimal polynomial P ( s , t ) = 0 . Suppose P ∈ Q [ x, y ] is given by (4.3) P ( x, y ) = x k y m − k + X i + j The proof follows the previous pattern. In the first case, note that the maximal degree condi-tion (4.3) is chosen in such a way that the leading term in (4.2) remains non-zero. This implies theresult. In the second, more general case, maximal degree condition (4.3) is replaced with (4.4) whichalso remains nonzero since the signs do not allow cancellations. The details are straightforward. (cid:3) Remark 4.11. The approach in the corollaries fails in two notable cases we cover in the next section.First, by Proposition 5.2 below, an isosceles triangle ∆ with side lengths (2 , , 1) can be domed if andonly if a unit rhombus ρ ( , √ 3) can be domed. Since the argument above is not applicable for ρ ( s, t )for which s , t ∈ Q , we cannot conclude that ∆ cannot be domed, cf. Conjecture 5.1.The second example where the above approach is inapplicable is the case of planar rhombi ρ (cid:0) s, t (cid:1) ,where s + t = 4, see § k j s + m j t = ( k j − m j ) d + 4 m j of theleading degree term in P can be equal to 4 when k j = ± m j = ± Big picture Integer-sided triangles. It may seem from the proof of Theorem 1.2, that only integral curveswith non-algebraic diagonals cannot be domed. In fact, we believe that only very few curves can bedomed. Conjecture 5.1. An isosceles triangle ∆ with side lengths (2 , , cannot be domed. As many other domes on curves problems, this conjecture turned out to be equivalent to that over acertain unit rhombus. Proposition 5.2. Let ∆ be an isosceles triangle with side lengths (2 , , , and let ρ ⋄ = ρ (cid:0) , √ (cid:1) . Then ∆ can be domed if and only if ρ ⋄ can be domed.Proof. Attach three unit triangles to ∆ as in Figure 8. Observe that the boundary of the resultingsurface is exactly ρ ⋄ . (cid:3) ∆ ρ Figure 8. Proof of Proposition 5.2: ∆ ∈ D if and only if ρ ⋄ ∈ D .In a contrapositive fashion, let us show that if ∆ ∈ D , then all integral triangles can be domed. Proposition 5.3. Let ∆ be an isosceles triangle with side lengths (2 , , . If ∆ can be domed, then socan every integer-sided triangle.Proof. Whenever clear, we denote polygons with their edge length sequence. Observe that all triangles( k, k, k ) and all trapezoids (1 , ℓ, , ℓ + 1) can be domed by a plane triangulation. To construct domesover all integer-sided triangles, we use the following rules:(1) for integer k > 1, 1 ≤ ℓ < √ k , two copies of ( k, k, k, k, ℓ ), and a (1 , ℓ, , ℓ + 1)trapezoid, give a triangle ( k, k, ℓ + 1) via a construction as above,(2) for integer k > ℓ < √ k , two copies of ( k, k, ℓ ), and one ( k, k, ℓ, ℓ, 1) via a tetrahedron.We now construct all triangles ( k, k, 1) one by one, alternating the rules above in the following order:∆ = (2 , , → (1) (2 , , → (2) (3 , , → (1) (3 , , → (2) (4 , , → . . . Next, we construct domes over general isosceles triangles ( k, k, ℓ ), for all 1 ≤ ℓ < k , as follows:( k, k, → (1) ( k, k, → (1) ( k, k, → (1) . . . Finally, we can span ( a, b, c ) using triangles ( k, k, a ), ( k, k, b ) and ( k, k, c ), for k ≥ max { a, b, c } largeenough. This completes the proof. (cid:3) Remark 5.4. Suppose, contrary to Conjecture 5.1, that a triangle (2 , , 1) can be domed. That wouldeasily imply Theorem 1.4. Indeed, let ℓ be an integer greater than the radius of rQ n . By Proposition 5.3,triangle ( ℓ, ℓ, r ) can also be domed. Symmetrically attach these triangles to all edges in rQ n , to form apyramid over rQ n .5.2. Flexible surfaces. Let S ⊂ R be a PL-surface homeomorphic to a sphere, and whose faces areunit triangles. We say that S is a closed dome . Such S is called flexible , if there is a continuous family { S t , t ∈ [0 , } of (intrinsically) isometric but non-congruent closed domes; closed dome S is called rigid otherwise. Conjecture 5.5. Every closed dome S ⊂ R is rigid. Curiously, this general conjecture implies Conjecture 5.1, which at first glance might seem unrelated. Proposition 5.6. Conjecture 5.5 implies Conjecture 5.1.Proof. By contradiction, suppose Conjecture 5.1 is false. In other words, suppose triangle ∆ with sidelengths (2 , , 1) can be domed. By Proposition 5.3, then so can every integer-sided triangle, includingtriangles with sides (3 , , 7) and (4 , , Bricard octahedron (see e.g. [25, § (cid:3) Remark 5.7. We believe that the rigidity claim in the conjecture can be replaced with infinitesimalrigidity , see e.g. [25, § Planar unit rhombi. Denote by A the set of all a ≥ 0, such that the planar rhombus ρ ( a, √ − a )can be domed. It follows from Lemma 2.1 that X ⊆ A , so A is infinite. Conjecture 5.8. Set A is countable. The following result is our only evidence in favor of this conjecture. Proposition 5.9. Conjecture 5.5 implies Conjecture 5.8. OMES OVER CURVES 15 Proof. By contradiction, suppose Conjecture 5.8 is false. Since there is a countable number of combina-torial types of triangulated surfaces with quadrilateral boundary, there is one type that with uncountablemany boundary rhombi ρ ( a, √ − a ), a ∈ A . Pick one for each rhombus: S a , a ∈ A . Denote by M the space of surfaces with this combinatorial type (modulo rigid motions, as always). Since the limitof realizations is a realization, the space M is compact. Thus there is a infinite converging sequence S a n → S a , where a n → a , a n ∈ A , n ∈ N . Since M is closed and semi-algebraic, this implies that a ∈ A , and defines a continuous deformation { S x , x ∈ [ a − ǫ, a ] } or { S x , x ∈ [ a, a + ǫ ] } . Attaching twocopies of S x along the rhombus boundary, gives a nontrivial deformation of the closed dome. This refutesConjecture 5.5. (cid:3) Finally, by analogy with Conjecture 5.1, we believe the following claim. Conjecture 5.10. We have: / / ∈ A . In other words, the planar unit rhombus ρ ♦ := ρ (1 / , √ / cannot be domed. There is a nice connection between these two conjectures. Proposition 5.11. If the planar integral rhombus ρ ♦ cannot be domed, then both Conjecture 5.1 andConjecture 5.10 are true.Proof. Observe that 2 ρ ♦ can be tiled with four copies of ρ ♦ . Similarly, 2 ρ ♦ can be tiled with four copiesof the triangle ∆ with sides (2 , , (cid:3) Space colorings. Denote by R a unit distance graph of R , i.e. a graph with vertices points in R and edges pairs ( x, y ) ∈ R such that | x, y | = 1. Questions about colorings of R avoiding certainsubgraphs are the main subject of the Euclidean Ramsey Theory , see e.g. [13, 29].Denote by χ : R → { , , } a coloring of R . We say that [ xyz ] ⊂ R is a rainbow triangle in χ , if[ xyz ] is a unit triangle, and χ ( x ) = 1, χ ( y ) = 2, χ ( z ) = 3. Proposition 5.12. Let ρ = [ uvwx ] ⊂ R be a unit rhombus. Suppose there is a coloring χ : R →{ , , } with no rainbow triangles, and such that χ ( u ) = χ ( v ) = 1 , χ ( w ) = 2 , χ ( x ) = 3 . Then ρ cannotbe domed.Proof. By contradiction, suppose S is a 2-dimensional triangulated surface with the boundary ∂S = ρ .Consider a closed 2-manifold M := S ∪ [ uvx ] ∪ [ vwx ]. By Sperner’s Lemma for closed 2-manifolds,see [24, Cor. 3.1], the number of rainbow triangles in M is even. Note that triangle [ vwx ] is rainbow,while triangle [ uvx ] is not. Thus S has at least one rainbow triangle, a contradiction. (cid:3) The idea to use the coloring to prove that some curves cannot be domed can be illustrated with thefollowing conjecture: Conjecture 5.13. Let ρ ♦ = [ uvwx ] ⊂ R be a unit rhombus defined above. Then there is a coloring asin Proposition 5.12. Domes over multi-curves. One can generalize Kenyon’s Question 1.1 to a disjoint union ofintegral curves Υ = γ ∪ . . . ∪ γ k , and ask for a dome over Υ. A special case of this, when Υ is union oftwo triangles is especially important in view of the Steinhaus problem , see § Conjecture 5.14. There are unit triangles ∆ , ∆ ⊂ R , such that Υ = ∆ ∪ ∆ cannot be domed. In the spirit of Proposition 5.2 and the proof of Theorem 1.3, we conjecture that for every integralcurve γ ⊂ R , whether it can be domed can be reduced to a single rhombus. This is the analogue of“cobordism for domes”. Formally, in the notation above, we believe the following holds: Conjecture 5.15. For every integral curve γ ∈ M n , there is a unit rhombus ρ ∈ M , and a dome over γ ∪ ρ . In the spirit of the proof Theorem 1.3, there is a natural way to split Conjecture 5.15 into two parts. Conjecture 5.16. For every integral curve γ ∈ M n , there is a finite set of unit rhombi ρ , . . . , ρ k ∈ M ,and a dome over γ ∪ ρ ∪ · · · ∪ ρ k . Conjecture 5.16 is of independent interest. If true, it reduces Conjecture 5.15 to the following claim: Conjecture 5.17. For every two unit rhombi ρ , ρ ∈ M , there is a unit rhombus ρ ∈ M and adome over ρ ∪ ρ ∪ ρ . General algebraic dependence. While much of the paper and earlier conjectures are largelyconcerned with reducing the problem to domes over rhombi, there is another direction one can explore.Namely, one can ask if Theorem 1.2 can be generalized to all integral curves.Let γ = [ v . . . v n ] ∈ M n be an integral curve. Denote by L n = Q [ x , , x , , . . . , x n − ,n ] the ringof rational polynomials with variables corresponding to diagonals of γ . Let CM n ⊂ L n be the idealspanned by all Cayley–Menger determinants on vertices { v , . . . , v n } , see [7] and [25, § Conjecture 5.18. Let γ = [ v . . . v n ] ∈ D n be an integral curve which can be domed, where n ≥ .Denote by d ij = | v i v j | the diagonals of γ , where ≤ i < j ≤ n . Then there is a nonzero polynomial P ∈ L n , P / ∈ CM n , such that P (cid:0) d , , d , , . . . , d n − ,n (cid:1) = 0 . This conjecture can be viewed as a direct analogue of Sabitov’s theory of volume being algebraicover squared diagonal lengths, see § Final remarks γ ” owes much to the architectural style of the iconic geodesic domes popularized by Buckminster Fuller, and his ill-fated 1960 proposal of a Dome overManhattan , see e.g. [5, pp. 321–324].6.2. Kenyon formulated Question 1.1 in [19, Problem 2], in an undated webpage going back to at leastApril 2005. It is best understood in the context of regular polygonal surfaces (see e.g. [1, 8]). Whilewe study only the weaker notion (realizations), both the immersed and the embedded surfaces can beconsidered, as they add further constraints to the domes. Note also that combinatorially, a dome is a unit distance complex of dimension two [17], a notion generalizing the unit distance graphs in § § B ≥ p / k, k, 1) triangle, while the matchingupper bound is based on elementary arguments in plane geometry. For general d , the best known bound B d ≤ d is due to Grinberg and Sevast’janov [14]. B´ar´any and others conjecture that B d = O ( √ d ), whichwould match the Bergstr¨om–type lower bound B d ≥ p ( d + 3) / 4. We refer to an interesting survey [3]for these results and further references.6.6. Building on his earlier work and on [7], Sabitov in [26] and [27, § OMES OVER CURVES 17 Scottish book , Steinhaus introduced the tetrahedral chains , which are polyhedra with a chain-likepartition into regular unit tetrahedra. They can be viewed as special types of domes over two triangles,see § R .While the former was given a negative answer in 1959 by ´Swierczkowski, the latter was partially resolvedonly recently by Elgersma and Wagon [11]. A somewhat stronger version was later proved by Stewart [32].Stewart’s paper is especially notable. He uses the ergodic theory of non-amenable group actions, andreproves the (previously known) fact that as a subgroup of O (3 , R ), the group G of face reflections of aregular tetrahedron is isomorphic to a free product: G ≃ Z ∗ Z ∗ Z ∗ Z . From there, Stewart showedthat G is dense in O (3 , R ). One can view this result as an advanced generalization of our Lemma 2.2.The original Steinhaus problem about the group of reflections being dense in the full group O (3 , R ) ⋉ R of rigid motions remains open. Acknowledgements. We are grateful to Arseniy Akopyan, Nikolay Dolbilin, Oleg Musin, Bruce Roth-schild and Ian Stewart, for interesting comments and help with the references. We are especially thankfulto Rick Kenyon for encouragement. The authors were partially supported by the NSF. References [1] I. M. Alevy, Regular polygon surfaces, 25 pp.; arXiv:1804.05452 .[2] M. F. Atiyah and I. G. Macdonald, Introduction to commutative algebra , Addison-Wesley, Reading, MA, 1969.[3] I. B´ar´any, On the power of linear dependencies, in Building bridges , Springer, Berlin, 2008, 31–45.[4] V. Bergstr¨om, Zwei S¨atze ¨uber ebene Vectorpolygone (in German), Abh. Math. Sem. Univ. Hamburg (1931),148–152.[5] E. Bookstein, The Smith Tapes , Princeton Arch. Press, New York, 2015, 416 pp.[6] R. Connelly, A counterexample to the rigidity conjecture for polyhedra, IH ´ES Publ. Math. (1977), 333–338.[7] R. Connelly, I. Sabitov and A. Walz, The bellows conjecture, Beitr. Algebra Geom. (1997), 1–10.[8] H. S. M. Coxeter, Regular polytopes (Second ed.), Macmillan, New York, 1963.[9] P. R. Cromwell, Polyhedra , Cambridge Univ. Press, New York, 1997.[10] N. P. Dolbilin, M. A. Shtan’ko and M. I. Shtogrin, Nonbendability of the covering of a sphere by squares, Dokl. Akad.Nauk (1997), no. 4, 443–445.[11] M. Elgersma and S. Wagon, Closing a platonic gap, Math. Intelligencer (2015), no. 1, 54–61.[12] A. A. Gaifullin and S. A. Gaifullin, Deformations of period lattices of flexible polyhedral surfaces, Discrete Comput.Geom. (2014), 650–665.[13] R. L. Graham, Euclidean Ramsey theory, in Handbook of discrete and computational geometry (Second ed.), CRCPress, Boca Raton, FL, 2004, 239–254.[14] V. S. Grinberg and S. V. Sevast’janov, Value of the Steinitz constant, Functional Anal. Appl. (1980), 125–126.[15] I. Izmestiev, Classification of flexible Kokotsakis polyhedra with quadrangular base, IMRN (2017), 715–808.[16] N. W. Johnson, Convex polyhedra with regular faces, Canadian J. Math. (1966), 169–200.[17] G. Kalai, Some old and new problems in combinatorial geometry I: around Borsuk’s problem, in Surveys in combi-natorics , Cambridge Univ. Press, Cambridge, 2015, 147–174.[18] O. N. Karpenkov, On the flexibility of Kokotsakis meshes, Geom. Dedicata (2010), 15–28.[19] R. Kenyon, Open problems , available at https://gauss.math.yale.edu/˜rwk25/openprobs/index.html; an April 5, 2005version is cached by the Wayback Machine .[20] S. Lang, Introduction to algebraic geometry , Addison-Wesley, Reading, MA, 1972.[21] J. Li, Flexible Polyhedra: Exploring finite mechanisms of triangulated polyhedra Amer. Math. Monthly (1970), 161–164.[23] P. Monsky, A conjecture of Stein on plane dissections, Math. Z. (1990), 583–592.[24] O. R. Musin, Extensions of Sperner and Tucker’s lemma for manifolds, J. Combin. Theory, Ser. A (2015),172–187.[25] I. Pak, Lectures on discrete and polyhedral geometry Izv. Math. (2002), 377–391.[27] I. Kh. Sabitov, Algebraic methods for the solution of polyhedra, Russian Math. Surveys (2011), 445–505.[28] B. Schulze, Combinatorial rigidity of symmetric and periodic frameworks, in Handbook of geometric constraint systemsprinciples , CRC Press, Boca Raton, FL, 2019, 543–565.[29] A. Soifer, The mathematical coloring book , Springer, New York, 2009.[30] H. Stachel, Flexible polyhedral surfaces with two flat poses, Symmetry (2015), 774–787.[31] R. P. Stanley, Enumerative combinatorics , vol. 1 (second ed.), Cambridge Univ. Press, 2012.[32] I. Stewart, Tetrahedral chains and a curious semigroup, Extracta Math. (2019), 99–122. Appendix A. Doubly periodic surface with a three-dimensional flex A.1. Flex dimension. In [12], Gaifullin and Gaifullin studied the case of doubly periodic surfaceshomeomorphic to the plane. In this case they proved a stronger result than Theorem 4.3, that there aretwo primitive vectors λ, µ ∈ Λ ∗ , such that the place φ is finite both on ( λ, λ ) and ( λ, µ ). They concludedthe following result: Theorem A.1 ([12, Thm 1.4]) . Every embedded doubly periodic triangular surface homeomorphic to aplane has at most one-dimensional doubly periodic flex. By a doubly periodic flex of the triangular surface S we mean a continuous rigid deformation { S t , t ∈ [0 , δ ) } for some δ > 0, which preserves double periodicity, i.e. invariant under the action of G = Z ⊕ Z (the action of G can also depend on t ). The continuity of S is meant with respect to all dihedral angles.Here we identify deformations modulo changes of parameter t and ask for the dimension of the space offlexing at t = 0, i.e. when S = S . For example, the surface in Figure 6, when triangulated along theshadow lines has only one doubly periodic flex along these lines.Let us mention that flexible doubly periodic surfaces is an important phenomenon in Rigidity Theory,with Kokotsakis surfaces introduced in 1933, giving classical examples, see e.g. [15, 18]. Note that thereare doubly periodic polyhedral surfaces whose flexes are not doubly periodic, see [30]. We refer to [28, § New construction. In [12, Question 1.5], the authors asked if Theorem A.1 can be extendedto surfaces which are not homeomorphic to a plane. In this section we give a negative answer to thisquestion by an explicit construction. Theorem A.2. There is a doubly periodic triangular surface whose doubly periodic flex is three-dimensional.Proof. First, consider a flexible polyhedron F satisfying the following conditions. Polyhedron F musthave two faces f and f , both of them centrally symmetric, such that the distance between their centerschanges during the flexing and all other faces of F are triangles. To construct such F , take, for example,one of Bricard’s flexible octahedra B (see e.g. [21, § § B is replaced with theusual octahedron for clarity. B F f f Figure 9. Illustration of the first step of the construction.We call the axis of F , the line segment connecting the centers of f and f (see Figure 9). Let H be a flexible polyhedron that has two faces congruent to one of the triangular faces of F . We attach H F F Fg g g g g g σ H Figure 10. Illustration of the second step of the construction. OMES OVER CURVES 19 two copies of F , which we call F and F , to each of these two faces of H (see Figure 10). For ourconstruction we are interested in such H that, when flexing H , the angle σ between the axes of F and F changes and is never zero. Again, a suitable Bricard’s octahedron satisfies this condition. We thenhave a three-dimensional flexing of the whole structure: flexing of F changes the length of the axis of F , flexing of F changes the length of the axis of F , flexing of H changes the angle σ between the axes.For the next step of the construction we consider F ′ , the image of F under the central symmetry withrespect to the center of f . By b F we denote the union of F and F ′ attached by f (see Figure 11). Fromnow on, we consider only flexes of b F such that it stays centrally symmetric with respect to the centerof f . Note that during the flex, face f and its counterpart in F ′ , face f ′ , are translates of each otherand the distance between their centers changes during the flex. One can think of b F as a polyhedralversion of accordion with bellows such that the sturdy parts of the accordion always stay parallel butthe distance between them may change. F' f ' f ' F f f g g H g g F F'F f f ' Figure 11. Illustration of the pieces of the infinite accordion.Using infinitely many copies of b F , we construct a periodic flexible surface S ( infinite accordion ) byattaching f of one copy of b F to f ′ of the next copy of b F . See Figure 12 for an illustration (cf. Figure 6).The space of periodic flexes of S is one-dimensional.Consider a flex { S t } of S = S with periodicity vector α . At one of the flexed copies of b F we attach H along g . Then attach to H along g to another copy of b F ′ which forms its own copy S ′ of the infiniteaccordion. Assume the vector of periodicity of S ′ is β . Now we attach all translates H + kα , k ∈ Z , tosurface S , and attach S ′ + kα to each of them. Then attach H + kα + mβ , m ∈ Z , to all S ′ + kα , andattach all translates S + mβ to all translates of H . The resulting surface F is doubly periodic. It canbe flexed in the following two ways: ◦ by changing lengths of both α and β when flexing { S t } and { S ′ z } , respectively, and ◦ by changing the angle σ between the axes of S and S ′ , by simultaneously flexing all copies of H. Therefore, the space of doubly periodic flexes of F is three-dimensional. (cid:3) SF FF' F'F FF'F' S' S+ α S'+ β Figure 12. Illustration of the infinite accordion surface S in the proof, and the threeother copies S + α , S ′ and S ′ + β , of the doubly periodic surface F . Remark A.3. Note that the surface S in the proof is not necessarily embedded. One can similarlyconstruct the analogous embedded surface, by a more careful choice of a flexible polyhedron, cf. [6].It would be interesting to see if such constructions can have engineering applications. We refer to arecent thesis [21], which reviews several new constructions of embedded flexible polyhedra with largerflex dimensions, and discusses various applications. Remark A.4. This surface F is a counterexample to a natural generalization of the Main Lemma 4.6.Let us mention why the proof of the Main Lemma fails for F . Note that when the initial polyhedra F and H are homeomorphic to a sphere, we have K/ Λ is a surface of genus 2 (in the notation of theproof of the Main Lemma), where the elements of the fundamental group corresponding to α and β donot commute, i.e. stand for two different handles of the surface. In particular, the inductive step in theFirst Case of the proof of the Main Lemma would not work for surgeries since cutting along t woulddisconnect all translates of S and all translates of S2