# The Configuration Space ofn-Tuples of Equiangular Unit Vectors for n=3, 4, and 5

Authors: Yasuhiko Kamiyama
Journal: Chinese Journal of Mathematics
 (2018)

---

## Abstract

LetMn(θ) be the configuration space of n-tuples of unit vectors in R3 such that all interior angles are θ. The space Mn(θ) is an (n-3)-dimensional space. This paper determines the topological type of Mn(θ) for n=3, 4, and 5.


---

## Body

## 1. Introduction

Recently, starting in [1], the topology of the configuration space of spatial polygons of arbitrary edge lengths has been considered by many authors. In the equilateral case, the definition is given as follows. For l>0, we set(1)Pnl=a1,…,an∈S2n∣∑i=1nlai=0SO3.Here ai∈S2 denote the unit vectors in the directions of the edges of a polygon; the group SO(3) acts diagonally on (a1,…,an).Many topological properties ofPn(l) are already known: First, it is clear that there is a homeomorphism(2)Pnl≅Pn1∀l.Second, it is proved in [2] that P5(1) is homeomorphic to del Pezzo surface of degree 5.Third, whenn is odd, the integral cohomology ring H∗(Pn(1);Z) was determined in [3]. We refer to [4] for other properties of Pn(1), which is an excellent survey of linkages.In another direction, we consider the space ofn-tuples of equiangular unit vectors in R3. More precisely, we define the following: We fix θ∈[0,π] and set(3)Anθ=a1,…,an∈S2n∣ai,ai+1=cos⁡θfor21≤i≤n-1,an,a1=cos⁡θ,where , denotes the standard inner product on R3. Using (3), we define(4)Mnθ=AnθSO3.It is expected that the spaceMn(θ) is much more difficult than Pn(l). For example, the following trivial observation shows that Mn(θ) does not admit a similar property to (2): when n is odd, we have Mn(0)={one  point} but Mnπ=⌀.We claim thatMn(θ) is a hypersurface of the torus Tn-2. In fact, if we forget the condition an,a1=cos⁡θ in (3), the space corresponding to (4) is Tn-2 as observed in [5, 6]. Hence the claim follows.We recall previous results onMn(θ). First, [7] considered the case for θ=π/2. The main result is that, realizing Mn(π/2) as a homotopy colimit of a diagram involving Mn-2(π/2) and Mn-1(π/2), we inductively computed χ(Mn(π/2)). In particular, we obtained a homeomorphism M5(π/2)≅Σ5, where Σ5 denotes a connected closed orientable surface of genus 5.Second, we set(5)Xnθ≔Pn1∩Mnθ.Note that Xn(θ) is the configuration space of equilateral and equiangular n-gons. Crippen [8] studied the topological type of Xn(θ) for n=3,4, and 5. The result is that Xn(θ) is either ⌀, one point, or two points depending on θ. Later, O’Hara [9] studied the topological type of X6(θ). The result is that X6(θ) is disjoint union of a certain number of S1’s and points.The purpose of this paper is to determine the topological type ofMn(θ) for n=3,4, and 5. In contrast to the fact that at most one-dimensional spaces appear in the results of [8, 9], surfaces appear in our results.This paper is organized as follows. In Section2, we state our main results and in Section 3 we prove them.


## 2. Main Results

Theorem A.
The topological type ofM3(θ) is given in Table 1.Table 1
The topological type ofM3(θ).
θ Topological type 2π/3<θ≤π ⌀ 2π/3 { o n e   p o i n t } 0<θ<2π/3 { t w o   p o i n t s } 0 { o n e   p o i n t }Theorem B.
(i) The topological type ofM4(θ) is given in Table 2.
(ii) Asθ approaches π/2, point A in Figure 1(a) approaches point B.Table 2
The topological type ofM4(θ).
θ Topological type π { o n e   p o i n t } π/2<θ<π Figure1(a) π/2 Figure1(b) 0<θ<π/2 Figure1(a) 0 { o n e   p o i n t }Figure 1
(a)M4(θ) for 0<θ<π/2 or π/2<θ<π. (b) M4(π/2).
(a) (b)Theorem C.
(i) The topological type ofM5(θ) is given in Table 3. Let Σg be a connected closed orientable surface of genus g.
(ii) (a) Letθ satisfy that 2π/5<θ<2π/3. We study the situation where θ approaches 2π/3. We identify the torus Σ1 with the Dupin cyclide, which we denote by D. (See Figure 2.)
Using this, we identifyΣ5 with #5⁡D, where the connected sum is formed by cutting a small circular hole away from the narrow part of D. As θ approaches 2π/3, the center of each narrow part pinches to a point. Thus the five singular points appear.
(b) We consider the situation whereθ increases from 2π/3. Then each pinched point of M5(2π/3) separates. Thus we obtain S2.
(c) Letθ satisfy that 2π/5<θ<2π/3. We consider the situation where θ approaches 2π/5. In contrast to (a), the center of exactly one narrow part pinches to a point. Thus one singular point appears.Table 3
The topological type ofM5(θ).
θ Topological type 4π/5<θ≤π ⌀ 4π/5 { o n e   p o i n t } 2π/3<θ<4π/5 S 2 2π/3 Contains five singular points 2π/5<θ<2π/3 Σ 5 2π/5 Contains one singular point 0<θ<2π/5 Σ 4 0 { o n e   p o i n t }Figure 2
The Dupin cyclide.Corollary D.
As a subspace of((S2)5×[0,π])/SO(3), we define the space(6)Y≔⋃0≤θ≤πM5θ.Then M5(0) is a singular point of Y and has a neighborhood CΣ4, where C denotes the cone.Remark 1.
Cone-type singularities appear in Theorems B and C and Corollary D. We note that singularities of configuration spaces of mechanical linkages have been studied extensively by Blanc and Shvalb [10].


## 3. Proofs of the Main Results

We fixθ∈[0,π] and set (7)e1=1,0,0,p=cos⁡θ,sin⁡θ,0.Normalizing a1 and a2 to be e1 and p, respectively, we have the following description:(8)Mnθ=a1,…,an∈S2n∣a1=e1,2a2=p,ai,ai+1=cos⁡θfor2≤i≤2n-1,an,a1=cos⁡θ.Hereafter we use (8).In order to prove our main results, we use the following fact, whose proof is left to the reader.Fact 2.
Let(α,β,γ)∈(S2)3 satisfy that(9)α,β=0,α,γ=cos⁡θ.Then, there exists ϕ∈R such that (10)γ=cos⁡θα+sin⁡θcos⁡ϕβ+sin⁡θsin⁡ϕα×β.Now we first consider the casen=5. Consider Fact 2 for α=p,β=(-sin⁡θ,cos⁡θ,0), and γ=a3. Then there exists x∈R such that(11)a3=cos⁡θp+sin⁡θcos⁡x-sin⁡θ,cos⁡θ,0+sin⁡θsin⁡x0,0,1.Next, we consider Fact2 for α=e1,β=(0,1,0), and γ=a5. Then there exists z∈R such that(12)a5=cos⁡θ,sin⁡θcos⁡z,sin⁡θsin⁡z.Finally, we consider Fact2 for α=a5 in (12), (13)β=-sin⁡θ,cos⁡θcos⁡z,cos⁡θsin⁡z,and γ=a4. Then there exists y∈R such that(14)a4=cos⁡θα+sin⁡θcos⁡yβ+sin⁡θsin⁡yα×β.Now we define the functionf:(R/2πZ)3×[0,π]→R by(15)fx,y,z,θ≔11,14-cos⁡θ.We can understand M5(θ) as a level set. More precisely, we define the function(16)h:f-10⟶Rby h(x,y,z,θ)=θ. Then we have(17)M5θ=h-1θif 0<θ≤π.Remark 3.
Sincef(x,y,z,0)=0 for all x,y, and z, we have h-1(0)=(R/2πZ)3. On the other hand, it is clear that M5(0)={onepoint}. Hence (17) does not hold for θ=0. Apart from this point, there is an identification (18)Y∖M50=f-10∖h-10,where Y is defined in (6).Lemma 4.
We set(19)S≔x,y,z,θ∈R2πZ3×0,π∣fx,y,z,θ=0,∂f∂xx,y,z,θ,∂f∂yx,y,z,θ,∂f∂zx,y,z,θ=0,0,0.Then S is given in Table 4.Table 4
The setS.
θ ( x , y , z ) 4π/5 ( 0,0 , π ) 2π/3 ( π , 0,0 ) ,   ( 0 , π , 0 ) ,   ( 0,0 , π ) ,   ( π , 0 , π ) ,   ( 0 , π , π ) 2π/5 ( 0,0 , π )Proof.
The lemma is proved by direct computations.Proof of Theorem C.
We considerh in (16) as a Morse function on f-1(0). First, Table 4 and (17) show that M5(4π/5)={onepoint}.
Second, direct computation shows that(20)∂f∂θ0,0,π,4π5=525-52.Since this is nonzero, the space f-1(0) is smooth at (0,0,π,4π/5). Actually, we can prove that the point is a nondegenerate critical point of the function h. Hence Morse lemma shows that there is a homeomorphism M5(θ)≅S2 for 2π/3<θ<4π/5. But if we use [11, Corollary B], we need not check that h is nondegenerate at (0,0,π,4π/5). For our reference, we draw the figure of M5(4π/5-0.1) in Figure 3.
Third, the other parts of Table3 follow from Table 4. This completes the proof of Theorem C.Figure 3
M 5 ( 4 π / 5 - 0.1 ).Proof of Corollary D.
The corollary is an immediate consequence of Theorem C.Proof of Theorem B.
We definea3 as in (11). We also define a4 to be the right-hand side of (12). We define the function f:(R/2πZ)2×[0,π]→R by (21)fx,z,θ≔a3,a4-cos⁡θ.Similarly to (17), we have M4(θ)=h-1(θ). Since h-1(θ) is one-dimensional, it is easy to draw its figure. Thus Theorem B follows.Proof of Theorem A.
We define the functionf:(R/2πZ)×[0,π]→R by f(x,θ)=a3,e1-cos⁡θ. Since M3(θ)=h-1(θ), Theorem A follows.



---

*Source: 9842324-2018-03-01.xml*