Let Mn(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.
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 of Pn(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, when n 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 of n-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=cosθ for21≤i≤n-1,an,a1=cosθ,where , denotes the standard inner product on R3. Using (3), we define(4)Mnθ=AnθSO3.
It is expected that the space Mn(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 that Mn(θ) is a hypersurface of the torus Tn-2. In fact, if we forget the condition an,a1=cosθ in (3), the space corresponding to (4) is Tn-2 as observed in [5, 6]. Hence the claim follows.
We recall previous results on Mn(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 of Mn(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 Section 2, we state our main results and in Section 3 we prove them.
2. Main ResultsTheorem A.
The topological type of M3(θ) is given in Table 1.
The topological type of M3(θ).
θ
Topological type
2π/3<θ≤π
⌀
2π/3
{onepoint}
0<θ<2π/3
{twopoints}
0
{onepoint}
Theorem B.
(i) The topological type of M4(θ) is given in Table 2.
(ii) As θ approaches π/2, point A in Figure 1(a) approaches point B.
The topological type of M4(θ).
θ
Topological type
π
{onepoint}
π/2<θ<π
Figure 1(a)
π/2
Figure 1(b)
0<θ<π/2
Figure 1(a)
0
{onepoint}
(a) M4(θ) for 0<θ<π/2 or π/2<θ<π. (b) M4(π/2).
Theorem C.
(i) The topological type of M5(θ) is given in Table 3. Let Σg be a connected closed orientable surface of genus g.
(ii) (a) Let θ 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 identify Σ5 with #5D, 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 where θ increases from 2π/3. Then each pinched point of M5(2π/3) separates. Thus we obtain S2.
(c) Let θ 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.
The topological type of M5(θ).
θ
Topological type
4π/5<θ≤π
⌀
4π/5
{onepoint}
2π/3<θ<4π/5
S2
2π/3
Contains five singular points
2π/5<θ<2π/3
Σ5
2π/5
Contains one singular point
0<θ<2π/5
Σ4
0
{onepoint}
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 fix θ∈[0,π] and set (7)e1=1,0,0,p=cosθ,sinθ,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=cosθ for 2≤i≤2n-1,an,a1=cosθ.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=cosθ.Then, there exists ϕ∈R such that (10)γ=cosθα+sinθcosϕβ+sinθsinϕα×β.
Now we first consider the case n=5. Consider Fact 2 for α=p,β=(-sinθ,cosθ,0), and γ=a3. Then there exists x∈R such that(11)a3=cosθp+sinθcosx-sinθ,cosθ,0+sinθsinx0,0,1.
Next, we consider Fact 2 for α=e1,β=(0,1,0), and γ=a5. Then there exists z∈R such that(12)a5=cosθ,sinθcosz,sinθsinz.
Finally, we consider Fact 2 for α=a5 in (12), (13)β=-sinθ,cosθcosz,cosθsinz,and γ=a4. Then there exists y∈R such that(14)a4=cosθα+sinθcosyβ+sinθsinyα×β.
Now we define the function f:(R/2πZ)3×[0,π]→R by(15)fx,y,z,θ≔11,14-cosθ.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.
Since f(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.
The set S.
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 consider h 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 Table 3 follow from Table 4. This completes the proof of Theorem C.
M5(4π/5-0.1).
Proof of Corollary D.
The corollary is an immediate consequence of Theorem C.
Proof of Theorem B.
We define a3 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-cosθ.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 function f:(R/2πZ)×[0,π]→R by f(x,θ)=a3,e1-cosθ. Since M3(θ)=h-1(θ), Theorem A follows.
Conflicts of Interest
The author declares that there are no conflicts of interest regarding the publication of this paper.
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