# The Topological Type of Equilateral and Almost Equiangular Polygon Spaces

Authors: Yasuhiko Kamiyama
Journal: Journal of Mathematics
 (2019)

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## Abstract

LetPn be the configuration space of equilateral spatial n-gons. For θ∈0,π and k∈0,1,…,n, let Pnkθ be the subspace of Pn consisting of elements whose first k bond angles are θ. Recently, the topological type of Pnkθ was determined for small n, special θ, and k=n or n−2. In this paper, we determine the topological type of Pnn−3θ for general n and θ.


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## Body

## 1. Introduction

Starting in [1], the topology of polygon spaces in the Euclidean space of dimension three has been studied extensively by many authors. For example, Milgram and Trinkle [2] obtained results by making excellent use of Morse surgery. We refer to [3] for an excellent exposition with emphasis on Morse theory.LetPn be the configuration space of equilateral spatial n-gons. Recently, in order to construct a mathematical model of chemical objects, attempts are made to impose restrictions on the bond angles of an element of Pn. For example, an equilateral and equiangular polygon can be considered as a mathematical model of a cycloalkane. We fix the number n of the vertices and the bond angle θ. Then, the set of all the possible shapes of a model is called conformations in chemistry and configuration spaces in mathematics (for more about chemistry, see Example 1 and the papers therein).Generalizing the configuration space which was already constructed, we give the following filtration ofPn.Definition 1.
Forθ∈0,π and k∈0,1,…,n, we set(1)Ankθ≔P=γ1,…,γn∈S2n∣the followingiandiihold.(1)
∑i=1nγi=0.(2)
γi,γi+1=−cosθ, for 1≤i≤k.
Here, denotes the standard inner product on ℝ3. Moreover, when k=n, we understand γn,γn+1 to be γn,γ1.
LetSO3 act on Ankθ diagonally, and we set Pnkθ≔Ankθ/SO3. Note that Pnkθ is the subspace of Pn consisting of elements whose first k bond angles are θ. In particular, we have Pn0θ=Pn.Example 1.
(i) An equilateral and equiangularn-gon can be considered as a mathematical model of a cycloalkane. Such a cycloalkane is called cyclobutanes, cyclopentanes, or cyclohexanes according as n=4,5, or 6. The space Pnnθ is the configuration space of such a model. The topological type of Pnnθ for various θ was determined by Crippen [4] for n=3,4, and 5 and by O’Hara [5] for n=6.
(ii)
Consider an equilateral polygon whose bond angles are the same except for the last two ones. It can be considered as a mathematical model of a ringed hydrocarbon molecule. The spacePnn−2θ is the configuration space of such a model. Goto et al. [6] studied the topological type of Pnn−2θ for the case that n is small and θ is slightly smaller than n−2/nπ. Here, note that n−2/nπ is the bond angle of the regular n-gon in ℝ2 (see Theorem 1).
The spacePnn−2θ is tough, and the only technical tools to analyze it are the implicit function theorem and Reeb’s theorem. For this reason, the known result is limited to the case that Pnn−2θ is homeomorphic to a sphere (see Theorem 1).
The purpose of this paper is to study the topology ofPnn−3θ. Since we can use usual techniques in topology, for example, fiber bundles, we can determine the topological type of Pnn−3θ for general n and θ.
This paper is organized as follows. In Section2, we prove preliminary results. In Section 3, we state our main theorem. In Section 4, we prove auxiliary lemmas. In Section 5, we prove our main theorem. Finally, in Section 6, we state our conclusions. (Note: throughout this paper, the notation X≅Y means that X is homeomorphic to Y.)


## 2. Preliminary Results

We first recall the results aboutPnn−2θ.Theorem 1  ([6]).
(i)
There is a homeomorphism:
(2)
P
n
n
−
2
θ
=
∅
,
if
n
−
2
n
π
<
θ
,
one point
,
if
θ
=
n
−
2
n
π
.
Note thatPnn−2n−2/nπ consists of the regular n-gon in ℝ2.(ii)
Letε be a sufficiently small positive real number. Then, for n≤8, we have Pnn−2n−2/nπ−ε≅Sn−4.In the definition ofPnkθ, we may normalize that γn−4=−cosθ,−sinθ,0 and γn−3=1,0,0. Next, we set(3)ui≔γn−3+i,for0≤i≤1,vi≔−γn−4−i,if0≤i≤n−5,−γ2n−4−i,ifn−4≤i≤n−3.Then, we obtain the following description ofPnn−3θ:(4)Pnn−3θ≔P=u0,u1,v0,v1,…,vn−3∈S2n∣the  followingi,ii, andiiihold.(1)
u0=1,0,0 and v0=cosθ,sinθ,0.(2)
u0+u1=∑i=0n−3vi.(3)
u0,u1=−cosθ and vi,vi+1=−cosθ for 0≤i≤n−6.Hereafter, we use description (4).The following definition and lemma will be used in Section3.Definition 2.
Letn be an integer greater than or equal to 5.(i)
We define the functionfnθ by(5)fnθ=3+4cosθ+−1ncosn−2θ.
Then, we defineαn to be largest θ which satisfies the following equation:(6)fnθ=0,0<θ<π.(ii)
We define the functiongnθ by(7)gnθ=2+4cosθ+cos2θ+−1ncosn−5θ+cosn−4θ−cosn−3θ.
Then, we defineβn as follows:(a)
β5≔arccos1/4.(b)
Whenn≥6, we define βn to be largest θ which satisfies the following equation:(8)gnθ=0,0<θ<π.Example 2.
(i)
α
5
=
2
π
/
3
=
2.09, α6=2.17 and α7=2.28.
(ii)
β
5
=
arccos
1
/
4
=
1.32, β6=π/2=1.57 and β7=1.91.Lemma 1.
We defineP1,P2, and P3 to be the elements of Pnn−3θ, which satisfy the following conditions:(a)
The linkagesPi are planar polygons such that vn−4=vn−3.(b)
The linkagesPi do not have self-intersection points.(c)
We sets≔detu0,u1 and ti≔detvi,vi+1. Then,(i)
ForP1, we require that s>0 and ti<00≤i≤n−6.(ii)
ForP2, we require that s<0 and ti<00≤i≤n−6.(iii)
ForP3, we require that s>0, ti<00≤i≤n−7 and tn−6>0.
Figures1(a)–1(c) are P1, P2, and P3 for n=7, respectively. Then, the following assertions hold:(i)
Forn≥5, the value θ=αn realizes P1(ii)
Forn≥5, the value θ=βn realizes P2(iii)
Forn≥6, the value θ=βn realizes P3Figure 1
P
1, P2, and P3 for n=7.
(a)
(b)
(c)Proof.
Since the proofs are similar, we only prove (i). We setRϕ≔cosϕ−sinϕsinϕcosϕ. Regarding P1, the coordinate of the endpoint of u1 is given by(9)10+Rπ−θ10.
On the other hand, the coordinate of the endpoint ofvn−5 is given by(10)∑i=0n−5Rπ+θicosθsinθ.
It is easy to see that the equation2−3=2 is equivalent to −fnθ/1+cosθ. Hence, (i) holds.


## 3. Main Theorem

First, it is easy to see that there is a homeomorphismP41θ≅S1 for all θ∈0,π.Main Theorem.
Letn be an integer greater than or equal to 5. Then, the following assertions hold:(i)
We have(11)Pnn−3θ=∅,ifαn<θ,one point,ifθ=αn.
Moreover, the element ofPnn−3αn is given by P1 in Lemma 1.(ii)
Whenβn<θ<αn, we have Pnn−3θ≅Sn−3.(iii)
We have(12)Pnn−3βn≅S2S0,ifn=5,S3S0#S3S0∪V,ifn=6,Sn−3S0#Sn−3S0,ifn≥7,where S2/S0 denotes the sphere with two points identified. Moreover, V is homeomorphic to S2 such that S3/S0#S3/S0∩V≅S1. Finally, for n≥6, the singular point of Sn−3/S0#Sn−3/S0 is given by P2 and P3 in Lemma 1.(iv)
Whenn−4/n−2π<θ<βn, we have(13)Pnn−3θ≅S12,ifn=5,S1×Sn−4#S1×Sn−4,ifn≥7.
Case (iv) does not occur forn=6 because n−4/n−2π=βn.(v)
Assume thatθ satisfies the following condition:(14)πn−2≤θ≤n−4n−2π,ifnis  odd,0<θ≤n−4n−2π,ifnis even.
Then, the topological type ofPnn−3θ does not depend on θ. Moreover, the topological type is given as follows:(a)
Whenn=6, the topological type is given by the above P63β6. In fact, we have n−4/n−2π=βn when n=6.(b)
Whenn≠6, there is a homeomorphism:(15)Pnn−3θ≅V1⋃V2,where V1 is homeomorphic to the space in the right-hand side of (13) and V2 is homeomorphic to Pn−2n−3θ×S2. Moreover, if we set W≔V1∩V2, then there is a fiber bundle:(16)S1⟶W⟶Pn−2n−3θ.(vi)
Whenn is odd and 0<θ<π/n−2, we have Pnn−3θ≅S1n−3.Example 3.
(i)
Regarding the Main Theorem (iii) forn=6, the space V means P43π/2×S2.
(ii)
Regarding the Main Theorem (v) (b) forn=5, we have P53π/3≅V1∪V2, where V1≅S12, V2≅S2 such that V1∩V2≅S1.


## 4. Auxiliary Lemmas

We fixθ and set(17)Aθ≔w∈S2∣1,0,0,w=−cosθ.We define the projectionpn,θ:Pnn−3θ⟶Aθ by pn,θP=u1. We study the topological type of the fiber pn,θ−1w for each w∈Aθ. In the following lemma, we assume that n≥7. The case for n=5 or 6 can be obtained by slight modifications.Lemma 2.
Letn be an integer greater than or equal to 7. Assume that θ satisfies the following condition:(18)0<θ<πn−2orn−4n−2π<θ<αn,ifnis odd,n−4n−2π<θ<αn,ifnis even.
We write an element ofAθ by(19)wϕ≔−cosθ,cosϕsinθ,sinϕsinθ.
Then, the following assertions hold:(i)
Whenβn<θ<αn, there exists a∈0,π such that(20)pn,θ−1wϕ≅Sn−4,if−a<ϕ<a,one point,ifϕ∈−a,a,∅,otherwise.(ii)
Whenθ=βn, we have(21)pn,θ−1wϕ≅Sn−4S0,ifϕ=0,one point,ifϕ=π,Sn−4,otherwise.(iii)
Whenn−4/n−2π<θ<βn, there exists a∈0,π such that(22)pn,θ−1wϕ≅S1×Sn−5,if−a<ϕ<a,Sn−4S0,ifϕ∈−a,a,Sn−4,otherwise.(iv)
Whenn is odd and 0<θ<π/n−2, we have pn,θ−1wϕ≅S1n−4.Proof.
We construct the following commutative diagram:(23)
First, we set(24)Xnθ≔Q=u0,u1,v0,v1,…,vn−5∈S2n−2∣the followinga,b,andchold,Ynθ≔Q=u0,u1,v0,v1,…,vn−5∈S2n−2∣the followingaandchold.
Here, we define the following:(a)
u0=1,0,0 and v0=cosθ,sinθ,0(b)
u0,u1=−cosθ and vi,vi+1=−cosθ for 0≤i≤n−6(c)
If we define the mapρ by ρQ≔∑i=0n−5vi−u0+u1, then we require that ρQ≤2
Second, we define the mapi to be the inclusion. Third, we define the map π as follows: for P=u0,u1,v0,v1,…,vn−3∈Pnn−3θ, we set(25)πP≔u0,u1,v0,v1,…,vn−6,vn−5.
Fourth, we define the mapsqn,θ and rn,θ to be the natural extensions of pn,θ.
We claim that forQ∈Xnθ, we have(26)π−1Q=one point,ifρQ=2,S1,if0<ρQ<2,S2,ifρQ=0.
In fact, we can identifyπ−1Q with the locus of the endpoint of vn−4. Note that the last case of (26) does not occur because such a case would imply that Pn−2n−3θ≠∅. But a result in [7] tells us that Pn−2n−3θ≠∅ if and only if (14) holds. Hence if θ satisfies (18), then the last case of (26) does not occur.
Now since the proofs of Lemma 2(i), (ii), and (iii) are similar, we prove only (iii). Note that for eachw∈Aθ, we have(27)rn,θ−1w≅S1n−5.
Using this fact, it is easy to see that(28)qn,θ−1wϕ≅S1×Dn−6,if−a<ϕ<a,Dn−5S0,ifϕ∈−a,a,Dn−5,otherwise.
Combining (23), (26) and (28), we obtain (iii).
(iv) By combining (23) and (27), we have qn,θ−1w⊂rn,θ−1w=S1n−5. We set(29)dn≔maxθ∈0,π∣qn,θ−1w=S1n−5,for allw∈Aθ.
We claim that whenn is odd, dn is given by arccosn2−4n−3/n2−4n+3. In fact, it is easy to see that if we fix θ, then maxρYnθ is attained by u0,u1,v0,v1,⋯,vn−5 such that u1=−cosθ,−sinθ,0 and(30)vi=−1,0,0,ifiis odd,cosθ,sinθ,0,ifiis even.
From this, we computemaxρYnθ. Next, solving the equation maxρYnθ=2 with respect to θ, we have θ=arccosn2−4n−3/n2−4n+3. Hence dn must be as in the claim. Now since π/n−2<arccosn2−4n−3/n2−4n+3, (iv) follows.
The following lemma will be used in Section5.Lemma 3.
LetM be a connected oriented k-dimensional manifold. For an embedding ϕ:S0×Dk⟶M, we define another k-manifold M′ to be(31)M′≔M\intimϕ⋃ϕS0×Sk−1I×Sk−1,where we glue I×Sk−1 so that M′ is an orientable manifold. Then, there is a homeomorphism: M′≅M#S1×Sk−1.Proof.
The lemma is well known (see, for example, ([9], Example 4.17)). For completeness, we give a proof. First, we consider the case that M=Sk. Since(32)Sk\intimϕ≅I×Sk−1,we see that Sk′≅S1×Sk−1. Using this, the case of general M is proved as follows:(33)M′≅M#Sk′≅M#Sk′≅M#S1×Sk−1.


## 5. Proof of the Main Theorem

The most difficult case for the proof of the Main Theorem is the item (iv) forn≥7. Hence, we first prove this. We begin with the following.Lemma 4.
Letn be an integer greater than or equal to 7. Assume that θ satisfies the condition n−4/n−2π<θ<βn. Let N denote ∪0≤ϕ≤πpn,θ−1wϕ or ∪π≤ϕ≤2πpn,θ−1wϕ. Then, we have the following homeomorphism:(34)N≅S1×Dn−4/intDn−3.Proof.
We define the functionνm:Sm⟶ℝ by νmx1,…,xm+1=xm+1. We fix real numbers A and B which satisfy 1<A<B. We define the function μ by(35)μ=A+νn−4B+ν1:Sn−4×S1⟶ℝ.
The functionμ is a Morse function with four critical points, which are pairs of the critical points of νn−4 and ν1. Let p be the critical point of μ such that μp=A−1B+1. It is well known that the index of μ at p is 1. Using this fact, we check how a level set of μ changes when it crosses through p. The answer is that the change coincides with that in Lemma 2 (iii). Hence, we have(36)N≅μ−1A−1B+1−ε,A−1B+1+ε.
Now sinceμp is the second smallest critical value, we obtain Lemma 4.Proof of the Main Theorem (iv).
Note that(37)Pnn−3θ=⋃0≤ϕ≤πpn,θ−1wϕ⋃⋃π≤ϕ≤2πpn,θ−1wϕ,where we identify the corresponding boundaries. First, we identify only the boundaries pn,θ−1w0. Using Lemma 4, we obtain the space S1×Sn−4\∐2intDn−3. Second, we identify the remaining boundaries. Using Lemma 3, we obtain S1×Sn−4#S1×Sn−4.Proof of the Main Theorem.
Since the case forn=5 or 6 can be proved by sight modifications, we assume that n≥7.(i)
It is easy to prove the item (see also [7]).(ii)
From Lemma 2(i), we havePnn−3θ≅∑Sn−4, where ∑ denotes the suspension.(iii)
The item follows from Lemma 2(ii).(iv)
The case forn≥7 is proved above.(v)
Since the arguments are easy, we prove only the item (b). From the continuity of the deformation ofPnn−3θ with respect to θ, the space V1 appears.In addition toV1, the new space V2 appears, where we set(38)V2≔P∈Pnn−3θ∣vn−4+vn−3=0.It is clear that V2≅Pn−2n−3θ×S2. Moreover, as recalled in the proof of Lemma 2, Pn−2n−3θ≠∅ if and only if (14) holds.Next, we determineW≔V1∩V2. We set(39)Znn−3θ≔P=u0,u1,v0,v1,…,vn−3∈Pnn−3θ∣u0+u1−∑i=0n−6vi=1.We fix P∈Znn−3θ and set P¯≔u0,u1,v0,…,vn−6. Then, the endpoint of vn−5 moves on the circle which satisfies the equation vn−6,vn−5=−cosθ and the center being the point ∑i=0n−6vi. Along the circle, we let the endpoint of vn−5 approach the point u0+u1. That is, we let the vector vn−5 approach v˜n−5≔u0+u1−∑i=0n−6vi. Then, the pair of vectors vn−4,vn−3 converges to the pair of vectors v˜n−4,v˜n−3 such that the configuration(40)u0,u1,v0,v1,…,vn−6,v˜n−5,v˜n−4,v˜n−3,is an element of W. If we do this procedure for all elements P∈Znn−3θ, then we obtain all elements of W.About (40), we regard u0,u1,v0,v1,…,vn−6,v˜n−5 as an element of Pn−2n−3θ.In order to emphasize P, we writev˜n−4 in (40) by v˜n−4P. Assume that we also take Q∈Znn−3θ such that P¯=Q¯. Then, the construction of (40) shows that the endpoints of v˜n−4P and v˜n−4Q lie on a circle which is determined by P¯. Hence, have the fiber bundle (16).(vi)
By Lemma 2(iv), there is a fiber bundle:(41)S1n−4⟶Pnn−3θ⟶S1.
We show that (41) is a trivial bundle. Let Drp be the closed three-dimensional ball of radius r>0 centered at p∈ℝ3. Since θ is small, there exists a sufficiently small positive real number ε such that for all P∈Pnn−3θ, the endpoint of u0+u1 is contained in DεO and that of ∑i=0n−5vi is contained in Dε1,0,0.
Consider the space(42)E≔ξ1,ξ2,w∈DεO×Dε1,0,0×S2∣w−ξ1=w−ξ2=1,where for P∈Pnn−3θ, we are denoting u0+u1 by ξ1, ∑i=0n−5vi by ξ2 and ∑i=0n−4vi by w. Since ε is small, there is a fiber bundle:(43)S1⟶E⟶DεO×Dε1,0,0.
Moreover, since the base space is contractible, (43) is a trivial bundle. Hence, bundle (41) is also trivial.


## 6. Conclusions

We have determined the topological type ofPnn−3θ for general n and θ. We try to study the other spaces Pnkθ. As stated in Section 1, only limited information is known for the case k=n−2,n−1 or n. In this direction, the necessary and sufficient condition on θ for Pnkθ to be nonempty was determined recently (see [7]).On the other hand, for the casek≤n−4, the following observation holds: let Bm,r be the space of spatial m-gons having edge lengths 1,1,…,1,r, where we do not consider the orbit space by the SO3-action. Note that the right-hand side of (26) coincides with the topological type of B3,r. Hence, in order to study Pnkθ, we will need information on Bn−k,r (see [1, 8] for information on Bm,r).



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