# The Configuration Space of Regular Spherical Even Polygons

Authors: Yasuhiko Kamiyama
Journal: International Journal of Mathematics and Mathematical Sciences
 (2019)

---

## Abstract

Leta be a real number satisfying 0<a<π. We denote by Mna the configuration space of regular spherical n-gons with side lengths a. In our previous paper, we determined χMna for odd n. In this paper, we determine it for even n. The main difference from our previous paper is given as follows. When n is odd, we prove that Mna is obtained from Sn−3 by successive Morse surgeries. On the other hand, when n is even, we show that Mna is obtained from Pn by successive Morse surgeries. Here, Pn denotes the configuration space of equilateral n-gons in ℝ2, which has singular points when n is even.


---

## Body

## 1. Introduction and Statement of the Main Result

Recently, the topology of polygon spaces in the Euclidean space of dimension two or three has been considered by many authors. The study of planar polygon spaces started in [1–3]. For example, the homology groups were determined in [4]. On the other hand, the study of spatial polygon spaces started in [5].Morse theory plays a key role in the study of polygon spaces. [6] is an excellent exposition about polygon spaces with emphasis on Morse theory. In [7], Milgram and Trinkle obtained results by making excellent use of Morse surgery.Later, Kapovich and Millson [8] studied spherical polygon spaces. They proved a remarkable and very useful theorem about a Morse function. We first recall their result.Letr=r1,r2,…,rn be an n-tuple of real numbers satisfying 0<ri<π. We set r′≔r1,…,rn−1 and define(1)N˜r′=u=u1,…,un∈S2ndui,ui+1=ri,1≤i≤n−1,where d is the spherical distance. Let SO3 act on N˜r′ diagonally and we set Nr′≔N˜r′/SO3. Kapovich and Millson defined the function ρn:Nr′⟶ℝ by(2)ρnu=du1,un.We will restrict tou’s such that 0<ρnu<π so that ρn is differentiable. Note that(3)Mr≔ρn−1rn⊂Nr′,is the moduli space of closed polygonal linkages in S2 with side lengths r1,…,rn.It is proved in [8] that an element of Nr′ is a critical point of ρn if and only if it is degenerate, i.e., it lies in a great circle γ in S2. In order to describe the signature, we give the following.Definition 1 (orientingγ).
Supposeu=u1,u2,…,un is a closed degenerate linkage contained in a great circle γ. Orient γ so that the arc joining u1 to un is positively directed. Thus, an edge ei is a back-track if it has the same direction as en=un,u1.Theorem 1 ([8], Main Theorem).
LetT∈Nr′ be a degenerate free n−1 linkage and P be the associated degenerate closed n-gon linkage. Then, the signature of D2ρnT is given by(4)bP+2wP−1,fP−2wP−1,where f=fP denotes the number of forward-tracks and b=bP denotes the number of back-tracks so f+b=n. Moreover, wP denotes the winding number of the degenerate configuration P, which is given by the formula ∑i=1nεiri=2πwP, where εi∈±1 is defined as follows: if ei is the forward-track, the we set ei=1, and if ei is the back-track, then we set ei=−1.Leta be a real number satisfying 0<a<π. We consider the case that r=a,a,…,a, that is, ri=a for 1≤i≤n. We set(5)Mna≔ρn−1a.Note thatMna is the moduli space of regular spherical polygons with side lengths a.We realizeMna as a level set of a Morse function, which is different from the above ρn. We first set(6)An≔P,a=u1,…,un,a∈S2n×0,πdui,ui+1=afor1≤i≤n−1anddun,u1=a.LetSO3 act on An by(7)P,ag≔Pg,a,where g∈SO3. Then, we set(8)Xn≔AnSO3.We define the functionμ:Xn⟶ℝ by(9)μP,a=a.Note that for alla∈0,π, we have(10)μ−1a=Mna.In [9], we studied the case of odd n and proved that μ is a Morse function. Since a level set is obtained by successive Morse surgeries, we could determine χMna for all a.The purpose of this paper is to study the case of evenn. Hereafter, we will always assume n to be even and set n=2m. When n is even, Mna has the following two characteristics, which do not hold for odd n. One is that there is a homeomorphism Mna≅Mnπ−a for all a (see Theorem 5 (i)). The other is that Mna has singular points for all a. In fact, we set(11)Σna≔P∈MnaPis degenerate such thatwP=0,where wP is defined in Theorem 1. Then, an element of Σna is a singular point of Mna.AvoidingΣna, we give the following definitions. First, we set(12)M¯na≔Mna\Σna.Second, we set(13)X¯n≔P,a∈XnP∈M¯na.Third, we denoteμ¯:X¯n⟶ℝ to be the restriction of μ to X¯n.Theorems 2 and 3, which are main theorems of this paper, assert thatμ¯ is a Morse function.Theorem 2.
An elementP,a∈X¯n is a critical point of μ¯ if and only if P is degenerate.Theorem 2 implies the following two results:(i)
In contrast to the fact that the critical points ofρn are the degenerate linkages, the critical points of μ¯ are the degenerate linkages P such that wP≠0.(ii)
Alla’s which are critical values for μ¯ are rational multiples of π. Moreover, the smallest critical value is π/m and the largest one is m−1π/m.From the definition off, b, and w in Theorem 1, for P∈Mna a degenerate linkage, the following relations hold:(14)f+b=n,af−b=2πw.Then, we have the following:Theorem 3.
LetP∈M¯na be a degenerate polygon. Then, the following hold:(i)
WhenwP>0, the signature of D2μ¯P,a is given by fP−2wP−1,bP+2wP−1(ii)
WhenwP<0, the signature of D2μ¯P,a is given by bP+2wP−1,fP−2wP−1This paper is organized as follows. In Section2, we first combine Theorems 2 and 3 into Theorem 4. Then, using the theorem, we determine χMna in Theorem 6. In Section 3, we prove Theorems 2 and 3. Theorem 1 is a key to proving Theorem 3. In Section 4, we give several examples about χMna.


## 2. Conclusions from Theorems 2 and 3

In Theorem 3, we set(15)s,t≔fP−bP2,wP,ifwP>0,−fP+bP2,−wP,ifwP<0.Then, we obtain the following theorem. We denote byℕ=1,2,3,…, the set of natural numbers.Theorem 4.
We set(16)Γn≔s,t∈ℕ×ℕt<s≤m.
Then, the following assertions hold:(i)
To eachs,t∈Γn, there corresponds a certain number of critical points of μ¯. All critical points are nondegenerate such that their information is given by Table 1.
(ii)
Conversely, a critical point ofμ¯ is attained by a unique s,t∈Γn.Table 1
The information onμ¯ at s,t.
Critical value
The number of critical points
The index
t
/
s
π
2
m
m
−
s
m
−
s
+
2
t
−
1Remark 1.
(i)
It is not true that critical points of the same critical value have the same index. For example,Γ12 contains elements 3,1 and 6,2 such that their critical values are π/3. On the other hand, the index of the former is 4 but that of the latter is 3.(ii)
The number of critical points does not depend ont. Moreover, the index of μ¯ has the same parity as m−s−1, which also does not depend on t.Proof of Theorem 4.
(i)
LetUn be the set of critical points of μ¯. We define the map(17)F:Un⟶Γn,by makingP,a∈Un correspond with s,t in (15). We check that F is certainly a map to Γn. First, we check that t>0. In fact, this is clear from (15) for wP positive or negative. Second, we check that t<s. In fact, we have from the second equation of (14) that(18)as=πt.Since0<a<π, (18) implies that t<s. Third, we check that s≤m. In fact, we have from the second equation of (15) that if wP>0, then s≤f/2≤m, and if wP<0, then s≤b/2≤m.Next, we check Table1. First, (18) tells us that the critical value in Table 1 is true.Second, we compute the index ofμ¯ at s,t. From the first equation of (14) and (15), we have(19)f,b=m+s,m−s,ifw>0,m−s,m+s,ifw<0.Then, using Theorem 3, (15), and (19), we have the index of μ¯ at s,t is m−s+2t−1 for w positive or negative. Hence, the index of μ¯ in Table 1 is true.Third, we compute the number of critical points. (15) tells us that critical points P,a and Q,a satisfy FP,a=FQ,a if and only if fP,bP=fQ,bQ or fP,bP=bQ,fQ. The description of f in terms of s is computed in (19). Since en is always back-track, we need to choose f elements from e1,e2,…,e2m−1. The total number of such choices is(20)2m−1m+s+2m−1m−s=2m−1m−s−1+2m−1m−s=2mm−s.Hence, the number of critical points in Table1 is true. This completes the proof of Theorem 4 (i).(ii)
The item is clear from (15).Theorem 5.
(i)
Forr=r1,r2,…,rn, we set r′≔r1′,r2′,…,rn′, where we put ri′≔π−ri for 1≤i≤n. Then, there is a homeomorphism Mr≅Mr′, where the space Mr is defined in (3). In particular, there is a homeomorphism Mna≅Mnπ−a for all a∈0,π.(ii)
LetPn be the configuration space of equilateral n-gons in ℝ2. Then, if 0<a<π/m or m−1π/m<a<π, there is a homeomorphism Mna≅Pn.Proof.
(i)
The item is proved in ([10], Theorem 2.2) and ([8], Section 4). More precisely, we define the map G:Mr⟶Mr′ by(21)Gu1,…,un≔v1,…,vn,where we setvi≔−1iui. Then, G is a homeomorphism.(ii)
It is clear that ifε is a sufficiently small positive real number, then Mnε is homeomorphic to Pn. Moreover, since any element of 0,π/m is a regular value of μ¯ by Theorem 4, we have Mna≅Pn for 0<a<π/m. Using the above item (i), we also have Mna≅Pn for m−1π/m<a<π.Theorem 6.
LetVn be the set of critical values of μ¯ (note that we can write the elements of Vn using Table 1). Then, the following assertions hold:(i)
Assume thata∉Vn. Then, we have(22)χMna=−1m+12m−1m.(ii)
Assume thata∈Vn. Let s0,t0 be the unique element of Γn, which satisfies the following two conditions:(23)gcds0,t0=1,t0s0π=a.
Then, we have(24)χMna=−1m+12m−1m+∑i=1m/s0−1s0i2mm−s0i.
Note that the right-hand side of (24) does not depend on t0.
In order to prove Theorem 6, we need the following.Theorem 7.
Leth:M⟶ℝ be a smooth function on a d-dimensional manifold M. For numbers ξ1<ξ2, we assume that h−1ξ1,ξ2 is compact and contains a unique nondegenerate critical point p of index r. Then, the following results hold:(i)
The level seth−1ξ1 is obtained from h−1ξ2 by removing Int∂Dd−r×Dr and attaching Dd−r×∂Dr along the boundary. We call this construction a surgery of type d−r,r.(ii)
Ifd is even, then we have(25)χh−1ξ1=χh−1ξ2.(iii)
We setξ3≔hp. Then, level set h−1ξ3 is obtained from h−1ξ2 by removing Int∂Dd−r×Dr and attaching CSd−r−1×Sr−1 along the boundary, where C denotes the cone. In particular, if d is even, then we have(26)χh−1ξ3=χh−1ξ2+−1r.Proof.
The theorem is well known in Morse surgery (see [11]).Proof of Theorem 6.
We apply Theorem 7 to the functionμ in (9). Every level set Mna contains the singular point set Σna and the information on μ¯ is determined in Theorem 4.(i)
Ifa∉Vn, then Table 1 tells us that a is a regular value of μ¯. Combining Theorem 7 (ii) and Theorem 5 (ii), we have χMna=χPn. Recall that χPn was determined in ([12], Theorem A). Hence, (i) follows.(ii)
Ifa∈Vn, then Table 1 tells us that the elements of Γn whose critical value equals to a are given by s0i,t0i for 1≤i≤m/s0. Then, combining Table 1 and Theorem 7 (iii), we obtain (ii).


## 3. Proofs of Theorems 2 and 3

Proof of Theorem 2.
We prove the theorem along the lines of ([8], Theorem 2.9) (see also the proof of ([9], Theorem 1.3)). Combining the following three assertions, we obtain Theorem 2:(i)
An analogue of ([8], Lemma 2.7 (ii)): by (10), the Zariski tangent space TPM¯na is given by(27)TPM¯na=kerdμ¯P,a.(ii)
An analogue of ([8], Corollary 2.8): we see from (i) that a point P is a singular point of M¯na if and only if P,a is a critical point of μ¯.(iii)
By ([13], Theorem 1.1), P is a singular point of M¯na if and only if P is degenerate.
In order to prove Theorem 3, we need the following.Lemma 1.
We consider the functionρn in (2) for the case that r′=a,a,…,a, that is, ri=a for 1≤i≤n−1. Let O be an open neighborhood of P,a in X¯n such that O contains no other degenerate polygons than P,a. We fix a sufficiently small positive real number ε. Then, the following assertions hold:(i)
Assume thatwP>0. Then, for all j∈−1,1, μ¯−1a+jε∩O is diffeomorphic to an open set of ρn−1a−jε.(ii)
Assume thatwP<0. Then, for all j∈−1,1, μ¯−1a+jε∩O is diffeomorphic to an open set of ρn−1a+jε.Proof of Lemma 1.
(i)
We write the side lengths of a spherical polygon asr1,r2,… …,rn. For δ∈0,ε, we deform the side lengths of an element of μ¯−1a+ε∩O by(28)a+jδ,a+jδ,…,a+jδ,a+j−ε+2δ.We need to check that this deformation is indeed possible. To see this, it will suffice to see that (28) does not cross a wall for any δ, where a wall is defined in ([8], p. 311).Recall thaten is always back-track (see Definition 1). Hence, we have about (28) that(29)the sum of forward‐track side‐lengths−the sum of back‐track side‐lengths=2πwP+jε+u−2δ,for some integeru. We claim that u≥3. In fact, the assumption wP>0 tells us that at least m+1 of the first 2m−1 components of (28) are forward-track. Hence, we have u≥m+1−m−2=3 and the claim follows.Now, since the termjε+u−2δ in (2) is near zero but not zero for all δ∈0,ε, (28) does not cross a wall for any δ. Finally, setting δ=0 or ε in (28), we complete the proof of (i).(ii)
Instead of (28), we deform the side lengths of an element of μ¯−1a+ε∩O by(30)a+jδ,a+jδ,…,a+jδ,a+jε.
We have about (30) that(31)the sum of forward‐track side‐lengths−the sum  of back‐track side‐lengths=2πwP−jε+vδ,for some integer v. We claim that v≥1. In fact, the assumption wP<0 tells us that at most m−1 of the first 2m−1 components of (30) are forward-track. Hence, we have −v≤m−1−m=−1 and the claim follows.
Now, since the termjε+vδ in (9) is near zero but not zero for all δ∈0,ε, (30) does not cross a wall for any δ. Finally, setting δ=0 or ε in (30), we complete the proof of (ii). This completes the proof of Lemma 1.Proof of Theorem 3.
We shall deduce Theorem 3 from Theorem 1. We apply Theorem 7 (i) to the mapμ¯. Note that dimX¯n=n−2. We denote by r the index of P. When the level set μ¯−1a+ε descends to μ¯−1a−ε, the critical point P gives a surgery of type(32)n−2−r,r.
We setλ≔fP−2wP−1.(i)
IfwP>0, then Lemma 1 (i) tells us that the above descent is equivalent to the ascent from ρn−1a−ε to ρn−1a+ε. Combining Theorem 7 (i) and Theorem 1, when we cross through P, a surgery of type(33)λ,n−2−λ.occurs. Comparing (32) and (33), we have(34)r=n−2−λ=bP+2wP−1.(ii)
IfwP<0, then the descent from μ¯−1a+ε to μ¯−1a−ε is equivalent to the descent from ρn−1a+ε to ρn−1a−ε. When we cross through P, a surgery of type(35)n−2−λ,λ,occurs. Comparing (32) and (35), we have(36)r=λ=fP−2wP−1.This completes the proof of Theorem 3.


## 4. Examples

Proposition 1.
We have the following examples:(i)
χMnπ/2=−1m+1⋅22m−2(ii)
χMnπ/3=−3m−1(iii)
χMnπ/4=−1m+1⋅2m−2⋅2m−1+1(iv)
χMnπ/6=−1m+1⋅4m−1/3+3m−1/2+1/6.Proof.
Computing the right-hand side of (24) explicitly, we obtain the proposition.Remark 2.
Proposition 1 (i) was obtained in ([14], Theorem 2) by a different method.



---

*Source: 7148538-2019-11-14.xml*