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The Topology of the Configuration Space of a Mathematical Model for Cycloalkenes

By Yasuhiko Kamiyama

Submitted: August 24th 2021Reviewed: September 27th 2021Published: November 11th 2021

DOI: 10.5772/intechopen.100723

Downloaded: 39

Abstract

As a mathematical model for cycloalkenes, we consider equilateral polygons whose interior angles are the same except for those of the both ends of the specified edge. We study the configuration space of such polygons. It is known that for some case, the space is homeomorphic to a sphere. The purpose of this chapter is threefold: First, using the h-cobordism theorem, we prove that the above homeomorphism is in fact a diffeomorphism. Second, we study the best possible condition for the space to be a sphere. At present, only a sphere appears as a topological type of the space. Then our third purpose is to show the case when a closed surface of positive genus appears as a topological type.

Keywords

  • cycloalkene
  • polygon
  • configuration space
  • h-cobordism theorem
  • closed surface

1. Introduction

The configuration space of mechanical linkages in the Euclidean space of dimension three, also known as polygon space, is the central objective in topological robotics. The linkage consists of nbars of length l1,,lnconnected by revolving joints forming a closed spatial polygonal chain.

The polygon space is quite important in various engineering applications: In molecular biology they describe varieties of molecular shapes, in robotics they appear as spaces of all possible configurations of some mechanisms, and they play a central role in statistical shape theory.

Mathematically, these spaces are also very interesting: The symplectic structure on the polygon space was studied in the seminal paper [1]. The integral cohomology ring was determined in [2] applying methods of toric topology. We refer to [3] for an excellent exposition with emphasis on Morse theory.

Recently, mathematicians are interested in a mathematical model for monocyclic hydrocarbons. The model is defined by imposing conditions on the interior angles of a polygon. The configuration space of such polygons corresponds in chemistry to the conformations of all possible shapes of a monocyclic hydrocarbon. Hence the configuration space is interesting both in mathematics and chemistry.

In order to give more detailed account, recall that monocyclic hydrocarbons are classified into two types: One is saturated type, and the other is unsaturated type. Mathematicians constructed a mathematical model for each type. We summarize the correspondence between chemical and mathematical terminologies in the following Table 1.

ChemistryMathematics
Monocyclic hydrocarbonEquilateral polygon in R3
Bond of a monocyclic hydrocarbonEdge of a polygon
Bond angle of a monocyclic hydrocarbonInterior angle of an equilateral polygon
ConformationsConfiguration space
CycloalkanesEquilateral and equiangular polygons. Their configuration space is denoted by Mnθ.
CycloalkenesEquilateral polygons whose interior angles are the same except for those of the both ends of the specified edge. Their configuration space is denoted by Cnθ.

Table 1.

The correspondence between chemical and mathematical terminologies.

Below we explain Table 1.

  1. Monocyclic saturated hydrocarbons are monocyclic hydrocarbons that contain only single bonds between carbon atoms. Monocyclic saturated hydrocarbons are called cycloalkanes. (See Figure 1 for the 6-membered cycloalkane.)

    • The mathematical model for cycloalkanes is the equilateral and equiangular polygons. Let Mnθbe the configurations of such n-gons with interior angle θ. The study of the topological type of Mnθoriginated in [4]. See the next item for more details.

    • The topological type of M4θand M5θwas determined in [4] for arbitrary θ, and that of M6θwas determined in [5] for arbitrary θ. The paper [4] also determined the topological type of M7θfor the case that θis the ideal tetrahedral bond angle, i.e. θ=arccos13109.47°. The result was generalized in [6] for generic θ.

  2. Monocyclic unsaturated hydrocarbons are monocyclic hydrocarbons with at least one double or triple bond between carbon atoms.

    • Hereafter, for simplicity, we consider only the monocyclic unsaturated hydrocarbons that contain exactly one multiple bond.

    • It is not mathematically important whether the multiple bond is a double or triple bond. Hence we assume that the multiple bond is a double bond.

    • Such monocyclic unsaturated hydrocarbons are called cycloalkenes. (See Figure 2 for the 6-membered cycloalkene.)

    • The mathematical model for cycloalkenes is the equilateral polygons whose interior angles are the same except for those of the both sides of the specified edge. Here the specified edge corresponds to the double bond. Let Cnθbe the configurations of such n-gons with interior angle θ. (See (1) for more precise definition of Cnθ.) The study of the topological type of Cnθoriginated in [7] and the result was generalized in [8]. See the next item for more details.

    • The following result was proved in [8] (see Theorem 6): There exists θ0such that for all θθ0n2nπ, Cnθis homeomorphicto Sn4.

    • Except for the above result in [8], we do not know strong results about the topology of Cnθ.

Figure 1.

Cyclohexane (6-membered cycloalkane).

Figure 2.

Cyclohexene (6-membered cycloalkene).

On the other hand, as a combinatorial result, the necessary and sufficient condition for Mnθand Cnθto be non-empty was proved in [9]. (See Theorem 3 about the result for Cnθ.)

As stated in the last item of the above ii, we do not have enough information about the topology of Cnθ. The purpose of this chapter is to obtain systematic information about Cnθ. More precisely, we study the following:

Problem 1.(i) We prove that the above homeomorphism in [8] is in fact a diffeomorphism.

(ii) We study the best possible value about the above θ0in [8].

(iii) At present, only a sphere appears as a topological type of Cnθ. We determine the topological type of C6θfor all θ. The result shows that for some θ, C6θis a closed surface of positive genus.

This chapter is organized as follows. In §2, we state our main results. In §3-§5, we prove them. In §6, we state the conclusions.

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2. Main results

We give the definition of the configuration space. Let θbe a real number satisfying 0θπ. We set

CnθP=u1unS2nthe followingiiiandiiihold.E1

  1. u1=1,0,0and un=cosθsinθ0.

  2. i=1nui=0.

  3. uiui+1=cosθfor 1in3, where denotes the standard inner product on R3.

About the conditions in (1), the following explanations are in order. (See Table 1 for chemical terminologies.)

  • The element uidenotes the unit vector in the direction of the edge of a polygon. Then the condition ii requires the fact that u1unis in fact a polygon.

  • We specify un1to be the special edge, which corresponds to the double bond of a cycloalkene. Then the condition iii requires the fact that the interior angles of an n-gon are θexcept for those of the both ends of un1.

Remark 2.In some papers, Cnθis defined as

Anθ/SO3,E2

where we set

Anθu1unS2n1iiiiiand the conditionunu1=cosθhold.

Let SO3act on Anθdiagonally. Then for an element u1unAnθ, we may normalize u1and unto be as in (1) i. Hence (2) in fact coincides with (1).

The following result is known.

Theorem 3([9], Theorems A and B). (i) Forn4, we haveCnθif and only ifθbelongs to the following interval:

2arcsin1n1n2nπ,ifnisodd,0n2nπ,ifnis even.E3

(ii). Letabe an endpoint of the intervals in(3). Then we haveCna=onepoint.

Example 4.For n=4or 5, the following results hold, where we omit the cases which can be read from Theorem 3.

  1. For 0<θ<π2, we have C4θ=twopoints.

  2. The topological type of C5θis given by the following Table 2.

Range of θ2 arcsin 14<θ<π5π5π5<θ<π3π3π3<θ<35π
Topological type of C5θS1η1S1S1η2S1

Table 2.

The topological type of C5θ.

Here we define η1and η2to be the following Figures 3 and 4, respectively.

Figure 3.

The spaceη1.

Figure 4.

The spaceη2.

The proof of the example will be given at the end of §5.

In [8], the following proposition is proved using the implicit function theorem.

Proposition 5([8], Proposition 1). There existsθ0such that for allθθ0n2nπ, the system of equations defined by(1) i, ii andiii intersect transversely. Hence for suchθ, Cnθcarries a natural differential structure.

The main result in [8] is the following:

Theorem 6([8], Theorem 1). Letθ0be as in Proposition 5. Then for allθθ0n2nπ, Cnθis homeomorphic toSn4.

Remark 7.In [8], Theorem 6 is proved by the following method: We construct a function f:CnθRand show that fhas exactly two critical points. Then Reeb’s theorem implies Theorem 6. Note that with this method, we cannot improve the assertion from homeomorphism to diffeomorphism. (See ([10], p. 25) for Reeb’s theorem and remarks about it.)

The following theorem is the answer to Problem 1 (i).

Theorem A.We equipSn4with the standard differential structure. Letθ0be as in Proposition 5. Then for allθθ0n2nπ, Cnθis diffeomorphic toSn4.

Next we consider Problem 1 (ii). We set

αninfθ00πCnθSn4holds forallθ(θ0n2nπ).

Here in what follows, the notation XYmeans that Xis homeomorphic to Y. Note that among the values of θ0in Theorem 6, αnis the best possible one.

The following result is known.

Theorem 8([11]). (i) We haveαn=n4n2πfor4n7.

(ii). We haveα8<57π.

Remark 9.About Theorem 8 (i), we can read α4and α5from the above Example 4, and α6from Table 4 in Theorem D below.

From Theorem 8, we naturally encounter the following:

Question 10.(i) Is it true that αn=n4n2πholds for n4?

(ii) Is it true that αn<n3n1πholds for n4? Note that if (i) is true then (ii) holds automatically.

The following theorem is the answer to Problem 1 (ii).

Theorem B.For4n14, the followingTable 3 holds.

nαnn4n2πn3n1πn2nπ
4000.333π0.500π
50.333π0.333π0.500π0.600π
60.500π0.500π0.600π0.667π
70.600π0.600π0.667π0.714π
80.676π0.667π0.714π0.750π
90.729π0.714π0.750π0.778π
100.767π0.750π0.778π0.800π
110.795π0.778π0.800π0.818π
120.817π0.800π0.818π0.833π
130.834π0.818π0.833π0.846π
140.848π0.833π0.846π0.857π

Table 3.

The value of αnfor 4n14.

The following theorem is the answer to Question 10.

Theorem C.(i) The statement in Question 10(i) is false forn8. In fact, we haven4n2π<αnforn8.

(ii) The statement in Question 10(ii) is false forn13. In fact, we haven3n1π<αnforn13.

The following theorem is the answer to Problem 1 (iii).

Theorem D.The topological type ofC6θis given by the followingTable 4, where we omit the cases which can be read from Theorem 3.

Range of θ0<θ<π3π3<θ<π2π2<θ<23π
Topological type ofC6θ#3S1×S1#3S1×S1S2

Table 4.

The topological type of C6θ.

Remark 11.(i) As indicated in Problem 1 (iii), not only S2but also #3S1×S1appears in Table 4 as a topological type of C6θ

(ii) We can also determine the topological type of C6π2and C6π3. (See Remark 18 in §5.) In particular, they have singular points.

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3. Proof of Theorem A

Following the method of [12], we set

XnPθS2n×0n2nπPCnθ.E4

We define the function μ:XnRby

μPθ=θ.E5

Note that for all θ0π, we have

μ1θ=Cnθ.E6

The following proposition holds:

Proposition 12.(i) Letθ0be as in Proposition 5. Then the spaceμ1θ0n2nπis a manifold, whereμ1θ0n2nπdenotes the inverse image of the interval.

(ii) Any element ofμ1θ0n2nπis a regular point ofμ.

(iii) Consider the casen=8. ThenC868πis a non-degenerate critical point ofμ.

In order to prove the proposition, we need a lemma. We set

Dnu1un2θS2n2×0πthe followingiandiihold.

  1. u1=1,0,0.

  2. uiui+1=cosθfor 1in3.

Lemma 13.(i) There is a diffeomorphism

f:S1n3×0πDn.

(ii) We define the mapL:DnRby

Lu1un2θ=cosθsinθ0+i=1n2ui2.

Then we have the following commutative diagram:

E7

Here the mapspandgare defined as follows.

  • Let

Pr:S1n3×0πRE8

be the projection to the0π-component and we denote bypthe restriction ofPrtoLf11:

pPrLf11.

  • The mapgis a homeomorphism which will be defined in(13).

(iii) For allθ0π, the restriction of the mapgin(7) naturally induces a homeomorphism

gCnθ:Cnθp1θ.E9

Proof of Lemma 13:(i) From an element

eiϕ1eiϕn3θS1n3×0π,

we construct the element u1un2θDnas follows: In the process of constructing ui, we also construct the elements viS2such that uivi=0. We set

ui+1cosθui+sinθcosϕivi+sinθsinϕiui×viE10

and

vi+1sinθuicosθcosϕivicosθsinϕiui×vi,E11

where ui×videnotes the cross product.

In (10) and (11) for i=1, we set u11,0,0and v10,1,0. Then we obtain u2and v2. Next using (10) and (11) for i=2, we obtain u3and v3. Repeating this process, we obtain uiand vifor 1in2. Now we define fby

feiϕ1eiϕn3θu1un2θ.

From the construction, fis a diffeomorphism.

(ii) We define the map h:XnL11by

hu1unθu1un2θ.E12

Since u1unis an element of Cnθ, the right-hand side of (12) is certainly an element of L11. It is clear that his a homeomorphism. Hence if we define the map gby

gf1h,E13

then gis also a homeomorphism. From the construction, it is clear that the diagram (7) is commutative.

(iii) The item is clear from (6) and the diagram (7).

Proof of Proposition 12:Recall that Lf11in (7) is a subspace of S1n3×0π. In order to prove Proposition 12, we calculate in the universal covering space. Let

q:Rn3×0πS1n3×0π

be the universal covering space and we define the map

Pr˜:Rn3×0πR

by Pr˜Prq,where the map Pris defined in (8). Then in addition to (7), we have the following commutative diagram:

E14

(i) Let xRn3×0πbe any element which satisfies the condition

qxp1θ0n2nπ.E15

Note that if we use the diagram (14), then (15) is equivalent to saying that

Pr˜xθ0n2nπandLfqx=1.

We set

gradxLfqLfqϕ1xLfqϕn3xLfqθx.E16

In order to prove Proposition 12 (i), it will suffice to prove that

gradxLfq00E17

(a) The case whenPr˜x=n2nπ.

We claim that xhas the form

x=00n2nπ.E18

To prove this, recall the homeomorphism gCnθwas defined in (9). Since gCnθ1qxis the regular n-gon, (18) follows.

We shall prove that

gradxLfq=00rE19

for some positive real number r.

First, note that the real-valued function Lfqϕ1ϕn3n2nπtakes the minimum value 0at ϕ1ϕn3=00. Hence the first n3-terms of the both sides of (19) coincide.

Second, direct computations show that

Lfq00θ=4i=1m1isin2i12θ2,ifn=2m+1,1+2i=1m11icosiθ2,ifn=2m.E20

The number rin (19) equals to the derivative of (20) at θ=n2nπ. It is easy to see that

1icos2i12m14m+2π<0,for1im,1i+1sini2m22mπ>0,for1im1E21

and

i=1m1isin2i12m14m+2π=12,1+2i=1m11icosi2m22mπ=1.E22

Using (21) and (22), we can check that the derivative of (20) at θ=n2nπis positive, i.e., ris positive. Thus we have obtained (19). This completes the proof of (17) for the case (a).

(b) The case whenPr˜xθ0n2nπ.

By Proposition 5, we have

Lfqϕ1xLfqϕn3x00.E23

Then using (16), we obtain (17). This completes the proof of (17) for the case (b), and hence also that of (i).

(ii) In order to prove by contradiction, assume that μ1θ0n2nπcontains a critical point of μ. Then using (7), p1θ0n2nπcontains a critical point of p. Lifting to the universal covering space using (14), there exists an element xRn3×0πwhich satisfies the following two items:

  • We have Pr˜xθ0n2nπ.

  • The point xis a critical point of the function Pr˜under the constraint

Lfqϕ1ϕn3θ=1.E24

We apply the Lagrange multiplier method to (24). Since Pr˜ϕ1ϕn3θ=θ, there exists λRsuch that

0,0,1=λgradxLfq.E25

We compare the first n3-components of the both sides of (25). Then by (23), we have λ=0. But this contradicts the last component of (25). Hence (25) cannot occur. This completes the proof of (ii).

(iii) Consider the Eq. (24). Using the implicit function theorem, we may assume that θis a function with variables ϕi1in3: θ=θϕ1ϕn3. Note that

Pr˜ϕ1ϕn3θϕ1ϕn3=θϕ1ϕn3.

Hence it will suffice to prove the following result for n=8:

2θϕ1ϕn3ϕiϕj001i,jn30,E26

where denotes the determinant. Computing by the method of second implicit derivative, we see that the value of the left-hand side of (26) for n=8is 1216384. Hence (26) holds for n=8. This completes the proof of (iii), and hence also that of Proposition 12.

In order to prove Theorem A, we recall the following:

Theorem 14([13], Corollary B). Ford2, letMbe ad-dimensional smooth manifold without boundary andF:MRa smooth function. We setmaxFM=mand assume thatmis attained by unique pointzM. LetaRsatisfy the following four conditions:

  1. a<m.

  2. F1amis compact.

  3. There are no critical points inF1am.

  4. d5.

Then there is a diffeomorphismF1aSd1.

Remark 15.For the proof of Theorem 14, the h-cobordism theorem (see [14], p. 108, Proposition A) is crucial. Hence we cannot drop the condition d5.

Proof of Theorem A:First, we consider the case n8.

  • For Fin Theorem 14, we consider

μ:μ1θ0n2nπR.

More precisely, we denote the restriction of μin (5) to μ1θ0n2nπby the same symbol μ. Note that from the definition of F, zin Theorem 14 is Cnn2nπ.

  • By Proposition 12 (i), μ1θ0n2nπis in fact a manifold.

  • For ain Theorem 14, we consider any element θin θ0n2nπ.

Below we check that the conditions i, ii, iii and iv in Theorem 14 are satisfied. The items i and ii are clear. The item iii follows from Proposition 12 ii. The item iv follows from the following argument: Since dimXn=n3, we have dimμ1θ0n2nπ5if and only if n8.

Now we can apply Theorem 14 and obtain that μ1θis diffeomorphic to Sn4if θsatisfies that θ0<θ<n2nπ. By (6), this is equivalent to saying that Cnθis also diffeomorphic to Sn4. This completes the proof of Theorem A for n8.

Second, we consider the case n=8. If we apply the Morse lemma to Proposition 12 (iii), then we obtain that μ1θis diffeomorphic to S4if θsatisfies that θ0<θ<68π. Hence C8θis also diffeomorphic to S4. This completes the proof of Theorem A for n=8.

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4. Proofs of Theorems B and C

Proof of Theorem B:In ([8], Lemma 1), certain conditions on an element u1unof Cnθare listed. For example, a condition is given by

u2=un.E27

Let Λbe the set of the conditions. For λΛ, we set

Θλinf{θ00n2nπforallθθ0n2nπ,Cnθdoes notcontainan element which satisfies the conditionλ}.E28

Using this, we set

βnmaxΘλλΛ.E29

Then it is proved in ([8], Proposition 1) that

αn=βn.E30

We explain how to compute Θλ. As an example of λ, we consider the condition (27). We construct the continuous function

Rλ:0πRE31

which satisfies the following two properties:

  1. We have Rλθ0for all θ.

  2. An element θ0πsatisfies Rλθ=0if and only if Cnθcontains an element which satisfies the condition (27).

In order to construct Rλin (31), we first fix θand define the space Ynθas follows:

Ynθu1unS2nthe followingiandiihold.

  1. u1=1,0,0and u2=un=cosθsinθ0.

  2. uiui+1=cosθfor 2in3.

Second, we define the function rλ:YnθRas follows: For u1unYnθ, we set

rλu1uni=1nui.E32

Third, we define Rλin (31) by

RλθminrλYnθ.

Below we check the above properties a and b of Rλ.

The item a is clear.

In order to prove the item b, we claim the following identification holds:

rλ10=u1unCnθu2=un.E33

In fact, an element u1unYnθbelongs to rλ10if and only if (1) ii holds. Hence (33) follows.

Now the item b is clear from (33). Thus we have checked the above properties a and b.

Next using the properties a and b, we can describe Θλin (28) as

Θλ=maxθ0πRλθ=0.E34

From the constructions in (10) and (11), we have

YnθS1n4×S2.

Using this fact, we can compute the right-hand side of (34) for n14.

By a similar method, we compute Θλfor each λΛ. Then from the definition of βnin (29), we can determine βn. Finally, using (30), we obtain αn. This completes the proof of Theorem B.

Remark 16.In the above proof of Theorem B, the identification (33) is crucial. Although rλ10is a critical submanifold of the function rλin (32), this fact allows us to compute the right-hand side of (34) for n14. See §6 (ii) for further remarks.

Proof of Theorem C:The theorem is clear from Table 3.

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5. Proof of Theorem D

The following proposition is a refinement of Proposition 12 for n=6.

Proposition 17.(i) The spaceX6is a manifold, whereXnis defined in(4).

(ii) The interior angleθis a critical point ofμif and only ifθequals toπ3, π2or23π.

Proof:We can prove the proposition is the same way as in Proposition 12. Since the dimension is low, we can perform direct computations.

We apply the first fundamental theorem of Morse theory to Proposition 17. Then we obtain the following assertion: If θ1and θ2belong to the same interval from the three intervals 0π3, π3π2and π223π, then we have C6θ1C6θ2.

The homeomorphism (9) tells us that in order to determine the topological type of C6θ, it will suffice to determine the topological type of p1θfor n=6. For a fixed ψ02π, we set

Mθψeiϕ1eiϕ2eiϕ3θp1θϕ1=ψ.

Since Mθψis a one dimensional object, it is not to difficult to draw its figure. The results are given as follows.

(i) The case whenπ2<θ<23π.

There exists ωin 0πsuch that the following homeomorphism holds:

Mθψonepoint,ifψ=ωor2πω,,ifω<ψ<2πω,S1,otherwise.

From this, we have p1θS2. The figure of C6θis given by the following Figure 6.

(ii) The case whenπ3<θ<π2.

There exists ωin 0πsuch that the following homeomorphism holds:

Mθψσ,ifψ=ωor2πω.S1,otherwise,E35

Here we set

σi=12xyR2x2+i2y2=1.

(The figure of σis given by the following Figure 5(b).)

Figure 5.

(a)Mθωε; (b)Mθω; (c)Mθω+ε.

We claim that the four intersection points in MθωMθ2πωare saddle points of p1θ. In fact, for a sufficiently small positive real number ε, the following Figure 5(a)(c) give the shape of Mθωε, Mθωand Mθω+ε, respectively. (The deformation of the shape of Mθψwhen ψis near 2πωis also given by Figure 5.) Now from Figure 5, we see that the four intersection points are in fact saddle points.

Since we identify Mθ0with Mθ2π, (35) and Figure 5 give the homeomorphism p1θ#3S1×S1. The figure of C6θis given by the following Figure 7.

(iii) The case when0<θ<π3.

The topological type of Mθψis the same as (35). Hence the argument in (ii) remains valid.

Remark 18.We determine the topological type of C6π2and C6π3.

(i) The figure of C6π2is given by the following Figure 8.

Thus C6π2is homeomorphic to the orbit space S2/, where the equivalence relation is generated by

1,0,01,0,0,0100,1,0and0010,0,1.

In particular, C6π2has three singular pints.

As θapproaches π2from below, each center of the three handles in Figure 7 shrinks. And when θ=π2, each center pinches to a point and we obtain Figure 8. If θincreases further from π2, then the pinched point separates and we obtain Figure 6.

Figure 6.

The spaceC6θforπ2<θ<23π.

(ii) The figure of C6π3is given by the following Figure 9.

The space C6π3contains subspaces

N1,N2andN3E36

which satisfy the following three properties:

  • N1S1×S1,.N2S1×S1andN3#2S1×S1

  • i=13Ni=C6π3.

  • i<jNiNji=13xyR2x2+i2y2=1.

The figure of C6π3±εis given by Figure 10 above.

As θapproaches π3, a cross-section of the four tubes in Figure 7 becomes a union of two circles: In the notation of (36), one circle becomes a handle of N3. And the other circle is a subspace of N1N2.

Figure 7.

The spaceC6θforπ3<θ<π2, where we identify the opposite boundaries.

Figure 8.

The spaceC6π2, where we identify the opposite vertices.

Figure 9.

The spaceC6π3.

Figure 10.

The spaceC6π3±ε.

On the other hand, the hole of the center of Figure 7 becomes a subspace of N1N2.

Proof of Example 4:We can prove the example in the same way as in Theorem D. We can also prove by the following method. Recall that the space Xnwas defined in (4). The figures of X4and X5are given by Figures 11 and 12 above, respectively.

Figure 11.

The spaceX4.

Figure 12.

The spaceX5.

The identification (6) tells us that each level set of Figure 11 gives C4θ, and that of Figure 12 gives C5θ. Thus we obtain Example 4.

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6. Conclusions

  1. We have the following comments about the proof of Theorem A. Recall that for the proof of Theorem A given in §3, we used Proposition 5 but we did not use Theorem 6. In other words, we did not use Reeb’s theorem. Instead, we used Theorem 14, for which the h-cobordism theorem is crucial. From the computations for small n, it seems that (26) holds for all n. If we could prove this, then we obtain a proof which uses only the Morse lemma. We pose the following question: Is it possible to prove (26) for all n?

  2. We have the following comments about the proof of Theorem B. Recalling (23) and (24), we consider the following system of equations:

Lfqϕ1xLfqϕn3x=00E37

and

Lfqϕ1ϕn3θ=1.E38

If we could solve the system of Eqs. (37) and (38) with respect to the variables ϕ1,,ϕn3and θ, then we could determine for which θ, Cnθhas a singular point and the set of singular points of Cnθ. In particular, we obtain Proposition 5. But it is not easy to solve a system of equations even if we can use a computer. Hence, as we remarked in Remark 16, we have given the proof of Theorem B such as in §4. We pose the following question: Is it possible to solve the system of Eqs. (37) and (38) with respect to the variables ϕ1,,ϕn3and θ?

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Yasuhiko Kamiyama (November 11th 2021). The Topology of the Configuration Space of a Mathematical Model for Cycloalkenes [Online First], IntechOpen, DOI: 10.5772/intechopen.100723. Available from:

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