Open access peer-reviewed chapter

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

Written By

Yasuhiko Kamiyama

Reviewed: 27 September 2021 Published: 11 November 2021

DOI: 10.5772/intechopen.100723

From the Edited Volume

Advanced Topics of Topology

Edited by Francisco Bulnes

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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 n bars of length l1,,ln connected 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 Mnθ be the configurations of such n-gons with interior angle θ. The study of the topological type of Mnθ originated in [4]. See the next item for more details.

    • The topological type of M4θ and M5θ was determined in [4] for arbitrary θ, and that of M6θ was determined in [5] for arbitrary θ. The paper [4] also determined the topological type of M7θ for the case that θ is 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 Cnθ be 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 Cnθ originated 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 θ0 such that for all θθ0n2nπ, Cnθ is homeomorphic to 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 Mnθ and Cnθ to 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 θ0 in [8].

(iii) At present, only a sphere appears as a topological type of Cnθ. We determine the topological type of C6θ for all θ. The result shows that for some θ, C6θ is 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 θ be a real number satisfying 0θπ. We set

CnθP=u1unS2nthe followingiiiandiiihold.E1

  1. u1=1,0,0 and un=cosθsinθ0.

  2. i=1nui=0.

  3. uiui+1=cosθ for 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 ui denotes the unit vector in the direction of the edge of a polygon. Then the condition ii requires the fact that u1un is in fact a polygon.

  • We specify un1 to 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 θ except for those of the both ends of un1.

Remark 2. In some papers, Cnθ is defined as

Anθ/SO3,E2

where we set

Anθu1unS2n1iiiiiand the conditionunu1=cosθhold.

Let SO3 act on Anθ diagonally. Then for an element u1unAnθ, we may normalize u1 and un to 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) For n4, we have Cnθ if and only if θ belongs to the following interval:

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

(ii). Let a be an endpoint of the intervals in (3). Then we have Cna=onepoint.

Example 4. For n=4 or 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 C5θ is 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 η1 and η2 to be the following Figures 3 and 4, respectively.

Figure 3.

The space η1.

Figure 4.

The space η2.

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 exists θ0 such that for all θθ0n2nπ, the system of equations defined by (1) i, ii and iii intersect transversely. Hence for such θ, Cnθ carries a natural differential structure.

The main result in [8] is the following:

Theorem 6 ([8], Theorem 1). Let θ0 be as in Proposition 5. Then for all θθ0n2nπ, Cnθ is homeomorphic to Sn4.

Remark 7. In [8], Theorem 6 is proved by the following method: We construct a function f:CnθR and show that f has 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 equip Sn4 with the standard differential structure. Let θ0 be as in Proposition 5. Then for all θθ0n2nπ, Cnθ is diffeomorphic to Sn4.

Next we consider Problem 1 (ii). We set

αninfθ00πCnθSn4holds forallθ(θ0n2nπ).

Here in what follows, the notation XY means that X is homeomorphic to Y. Note that among the values of θ0 in Theorem 6, αn is the best possible one.

The following result is known.

Theorem 8 ([11]). (i) We have αn=n4n2π for 4n7.

(ii). We have α8<57π.

Remark 9. About Theorem 8 (i), we can read α4 and α5 from the above Example 4, and α6 from Table 4 in Theorem D below.

From Theorem 8, we naturally encounter the following:

Question 10. (i) Is it true that αn=n4n2π holds for n4?

(ii) Is it true that αn<n3n1π holds for n4? Note that if (i) is true then (ii) holds automatically.

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

Theorem B. For 4n14, the following Table 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 αn for 4n14.

The following theorem is the answer to Question 10.

Theorem C. (i) The statement in Question 10 (i) is false for n8. In fact, we have n4n2π<αn for n8.

(ii) The statement in Question 10 (ii) is false for n13. In fact, we have n3n1π<αn for n13.

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

Theorem D. The topological type of C6θ is given by the following Table 4, where we omit the cases which can be read from Theorem 3.

Range of θ0<θ<π3π3<θ<π2π2<θ<23π
Topological type of C6θ#3S1×S1#3S1×S1S2

Table 4.

The topological type of C6θ.

Remark 11. (i) As indicated in Problem 1 (iii), not only S2 but also #3S1×S1 appears in Table 4 as a topological type of C6θ

(ii) We can also determine the topological type of C6π2 and 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:XnR by

μPθ=θ.E5

Note that for all θ0π, we have

μ1θ=Cnθ.E6

The following proposition holds:

Proposition 12. (i) Let θ0 be as in Proposition 5. Then the space μ1θ0n2nπ is a manifold, where μ1θ0n2nπ denotes the inverse image of the interval.

(ii) Any element of μ1θ0n2nπ is a regular point of μ.

(iii) Consider the case n=8. Then C868π is a non-degenerate critical point of μ.

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

Dnu1un2θS2n2×0πthe followingiandiihold.

  1. u1=1,0,0.

  2. uiui+1=cosθ for 1in3.

Lemma 13. (i) There is a diffeomorphism

f:S1n3×0πDn.

(ii) We define the map L:DnR by

Lu1un2θ=cosθsinθ0+i=1n2ui2.

Then we have the following commutative diagram:

E7

Here the maps p and g are defined as follows.

  • Let

Pr:S1n3×0πRE8

be the projection to the 0π-component and we denote by p the restriction of Pr to Lf11:

pPrLf11.

  • The map g is a homeomorphism which will be defined in (13).

(iii) For all θ0π, the restriction of the map g in (7) naturally induces a homeomorphism

gCnθ:Cnθp1θ.E9

Proof of Lemma 13: (i) From an element

eiϕ1eiϕn3θS1n3×0π,

we construct the element u1un2θDn as follows: In the process of constructing ui, we also construct the elements viS2 such that uivi=0. We set

ui+1cosθui+sinθcosϕivi+sinθsinϕiui×viE10

and

vi+1sinθuicosθcosϕivicosθsinϕiui×vi,E11

where ui×vi denotes the cross product.

In (10) and (11) for i=1, we set u11,0,0 and v10,1,0. Then we obtain u2 and v2. Next using (10) and (11) for i=2, we obtain u3 and v3. Repeating this process, we obtain ui and vi for 1in2. Now we define f by

feiϕ1eiϕn3θu1un2θ.

From the construction, f is a diffeomorphism.

(ii) We define the map h:XnL11 by

hu1unθu1un2θ.E12

Since u1un is an element of Cnθ, the right-hand side of (12) is certainly an element of L11. It is clear that h is a homeomorphism. Hence if we define the map g by

gf1h,E13

then g is 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 Lf11 in (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 Pr is defined in (8). Then in addition to (7), we have the following commutative diagram:

E14

(i) Let xRn3×0π be 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 when Pr˜x=n2nπ.

We claim that x has the form

x=00n2nπ.E18

To prove this, recall the homeomorphism gCnθ was defined in (9). Since gCnθ1qx is 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 Lfqϕ1ϕn3n2nπ takes the minimum value 0 at 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 r in (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=n2nπ is positive, i.e., r is positive. Thus we have obtained (19). This completes the proof of (17) for the case (a).

(b) The case when Pr˜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 μ1θ0n2nπ contains a critical point of μ. Then using (7), p1θ0n2nπ contains a critical point of p. Lifting to the universal covering space using (14), there exists an element xRn3×0π which satisfies the following two items:

  • We have Pr˜xθ0n2nπ.

  • The point x is 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 λR such 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 θ is 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=8 is 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). For d2, let M be a d-dimensional smooth manifold without boundary and F:MR a smooth function. We set maxFM=m and assume that m is attained by unique point zM. Let aR satisfy the following four conditions:

  1. a<m.

  2. F1am is compact.

  3. There are no critical points in F1am.

  4. d5.

Then there is a diffeomorphism F1aSd1.

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 F in Theorem 14, we consider

μ:μ1θ0n2nπR.

More precisely, we denote the restriction of μ in (5) to μ1θ0n2nπ by the same symbol μ. Note that from the definition of F, z in Theorem 14 is Cnn2nπ.

  • By Proposition 12 (i), μ1θ0n2nπ is in fact a manifold.

  • For a in Theorem 14, we consider any element θ in 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 dimμ1θ0n2nπ5 if and only if n8.

Now we can apply Theorem 14 and obtain that μ1θ is diffeomorphic to Sn4 if θ satisfies that θ0<θ<n2nπ. By (6), this is equivalent to saying that Cnθ is 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 μ1θ is diffeomorphic to S4 if θ satisfies that θ0<θ<68π. Hence C8θ is 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 u1un of Cnθ are listed. For example, a condition is given by

u2=un.E27

Let Λ be 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 Rλθ0 for all θ.

  2. An element θ0π satisfies Rλθ=0 if and only if Cnθ contains an element which satisfies the condition (27).

In order to construct Rλ in (31), we first fix θ and define the space Ynθ as follows:

Ynθu1unS2nthe followingiandiihold.

  1. u1=1,0,0 and u2=un=cosθsinθ0.

  2. uiui+1=cosθ for 2in3.

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

rλu1uni=1nui.E32

Third, we define Rλ in (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 u1unYnθ belongs to rλ10 if 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 Θλ in (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 Θλ for each λΛ. Then from the definition of βn in (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 rλ10 is a critical submanifold of the function rλ in (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 space X6 is a manifold, where Xn is defined in (4).

(ii) The interior angle θ is a critical point of μ if and only if θ equals to π3, π2 or 23π.

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 θ1 and θ2 belong to the same interval from the three intervals 0π3, π3π2 and 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 p1θ for n=6. For a fixed ψ02π, we set

Mθψeiϕ1eiϕ2eiϕ3θp1θϕ1=ψ.

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

(i) The case when π2<θ<23π.

There exists ω in 0π such that the following homeomorphism holds:

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

From this, we have p1θS2. The figure of C6θ is given by the following Figure 6.

(ii) The case when π3<θ<π2.

There exists ω in 0π such that the following homeomorphism holds:

Mθψσ,ifψ=ωor2πω.S1,otherwise,E35

Here we set

σi=12xyR2x2+i2y2=1.

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

Figure 5.

(a) Mθωε; (b) Mθω; (c) Mθω+ε.

We claim that the four intersection points in MθωMθ2πω are 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, Mθω and Mθω+ε, respectively. (The deformation of the shape of Mθψ when ψ is near 2πω is also given by Figure 5.) Now from Figure 5, we see that the four intersection points are in fact saddle points.

Since we identify Mθ0 with Mθ2π, (35) and Figure 5 give the homeomorphism p1θ#3S1×S1. The figure of C6θ is given by the following Figure 7.

(iii) The case when 0<θ<π3.

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

Remark 18. We determine the topological type of C6π2 and C6π3.

(i) The figure of C6π2 is given by the following Figure 8.

Thus C6π2 is 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, C6π2 has three singular pints.

As θ approaches π2 from 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 θ increases further from π2, then the pinched point separates and we obtain Figure 6.

Figure 6.

The space C6θ for π2<θ<23π.

(ii) The figure of C6π3 is given by the following Figure 9.

The space C6π3 contains subspaces

N1,N2andN3E36

which satisfy the following three properties:

  • N1S1×S1.N2S1×S1 and N3#2S1×S1

  • i=13Ni=C6π3.

  • i<jNiNji=13xyR2x2+i2y2=1.

The figure of C6π3±ε is given by Figure 10 above.

As θ approaches 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 space C6θ for π3<θ<π2, where we identify the opposite boundaries.

Figure 8.

The space C6π2, where we identify the opposite vertices.

Figure 9.

The space C6π3.

Figure 10.

The space C6π3±ε.

Figure 11.

The space X4.

Figure 12.

The space X5.

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 Xn was defined in (4). The figures of X4 and X5 are given by Figures 11 and 12 above, respectively.

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,,ϕn3 and θ, then we could determine for which θ, Cnθ has 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,,ϕn3 and θ?

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Written By

Yasuhiko Kamiyama

Reviewed: 27 September 2021 Published: 11 November 2021