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

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 three-fold: 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.


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 l 1 , ⋯, l n 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.
Below we explain Table 1.
i. 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 M n θ ð Þ be the configurations of such n-gons with interior angle θ. The study of the topological type of M n θ ð Þ originated in [4]. See the next item for more details.
ii. 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. Table 1.
The correspondence between chemical and mathematical terminologies. • 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 C n θ ð Þ be the configurations of such n-gons with interior angle θ. (See (1) for more precise definition of C n θ ð Þ.) The study of the topological type of C n θ ð Þ originated in [7] and the result was generalized in [8]. See the next item for more details.
• Except for the above result in [8], we do not know strong results about the topology of C n θ ð Þ.
On the other hand, as a combinatorial result, the necessary and sufficient condition for M n θ ð Þ and C n θ ð Þ to be non-empty was proved in [9]. (See Theorem 3 about the result for C n θ ð Þ.) As stated in the last item of the above ii, we do not have enough information about the topology of C n θ ð Þ. The purpose of this chapter is to obtain systematic information about C n θ ð Þ. 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 θ 0 in [8].
(iii) At present, only a sphere appears as a topological type of C n θ ð Þ. We determine the topological type of C 6 θ ð Þ for all θ. The result shows that for some θ, C 6 θ ð Þ 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.
ii. P n i¼1 u i ¼ 0: iii. u i , u iþ1 h i¼ À cos θ for 1 ≤ i ≤ n À 3, where , h i denotes the standard inner product on  3 .
About the conditions in (1), the following explanations are in order. (See Table 1 for chemical terminologies.) • The element u i denotes the unit vector in the direction of the edge of a polygon.
Then the condition ii requires the fact that u 1 , ⋯, u n ð Þis in fact a polygon.
• We specify u nÀ1 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 u nÀ1 .

Remark 2.
In some papers, C n θ ð Þ is defined as where we set iii and the condition u n , u 1 h i¼ À cos θ hold È É : Let SO 3 ð Þ act on A n θ ð Þ diagonally. Then for an element u 1 , ⋯, u n ð Þ∈ A n θ ð Þ, we may normalize u 1 and u n 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 n ≥ 4, we have C n θ ð Þ 6 ¼ ∅ if and only if θ belongs to the following interval: if n is even: (ii). Let a be an endpoint of the intervals in (3). Then we have C n a ð Þ ¼ one point f g : Example 4. For n ¼ 4 or 5, the following results hold, where we omit the cases which can be read from Theorem 3.
ii. The topological type of C 5 θ ð Þ is given by the following Table 2.
Here we define η 1 and η 2 to be the following Figures 3 and 4, respectively. 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 θ 0 such that for all θ ∈ θ 0 , nÀ2 n π À Á , the system of equations defined by (1) i , ii and iii intersect transversely. Hence for such θ, C n θ ð Þ carries a natural differential structure. The main result in [8] is the following: Theorem 6 ([8], Theorem 1). Let θ 0 be as in Proposition 5. Then for all θ ∈ θ 0 , nÀ2 n π À Á , C n θ ð Þ is homeomorphic to S nÀ4 . Remark 7. In [8], Theorem 6 is proved by the following method: We construct a function f : C n θ ð Þ !  and show that f has exactly two critical points. Then Reeb's  5 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 S nÀ4 with the standard differential structure. Let θ 0 be as in Proposition 5. Then for all θ ∈ θ 0 , nÀ2 n π À Á , C n θ ð Þ is diffeomorphic to S nÀ4 . Next we consider Problem 1 (ii). We set Here in what follows, the notation X ffi Y means that X is homeomorphic to Y. Note that among the values of θ 0 in Theorem 6, α n is the best possible one.
The following result is known.
From Theorem 8, we naturally encounter the following: Question 10. (i) Is it true that α n ¼ nÀ4 nÀ2 π holds for n ≥ 4? (ii) Is it true that α n < nÀ3 nÀ1 π holds for n ≥ 4? Note that if (i) is true then (ii) holds automatically.
The following theorem is the answer to Problem 1 (ii). Theorem B. For 4 ≤ n ≤ 14, the following Table 3 holds.
The following theorem is the answer to Question 10. Theorem C. (i) The statement in Question 10 (i) is false for n ≥ 8. In fact, we have nÀ4 nÀ2 π < α n for n ≥ 8.
(ii) The statement in Question 10 (ii) is false for n ≥ 13. In fact, we have nÀ3 nÀ1 π < α n for n ≥ 13.
The following theorem is the answer to Problem 1 (iii). Theorem D. The topological type of C 6 θ ð Þ is given by the following Table 4, where we omit the cases which can be read from Theorem 3.  Remark 11. (i) As indicated in Problem 1 (iii), not only S 2 but also # 3 S 1 Â S 1 À Á appears in Table 4 as a topological type of C 6 θ ð Þ (ii) We can also determine the topological type of C 6 π 2 À Á and C 6 π 3 À Á . (See Remark 18 in §5.) In particular, they have singular points.

Proof of Theorem A
Following the method of [12], we set We define the function μ : X n !  by Note that for all θ ∈ 0, π ð , we have The following proposition holds: is a non-degenerate critical point of μ. In order to prove the proposition, we need a lemma. We set Þ j the following i and ii hold n o :

Lemma 13. (i) There is a diffeomorphism
(ii) We define the map L : D n !  by Then we have the following commutative diagram: Here the maps p and g are defined as follows.
• Let be the projection to the 0, π ð Þ-component and we denote by p the restriction of Pr to L ∘ f ð Þ À1 1 ð Þ: ð Þ : • The map g is a homeomorphism which will be defined in (13).
(iii) For all θ ∈ 0, π ð Þ, the restriction of the map g in (7) naturally induces a homeomorphism Proof of Lemma 13: (i) From an element we construct the element u 1 , ⋯, u nÀ2 , θ ð Þ ∈ D n as follows: In the process of constructing u i , we also construct the elements v i ∈ S 2 such that u i , v i h i¼0. We set and where u i Â v i denotes the cross product. In (10) and (11) for i ¼ 1, we set u 1 ≔ 1, 0, 0 ð Þand v 1 ≔ 0, 1, 0 ð Þ. Then we obtain u 2 and v 2 . Next using (10) and (11) for i ¼ 2, we obtain u 3 and v 3 . Repeating this process, we obtain u i and v i for 1 ≤ i ≤ n À 2. Now we define f by From the construction, f is a diffeomorphism.
(ii) We define the map h : X n ! L À1 1 ð Þ by Since u 1 , ⋯, u n ð Þis an element of C n θ ð Þ, the right-hand side of (12) is certainly an element of L À1 1 ð Þ. It is clear that h is a homeomorphism. Hence if we define the map g by 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 L ∘ f ð Þ À1 1 ð Þ in (7) is a subspace of In order to prove Proposition 12, we calculate in the universal covering space. Let be the universal covering space and we define the map e Pr :  nÀ3 Â 0, π ð Þ !  by e Pr ≔ Pr ∘ q, where the map Pr is defined in (8). Then in addition to (7), we have the following commutative diagram: Þbe any element which satisfies the condition q x ð Þ ∈ p À1 θ 0 , n À 2 n π ! : Note that if we use the diagram (14), then (15) is equivalent to saying that e Pr x ð Þ ∈ θ 0 , n À 2 n π ! and L ∘ f ∘ q ð Þx ð Þ ¼ 1: We set In order to prove Proposition 12 (i), it will suffice to prove that (a) The case when e Pr x ð Þ ¼ nÀ2 n π. We claim that x has the form To prove this, recall the homeomorphism gj C n θ ð Þ was defined in (9). Since Þis the regular n-gon, (18) follows.
Second, direct computations show that The number r in (19) equals to the derivative of (20) at θ ¼ nÀ2 n π. It is easy to see that and Using (21) and (22), we can check that the derivative of (20) at θ ¼ nÀ2 n π 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 e Pr x ð Þ ∈ θ 0 , nÀ2 n π À Á . By Proposition 5, we have Then using (16), we obtain (17). This completes the proof of (17) for the case (b), and hence also that of (i).
• The point x is a critical point of the function e Pr under the constraint We apply the Lagrange multiplier method to (24). Since e Pr ϕ 1 , ⋯, ϕ nÀ3 , θ ð Þ¼θ, there exists λ ∈  such that We compare the first n À 3 ð Þ-components of the both sides of (25). Then by (23), we have λ ¼ 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 Hence it will suffice to prove the following result for n ¼ 8: where jj 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 ¼ 8 is 1À ffiffi 2 p 16384 . Hence (26) holds for n ¼ 8. This completes the proof of (iii), and hence also that of Proposition 12.
In order to prove Theorem A, we recall the following: ii. F À1 a, m ½ is compact.
iii. There are no critical points in F À1 a, m ½ Þ.
Then there is a diffeomorphism F À1 a ð Þ ffi S dÀ1 . 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 d 6 ¼ 5.
Proof of Theorem A: First, we consider the case n 6 ¼ 8.
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 dim X n ¼ n À 3, we have dim μ À1 θ 0 , nÀ2 n π À Ã 6 ¼ 5 if and only if n 6 ¼ 8. Now we can apply Theorem 14 and obtain that μ À1 θ ð Þ is diffeomorphic to S nÀ4 if θ satisfies that θ 0 < θ < nÀ2 n π. By (6), this is equivalent to saying that C n θ ð Þ is also diffeomorphic to S nÀ4 . This completes the proof of Theorem A for n 6 ¼ 8.
Second, we consider the case n ¼ 8. If we apply the Morse lemma to Proposition 12 (iii), then we obtain that μ À1 θ ð Þ is diffeomorphic to S 4 if θ satisfies that θ 0 < θ < 6 8 π. Hence C 8 θ ð Þ is also diffeomorphic to S 4 . This completes the proof of Theorem A for n ¼ 8.

Proofs of Theorems B and C
Proof of Theorem B: In ([8], Lemma 1), certain conditions on an element u 1 , ⋯, u n ð Þof C n θ ð Þ are listed. For example, a condition is given by Let Λ be the set of the conditions. For λ ∈ Λ, we set Θ λ ≔ inf fθ 0 ∈ 0, n À 2 n π | for all θ ∈ θ 0 , n À 2 n π , C n θ ð Þdoes not containan element which satisfies the condition λg: Using this, we set Then it is proved in ([8], Proposition 1) that We explain how to compute Θ λ . As an example of λ, we consider the condition (27). We construct the continuous function which satisfies the following two properties: a. We have R λ θ ð Þ ≥ 0 for all θ.
In order to construct R λ in (31), we first fix θ and define the space Y n θ ð Þ as follows: Y n θ ð Þ ≔ u 1 , ⋯, u n ð Þ∈ S 2 À Á n j the following i and ii hold È É : ii. u i , u iþ1 h i¼ À cos θ for 2 ≤ i ≤ n À 3.
Second, we define the function r λ : Y n θ ð Þ !  as follows: For u 1 , ⋯, u n ð Þ∈ Y n θ ð Þ, we set Third, we define R λ in (31) by Below we check the above properties a and b of R λ . The item a is clear.
In order to prove the item b, we claim the following identification holds: In fact, an element u 1 , ⋯, u n ð Þ∈ Y n θ ð Þ belongs to r À1 λ 0 ð Þ 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 Θ λ in (28) as From the constructions in (10) and (11), we have Using this fact, we can compute the right-hand side of (34) for n ≤ 14.
By a similar method, we compute Θ λ for each λ ∈ Λ. Then from the definition of β n in (29), we can determine β n . Finally, using (30), we obtain α n . This completes the proof of Theorem B.
Remark 16. In the above proof of Theorem B, the identification (33) is crucial. Although r À1 λ 0 ð Þ is a critical submanifold of the function r λ in (32), this fact allows us to compute the right-hand side of (34) for n ≤ 14. See §6 (ii) for further remarks.
Proof of Theorem C: The theorem is clear from Table 3.

Proof of Theorem D
The following proposition is a refinement of Proposition 12 for n ¼ 6. Proposition 17. (i) The space X 6 is a manifold, where X n is defined in (4).
(ii) The interior angle θ is a critical point of μ if and only if θ equals to π 3 , π 2 or 2 3 π. Proof: We can prove the proposition is the same way as in Proposition 12. Since the dimension is low, we can perform direct computations.

17
The Topology of the The figure of C 6 π 3 AE ε À Á is given by Figure 10 above. As θ approaches π 3 , 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 N 3 . And the other circle is a subspace of N 1 ∪ N 2 .
On the other hand, the hole of the center of Figure 7 becomes a subspace of N 1 ∪ N 2 .  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 X n was defined in (4). The figures of X 4 and X 5 are given by Figures 11 and 12 above, respectively.
The identification (6) tells us that each level set of Figure 11 gives C 4 θ ð Þ, and that of Figure 12 gives C 5 θ ð Þ. Thus we obtain Example 4.

Conclusions
i. 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?
ii. We have the following comments about the proof of Theorem B. Recalling (23) and (24), we consider the following system of equations: and If we could solve the system of Eqs. (37) and (38) with respect to the variables ϕ 1 , ⋯, ϕ nÀ3 and θ, then we could determine for which θ, C n θ ð Þ has a singular point and the set of singular points of C n θ ð Þ. 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 ϕ 1 , ⋯, ϕ nÀ3 and θ?

Author details
Yasuhiko Kamiyama Faculty of Science, Department of Mathematical Sciences, University of the Ryukyus, Okinawa, Japan *Address all correspondence to: kamiyama@sci.u-ryukyu.ac.jp © 2021 The Author(s). Licensee IntechOpen. This chapter is distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/ by/3.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.