Research Article | Open Access
Yasuhiko Kamiyama, "The Topological Type of Equilateral and Almost Equiangular Polygon Spaces", Journal of Mathematics, vol. 2019, Article ID 6958218, 6 pages, 2019. https://doi.org/10.1155/2019/6958218
The Topological Type of Equilateral and Almost Equiangular Polygon Spaces
Let be the configuration space of equilateral spatial n-gons. For and , let be the subspace of consisting of elements whose first k bond angles are θ. Recently, the topological type of was determined for small n, special θ, and or . In this paper, we determine the topological type of for general n and θ.
Starting in , the topology of polygon spaces in the Euclidean space of dimension three has been studied extensively by many authors. For example, Milgram and Trinkle  obtained results by making excellent use of Morse surgery. We refer to  for an excellent exposition with emphasis on Morse theory.
Let 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 . 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 of .
Definition 1. For and , we set(1).(2), for .Here denotes the standard inner product on . Moreover, when , we understand to be .
Let act on diagonally, and we set . Note that is the subspace of consisting of elements whose first k bond angles are θ. In particular, we have .
Example 1. (i) An equilateral and equiangular n-gon can be considered as a mathematical model of a cycloalkane. Such a cycloalkane is called cyclobutanes, cyclopentanes, or cyclohexanes according as or 6. The space is the configuration space of such a model. The topological type of for various θ was determined by Crippen  for and 5 and by O’Hara  for .(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 space is the configuration space of such a model. Goto et al.  studied the topological type of for the case that n is small and θ is slightly smaller than . Here, note that is the bond angle of the regular n-gon in (see Theorem 1).The space 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 is homeomorphic to a sphere (see Theorem 1).
The purpose of this paper is to study the topology of . Since we can use usual techniques in topology, for example, fiber bundles, we can determine the topological type of for general n and θ.
This paper is organized as follows. In Section 2, 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 means that X is homeomorphic to Y.)
2. Preliminary Results
We first recall the results about .
Theorem 1 (). (i)There is a homeomorphism:Note that consists of the regular n-gon in .(ii)Let ε be a sufficiently small positive real number. Then, for , we have .
In the definition of , we may normalize that and . Next, we set
Then, we obtain the following description of :(1) and .(2).(3) and for .
Hereafter, we use description (4).
The following definition and lemma will be used in Section 3.
Definition 2. Let n be an integer greater than or equal to 5.(i)We define the function byThen, we define to be largest θ which satisfies the following equation:(ii)We define the function byThen, we define as follows:(a).(b)When , we define to be largest θ which satisfies the following equation:
Example 2. (i), and .(ii), and .
Lemma 1. We define and to be the elements of , which satisfy the following conditions:(a)The linkages are planar polygons such that .(b)The linkages do not have self-intersection points.(c)We set and . Then,(i)For , we require that and .(ii)For , we require that and .(iii)For , we require that , and .Figures 1(a)–1(c) are , , and for , respectively. Then, the following assertions hold:(i)For , the value realizes (ii)For , the value realizes (iii)For , the value realizes
Proof. Since the proofs are similar, we only prove (i). We set . Regarding , the coordinate of the endpoint of is given byOn the other hand, the coordinate of the endpoint of is given byIt is easy to see that the equation is equivalent to . Hence, (i) holds.
3. Main Theorem
First, it is easy to see that there is a homeomorphism for all .
Main Theorem. Let n be an integer greater than or equal to 5. Then, the following assertions hold:(i)We haveMoreover, the element of is given by in Lemma 1.(ii)When , we have .(iii)We havewhere denotes the sphere with two points identified. Moreover, V is homeomorphic to such that . Finally, for , the singular point of is given by and in Lemma 1.(iv)When , we haveCase (iv) does not occur for because .(v)Assume that θ satisfies the following condition:Then, the topological type of does not depend on θ. Moreover, the topological type is given as follows:(a)When , the topological type is given by the above . In fact, we have when .(b)When , there is a homeomorphism:where is homeomorphic to the space in the right-hand side of (13) and is homeomorphic to . Moreover, if we set , then there is a fiber bundle:(vi)When n is odd and , we have .
Example 3. (i)Regarding the Main Theorem (iii) for , the space V means .(ii)Regarding the Main Theorem (v) (b) for , we have , where , such that .
4. Auxiliary Lemmas
We fix θ and set
We define the projection by . We study the topological type of the fiber for each . In the following lemma, we assume that . The case for or 6 can be obtained by slight modifications.
Lemma 2. Let n be an integer greater than or equal to 7. Assume that θ satisfies the following condition:We write an element of byThen, the following assertions hold:(i)When , there exists such that(ii)When , we have(iii)When , there exists such that(iv)When n is odd and , we have .
Proof. We construct the following commutative diagram:First, we setHere, we define the following:(a) and (b) and for (c)If we define the map ρ by , then we require that Second, we define the map i to be the inclusion. Third, we define the map π as follows: for , we setFourth, we define the maps and to be the natural extensions of .
We claim that for , we haveIn fact, we can identify with the locus of the endpoint of . Note that the last case of (26) does not occur because such a case would imply that . But a result in  tells us that 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 each , we haveUsing this fact, it is easy to see thatCombining (23), (26) and (28), we obtain (iii).
(iv) By combining (23) and (27), we have . We setWe claim that when n is odd, is given by . In fact, it is easy to see that if we fix θ, then is attained by such that andFrom this, we compute . Next, solving the equation with respect to θ, we have . Hence must be as in the claim. Now since , (iv) follows.
The following lemma will be used in Section 5.
Lemma 3. Let M be a connected oriented k-dimensional manifold. For an embedding , we define another k-manifold to bewhere we glue so that is an orientable manifold. Then, there is a homeomorphism:
Proof. The lemma is well known (see, for example, (, Example 4.17)). For completeness, we give a proof. First, we consider the case that . Sincewe see that . Using this, the case of general M is proved as follows:
5. Proof of the Main Theorem
The most difficult case for the proof of the Main Theorem is the item (iv) for . Hence, we first prove this. We begin with the following.
Lemma 4. Let n be an integer greater than or equal to 7. Assume that θ satisfies the condition . Let N denote or . Then, we have the following homeomorphism:
Proof. We define the function by . We fix real numbers A and B which satisfy . We define the function μ byThe function μ is a Morse function with four critical points, which are pairs of the critical points of and . Let p be the critical point of μ such that . 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 haveNow since is the second smallest critical value, we obtain Lemma 4.
Proof of the Main Theorem (iv). Note thatwhere we identify the corresponding boundaries. First, we identify only the boundaries . Using Lemma 4, we obtain the space Second, we identify the remaining boundaries. Using Lemma 3, we obtain .
Proof of the Main Theorem. Since the case for or 6 can be proved by sight modifications, we assume that .(i)It is easy to prove the item (see also ).(ii)From Lemma 2(i), we have , where denotes the suspension.(iii)The item follows from Lemma 2(ii).(iv)The case for is proved above.(v)Since the arguments are easy, we prove only the item (b). From the continuity of the deformation of with respect to θ, the space appears. In addition to , the new space appears, where we setIt is clear that . Moreover, as recalled in the proof of Lemma 2, if and only if (14) holds. Next, we determine . We setWe fix and set . Then, the endpoint of moves on the circle which satisfies the equation and the center being the point . Along the circle, we let the endpoint of approach the point . That is, we let the vector approach . Then, the pair of vectors converges to the pair of vectors such that the configurationis an element of W. If we do this procedure for all elements , then we obtain all elements of W. About (40), we regard as an element of . In order to emphasize P, we write in (40) by . Assume that we also take such that . Then, the construction of (40) shows that the endpoints of and lie on a circle which is determined by . Hence, have the fiber bundle (16).(vi)By Lemma 2(iv), there is a fiber bundle:We show that (41) is a trivial bundle. Let be the closed three-dimensional ball of radius centered at . Since θ is small, there exists a sufficiently small positive real number ε such that for all , the endpoint of is contained in and that of is contained in .
Consider the spacewhere for , we are denoting by , by and by . Since ε is small, there is a fiber bundle:Moreover, since the base space is contractible, (43) is a trivial bundle. Hence, bundle (41) is also trivial.
We have determined the topological type of for general n and θ. We try to study the other spaces . As stated in Section 1, only limited information is known for the case or n. In this direction, the necessary and sufficient condition on θ for to be nonempty was determined recently (see ).
On the other hand, for the case , the following observation holds: let be the space of spatial m-gons having edge lengths , where we do not consider the orbit space by the -action. Note that the right-hand side of (26) coincides with the topological type of . Hence, in order to study , we will need information on (see [1, 8] for information on ).
No data were used to support this study.
Conflicts of Interest
The author declares that there are no conflicts of interest regarding the publication of this paper.
This study was supported by JSPS KAKENHI (grant no. 15K04877).
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Copyright © 2019 Yasuhiko Kamiyama. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.