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`International Journal of Mathematics and Mathematical SciencesVolume 2013 (2013), Article ID 305926, 7 pageshttp://dx.doi.org/10.1155/2013/305926`
Research Article

## The Height of a Class in the Cohomology Ring of Polygon Spaces

Department of Mathematics, University of the Ryukyus, Nishihara-Cho, Okinawa 903-0213, Japan

Received 26 September 2013; Accepted 9 December 2013

Copyright © 2013 Yasuhiko Kamiyama and Kazufumi Kimoto. 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.

#### Abstract

Let be the configuration space of planar -gons having side lengths and modulo isometry group. For generic , the cohomology ring has a form where is the first Stiefel-Whitney class of a certain regular -cover and the ideal is in general big. For generic , we determine the number such that but .

#### 1. Introduction

Given a string of positive real numbers, one considers the configuration space of planar polygon linkages having side lengths modulo isometry group. Starting in [13], the topology of has been considered by many authors. A notable achievement is [4] which determined the cohomology ring for generic . The study culminated in the proof by [5] of a conjecture by Kevin Walker which states that one can recover relative lengths of edges from the cohomology ring of the configuration space.

By [4], the cohomology ring has a form where is the first Stiefel-Whitney class of a certain regular -cover and the ideal is in general big. For example, when ,   is generated by polynomials whose number is approximately . While [5] made clever arguments to distinguish the cohomology rings, it is difficult to extract more concrete information from the ring. The reason is that the polynomials do not form a Gröbner basis.

In this direction, one natural problem is to compute the height of , that is, the unique such that but . The purpose of this paper is to determine for the case . Note that such is a string which comes next to the equilateral case. For example, the homology groups were determined in [6].

Finally, we note that the height of an element in the cohomology ring of a space has been studied in order to give nice lower bounds on the Lusternik-Schnirelmann category of . For example, [7] studied the problem for the case that and the element is the first Stiefel-Whitney class of the universal bundle.

This paper is organized as follows. In Section 2 we state our main results. Theorem A determines the height of . Theorem B determines which element represents the fundamental class of . In Section 3 we prove auxiliary results. In Section 4 we prove Theorem B and in Section 5 we prove Theorem A.

#### 2. Main Results

For and , we define the configuration space of planar -gons having side lengths and by

Here denotes the unit vectors in the direction of the sides of a polygon. Note that comes with a natural involution

We set

It is clear that if and . In [13] it is proved that if is a natural number which satisfies and has the same parity as , then there is a diffeomorphism

Moreover, if is even and , then we have , where acts on by complex conjugate. On the other hand, if and has the different parity as , then has singular points.

Hereafter we assume that is a natural number which satisfies and has the same parity as . In this case, is a connected closed manifold of dimension . Moreover, is orientable if and only if is even.

The following examples are well known:

The following theorem is crucial in this paper.

Theorem 1 (see [4, Corollary 9.2]). The cohomology ring of is where and are of degree and is the ideal generated by the three families of elements:(R1)(R2)(R3)
The symbol in runs over all subsets of including the empty set. By a term of the sum in vanishes if .

The class coincides with the first Stiefel-Whitney class of the regular -cover . We define the height of is the unique such that but .

In order to state our main results, we prepare notations. Throughout this paper, the notation means that . We set

Moreover, we set

Now our first result is the following.

Theorem A. For all and which satisfies and has the same parity as , are has .

We study the generator of . Note that if then we have in that

We set

Our second result is the following.

Theorem B. Let and be as in Theorem A. Then for , the following equality holds in :

Theorem A implies that is a generator if and only if . This is in agreement with Theorem B for .

For the case of almost equilateral polygons, that is, the case for or , we can write Theorem A more explicitly.

Proposition C. (i) About for odd , one has the following.(a)One has if and only if is of the form .(b)If satisfies that then .(c)One has for all .
(ii) About for even , one has the following.(a)One has if and only if is of the form .(b)If , then . On the other hand, if satisfies that then .(c)One has for all .

One deduces the proposition from Theorem A in Section 5.

We give two examples of Theorem A. The first example is about the case when is small.

Example 2. We consider Table 1 for the case . Then (where is defined in (15)) in if and only if satisfies the case which is given in Table 2.

Table 1: for .
Table 2: The cases for .

Our second example is about the case when is large.

Example 3. (i) We have
(ii) We have.
In fact, (i) for odd and (ii) correspond to the case in Theorem A. But we know a stronger result as in (6).

#### 3. Auxiliary Results

Proposition 4. Let be indeterminate and . Let be an matrix over defined as follows:
Then one has the following results. (i)One has. (ii)One has if and only if in .(iii)When , the following results hold.(a)Let be the unique nontrivial element of . Then one has(b)Letbe as in (a). Then one has

Proof. We define 4 matrices as follows:
Since , they are regular matrices.
We write and as follows:
Using the Chu-Vandermonde identity we have
(i) Since and is a regular matrix, we have .
(ii) Note that In fact, both the sides coincide for and vanish for such that .
Now (27) tells us that if and only if .
(iii)(a) Since , we have . We must prove that for . Since we are assuming that , it will suffice to prove that
We use the formula which holds over . Setting , , and , we have in that the left-hand side of (29) is . Since , this is .
(b) The left-hand side of (b) is that of (29) for . Then this is .

Corollary 5. Assume that and let be the element in Proposition 4(iii)(a). One sets If one has also that , then one has

Proof. The unique nontrivial element of is given by The corollary is clear from this.

The following property of will be used in Section 5.

Lemma 6. (i) If and , one has
(ii) If is odd and satisfies that then the following results hold.(a)One has if and only if is of the form .(b).

Proof. (i) We need to consider two cases according to or .
Case (a) .
Note that
We shall prove that
We consider Proposition 4 for and . For in the proposition, we set and put . Note that is the sequence which appears in the definition of . This and the definition of in Corollary 5 tell us that
Using Corollary 5, (36), and (37), we have
Combining (37), (39), and (40), we obtain (38).
Now (i) follows from the following computation: Case (b) .
By definition, we have . Hence it will suffice to prove . This is equivalent to the assertion that and also equivalent to the assertion that . But the last assertion is clear if we apply (36) and (37) to our assumptions that
(ii) If we prove (b), then (a) is clear from this. Since (b) clearly holds for , we assume that . We set . By definition,
By Kummer, is odd if and only if there are no carries when adding and in base . Hence if we write (where ), the greatest (where ) such that is
Hence and

Modifying Proposition 4 slightly, we have the following.

Lemma 7. Let be an matrix over defined as follows: Then one has the following results. (i)One has .(ii)Let be the unique nontrivial element of . Then one has

Proof. (i) Let be the matrix in the proof of Proposition 4 and we define
Since , , and , the result follows;
(ii) follows from (29).

#### 4. Proof of Theorem B

For , let be the polynomial in Theorem 1   (R3) for . Note that the coefficient of in is , where is defined in (15). We list the coefficient in the following matrix : Theorem 1 tells us that

In the notation of Lemma 7, we have

Since generate , (51) tells us that

We compute as follows:

#### 5. Proof of Theorem A

Theorem A is proved by showing the upper bound for those with coinciding with the lower bound. About the upper bound, we have the following.

Proposition 8. In , one has .

Proof. Since , the case for is clear. Hence we may assume that . For in , we denote by the sum of polynomials in Theorem 1   (R3) for all with . Note that . We use the notation in the proof of Lemma 6. We claim the following:
In order to prove (55), we define
Then the following equality holds:
In fact, the left equality follows from the fact that for . On the other hand, the right equality is proved as follows. Note that we can write as where is the first row of and is the matrix defined from by shifting the -element to the -element. Since , we have . Now the right equality of (57) follows from Proposition 4(iii)(b).
Let be the th symmetric polynomial in variables . Then we have
In fact, the binomial coefficient is computed from for . Now we complete the proof of (55) as follows:
Hence (55) holds.
Now by Theorem 1 (R3), we have in . Since the left-hand side of (55) vanishes, so does the right-hand side.

Proof of Theorem A. We prove the theorem by induction on and for all .
(I) The case for .
Since , Theorem A holds.
(II) Fix and assume that Theorem A holds for all with .
We shall prove Theorem A for .
Case (i) .
We need to consider the cases according to or . Since the case of is proved in Theorem B, we may assume that . We define an inclusion by
Note that the map naturally induces a map and we have the following diagram of regular 2-covers:(63)
Since is the first Stiefel-Whitney class of the regular -cover , we have
By inductive hypothesis, we have in that
Using (64), we have in that
Using Lemma 6(i), we have .
On the other hand, we have by Proposition 8 that . Hence we have proved that .
Case (ii) .
As in (i), we may assume that . Lemma 6(ii) tells us that if , then . Moreover, Proposition 8 tells us that . Hence it will suffice to prove that in . We define an inclusion by
The map induces a map . Using a similar diagram to (63) and the fact that in (see Theorem B), we have in that .

Proof of Proposition C. Proposition C(i)(a) and (b) follow from Theorem A and Lemma 6(ii), and (i)(c) follows from Theorem B. If we modify Lemma 6(ii) slightly for even , then we can prove Proposition C(ii) similarly.

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