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

Authors: Yasuhiko Kamiyama; Kazufumi Kimoto
Journal: International Journal of Mathematics and Mathematical Sciences
 (2013)

---

## Abstract

LetM-n,r be the configuration space of planar n-gons having side lengths 1,…,1 and r modulo isometry group. For generic r, the cohomology ring H*(M-n,r;ℤ2) has a form H*(M-n,r;ℤ2)=ℤ2[R(n,r),V1,…,Vn-1]/ℐn,r, where R(n,r) is the first Stiefel-Whitney class of a certain regular 2-cover π:Mn,r⟶M-n,r and the ideal ℐn,r is in general big. For generic r, we determine the number h(n,r) such that R(n,r)h(n,r)≠0 but R(n,r)h(n,r)+1=0.


---

## Body

## 1. Introduction

Given a stringr=(r1,…,rn) of n positive real numbers, one considers the configuration space M-r of planar polygon linkages having side lengths ri modulo isometry group. Starting in [1–3], the topology of M-r has been considered by many authors. A notable achievement is [4] which determined the mod2 cohomology ring H*(M-r;ℤ2) for generic r. 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 mod2 cohomology ring of the configuration space.By [4], the cohomology ring H*(M-r;ℤ2) has a form
(1)H*(M-r;ℤ2)=ℤ2[Rr,V1,…,Vn-1]ℐr,
where Rr is the first Stiefel-Whitney class of a certain regular 2-cover π:Mr→M-r and the ideal ℐr is in general big. For example, when r=(1,…,1),  ℐr is generated by polynomials whose number is approximately 2n-2. 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 ofRr, that is, the unique h(r) such that Rrh(r)≠0 but Rrh(r)+1=0. The purpose of this paper is to determine h(r) for the case r=(1,…,1,r). Note that such r is a string which comes next to the equilateral case. For example, the homology groups H*(Mr;ℤ) were determined in [6].Finally, we note that the height of an element in the cohomology ring of a spaceX has been studied in order to give nice lower bounds on the Lusternik-Schnirelmann category of X. For example, [7] studied the problem for the case that X=Gk(ℝn+k) and the element is the first Stiefel-Whitney class of the universal bundle.This paper is organized as follows. In Section2 we state our main results. Theorem A determines the height of Rr. Theorem B determines which element represents the mod2 fundamental class of M-r. 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

For4≤n∈ℕ and 0<r∈ℝ, we define the configuration space of planar n-gons having side lengths 1,…,1 and r by
(2)Mn,r={(z1,…,zn-1)∈(S1)n-1∣∑i=1n-1‍zi=r}.Herezi∈S1⊂ℂ denotes the unit vectors in the direction of the sides of a polygon. Note that Mn comes with a natural involution
(3)τ:Mn,r⟶Mn,r,τ(z1,…,zn-1)=(z-1,…,z-n-1).We set(4)M-n,r=Mn,rτ.It is clear thatM-n,r=∅ if r>n-1 and M-n,n-1={onepoint}. In [1–3] it is proved that if r is a natural number which satisfies r≤n-2 and has the same parity as n, then there is a diffeomorphism
(5)M-n,r′≅M-n,r,∀r′∈(r-1,r+1).Moreover, ifn is even and r′∈(0,1), then we have M-n,r′≅Mn-1,1×τS1, where τ acts on S1 by complex conjugate. On the other hand, if n>4 and r∈ℕ has the different parity as n, then M-n,r has singular points.Hereafter we assume thatr is a natural number which satisfies r≤n-2 and has the same parity as n. In this case, M-n,r is a connected closed manifold of dimension n-3. Moreover, M-n,r is orientable if and only if n is even.The following examples are well known:(6)M-n,n-4≅#nℝPn-3,M-n,n-2≅ℝPn-3.The following theorem is crucial in this paper.Theorem 1 (see [4, Corollary 9.2]).
Themod2 cohomology ring of M-n,r is
(7)H*(M-n,r;ℤ2)=ℤ2[R(n,r),V1,…,Vn-1]ℐn,r,
where R(n,r) and V1,…,Vn-1 are of degree 1 and ℐn,r is the ideal generated by the three families of elements:(R1)
(8)Vi2+R(n,r)Vi,wherei=1,…,n-1,(R2)
(9)∏i∈S‍Vi,whereS⊂{1,…,n-1}issuchthat|S|≥n-r2,(R3)
(10)∑‍S⊂LR(n,r)|L-S|-1∏‍i∈SVi,whereL⊂{1,…,n-1}issuchthat|L|≥n+r2.
The symbolS in (R3) runs over all subsets of L including the empty set. By (R2) a term of the sum in (R3) vanishes if |S|≥(n-r)/2.The classR(n,r) coincides with the first Stiefel-Whitney class of the regular 2-cover π:Mn,r→M-n,r. We define the height of R(n,r)is the unique h(n,r) such that R(n,r)h(n,r)≠0 but R(n,r)h(n,r)+1=0.In order to state our main results, we prepare notations. Throughout this paper, the notationa≡b means that a≡b(mod2). We set
(11)D(n)=n-2,e(n,r)=n-r2-1,k(n,r)=max{(D(n)-e(n,r)+ii)≡1}i∣0≤i≤e(n,r)-1,(D(n)-e(n,r)+ii)≡1}.Moreover, we set(12)ϕ(n,r)={n-3,if(D(n)e(n,r))≡1,n+r2+k(n,r)-2,if(D(n)e(n,r))≡0.Now our first result is the following.Theorem A.
For alln≥4 and r∈ℕ which satisfies r≤n-2 and has the same parity as n, are has h(n,r)=ϕ(n,r).We study the generator ofHn-3(M-n,r;ℤ2)=ℤ2. Note that if
(13)1≤ν1<⋯<νi≤n-1,
then we have in Hn-3(M-n,r;ℤ2) that
(14)R(n,r)n-3-iV1⋯Vi=R(n,r)n-3-iVν1⋯Vνi.We set(15)p(n,r,i)=R(n,r)n-3-iV1⋯Vi.Our second result is the following.Theorem B.
Letn and r be as in Theorem A. Then for 0≤i≤e(n,r), the following equality holds in Hn-3(M-n,r;ℤ2):
(16)p(n,r,i)=(D(n)-ie(n,r)-i).Theorem A implies thatR(n,r)n-3 is a generator if and only if (D(n)e(n,r))≡1. This is in agreement with Theorem B for i=0.For the case of almost equilateral polygons, that is, the case forr=1 or 2, we can write Theorem A more explicitly.Proposition C.
(i) AboutM-n,1 for odd n, one has the following.(a)
One has(D(n)e(n,r))≡1 if and only if n is of the form n=2s+1.(b)
Ifn satisfies that
(17)2s+1≤n≤2s+1-1,
then h(n,1)=2s-2.(c)
One hasp(2s+1,1,i)≠0 for all i.
(ii) AboutM-n,2 for even n, one has the following.(a)
One has(D(n)e(n,r))≡1 if and only if n is of the form n=2s.(b)
Ifn=2s, then h(n,2)=2s-3. On the other hand, if n satisfies that
(18)2s+2≤n≤2s+1-2,
then h(n,2)=2s-2.(c)
One hasp(2s,2,i)≠0 for all i.One deduces the proposition from Theorem A in Section5.We give two examples of Theorem A. The first example is about the case whenn is small.Example 2.
We consider Table1 for the case h(n,r)=n-3. Then p(n,r,i)=0 (where p(n,r,i) is defined in  (15)) in Hn-3(M-n,r;ℤ2) if and only if (n,r,i) satisfies the case which is given in  Table 2.Table 1
h
(
n
,
r
) for 4≤n≤20.
r
n
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
1
—
2
—
2
—
6
—
6
—
6
—
6
—
14
—
14
—
2
1
—
2
—
5
—
6
—
6
—
6
—
13
—
14
—
14
3
—
2
—
4
—
6
—
6
—
6
—
12
—
14
—
14
—
4
—
—
3
—
4
—
6
—
6
—
11
—
12
—
14
—
14
5
—
—
—
4
—
6
—
6
—
10
—
12
—
14
—
14
—
6
—
—
—
—
5
—
6
—
9
—
10
—
13
—
14
—
14
7
—
—
—
—
—
6
—
8
—
10
—
10
—
14
—
14
—
8
—
—
—
—
—
—
7
—
8
—
10
—
10
—
14
—
14
9
—
—
—
—
—
—
—
8
—
10
—
10
—
14
—
14
—
10
—
—
—
—
—
—
—
—
9
—
10
—
13
—
14
—
14
11
—
—
—
—
—
—
—
—
—
10
—
12
—
14
—
14
—
12
—
—
—
—
—
—
—
—
—
—
11
—
12
—
14
—
14
13
—
—
—
—
—
—
—
—
—
—
—
12
—
14
—
14
—
14
—
—
—
—
—
—
—
—
—
—
—
—
13
—
14
—
17
15
—
—
—
—
—
—
—
—
—
—
—
—
—
14
—
16
—
16
—
—
—
—
—
—
—
—
—
—
—
—
—
—
15
—
16
17
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
16
—
18
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
17Table 2
The cases forp(n,r,i)=0.
(
n
,
r
)
i
(
9,3
)
1
(
13,7
)
1
(
15,5
)
1,3
(
17,3
)
1,3
,
5
(
17,5
)
2,3
(
17,7
)
1,2
,
3
(
17,11
)
1Our second example is about the case whenr is large.Example 3.
(i) We have(19)h(n,n-4)={n-3nisodd,n-4niseven.
(ii) We haveh(n,n-2)=n-3.
In fact, (i) for oddn and (ii) correspond to the case (D(n)e(n,r))≡1 in Theorem A. But we know a stronger result as in (6).


## 3. Auxiliary Results

Proposition 4.
Letx be indeterminate and e∈ℕ. Let Ae(x) be an e×e matrix over ℤ2[x] defined as follows:
(20)Ae(x)=((x+ij))1≤i≤e1≤j≤e.
Then one has the following results.(i)
One hasrankAe(x)≥e-1.(ii)
One hasrankAe(x)=eif and only if(x+ee)≠0inℤ2[x].(iii)
WhenrankAe(x)=e-1, the following results hold.
(a)
Letyt=(y1,…,ye)tbe the unique nontrivial element ofKerAe(x). Then one has(21)yj=(x+e-je-j).(b)
Letyt=(y1,…,ye)tbe as in (a). Then one has(22)∑j=1e‍(x+e+1j)yj=1.Proof.
We define 4 matrices as follows:(23)Pe=((ij))1≤i≤e0≤j≤e,Qe(x)=((xj-i))0≤i≤e1≤j≤e,P^e=((ij))1≤i,j≤e,Q^e(x)=((xj-i))1≤i,j≤e.
SincedetP^e=detQ^e(x)=1, they are regular matrices.
We writePe and Qe(x) as follows:
(24)Pe=(1P^e),Qe(x)=(qQ^e(x)).
Using the Chu-Vandermonde identity(25)(x+ij)=∑k=0j‍(ik)(xj-k),
we have
(26)Ae(x)=PeQe(x)=1q+P^eQ^e(x).
(i) Sincerank1q=1 and P^eQ^e(x) is a regular matrix, we have rankAe(x)≥e-1.
(ii) Note that(27)detAe(x)=(x+ee).In fact, both the sides coincide for x=0 and vanish for x∈ℤ such that -e≤x≤-1.
Now (27) tells us that rankAe(x)=e if and only if (x+ee)≠0.
(iii)(a) Sinceye=1, we have y≠0t=. We must prove that
(28)∑j=1e‍(x+ij)(x+e-je-j)=0
for 1≤i≤e. Since we are assuming that rankAe(x)=e-1, it will suffice to prove that
(29)∑j=0e‍(x+ij)(x+e-jx)=0.
We use the formula(30)∑j=0c‍(-1)j(aj)(b-jb-c)=(b-ac)
which holds over ℤ. Setting a=x+i, b=x+e, and c=e, we have in ℤ2 that the left-hand side of (29) is (e-ie). Since 1≤i≤e, this is 0.
(b) The left-hand side of (b) is that of (29) for i=e+1. Then this is (e-ie)=(-1e)=1.Corollary 5.
Assume thatrankAe(x)=e-1 and let yt be the element in Proposition 4(iii)(a). One sets
(31)ℓ(x,e)=min{j∣1≤j≤e,yj≡1}.
If one has also that rankAe(x-1)=e-1, then one has
(32)ℓ(x-1,e)=ℓ(x,e)-1.Proof.
The unique nontrivial element ofKerAe(x-1) is given by
(33)(y1+y2,y2+y3,…,ye-1+ye,ye)t.
The corollary is clear from this.The following property ofϕ(n,r) will be used in Section 5.Lemma 6.
(i) Ifr≥2 and (D(n)e(n,r))≡0, one has
(34)ϕ(n-1,r-1)=ϕ(n,r).
(ii) Ifn is odd and satisfies that
(35)2s+1≤n≤2s+1-1,
then the following results hold.(a)
One has(D(n)e(n,r))≡1if and only if n is of the form n=2s+1.(b)
ϕ(n,1)=2s-2.Proof.
(i) We need to consider two cases according to(D(n-1)e(n-1,r-1))≡0 or 1.
Case (a)((D(n-1)e(n-1,r-1))≡0).
Note that(36)D(n-1)=D(n)-1=n-3,(37)e(n-1,r-1)=e(n,r)=n-r2-1.
We shall prove that(38)k(n-1,r-1)=k(n,r)+1.
We consider Proposition4 for x=D(n)-e(n,r) and e=e(n,r). For y in the proposition, we set y~j=ye-j and put y~=(y~0,…,y~e-1). Note that y~ is the sequence which appears in the definition of k(n,r). This and the definition of ℓ(x,e) in Corollary 5 tell us that
(39)k(n,r)=e(n,r)-ℓ(D(n)-e(n,r),e(n,r)).
Using Corollary5, (36), and (37), we have
(40)ℓ(D(n-1)-e(n-1,r-1),e(n-1,r-1))=ℓ(D(n)-e(n,r),e(n,r))-1.
Combining (37), (39), and (40), we obtain (38).
Now (i) follows from the following computation:(41)ϕ(n-1,r-1)=n-1+r-12+(k(n,r)+1)-2=ϕ(n,r).
Case (b) ((D(n-1)e(n-1,r-1)))≡1.
By definition, we haveϕ(n-1,r-1)=n-4. Hence it will suffice to prove ϕ(n,r)=n-4. This is equivalent to the assertion that k(n,r)=e(n,r)-1 and also equivalent to the assertion that (D(n)-1e(n,r)-1)≡1. But the last assertion is clear if we apply (36) and (37) to our assumptions that
(42)(D(n)e(n,r))≡0,(D(n-1)e(n-1,r-1))≡1.
(ii) If we prove (b), then (a) is clear from this. Since (b) clearly holds forn=2s+1, we assume that 2s+3≤n≤2s+1-1. We set n=2m+1. By definition,
(43)k(n,r)=max{i∣0≤i≤m-2,(m+ii)≡1}.
By Kummer,(a+bb) is odd if and only if there are no carries when adding a and b in base 2. Hence if we write m=2s-1+u (where 1≤u≤2s-1), the greatest i (where i≤m-2) such that (m+ii)≡1 is
(44)i=2s-2+2s-3+⋯+1-u=2s-1-1-u.
Hencek(n,r)=2s-1-1-u and
(45)ϕ(n,1)=(m+1)+k(n,r)-2=(2s-1+u+1)+(2s-1-1-u)-2=2s-2.Modifying Proposition4 slightly, we have the following.Lemma 7.
LetAˇe(x) be an e×(e+1) matrix over ℤ2[x] defined as follows:
(46)Aˇe(x)=(x+ij)1≤i≤e0≤j≤e.
Then one has the following results.  (i)
One hasrankAˇe(x)=e.(ii)
Letyˇt=t(yˇ0,…,yˇe) be the unique nontrivial element of KerA˘e(x). Then one has
(47)yˇj=(x+e-je-j).Proof.
(i) LetPe be the matrix in the proof of Proposition 4 and we define
(48)Qˇe(x)=((xj-i))0≤i≤e0≤j≤e.
SinceAˇe(x)=PeQ˘e(x), rankPe=e, and detQ˘e(x)=1, the result follows;
(ii) follows from (29).


## 4. Proof of Theorem B

For(n+r)/2≤i≤n-2, let fi(R(n,r),V1,…,Vn-1) be the polynomial in Theorem 1   (R3) for L={1,…,i}. Note that the coefficient of p(n,r,j) in
(49)R(n,r)n-2-ifi(R(n,r),V1,…,Vn-1)
is (ij), where p(n,r,j) is defined in (15). We list the coefficient in the following e(n,r)×(e(n,r)+1) matrix Bn,r:
(50)Bn,r=((ij))((n+r)/2)≤i≤n-20≤j≤e(n,r).
Theorem 1 tells us that
(51)Bn,r(p(n,r,0),…,p(n,r,j),…,p(n,r,e(n,r)))t=(0,…,0,…,0)t.In the notation of Lemma7, we have
(52)Bn,r=Aˇe(n,r)(n+r2-1).Sincep(n,r,j)(0≤j≤e(n,r)) generate Hn-3(M-n,r;ℤ2), (51) tells us that
(53)p(n,r,j)=yˇj.We computeyˇj as follows:
(54)yˇj=((n+r2-1)+(n-r2-1)-je(n,r)-j)=(n-2-je(n,r)-j)=(D(n)-je(n,r)-j).


## 5. Proof of Theorem A

Theorem A is proved by showing the upper bound for thoseλ with R(n,r)λ≠0coinciding with the lower bound. About the upper bound, we have the following.Proposition 8.
InH*(M-n,r;ℤ2), one has R(n,r)ϕ(n,r)+1=0.Proof.
SincedimM-n,r=n-3, the case for (D(n)e(n,r))≡1 is clear. Hence we may assume that (D(n)e(n,r))≡0. For p in ((n+r)/2)≤p≤n-2, we denote by gp(R(n,r),V1,…,Vn-1) the sum of polynomials in Theorem 1   (R3) for all L with |L|=p. Note that deggp(R(n,r),V1,…,Vn-1)=p-1. We use the notation y~ in the proof of Lemma 6. We claim the following:
(55)∑j=0k(n,r)‍R(n,r)k(n,r)-jg((n+r)/2)+j(R(n,r),V1,…,Vn-1)y~j=R(n,r)ϕ(n,r)+1.
In order to prove (55), we define
(56)Cn,r=((D(n)+1-ie(n,r)-j))0≤i≤e(n,r)0≤j≤e(n,r)-1,C~n,r=((D(n)+1-ie(n,r)-j))0≤i≤e(n,r)0≤j≤k(n,r).
Then the following equality holds:(57)C~n,r(y~0,…,y~k(n,r))=Cn,rty~=t(1,0,…,0)t.
In fact, the left equality follows from the fact thaty~i=0 for k(n,r)+1≤i≤e(n,r)-1. On the other hand, the right equality is proved as follows. Note that we can write Cn,r as
(58)Cn,r=(cA~e(n,r)(D(n)-e(n,r))),
where c is the first row of Cn,r and A~e(n,r)(D(n)-e(n,r)) is the matrix defined from Ae(n,r)(D(n)-e(n,r)) by shifting the (i,j)-element to the (e(n,r)+1-i,e(n,r)+1-j)-element. Since ty∈KerAe(n,r)(D(n)-e(n,r)), we have y~t∈KerA~e(n,r)(D(n)-e(n,r)). Now the right equality of (57) follows from Proposition 4(iii)(b).
Letσi be the ith symmetric polynomial in variables V1,…,Vn-1. Then we have
(59)g((n+r)/2)+j=∑i=0e(n,r)‍(D(n)+1-ie(n,r)-j)R(n,r)((n+r)/2)+j-i-1σi.
In fact, the binomial coefficient is computed from((n-1p)(pi))/(n-1i) for p=((n+r)/2)+j. Now we complete the proof of (55) as follows:
(60)thelefthandsideof(55)=(R(n,r)((n+r)/2)+k(n,r)-1σ0,…,R(n,r)((n+r)/2)+k(n,r)-e(n,r)-1σe(n,r))×C~n,r(y~0,…,y~k(n,r))t(by(59))=(R(n,r)((n+r)/2)+k(n,r)-1σ0,…,R(n,r)((n+r)/2)+k(n,r)-e(n,r)-1σe(n,r))×(1,0,…,0)t(by(57))=R(n,r)((n+r)/2)k(n,r)-1=R(n,r)ϕ(n,r)+1.
Hence (55) holds.
Now by Theorem1 (R3), we have
(61)gp(R(n,r),V1,…,Vn-1)=0
in Hp-1(M-n,r;ℤ2). Since the left-hand side of (55) vanishes, so does the right-hand side.Proof of Theorem A.
We prove the theorem  by induction onn and for all r.
(I) The case forn=4.
SinceM-4,2=S1, Theorem A holds.
(II) Fixn and assume that Theorem A holds for all k with k<n.
We shall prove Theorem A forM-n,r.
Case (i)(2≤r≤n-2).
We need to consider the cases according to(D(n)e(n,r))≡0 or 1. Since the case of 1 is proved in Theorem B, we may assume that (D(n)e(n,r))≡0. We define an inclusion i:Mn-1,r-1→Mn,r by
(62)i(z1,…,zn-2)=(z1,…,zn-2,1).
Note that the mapi naturally induces a map ı-:M-n-1,r-1→M-n,r and we have the following diagram of regular 2-covers:(63)
SinceR(n,r) is the first Stiefel-Whitney class of the regular 2-cover π:Mn,r→M-n,r, we have
(64)ı-*(R(n,r))=R(n-1,r-1).
By inductive hypothesis, we have inH*(M-n-1,r-1;ℤ2) that
(65)R(n-1,n-1)ϕ(n-1,r-1)≠0.
Using (64), we have in H*(M-n,r;ℤ2) that
(66)R(n,r)ϕ(n-1,r-1)≠0.
Using Lemma6(i), we have R(n,r)ϕ(n,r)≠0.
On the other hand, we have by Proposition8 that R(n,r)ϕ(n,r)+1=0. Hence we have proved that h(n,r)=ϕ(n,r).
Case (ii)(r=1).
As in (i), we may assume that(D(n)e(n,r))≡0. Lemma 6(ii) tells us that if 2s+3≤n≤2s+1-1, then ϕ(n,1)=2s-2. Moreover, Proposition 8 tells us that R(n,1)2s-1=0. Hence it will suffice to prove that R(n,1)2s-2≠0 in H*(M-n,1;ℤ2). We define an inclusion j:M2s+1,1→Mn,1 by
(67)i(z1,…,z2s)=(z1,…,z2s,1,…,1︸d,-1,…,-1︸d).
The map induces a mapȷ-:M-2s+1,1→M-n,1. Using a similar diagram to (63) and the fact that R(2s+1,1)2s-2≠0 in H*(M-2s+1,1;ℤ2) (see Theorem B), we have in H*(M-n,1;ℤ2) that R(n,1)2s-2≠0.Proof of Proposition C.
Proposition C(i)(a) and (b) follow from Theorem A and Lemma6(ii), and (i)(c) follows from Theorem B. If we modify Lemma 6(ii) slightly for even n, then we can prove Proposition C(ii) similarly.



---

*Source: 305926-2013-12-26.xml*