Let M-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.
1. Introduction
Given a string r=(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 of Rr, 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 space X 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 Section 2 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
For 4≤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}.
Here zi∈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 that M-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, if n 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 that r 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]).
The mod2 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:
The symbol S 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 class R(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 notation a≡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 all n≥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 of Hn-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.
Let n 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 that R(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 for r=1 or 2, we can write Theorem A more explicitly.
Proposition C.
(i) About M-n,1 for odd n, one has the following.
One has (D(n)e(n,r))≡1 if and only if n is of the form n=2s+1.
If n satisfies that
(17)2s+1≤n≤2s+1-1,
then h(n,1)=2s-2.
One has p(2s+1,1,i)≠0 for all i.
(ii) About M-n,2 for even n, one has the following.
One has (D(n)e(n,r))≡1 if and only if n is of the form n=2s.
If n=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.
One has p(2s,2,i)≠0 for all i.
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 n is small.
Example 2.
We consider Table 1 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.
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
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6
—
10
—
12
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14
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14
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6
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5
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6
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9
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10
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13
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14
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14
7
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6
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8
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10
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10
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14
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14
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8
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7
—
8
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10
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10
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14
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14
9
—
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—
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8
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10
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10
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14
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14
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10
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9
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10
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13
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14
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14
11
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10
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12
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14
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14
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12
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11
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12
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14
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14
13
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12
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14
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14
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14
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13
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14
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17
15
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14
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16
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16
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15
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17
The cases for p(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)
1
Our second example is about the case when r 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 odd n 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 ResultsProposition 4.
Let x 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.
One hasrankAe(x)≥e-1.
One has rankAe(x)=eif and only if (x+ee)≠0in ℤ2[x].
When rankAe(x)=e-1, the following results hold.
Let yt=(y1,…,ye)tbe the unique nontrivial element of KerAe(x). Then one has(21)yj=(x+e-je-j).
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.
Since detP^e=detQ^e(x)=1, they are regular matrices.
We write Pe 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) Since rank1q=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) Since ye=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 that rankAe(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 of KerAe(x-1) is given by
(33)(y1+y2,y2+y3,…,ye-1+ye,ye)t.
The corollary is clear from this.
The following property of ϕ(n,r) will be used in Section 5.
Lemma 6.
(i) If r≥2 and (D(n)e(n,r))≡0, one has
(34)ϕ(n-1,r-1)=ϕ(n,r).
(ii) If n is odd and satisfies that
(35)2s+1≤n≤2s+1-1,
then the following results hold.
One has (D(n)e(n,r))≡1if and only if n is of the form n=2s+1.
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 Proposition 4 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 Corollary 5, (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 have ϕ(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 for n=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.
Hence k(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 Proposition 4 slightly, we have the following.
Lemma 7.
Let Aˇe(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.
One has rankAˇe(x)=e.
Let yˇt=t(yˇ0,…,yˇe) be the unique nontrivial element of KerA˘e(x). Then one has
(47)yˇj=(x+e-je-j).
Proof.
(i) Let Pe be the matrix in the proof of Proposition 4 and we define
(48)Qˇe(x)=((xj-i))0≤i≤e0≤j≤e.
Since Aˇe(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 Lemma 7, we have
(52)Bn,r=Aˇe(n,r)(n+r2-1).
Since p(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 compute yˇj 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 those λ with R(n,r)λ≠0coinciding with the lower bound. About the upper bound, we have the following.
Proposition 8.
In H*(M-n,r;ℤ2), one has R(n,r)ϕ(n,r)+1=0.
Proof.
Since dimM-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 that y~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).
Let σi 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 Theorem 1 (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 on n and for all r.
(I) The case for n=4.
Since M-4,2=S1, Theorem A holds.
(II) Fix n and assume that Theorem A holds for all k with k<n.
We shall prove Theorem A for M-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 map i naturally induces a map ı-:M-n-1,r-1→M-n,r and we have the following diagram of regular 2-covers:(63)
Since R(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 in H*(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 Lemma 6(i), we have R(n,r)ϕ(n,r)≠0.
On the other hand, we have by Proposition 8 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 map ȷ-: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 Lemma 6(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.
HausmannJ. C.Sur la topologie des bras articules19891474Berlin, GermanySpringer146159Lecture Notes in MathematicsKapovichM.MillsonJ.On the moduli space of polygons in the Euclidean plane19954224304642-s2.0-0029679793WalkerK.1985Princeton UniversityHausmannJ.-C.KnutsonA.The cohomology ring of polygon spaces19984812813212-s2.0-0032391132FarberM.HausmannJ.SchützD.On the conjecture of Kevin Walker20091165862-s2.0-7795488225610.1142/S1793525309000023KamiyamaY.TezukaM.TomaT.Homolog y of the configuration spaces of quasi-equilateral polygon linkages199835012486948962-s2.0-22444455389HillerH.On the cohomology of real Grassmannians1980257521533