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

Abstract

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,…,𝑉_𝑛−1]/ℐ_𝑛,𝑟, where 𝑅(𝑛,𝑟) is the first Stiefel-Whitney class of a certain regular 2-cover 𝜋∶𝑀_𝑛,𝑟⟶𝑀_𝑛,𝑟 and the ideal ℐ_𝑛,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,…,𝑟_𝑛) of 𝑛 positive real numbers, one considers the configuration space 𝑀_𝐫 of planar polygon linkages having side lengths 𝑟_𝑖 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 𝐻^∗𝑀_𝐫;ℤ₂=ℤ₂𝑅_𝐫,𝑉₁,…,𝑉_𝑛−1ℐ_𝐫, (1) where 𝑅_𝐫 is the first Stiefel-Whitney class of a certain regular 2-cover 𝜋∶𝑀_𝐫→𝑀_𝐫 and the ideal ℐ_𝐫 is in general big. For example, when 𝐫=(1,…,1),  ℐ_𝐫 is generated by polynomials whose number is approximately 2^𝑛−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 𝑅_𝐫, that is, the unique ℎ(𝐫) such that 𝑅_𝐫^ℎ(𝐫)≠0 but 𝑅_𝐫^ℎ(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_∗(𝑀_𝐫;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=𝐺_𝑘(ℝ^𝑛+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 𝑅_𝐫. 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 𝑀_𝑛,𝑟=𝑧₁,…,𝑧_𝑛−1∈𝑆^1𝑛−1∣^𝑛−1_𝑖=1𝑧_𝑖=𝑟. (2)

Here 𝑧_𝑖∈𝑆¹⊂ℂ denotes the unit vectors in the direction of the sides of a polygon. Note that 𝑀_𝑛 comes with a natural involution 𝜏∶𝑀_𝑛,𝑟⟶𝑀_𝑛,𝑟,𝜏𝑧₁,…,𝑧_𝑛−1=𝑧₁,…,𝑧_𝑛−1. (3)

We set 𝑀_𝑛,𝑟=𝑀_𝑛,𝑟𝜏. (4)

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 𝑀_𝑛,𝑟′≅𝑀_𝑛,𝑟,∀𝑟^′∈(𝑟−1,𝑟+1). (5)

Moreover, if 𝑛 is even and 𝑟^′∈(0,1), then we have 𝑀_𝑛,𝑟′≅𝑀_𝑛−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: 𝑀_{𝑛,𝑛−4}≅#_𝑛ℝ𝑃^𝑛−3,𝑀_{𝑛,𝑛−2}≅ℝ𝑃^𝑛−3. (6)

The following theorem is crucial in this paper.

Theorem 1 (see [4, Corollary 9.2]). The mod2 cohomology ring of 𝑀_𝑛,𝑟 is 𝐻^∗𝑀_𝑛,𝑟;ℤ₂=ℤ₂𝑅(𝑛,𝑟),𝑉₁,…,𝑉_𝑛−1ℐ_𝑛,𝑟, (7) where 𝑅(𝑛,𝑟) and 𝑉₁,…,𝑉_𝑛−1 are of degree 1 and ℐ_𝑛,𝑟 is the ideal generated by the three families of elements:(R1)  𝑉²_𝑖+𝑅(𝑛,𝑟)𝑉_𝑖,𝑤ℎ𝑒𝑟𝑒𝑖=1,…,𝑛−1, (8)(R2)_𝑖∈𝑆𝑉_𝑖,𝑤ℎ𝑒𝑟𝑒𝑆⊂{1,…,𝑛−1}𝑖𝑠𝑠𝑢𝑐ℎ𝑡ℎ𝑎𝑡|𝑆|≥𝑛−𝑟2, (9)(R3)_𝑆⊂𝐿𝑅(𝑛,𝑟)^{|𝐿−𝑆|−1}_𝑖∈𝑆𝑉_𝑖{}|𝐿|≥,𝑤ℎ𝑒𝑟𝑒𝐿⊂1,…,𝑛−1𝑖𝑠𝑠𝑢𝑐ℎ𝑡ℎ𝑎𝑡𝑛+𝑟2. (10)
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∶𝑀_𝑛,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 𝐷(𝑛)=𝑛−2,𝑒(𝑛,𝑟)=𝑛−𝑟2𝑘()()𝐷(𝑛)()𝑖.−1,𝑛,𝑟=max𝑖∣0≤𝑖≤𝑒𝑛,𝑟−1,−𝑒𝑛,𝑟+𝑖≡1 (11)

Moreover, we set 𝜙()=𝐷(𝑛)𝑒()𝑛,𝑟𝑛−3,if𝑛,𝑟≡1,𝑛+𝑟2()𝐷(𝑛)𝑒()+𝑘𝑛,𝑟−2,if𝑛,𝑟≡0. (12)

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 𝐻^𝑛−3(𝑀_𝑛,𝑟;ℤ₂)=ℤ₂. Note that if 1≤𝜈₁<⋯<𝜈_𝑖≤𝑛−1, (13) then we have in 𝐻^𝑛−3(𝑀_𝑛,𝑟;ℤ₂) that 𝑅(𝑛,𝑟)^{𝑛−3−𝑖}𝑉₁⋯𝑉_𝑖=𝑅(𝑛,𝑟)^{𝑛−3−𝑖}𝑉_𝜈1⋯𝑉_𝜈𝑖. (14)

We set 𝑝(𝑛,𝑟,𝑖)=𝑅(𝑛,𝑟)^{𝑛−3−𝑖}𝑉₁⋯𝑉_𝑖. (15)

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 𝐻^𝑛−3(𝑀_𝑛,𝑟;ℤ₂): 𝑝()=𝐷(𝑛)𝑒()𝑛,𝑟,𝑖−𝑖𝑛,𝑟−𝑖. (16)

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.(a)One has ^𝐷(𝑛)_{𝑒(𝑛,𝑟)}≡1 if and only if 𝑛 is of the form 𝑛=2^𝑠+1.(b)If 𝑛 satisfies that 2^𝑠+1≤𝑛≤2^𝑠+1−1, (17) then ℎ(𝑛,1)=2^𝑠−2.(c)One has 𝑝(2^𝑠+1,1,𝑖)≠0 for all 𝑖.
(ii) About 𝑀_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=2^𝑠.(b)If 𝑛=2^𝑠, then ℎ(𝑛,2)=2^𝑠−3. On the other hand, if 𝑛 satisfies that 2^𝑠+2≤𝑛≤2^𝑠+1−2, (18) then ℎ(𝑛,2)=2^𝑠−2.(c)One has 𝑝(2^𝑠,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 𝐻^𝑛−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.

Table 2 

The cases for 𝑝(𝑛,𝑟,𝑖)=0.

Our second example is about the case when 𝑟 is large.

Example 3. (i) We haveℎ(𝑛,𝑛−4)=𝑛−3𝑛isodd,𝑛−4𝑛iseven. (19)
(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 Results

Proposition 4. Let 𝑥 be indeterminate and 𝑒∈ℕ. Let 𝐴_𝑒(𝑥) be an 𝑒×𝑒 matrix over ℤ₂[𝑥] defined as follows: 𝐴_𝑒(𝑥)=𝑗𝑥+𝑖_{1≤𝑖≤𝑒1≤𝑗≤𝑒}. (20)
Then one has the following results. (i)One hasrank𝐴_𝑒(𝑥)≥𝑒−1. (ii)One has rank𝐴_𝑒(𝑥)=𝑒if and only if (_𝑒^𝑥+𝑒)≠0in ℤ₂[𝑥].(iii)When rank𝐴_𝑒(𝑥)=𝑒−1, the following results hold.(a)Let ^𝑡𝐲=^𝑡(𝑦₁,…,𝑦_𝑒)be the unique nontrivial element of Ker𝐴_𝑒(𝑥). Then one has𝑦_𝑗=𝑥+𝑒−𝑗𝑒−𝑗. (21)(b)Let^𝑡𝐲=^𝑡(𝑦₁,…,𝑦_𝑒)be as in (a). Then one has^𝑒𝑗=1𝑗𝑦𝑥+𝑒+1_𝑗=1. (22)

Proof. We define 4 matrices as follows: 𝑃_𝑒=𝑖𝑗_{1≤𝑖≤𝑒0≤𝑗≤𝑒},𝑄_𝑒(𝑥)=𝑥𝑗−𝑖_{0≤𝑖≤𝑒1≤𝑗≤𝑒},𝑃_𝑒=𝑖𝑗_{1≤𝑖,𝑗≤𝑒},𝑄_𝑒(𝑥)=𝑥𝑗−𝑖_{1≤𝑖,𝑗≤𝑒}. (23)
Since det𝑃_𝑒=det𝑄_𝑒(𝑥)=1, they are regular matrices.
We write 𝑃_𝑒 and 𝑄_𝑒(𝑥) as follows: 𝑃_𝑒=𝟏𝑃_𝑒,𝑄_𝑒(𝑥)=𝐪𝑄_𝑒(𝑥). (24)
Using the Chu-Vandermonde identity 𝑗=𝑥+𝑖^𝑗_𝑘=0𝑖𝑘𝑥𝑗−𝑘, (25) we have 𝐴_𝑒(𝑥)=𝑃_𝑒𝑄_𝑒(𝑥)=𝟏𝐪+𝑃_𝑒𝑄_𝑒(𝑥). (26)
(i) Since rank𝟏𝐪=1 and 𝑃_𝑒𝑄_𝑒(𝑥) is a regular matrix, we have rank𝐴_𝑒(𝑥)≥𝑒−1.
(ii) Note that det𝐴_𝑒(𝑥)=𝑒𝑥+𝑒. (27)In fact, both the sides coincide for 𝑥=0 and vanish for 𝑥∈ℤ such that −𝑒≤𝑥≤−1.
Now (27) tells us that rank𝐴_𝑒(𝑥)=𝑒 if and only if (_𝑒^𝑥+𝑒)≠0.
(iii)(a) Since 𝑦_𝑒=1, we have ^𝑡𝐲≠0=. We must prove that ^𝑒𝑗=1𝑗𝑥+𝑖𝑥+𝑒−𝑗𝑒−𝑗=0 (28) for 1≤𝑖≤𝑒. Since we are assuming that rank𝐴_𝑒(𝑥)=𝑒−1, it will suffice to prove that ^𝑒𝑗=0𝑗𝑥𝑥+𝑖𝑥+𝑒−𝑗=0. (29)
We use the formula ^𝑐_𝑗=0(−1)^𝑗𝑎𝑗=𝑐𝑏−𝑗𝑏−𝑐𝑏−𝑎 (30) 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 𝑒𝑒−𝑖. 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^𝑒−𝑖=_𝑒⁻¹=1.

Corollary 5. Assume that rank𝐴_𝑒(𝑥)=𝑒−1 and let ^𝑡𝐲 be the element in Proposition 4(iii)(a). One sets ℓ(𝑥,𝑒)=min𝑗∣1≤𝑗≤𝑒,𝑦_𝑗≡1. (31) If one has also that rank𝐴_𝑒(𝑥−1)=𝑒−1, then one has ℓ(𝑥−1,𝑒)=ℓ(𝑥,𝑒)−1. (32)

Proof. The unique nontrivial element of Ker𝐴_𝑒(𝑥−1) is given by ^𝑡𝑦₁+𝑦₂,𝑦₂+𝑦₃,…,𝑦_𝑒−1+𝑦_𝑒,𝑦_𝑒. (33) 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 𝜙(𝑛−1,𝑟−1)=𝜙(𝑛,𝑟). (34)
(ii) If 𝑛 is odd and satisfies that 2^𝑠+1≤𝑛≤2^𝑠+1−1, (35) then the following results hold.(a)One has ^𝐷(𝑛)_{𝑒(𝑛,𝑟)}≡1if and only if 𝑛 is of the form 𝑛=2^𝑠+1.(b)𝜙(𝑛,1)=2^𝑠−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 𝐷()(𝑛)𝑒()()=𝑛−1=𝐷−1=𝑛−3,𝑛−1,𝑟−1=𝑒𝑛,𝑟𝑛−𝑟2−1. (36)(37)
We shall prove that 𝑘(𝑛−1,𝑟−1)=𝑘(𝑛,𝑟)+1. (38)
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=𝑦_𝑒−𝑗 and put 𝐲=(𝑦₀,…,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 𝑘(𝑛,𝑟)=𝑒(𝑛,𝑟)−ℓ(𝐷(𝑛)−𝑒(𝑛,𝑟),𝑒(𝑛,𝑟)). (39)
Using Corollary 5, (36), and (37), we have ℓ(𝐷()()((𝐷(𝑛)()(𝑛−1−𝑒𝑛−1,𝑟−1,𝑒𝑛−1,𝑟−1))=ℓ−𝑒𝑛,𝑟,𝑒𝑛,𝑟))−1. (40)
Combining (37), (39), and (40), we obtain (38).
Now (i) follows from the following computation: 𝜙(𝑛−1,𝑟−1)=𝑛−1+𝑟−12+(𝑘())().𝑛,𝑟+1−2=𝜙𝑛,𝑟 (41) 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)−1}_{𝑒(𝑛,𝑟)−1}≡1. But the last assertion is clear if we apply (36) and (37) to our assumptions that 𝐷(𝑛)𝑒()𝐷()𝑛,𝑟𝑒()≡0,𝑛−1𝑛−1,𝑟−1≡1. (42)
(ii) If we prove (b), then (a) is clear from this. Since (b) clearly holds for 𝑛=2^𝑠+1, we assume that 2^𝑠+3≤𝑛≤2^𝑠+1−1. We set 𝑛=2𝑚+1. By definition, 𝑘()𝑖𝑛,𝑟=max𝑖∣0≤𝑖≤𝑚−2,𝑚+𝑖≡1. (43)
By Kummer, 𝑏𝑎+𝑏 is odd if and only if there are no carries when adding 𝑎 and 𝑏 in base 2. Hence if we write 𝑚=2^𝑠−1+𝑢 (where 1≤𝑢≤2^𝑠−1), the greatest 𝑖 (where 𝑖≤𝑚−2) such that _𝑖^𝑚+𝑖≡1 is 𝑖=2^𝑠−2+2^𝑠−3+⋯+1−𝑢=2^𝑠−1−1−𝑢. (44)
Hence 𝑘(𝑛,𝑟)=2^𝑠−1−1−𝑢 and 𝜙()=()()=2𝑛,1𝑚+1+𝑘𝑛,𝑟−2_𝑠−1+𝑢+1+2^𝑠−1−1−𝑢−2=2^𝑠−2. (45)

Modifying Proposition 4 slightly, we have the following.

Lemma 7. Let 𝐴̌_𝑒(𝑥) be an 𝑒×(𝑒+1) matrix over ℤ₂[𝑥] defined as follows: 𝐴̌_𝑒(𝑥)=𝑗𝑥+𝑖_{1≤𝑖≤𝑒0≤𝑗≤𝑒}. (46) Then one has the following results. (i)One has rank𝐴̌_𝑒(𝑥)=𝑒.(ii)Let 𝐲̌^𝑡=^𝑡(𝑦̌₀,…,𝑦̌_𝑒) be the unique nontrivial element of Ker𝐴̆_𝑒(𝑥). Then one has 𝑦̌_𝑗=𝑥+𝑒−𝑗𝑒−𝑗. (47)

Proof. (i) Let 𝑃_𝑒 be the matrix in the proof of Proposition 4 and we define 𝑄̌_𝑒(𝑥)=𝑥𝑗−𝑖_{0≤𝑖≤𝑒0≤𝑗≤𝑒}. (48)
Since 𝐴̌_𝑒(𝑥)=𝑃_𝑒𝑄̆_𝑒(𝑥), rank𝑃_𝑒=𝑒, and det𝑄̆_𝑒(𝑥)=1, the result follows;
(ii) follows from (29).

4. Proof of Theorem B

For (𝑛+𝑟)/2≤𝑖≤𝑛−2, let 𝑓_𝑖(𝑅(𝑛,𝑟),𝑉₁,…,𝑉_𝑛−1) be the polynomial in Theorem 1   (R3) for 𝐿={1,…,𝑖}. Note that the coefficient of 𝑝(𝑛,𝑟,𝑗) in 𝑅(𝑛,𝑟)^{𝑛−2−𝑖}𝑓_𝑖𝑅(𝑛,𝑟),𝑉₁,…,𝑉_𝑛−1 (49) 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 𝐵_𝑛,𝑟: 𝐵_𝑛,𝑟=𝑖𝑗_{((𝑛+𝑟)/2)≤𝑖≤𝑛−20≤𝑗≤𝑒(𝑛,𝑟)}. (50) Theorem 1 tells us that 𝐵^𝑡_𝑛,𝑟𝑝()((=𝑛,𝑟,0,…,𝑝𝑛,𝑟,𝑗,…,𝑝𝑛,𝑟,𝑒𝑛,𝑟))_𝑡(0,…,0,…,0). (51)

In the notation of Lemma 7, we have 𝐵_𝑛,𝑟=𝐴̌_{𝑒(𝑛,𝑟)}𝑛+𝑟2−1. (52)

Since 𝑝(𝑛,𝑟,𝑗)(0≤𝑗≤𝑒(𝑛,𝑟)) generate 𝐻^𝑛−3(𝑀_𝑛,𝑟;ℤ₂), (51) tells us that 𝑝𝑛,𝑟,𝑗=𝑦̌_𝑗. (53)

We compute 𝑦̌_𝑗 as follows: 𝑦̌_𝑗=𝑛+𝑟2−1+𝑛−𝑟2𝑒()=𝑒()=𝐷(𝑛)𝑒().−1−𝑗𝑛,𝑟−𝑗𝑛−2−𝑗𝑛,𝑟−𝑗−𝑗𝑛,𝑟−𝑗 (54)

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 𝑔_𝑝(𝑅(𝑛,𝑟),𝑉₁,…,𝑉_𝑛−1) the sum of polynomials in Theorem 1   (R3) for all 𝐿 with |𝐿|=𝑝. Note that deg𝑔_𝑝(𝑅(𝑛,𝑟),𝑉₁,…,𝑉_𝑛−1)=𝑝−1. We use the notation 𝐲 in the proof of Lemma 6. We claim the following: ^{𝑘(𝑛,𝑟)}_𝑗=0𝑅(𝑛,𝑟)^{𝑘(𝑛,𝑟)−𝑗}𝑔_{((𝑛+𝑟)/2)+𝑗}𝑅(𝑛,𝑟),𝑉₁,…,𝑉_𝑛−1𝑦_𝑗=𝑅(𝑛,𝑟)^{𝜙(𝑛,𝑟)+1}. (55)
In order to prove (55), we define 𝐶_𝑛,𝑟=𝐷(𝑛)𝑒()+1−𝑖𝑛,𝑟−𝑗_{0≤𝑖≤𝑒(𝑛,𝑟)0≤𝑗≤𝑒(𝑛,𝑟)−1},𝐶_𝑛,𝑟=𝐷(𝑛)𝑒()+1−𝑖𝑛,𝑟−𝑗_{0≤𝑖≤𝑒(𝑛,𝑟)0≤𝑗≤𝑘(𝑛,𝑟)}. (56)
Then the following equality holds: 𝐶^𝑡_𝑛,𝑟𝑦₀,…,𝑦_{𝑘(𝑛,𝑟)}=𝐶^𝑡_𝑛,𝑟𝐲=^𝑡(1,0,…,0). (57)
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 𝐶_𝑛,𝑟 as 𝐶_𝑛,𝑟=𝐜𝐴_{𝑒(𝑛,𝑟)}(𝐷(𝑛)−𝑒(𝑛,𝑟)), (58) where 𝐜 is the first row of 𝐶_𝑛,𝑟 and 𝐴_{𝑒(𝑛,𝑟)}(𝐷(𝑛)−𝑒(𝑛,𝑟)) is the matrix defined from 𝐴_{𝑒(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∈Ker𝐴_{𝑒(𝑛,𝑟)}(𝐷(𝑛)−𝑒(𝑛,𝑟)), we have 𝐲∈Ker𝐴^𝑡_{𝑒(𝑛,𝑟)}(𝐷(𝑛)−𝑒(𝑛,𝑟)). Now the right equality of (57) follows from Proposition 4(iii)(b).
Let 𝜎_𝑖 be the 𝑖th symmetric polynomial in variables 𝑉₁,…,𝑉_𝑛−1. Then we have 𝑔_{((𝑛+𝑟)/2)+𝑗}=^{𝑒(𝑛,𝑟)}_𝑖=0𝐷(𝑛)𝑒()𝑅()+1−𝑖𝑛,𝑟−𝑗𝑛,𝑟^{((𝑛+𝑟)/2)+𝑗−𝑖−1}𝜎_𝑖. (59)
In fact, the binomial coefficient is computed from (_𝑝^𝑛−1𝑝_𝑖)/_𝑖^𝑛−1 for 𝑝=((𝑛+𝑟)/2)+𝑗. Now we complete the proof of (55) as follows: ()=𝑅()thelefthandsideof55𝑛,𝑟^{((𝑛+𝑟)/2)+𝑘(𝑛,𝑟)−1}𝜎₀𝑅(),…,𝑛,𝑟^{((𝑛+𝑟)/2)+𝑘(𝑛,𝑟)−𝑒(𝑛,𝑟)−1}𝜎_{𝑒(𝑛,𝑟)}×𝐶^𝑡_𝑛,𝑟𝑦₀,…,𝑦_{𝑘(𝑛,𝑟)}()=𝑅()by59𝑛,𝑟^{((𝑛+𝑟)/2)+𝑘(𝑛,𝑟)−1}𝜎₀𝑅(),…,𝑛,𝑟^{((𝑛+𝑟)/2)+𝑘(𝑛,𝑟)−𝑒(𝑛,𝑟)−1}𝜎_{𝑒(𝑛,𝑟)}×^𝑡()()()1,0,…,0by57=𝑅𝑛,𝑟^{((𝑛+𝑟)/2)𝑘(𝑛,𝑟)−1}=𝑅(𝑛,𝑟)^{𝜙(𝑛,𝑟)+1}. (60)
Hence (55) holds.
Now by Theorem 1 (R3), we have 𝑔_𝑝𝑅(𝑛,𝑟),𝑉₁,…,𝑉_𝑛−1=0 (61) in 𝐻^𝑝−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∶𝑀_{𝑛−1,𝑟−1}→𝑀_𝑛,𝑟 by 𝑖𝑧₁,…,𝑧_𝑛−2=𝑧₁,…,𝑧_𝑛−2,1. (62)
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∶𝑀_𝑛,𝑟→𝑀_𝑛,𝑟, we have 𝚤^∗(𝑅(𝑛,𝑟))=𝑅(𝑛−1,𝑟−1). (64)
By inductive hypothesis, we have in 𝐻^∗(𝑀_{𝑛−1,𝑟−1};ℤ₂) that 𝑅(𝑛−1,𝑛−1)^{𝜙(𝑛−1,𝑟−1)}≠0. (65)
Using (64), we have in 𝐻^∗(𝑀_𝑛,𝑟;ℤ₂) that 𝑅(𝑛,𝑟)^{𝜙(𝑛−1,𝑟−1)}≠0. (66)
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 2^𝑠+3≤𝑛≤2^𝑠+1−1, then 𝜙(𝑛,1)=2^𝑠−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∶𝑀₂^𝑠_+1,1→𝑀_𝑛,1 by 𝑖𝑧₁,…,𝑧₂^𝑠=𝑧₁,…,𝑧₂^𝑠,1,…,1_𝑑,−1,…,−1_𝑑. (67)
The map induces a map 𝚥∶𝑀₂^𝑠_+1,1→𝑀_𝑛,1. Using a similar diagram to (63) and the fact that 𝑅(2^𝑠+1,1)^2𝑠−2≠0 in 𝐻^∗(𝑀₂^𝑠_+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.