Relations in the cohomology ring of the moduli space of flat SO(2n + 1)-connections on a Riemann surface

Abstract

We consider the moduli space of flat SO(2n + 1)-connections (up to gauge transformations) on a Riemann surface, with fixed holonomy around a marked point. There are natural line bundles over this moduli space; we construct geometric representatives for the Chern classes of these line bundles, and prove that the ring generated by these Chern classes vanishes below the dimension of the moduli space, generalizing a conjecture of Newstead.

This article is part of the theme issue ‘Finite dimensional integrable systems: new trends and methods’.

1. Introduction

Let G be a compact Lie group with Lie algebra , and pick a maximal torus T⊂G. Spaces of the form (Hom π₁(Σ), G)/G, where Σ is a Riemann surface of genus g≥2, arise in various branches of geometry. Such spaces have interpretations as the moduli space of flat connections up to gauge transformations [1–3]; they also occur as a building block in the topological quantum field theoretical construction of invariants of 3-manifolds with boundary Σ. See also the related discussions of character varieties [4,5], of parabolic Higgs bundles [6] and of polygon spaces [7].

A related space is obtained by marking a point p∈Σ and prescribing the holonomy of the connections around that point. That is, fix a generator c∈π₁(Σ\p) that represents a small curve around p, and for t∈T, let S_g(t): = {ρ∈Hom(π₁(Σ\p), G|ρ(c)∼t}/G, where ∼ denotes conjugacy in G. In the case where the holonomy ζ lies in the centre of G, the spaces S_g(ζ) are moduli spaces of stable holomorphic vector bundles over Σ (e.g. [8–13]).

We will consider the space S_g(t), where t is a generic torus element with Stabt = T. In this paper, we will take G = SO(2n + 1). Consider the torus bundle V_g(t) → S_g(t) given by V_g(t): = {ρ∈Hom(π₁(Σ\p), G)|ρ(c) = t}. If is a root of , let L_ϕ be the line bundle associated with V_g(t) via the torus representation with weight ϕ. We will construct geometric representatives for the first Chern classes of these line bundles. By considering these geometric representatives, we are able to identify several particular products of Chern classes which vanish in H*(S_g(t)); we also give a combinatorial proof that any monomial in the c₁(L_ϕ) is equivalent in H*(S_g(t)) to a combination of those particular monomials and hence also vanishes. Our geometric representatives are analogous to Schubert cycles for flag manifolds; however, a key difference is that there is no canonical complex structure on S_g(t), and our geometric representatives for Chern classes will not generally be complex subvarieties of S_g(t) with respect to an arbitrary choice of Kähler structure. Thus, a feature of our topological approach is that it enables us to make use of these particular geometric representatives which would not show up under an algebraic geometric treatment of the subject.

This paper builds on our earlier work [14], where we used a similar approach for the case G = SU(n), making S_g(t) the moduli space of parabolic holomorphic vector bundles over Σ. (For a different approach to finding generators and relations for the cohomology of this moduli space, see [11].) This itself was based on the earlier paper [15], which used this geometric approach in the case G = SU(2) to provide a geometric proof to a conjecture of Newstead [16].

Let us now take G = SO(2n + 1), and fix generators a₁, …, a_g, b₁, …, b_g, c for the fundamental group π₁(Σ\{p}), such that c represents the boundary of Σ\{p} and . Recall that the set of roots of is . Choose the maximal torus T⊂G consisting of elements

for ease of notation such elements will be denoted (θ₁, …, θ_n).

Definition 1.1 —

We say an element t = (θ₁, …, θ_n) is generic if Stabt = T, and if the only relation with λ_i∈{ ± 1, 0} is the trivial relation λ₁ =  · s = λ_n = 0.

Let t∈T be generic, and set

Let denote the one-dimensional torus representation

For a root of , we consider line bundles , where the quotient is by the diagonal T-action. We will also denote L_{(ηi ± ηj)} by L _ij± ; observe that L_{−(ηi ± ηj)}≅(L_{ηi ± ηj})*.

Theorem 1.2 —

The product vanishes whenever .

The dimension of S_n,g(t) is 2gn(2n + 1) − 2n(n + 1), so when g≥3n/2, our theorem shows that the top cohomology classes of S_n,g(t) must be generated by elements other than the c₁(L_ϕ).

We begin in the case n = 1, the proof for which is analogous to Weitsman's proof [15] of the Newstead conjecture. When n = 1, we have just the one pair of roots ± ϕ of , and the corresponding pair of line bundles L_ϕ and L_−ϕ = L^*_ϕ; of course c₁(L_ϕ) = − c₁(L^*_ϕ) = − c₁(L_−ϕ). In this case, the theorem is that c₁(L_ϕ)^2g = 0, as follows:

Proposition 1.3 —

Let G = SO(3). Let t∈T be generic, and let L → S_1,g(t) be the line bundle associated with V_g(t) by the representation . Then (c₁(L))^2g = 0 in .

Proof. —

For x∈{a₁, …, a_g, b₁, …, b_g}, consider the section s_x of L induced by the T-equivariant map

(where the subscript ij denotes the (i, j)th matrix entry). If s_x = 0, then

But since ρ(x)∈SO(3), we also have |c| = |d| and cd = 0, thus c = d = 0. Similarly e = f = 0, and so a² + b² = 1 and g = 1; that is, ρ(x)∈T. Suppose ρ(x) = 0 for all x∈{a₁, …, a_g, b₁, …, b_g}. Then ρ(a₁), …, ρ(a_g), ρ(b₁), …, ρ(b_g)∈T. But then , so in fact there are no such ρ in V_g(t). Hence the section (s_a1, …, s_ag, s_b1, …, s_bg) of L^⊕2g is nowhere zero, so c_2g(L^⊕2g) = (c₁(L))^2g = 0. ▪

When n > 1, the combinatorics of the vanishing loci of the relevant sections becomes more complicated. The key idea in the proof of theorem 1.2 is similar to that in our earlier paper [14], but the combinatorics required in this case is more intricate.

The outline of the rest of this paper is as follows.

In §2, we prove the theorem for the case G = SO(5). As in [14], the proof begins by identifying specific products of Chern classes which must vanish because there are sections (constructed explicitly) of the relevant line bundles with no common zeros. The proof is completed by showing that any monomial of sufficiently high degree is equivalent to a combination of these products of Chern classes which have already been shown to vanish. In §3, we allow n to be an arbitrary positive integer, exhibit some collections of sections of the appropriate line bundles with no common zeros, and conclude that the corresponding products of Chern classes vanish. In §4 we turn to the combinatorial heart of the argument, showing that every product of Chern classes of sufficient degree is equivalent in H*(S_n,g(t)) to a combination of those shown to vanish in §3.

2. G = SO(5)

We have chosen to devote a section to proving our result in the case G = SO(5). This is intended to provide intuition for the general case in a more tractable context. Let G = SO(5) and let T⊂G be the maximal torus

to simplify notation we will also denote elements of T by (θ₁, θ₂). Choose an element t∈T that is generic in the sense that . Recall that the roots of are ± (2η₁, 2η₂, η₁ + η₂, η₁ − η₂). If ϕ =  ± (η_j ± η_k) we denote by the one-dimensional torus representation (θ₁, θ₂) · z = e^{± i(θ_j ± θ_k)}z. Consider the line bundles for ϕ∈Φ(G). Observe that L_−ϕ≅L^*_ϕ, and L_ϕ+ψ≅L_ϕ⊗L_ψ. Hence c₁(L^*_ϕ) = − c₁(L_ϕ), and c₁(L_ϕ+ψ) = c₁(L_ϕ) + c₁(L_ψ). In this case, theorem 1.2 says

Proposition 2.1 —

The product vanishes whenever .

As in the case of G = SO(3), and following the strategy for G = SU(n) in [14], we prove this by identifying collections of sections of the L_ϕ with no common zeros, thus observing that the corresponding products of first Chern classes c₁(L_ϕ) vanish. We will identify three such vanishing products, and then prove that any other monomial of degree at least 8g is equivalent in H*(S_2,g(t)) to a combination of those three.

Definition 2.2 —

For x∈{a_l, b_l|1 ≤ l ≤ g}, let s ± _ij(x) be the section of L_{ηi ± ηj} induced by the T-equivariant map

(Note that s⁻_ii(x) is a section of the trivial bundle L₀.)

Observe that if s⁺_ij(x) vanishes, then the (i, j)^th 2-by-2 block in ρ(x) takes the form ; if for any given i, j, both s⁺_ij(x) and s⁻_ij(x) vanish, then the (i, j)^th 2-by-2 block in ρ(x) is zero.

To simplify notation, will write c ± _ij: = c₁(L ± _ij)

Lemma 2.3 —

The monomial z₁: = (c⁺₁₁c⁺₁₂c⁻₁₂)^2g vanishes in H^12g(S_2,g(t)).

Proof. —

Let x∈{a_l, b_l}, and consider the sections s⁺₁₁(x), s⁺₁₂(x), s⁻₁₂(x). If these all vanish, then

for some . Since ρ(x)∈SO(5), c = d = 0, so in fact

for some with a² + b² = 1. If all 6g sections s⁺₁₁(x), s⁺₁₂(x), s⁻₁₂(x), for x = a₁, …, a_g, b₁, …, b_g vanish, then

(since t was chosen to be generic). But ρ(c) = t for ρ∈V_g(t), and so these 6g sections have no common zeros; thus (c⁺₁₁c⁺₁₂c⁻₁₂)^2g = 0 as claimed. ▪

Similarly, z₂: = (c⁺₂₂c⁺₁₂c⁻₁₂)^2g = 0.

Lemma 2.4 —

Suppose A_l∈SO(5) for 1 ≤ l ≤ g, and each A_l has the form

where R_θ represents the 2-by-2 block . Then cannot be a generic torus element.

Proof. —

Consider the map

This is an injective homomorphism, and κ(M)∈T⇔M is diagonal in U(2). Each A_l is in the image of κ; say A_l = κ(M_l), for M_l∈U(2). Then

Observe that has determinant 1. If M is not diagonal, then κ(M)≠t. If M is diagonal, then it must be for some θ, so is not generic. Thus, in particular, . ▪

Corollary 2.5 —

The monomial p = (c⁺₁₁c⁺₁₂c⁺₂₁c⁺₂₂)^2g vanishes in H^16g(S_2,g(t)).

Proof. —

For each x∈{a_l, b_l|1 ≤ l ≤ g}, consider the sections s⁺₁₁(x), s⁺₁₂(x), s⁺₂₁(x), and s⁺₂₂(x). If these were all to vanish, then each ρ(x) would have the form

where . By lemma 2.4, if these 8g sections all vanish, then . Thus these 8g sections have no common zeros, so (c⁺₁₁c⁺₁₂c⁺₂₁c⁺₂₂)^2g = 0. ▪

Lemma 2.6 —

Suppose B_k∈SO(5) for 1 ≤ k ≤ 2g, and each B_k has the form

Then .

Proof. —

Let

Observe that EB_kE⁻¹ has the form described in lemma 2.4. So by lemma 2.4, cannot be a generic torus element. If , then . Note that

so EtE⁻¹ is a generic torus element. This is impossible by lemma 2.4, hence . ▪

Corollary 2.7 —

The monomial q = (c⁺₁₁c⁺₂₂c⁻₁₂c⁻₂₁)^2g vanishes in H^16g(S_2,g(t)).

Proof. —

Consider the sections s⁺₁₁(x), s⁺₂₂(x), s⁻₁₂(x) and s⁻₂₁(x) for each x∈{a_l, b_l|1 ≤ l ≤ g}. If these all vanish, then each ρ(x) has the form described in lemma 2.6, and so . So these 8g sections have no common zeros, so (c⁺₁₁c⁺₂₂c⁻₁₂c⁻₂₁)^2g = 0. ▪

Lemma 2.8 —

Let 0 ≤ m ≤ 2g. The monomials

2.1

and

2.2

vanish in H^16g(S_2,g(t)).

Proof. —

For fixed x, we have seen above that the sections s⁺₁₁(x), s⁺₁₂(x), s⁺₂₁(x) and s⁺₂₂(x) all vanish when ρ(x) has the form

2.3

Observe that if the sections s⁺₁₁(x), s⁻₁₂(x), s⁺₁₂(x) and s⁺₂₂(x) all vanish, then ρ(x) has the form

which is in particular also in the form (2.3). Thus if X_m is any m-element subset of {a₁, …, a_g, b₁, …, b_g}, then the collection of sections

has no common zeros, so the monomial (2.1) vanishes.

Similarly, the collection of sections s⁺₁₁(x), s⁺₂₂(x), s⁻₁₂(x) and s⁻₂₁(x) vanishes when ρ(x) has the form

and the collection s⁺₁₁(x), s⁺₂₂(x), s⁻₁₂(x) and s⁺₁₂(x) vanishes when ρ(x) has the form

so by lemma 2.6 the collection of sections

has no common zeros, so the monomial (2.2) vanishes. ▪

Proposition 2.9 —

Suppose ζ is a monomial in the c ± _ij with degree at least 8g. Then ζ is a combination of monomials z₁, z₂, y⁺_m and y⁻_m.

Proof. —

Use the relations

to write ζ as a sum of terms (c⁺₁₁)^a(c⁺₂₂)^b, where a + b≥8g, and consider each term separately. We must have a≥2g or b≥2g; without loss of generality assume a≥2g. Rewrite this term as (c⁺₁₁)^2g(2c⁺₁₂ − c⁺₂₂)^a−2g(c⁺₂₂)^b, and consider each term λ(c⁺₁₁)^2g(c⁺₁₂)^a−2g−m(c⁺₂₂)^b+m. Note that a − 2g + b≥6g, so either a − 2g − m≥2g or b + m≥2g.

• — If a − 2g − m≥2g, rewrite this term as λ′(c⁺₁₁)^2g(c⁺₁₂)^2g(c⁻₁₂ + c⁺₂₂)^a−m−4g(c⁺₂₂)^b+m, and consider each term in the resulting expansion. Note that a + b − 4g≥4g, so either a − m − 4g − s≥2g or b + m + s≥2g.

  • (i) If a − m − 4g − s≥ 2g, this term is a multiple of z₁.

  • (ii) If b + m + s≥2g, rewrite this term as . Each term in the expansion of this polynomial is a multiple of some y ± _m.

• — If b + m≥2g, rewrite the term as λ(c⁺₁₁)^2g(c⁺₂₂)^2g(c⁺₁₂)^a−2g−m(c⁺₁₂ − c⁻₁₂)^b+m−2g. Each term in the expansion of this polynomial is a multiple of some y ± _m.

Hence ζ is equal to a sum of multiples of the monomials z₁, z₂ and y ± _m as claimed. ▪

Corollary 2.10 —

vanishes whenever

3. The general case

We now begin our study of the general case G = SO(2n + 1). As we did for SO(5) in the previous section, we will show that some particular products of the c₁(L_ϕ), for roots ϕ of , vanish, by finding sections of the L_ϕ with no common zeros.

A T-equivariant map induces a section of . Let x∈{a₁, …, a_g, b₁, …, b_g} be one of our chosen generators of π₁(Σ\{p}) other than c, and consider the maps

These maps are T-equivariant and induce sections s⁺_ij(x) of L ± _ij. The nature of the T action on SO(2n + 1) is such that top left 2n-by-2n corner of elements of SO(2n + 1) can be divided into 2-by-2 matrices on which the behaviour of the torus action can be considered separately. Thus, it will often be convenient to write elements of SO(2n + 1) in the form

adopting the notational convention that capital letters denote 2-by-2 arrays whereas lowercase letters represent real numbers.

The next lemma is the direct generalization of lemma 2.4 to the general case.

Lemma 3.1 —

Suppose M_l∈SO(2n + 1) for 1 ≤ l ≤ 2g, and each M_l has the form

3.1

where each R^ij_l is a 2-by-2 block of the form , with . Then cannot be a generic torus element.

Proof. —

Consider the map

This is an injective homomorphism, and κ(M)∈T⇔M is diagonal in U(n). Under the hypothesis of this lemma, each M_l is in the image of κ; say M_l = κ(N_l), for N_l∈U(n). Then

Observe that if has determinant 1. If N is not diagonal, then κ(N) is not a torus element (so certainly not a generic torus element). Assume κ(N) is a torus element. Then N is a diagonal matrix in U(n) of determinant 1, so is

where . Thus κ(N) is (θ₁, …, θ_n)∈T, which is not generic in the sense of definition 1.1 because . ▪

Corollary 3.2 —

The sections {s⁺_ij(x)|1 ≤ i, j ≤ n, x∈ {a₁, …, a_g, b₁, …, b_g}} of the line bundles L⁺_ij have no common zeros.

Proof. —

Fix x∈{a₁, …, a_g, b₁, …, b_g} and consider the sections {s⁺_ij(x)|1 ≤ i, j,  ≤ n}. These sections all vanish when

Since ρ(x)∈SO(2n + 1), we must have z²_2l−1 = z²_2l, and z_2l−1z_2l = 0, so z₁ =  · s = z_2n = 0. Similarly, w₁ =  · s = w_2n = 0, hence u = 1, so ρ(x) is in the form (3.1) from lemma 3.1. At points in S_n,g(t) where all of our sections vanish, therefore, ρ(x) is in this form for every x∈{a₁, …, a_g, b₁, …, b_g}, so by lemma 3.1, cannot be a generic torus element. However, which was chosen to be a generic torus element. Thus, there are no such points in S_n,g(t), that is, the locus on which every one of these sections vanishes is empty. ▪

In the coming discussion, we will make use of the following definition from our earlier paper [14].

Definition 3.3 —

Let X be a finite set. A block B in X × X is a subset of X × X of the form V × V^c, where is a proper non-empty subset of X. We denote the set of all blocks in X × X by . If B = V × V^c, let . We will also make use of the indicator functions

and

Notation —

For a positive integer m, we will denote by [m] the set {1, …, m}.

For l∈[n], let E_l∈O(2n + 1) be the matrix

(so conjugation by E_l switches the (2l − 1)th and (2l)th rows, and the (2l − 1)th and (2l)th columns).

Lemma 3.4 —

Let . Suppose M_l∈SO(2n + 1) for each 1 ≤ l ≤ 2g, and each M_l has the form

3.2

where S^ij_l takes the form

Then cannot be a generic torus element.

Proof. —

Let , and observe that if M_l is in the form (3.2) then EME⁻¹ is in the form (3.1). Thus is not a generic torus element. Note that if h = (θ₁, …, θ_n)∈T is a generic torus element, then EhE⁻¹ = (ϵ_V(1)θ₁, …, ϵ_V(n)θ_n) is also generic. Hence is not a generic torus element. ▪

Corollary 3.5 —

Let . The sections

have no common zeros.

Proof. —

Consider the sections {s^−ϵ_B(i,j)_ij(x)|1 ≤ i, j ≤ n, x = a₁, …, a_g, b₁, …, b_g}. These sections all vanish when each

where R^ij_x takes the form

Since ρ(x)∈SO(2n + 1), this forces z₁ =  · s = z_2n = w₁ =  · s = w_2n = 0, and thus u = 1. So the sections we are considering all vanish when every ρ(x) is of the form (3.2) described in lemma 3.4, in which case cannot be a generic torus element. But again, was chosen to be generic, so these sections have no common zeros. ▪

Now fix x∈{a₁, …, a_g, b₁, …, b_g}, and let be another block (not necessarily distinct from B). Consider the sections

3.3

If these sections all vanish, then

where T^ij_x takes the form

Since ρ(x)∈SO(2n + 1), this forces z_2i−1 = z_2i = 0 for all i∈U; thus ρ(x) is block diagonal up to reordering of basis elements, so also for . Thus again, z₁ =  · s = z_2n = w₁ =  · s = w_2n = 0, and u = 1. Observe that in particular, ρ(x) is in the form (3.2) described in lemma 3.4, and further, that this form (3.2) is independent of C. More generally, ρ(x) takes the same form (3.2) when the sections obtained from those in (3.3) by replacing the block C with a union of blocks all vanish. If we instead take B to be the empty set, then ρ(x) is in the form (3.1) described in lemma 3.1. This leads to the following lemma:

Lemma 3.6 —

Let . For 1 ≤ l ≤ 2g, let D_l be a (possibly empty) union of blocks in . Then the following collection of sections has no common zeros:

3.4

Proof. —

As discussed above, if these sections all vanish, then ρ(x) has the form (3.2) for every x∈{a₁, …, a_g, b₁, …, b_g}. Thus cannot be a generic torus element by lemma 3.4. But was chosen to be a generic torus element, so these sections have no common zeros. ▪

Remark 3.7 —

In lemma 3.4, we proved that if each ρ(x) is of the form (3.2), then cannot be a generic torus element. Observe that if we have 2g block diagonal matrices A₁, …, A_g, B₁, …, B_g, where the first block in each is a k-by-k matrix of the form (3.2) , then their product of commutators also cannot be a generic torus element.

We will use this idea to find collections of sections with no common zeros by extending collections that worked for lower-rank cases. Note that in the example above, the matrices only need to be block diagonal up to reordering of the basis elements. Thus it will be useful to introduce the following notation:

Notation —

Let G = SO(2n + 1), let X be a non-empty finite subset of [n], let , and let D be a 2g-tuple of unions of blocks in . Let

Define the collection of sections

When the sections in Q_n(X, B, D) all vanish, the |X|-by-|X| submatrix of each ρ(x) induced by considering only the rows and columns 2i − 1 and 2i for elements i in X must take the form (3.2). Lemma 3.6 says that the sections in Q_n([n], B, D) have no common zeros.

Definition 3.8 —

Consider sets , for , defined recursively as follows:

• — if and 0 ≤ r ≤ 2g, then .

• — if X⊔Y = [n] is a partition, ψ:[|X|] → X is a bijection, and , then .

This induced map ψ_* sends a section s ± _ij(x) of the line bundle L ± _ij over S_|X|,g(t) to the section s ± _ψ(i),ψ(j)(x) of the line bundle L ± _ψ(i),ψ(j) over S_n,g(t).

Proposition 3.9 —

If is a collection of sections in the set just described, then the sections in P have no common zeros.

Proof. —

An element takes the form

for some 0 ≤ k ≤ n. Observe that this collection contains the collection of sections

When these sections all vanish, each ρ(x) is (up to reordering) of the form

where each A_x is of the form (3.2). But the form of A_x forces z_2l−1 = z_2l = 0 for each l∈X, so each ρ(x) is in fact block diagonal (up to reordering), of the form discussed in remark 3.7. Hence these sections have no common zeros. ▪

Corollary 3.10 —

Suppose

where k⁺_ij(x) and for 1 ≤ i, j ≤ n and x∈{a₁, …, a_g, b₁, …, b_g}, and k⁻_ij(x) = 0 whenever i = j. Then the cohomology class

vanishes in .

4. The combinatorics

So far we have found a class of ‘good’ products of the c₁(L ± _ij) which vanish. The rest of this paper is devoted to proving that any product of at least of the c₁(L ± _ij) is equivalent in to a combination of these ‘good’ products, and hence also vanishes.

The proof is combinatorial, and makes extensive use of the relations

4.1

4.2

4.3

For the terms appearing in these relations to be defined, we must have fixed the rank n, but the same relations hold no matter which n is chosen. In order to use inductive arguments, we wish to be able to make statements that do not rely on having fixed the rank. We thus move away from considering the Chern classes themselves and instead study the combinatorial properties of another ring R whose elements satisfy the same relations (4.1) as our Chern classes.

Definition 4.1 —

Let X be a finite subset of . Define the associated auxiliary sets

• — Y⁺(X): = {y⁺_ij|i, j∈X}, 

• — Y⁻(X): = {y⁻_ij|i, j∈X;i≠j}

• — Y (X): = Y⁺(X)⊔Y⁻(X),

as well as subsets Y_z(X): = {y ± _ij∈Y (X)|i = zor j = z} for each x∈X. If is a block as defined in definition 3.3, then we also introduce

• — Y⁺_B(X): = {y^−ϵ_B(i,j)_ij|i, j∈X}

• — Y⁻_B(X): = {y^ϵ_B(i,j)_ij|i, j∈X;i≠j}.

We would like to form a quotient of by relations corresponding to those satisfied by the c₁(L ± _ij) listed above (4.1).

Definition 4.2 —

Let be the ideal generated by the elements

• — y⁻_ij + y⁻_ji

• — y⁺_ij − y⁺_ji

• — y⁻_ij + y⁻_jk + y⁻_ki

• — y⁺_ij − y⁺_jk + y⁻_ki

for all triples of elements i, j, k∈X. Let be the quotient of by this ideal; if we will denote by [p] its image in R. Note that this quotient preserves the grading by degree of .

In the previous section, we found sets of collections of sections with no common zeros, and concluded that the corresponding products of Chern classes vanished in H*(S_n,g(t)). We wish now to work in the ring R and not in H*(S_n,g(t)), so we introduce the map

Note that α is a map of sets, not a ring homomorphism; we will use the same notation for the map that takes a set of sections to the product of their images in . Observe that α forgets both n and x. The main goal of the remainder of this paper is to prove that for any monomial of degree at least , there exist some and monomials such that in R. Recall that the sets were defined recursively. We make this more explicit in the statement of the proposition:

Proposition 4.3 —

Let with |X| = n. Let be a homogeneous polynomial of degree at least . Then for each and , we can find

• — a finite set whose elements are 2g-tuples of unions of blocks in

• — homogeneous polynomials , as well as polynomials χ_D for each

• — elements and , and

• — bijections f_V:[|V |] → V and f_V^c:[|V^c|] → V^c such that

  

There are several points during the course of the proof of this proposition when we apply the pigeonhole principle, generally in order to show that the restriction of a polynomial to some subring of has high enough degree to apply an inductive hypothesis. In order to reduce clutter, we collect the relevant calculations into the following five lemmas.

Lemma 4.4 —

Let X be a finite subset of with |X| = m > 3, and let be a monomial of degree at least 2gm(m − 1) − m + 1. Then there exists some z∈X such that if we factorize p as p = q_zr_z, where and are monomials, then q_z has degree at least 2g(m − 1)(m − 2) − m + 2.

Proof. —

Write

(where d⁻_ii = 0 for all i). Given z∈X,

Note that each factor y⁺_ij of p appears in q_z precisely when i, j≠z, and thus appears in at least m − 2 of the q_z as z ranges over X (exactly m − 2 except for the y⁺_ii which appear in m − 1). Hence has degree ≥(m − 2)(2gm(m − 1) − m + 1). Suppose by the way of contradiction that each q_z has degree at most 2g(m − 1)(m − 2) − m + 1. Then has degree at most

This is a contradiction, so the desired z must exist. ▪

Lemma 4.5 —

Let X = {e₁, …, e_h, f₁, …, f_w, z} be a subset of with |X| = m. Let be a monomial of degree at least 2gm(m − 1) − m + 1 − 4gwh that factorizes as p = p_hp_w, where and are monomials. Then either , or .

Proof. —

Suppose . Then

 ▪

Lemma 4.6 —

Let X = H⊔W be a finite subset of with |H| = h > 0, |W| = w > 0, and |X| = w + h = n. Let be a monomial of degree at least that factorizes as p = p_wp_h, where and . Then either or .

Proof. —

Suppose . Then

 ▪

Lemma 4.7 —

Let p = qr be a monomial of degree at least . Then either or .

Proof. —

Suppose . Then

 ▪

Lemma 4.8 —

Let with w + h = m − 1, and let p = qr be a monomial of degree at least 2gm(m − 1) − m + 1 − 4gwh − h(h + 1)g. Then either or .

Proof. —

Suppose . Then

 ▪

We will also need the following results about interactions between different blocks.

Lemma 4.9 —

Suppose X = {e₁, …, e_h, f₁, …, f_w, z}. Let be the block B = {e₁, …, e_h} × {f₁, …, f_w}, and let C be a block in . Then the union contains a block .

Proof. —

Suppose without loss of generality C = {f₁, …, f_d} × {f_d+1, …, f_w, z}, for some 1 ≤ d ≤ w. Then . ▪

Lemma 4.10 —

Again, suppose X = {e₁, …, e_h, f₁, …, f_w, z}, and let . Let and . Then there exists a block such that either , or .

Proof. —

Suppose without loss of generality C = {e₁, …, e_k} × {e_k+1, …, e_h, z}, for some 1 ≤ k ≤ h. Either E or has the form {f₁, …, f_l} × {f_l+1, …, f_w, z} (up to relabelling), so take A_CE = {e₁, …, e_k, f₁, …, f_l} × {e_k+1, …, e_h, f_l+1, …, f_w, z}. ▪

Lemma 4.11 (The symmetric difference of two blocks is a block) —

Let . Then there exists a block such that .

Proof. —

Suppose B = {e₁, …, e_h} × {f₁, …, f_w}, and

Then take D = {e₁, …, e_s, f_t+1, …, f_w} × {e_s+1, …, e_h, f₁, …, f_t}. ▪

Remark 4.12 —

Let and z∈X. Then .

We are now ready to study the quotient ring R. We begin by observing that the polynomial ring obtained by adjoining the elements in the subset Y⁻(X)⊂Y (X) to is isomorphic to the polynomial ring introduced in [14] to study the case G = SU(n); we can therefore use the following result from our earlier work:

Lemma 4.13 —

Let X be a finite set. Let be a monomial of degree at least , for some . Then for each we can find a monomial such that

Proof. —

This follows from replacing 2g by a in the proof of Proposition 3.6 in [14]. ▪

Remark 4.14 —

Let X be a finite set and let B = V × V^c be a block in . Consider the isomorphism given by

or more concisely, y⁺_ij↦±ϵ_V(i)y^−ϵ_B(i,j)_ij. This map restricts to an isomorphism ; it is straightforward to check that f_B descends to an isomorphism on the quotient R.

Corollary 4.15 —

Let X be a finite set, let be a block, let , and let be a monomial of degree at least . Then for each we can find a monomial such that

Proof. —

Consider the isomorphism defined above. Applying lemma 4.13 to f⁻¹_B(p), we can write

Thus

where λ_C = |{(i, j)∈C|i∈V }|. ▪

Lemma 4.16 —

Suppose X = {e₁, …, e_h, f₁, …, f_w, z}, where both w and h are greater than zero. Let D = {e₁, …, e_d} × {e_d+1, …, e_h, z}, where 0 ≤ d ≤ h, be either a block in {e₁, …, e_h, z} × {e₁, …, e_h, z} or the empty set. Then a polynomial is equivalent in R to a sum of terms of the form α_hα_w, where and .

Proof. —

First use the relations [y⁻_eifj] = [y⁻_eiz + y⁻_zfj] and [y⁺_eifj] = [y⁺_fjz − y⁻_zei] to rewrite [p] using only elements of and . We must show that the images in R of elements of have representatives that are sums of elements in and in .

Suppose i, j∈{e₁, …, e_h}, and observe that [y⁺_ij] = [y⁻_jz + y⁻_zk + y⁺_zk − y⁻_zi], where we may choose k∈{f₁, …, f_w}. Observe further that [y⁺_iz] = [y⁻_iz + y⁻_zk + y⁺_zk], where again we may choose k∈{f₁, …, f_w}. This proves the lemma in the case where D is empty. If D is non-empty, let be a block with D⊆B, an deconsider the images of the above equations under the isomorphism f_B defined in remark 4.14:

Since D⊆B, we know that ϵ_B(i, j) = ϵ_D(i, j) whenever i, j∈{e₁, …, e_h, z}. Thus, these relations allow us to rewrite [y^−ϵ_D(i,j)_ij] using only elements of and . ▪

Lemma 4.17 —

Let X be a finite subset of with |X|≥2, and let . Let z∈V^c. Let be a homogeneous polynomial of degree at least 2gm(m − 1) − m + 1 − (m − 1)(m − 2)g. Then [p] has a representative in ; furthermore, this representative can be chosen to be a linear combination of monomials, each of which, when factorized as qr with and , either satisfies , or for some monomial θ.

Proof. —

We first claim that for any i∈X\{z}, the element [y^−ϵ_V(i)_iz] together with the images in R of the elements of Y⁻_B(X) form a set of generators for R. To see this, fix i∈X\{z}. Since z∈V^c, we know that y^ϵ_V(i)_iz∈Y⁻_B(X), and so both y⁺_iz and y⁻_iz are in Y⁻_B(X)∪{y^−ϵ_V(i)_iz}. Observe that [y⁺_ij + y⁻_ij] = [y⁺_iz + y⁻_iz], and for any j∈X\{z, i}, either y ± _ij∈Y⁻_B(X) or y⁻_ij∈Y⁻_B(X). Thus both [y⁺_ij] and [y⁻_ij] are elements of 〈[Y⁻_B(X)], [y^−ϵ_V(i)_iz]〉⊂R. Similarly, given any l≠j∈X, either y⁺_lj∈Y⁻_B(X) or y⁻_lj∈Y⁻_B(X). But [y⁺_lj + y⁻_lj] = [y⁺_ij + y⁻_ij], so [y⁺_lj] and [y⁻_lj] are both in 〈[Y⁻_B(X)], [y^−ϵ_V(i)_iz]〉. Finally, observe that [y⁺_ll] = [y⁺_lj + y⁻_lj], so each [y⁺_ll] is also in 〈[Y⁻_B(X)], [y^−ϵ_V(i)_iz]〉.

Without loss of generality, suppose X = [m] and z = m. Pick a representative for [p] in , and factorize each term as [q₁(y^−ϵ_V(1)_1z)^d₁], with . If is at least m(m − 1)g − m + 2, this term has the desired form; else

In this case, pick a representative for [q₁(y^−ϵ_V(1)_1z)^d₁] that is a sum of terms of the form (y^−ϵ_V(1)_1z)^2gq₂(y^−ϵ_V(2)_2z)^d₂, where , and consider each term separately. Any term where now has the desired form; else d₂≥2g(m − 1) − 2g = 2g(m − 2)≥2g. Repeat this process: for each term, either , or d_i≥2g(m − i). Thus any term with has d_i≥2g ∀1 ≤ i ≤ m − 1, and so is a multiple of . ▪

Lemma 4.18 —

Let X be a finite subset of with |X|≥2, and suppose is a monomial of degree at least 2g|X|(|X| − 1) − |X| + 1. Then we can find a homogeneous polynomial , together with polynomials for each block , such that

Illustration. Our goal is to use the relations between the c₁(L ± _ij) to express any monomial [p]∈R as a sum of images of ‘good’ collections of sections defined at the end of §3. While we are working in R, we will illustrate monomials appearing in our argument with pictures of the form ρ(x) takes in the vanishing loci of the corresponding sections s⁺_ij(x). Though ρ(x)∈SO(2n + 1), our illustrations will be n × n grids, in which we represent 2 × 2 blocks of the form by , the 2 × 2 blocks of the form by , and the 2 × 2 blocks of zeros (in the vanishing loci of (s⁺_ij(x), s⁻_ij(x)) by . We have chosen to illustrate our argument using blocks B with the property that i < j ∀(i, j)∈B, since these produce the simplest visualizations; note that this is not an assumption we make in the proof. Lemma 4.18 says we can express any monomial of degree ≥2gn(n − 1) − n + 1 as a sum of terms whose corresponding sections have vanishing loci taking one of the forms shown in figure 1.

Figure 1.

(a) A block of , otherwise , above the diagonal. (Or all, if B = ø.) and (b) a block of zeros. Note that this alone does not make ρ(x)∈SO(2n + 1) block diagonal, since this is only a 2h-by-2(n − h) block of zeros. (Online version in colour.)

Proof. —

By induction on |X|.

For |X| = 2, without loss of generality take X = {1, 2}. Then

Since [y⁺₁₂] = [y⁺₂₁] and [y⁻₁₂] = [ − y⁻₂₁] in R, any monomial of degree at least 4g − 1 is equivalent in R to λ(y⁺₁₂)^a(y⁻₁₂)^b, where and a + b≥4g − 1. So either a≥2g or b≥2g. If a≥2g, take , and ψ_B = χ_B = 0 for all . If b≥2g, take χ_(1,2) = (y⁺₁₂)^a(y⁻₁₂)^b−2g, and for all . (✓)

Now suppose |X| = m≥3. By lemma 4.4, there is some z∈X for which when we factorize p as q_zr_z, with monomials and , the degree of q_z is at least 2g(m − 1)(m − 2) − m + 2. By the inductive hypothesis, there exist homogeneous polynomials , for all , such that

We will consider these terms separately, as it suffices to show each term has the desired form.

Fix a block , say C = {e₁, …, e_h} × {f₁, …, f_w} (so h + w = m − 1). Consider the term

4.4

in [p] (figure 2). Observe that . Since [y⁺_ij] = [y⁺_iz + y⁻_zj] and [y⁻_ij] = [y⁻_iz + y⁻_zj], each term of can be expressed as a sum of terms of the form [p_hp_w], where and are monomials. By lemma 4.5, either , or (figure 3). Consider one such monomial in the sum, and without loss of generality assume the former. By the inductive hypothesis, we can find homogeneous polynomials ψ_D, χ_D for each , together with , with , such that

(see figure 4).

Fix a block D = {e₁, …, e_d} × {e_d+1, …, e_h, z} (some 1 ≤ d ≤ h), and consider the term

of [p]. By lemma 4.9, there is a block with , so our term is a multiple of and hence has the desired form (figure 4a).

Now fix D = {e₁, …, e_d} × {e_d+1, …, e_h, z} where 0 ≤ d ≤ h (that is, allow D to be empty), and consider the term

of [p] (figure 4b). By lemma 4.16, we may express [p_wχ_D] as a sum of terms of the form [α_wα_h], where and . By lemma 4.8, either or (figure 5). If , then by corollary 4.15, for each block we can find a monomial β_E such that

Then for fixed E, the term

of [p] contains the factor

(up to replacing y⁺_ji with y⁺_ij whenever (i, j)∈E, , and i < j). By lemma 4.9, contains a block , so this term has the desired form (figure 6).

Suppose instead that . Then by the inductive hypothesis,

for some homogeneous polynomials θ_F, .

Fix a block and consider each term separately. By lemma 4.9, contains a block , and so the term

of [p] contains the factor and hence is of the desired form (figure 7a).

By lemma 4.10, there exists a block with either , or , and thus the term

of [p] contains the factor , and hence is of the desired form (figure 7b). Hence each term has a representative in of the desired form.

It remains to show that

(figure 8) has a representative in of the desired form; again it suffices to show this for each term separately.

Fix and consider

4.5

Observe that . By lemma 4.17, has a representative in that is a sum of terms of the form , where either , or d_i≥2g ∀i∈X\{z} (figure 9).

In the latter case, the corresponding terms of [p] have the factor

where is the extension of the block C obtained by adding z to the second factor. Hence these terms have the desired form (figure 9a).

If a term has , then by corollary 4.15 there exist homogeneous polynomials γ_G, for each , such that

(figure 9b). Then each such term

in (4.5) contains the factor (up to replacing y⁺_ji by y⁺_ij whenever , (i, j)∈G and i > j). By remark 4.12, , and thus we are reduced to a monomial of the form (4.4) (figure 10), which we dealt with above.

Thus we have found a representative for [p] in that has the desired form. ▪

Figure 2.

The inductive hypothesis gave us a block of zeros in , using a monomial of degree 4gwh. (Online version in colour.)

Figure 3.

We can apply the inductive hypothesis to one of these triangles T_h, T_w. (Online version in colour.)

Figure 4.

Applying the inductive hypothesis to T_h, we find either a block of zeros, or a block of with the rest . (a) A block D of zeros in . Observe that the union C∪D contains a full block B_CD∈B[X] and (b) A block of , the rest in T_h. (Online version in colour.)

Figure 5.

We can apply either corollary 4.15 to (left), or the inductive hypothesis to T_w. (Online version in colour.)

Figure 6.

Applying corollary 4.15 to gives a block E of or as shown in , which combines with the and in T_h to get zeros everywhere in E. The union C∪E contains a full block . (Online version in colour.)

Figure 7.

Applying the inductive hypothesis in T_w results in one of these two situations. (a) A block F of zeros in T_w; the union C∪F contains a full block and (b) A block of , the rest , in T_w. There is now a full block with (or zeros) in every position inside and (or zero) in every position outside. (Online version in colour.)

Figure 8.

The inductive hypothesis gave us a block of , and everywhere else above the diagonal and away from z. (Online version in colour.)

Figure 9.

Lemma 4.17 puts us in one of these situations. (a) We can either get the appropriate pattern of and along z to extend C to a full block of with everything else and (b) or we can apply corollary 4.15 to find a full block G in the subring . (Online version in colour.)

Figure 10.

Combining the block G (figure 9b) with everywhere in C and elsewhere away from z (figure 8) gives a block of zeros in , reducing us to the earlier case (figure 2). (Online version in colour.)

Lemma 4.19 —

Let be a block (or the empty set), let be a monomial, and fix 1 ≤ i ≤ n. Then [p] is equivalent in R to a sum of terms of the form [qr], where and .

Proof. —

We must check that the images in R of the elements of Y⁻_B(X) together with [y⁺_ii] generate R. We showed in the proof of lemma 4.17 that the images of elements in Y⁻_B(X) together with [y^−ϵ_B(i,z)_iz] (for some z≠i generate R, so it suffices to show that [y^−ϵ_B(i,z)_iz]∈ 〈[Y⁻_B(X)], [y⁺_ii]〉. But this is clear, because y^ϵ_B(i,z)_iz∈Y⁻_B(X), and [y^ϵ_B(i,z)_iz + y^−ϵ_B(i,z)_iz] = [y⁺_ii]. ▪

Notation —

If is a monomial, let d ± _ij (for i, j∈X) be the integers such that (where ), and define d_ij(p): = d⁺_ij + d⁻_ij + d⁺_ji + d⁻_ji.

Lemma 4.20 —

Let X = [n] and let be a monomial of degree at least . Let be a fixed block (or the empty set). Then we can find a (finite) set whose elements are 2g-tuples of (possibly empty) unions of blocks in , and homogeneous polynomials ϕ_D for each and θ_B for each , such that

Proof. —

First factorize p as p − p + , where and . If , we are done by corollary 4.15. Otherwise, let a < 2g be the largest integer such that , and write

(which is possible by corollary 4.15). Each monomial in the resulting polynomial has the form αβγ, where , for some 0 ≤ s ≤ 2g and unions D_m of blocks in , and . Consider monomials of this form.

Case 1: If , then [αβγ] has a representative of the desired form by corollary 4.15.

Case 2: If s = 2g and there is a block such that B⊆D_m for every 1 ≤ m ≤ 2g, then β contains the factor , so αβγ is of the desired form.

Case 3: Observe further that for each i < j, we know and . Thus, if d_ij(βγ)≥2g ∀i≠j, then swapping y⁺_ij with y⁺_ji or y⁻_ij with −y⁻_ji as necessary and taking for s + 1 ≤ m ≤ 2g, we may write

thus finding a representative for [αβγ] of the desired form.

The following procedure takes monomials of degree at least , and finds representatives for them as sums of monomials, each of which falls into one of the three cases above.

Start with a monomial αβγ of degree , where , for some 0 ≤ s ≤ 2g and unions D_m of blocks in , and .

• (1) If , we are in Case 1. Stop. Otherwise, go to Step 2.

• (2) If , go to Step 3. Otherwise, go to Step 4.

• (3) Apply corollary 4.15 to write . Consider each term in the resulting polynomial separately.

  • (a) If s < 2g: Replace α with ϕ_B. Set D_s+1: = B. Replace β with . Go to Step 2.

  • (b) If s = 2g:

    • (i) If B⊆D_m for all 1 ≤ m ≤ 2g, then we are in Case 2. Stop.

    • (ii) Otherwise, pick 1 ≤ m ≤ 2g with B⊆D_m. Replace D_m with B∪D_m in β. Go to Step 2.

• • (i) If d_ij(βγ)≥2g for all i < j, we are in Case 3. Stop.

  • (ii) Otherwise, pick i < j with d_ij(βγ) < 2g. We know , whilst , so . Thus, there must be some k, l∈X with and y^−ϵ_C(k,l)_kl must be a factor of γ. Replace one instance of y^−ϵ_C(k,l)_kl in γ with y^ϵ_C(k,i)_ki + y^−ϵ_C(i,j)_ij + y^ϵ_C(l,j)_l,j (the two are equivalent in R), and treat each term in the resulting polynomial separately. Go to Step 1.

Observe that Step 3 increases the number b of blocks (counted with multiplicity) in ∪^s_r=1D_r, and Step 4b either increases or decreases , for each term in the polynomials these steps create. None of the steps decrease b or , or increase d. If , or d = 0, or b gets large enough to force every D_m to contain some common block B ( will do), then the algorithm terminates. So even when a step results in a polynomial with more than one term, there are still only finitely many terms, each of which are also closer to a terminating condition than the previous monomial. So this procedure terminates in finite time, giving us a representative of the desired form for any monomial of sufficient degree. ▪

Proposition 4.21 —

Let with |X| = n. Let be a homogeneous polynomial of degree at least . Then for each and each , we can find

• — homogeneous polynomials

• — elements and

• — bijections f_V:[|V |] → V and f_V^c:[|V^c|] → V^c

• — a finite set whose elements are 2g-tuples of unions of blocks in , and

• — homogeneous polynomials for each

such that

4.6

Remark 4.22 —

By the definition of the sets (see definition 3.8), if this relation holds, then there exist , monomials , and a bijection f_X:[n] → X such that .

Proof. —

By induction on n.

For n = 1, if is a monomial of degree ≥2g, it is in the desired form. (✓)

Suppose n≥2. By lemma 4.18, we can write

As usual, we treat each term separately. Fix a block and consider a monomial . Suppose B = {e₁, …, e_h} × {f₁, …, f_w}; then . By using the relations , we can express [ψ_B] as a sum of terms of the form [ψ_hψ_w], where and are monomials of total degree . By lemma 4.6, in each of these terms either or , and we can apply the inductive hypothesis to the appropriate monomial. Applying the inductive hypothesis to ψ_h, we find that there exist , monomials and a bijection f_V:[h] → V such that ; applied to ψ_w the inductive hypothesis gives , where , , and f_V^c:[w] → V^c is a bijection. Applying the inductive hypothesis to the appropriate factor in each term of ψ_B and summing the resulting representatives, we obtain

for homogeneous polynomials and elements and . This is in the desired form (4.6) (taking ψ_C = 0 ).

Now fix and consider the term . Observe that . By repeated application of lemmas 4.7 and 4.19, we can write , where is a homogeneous polynomial with , and . By corollary 4.15, , for homogeneous polynomials . Thus

(for some ), which is of the form considered above.

Finally, consider the monomial . Note that . Applying lemma 4.20,

As usual, consider each monomial individually. Fix F, and consider

This contains the factor , and so is of the form discussed above. The remaining term

is already in the form (4.6). ▪

(a). Proof of the main theorem

Theorem 4.23 —

For each ϕ∈Φ(G), let k_ϕ be a non-negative integer. Then the cohomology class vanishes whenever .

Proof. —

Let X = [n] and consider the rings and . Let be the subring generated by the c₁(L_ϕ) for ϕ∈Φ(G). Since the relations (4.1) hold in J, the map

defines a ring homomorphism. Consider the element . It has a representative in R. Suppose . Then by proposition 4.21, [p] is equivalent in R to an expression of the form (4.6). So π(p) vanishes in by corollary 3.10. ▪

Data accessibility

This article has no additional data.

Author's contributions

E.A.G. and J.W. proposed the project, carried out research and approved the final version. E.A.G. wrote the draft of the manuscript.

Competing interests

We declare we have no competing interests.

Funding

The first author was partially supported by NSERC PostDoctoral Fellowship 488168. Both authors were partially supported by NSF grant no. DMS 12-11819.