THE 1-LINE OF THE K-THEORY BOUSFIELD-KAN SPECTRAL SEQUENCE FOR Spin(2n + 1)

MARTIN BENDERSKY AND DONALD M. DAVIS

1. Statement of results

The p-primary v_{1}-periodic homotopy groups, v_{1}^{−1}π_{∗}(X; p), of a topological space X, as defined in [13], are a localization of the portion of the actual homotopy groups of X detected by K-theory. In [10, 4, 6], v_{1}^{−1}π_{∗}(X; p) was calculated for classical groups X and primes p in all cases except (SO(n), 2). In this paper we make a first step toward the calculation of v_{1}^{−1}π_{∗}(SO(n); 2), which is of course isomorphic to v_{1}^{−1}π_{∗}(Spin(n); 2).

In [7], a Bousfield-Kan-type spectral sequence E_{r}^{s,t}(X) based on periodic K-theory was introduced. It converges to the homotopy groups of the K-completion X_{K}^{∧} of the space X. In this paper, we will deal exclusively with the localization of this spectral sequence at the prime 2. We expect to prove in a subsequent paper that if X = Spin(n) then π_{i}(X_{K}^{∧}) ≈ v_{1}^{−1}π_{i}(X; 2) for sufficiently large values of i. Thus, since v_{1}-periodic homotopy groups are periodic, a computation of this spectral sequence for X = Spin(n) and t − s large would yield a complete computation of v_{1}^{−1}π_{∗}(Spin(n)).

If X is an H-space such that K_{∗}(X) is a free commutative algebra on odd-dimensional classes (e.g. X = Spin(n)), then ([7, 4.9]) E_{2}(X) is the homology of an unstable cobar complex determined by these classes and their K_{∗}K-coaction. The main result of this paper is an explicit computation of the 1-line E_{2}^{1,t}(Spin(2n + 1)) of this spectral sequence when p = 2 and X = Spin(2n + 1), .

The first step toward this computation is the following general result, which was inspired by the results of Bousfield ([9]) at the odd primes. This result will be proved

Date: September 20, 1999.

1991 Mathematics Subject Classification. 55T15,55Q52.

Key words and phrases. homotopy groups, Adams operations, Spinor groups.

2 BENDERSKY AND DAVIS

in Section 2. In it, (−)^{#} denotes the Pontryagin dual, and P K^{1}(X) the primitives in K^{1}(X).

Theorem 1.1. If X is a simply-connected finite H-space with K_{∗}X an exterior algebra on odd-dimensional classes of degree ≤ 2M + 1, then E_{2}^{1,2m}(X) = 0 for all m, and, if m > M, then

E_{2}^{1,2m+1}(X) ≈ (P K^{1}(X)/ im(ψ^{r} − r^{m} : r = 2, 3, −1))^{#}.

For simplicity of notation, we define

ṽ^{2m}(X) := P K^{1}(X)/ im(ψ^{r} − r^{m} : r = 2, 3, −1)

and the closely-related functor

v^{2m}(X) := P K^{1}(X)/ im(ψ^{2}, ψ^{3} − 3^{m}, ψ^{−1} − (−1)^{m}). The advantage of ṽ^{∗}(−) is that it is more closely related to E_{2}^{1,∗+1}(−), while the advantage of v^{∗}(−) is that it is periodic and has simpler formulas. They are related by the following result, which will be proved at the end of this section.

Proposition 1.2. If X is a simply-connected finite H-space with K_{∗}X an exterior algebra on classes of degrees 2d_{1} +1, …, 2d_{r} +1, then v^{2m}(X) ≈ ṽ^{2m}(X) if m > ∑ d_{i}.

The second purpose of this note is to compute v^{2m}(Spin(2n + 1)), which, by Theorem 1.1 and Proposition 1.2, is isomorphic to E_{2}^{1,2m+1}(Spin(2n + 1)) if m > n^{2}. Here we use the rational equivalence of Spin(2n + 1) with ∏_{i=1}^{n} S^{4i−1}, and that the sum of the odd integers up to 2n − 1 is n^{2}.

We will show in Proposition 3.1 that there is an injection of Adams-modules

P K^{1}(Sp(n)) → P K^{1}(Spin(2n + 1)),

and hence a morphism

v^{2m}(Sp(n)) → v^{2m}(Spin(2n + 1)),  (1.3)

which is dual to a morphism of E_{2}^{1,2m+1}(−). The following result was proved in [5]. Here and throughout, ν(−) denotes the exponent of 2 in a number.

Theorem 1.4. ([5]) If m is odd and m ≥ 2n, then E_{2}^{1,2m+1}(Sp(n)) ≈ Z/2^{eSp(m,n)}, where

eSp(m, n) := min(ν(S_{m,j}) : j > 2n),

1-LINE FOR SPIN(2N+1) 3

with S_{m,j} = ∑(−1)^{k}\binom{j}{k}k^{m}. If m is even, then E_{2}^{1,2m+1}(Sp(n)) = Z/2.

Actually, the result in [5] was proved for the BP -based unstable Novikov spectral sequence, but one might use the change of rings theorem in [7], or mimic the calculation in [5] to yield the result for the K-based spectral sequence.

Our second main result is as follows. In light of 1.1, it gives E_{2}^{1,2m+1}(Spin(2n + 1)) when m > n^{2}.

Theorem 1.5. If n = 3 or n ≥ 5, and m is odd, then v^{2m}(Spin(2n+1)) ≈ Z/2^{min(eSp(m,n),ν(R_{1}(m,n)),ν(R_{2}(m,n)))}⊕Z/2^{min(2+ν(m+1),n)}, where

R_{1}(m, n) = ∑_{odd k≥1} k^{m}(∑_{i=0}^{n−k} \binom{2n+1}{i} − 2 ∑_{t≥0} \binom{2n+2}{n−1−k−4t})

and R_{2}(m, n) =

((2^{2n+1}−3^{m+1}+1) ∑_{odd k≥1} k^{m} ∑_{t≥0} \binom{2n+2}{n−1−k−4t}−3·2^{2n} ∑_{odd k≥1} k^{m} ∑_{t≥0} \binom{2n+1}{n−1−k−3t}) / 2^{min(2+ν(m+1),n)}.

If m is even, then v^{2m}(Spin(2n + 1)) ≈ Z/2 ⊕ Z/2.

The sums over odd k could include even values as well, as long as m > n^{2}, since this is greater than the largest possible exponent of v^{∗}(Spin(2n + 1)). We will discuss the slight anomaly when n = 4 in Proposition 4.21. Since Spin(5) ≈ Sp(2), the case n = 2 has been covered in [6] (and is quite different).

The morphism (1.3) sends a generator to g_{1} + ag_{2}, for some integer a. We could certainly determine a, but shall not bother to do so here.

We illustrate with concrete calculations when n = 5 and 6. These results are obtained by computer calculation, although we shall show after (4.20) how to obtain some of them by hand.

Proposition 1.6. For n = 5 and 6, v^{2m}(Spin(2n+1)) ≈ Z^{e_{1}(2n+1)} ⊕Z^{e_{2}(2n+1)}, where e_{1}(2n + 1) and e_{2}(2n + 1) are as in the following table, which also presents eSp(m, n) for comparison.

4 BENDERSKY AND DAVIS

n  m  eSp(m, n)  e_{1}(2n + 1)  e_{2}(2n + 1)
5 3 mod 8  8  5  4
5 7 mod 8 min(11, ν(m − 7) + 6) min(11, ν(m − 103) + 2)  5
5 1 mod 4 min(14, ν(m − 73) + 6) min(12, ν(m − 73) + 4)  3
6 15 mod 16  11  6  6
6 7 mod 16  11  min(9, ν(m − 23) + 4)  5
6 3 mod 8 min(15, ν(m − 75) + 9) min(15, ν(m − 523) + 5)  4
6 1 mod 4 min(14, ν(m − 9) + 8) min(13, ν(m − 201) + 5)  3

Proof of Proposition 1.2. There is a rational equivalence X ≃ ∏ S^{2d_{i}+1}. Thus there are elements x_{i} ∈ P K^{1}(X) for which ψ^{k}(x_{i}) = k^{d_{i}}x_{i} for all k. Similarly to the procedure in [11], we can find a basis {w_{1}, …, w_{r}} of P K^{1}(X)_{(2)} such that each w_{i} is a rational linear combination of {x_{j} : j ≥ i}, and hence the matrix of each ψ^{k} is triangular with {k^{d_{1}}, …, k^{d_{r}}} along the diagonal. Thus the Adams module P K^{1}(X) can be built from P K^{1}(S^{2d_{1}+1}), …, P K^{1}(S^{2d_{r}+1}) by short exact sequences, and by the Snake Lemma this implies the same of v^{2m}(X) (built from v^{2m}(S^{2d_{i}+1})). Since 2^{d_{i}}v^{2m}(S^{2d_{i}+1}) = 0 by [7], we deduce 2^{∑ d_{i}}v^{2m}(X) = 0 (a very conservative estimate). If m ≥ ∑ d_{i}, then ·2^{m} is 0 in v^{2m}(X). Since the only difference between v^{2m}(X) and ṽ^{2m}(X) is multiples of 2^{m}, we conclude that they are equal.

2. Proof of Theorem 1.1

In this section, we prove Theorem 1.1. We begin by observing that the exact sequence in E_{2} induced by the coefficient sequence

0 → Z → Q → Q/Z → 0

induces isomorphisms E_{2}^{s,t}(X; Q/Z) → E_{2}^{s+1,t}(X) except when s = 0, t = 2d_{i} + 1. Hence there is an isomorphism when s = 0, t > 2d_{r} + 1. Everything is localized at 2. We have E_{2}^{0,2m+1}(X; Q/Z) ≈ ker(d̄: K_{2m+1}(X) ⊗ Q/Z → (K_{∗}K ⊗ K_{∗}X ⊗ Q)/U(X)),

(2.1)

where d̄ is induced from d : (K_{∗}(X) ⊗ Q, K_{∗}(X)) → (K_{∗}K ⊗ K_{∗}X ⊗ Q, U(X)),

1-LINE FOR SPIN(2N+1) 5

the boundary in the unstable cobar complex described in [7, §4]. The following result, whose proof is deferred until the end of this section, is crucial. Here QK_{∗}(−) denotes the indecomposables.

Proposition 2.2. Suppose QK_{∗}(X) has basis {y_{1}, …, y_{r}} as a K_{∗}-module with y_{j} ∈ K_{2d_{j}+1}(X), and x ∈ K_{∗}(X) ⊗ Q has d(x) = ∑ s_{j} ⊗ y_{j}, with s_{j} ∈ K_{∗}K ⊗ Q. Then x ∈ ker(d̄) (of (2.1)) if and only if for all positive integers k and all subscripts j

k^{d_{j}}⟨ψ^{k}, s_{j}⟩ ∈ K_{∗}_{(2)}.  (2.3)

The condition (2.3) says that each k^{d_{j}}⟨ψ^{k}, s_{j}⟩ is a 2-local integer times the generator of K_{|s_{j}|}.

Now we prove Theorem 1.1. Suppose {y_{1}, …, y_{r}} is as in Proposition 2.2 and

ψ : K_{∗}X → K_{∗}K ⊗ K_{∗}X

satisfies

ψ(y_{j}) = ∑_{i} s_{i,j} ⊗ y_{i}.  (2.4)

Let u ∈ K_{2} denote the Bott class, so that K_{∗} = Z_{(2)}[u], and let v = η_{R}(u). Then

d(u^{e}y_{j}) = (v^{e} − u^{e}) ⊗ y_{j} − ∑_{i≠j} u^{e}s_{i,j} ⊗ y_{i}.

Since, by [8, p.676], ⟨ψ^{k}, v⟩ = ku and ⟨ψ^{k}, u⟩ = u, the expression whose coefficients correspond to ⟨ψ^{k}, s_{j}⟩ in (2.3) equals

k^{e}u^{e}y_{j} − ∑_{all i}⟨ψ^{k}, s_{i,j}⟩u^{e}y_{i}.

A basis for QK_{2m+1}(X) is given by {u^{m−d_{1}}y_{1}, …, u^{m−d_{r}}y_{r}}, and, by Proposition 2.2, E_{2}^{0,2m+1}(X; Q/Z) is the intersection, over all integers k > 1, of the kernel of the morphism

δ_{k} : K_{2m+1}(X) ⊗ Q/Z → K_{2m+1}(X) ⊗ Q/Z

defined by

δ_{k}(u^{m−d_{j}}y_{j}) = −k^{m−d_{j}}u^{m−d_{j}}k^{d_{j}}y_{j} + u^{m−d_{j}} ∑_{i}⟨ψ^{k}, s_{i,j}⟩k^{d_{i}}y_{i}.

6 BENDERSKY AND DAVIS

Here we have negated the differential d for later convenience. Define α_{i,j,k} ∈ Z by ⟨ψ^{k}, s_{i,j}⟩k^{d_{i}} = α_{i,j,k}u^{d_{j}−d_{i}}, and define an r ×r-matrix A_{k} = (α_{i,j,k}). Then δ_{k} is a linear transformation of free Q/Z-modules with matrix A_{k} − k^{m}I.

Now let w_{i} ∈ K^{2d_{i}+1}(X) be dual to y_{i}, and let z_{i} = u^{d_{i}}w_{i} ∈ K^{1}(X). By [8, 11.19] applied to (2.4), we have

ψ^{k}(w_{i}) = ∑_{j}⟨ψ^{k}, s_{i,j}⟩w_{j}.

Thus

ψ^{k}(z_{i}) = k^{d_{i}}u^{d_{i}} ∑_{j}⟨ψ^{k}, s_{i,j}⟩u^{−d_{j}}z_{j} = ∑_{j} α_{i,j,k}z_{j}.

Thus A_{k} − k^{m}I is the matrix whose ith row gives (ψ^{k} − k^{m})(z_{i}) expressed in terms of the basis {z_{1}, …, z_{r}}. Then ṽ^{2m}(X) is the abelian group presented by a matrix obtained by stacking the matrices A_{k} − k^{m}I for all k. Thus we see that the same matrix A, defined to be all matrices A_{k} − k^{m}I stacked, yields E_{2}^{0,2m+1}(X; Q/Z) if we take the kernel of the linear transformation of free Q/Z-modules whose matrix is A, and it yields ṽ^{2m}(X) if we think of it as the presentation of an abelian group. Theorem 1.1 follows now from the following proposition.

Proposition 2.5. If A is an integer matrix, let K(A) denote the kernel of the linear transformation of free Q/Z-modules whose matrix is A, and let G(A) denote the abelian group presented by A. Then there is an isomorphism of abelian groups K(A) ≈ (G(A))^{#}. The same is true if we are localized at a prime p.

Proof. The matrix A can be brought to diagonal form (with rows and columns of 0’s adjoined) by integer row and column operations. These operations consist of interchanging, multiplying by −1, and adding a multiple of one to another. For the first interpretation of the matrix, they correspond to change-of-basis in the domain and range, and hence do not affect the kernel (up to isomorphism), while in the second interpretation they correspond to invertible change of generating set and to simplification of relations, and hence do not affect the isomorphism class of the group presented. Let D be a diagonal matrix with entries d_{i} > 1, m 1’s, and n 0’s on the diagonal, and possibly many additional rows of 0’s. Then K(D) ≈ ⨁ Z/d_{i} ⊕ (Q/Z)^{n}, while G(D) ≈ ⨁ Z/d_{i} ⊕(Z)^{n}. Since (Z)^{#} ≈ Q/Z and (Z/d)^{#} ≈ Z/d, the proposition follows.

1-LINE FOR SPIN(2N+1) 7

Proof of Proposition 2.2. We must show that if d(x) = ∑ s_{j} ⊗ y_{j}, then d̄(x) ∈ U(X) if and only if k^{d_{j}}⟨ψ^{k}, s_{j}⟩ is 2-integral for all j.

The following analysis of the unstable condition (i.e. the condition for being in U(X)) is based on [7, §7], which derived from [3]. Let K_{2n} denote the 2nth space in the Ω-spectrum for K. There is a homotopy equivalence K_{2n} → BU, and the structure maps Σ^{2}K_{2n} → K_{2n+2} are the Bott maps B. These combine to a tower

K_{2e}(BU)⟶^{B_{∗}}K_{2e+2}(BU)⟶^{B_{∗}} ⋯ ⟶^{B_{∗}}K_{2e+2d_{j}}(BU)⟶^{B_{∗}} ⋯ → K_{2e}(K),

with K_{2e}(K) the direct limit of the tower. There is a similar tower after tensoring with Q. Suppose s_{j} ∈ K_{2e}(K) ⊗ Q. Then s_{j} ⊗ y_{j} ∈ U(X) if and only if s_{j} pulls back to an element of K_{2e+2d_{j}}(BU). This follows from the description of unstable comodules in the fourth paragraph after [7, 4.3].

For purposes of calculation, it is convenient to write the above tower in the equivalent form

K_{0}(BU)⟶^{B_{∗}}K_{0}(BU)⟶^{B_{∗}} ⋯ ⟶^{B_{∗}}K_{0}(BU)⟶^{B_{∗}} ⋯ → K_{0}(K).

In [3, 1.2-1.4], it is observed that K_{0}(BU) ⊂ Q[w] is a polynomial algebra generated by \binom{w}{n} = w(w − 1) ⋯ (w − n + 1)/n! ∈ Q[w] for n ≥ 1. Let A denote the free abelian group on the rational polynomials \binom{w}{n}. Then A consists of all rational polynomials f(w) such that f(n) is an integer for every integer n, so-called numerical polynomials. Then, still following [3], B_{∗}(f)−wf is decomposable for all f ∈ A, and B_{∗} annihilates decomposables, from which it is deduced that K_{0}(K) ≈ w^{−1}A, with an element f in the dth factor corresponding to w^{−d}f in the limit.

It is also noted in [3, p.390] that w = u^{−1}v ∈ K_{0}(K). This implies that ⟨ψ^{k}, w^{n}⟩ = k^{n}. Thus ⟨ψ^{k}, f(w)⟩ = f(k), and so A consists of those polynomials f(w) for which ⟨ψ^{k}, f(w)⟩ is integral for all k.

Now s_{j} ∈ K_{0}(K) ⊗ Q is represented by a finite Laurent series s_{j}(w) and pulls back to an element in the d_{j}th copy of K_{0}(BU) if and only if w^{d_{j}}s_{j}(w) is a numerical polynomial if and only if ⟨ψ^{k}, w^{d_{j}}s_{j}(w)⟩ is integral for all k. Since ⟨ψ^{k}, w^{d_{j}}s_{j}⟩ = k^{d_{j}}⟨ψ^{k}, s_{j}⟩, our desired conclusion follows.

8 BENDERSKY AND DAVIS

3. Proof of Theorem 1.5

We begin with a result culled from [16].

Proposition 3.1. There are morphisms of Adams modules

P K^{1}(SU(2n + 1))⟶^{j^{∗}} P K^{1}(Sp(n)) ⟶^{θ}P K^{1}(Spin(2n + 1))
⟨B_{1}, …, B_{2n}⟩ ⟶ ⟨B′_{1}, …, B′_{n}⟩ ⟶ ⟨B′′_{1}, …, B′′_{n−1}, D⟩,

where ⟨−, …, −⟩ denotes the free abelian group on the indicated classes, θ(B′_{i}) = B′′_{i} if i < n, θ(B′_{1} + ⋯ + B′_{n}) = 2^{n+1}D, and

j^{∗}(B_{i}) = \begin{cases}B′_{i} & i ≤ n \\ B′_{2n+1−i} & i > n. \end{cases}

Proof. We use Hodgkin’s result ([14]) that P K^{1}(G) ≈ ⟨β(ρ_{1}), …, β(ρ_{l})⟩, where ρ_{1}, …, ρ_{l} are the fundamental representations of G, and β(ρ) is the virtual vector bundle over ΣG associated to ρ − dim(ρ). Naylor ([16, p.151]) notes that the maps

Sp(n)⟶^{k}SU(2n)⟶^{i}SU(2n + 1)

induce morphisms of representation rings

R(SU(2n + 1))⟶^{i^{∗}}R(SU(2n))⟶^{k^{∗}}R(Sp(n))

satisfying

k^{∗}(μ′_{i}) = \begin{cases}η_{i} & i ≤ n \\ η_{2n−i} & i > n \end{cases}

and i^{∗}(μ_{i}) = μ′_{i} +μ′_{i−1}, where η_{i}, 1 ≤ i ≤ n, μ′_{i}, 1 ≤ i ≤ 2n−1, and μ_{i}, 1 ≤ i ≤ 2n, are the representations given by exterior power operations on a canonical representation. We have μ′_{0} = μ′_{2n} = 1. Letting B′_{i} = β(η_{i})+β(η_{i−1}) and B_{i} = β(μ_{i}) yields the desired result about j^{∗}, which is the composite k^{∗} ∘ i^{∗}.

Naylor also considers the composite

j′ : Spin(2n + 1) → SO(2n + 1) → SU(2n + 1).

We let λ_{i} denote the exterior powers of the canonical unitary representation of Spin(2n + 1), Δ the Spin representation, B′′_{i} = β(λ_{i}), and D = β(Δ). Naylor shows that

j′^{∗}(B_{i}) = \begin{cases}B′′_{i} & i ≤ n \\ B′′_{2n+1−i} & i > n \end{cases}

1-LINE FOR SPIN(2N+1) 9

and 2^{n+1}D = B′′_{1} + ⋯ + B′′_{n}. Since ker(j^{∗}) = ker(j′^{∗}), there is a homomorphism of abelian groups

P K^{1}(Sp(n))⟶^{θ}P K^{1}(Spin(2n + 1))

defined by θ(B′_{i}) = B′′_{i} for i ≤ n, and it will be an Adams-module homomorphism since j^{∗} and j′^{∗} are Adams-module homomorphisms with j^{∗} surjective.

Next we relate the above basis of P K^{1}(SU(n)) with one used by Bousfield ([9]).

Proposition 3.2. There is an isomorphism of Adams modules

P K^{1}(SU(n)) ≈ K̃^{0}(CP^{n−1}).  (3.3)

Let ξ_{k} denote the element of P K^{1}(SU(n)) which corresponds to ξ^{k} − 1 under this isomorphism. Here ξ denotes the Hopf bundle over CP^{n−1}. Then

B_{j} = ∑(−1)^{k+1}\binom{n}{j−k}ξ_{k}.

Proof. If I denotes the augmentation ideal of R(G), then there is an isomorphism I/I^{2} ≈ P K^{1}(G) under which (−1)^{k+1}λ^{k} corresponds to ψ^{k}. (See, e.g., [11, 2.1ff].) We first claim that under this isomorphism and (3.3) (which is well-known) {θ_{n}−n} ∈ I/I^{2} corresponds to {ξ − 1} ∈ K̃^{0}(CP^{n−1}). To see this, we use a result in [15, p.206] which states that there is a commutative diagram

CP^{n−1} ⟶^{f} BU
j↓  ↓^{B}
ΩSU(n) ⟶^{Ωi} ΩSU

in which j is a canonical map, i is the inclusion, B the Bott map, and f satisfies f^{∗}(c_{i}) = 0 unless i = 1, and f^{∗}(c_{1}) is the usual generator of H^{2}(CP^{n−1}). This implies that f classifies ξ. The isomorphism (3.3) is defined using the maps j and B. Hence ξ corresponds under this isomorphism to the map i, which is the canonical representation θ_{n}.

Since ψ^{k}(ξ) = ξ^{k}, we obtain that

ξ_{k} = (−1)^{k+1}λ^{k}(θ_{n} − n) = (−1)^{k+1} ∑_{j=0}^{k} \binom{−n}{k−j}λ^{j}(θ_{n}).

But λ^{j}(θ_{n}) = B_{j}. To invert this equation, we take a formal sum of the equations for all values of k, using powers of an indeterminate x. We have

∑(−1)^{k+1}ξ_{k}x^{k} = ∑ x^{k} ∑_{j=0}^{k} \binom{−n}{k−j}B_{j} = (1 + x)^{−n} ∑ x^{j}B_{j}

and hence

∑ x^{j}B_{j} = ∑_{i} \binom{n}{i}x^{i} ∑_{k}(−1)^{k+1}ξ_{k}x^{k} = ∑ x^{j} ∑_{k} \binom{n}{j−k}(−1)^{k+1}ξ_{k},

as desired.

The following corollary is immediate from the description of j^{∗} in Proposition 3.1 along with Proposition 3.2.

Corollary 3.4. P K^{1}(Sp(n)) ≈ ⟨ξ_{1}, ξ_{2}, …: R_{n+1}, …, R_{2n}, S_{j}, j > 2n⟩, where R_{j} = ∑(−1)^{k+1}\binom{2n+1}{j−k}ξ_{k} − ∑(−1)^{k+1}\binom{2n+1}{2n+1−j−k}ξ_{k},

S_{j} = ∑(−1)^{k}\binom{j}{k}ξ_{k}, and ψ^{t}ξ_{k} = ξ_{kt}.

Note that the relations allow one to express each ξ_{i} with i > n in terms of ξ_{1}, …, ξ_{n}. From Corollary 3.4, we easily deduce the following result.

Corollary 3.5. v^{2m}(Sp(n)) ≈ Z/2^{eSp′(m,n)}, where

eSp′(m, n) := min(ν(R_{m,n+1}), …, ν(R_{m,2n}), ν(S′_{m,j}), j > 2n), with R_{m,j} and S′_{m,j} defined by

R_{m,j} = ∑_{k odd} \binom{2n+1}{j−k}k^{m} − ∑_{k odd} \binom{2n+1}{2n+1−j−k}k^{m},

S′_{m,j} = ∑_{k odd} \binom{j}{k}k^{m}.

Proof. The sums are taken over all odd values of k for which the binomial coefficients are nonzero. The generator of the group is ξ_{1}, and the relations are obtained from those in 3.4 by ξ_{k} = ψ^{2}(ξ_{k/2}) ≡ 0 if k is even, while if k is odd, ξ_{k} = ψ^{k}(ξ_{1}) ≡ k^{m}ξ_{1}. Here we are using Corollary 3.9 to imply that ψ^{k} ≡ k^{m} in v^{2m}(−) for all odd k.

Because of Theorems 1.1 and 1.4 and Corollary 3.5, we have

1-LINE FOR SPIN(2N+1) 11

Proposition 3.6. eSp′(m, n) = eSp(m, n); i.e., min(ν(R_{m,n+1}), …, ν(R_{m,2n}), ν(S_{m,j}), j > 2n) = min(ν(S_{m,j}), j > 2n), and so the relations R_{m,j} are superfluous in eSp′(m, n)

We should remark on the minor difference between S_{m,j} in 1.4 and S′_{m,j} in 3.5, in that one involves a sum over all values of k, while the other involves a sum only over odd values of k. As remarked in Section 1, we only consider values of m large enough that summands which are multiples of 2^{m} would not affect divisibility by 2 of the sum. Thus ν(S_{m,j}) = ν(S′_{m,j}) for such values of m.

Next we explain why the relations ψ^{3}−3^{m} and ψ^{−1}−(−1)^{m} imply relations ψ^{k} −k^{m} for all odd integers k. We contrast with the situation when we are localized at an odd prime. The difference is due to the following result, which is stated in [1, 2.9].

Proposition 3.7. If p is an odd prime, then the multiplicative group of units (Z/p^{n})^{×} is cyclic of order (p − 1)p^{n−1}. If r generates (Z/p^{2})^{×}, then any integer congruent to r mod p^{2} generates (Z/p^{n})^{×}. The group (Z/2^{n})^{×} is the direct product of the subgroup consisting of ±1 and the subgroup of classes congruent to 1 mod 4; the latter subgroup is cyclic of order 2^{n−2} generated by any integer congruent to 5 mod 8.

This implies that, for any n, any odd integer is congruent mod p^{n} to ±3^{e} for some e. We also need the following result of Adams.

Proposition 3.8. ([2, 5.1]) If X is a finite complex, there exists an integer e so that for all k and m, ψ^{k} ≡ ψ^{k+m^{e}} in K(X)/mK(X).

We will use the following consequence of these results.

Corollary 3.9. Let X be a finite complex, and if p is odd let r generate (Z/p^{2})^{×}. Let

Q_{m} = \begin{cases}P K^{1}(X; Z_{(p)})/(im(ψ^{r} − r^{m})) & if p is odd \\ P K^{1}(X; Z_{(2)})/(im(ψ^{3} − 3^{m}, ψ^{−1} − (−1)^{m})) & if p = 2. \end{cases}

Then ψ^{s} = s^{m} in Q_{m} for all s ≢ 0 mod p.

Proof. We consider just the case p = 2. The case when p is odd is similar and slightly easier. We show ψ^{s} ≡ s^{m} mod 2^{n} for all positive integers n. This implies that they are equal in the finitely generated Z_{(2)}-module.

12 BENDERSKY AND DAVIS

We first show that Proposition 3.8 is valid for K^{1}(−) provided k and m are relatively prime. Adams’ result dealt with K^{0}(−). The validity in K^{−1}(−) is immediate since K^{−1}(X) = K^{0}(ΣX). Since ψ^{k} in K^{−1}(−) corresponds to kψ^{k} in K^{1}(−), the periodicity in K^{1}(−) follows, as long as k is a unit mod m.

Choose an integer e that works for X in Proposition 3.8. Use Proposition 3.7 to find an integer ℓ and ϵ = 0 or 1 so that s ≡ (−1)^{ϵ}3^{ℓ} mod 2^{ne}. Then, for some k, we have, in Q_{m} mod 2^{n},

ψ^{s} = ψ^{(−1)^{ϵ}3^{ℓ}+k2^{ne}} ≡ ψ^{(−1)^{ϵ}}ψ^{3^{ℓ}} ≡ ((−1)^{ϵ})^{m}(3^{ℓ})^{m} ≡ s^{m}.

The following result is immediate from Propositions 3.1 and 3.2.

Proposition 3.10. The group P K^{1}(Spin(2n + 1)) has a subgroup isomorphic to P K^{1}(Sp(n)), as described in Corollary 3.4, and an additional generator D satisfying

2^{n+1}D = ∑_{j=1}^{n} ∑_{k=1}^{j}(−1)^{k+1}\binom{2n+1}{j−k}ξ_{k}.  (3.11)

The Adams operations are determined by ψ^{t}ξ_{k} = ξ_{kt}.

The next two theorems, 3.15 and 3.18, determine v^{2m}(Spin(2n + 1)). Two of the relations, R_{2} and R_{3}, are defined in terms of the following algorithms. The algorithms will be justified in the proof of 3.15 and computed in 3.18. We use the relations R_{j}, n + 1 ≤ j ≤ 2n, and S_{j}, 2n < j ≤ 3n, of Corollary 3.4. Note that each equals ±ξ_{j} mod terms with smaller subscripts of ξ.

For R_{2}, begin with

∑_{j=1}^{n} ∑_{k=1}^{j}(−1)^{k+1}\binom{2n+1}{j−k}ξ_{2k},  (3.12)

and subtract multiples of R_{2n}, …, R_{n+1} to eliminate ξ_{2n}, …, ξ_{n+1}. This results in a linear expression E in ξ_{1}, …, ξ_{n} which, we will show in (4.2), equals (−1)^{n+1}2^{n}ξ_{n} plus lower terms. Write (3.11) in the form

2^{n+1}D + (−1)^{n}ξ_{n} + ∑_{k=1}^{n−1}(−1)^{k}ξ_{k} ∑_{i=0}^{n−k} \binom{2n+1}{i}.  (3.13)

1-LINE FOR SPIN(2N+1) 13

Add 2^{n} times (3.13) to E to eliminate ξ_{n}. The resulting linear expression in ξ_{1}, …, ξ_{n−1}, and D is divisible by 2^{n+1}. Divide it by 2^{n+1}, and then replace ξ_{k} by k^{m} (resp. 0) if k is odd (resp. even) to get R_{2}.

For R_{3}, begin with

∑_{j=1}^{n} ∑_{k=1}^{j}(−1)^{k+1}\binom{2n+1}{j−k}ξ_{3k},  (3.14)

and subtract multiples of S_{3n}, …, S_{2n+1}, R_{2n}, …, R_{n+1}, and of (3.13) to eliminate ξ_{3n}, …, ξ_{n+1}, ξ_{n}. This results in a linear expression in ξ_{1}, …, ξ_{n−1}, and D which is divisible by 2^{n+1}. Divide it by 2^{n+1} and replace ξ_{k} by k^{m} or 0 as before to get an expression E′. The relation R_{3} is 3^{m}D − E′.

Theorem 3.15. The abelian group v^{2m}(Spin(2n + 1)) has generators ξ_{1} and D with relations

2^{n+1}D = (∑_{j=1}^{n} ∑_{odd k} \binom{2n+1}{j−k}k^{m})ξ_{1},  (3.16)

2^{eSp(m,n)}ξ_{1}, R_{2} and R_{3}.

Proof. The second relation is the one obtained in Propositions 3.5 and 3.6. This relation due to Adams operations on ξ_{i}-classes is still present in v^{2m}(Spin(2n + 1)). The first relation is the definition of D, together with ξ_{k} ∼ ψ^{k}ξ_{1} ∼ k^{m}ξ_{1} if k is odd, and ξ_{k} ∼ ψ^{k}ξ_{1} ∼ 0 if k is even.

The relations R_{2} and R_{3} are consequences of ψ^{2}D ∼ 0 and ψ^{3}D ∼ 3^{m}D, respectively. We evaluate ψ^{2}(2^{n+1}D) and ψ^{3}(2^{n+1}D) using (3.11) and ψ^{ℓ}ξ_{k} ∼ ξ_{kℓ}. The relations are reduced using R_{j} and S_{j}. Since ψ^{k}D must be an integral combination of the generators ξ_{1}, …, ξ_{n−1}, and D, it follows that ψ^{2}(2^{n+1}D) and ψ^{3}(2^{n+1}D), when expressed in terms of these classes, must be divisible by 2^{n+1}. Performing the division yields the desired relations for ψ^{2}(D) and ψ^{3}(D).

By Corollary 3.9, we need only to append the relation ψ^{−1}(D) ∼ (−1)^{m}D in order to achieve all relations ψ^{k}(D) ∼ k^{m}D for all odd k. However, by Proposition 3.17, this relation adds no information if m is odd, while if m is even, which will be treated at the end of this section, it implies 2D ∼ 0.

Proposition 3.17. ψ^{−1} = −1 in P K^{1}(Spin(2n + 1)).

14 BENDERSKY AND DAVIS

Proof. First note that ψ^{−1} = (−1)^{m} in P K^{1}(S^{2m+1}). We use this to prove, by induction on n, that ψ^{−1} = −1 in P K^{1}(Sp(n)).

Indeed, consider the short exact sequence

0 → P K^{1}(S^{4n−1})⟶^{p^{∗}}P K^{1}(Sp(n))⟶^{i^{∗}}P K^{1}(Sp(n − 1)) → 0,

and assume ψ^{−1} = −1 in P K^{1}(Sp(n−1)). Let y generate im(p^{∗}). Note ψ^{−1}(y) = −y. Let x ∈ P K^{1}(Sp(n)). By naturality of ψ^{−1} and the induction hypothesis, we have i^{∗}(ψ^{−1}(x) + x) = 0, and hence ψ^{−1}(x) + x = αy for some integer α. Apply ψ^{−1} and use ψ^{−1}ψ^{−1} = 1 to obtain αy = −αy, and hence α = 0.

Now we use the morphism P K^{1}(Sp(n))⟶^{θ}P K^{1}(Spin(2n+1)) considered in Proposition 3.1 to deduce the proposition. Since 2^{n+1}D ∈ im(θ), we obtain ψ^{−1}(2^{n+1}D) = −2^{n+1}D, and hence ψ^{−1}(D) = −D since there is no torsion in P K^{1}(Spin(2n + 1)). All other elements are in the image from P K^{1}(Sp(n))).

The algorithm distills down to the following closed form for the relations in Theorem 3.15, which was observed from extensive computer calculations by the second author and then proved (see Section 4) utilizing a number of elaborate combinatorial arguments.

Theorem 3.18. v^{2m}(Spin(2n + 1)) is the abelian group with generators ξ_{1} and D and four relations as below:

2^{eSp(m,n)}ξ_{1},
( ∑_{odd k} k^{m} ∑_{i=0}^{n−k} \binom{2n+1}{i})ξ_{1} − 2^{n+1}D,  (3.19)
( ∑_{odd k} k^{m} ∑_{t≥0} \binom{2n+2}{n−1−k−4t})ξ_{1} − 2^{n}D,  (3.20)

and

(2^{n} ∑_{odd k} k^{m} ∑_{t≥0} \binom{2n+1}{n−1−k−3t})ξ_{1} − \frac{1}{3}(2^{2n+1} + 1 − 3^{m+1})D. (3.21)

If m is odd and n ≠ 4, the coefficients of ξ_{1} in (3.19) and (3.20) are divisible by 2^{n}.

1-LINE FOR SPIN(2N+1) 15

¿From this, we easily deduce Theorem 1.5. Subtract 2 times (3.20) from (3.19) to obtain the relation R_{1}(m, n) of 1.5 on the generator ξ_{1}. Note that the exponent of 2 in the coefficient of D in (3.21) is min(2n + 1, ν(m + 1) + 2) if m is odd. If n ≤ ν(m + 1) + 2, divide (3.20) by 2^{n} to split off Z/2^{n}, the second summand in 1.5, and then add an appropriate multiple of (3.20) to (3.21) to eliminate D, yielding the relation R_{2}(m, n) of 1.5 on ξ_{1}. The last part of Theorem 3.18 is necessary in order to know that we can split off Z/2^{n}.

Similarly, if ν(m + 1) + 2 < n, use (3.21) to split off Z/2^{ν(m+1)+2}, and then subtract an appropriate multiple of (3.21) from (3.20) to obtain R_{2}(m, n). This completes the proof of Theorem 1.5 when m is odd.

To prove that v^{2m}(Spin(2n + 1)) ≈ Z/2 ⊕ Z/2 when m is even, we begin by showing that eSp′(m, n) = 1 in Corollary 3.5 when m is even. It suffices to show that R_{m,2n} ≡ 2 mod 4 and that R_{m,j} and S′_{m,j} are even for all relevant j. Since k^{m} ≡ 1 mod 4 if k is odd and m even, we have, mod 4,

S′_{m,j} ≡ ∑_{k odd} \binom{j}{k} = 2^{j−1},

R_{m,2n} ≡ ∑_{k odd} \binom{2n+1}{k} − 2\binom{2n+1}{2n+1} = 2^{2n} − 2,

and other R_{m,j} are congruent to 2^{2n} minus 2 times a sum of binomial coefficients.

As observed in the proof of 3.15, ψ^{−1} implies 2D ∼ 0 when m is even. Thus it remains to show that the coefficients of relations in Theorem 3.18 are even when m is even, which reduces to showing

∑_{k odd} ∑_{i=0}^{n−k} \binom{2n+1}{i} ≡ 0 mod 2  (3.22)

and

∑_{k odd} ∑_{t≥0} \binom{2n+2}{n−k−1−4t} ≡ 0 mod 2.  (3.23)

The sum in (3.22) is ∑_{i=0}^{n−1} \binom{2n+1}{i}[\frac{n−i+1}{2}]. If n is even, terms 2j and 2j +1 in this sum will have the same parity. If n ≡ 3 mod 4, terms 2j + 1 and 2j + 2 will have the same parity. If n ≡ 1 mod 4, the only odd terms will occur in pairs i = 8j and i = 8j + 3.

16 BENDERSKY AND DAVIS

The sum in (3.23) is ∑ \binom{2n+2}{n−2j}[\frac{j+1}{2}]. If n is odd, all coefficients \binom{2n+2}{n−2j} are even, while if n = 2N, this becomes ∑ \binom{2N+1}{N−j}[\frac{j+1}{2}] mod 2, which equals the first sum, and hence is even. This completes the proof of Theorem 1.5.

4. Proof of Theorem 3.18

In this section, we perform the algorithm in Theorem 3.15 and obtain Theorem 3.18. First note that (3.19) is an elementary manipulation of (3.16).

We begin by deriving (3.20). Let M be a matrix whose rows present the relations R_{2n}, …, R_{n+1} of 3.4 expressed in terms of ξ_{2n}, …, ξ_{1}. Then

M_{i,j} = (−1)^{j−i}\binom{2n+1}{j−i} + (−1)^{i+j}\binom{2n+1}{j+i−2n−1}.

Let M_{L} (resp. M_{R}) denote the left (resp. right) half of M.

Let σ_{j} := ∑_{k=0}^{j} \binom{2n+1}{k}. Let P denote a row vector of length 2n whose (2j + 1)st entry is (−1)^{n+1+j}σ_{j} for 0 ≤ j < n, with other entries 0. This represents (3.12). Write P = (P_{L}|P_{R}), with P_{L} and P_{R} of length n. We wish to add multiples of the rows of M to P to annihilate P_{L}. This is facilitated by first replacing M by the equivalent set of relations M_{L}^{−1}M = [I|M_{L}^{−1}M_{R}]. The linear expression E in the algorithm for R_{2} described before Theorem 3.15 corresponds to P_{R} − P_{L}M_{L}^{−1}M_{R}. We will show that this equals

(−1)^{n+1}2^{n} ∑_{j=0}^{n−1}(−1)^{j}\binom{2n+1}{j}V_{j},  (4.1)

where V_{j} is a vector of length n which begins with j 0’s followed by

1, 1, −1, −1, 1, 1, −1, ….

The expression corresponding to (4.1) is

2^{n}(∑_{k=1}^{n}(−1)^{k+1}ξ_{k}(∑_{j=0}^{n−k} \binom{2n+1}{j} − 2 ∑_{t≥0} \binom{2n+2}{n−1−k−4t})).  (4.2)

We have used

\binom{2n+2}{a} = \binom{2n+1}{a} + \binom{2n+1}{a−1}  (4.3)

to show that this corresponds to (4.1). Adding 2^{n} times (3.13) to (4.2) and replacing ξ_{k} by k^{m} or 0 yields −2^{n+1} times (3.20), as desired.

1-LINE FOR SPIN(2N+1) 17

To complete the proof of (3.20), it remains to prove (4.1). Let c_{i} = \binom{2n+1}{i} and d_{i} = \binom{2n+i}{i}. Consideration of generating functions implies that for any positive integer m

∑_{j=0}^{m}(−1)^{j}d_{j}c_{m−j} = 0  (4.4)

and that the (i, j)th entry of M_{L}^{−1} is d_{j−i}. Using these facts, one can easily prove the following lemma.

Lemma 4.5. M_{L}^{−1}M_{R} = ∑_{j=1}^{n} D_{j}C_{j}, where D_{j} is a column vector of length n with t^{th} entry d_{n+1−t−j} − d_{n+j−t}, and C_{j} is a row vector of length n with t^{th} entry (−1)^{t−j}c_{t−j}.

To simplify notation, we shall complete the proof of (4.1) when n is odd. The proof when n is even is virtually the same. Let n = 2e + 1. We shall prove the following lemma.

Lemma 4.6. For 1 ≤ j ≤ n,

−P_{L}D_{j} = (−1)^{[(j−1)/2]}2^{n} + ∑_{u=1}^{[j/2]}(−1)^{e+u+1}σ_{e+u}d_{j−2u}.

Then the entry in the t^{th} position of −P_{L}M_{L}^{−1}M_{R} is

(−P_{L} ∑_{j} D_{j}C_{j})_{t} = A_{1,t} + A_{2,t},

where

A_{1,t} = 2^{n} ∑_{j=1}^{t}(−1)^{[(j−1)/2]}(−1)^{t−j}c_{t−j},

which equals the t^{th} entry of (4.1), and

A_{2,t} = ∑_{j=1}^{t}(−1)^{t−j}c_{t−j} ∑_{u=1}^{[j/2]}(−1)^{e+u+1}σ_{e+u}d_{j−2u}.

Using (4.4), one can check that

A_{2,t} = \begin{cases}0 & t odd \\ (−1)^{e+T +1}σ_{e+T} & t = 2T. \end{cases}

Thus P_{R} + A_{2} = 0, and hence P_{R} − P_{L}M_{L}^{−1}M_{R} equals (4.1), as claimed. Once we have given the proof of Lemma 4.6, we will be done with the proof of the R_{2}-part of Theorem 3.18.

18 BENDERSKY AND DAVIS

Proof of Lemma 4.6. We use generating functions. The coefficients of x^{0}, …, x^{n−1} in p(x) := (1 − x^{2})^{2n+1}/(1 + x^{2}) give the entries of P_{L}, while q(x) := (1 − x)^{−(2n+1)} = ∑ d_{i}x^{i}. Note that p(x)q(x) = (1 + x)^{2n+1}/(1 + x^{2}). We obtain

−P_{L}D_{j} = ∑_{t=1}^{n} coef(x^{t−1}, p(x)) · (coef(x^{n+j−t}, q(x)) − coef(x^{n−j+1−t}, q(x))
= coef(x^{n+j−1}, p(x)q(x)) − coef(x^{n−j}, p(x)q(x))  (4.7)

− ∑_{t>n} coef(x^{t−1}, p(x)) · d_{n+j−t}.

The sum subtracted at the end is due to the truncation of D_{j}; one easily checks that this equals the u-sum in Lemma 4.6. Finally we observe that, if we employ the symmetry property of binomial coefficients, then the expression in line (4.7) equals the following expression if j ≡ 1 or 2 mod 4, and equals the negative of this expression otherwise.

\binom{2n+1}{n} − \binom{2n+1}{n−1} − \binom{2n+1}{n−2} + \binom{2n+1}{n−3} + \binom{2n+1}{n−4} − \binom{2n+1}{n−5} − ⋯ This expression equals 2^{n}, as can be seen by expanding (1 − i)^{2n+1} or by induction on n.

The derivation of (3.21) is similar to that of (3.20) just performed, although somewhat more complicated. Replacing the matrix M used in (3.20) is a matrix N whose rows present the relations S_{3n}, …, S_{2n+1}, R_{2n}, …, R_{n+1} of 3.4 in terms of ξ_{3n}, …, ξ_{1}. Then

N_{i,j} = \begin{cases}(−1)^{j−i}\binom{3n+1−i}{j−i} & 1 ≤ i ≤ n \\ (−1)^{j−i}\binom{2n+1}{j−i} + (−1)^{i+j}\binom{2n+1}{j+i−4n−1} & n < i ≤ 2n. \end{cases}

Divide N into n × n submatrices which we name as follows:

N = (\begin{matrix}T_{1} & U_{1} & U_{2} \\ 0 & T_{2} & U_{3} \end{matrix}) .

We perform row operations on N to reduce it to

N′ = (\begin{matrix}I & 0 & B_{2} − B_{1}W \\ 0 & I & W \end{matrix} ) ,

where B_{i} = T_{1}^{−1}U_{i}, and W = T_{2}^{−1}U_{3} = ∑_{j=1}^{n} D_{j}C_{j}, exactly as in Lemma 4.5.

Let σ_{j}, c_{i}, d_{i}, C_{j}, and D_{j} be as in the preceding proof, with σ_{j} = 0 if j is not an integer. Let Q denote a row vector of length 3n whose (3j + 1)st entry is (−1)^{n+1+j}σ_{j}

1-LINE FOR SPIN(2N+1) 19

for 0 ≤ j < n, with other entries 0. This represents (3.14). Write Q = (Q_{0}|Q_{1}|Q_{2}) with each Q_{i} of length n. We use N′ to reduce Q to (0|0|Q′_{2}), where

Q′_{2} = Q_{2} − Q_{0}(B_{2} − B_{1}W) − Q_{1}W.  (4.8)

We will show this equals

(−1)^{n}(\frac{1}{3}(2^{2n+1} + 1)(−σ_{0}, σ_{1}, −σ_{2}, σ_{3}, …, σ_{n−1})

+ 2^{2n+1}(0, −c_{0}, c_{1}, −c_{2}, c_{3} + c_{0}, −(c_{4} + c_{1}), …)). (4.9)

Thus the vector (4.9) expresses ψ^{3}(2^{n+1}D) in terms of ξ_{n}, …, ξ_{1} as

\frac{1}{3}(2^{2n+1} + 1) ∑_{k=1}^{n}(−1)^{k+1}ξ_{k} ∑_{i=0}^{n−k} \binom{2n+1}{i} + 2^{2n+1} ∑_{k=1}^{n−1}(−1)^{k}ξ_{k} ∑_{t≥0} \binom{2n+1}{n−1−k−3t}. (4.10)

We add \frac{1}{3}(2^{2n+1} + 1) times (3.13) to eliminate ξ_{n} and get

\frac{1}{3}(2^{2n+1} + 1)2^{n+1}D + 2^{2n+1} ∑_{k=1}^{n−1}(−1)^{k}ξ_{k} ∑_{t≥0} \binom{2n+1}{n−1−k−3t}.

Divide by 2^{n+1}, replace ξ_{k} by k^{m} or 0, and subtract from 3^{m}D; this yields exactly (3.21).

It remains to show that Q′_{2} equals (4.9). We show this when n ≡ 0 mod 6; other congruences can be handled similarly. Let r = n/3. It is not difficult to prove that (B_{1})_{i,j} = (−1)^{j}\binom{n+j−i−1}{n−i}\binom{3n+1−i}{n+j−i} and (B_{2})_{i,j} = (−1)^{j}\binom{2n+j−i−1}{n−i}\binom{3n+1−i}{2n+j−i}. We obtain that, for 0 ≤ t < n, the (t + 1)st entry of Q′_{2} is

(−1)^{t+1}σ_{2r+t/3} − ∑_{u=0}^{r−1}(−1)^{u+t}σ_{u}\binom{2n+t−3u−1}{n+t}\binom{3n−3u}{n−t} (4.11)
+ ∑_{s=0}^{n−1} ∑_{u=0}^{r−1}(−1)^{u+s}σ_{u}\binom{n+s−3u−1}{s}\binom{3n−3u}{2n−s}  (4.12)

· ∑_{j=0}^{t}(d_{n−s+j−t−1} − d_{n+t−j−s})(−1)^{j}c_{j}

+ ∑_{u=0}^{r−1}(−1)^{u}σ_{r+u} ∑_{j=0}^{t}(d_{n−3u−1+j−t} − d_{n+t−j−3u})(−1)^{j}c_{j}. (4.13)

20 BENDERSKY AND DAVIS

It follows from (4.4) that the j-sum in (4.12) is 0 for s ≥ n, except that it is −1 when s = n + t. Hence (4.12) can be made into a sum over all s ≥ 0 with the sum-part of (4.11) incorporated into it.

Next we note that if u = r + δ with δ ≥ 0 in (4.13), then, again using (4.4), the j-sum in (4.13) is nonzero only when δ = t/3, in which case it is −1. Thus (4.13) can be extended to u ≥ 0 if the initial term of (4.11) is included. Similarly we note that if u = r + Δ with Δ ≥ 0 in (4.12), then \binom{n+s−3u−1}{s}\binom{3n−3u}{2n−s} is nonzero only when s = 3Δ, in which case it is (−1)^{Δ}. Thus the modified (4.13) can be incorporated into (4.12) by extending the u-sum of (4.12) to u ≥ 0. Now the (t + 1)st entry of Q′_{2} has been simplified to

∑_{s≥0} ∑_{u≥0}(−1)^{u+s}σ_{u}\binom{n+s−3u−1}{s}\binom{3n−3u}{2n−s} ∑_{j=0}^{t}(d_{n−s+j−t−1} − d_{n+t−j−s})(−1)^{j}c_{j}. (4.14)

We will prove the following lemma at the end of this section.

Lemma 4.15. If n ≥ 0, and v and w are arbitrary integers, then

∑_{s≥0}(−1)^{s}\binom{n+s−1−w}{s}\binom{3n−w}{2n−s}\binom{3n−s+v}{2n} = \binom{4n−w+v}{2n}.

Using this with w = 3u, (4.14) reduces to

∑_{j=0}^{t}(−1)^{j}c_{j} ∑_{u≥0}(−1)^{u}σ_{u}(\binom{4n−3u+j−t−1}{2n} − \binom{4n−3u+t−j}{2n}). (4.16)

Viewing \binom{4n−3u+ϵ}{2n} as (−1)^{u+ϵ} coef(x^{2n−3u+ϵ}, (1 + x)^{−(2n+1)}), and σ_{u} as

coef(x^{3u}, (1 + x^{3})^{2n+1}/(1 − x^{3})),

we obtain a simplification of (4.16) to

(−1)^{t+1} ∑_{j=0}^{t} c_{j}(coef(x^{2n+j−t−1}, \frac{(1−x+x^{2})^{2n+1}}{1−x^{3}} ) + coef(x^{2n+t−j}, \frac{(1−x+x^{2})^{2n+1}}{1−x^{3}} )).

(4.17)

We will prove the following lemma at the end of this section.

Lemma 4.18. Let f_{m}(x) = (1 − x + x^{2})^{m+1}/(1 − x^{3}). Then for m ≥ 0 and j ≥ 0,

coef(x^{m−j−1}, f_{m}(x))+coef(x^{m+j}, f_{m}(x)) = \begin{cases}(1 + 2(−2)^{m})/3 & j ≡ 0, 2 mod 3 \\ (1 + 2(−2)^{m+1})/3 & j ≡ 1 mod 3. \end{cases}

We obtain that (4.17) equals

(−1)^{t+1}(\frac{1}{3}(2^{2n+1} + 1) ∑_{j=0}^{t} c_{j} − 2^{2n+1} ∑_{\begin{matrix}0≤j≤t \\ t−j≡1 (3) \end{matrix}} c_{j}),

which equals the component of (4.9) in position t+1, as desired, completing the proof of (3.21), as described surrounding (4.9).

Next we prove the portion of Theorem 3.18 which states that

∑_{odd k} k^{m} ∑_{t≥0} \binom{2n+2}{n−1−k−4t}  (4.19)

is divisible by 2^{n} if m is odd and n ≠ 4. This occurs in (3.20). We accomplish this by showing later in this section that, if m = 2a + 1, then (4.19) equals

\binom{n−1}{1}2^{2n−4} + ∑_{j=2}^{n/2} \binom{n−j}{j}2^{2n−4j} ∑_{i≥j−1} 8^{i}\binom{a}{i} ∑_{t=0}^{j−2}(−1)^{t}\binom{2j−1}{t}(2j − 2t − 1)\binom{j−t}{2}^{i}. (4.20)

Then we note that the first term of (4.20) is divisible by 2^{n} provided n ≥ 3. For each value of j ≥ 2, the j-summand is nonzero for n ≥ 2j and is clearly divisible by 2^{2n−4j+3(j−1)}, which is ≥ 2^{n} provided n ≥ j + 3. The case n = 4 and j = 2 is the only time that this gives a nonzero term which might not be divisible by 2^{n}; it is what causes the restriction n ≠ 4 in the last part of Theorem 3.18.

The expression (4.20), while not as elegant as (4.19), is often more useful in performing specific calculations of ν(−), such as in Proposition 1.6. Before proving (4.20), we handle the case n = 4, which was excluded from Theorem 1.5.

Proposition 4.21.

v^{2m}(Spin(9)) ≈ \begin{cases}Z/2 ⊕ Z/2 & m even \\ Z/8 ⊕ Z/2^{e(m)} & m odd, \end{cases}

where

e(m) = \begin{cases}min(ν(m − 5) + 2, 6) & if m ≡ 1 mod 4 \\ min(ν(m − 7) + 2, 8) & if m ≡ 3 mod 4. \end{cases}

Proof. The numbers eSp(m, n) are as in [6, 5.1] (suitably interpreted), and large enough as to not matter here. Let P = 3^{m} − 3. The other three relations of 3.18 are (160 + 10P)ξ_{1} − 32D, (48 + P)ξ_{1} − 16D, and 16(39 + P)ξ_{1} + (P − 168)D. If m ≡ 3

22 BENDERSKY AND DAVIS

mod 4, then P ≡ 8 mod 16, so we get a Z/8 due to (48+P)ξ_{1}. (This is the anomaly.) If m ≡ 1 mod 4, P ≡ 0 mod 16, and we get a Z/8 due to (P − 168)D.

If m ≡ 1 mod 4, let Q = 3(3^{m−5} − 1), so that ν(Q) = ν(m − 5) + 2. Replace P by 3^{4}Q + 3^{5} − 3. Subtract multiples of the third relation from the other two so as to eliminate the D-term from each. The relations obtained are, up to odd multiples, 2^{6} + 2Q + Q^{2}/4 and 2^{8} + Q + Q^{2}/8. For ν(Q) ≥ 4, the minimum of the exponent of 2 of these two expressions is min(6, ν(Q)), as claimed. The case m ≡ 3 mod 4 is handled similarly.

Now we establish the equivalence of (4.19) and (4.20). A key ingredient is the following lemma.

Lemma 4.22. If n ≥ 0 and A is an integer, then

∑_{j=0}^{n/2}(−1)^{j}\binom{n−j}{j}4^{n−2j}\binom{2j−1}{j−A} = (−1)^{A} ∑_{t≥0} \binom{2n+2}{n−2A−4t}.

Proof. Thinking of either expression as a function f(n, A), we establish that each satisfies

f(0, A) = \begin{cases}0 & A > 0 \\ 1 & A ≤ 0, even \\ −1 & A < 0, odd, \end{cases}

f(−1, A) = 0, and the recursive formula f(n, A) = 4f(n−1, A)−f(n−2, A−1)−2f(n−2, A)−f(n−2, A+1). Thus they are equal. The recursive formula for the LHS of 4.22 is proved using several applications of (4.3), while for the RHS one proves the closely related formula

\binom{2n+2}{m} = 4\binom{2n}{m−1} + \binom{2n−2}{m} − 2\binom{2n−2}{m−2} + \binom{2n−2}{m−4}

by multiplying both sides by x^{m} and summing over all values of m.

Replace k in (4.19) by 2s + 1, and m by 2a + 1. Using 4.22 with A = s + 1, (4.19) becomes

∑_{s≥0}(2s + 1)^{2a+1}(−1)^{s+1} ∑_{j=0}^{n/2}(−1)^{j}\binom{n−j}{j}4^{n−2j}\binom{2j−1}{j−s−1}

= ∑_{j=0}^{n/2}(−1)^{j}\binom{n−j}{j}4^{n−2j} ∑_{s≥0}(−1)^{s+1}(2s + 1)\binom{2j−1}{j−s−1} ∑_{i≥0} \binom{a}{i}(4s^{2} + 4s)^{i}.

Replacing s by j − t − 1 yields a formula which can easily be manipulated to (4.20), except that the i-sum is unrestricted. The restriction to i ≥ j − 1 follows from the following lemma.

Lemma 4.23. If 1 ≤ i ≤ j − 2, then

∑_{t=0}^{j−2}(−1)^{t}\binom{2j−1}{t}(2j − 2t − 1)\binom{j−t}{2}^{i} = 0.

Proof. We solve the system of equations

a_{1}\binom{3}{2} + a_{2}\binom{4}{2} + ⋯ + a_{j−2}\binom{j}{2} = 1
a_{1}\binom{3}{2}^{2} + a_{2}\binom{4}{2}^{2} + ⋯ + a_{j−2}\binom{j}{2}^{2} = 1
⋮
a_{1}\binom{3}{2}^{j−2} + a_{2}\binom{4}{2}^{j−2} + ⋯ + a_{j−2}\binom{j}{2}^{j−2} = 1

by Cramer’s rule and the Vandermonde evaluation of determinants, obtaining

a_{s−2} = \frac{1}{\binom{s}{2}} \frac{∏_{t>s}(\binom{t}{2} − 1) ∏_{t<s}(1 − \binom{t}{2})}{∏_{t>s}(\binom{t}{2} − \binom{s}{2}) ∏_{t<s}(\binom{s}{2} − \binom{t}{2}))}
= \frac{2}{s(s − 1)} \frac{∏_{t≠s}(t(t − 1) − 2)}{∏_{t≠s}(t(t − 1) − s(s − 1))}
= \frac{2}{s(s − 1)} \frac{∏_{t≠s}(t − 2)(t + 1)}{∏_{t≠s}(t − s)(t + s − 1)}
= (−1)^{s+1} \frac{2s − 1}{3} \begin{matrix}( & {2j−1 \atop j−s} & ) \\ ( & {2j−1 \atop j−2} & ) \end{matrix}.

The values of t in the products always run from 3 to j. Substituting the solutions into the equations yields equations which reduce to those that were to be proved.

24 BENDERSKY AND DAVIS

This completes the proof that the coefficient of ξ_{1} in (3.20) is divisible by 2^{n} if n ≠ 4. The proof that the coefficient of ξ_{1} in (3.19) is also divisible by 2^{n} is extremely similar. Using virtually identical methods, we show it equals

\binom{n+1}{1}2^{2n−3}+∑_{j≥2} 2^{2n+1−4j}(\binom{n+2−j}{j}−\binom{n−j}{j−2}) ∑_{i≥j−1} 8^{i}\binom{a}{i} ∑_{t=0}^{j−2}(−1)^{t}\binom{2j−1}{t}(2j−2t−1)\binom{j−t}{2}^{i}.

Each term of this is divisible by 2^{n+1} except when n = 4 and j = 2, in which case it is divisible by 2^{n}.

We complete the paper by giving deferred proofs of two lemmas.

Proof of Lemma 4.15. This lemma can probably be proved by some sort of generating function argument. However, we present a proof along the lines by which it was discovered, utilizing Maple and the method of [17].

As described in [17], one can often prove f(n) := ∑_{s} F (n, s) is constant (independent of n) by finding a function G such that

F (n + 1, s) − F (n, s) = G(n, s + 1) − G(n, s).  (4.24)

If such a function G can be found, then summing this equation over all values of s yields the conclusion f(n + 1) − f(n) = 0, and hence f is constant.

In our application, we run the software associated with the book [17] on the function

F (n, s) := (−1)^{s}\binom{n+s−1−w}{s}\binom{3n−w}{2n−s}\binom{3n−s+v}{2n} / \binom{4n−w+v}{2n}. (4.25)

Here n, w, and v are all free parameters, but we tell Maple that n is the induction parameter by running zeil(F(n,s),s,n,N), where F is as in (4.25). The second argument of zeil is the summation variable, the third is the parameter to be used for the recursion, and the fourth is the symbol to use for the operator which increases the value of the third parameter by 1. The output in this case is N-1, followed by G(n, s), an explicit polynomial, involving v and w in addition to n and s, for which (4.24) holds. In this case, the polynomial G involves approximately 500 monomials. The significance of the N-1 is that this operator acting on F equals G(n, s+1)−G(n, s). This says exactly (4.24). So we know that the ratio of the two expressions involved in the lemma is constant. To know that this ratio is 1, we note that when n = 0 each equals 1.

Proof of Lemma 4.18. The proof is by induction on m. The lemma is readily verified if m = 0. Let s_{m,j} denote the sum of the two coefficients with which the lemma deals. Since f_{m+1} = (1 − x + x^{2})f_{m}, we obtain

s_{m+1,j} = \begin{cases}s_{m,1} & if j = 0 \\ s_{m,j−1} − s_{m,j} + s_{m,j+1} & if j > 0. \end{cases}

The induction step follows easily.

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Hunter College, CUNY, NY, NY 10021

E-mail address: mbenders@shiva.hunter.cuny.edu

Lehigh University, Bethlehem, PA 18015

E-mail address: dmd1@lehigh.edu