September 1997.

Commentary on HSW.

The following is an annotated copy of HSW, with comments set in narrower paragraphs. The principal aim is to make the presentation less dependent on peculiarities of the complex numbers (especially the absolute value). We omit the introduction.

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Let F be a field of q elements, V the category of finite dimesional vector-spaces over F. Consider a functor Φ : V → L, where L is the category of lattices, i.e., of finitely generated torsion-free ZZ-modules. Note that every object V of V is acted on by G = Aut(F) = F^{×}, and so is every Φ(V). We shall assume that Φ is G-free, in the sense that Φ^{G} = {0}.

Theorem. Let V be an (n + 1)-dimensional F -space, L and H the sets of its subspaces of dimensions 1 and n, respectively. Consider a family of maps a_{l} : F → V, (l ∈ L) such that im(a_{l}) = l and another family b_{h} : V → F, (h ∈ H) such that ker(b_{h}) = h. Composing α = ∏_{l∈L} Φ(a_{l}) with β = ∏_{h∈H} Φ(b_{h}), we then obtain an injective map

∏_{l∈L} Φ(F) → Φ(V) → ∏_{h∈H} Φ(F)

whose cokernel has order q^{N}, where 2N = n · r · (1 + ⋯ + q^{n}), r denoting the rank of Φ(F). Proof. Choosing a non-degenerate bilinear form φ on V, we can produce a bijection H → L which allows us to identify the source of α with the target of β and to regard γ = β ∘ α as an endomorphism of Λ = ∏_{l∈L} Φ(F). G acts diagonally on this product, and γ is given by the matrix γ(l, h) = β_{h} ∘ α_{l} ∈ ZZG. We have to show that detZZ(γ) = ± q^{N}.

This means that we consider γ an endomorphism of the ZZ-module Λ and evaluate its determinant with respect to some ZZ-basis of the latter — the matrix γ(l, h) ∈ ZZG is a bit of a red herring at this point. The desired determinant can, of course, be computed in K ⊗ Λ, where K is any field of characteristic 0, for instance K = C.

C ⊗ Φ(F) is a direct sum of 1-dimensional subspaces W_{i} (i = 1, …, r), on each of which G acts via a character μ_{i} : G → C^{×}, and γ is a direct sum of endomorphisms γ_{i} : ∑_{l∈L} W_{i} → ∑_{l∈L} W_{i}. Hence | det C(γ)| = ∏_{i} | det(γ_{i})|, where γ_{i} is the matrix μ_{i}(β_{h} ∘ α_{l}).

Since Φ(F)^{G} = {0}, all the characters μ_{i} are non-trivial. From this it will follow that | det(γ_{i})| = q^{N/r}, which is what we want.

Strictly speaking, it is C ⊗ γ which splits into ∑_{i} γ_{i}. Nota bene: the indices i label the simple components W_{i} of C ⊗ Φ(F), not the characters μ_{i} (which may not be distinct). The first sentence of the passage works just as well if C is replaced by a K admitting an injection χ : G → K^{×}. In the second one, the absolute values could be removed, giving det_{K}(γ) = ∏_{i} det(γ_{i}). The last sentence is trickier. What will actually be proved is: μ_{j} = μ_{i}^{−1} ⟹ det(γ_{i}) det(γ_{j}) = q^{2N/r} . To conclude that this is “what we want”, without absolute values, we would need additional information …

In fact, let Q be the set of x ∈ V such that x = a_{l}(1) for some l ∈ L, and let R be the set of y ∈ V such that φ(−, y) = b_{h} for some h ∈ H. Then γ_{i} produces the matrix M_{i}(x, y) = μ_{i}(φ(x, y)) defined on the index set Q × R. For the sake of tidiness, let us make R = Q by adjusting each y by a suitable factor c ∈ F; this changes the corresponding column of the matrix by the unimodular complex number μ_{i}(c), hence has no effect on | det(M_{i})|. Our theorem is now implied by the following lemma.

This adjustment could have been made by simply using cb_{h} instead of b_{h} in our family of maps V → F. Or we could say: det_{K}(γ) changes by ∏_{i} μ_{i}(c), which equals the determinant of the action of c ∈ G on the lattice Φ(C), hence must be ±1.

Lemma. With V, φ, Q as above, let μ be a non-trivial multiplicative character on F. On the index set Q × Q consider the matrix M(x, y) = μ(φ(x, y)). Then q^{−n/2}M is a unitary matrix. Proof. We must show that, for x, y ∈ Q,

(∗)  ∑_{z∈Q} μ(φ(x, z)) μ^{−1}(φ(y, z)) = q^{n}δ(x, y) ,

where the δ is Kronecker’s. To prove this identity, consider linear functionals ξ, η : V → F and form the sum

S(ξ, η) = ∑_{z∈Q} μ(ξ(z)) μ^{−1}(η(z)) .

We shall see that: (1) S(ξ, ξ) = q^{n} if ξ is non-trivial, and (2) S(ξ, η) = 0 if ξ, η are linearly independent. Actually it is easier to work with the wider sum

T (ξ, η) = ∑_{v∈V} μ(ξ(v)) μ^{−1}(η(v)) .

Since its summands are constant on projective points, we have T (ξ, η) = (q − 1)S(ξ, η). (1) Writing v = (u, a) with u ∈ U = ker(ξ) and a = ξ(v), we get

T (ξ, ξ) = ∑_{u∈U} ∑_{a∈F^{×}} μ(a) μ^{−1}(a) = q^{n}(q − 1) .

(2) Writing v = (u, a, b) with u ∈ U = ker(ξ, η), a = ξ(v), and b = η(v), we get

T (ξ, η) = ∑_{u∈U} ∑_{a,b∈F^{×}} μ(a) μ^{−1}(b) = q^{n−1} ∑_{a∈F^{×}} μ(a) ∑_{b∈F^{×}} μ^{−1}(b) .

Since μ is not trivial, these sums are.

What is being proved here is (∗), i.e., M_{μ}M_{ν}^{t} = q^{n}I_{m} for a non-trivial μ : G → K^{×} and ν = μ^{−1}. Taking K = C, we can deduce the theorem as in this text from HSW. However, the lemma yields more information. For instance, for K = Q_{p} and q = p, all characters are of the form μ = τ^{d}, where τ is the Teichmüller character. Writing M_{d} instead of M_{μ}, we then get M_{p−1−d}^{t} = p^{n}M_{d}^{−1}, with all entries in ZZ_{p}. If LM_{d}R = Δ_{d} is diagonal (with L and R invertible over ZZ_{p}), this implies L^{−t}M_{p−1−d}R^{−t} = p^{n}Δ_{d}^{−1}. Hence, if the elementary divisors of M^{d} are p^{r_{1}} ≥ ⋯ ≥ p^{r_{m}}, those of M_{p−1−d} will be p^{n−r_{m}} ≥ ⋯ ≥ p^{n−r_{1}}. Can this help us determine the structure of coker(γ_{d}) ?

Look at ”EL-ABA” for more (inconclusive) thoughts on this weighty matter, in particular, its relation to coker(α_{d}) — and don’t deny yourself the insights of the “Functor” paper in CanMathBull.