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Special classes in ker γ. The exact sequence of Galois modules

iaiηi to ((−1)a1, (−1)a2, ..., (−1)a2n+1) is an isomorphism of Gal(ks/k)-modules.

Then we complete the proof by noting that the F2-homomorphism mapping ηi to [dηi] identifies J [2](ks) with A/F2 which is isomorphic to ResL/kµ2(ks)/µ2(ks).

4.2. Special classes in ker γ. The exact sequence of Galois modules

0→ J[2](ks)→ J(ks)−→ J(k2 s)→ 0 induces the exact sequence

(14) 0→ J(k)/2J(k) → H1(k, J [2])→ H1(k, J )[2]→ 0 By the lemma above we have H1(k, J [2]) ∼= H1(k, GT).

Proposition 4.2.1. The subgroup J(k)/2J(k) of H1(k, J [2]) = H1(k, GT) lies in ker γ.

Proof. Recall that in §3.3.4, we associate to each class α∈ H1(k, GT) the orthogonal space Wα whose underlying space is L, with symmetric bilinear form

< λ, µ >α:= the coefficient of β2n in the product αλµ.

Now we use ”left multiplication by β” to construct another symmetric bilinear form

< βλ, µ >α on L. Let M = L

k, a 2n + 2-dimensional vector space over k. We define two bi-linear forms (or quadratic forms) on M : for (λ, a)∈ M,

Q(λ, a) =< λ, λ >α, Q(λ, a) =< βλ, λ >α+a2.

The pencil uQ− vQ with u, v ∈ ks has discriminant v2n+2f (u/v). Indeed, write lβ for the matrix associated to the left multiplication by β, and

( A 0

0 0 )

and

(

for those associate to Q and Q, respectively. Then

disc(uQ− vQ) = det(u

Thus, the pencil uQ− vQ is non-degenerate and contains exactly 2n + 2 singular elements over ks, which are the quadric Q, and the 2n + 1 quadrics η(β)Q − Q with f (η(β)) = 0. There is a Fano variety Fα whose ks-points corresponds to n-dimensional ks-subspaces Z of M which are isotropic for all of the quadrics in the pencil, and Fα is actually a principle homogeneous space of order 2 of the Jacobian J , corresponding to the image of α in H1(k, J )[2] [Donagi].

If α ∈ J(k)/2J(k), then by (14), the image of α in H1(k, J )[2] is trivial, so that the corresponding principle homogeneous space Fα has a k-rational point. Thus, there is a n-dimensional subspace Z of M over k which is isotropic for all quadrics in the pencil, hence isotropic for Q and Q. Since Z is isotropic for Q, Z contains no elements in M of the form (0, a), a ̸= 0. Hence, the projection of Z onto L is a n-dimensional k-subspace of L which is isotropic for Q. This implies the symmetric bilinear form <, >α is split. Therefore, α is in the kernel of γ.

5. Arithmetic fields

In this section, we consider the special cases where k is a finite field, a local field, or a global field. As before, the cases where V = W , V = Λ2(W ) and V = Sym2(W ) are respectively referred as the first, send and third case.

5.1. The finite field case. Let k be a finite field of odd order q. By Lang’s theorem (cf. [GC, III §2.3 Theorem 1]), H1(k, SO(W )) = 1 since SO(W ) is connected.

Thus by Galois descent theory, every orthogonal space of dimension 2n + 1 with determinant (−1)n is split so that H1(k, GT)⊆ ker γ.

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5.1.1. The first case. In the notation of §3.1, for each v∈ V with non-zero discrim-inant, Gv = SO(U ) which is also connected, hence H1(k, Gv) = 1. Thus for every non-zero element d∈ k, there is a unique orbit of vectors with△(v) = d.

5.1.2. The second case. For each S ∈ V with △(S) ̸= 0, GS is also connected because it is a maximal torus (Corollary 3.2.5) of the connected algebraic group SO(W ), hence has trivial cohomology. Thus for any separable polynomial f (x) = xg(x2), there is a unique orbit of S ∈ V with characteristic polynomial f(x).

5.1.3. The third case. Let T ∈ V be the special operator with non-zero discriminant so that GT = (ResL/kµ2)N =1. The exact sequence (9) induces the long exact sequence

(15)

· · · → µ2(L)−→ µN 2(k)→H1(k, GT)→H1(k, ResL/kµ2)−→HN 1(k, µ2)→ · · ·

(L/L∗2) −→ (kN /k∗2).

Let m + 1 denotes the number of irreducible factors of f (x)∈ k[x]. Then

|H1(k, ResL/kµ2)| = 2m+1.

Also, since 2n + 1 is an odd integer, the composition µ2(k) → µ2(L) −→ µN 2(k) is surjective. Hence H1(k, GT) → H1(k, ResL/kµ2) is injective. The lemma below asserts that

H1(k, ResL/kµ2)−→ HN 1(k, µ2) is actually surjective. These together imply

|H1(k, GT)| = 2m =|(µ2(L))N =1| = |GT(k)|.

This means that there are 2m distinct G(k)-orbits with characteristic polynomial f (x) that lie in the G(ks)-orbit with characteristic polynomial f (x). Since the order of each orbit equals to

|SO(W )(k)|

|GT(k)| = |SO(W )(k)|

2m ,

the number of self-adjoint operators S ∈ V associated to a fixed separable polyno-mial equals to |SO(W )(k)|.

Lemma 5.1.1. The map H1(k, ResL/kµ2)−→ HN 1(k, µ2) is surjective.

Proof. By (15), it is sufficient to show that L∗ N−→ k is surjective. By Chinese Remainder Theorem, we can write L =

m+1

i=1

Li, where each Li is a field extension of k, so that if t = (t1, ..., tm+1)∈ L, then N (t) =m+1

i=1 NLi/k(ti). Since k is a finite field, each norm map NLi/k is surjective. Hence the lemma follows. □ By Lang’s theorem, H1(k, J ) = 0,where J is the Jacobian of the smooth hyper-elliptic curve y2 = f (x) of genus n over k. Hence J (k)/2J (k) ∼= H1(k, GT) and every orbit associated to f (x) comes from a k-rational point on the Jacobian.

5.2. The non-archimedean local field case. Let k be a non-archimedean local field, with ring of integer O and finite residue field F := O/πO of odd order pα.

The simply-connected covering of SO is the spin group spin. We have an exact sequence

1→ µ2 → spin → SO → 1.

By Kneser’s theorem (cf. [GC, III §3.1 Conjecture II(a)]) we have H1(k, spin(W)) = 1,

and hence the injective connecting map:

δ : H1(k, SO(W)) ,→ H2(k, µ2) ∼=Z/2Z.

5.2.1. Surjectivity of δ. We’ll prove the surjectivity of δ by using induction on di-mension of W .

Lemma 5.2.1. Let V be a k-orthogonal space with dim(V ) = 2, then|H1(k, SO(V )| = 1 if det(V ) =−1(mod k∗2) and|H1(k, SO(V ))| = 2 if det(V ) = −d ̸= −1(mod k∗2).

Proof. (1) When det(V ) =−1(mod k∗2):

For every 2-dimensional orthogonal space (V, < ·, · >) with det(V) =

−1(mod k∗2), the case V has a nonzero isotropic vector is clearly split.

We may assume V contains no nonzero isotropic vectors. By using Gram-Schmidt’s orthogonal process, there is an orthogonal basisB = {e1, e2} such that

(2) When det(V ) =−d ̸= −1(mod k∗2):

For every 2-dimensional orthogonal space (V, < ·, · >) with det(V) ̸=

−1(mod k∗2), V contains no nonzero isotropic vectors. By using Gram-schmidt’s process , there is an orthogonal basis B = {e1, e2} and the qua-dratic form q associates to <·, · > under this basis is of the form

q(xe1+ ye2) = ax2 − by2 where xe1+ ye2 ∈ V and ab = d.

Consider the special case a = 1, b = d. This gives an orthogonal space (V0, <·, · >0) , and the associated quadratic form q0 is just the norm N from k(√

d) to k.

On the other hand,|k/N k(√

d)| = 2 by local class field theory, so we choose c∈ k not a norm from k(√

d), consider the orthogonal space

(V1, <·, · >1) with associated quadratic form q1 of the form cx2−cdy2. Then V0 and V1 have the same determinant but they are not isomorphic because q0 and q1 represent different numbers in k. Hence |H1(k, SO(V ))| = 2.

Lemma 5.2.2. Let (V, < ·, · >) be a k-orthogonal space with dim(V ) = 3, then

|H1(k, SO(V )| = 2.

Proof. If there is an isotropic vector v in V . Then the quadratic form q, which corresponds to < ·, · >, represents zero. Hence q represents every number. This means that △(u) =< u, u > can be every number as u varies in V . If det(V ) = a(mod k∗2), we choose v such that △(v) ̸= −a(mod k∗2) and consider V = (kv). We have det(V)̸= −1(mod k∗2), thus H1(k, SO(V)) has 2 elements by Lemma5.2.1, and so is H1(k, SO(V )).

If V contains no nonzero isotropic vector , then (kv) also contains no isotropic vector, for every v ∈ V , and hence H1(k, SO(V)) has 2 elements by Lemma5.2.1,

and so is H1(k, SO(V )).

Proposition 5.2.3. Let (V, <·, · >) be a k-orthogonal space with dim(V ) ⩾ 3, then

|H1(k, SO(V )| = 2. This proves the surjectivity of δ.

Proof. We prove by induction on dim(V ). When dim(V ) = 3, the proposition holds by Lemma 5.2.2. Assume the proposition holds for dim(V ) = l , we consider the case dim(V ) = l + 1. By Choosing a vector v ∈ V , we have the decomposition V = V⊕ kv where V = (kv).

Now V ,→ V induces H1(k, SO(V)) ,→ H1(k, SO(V ))(the proof is similar to The-orem 1), and the induction hypothesis asserts that |H1(k, SO(V))| = 2. Thus

|H1(k, SO(V )| = 2.

5.2.2. The first case. By Lemma 5.2.1, H1(k, Gv) = H1(k, SO(U )) ∼= Z/2Z except when dim(W ) = 3 and△(v) = 1(mod k∗2). Thus γ is bijection except in the special case.

5.2.3. The second case. Kottwitz has shown that γ is actually a group homomor-phism when we identify H1(k, Gv) and H1(k, SO(W )) as K/N E andZ/2Z respec-tively. (cf. [Kottwiz]). Let f (x) = xg(x2),K = k[x]/(g(x)) and E = k[x]/(g(x2)).

By local class field theory, |K/N E| = 2m where m is the number of irreducible factors gi(x) of g(x) such that gi(x2) still irreducible over k. Kottwitz also shows that γ is surjective when m ⩾ 1. Thus when m = 0, the number of orbits with characteristic polynomial f (x) is 1. And when m⩾ 1, the number of such orbits is 2m−1.

5.2.4. The third case. We can view H1(k, J [2]) as a F2-vector space. Let q denote the composition:

H1(k, Gv) = H1(k, J [2])−→

γ H1(k, SO(W ))−→

ρ H2(k, µ2) and consider the bilinear map

ϕ : H1(k, J [2])× H1(k, J [2])→ H2(k, µ2)

induced from the Weil pairing e2 : J [2]× J[2] → µ2 by using cup product. Then q is actually a quadratic refinement of ϕ [Wang, theorem 2.15] in the sense that

ϕ(a, b) = q(a + b)− q(a) − q(b) + q(0), for all a, b ∈ H1(k, J [2]).

This implies ϕ is an even bilinear form, because for all a∈ H1(k, J [2])

ϕ(a, a) = q(a + a)− q(a) − q(a) + q(0) = q(2a) − 2q(a) + q(0) = 2q(0) = 0.

Moreover, q is a quadratic form since

q(0) = 0 and q(cv) = c2q(v) for all c∈ F2 , v ∈ H1(k, J[2]).

Let m + 1 be the number of irreducible factors of f (x) in k[x]. Then L =

m+1

i=1

Ki

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where each Ki is a field extension of k. By Theorem 3 H1(k, Gv) = (L/L∗2)N=1 = (

m+1

i=1

Ki/K∗2i )N=1.

Each Ki is a local field of odd residue characteristic p. Hence, |Ki/Ki∗2| = 22 and

|(m+1

i=1

Ki/Ki∗2)N =1| = 22m. Therefore, which implies dimF2H1(k, J[2]) = 2m. More-over, the Arf invariant of the quadratic form q is 0 because there is a m dimensional F2-subspace J(k)/2J(k) which is isotropic under q (Proposition 4.2.1). Therefore, the cardinality of ker(γ) equals 2m−1(2m+ 1) (cf. [GALA, Theorem 6.2.38]) that, by Theorem 3, is the same as the number of orbits with characteristic polynomial equal to f (x).

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