We begin by recalling the definition of the signature of a non-degenerate real inner product space over R. Given a non-degenerate real inner product (non-degenerate symmetric R-bilinear map) space (V, <, >), we can choose a suitable basis under which the <, > correspond to the diagonal matrix D whose diagonal entries are±1.
Definition 5.3.1. We say (V, <, >) has signature (p, q) if there are p many 1’s and q many −1’s in the diagonal entries of D.
Thus, two real inner product spaces having the same signature are isomorphic as orthogonal spaces. In the second and third case,H1(R, GT) isomorphic to K∗/N E∗ and (L∗/L∗2)N =1 respectively, both are elementary abelian 2-group. Here we only consider the case where H1(k, GT) has maximal rank.
5.3.1. The first case. Recall that H1(R, SO(W)) represents the set of k-isomorphism classes of non-degenerate orthogonal spaces W′of dimension 2n+1 and determinant (−1)n ∈ R∗/R∗2. SinceR∗/R∗2 ={±1},
|H1(k, SO(W))| = n + 1 because the signature associated to each class must satisfy
p + q = 2n + 1 and q ≡ n (mod 2).
Now H1(R, Gv) = H1(R, SO(U)) represents the set of R-isomorphism classes of non-degenerate orthogonal spaces U′ of dimension 2n and discriminant △(v) over R. Because △(v) = ±1 , we separate our computation of |H1(R, Gv)| into two cases:
(1) △(v) = (−1)n:
This forces U′to have positive determinant so that its signature (p, q) satisfies p + q = 2n and q≡ 2 (mod 2).
Hence |H1(R, Gv)| = n + 1.
(2) △(v) = (−1)n+1:
Then U′ has negative determinant and its signature (p, q) satisfies p + q = 2n and q≡ 1 (mod 2).
Hence |H1(R, Gv)| = n.
Therefore, γ is bijective when△(v) = (−1)n and injective when △(v) = (−1)n+1. 5.3.2. The second case. To have K∗/N E∗ achieve the maximal rank, we need to have f (x) = x∏n
i=1
(x2+ ci) where ci ∈ R>0. Then
K∗/N E∗ ∼= (R∗/R>0)n ∼= (Z/2Z)n.
The real orthogonal space W then decompose into n orthogonal T -invariant planes and an orthogonal line kv′ with T v′ = 0. Indeed, by the strict real version of spectrum theorem, if S is a skew self adjoint operator on W , there is an orthogonal basis A of W such that We regard this decomposition as the spectral decomposition of S.
Note that for each i,
and
< vi1, vi2>= 1
√ci < vi1, Svi1 >= 1
√ci <−Svi1, Si1 >=− < vi2, vi1>,
and hence < vi1, vi2 >= 0, because <, > is symmetric. Therefore, the signature of each Wi is either (2, 0) or (0, 2). Let the ordered set
ω(S) :={ω(W0), ω(W1), ..., ω(Wn)}
denote the signatures of W0, W1,...,Wn and call it the signature of S.
By using the standard basis B, we see that the signature of W is (n + 1, n). Let m denote the positive integer such that n = 2m + 1 or n = 2m. If n is odd, then W0
has signature (1, 0) ad there are exactly m Wi having signature (0, 2); if n is even, then W0 has signature (0, 1) and there are exactly m Wi having signature (2, 0).
Lemma 5.3.2. Suppose S, S′ ∈ Λ2(W ) have the same characteristic polynomial f (x) = x
∏n i=1
(x2+ ci). Then S and S′ lie in the same SO(W)(R)-orbit if and only if they have the same signature.
Proof. For each g ∈ SO(W)(R) the spectral decomposition of S′ := gSg−1 is W = W0′⊕ W1′ ⊕ · · · ⊕ Wn′
where W0′ = span{gv0}, Wi′ = span{gvi1, gvi2} for 1 ⩽ i ⩽ n. Then since
< gu, gv >=< u, v >, for all u, v∈ B, we have
ω(S) := ω(S′).
Conversely , if S, S′ ∈ Λ2(W ) have the same signature and W = W0′⊕ W1′ ⊕ · · · ⊕ Wn′
is the special decomposition of S′, then because ω(Wj′) = ω(Wj), for j = 0, ..., n, we can arrange to have the corresponding basis
A′ ={v0′, v′11, v′12,· · · , vn1′ , vn2′ } satisfying
< v0, v0 >=< v0′, v0′ >, < vi1, vi1 >=< vi1′ , vi1′ > and < vi2, vi2>=< v′i2, vi2′ > .
Choose the linear map g such gv0 =±v′0, gvi1 = v′i1 and gvi2 = v′i2, 1≤ i ≤ n to have g ∈ SO(W ). Then S′ = gSg−1. Hence S and S′ are in the same SO(W )(
R)-orbit. □
Proposition 5.3.3. Let m denote the positive integer such that n = 2m + 1 or n = 2m. There are exactly (n
m
) elements in ker γ.
Proof. Let S = T . For every σ in the symmetric group Sn, let gi,σ(i) : Wi → Wσ(i)
be the isomorphism sending vi1, vi2 respectively to vσ(i)1, vσ(i)2 and let Tσ denote the linear transformation such that
Tσ |W0= T |W0 and Tσ |Wi= g−1iσ(i)◦ T |Wσ(i) ◦giσ(i). factors completely over R, say f(x) =2n+1∏
i=1
(x− ci). In this case,
H1(R, GT) ∼= (L∗/L∗2)N =1∼= ((R∗)2n+1/(R>0)2n+1)N =1 ∼= (Z/2Z)2n.
tBy spectrum theorem, S is a self-adjoint operator on W of characteristic poly-nomial f (x) if and only if there is an orthogonal basis A = {v1, v2,· · · , v2n+1,} of decomposition of S. We define the signature of S to be the ordered set
ω(S) :={ω(W1), ..., ω(W2n+1)} where ω(Wi) denote the signature of Wi.
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Lemma 5.3.4. Let S, S′ be elements of Sym2(W ) having f (x) =
2n+1∏
i=1
(x− ci) as their characteristic polynomial. Then S and S′ lie in the same SO(W )(R)-orbit if and only if ω(S) = ω(S′).
Proof. If g∈ SO(W )(R), then the decomposition of gSg−1 is W = W1′ ⊕ · · · ⊕ W2n+1′
The proof of the following is similar to that of Proposition 5.3.3.
Proposition 5.3.5. There are (2n+1
n
) elements in ker γ.
5.4. The Global field case. Here we only consider the case V = Sym2(W ).
The exact sequence where a, b, c are localization maps. Recall that the 2-Selmer group
Sel(J/k, 2) := ker(c◦ ι∗).
Proposition 5.4.1. The 2-Selmer group Sel(J/k, 2)⊂ H1(k, J [2]) lies in ker γ.
Proof. Consider the commutative diagram
H1(k, J [2]) γ > H1(k, SO(W ))
∏
v
H1(kv, J [2])
∨b
(γv)
> H1(kv, SO(W ))
∨d
where d is also the localization map. Suppose α∈ Sel(J/k, 2). Then b(α) is in the image of δ. Proposition 4.2.1 and the above diagram imply
d(γ(α)) = 0.
But Hasse-Minkowski theorem says d is injective. Hence γ(α) = 0 as desired.
□ In previous sections, we have seen that ker γ are finite in the cases where its order cam be estimated. However, in general ker γ is not always finite. We give a counterexample below.
Counterexample 5.4.2. Let k =Q, and f(x) = (x − 1)(x2n− 3).
For each d∈ Q∗/Q∗2, consider the twisted hyper elliptic curve with affine equation Cd: dy2 = f (x).
and let Jd denote the Jacobian variety. Then as Galois modules Jd[2](Qs) = J [2](Qs),
because both are generated by dη as described in Lemma 4.1.2. Furthermore, by Proposition 5.4.1, the 2-Selmer group Sel(Jd/k, 2) of H1(k, Jd[2]) = H1(k, J [2]) also lies in ker γ. The 2-Selmer groups of quadratic twists of such hyper-elliptic curve y2 = f (x) can be arbitrarily large [Chang, Theorem 5.5]. Hence ker γ does contain infinitely many elements.
6. Appendix 6.1. Special case of Galois descent.
Let V be a vector space over a field k with a fixed non-degenerate quadratic form x. Two pairs (V, x) and (V′, x′) are called k-isomorphic if there is a k-linear isomorphism
f : V → V′
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such that f (x) = x′.
Let K/k be a finite Galois extension with Galois group G,and VK = V ⊗kK be the vector space over K. The quadratic form x defines a quadratic form xK in the obvious way.We say (V, x) and (V′, x′) are K-isomorphic if (VK, xK) and (VK′ , x′K) are isomorphic. Denote by EV(K/k) the set of k-isomorphism classes of (V′, x′) that are K-isomorphic to (V, x).
Let AK be the group of K-automorphisms of (VK, xK). The group G acts on AK as follows: s∈ G acts on VK by s(x⊗ λ) = x ⊗ s(λ). Now if f : VK → VK is a linear map, put s(f )(x) = s(f (s−1(x))).
So let (V′, x′) ∈ EV(K, k) and f : VK → VK′ be a K-isomorphism.For each s∈ G,put
ps = f−1◦ s(f) = f−1◦ s ◦ f ◦ s−1.
We have ps ∈ AK. The map s 7→ ps is a 1-cocycle,and changing f gives another ps that differs from the original ps by a 1-coboundary. Hence we have defined a map
θ : EV(K/k) → H1(G, AK)
Also note that here AK is actually the orthogonal group OK(x) of the quadratic form x over K.
Proposition 6.1.1. The map θ is bijective.
Proof. To show θ is injective. Let (V1′, x′1) and (V2′, x′2) correspond to the same cocycle ps. And let f1,f2 be the corresponding K-isomorphisms. Then
f1−1◦ s(f1) = f2−1◦ s(f2).
Hence s(f2f1−1) = f2f1−1, and the map f2f1−1 is a k-isomorphism from (V1′, x′1) to (V2′, x′2).Thus θ is injective.
To show θ is surjective.Let ps be a 1-cocycle of G with values in AK. Because AK ⊂ GL(VK) and H1(G, GL(VK)) = 1, there is a K-automorphism f of VK such that
ps= f−1◦ s(f)
for all s∈ G.And put x′ = f (x),x′ is defined over k. Indeed, for all s∈ G,we have s(x′) = s(f )(s(x)) = s(f )(x) = f◦ ps(x) = f (x) = x′.
Hence (V, x′)∈ EV(K/k) and θ((V, x′)) is equal to the class of ps. □
Now we let EV′ (K/k) be the set of k-isomorphism classes of (V′, x′) that having the same discriminant as (V, x) which are K-isomorphic to (V, x) and AK be SOK(x).
Then for (V′, x′)∈ EV′ (K/k) and f : VK → VK′ be a K isomorphism.We have seen before that ps ∈ OK(x). In this case, since det(f ) is defined over k,
det(ps) = det(f−1) det(s(f )) = det(f−1)[s· det(f)] = det(f−1) det(f ) = 1.
Thus ps ∈ SOK(x) and we can define θ : EV′ (K/k) → H1(G, SOK(x)) as before.
Now the proof of bijectivity of θ is almost the same as the proof of Proposition 6.1.1.
We only need to replace the statement ”AK ⊂ GLK(V )” by ”AK ⊂ SLK(V )” In the proof of surjectivity of θ.
Hence When K = ks, U defined in Lemma 3.1.2, we have H1(k, SO(U ))={k-isomorphism classes of non-degenerate orthogonal spaces U′ of dimension 2n with discriminant d over k}.
And When K = ks, W defined in Section 2, we have H1(k, SO(W ))= {k-isomorphism classes of non-degenerate orthogonal spaces of dimension 2n + 1 with determinant (−1)n over k}.
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