The Arithmetic of Quaternion Algebra

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46 CHAPTER 3. QUATERNION ALGEBRA OVER A GLOBAL FIELD norm na : H × A → A× too. Suppose that G ′ is a group with unit 1, and for every place v ∈ V we have defined the homomorphism fv Gv → G′ such that fv(Cv) = 1 p.p.. We can then define in G ′ the product fA(x) = � fv(xv), if x = (xv) ∈ GA. v∈V EXAMPLE. We can define the norm NA, the modulus || · ||A in H × A , and A× too. NOTATIONS. It is convenient to consider Gv as that which has been embedded in GA and identifies canonically with � w�=v 1w × Gv, where 1w is the unit of Gw, w ∈ V . When GA is the adele group of an algebraic group defined over K, the group GK then is the group of points in G with its value in K. For every place v ∈ V we choose an inclusion of GK in Gv denoted by iv. For almost every place iv(GK) ⊃ Cv, then the mapping � v∈V iv defines an inclusion of GK in GA. We put X = XK = H or K, and Yv = Ov or Rv,p.p.. Quasi-characters. Recall that a quasi-character of a locally compact group is a continuous homomorphism of the group in C × . Let ψA be a quasi-character of GA. By restricting to Gv it defines a quasi-character ψv of Gv. We have naturally the relation ψA = � ψv(xv)if x = (xv) ∈ GA. v∈V For the convergence of the product in C × if and only if ψv(Cv) = 1, p.p.. In fact, if this property is not satisfied, we then could find cv ∈ Cv such that |ψv(cv)−1| > 1/2, p.p. and the product would not be convergent for the elements x such that xv = cvp.p.. Thus we have proved the following theorem. Lemma 3.1.2. The mapping ψA ↦→ (ψv) is an isomorphism of the group of quasi-characters of GA and the group {(ψv)} such that ψv is the quasi-character of Gv, and ψv(Cv) = 1, p.p. We can apply the local results of last chapter to the quasi-characters of XA. Let ψA = � v∈V ψv be the product of the canonical local character (exercise II.4.1); the product is well-defined because of ψv(Yv) = 1, p.p.. The above lemma shows that every character of XA is of the form x ↦→ ψa(ax), where a = (av) ∈ Xv, and av ∈ Ker(ψv), p.p.. Since Ker(ψv) = Yv, p.p., it follows that a ∈ A. Therefore, XA is self-dual. Let us turn firstly to the case where X = Q or Fp(T ) is a prime field, we shall verify that ψA is trivial on XK, and the dual of XA/XK is XK, cf. Weil [1]. Proposition 3.1.3. XA is self-dual, and XK is the dual of XA/XK. We are going now to give the principal theorems of adeles XA and X × A . These theorems are still valid if X is a central simple algebra over K. The proof in the special case treated by us gives a good idea of the proof in the general case (Weil [1]). Theorem 3.1.4. (Fundamental Theorem) Adeles. 1) XK is discrete in XA and XA/XK is compact.
3.1. ADELES 47 2)(theorem of approximation). For every place v, XK + Xv is dense in XA. Ideals. 1)X × K is discrete in X× A . 2)(product formula) The modulus equals 1 on X × K 3) (Fujisaki’s theorem [1]). If X is a field, the image in X × A /X× K of the set Y = {x ∈ X × A |0 < m ≤ ||x||A ≤ M, n, m is real} is compact. 4) For every place v or infinity if K is a number field, there exists a compact set C of XA such that X × A = X× KX× v C. Proof. Adeles.1) We prove that XK is discrete in XA. It suffices to verify that 0 is not an accumulative point of XK. In a sufficiently small neighborhood of 0, the only possible elements of XK are the integers for every finite places: hence a finite number of it if K is a function field, and belonging to Z if X = Q. In these two cases, it is clear, 0 is impossible to be an accumulative point. We have the same result for every X, since X is a vector space of finite dimension over Q or a function field. The dual group of a discrete group is compact, and hence XA/XK, dual to XK, is compact. 2)Theorem of approximation. We show that a character of XA being trivial on XK is determined by its restriction to Xv. In fact, a character being trivial on XK and on Xv has the form x ↦→ ψA(ax) where ψA is the canonical character with a in XK and ψv(axv) = 1 for every xv ∈ Xv. It implies a = 0, and the character ψA(ax) is trivial. Ideals. 1) Prove X × is discrete in X× it is sufficient to prove that 1 is not an K A accumulative point. A series of elements (xn) of X × converges to 1 if and only if K (xn) and x−1 n ) converge to 1. It suffices that (xn) converges to 1, hence that 1 is an accumulative point of XK in XA. It is impossible according to the theorem of adeles. Product formula. Let x be an element of XK; For proving the modulus of x equals equals 1, it is necessary and sufficient to verify the volume of an mea- surable set Y ⊂ XA equals the volume xY for an arbitrary Haar measure. We have � vol(xY ) = ϕ(x −1 � y)dy = � ϕ(zx −1 y)dy · XA � = � XK\XA z∈XK XK\XA z∈XK ϕ(zy)dy · = vol(Y ), where ϕ is the characteristic function of Y , and dy · is the measure on XK\XA induced by the compatibility with dy and the discrete measure on XK. has the form Fusijaki’s theorem. A compact set of X × A {x ∈ X × A |(x, x−1 ) ∈ C × C ′ } for two compact sets C and C ′ of XA. For element x of Y , i.e. 0 < m ≤ ||x|| ≤ M, we look for an element of X × K such that xa ∈ C and a−1 x −1 ∈ C ′ . We choose in XA a compact set C ′′ of volume sufficiently large, greater than vol(XA/XK)Sup(m −1 , M)
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The Arithmetic of Quaternion Algebra

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