NOTES ON TATE S THESIS

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1 NOTES ON TATE S THESIS YICHAO TIAN The aim of this short note is to explain Tate s thesis [Ta50] on the harmonic analysis on Adèles and Idèles, the functional equations of Dedekind Zeta functions and Hecke L-series. or general reference on adèles and idèles, we refer the reader to [We74]. 1. Local Theory 1.1. Let k be a local field of characteristic 0, i.e. R, C or a finite extension of Q p. If k is p-adic, we denote by O k the ring of integers in k, p O the maximal ideal, and ϖ p a uniformizer of O. If a is a fractional ideal of O, we denote by Na Q the norm of a. So if a O is an ideal, we have Na = O/a. Let : k R 0 be the normalized absolute value on k, i.e. for x k, we have x R if k = R; x = x 2 C if k = C; N(p) ordϖ(x) if k is p-adic and x = uϖ ordϖ(x) with u O. We denote by k + the additive group of k. Consider the unitary character ψ : k + C defined as follows: e 2πix if k = R; (1.1.1) ψ(x) = e 2πi(x+ x) if k = C; e 2πiλ(Tr k/qp (x)) if k is p-adic, where λ( ) means the decimal part of a p-adic number. or any ξ k, we note by ψ ξ the additive character x ψ(xξ) of k +. Note that if k is non-archimedean, ψ(x) = 1 if and only if x d 1, where d is the different of k over Q p, i.e. x d 1 Tr k/qp (xy) Z p y O. Proposition 1.2. The map Ψ : ξ ψ ξ defines an isomorphism of topological groups k + k +, where k + denotes the group of unitary characters of k +. Proof. If k = R or C, this is well known in classical ourier analysis. We assume here k is non-archimedean. (1) It s clear that Ψ is a homomorphism of groups. We show first that Ψ is continuous (at 0). If ξ p m, then ψ ξ is trivial on d 1 p m. Since the subsets U m = {χ k + χ is trivial on d 1 p m } form a fundamental system of open neighborhoods of 0 in k +, the continuity of Ψ follows immediately. 1

2 2 YICHAO TIAN (2) Next, we show that Ψ : k Ψ(k) k is homoemorphism of k onto its image. We need to check that if (x n ) n 1 k is a sequence such that ψ xn 1 uniformly for all compact subsets of k, then x n converges to 1 in k. Consider the compact open subgroup p m for m Z. Then for any 1/2 > ɛ > 0, there exists an integer N > 0 such that ψ(x n z) 1 < ɛ for all n > N and z p m. But x n p m is a subgroup and the open ball B(1, ɛ) C contains no subgroup of S 1. Hence we have ψ(x n z) = 1 for all z p m, so x n d 1 p m. (3) The image of Ψ is dense in k. Let H be the image of Ψ, and H k be its closure. Then we have H = {x k k χ(x) = 1, χ H} = {x k ψ(xξ) = 1, ξ k} = {0} Hence, we have H = k. (4) The proof of the Proposition will be complete by the Lemma 1.3 below. Lemma 1.3. Let G be a locally compact topological group, H G be a locally compact subgroup. Then H is closed in G. Proof. Let h n be a sequence in H that converges to g G. We need to prove that g H. Let (U r ) r 0 be a fundamental system of compact neighborhoods of 0. We have r 0 U r = {0}. Then for any r, there exists an integer N r > 0 such that h n g + U r for all n N r. Up to modifying U r, we may assume h n h m H U r 1 for any n, m N r. Note that H U r 1 is also compact by the local compactness of H. Up to replacing {U r } r 0 by a subsequence, we may choose m r for each integer r such that By compactness, the intersection h mr+1 + U r H h mr + U r 1 H. r 1 (h mr + U r 1 H) must contain an element h H. It s easy to see that h = g, since r 0 U r = {0} Now we choose a Haar measure dx on k as follows. If k = R, we take dx to be the usual Lebesgue measure on R; if k = C, we take dx to be twice of the usual Lebesgue measure on C; and if k is non-archimedean, we normalize the measure by O dx = (Nd) 1 2. Let L 1 (k, C) be the space of complex valued absolutely integrable functions on k. or f L 1 (k, C), we define the ourier transform of f to be (1.4.1) ˆf(ξ) = f(x)ψ(xξ)dx. Let S(k) be the space of Schwartz functions on k, i.e. k S(R) = {f C (R) n, m N, x n dm f is bounded}; dxm

3 NOTES ON TATE S THESIS 3 we have a similar definition for k = C; and if k is p-adic, S(k) consists of locally constant and compactly supported functions on k. In all these cases, the space S(k) is dense in L 1 (k, C). Proposition 1.5. The map f ˆf preserves S(k), and we have ˆf(x) = f( x) for any f S(k). The following lemma will be useful in the sequels. Lemma 1.6. Assume k is non-archimedean. The local ourier transform of f = 1 a+p l, the characteristic function of the set a + p l, is (1.6.1) ˆf(x) = ψ(ax)(nd) 1 2 (Np) l 1 d 1 p l. In particular, we have ˆf S(k). Proof. By definition, we have ˆf(x) = ψ(xy)dy = ψ(ax) ψ(xy)dy. a+p l p l The lemma follows immediately from { (Nd) 1 2 (Np) l if x d 1 p l ψ(xy)dy = p l 0 otherwise. Proof of 1.5. If k is archimedean, this is well-known in classical analysis. Consider here the non-archimedean case. Since any compactly supported locally constant function on k is a linear combination of functions 1 a+p l. We may assume thus f = 1 a+p l. The first part of the proposition follows from the previous lemma. or the second part, we have ˆf(x) = ˆf(y)ψ(xy)dy = (Nd) 1 2 (Np) l ψ((x + a)y)dy d 1 p l k = (Nd) 1 2 (Np) l (Nd) 1 2 (Np) ord ϖ(d)+l 1 a+p l = 1 a+p l. In the third equality above, we have used (1.6.1) with l replaced by ord ϖ (d) l and x replaced by x + a. Now it s clear that ˆf(x) = f( x) Now consider the multiplicative group k, and put U = {x k x = 1}. So U = {±1} if k = R, U = S 1 is the group of unit circle if k = C, and U = O if k is p-adic. We have { k R + if k = R, C; /U = Z if k is p-adic.

4 4 YICHAO TIAN Recall that a quasi-character of k is a continuous homomorphism χ : k C. We say χ is a (unitary) character if χ(x) = 1 for all x k, and χ is unramified if χ U is trivial. So χ is unramified if and only if there is s C such that χ(x) = x s. Note that such an s is determined by χ if k = R or C, and determined up to 2πi/log(Np) if k is p-adic. Lemma 1.8. or any quasi-character χ of k, there exists a unique unitary character χ 0 of k such that χ = χ 0 s. Proof. or any x k, one can write uniquely x = xρ where x U and ρ R + if k = R or C, and ρ ϖ Z if k is non-archimedean. We define χ 0 as χ 0 (x) = (χ U )( x). One checks easily that the quasi-character χ/χ 0 is unramified. Let χ be a quasi-character of k, and s C be the number appearing in the Lemma above. Note that σ(χ) = R(s) is uniquely determined by χ, and we call it the exponent of χ. Let ν Z 0 be the minimal integer such that χ 1+p ν is trivial. We call the ideal f χ = p ν conductor of χ. So the conductor of χ is O if and only if χ is unramified We choose the Haar measure on k to be d x = δ(k)dx/ x, where { 1 if k = R, C; (1.9.1) δ(k) = if k is non-archimedean. Np Np 1 If k is non-archimedean, the factor δ(k) is justified by the fact that dx = (Nd) 1 2. U Definition or f S(k), we put ζ(f, χ) = f(x)χ(x) d x, k which converges for any quasi-character χ with σ(χ) > 0. We call ζ(f, χ) the local zeta function associated with f (in quasi-characters). Proposition or any f, g S(k), we have ζ(f, χ)ζ(ĝ, ˆχ) = ζ( ˆf, ˆχ)ζ(g, χ), where ˆf, ĝ are ourier transforms of f and g, and ˆχ = χ 1 for any quasi-character χ with 0 < σ(χ) < 1. Proof. ( ) ζ(f, χ)ζ(ĝ, ˆχ) = f(x)ĝ(xy) x d x χ(y 1 ) y d y k k ( ) = δ(k) f(x)g(z)ψ(xyz)dzdx χ(y 1 ) y d y. k k k To finish the proof of the Proposition, it suffices to note that the expression above is symmetric for f and g.

5 NOTES ON TATE S THESIS 5 We endow the set of quasi-characters with a structure of complex manifold such that for any fixed quasi-character χ the map s χ s induces an isomorphism of complex manifolds from C to a connected component of the set of quasi-characters. Theorem or any f S(k), the function ζ(f, χ) can be continued to a meromorphic function on the space of all quasi-characters. Moreover, it satisfies the functional equation (1.12.1) ζ(f, χ) = ρ(χ)ζ( ˆf, ˆχ), where ρ(χ) is a meromorphic function of χ independent of f given as follows: (1) If k = R, then χ(x) = x s or χ(x) = sgn(x) x s for some s C. We have ρ( s ) = 2 1 s π s cos( πs 2 )Γ(s), ρ(sgn s ) = i2 1 s π s sin( πs 2 )Γ(s). (2) If k = C, then there exists n Z and s C such that χ = χ n s where χ n is the unitary character χ n (re iθ ) = e inθ. We have ρ(χ n s ) = i n (2π) 1 s Γ(s + n 2 ) (2π) s Γ(1 s + n 2 ). (3) Assume k is p-adic. If χ is unramified, then ρ( s ) = (Nd) s (Np) s 1 1 Np s. If χ = χ 0 s is ramified, where χ 0 is unitary with χ 0 (ϖ) = 1 as in Lemma 1.8, then one has ρ(χ 0 s ) = N(df χ ) s 1 2 ρ0 (χ 0 ) with ρ 0 (χ 0 ) = N(f χ ) 1 x 2 χ 0 ( x)ψ( ϖ ordϖ(dfχ) ) where x runs over a set of representatives of O /(1 + f χ ). x Proof. By Proposition 1.11, the function ρ(χ) = ζ(f,χ) is independent of f. This proves ζ( ˆf,ˆχ) the functional equation (1.12.1). Note that ζ(f, χ) is well defined if σ(χ) > 0, and ζ( ˆf, ˆχ) is well defined if σ(χ) < 1. Therefore, once we show that ρ(χ) is meromorphic as in the statement, it will follow from the functional equation (1.12.1) that ζ(f, χ) can be continued to a meromorphic function in χ. It remains to compute ρ(χ) by choosing special functions f S(k). (1) Assume k = R. If χ = s, we choose f = e πx2. We have ζ(f, s ) = On the other hand, (1.12.2) ˆf(y) = R e πx2 R x s d x = 2 + e π(x2 +2ixy) dx = e y2 0 e πx2 x s 1 dx = π s 2 Γ( s 2 ). R e π(x+yi)2 dx.

6 6 YICHAO TIAN Using the well-known fact that e π(x+yi)2 dx = R R e πx2 dx = 1, we get ˆf = f. Hence, we have ζ( ˆf, 1 s ) = π 1 s 2 Γ( 1 s 2 ) and ρ( s ) = π s 2 Γ( s 2 ) π 1 s 2 Γ( 1 s 2 ) = π s π Γ( s s+1 2 )Γ( Γ( 1 s 2 2 ) )Γ( 1+s 2 ) Now the formula for ρ( s ) follows from the properties of Gamma functions Γ( s 2 )Γ(s ) = 21 s πγ(s), Γ( 1 s 2 )Γ(1 + s 2 ) = π sin( π(1+s) If χ = sgn s, we take f = xe πx2. A similar computation shows that ζ(f, sgn s ) = π s+1 2 Γ( s ). 2 ). Taking derivatives with respect to y in (1.12.2), we get ˆf = if. So we have Therefore, we get ζ( ˆf, sgn 1 s ) = iπ s 2 1 Γ(1 s 2 ). s+1 ρ(sgn s π 2 Γ( s+1 2 ) = ) s+1 iπ s 2 1 Γ(1 s 2 ) = iπ s Γ( 2 π )Γ( s 2 ) Γ(1 s 2 )Γ( s 2 ) = i2 1 s π s sin( πs 2 )Γ(s). (2) Assume k = C. If χ = s, we take f(z) = e π(z z). The local zeta function associated with f is ζ(f, s ) = e πz z (z z) s d z C = 2π + = 4π θ=0 r= e πr2 2s 2rdrdθ r r 2 e πr2 r 2s 1 dr = 4π t s 1 2 e πt dt 0 2 t = 2π 1 s Γ(s). (set t = r 2 )

7 The ourier transform of f is (1.12.3) ˆf(z) = e πw w e 2πi(zw+ z w) dw = 2 C + + = 2e 4π(x2 +y 2 ) = 2f(2z). + NOTES ON TATE S THESIS 7 e π(u2 +v 2) e 4πi(ux vy) dudv e π(u+2ix)2 du + (put z = x + iy, w = u + iv) e π(v 2iy)2 dv Therefore, one has ζ( ˆf, 1 s ) = 2 2s 1 ζ(f, 1 s ) = 2 2s π s Γ(1 s), thus ρ( s ) = (2π) 1 2s Γ(s) Γ(1 s). Let n 1 and χ = χ n s. We put f n = z n e π(z z). We compute first the local zeta function of f n : (1.12.4) ζ(f n, χ n s ) = z n e π(z z) χ n (z)(z z) s d z C = 2π + = 4π = 4π θ=0 r= e πr2 2s+n 2rdrdθ r r 2 e πr2 r 2s+n 1 dr e πt n 1 s+ t 2 = 2π 1 (s+ n 2 ) Γ(s + n 2 ). dt 2 t To find the ourier transform of f n, we consider the equality (1.12.3) 2e 4π(z z) = e π(w w) e 2πi(zw+ z w) dw. Regarding z and z as independent variables and applying n z, we get n 2( 2i z) n e 4πz z = w n e π(w w) e 2πi(zw+ z w) dw, that is, ˆf n (z) = 2f n (2iz). A similar computation as (1.12.4) shows that ζ( f ˆ n (z), ˆχ) = ζ(2f n (2iz), χ n 1 s ) = ( i) n 2 2s π s n n 2 Γ(s + 2 ). Therefore, we get C C ρ(χ n s ) = 2π1 (s+ n 2 ) Γ(s + n 2 ) ( i) n 2 2s π s n 2 Γ(s + n 2 ) = in (2π) 1 2s Γ(s + n 2 ) Γ( n s). The formulae for ρ(χ n s ) can be proved in the same way by choosing f = f n.

8 8 YICHAO TIAN (3) Assume k is p-adic. Consider first the case χ = s. We take f = 1 O. In the proof of Proposition 1.5, we have seen that ˆf = (Nd) d 1. We have ζ(f, χ) = x s d x. As O {0} = + n=0 ϖn O, it follows that + ζ(f, χ) = (Np) ns n=0 O {0} d x = (Nd) 1 2 O Similarly, using d 1 {0} = + n= ord ϖ(d) ϖn O, one obtains ζ( ˆf, ˆχ) = (Nd) 1 2 x 1 s d x = (Nd) 1 2 d 1 {0} + n= ord ϖ(d) = (Nd) 1 (Np) ordϖ(d)(1 s) + = (Nd) s 1 1 Np s (Np) s (Np) n(s 1) O d x n=0 Np n(s 1) The formula for ρ( s ) follows immediately. Now consider the case χ = χ 0 s with χ 0 ramified, unitary and χ 0 (ϖ) = 1. We take x f(x) = ψ( ϖ ordϖ(dfχ) )1 O. The local zeta function of f is x ζ(f, χ) = ψ( ϖ ordϖ(dfχ) )χ 0(x) x s d x = O {0} + n=0 (Np) O ns xϖ n ψ( ϖ ordϖ(dfχ) )χ 0(x)d x We claim that xϖ n (1.12.5) ψ( O ϖ ordϖ(dfχ) )χ 0(x)d x = 0 for n 1. Consider first the case n ord ϖ (f χ ). We have xϖ n ψ( ϖ ordϖ(dfχ) ) = 1 as xϖ n ϖ ordϖ(dfχ) d 1. If S is a set of representatives of O /(1 + f χ ), the integral above is equal to χ 0 (x)d x = ( d x ) χ 0 (x) = 0. O 1+f χ x S

9 NOTES ON TATE S THESIS 9 Assume 0 n ord ϖ (f χ ) 1. or any y 1 + p n f χ, we have xyϖ n xϖ ψ( ) = ψ( ϖordϖ(dfχ) ϖ n ). ordϖ(dfχ) Therefore, if S n S denotes a subset of representatives of O /(1 + p n f χ ), we get xϖ n ψ( O ϖ ordϖ(dfχ) )χ 0(x)d x = ( d x ) xϖ n χ 0 (x)ψ( 1+f χ ϖ ordϖ(dfχ) ) x S = ( d x ) xϖ n χ 0 (x)ψ( 1+f χ ϖ ordϖ(dfχ) ) x S n y where y runs over a set of representatives of (1 + p n f χ )/(1 + f χ ). Note that { 0 if 1 n ord ϖ (f χ ), χ 0 (y) = 1 if n = 0. y x S χ 0 (y), This proves the claim. It follows that (1.12.6) ζ(f, χ) = ( d x ) x χ 0 (x)ψ( 1+f χ ϖ ordϖ(dfχ) ) = χ 0( 1)( d x)(nf χ ) 1 2 ρ0 (χ 0 ), 1+f χ where we have used the definition of ρ 0 in the last step. As in the proof of 1.5, the ourier transform of f is y ˆf(x) = ψ( )ψ(xy)dy O ϖordϖ(dfχ) 1 = ψ(y(x + ))dy ϖordϖ(dfχ) O = (Nd) ϖ ordϖ(dfχ) +d 1. We get the local zeta function of ˆf ζ( ˆf, ˆχ) = (Nd) 1 2 x 1 s χ 1 ϖ ordϖ(dfχ) 0 (x)d x +d 1 = (Nd) 1 2 (Np) ord ϖ(df χ)(1 s) ϖ ordϖ(dfχ) (1+f χ) Since χ 1 0 ( ϖordϖ(dfχ) (1 + y)) = χ 0 ( 1) for any y f χ, we get ζ( ˆf, ˆχ) = χ 0 ( 1)(Nd) 1 2 N(dfχ ) 1 s( 1+f χ d x ). χ 1 0 (x)d x It thus follows that ρ(χ 0 s ) = ζ(f, χ 0 s ) ζ( ˆf, χ s ) = N(df χ) s 1 2 ρ0 (χ 0 ).

10 10 YICHAO TIAN Remark The number ρ 0 (χ 0 ) in (3) is a generalization of (normalized) Gauss sum. By the same method as the classical case, we can show that ρ 0 (χ 0 ) = 1. In general, it s an interesting and difficult problem to find the exact argument of ρ 0 (χ 0 ). 2. Global Theory Let be a number field, O be its ring of integers. Let Σ be the set of all places of, and Σ f Σ (resp. Σ Σ) be the subset of non-archimedean (resp. archimedean) places. or v Σ, we denote by v the completion of at v. Let dx v be the self-dual Haar measure on v defined in 1.4. If v is finite, we denote by O v the ring of integers of v, by p v the maximal ideal of O v, and we fix a uniformizer ϖ v p v. Let A be the adèle ring of, i.e. the subring of v Σ v consisting of elements x = (x v ) v with x v O v for almost all v, and A,f be the ring of finite adèles. We choose the Haar measure on A as dx = v dx v. It induces a quotient Haar measure on A /. Lemma 2.1. Under the notation above, we have A / dx = 1. Proof. By Chinese reminders theorem, we have A = + v Σ f O v v Σ v. We get thus an isomorphism A / ( O v v )/O. v Σ f v Σ Hence we have dx = dx v A / v Σ O v ( dx v f v Σ v)/o v Σ = (Nd v ) 1 2 1/2, v Σ f where d v denotes the different of v and is the discriminant of. If d denotes the different of /Q, then the lemma follows easily from the product formula: = Nd = Nd v. v Σ f or v Σ, let ψ v be the additive character of the local field v defined in (1.1.1). It s easy to check that ψ = v Σ ψ v is trivial on additive group, therefore it defines a character of the quotient A /. We call it the basic character of A / (or A ). or any ξ A, let ψ ξ : A C be the character given by x ψ(xξ). Proposition 2.2. The map Ψ : ξ ψ ξ defines an isomorphism between A and its topological dual Â. Moreover ψ ξ is a character of A / if and only if ξ, i.e. ξ ψ ξ gives rise to an isomorphism of topological groups  /. Proof. The proof is similar to that of Proposition 1.2. One checks easily that Ψ is continuous and injective, and Ψ induces a homeomorphism of A onto its image. Conversely, let ψ : A C be a continuous character. The restriction ψ v = ψ v to the v-th local

11 NOTES ON TATE S THESIS 11 component defines a continuous character of v. By Proposition 1.2, there exists ξ v v such that ψ v = ψ v (ξ v ). Since ψ is continuous, there exists an open neighborhood v S U v v / S O v of 0 such that its image under ψ lies in B(1, 1/2) C. As B(1, 1/2) contains no non-trivial subgroups of S 1, we see that for any v / S, we have ξ v O v. This shows that ξ = (ξ v ) v Σ A, and ψ = ψ ξ. This shows that Ψ : A Â is a bijective continuous homomorphism of topological groups. To conclude that Ψ is an isomorphism, we need to show that if ξ n A is a sequence such that ψ ξn 1 in Â, we have ξ n 0 in A as n +. Actually, for any compact subset U v v with U v = O v for almost all v and any ɛ > 0, we have ψ ξn 1 < ɛ for n sufficiently large. By Proposition v Uv 1.2, for any finite subset S Σ containing Σ, we can take (U v ) v S sufficiently large and U v = O v for v / S such that ξ n v < ɛ for v S and ξ n O v for v / S. This means that ξ n 0 in A. or the second part, let Γ A be the subgroup such that Ψ(Γ) Â consists of all characters trivial on. It s clear that Γ since ψ is trivial on. To show that Γ =, we consider first the case = Q. Let γ Γ. Since A Q = Q + ( 1 2, 1 2 ] p Z p, we can write γ = b + c, where b Q, c ( 1/2, 1/2] and c p Z p for all primes p. Then we have 1 = ψ γ (1) = ψ(γ) = ψ(b + c) = ψ(c) = e 2πic. Hence we have c = 0. Moreover, for any prime p and any integer n 0, we deduce from 1 = ψ γ ( 1 p n ) = ψ( 1 cp (b + c)) = e2πiλ( p n ) pn that c p p n Z p, i.e. we have c p = 0. This shows γ = b, and hence Γ = Q. In the general case, we note that the basic character of A is the composition of that on A Q with the trace map Tr /Q : A A Q. The following lemma will conclude the proof. Lemma 2.3. Let x = (x v ) v Σ A such that Tr /Q (xy) Q A Q for all y. Then we have x. Proof. Let (e i ) 1 i d be a basis of /Q, and (e i ) 1 i d be the dual basis with respect to the perfect pairing Q given by (x, y) Tr /Q (xy). or any place p of Q, we have a canonical isomorphism of Q p -algebras Q p v p v. We put x p = (x v ) v p v p v. Then we can write x p = d i=1 a p,ie i with a p,i Q p. As Tr /Q (xe i ) Q A Q for any i, we deduce that a p,i Q and it s independent of p. This shows that x. Let S(A ) be the space of Schwartz functions on A, i.e. the space of finite linear combinations of functions on A of the form f = v f v, where f v S( v ) and f v = 1 Ov for almost all v. or any f S(A ), we define the ourier transform of f to be (2.3.1) ˆf(ξ) = A f(x)ψ(xξ)dx.

12 12 YICHAO TIAN Proposition 2.4. (a) The ourier transform f ˆf preserves the space S(A ), and ˆf(x) = f( x). (b) If f = v f v with f v S( v ) and f v = 1 Ov for almost all v. Then ˆf = v ˆfv, where ˆf v is the local ourier transform (1.4.1) of f v. (c) or any f S(A), the infinite sum x f(x) converges, and we have the Poisson formulae (2.4.1) f(x) = ˆf(ξ). x ξ Proof. Statement (a) is a direct consequence of (b), which in turn follows from the local computations in the proof of 1.5. Now we start to prove (c). We may assume f = v f v with f v S( v ) and f v = 1 Ov for almost all v. Then there exists an open compact subgroup U A f such that Supp(f) U v Σ v. Put O U = (U v Σ v ). This is a lattice in. Each individual term in the summation x f(x) is non-zero only if x O U. Write f = f f, where f = v Σf f v and f = v Σ f v. Then there exists a constant C > 0 such that f (x) < C for all x U. Hence, we have f(x) = f(x) < C f (x). x x O U x O U By classical analysis, the sum on the right hand side is convergent. This proves the first part of (c). It remains to show Poisson s summation formula (2.4.1). Consider the function g(x) = y f(x + y), which converges for any x A by the first part of (c). As g(x) is invariant under translation of, we regard g(x) as a function on A /. Its ourier transform of g(x) is ĝ(ξ) = g(x)ψ(xξ)dx (for ξ ) A / = f(x)ψ(xξ)dx = ˆf(ξ). A By the ourier inverse formulae (a), we have g(x) = ξ ĝ(ξ)ψ( xξ). The formulae (2.4.1) follows by setting x = Let I = A be the multiplicative group of idèles of, i.e. the subgroup of v Σ v consisting of elements x = (x v ) v with x v O v for almost all v, and I 1 be the subgroup of I of idèles with norm 1. The diagonal embedding I 1 identifies with a discrete subgroup of I 1 for the induced restricted product topology on I 1. A fundamental theorem in the theory of idèles says that the quotient I 1 / is compact [We74, IV 4 Thm.6]. We consider the Haar measure d x = v d x v on I, where d x v is the local Haar measure on v considered in 1.9. We use the same notation for the induced Haar measures on I 1 and I 1 /.

13 NOTES ON TATE S THESIS 13 Proposition 2.6. Under the notation above, we have Vol(I 1 / ) = d x = 2r1 (2π) r2 hr I 1 / 1/2 w, where r 1 (resp. r 2 ) is the number of real places (resp. complex places) of, h is the class number of, is the discriminant, R is the regulator, and w denotes the number of roots of unity in. Proof. Note first that Vol(I 1 / ) is finite, since I 1 / is compact. or each x = (x v ) v Σ I, we denote by Div(x) = v Σ f pv ordv(xv) be the fractional ideal associated with x. Then Div induces a short exact sequence 0 ( v Σ f O v (R ) r1 (C ) r2 ) I Cl 0, where Cl denotes the class group of. Let Ω be the subgroup of (R ) r 1 (C ) r 2 with product of absolute values r 1 i=1 x i r 2 i=1 z i C = 1. The the exact sequence above induces a similar exact sequence 0 ( O v Ω) I 1 Cl 0. v Σ f Therefore, one gets d x = h I 1 / ( v O v Ω)/( d x. v O v Ω) Let U denote the group of units of. We have ( v O v Ω) = U, and hence ( v O v Ω)/ ( = ( dx v O v Ω) O v v ) d x = Nd 1 2 v d x. Ω/U v Σ Ω/U f v Σ f In view of the product formula v Σ f Nd 1 2 = 1 2, to complete the proof, it suffices to prove that (2.6.1) d x = 2r1 (2π) r2 R. Ω/U w Consider the map Log : (R ) r 1 (C ) r 2 R r 1+r 2 ((x i ) 1 i r1, (z j ) 1 j r2 ) ((log x i ) 1 i r1, (log z j 2 ) 1 j r2 ). Let S 1 be the unit circle subgroup of C, and V be the subspace of R r 1+r 2 defined by the linear equation r 1 i=1 x i + r 2 j=1 y j = 0. Then the map Log induces a short exact sequence of abelian groups 0 {±1} r 1 (S 1 ) r 2 Ω Log V 0.

14 14 YICHAO TIAN If µ denotes the group of roots of unity in, we have ({±1} r 1 (S 1 ) r 2 ) U = µ. Therefore, one obtains d x = ( d x ) ( d x ). Ω/U {±1} r 1 (S 1 ) r 2 /µ V/Log(U ) By the definition of the Haar measure on I, the induced measure on Ω (R ) r 1 (C ) r 2 is determined as follows. On each copy of R, the measure is given by d x = dx x, where dx is the usual Lebesgue measure on R; on each copy of C, the measure is given by d dx dy z = 2 z 2 = d(r2 ) r 2 dθ, where z = x + iy = re iθ. Therefore, the Haar measure on I induces the usual Lebesgue measure on Log(Ω) = V, and the measure r 2 j=1 dθ j on (S 1 ) r 2. 1 It follows that d x = 2r1 (2π) r2. {±1} r 1 (S 1 ) r 2 /µ w By the definition of the regulator, we have R = V/Log(U ) dx. Now the formula (2.6.1) follows immediately. This finished the proof A Hecke character (or Grössencharacter) of is a continuous homomorphism χ : I / C. We say χ is unramified if there exists a complex number s C such that χ(x) = x s. We denote by X the set of Hecke characters of. We equip X with a structure of Riemann surface such that for each fixed character χ, the map s χ s is a local isomorphism of C into X. Now we choose a splitting I / = I 1 / R + of the norm map : I / R +. or every Hecke character χ, we put χ 0 = χ I 1 / and denote still by χ 0 its extension to I / by requiring χ 0 is trivial on the chosen complement R + of I1 /. Note that χ 0 is necessarily unitary since I 1 / is compact and χ/χ 0 is unramified, i.e. χ = χ 0 s with s C. We put σ(χ) = R(s), which is independent of the choice of the splitting. or v Σ, we put χ v = χ v. The local component χ v is unramified for almost all v. Definition 2.8. Let f S(A ) and χ be a Hecke character of. We define the zeta function of f at χ to be ζ(f, χ) = f(x)χ(x)d x. I Lemma 2.9. Let f S(A ) and χ X. absolutely for σ(χ) > 1. Then the zeta function ζ(f, χ) converges 1 Note that the finiteness of Vol(I 1 / ) implies that V /Log(U ) d x is finite, and hence Log(U ) V is a lattice. This actually gives another proof of Dirichlet s theorem that U has rank r 1 + r 2 1.

15 NOTES ON TATE S THESIS 15 Proof. We may assume f = v Σ f v with f v = 1 Ov for almost all v Σ f, and χ = χ 0 s where χ 0 : I 1 / S 1 unitary. By definition, we have an Euler product ζ(f, χ) = v Σ ζ(f v, χ v ), where ζ(f v, χ v ) is the local zeta function defined in We have seen in the proof of Theorem 1.12(3) that ζ(1 Ov, s v) = (Nd v ) Np s. v Thus there exists a finite subset S of places such that ζ(f, χ 0 s ) ζ(f v, χ 0,v s v) 1 1 Np σ. v S v v / S Since each ζ(f v, χ 0,v s v) converges for R(s) > 0, we are reduced to showing that the product 1 v / S converges absolutely for σ > 1. If = Q, this is a well-known 1 Np σ v theorem of Euler. In the general case, we have 1 1 Np σ 1 v 1 Np σ ( 1 p v 1 p σ )[ :Q]. p v / S v p Theorem 2.10 (Tate). Let f S(A ). The zeta function ζ(f, χ) can be analytically continued to a meromorphic function on the whole complex manifold X. It satisfies the functional equation (2.10.1) ζ(f, χ) = ζ( ˆf, ˆχ), where ˆf is the ourier transform of f (2.3.1), and ˆχ = χ 1. Moreover, ζ(f, χ) is holomorphic on the complex manifold X except for two simple poles at χ = 1 and χ =, with residues f(0)vol(i 1 / ) at χ = 1 and ˆf(0)Vol(I 1 / ) at χ =, where Proof. Let I 1 I 1 I 1 Vol(I 1 / ) = 2r 1 (2π) r 2 hr w. (resp. I 1 ) be the subset of I with norm 1 (resp. 1). Since I 1 = has Haar measure 0 in I, we have ζ(f, χ) = f(x)χ(x)d x = I f(x)χ(x)d x + f(x)χ(x)d x. I 1 Note that f is well-behaved when x, the first integral I 1 f(x)χ(x)d x converges absolutely for all χ X, thus defines a holomorphic function on the whole complex manifold X. or the second integral, we have I 1 f(x)χ(x)d x = ( I 1 / I 1 ξ f(ξx))χ(x)d x

16 16 YICHAO TIAN by the triviality of χ on. It s easy to check that the ourier transform of f(x ) is x 1 ˆf( x ). It follows from the Poisson formulae (2.4.1) that 1 f(xξ) = x ˆf( ξ x ) + 1 x ˆf(0) f(0). ξ ξ Therefore, we get I 1 / f(x)χ(x)d x = 1 ( I 1 / ξ x ˆf( ξ x ))χ(x)d x+ I 1 / ( 1 x ˆf(0) f(0))χ(x)d x. Making the change of variable y = 1 x, the first term on the right hand side above becomes ξ ˆf( x ))χ(x)d = ˆf(yξ))ˆχ(y)d y ( I 1 / ξ = ( I 1 / ξ I 1 ˆf(y)ˆχ(y)d y. We choose a splitting I / = I 1 / R + as in 2.7, and write that χ = χ 0 s, with χ 0 : I 1 / C is unitary and s C. We have ˆf(0) 1 f(0))χ(x)d x = ( χ 0 (x)d x)( ( ˆf(0) f(0))t s 1 dt. x I 1 / t=0 t We have I 1 / ( I 1 / χ 0 (x)d x = Vol(I 1 / )δ χ0,1 = or the second term, we have 1 ( ˆf(0) t=0 t Combining all the computations above, we get ζ(f, χ) = I 1 f(x)χ(x)d x + f(0))t s 1 dt = I 1 { 0 if χ 0 is non-trivial; Vol(I 1 / ) if χ 0 is trivial. ˆf(0) s 1 f(0) s. ˆf(x)ˆχ(x)d x + Vol(I 1 / )( ˆf(0) s 1 f(0) s )δ χ 0,1. Now it s clear that the right hand side of the equation above is invariant with f replaced by ˆf and χ replaced by ˆχ. Thus (2.10.1) follows immediately. The moreover part follows from the fact that the first two integrals above define holomorphic functions on X We indicate how to apply Tate s general theory to recover the classical results on the Dedekind Zeta function of a number field. Recall that Dedekind s zeta function is defined to be ζ (s) = 1 1 Np s = 1 v Σ v (Na) s, a O which converges absolutely for R(s) > 1. In the classical theory of Dedekind s zeta function, we have

17 NOTES ON TATE S THESIS 17 Theorem Let be a number field with r 1 real places and r 2 complex places. We put Z (s) = G 1 (s) r 1 G 2 (s) r 2 ζ (s), where G 1 (s) = π s 2 Γ( s 2 ), G 2(s) = (2π) 1 s Γ(s). Then Z (s) is a meromorphic function in the s-plan, holomorphic except for simple zeros at s = 0 and s = 1, and satisfies the functional equation Z (s) = 1 2 s Z (1 s). Its residues at s = 0 and s = 1 are respectively Vol(I 1 / ) and Vol(I 1 / ). Proof. We apply Tate s theorem 2.10 to χ = s, and f = f v with e πx2 v if v is real; πxv xv f v = e if v is complex; 1 Ov if v is non-archimedean. By the local computations in 1.12, we have π s 2 Γ( s ζ(f v, s 2 ) ) = 2π 1 s Γ(s) (Nd v ) Np s Therefore, we get if v is real; if v is complex; if v is non-archimedean. ζ(f, χ) = v Σ ζ(f v, s ) = 2 r 2s 1 2 Z (s). On the other hand, we have ˆf = v ˆfv with ˆf v = f v if v is real, ˆfv (z) = 2f v (2z) if v is complex, and ˆf v = (Nd v ) d 1 if v is non-archimedean. The local zeta functions are v π 1 s 2 Γ( 1 s ζ( ˆf v, 1 s 2 ) if v is real; ) = 2 s (2π) s Γ(1 s) if v is complex; (Nd v ) s 1 if v is non-archimedean. 1 Np s 1 Hence, we obtain ζ( ˆf, s ) = v Σ ζ( ˆf v, s ) = 2 r 2s s Z (1 s). The functional equation of Z (s) follows immediately from ζ(f, s ) = ζ( ˆf, 1 s ). The resides of Z (s) follows from the residues of ζ(f, s ) and the fact that f(0) = 1 and ˆf(0) = 2 r References [Ta50] Tate, John T. (1950), ourier analysis in number fields, and Hecke s zeta-functions, Algebraic Number Theory (Proc. Instructional Conf., Brighton, 1965), Thompson, Washington, D.C., pp [We74] A. Weil, Basic Number Theory, Third Edition, Springer-Verlag, (1974).