Notes on Tate's Thesis

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 Adèlesand Idèles,the functional equations of Dedekind Zeta functions and Hecke L-series. For general reference on adèlesand 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 Qp. If k is p-adic, we denote by O ⊂ k the ring of integers in k, p ⊂ O the maximal ideal, and $ ∈ 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| if k = ;  R R |x| = |x|2 if k = ; C C N(p)−ord$(x) if k is p-adic and x = u$ord$(x) with u ∈ O×. + + × We denote by k the additive group of k. Consider the unitary character ψ : k → C defined as follows:  −2πix e if k = R;  −2πi(x+¯x) (1.1.1) ψ(x) = e if k = C;  2πiλ(Tr (x)) e k/Qp if k is p-adic, where λ(·) means the decimal part of a p-adic number. For any ξ ∈ k, we note by ψξ the additive character x 7→ ψ(xξ) of k+. Note that if k is non-archimedean, ψ(x) = 1 if and −1 only if x ∈ d , where d is the different of k over Qp, i.e. −1 x ∈ d ⇔ Trk/Qp (xy) ∈ Zp ∀y ∈ O.

Proposition 1.2. The map Ψ: ξ 7→ ψξ deﬁnes an isomorphism of topological groups k+ ' kc+, where kc+ denotes the group of unitary characters of k+. Proof. If k = R or C, this is well known in classical Fourier 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 m −1 −m (at 0). If ξ ∈ p , then ψξ is trivial on d p . Since the subsets + −1 −m Um = {χ ∈ kc | χ is trivial on d p } form a fundamental system of open neighborhoods of 0 in kc+, the continuity of Ψ follows immediately. 1 2 YICHAO TIAN

(2) Next, we show that Ψ : k → Ψ(k) ⊂ bk is homoemorphism of k onto its image.

We need to check that if (xn)n≥1 ∈ k is a sequence such that ψxn → 1 uniformly for all compact subsets of k, then xn converges to 1 in k. Consider the compact open subgroup −m p for m ∈ Z. Then for any 1/2 > > 0, there exists an integer N > 0 such that −m −m |ψ(xnz) − 1| < for all n > N and z ∈ p . But xnp is a subgroup and the open ball × 1 −m B(1, ) ⊂ C contains no subgroup of S . Hence we have ψ(xnz) = 1 for all z ∈ p , so −1 m xn ∈ d p . (3) The image of Ψ is dense in k. Let H be the image of Ψ, and H¯ ⊂ bk be its closure. Then we have

H¯ ⊥ = {x ∈ bk ' k | χ(x) = 1, ∀χ ∈ H} = {x ∈ k | ψ(xξ) = 1, ∀ξ ∈ k} = {0}

Hence, we have H¯ = bk. (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 hn be a sequence in H that converges to g ∈ G. We need to prove that g ∈ H. Let (Ur)r≥0 be a fundamental system of compact neighborhoods of 0. We have ∩r≥0Ur = {0}. Then for any r, there exists an integer Nr > 0 such that hn ∈ g + Ur for all n ≥ Nr. Up to modifying Ur, we may assume hn −hm ∈ H ∩Ur−1 for any n, m ∈ Nr. Note that H ∩ Ur−1 is also compact by the local compactness of H. Up to replacing {Ur}r≥0 by a subsequence, we may choose mr for each integer r such that

hmr+1 + Ur ∩ H ⊂ hmr + Ur−1 ∩ H. By compactness, the intersection \ (hmr + Ur−1 ∩ H) r≥1 must contain an element h ∈ H. It’s easy to see that h = g, since ∩r≥0Ur = {0}.

1.4. 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 1 R − 2 measure on C; and if k is non-archimedean, we normalize the measure by O dx = (Nd) . 1 Let L (k, C) be the space of complex valued absolutely integrable functions on k. For 1 f ∈ L (k, C), we deﬁne the Fourier transform of f to be Z (1.4.1) fˆ(ξ) = f(x)ψ(xξ)dx. k Let S(k) be the space of Schwartz functions on k, i.e. dmf S( ) = {f ∈ C∞( ) | ∀n, m ∈ , |xn | is bounded}; R R N dxm 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 1 L (k, C). ˆ Proposition 1.5. The map f 7→ fˆ preserves S(k), and we have fˆ(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 Fourier transform of f = 1a+p` , the characteristic function of the set a + p`, is ˆ − 1 −` (1.6.1) f(x) = ψ(ax)(Nd) 2 (Np) 1d−1p−` . In particular, we have fˆ ∈ S(k). Proof. By deﬁnition, we have Z Z fˆ(x) = ψ(xy)dy = ψ(ax) ψ(xy)dy. a+p` p` The lemma follows immediately from

( − 1 −` −1 −` Z (Nd) 2 (Np) if x ∈ d p ψ(xy)dy = p` 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 1a+p` . We may assume thus f = 1a+p` . The ﬁrst part of the proposition follows from the previous lemma. For the second part, we have Z Z ˆ − 1 −` fˆ(x) = fˆ(y)ψ(xy)dy = (Nd) 2 (Np) ψ((x + a)y)dy k d−1p−` 1 1 − −` − ord$(d)+` = (Nd) 2 (Np) (Nd) 2 (Np) 1−a+p`

= 1−a+p` .

In the third equality above, we have used (1.6.1) with ` replaced by −ord$(d) − ` and x ˆ replaced by x + a. Now it’s clear that f(x) = f(−x). 1.7. Now consider the multiplicative group k×, and put U = {x ∈ k× | |x| = 1}. 1 × So U = {±1} if k = R, U = S is the group of unit circle if k = C, and U = O if k is p-adic. We have ( × if k = , ; k×/U = R+ R C Z if k is p-adic. 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. s So χ is unramified if and only if there is s ∈ C such that χ(x) = |x| . 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. For any quasi-character χ of k , there exists a unique unitary character χ0 × s of k such that χ = χ0| · | . × × Proof. For any x ∈ k , one can write uniquely x =xρ ˜ wherex ˜ ∈ U and ρ ∈ R+ if k = R or C, and ρ ∈ $Z if k is non-archimedean. We define χ0 as χ0(x) = (χ|U )(˜x). One checks easily that the quasi-character χ/χ0 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) = <(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+pν is trivial. We call the ideal ν fχ = p conductor of χ. So the conductor of χ is O if and only if χ is unramified. 1.9. 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) = Np Np−1 if k is non-archimedean. If k is non-archimedean, the factor δ(k) is justified by the fact that Z − 1 dx = (Nd) 2 . U Definition 1.10. For f ∈ S(k), we put Z 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 1.11. For any f, g ∈ S(k), we have ζ(f, χ)ζ(ˆg, χˆ) = ζ(f,ˆ χˆ)ζ(g, χ), where f,ˆ gˆ are Fourier transforms of f and g, and χˆ = | · |χ−1 for any quasi-character χ with 0 < σ(χ) < 1. Proof. Z Z ζ(f, χ)ζ(ˆg, χˆ) = f(x)ˆg(xy)|x|d×x χ(y−1)|y|d×y k× k× Z Z Z = 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. 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 7→ χ| · |s induces an isomorphism of complex manifolds from C to a connected component of the set of quasi-characters. Theorem 1.12. For 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(f,ˆ χˆ), where ρ(χ) is a meromorphic function of χ independent of f given as follows: s s (1) If k = R, then χ(x) = |x| or χ(x) = sgn(x)|x| for some s ∈ C. We have πs πs ρ(| · |s) = 21−sπ−s cos( )Γ(s), ρ(sgn| · |s) = i21−sπ−s sin( )Γ(s). 2 2 s (2) If k = C, then there exists n ∈ Z and s ∈ C such that χ = χn| · | where χn is the iθ inθ unitary character χn(re ) = e . We have 1−s |n| s |n| (2π) Γ(s + 2 ) ρ(χn| · | ) = i . s |n| (2π) Γ(1 − s + 2 ) (3) Assume k is p-adic. If χ is unramified, then s−1 s s− 1 1 − (Np) ρ(| · | ) = (Nd) 2 . 1 − Np−s s If χ = χ0| · | is ramified, where χ0 is unitary with χ0($) = 1 as in Lemma 1.8, then one has s s− 1 ρ(χ0| · | ) = N(dfχ) 2 ρ0(χ0) with − 1 X x ρ0(χ0) = N(fχ) 2 χ0(−x)ψ( ) ord$(dfχ) x $ × where x runs over a set of representatives of O /(1 + fχ).

Proof. By Proposition 1.11, the function ρ(χ) = ζ(f,χ) is independent of f. This proves ζ(f,ˆ χˆ) the functional equation (1.12.1). Note that ζ(f, χ) is well deﬁned if σ(χ) > 0, and ζ(f,ˆ χˆ) 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). s −πx2 (1) Assume k = R. If χ = | · | , we choose f = e . We have Z Z +∞ s −πx2 s × −πx2 s−1 − s s ζ(f, | · | ) = e |x| d x = 2 e x dx = π 2 Γ( ). R× 0 2 On the other hand,

Z 2 2 Z 2 (1.12.2) fˆ(y) = e−π(x +2ixy)dx = e−y e−π(x+yi) dx. R R 6 YICHAO TIAN

Using the well-known fact that

Z 2 Z 2 e−π(x+yi) dx = e−πx dx = 1, R R

1−s ˆ ˆ 1−s − 2 1−s we get f = f. Hence, we have ζ(f, | · | ) = π Γ( 2 ) and

s − 2 s s s+1 s π Γ( 2 ) −s√ Γ( 2 )Γ( 2 ) ρ(| · | ) = 1−s = π π 1−s 1+s − 2 1−s Γ( )Γ( ) π Γ( 2 ) 2 2 Now the formula for ρ(| · |s) follows from the properties of Gamma functions s s + 1 √ 1 − s 1 + s π Γ( )Γ( ) = 21−s πΓ(s), Γ( )Γ( ) = . 2 2 2 2 π(1+s) sin( 2 )

If χ = sgn| · |s, we take f = xe−πx2 . A similar computation shows that

s − s+1 s + 1 ζ(f, sgn| · | ) = π 2 Γ( ). 2

Taking derivatives with respect to y in (1.12.2), we get fˆ = −if. So we have

1−s s −1 s ζ(f,ˆ sgn| · | ) = −iπ 2 Γ(1 − ). 2 Therefore, we get

− s+1 s+1 s+1 s π 2 Γ( ) √ Γ( )Γ( ) ρ(sgn| · |s) = 2 = iπ−s π 2 2 s −1 s s s 2 Γ(1 − )Γ( ) −iπ Γ(1 − 2 ) 2 2 πs = i21−sπ−s sin( )Γ(s). 2

s −π(zz¯) (2) Assume k = C. If χ = | · | , we take f(z) = e . The local zeta function associated with f is Z ζ(f, | · |s) = e−πzz¯(zz¯)sd×z C× Z 2π Z +∞ −πr2 2s 2rdrdθ = e r 2 θ=0 r=0 r +∞ Z 2 = 4π e−πr r2s−1dr 0 Z +∞ s− 1 −πt dt 2 = 4π t 2 e √ (set t = r ) 0 2 t = 2π1−sΓ(s). NOTES ON TATE’S THESIS 7

The Fourier transform of f is Z (1.12.3) fˆ(z) = e−πww¯e−2πi(zw+¯zw¯)dw C +∞ +∞ Z Z 2 2 = 2 e−π(u +v )e−4πi(ux−vy)dudv (put z = x + iy, w = u + iv) −∞ −∞ +∞ +∞ 2 2 Z 2 Z 2 = 2e−4π(x +y ) e−π(u+2ix) du e−π(v−2iy) dv −∞ −∞ = 2f(2z). Therefore, one has ζ(f,ˆ | · |1−s) = 22s−1ζ(f, | · |1−s) = 22sπsΓ(1 − s), thus Γ(s) ρ(| · |s) = (2π)1−2s . Γ(1 − s) s n −π(zz¯) Let n ≥ 1 and χ = χ−n| · | . We put fn = z e . We compute ﬁrst the local zeta function of fn: Z s n −π(zz¯) s × (1.12.4) ζ(fn, χ−n| · | ) = z e χ−n(z)(zz¯) d z C× Z 2π Z +∞ −πr2 2s+n 2rdrdθ = e r 2 θ=0 r=0 r +∞ Z 2 = 4π e−πr r2s+n−1dr 0 Z +∞ −πt s+ n−1 dt = 4π e t 2 √ 0 2 t 1−(s+ n ) n = 2π 2 Γ(s + ). 2 To ﬁnd the Fourier transform of fn, we consider the equality (1.12.3) Z 2e−4π(zz¯) = e−π(ww¯)e−2πi(zw+¯zw¯)dw. C ∂n Regarding z andz ¯ as independent variables and applying ∂zn , we get Z 2(−2iz¯)ne−4πzz¯ = wne−π(ww¯)e−2πi(zw+¯zw¯)dw, C that is, fˆn(z) = 2f¯n(2iz). A similar computation as (1.12.4) shows that 1−s n 2s s− n n ζ(fˆ (z), χˆ) = ζ(2f¯ (2iz), χ | · | ) = (−i) 2 π 2 Γ(s + ). n n n 2 Therefore, we get 1−(s+ n ) n n 2π 2 Γ(s + ) Γ(s + ) ρ(χ | · |s) = 2 = in(2π)1−2s 2 . −n s− n n n n 2s 2 Γ( + 1 − s) (−i) 2 π Γ(s + 2 ) 2 s The formulae for ρ(χn| · | ) can be proved in the same way by choosing f = f¯n. 8 YICHAO TIAN

s (3) Assume k is p-adic. Consider ﬁrst the case χ = | · | . We take f = 1O. In the proof ˆ − 1 of Proposition 1.5, we have seen that f = (Nd) 2 1d−1 . We have Z ζ(f, χ) = |x|sd×x. O−{0} `+∞ n × As O − {0} = n=0 $ O , it follows that +∞ Z X 1 1 −ns × − 2 ζ(f, χ) = (Np) d x = (Nd) −s × 1 − (Np) n=0 O Similarly, using d−1 − {0} = `+∞ $nO×, one obtains n=−ord$(d) Z − 1 1−s × ζ(f,ˆ χˆ) = (Nd) 2 |x| d x d−1−{0} +∞ Z − 1 X n(s−1) × = (Nd) 2 (Np) d x O× n=−ord$(d) +∞ X = (Nd)−1(Np)ord$(d)(1−s) Npn(s−1) n=0 1 = (Nd)−s . 1 − Nps−1 The formula for ρ(| · |s) follows immediately. s Now consider the case χ = χ0| · | with χ0 ramiﬁed, unitary and χ0($) = 1. We take x f(x) = ψ( )1O. $ord$(dfχ) The local zeta function of f is Z x ζ(f, χ) = ψ( )χ (x)|x|sd×x ord (df ) 0 O−{0} $ $ χ +∞ X Z x$n = (Np)−ns ψ( )χ (x)d×x ord (df ) 0 × $ $ χ n=0 O We claim that Z x$n (1.12.5) ψ( )χ (x)d×x = 0 for n ≥ 1. ord (df ) 0 O× $ $ χ

Consider ﬁrst the case n ≥ ord$(fχ). We have x$n x$n ψ( ) = 1 as ∈ d−1. $ord$(dfχ) $ord$(dfχ) × If S is a set of representatives of O /(1 + fχ), the integral above is equal to Z Z × × X χ0(x)d x = d x χ0(x) = 0. × O 1+fχ x∈S NOTES ON TATE’S THESIS 9

−n Assume 0 ≤ n ≤ ord$(fχ) − 1. For any y ∈ 1 + p fχ, we have xy$n x$n ψ( ) = ψ( ). $ord$(dfχ) $ord$(dfχ) × −n Therefore, if Sn ⊂ S denotes a subset of representatives of O /(1 + p fχ), we get Z x$n Z X x$n ψ( )χ (x)d×x = d×x χ (x)ψ( ) ord (df ) 0 0 ord (df ) × $ $ χ $ $ χ O 1+fχ x∈S Z n × X x$ X = d x χ0(x)ψ( ) χ0(y), ord$(dfχ) 1+fχ $ x∈Sn y −n where y runs over a set of representatives of (1 + p fχ)/(1 + fχ). Note that ( X 0 if 1 ≤ n ≤ ord$(fχ), χ0(y) = y 1 if n = 0. This proves the claim. It follows that (1.12.6) Z Z × X x × 1 ζ(f, χ) = d x χ0(x)ψ( ) = χ0(−1)( d x)(Nfχ) 2 ρ0(χ0), $ord$(dfχ) 1+fχ x∈S 1+fχ where we have used the deﬁnition of ρ0 in the last step. As in the proof of 1.5, the Fourier transform of f is Z y fˆ(x) = ψ( )ψ(xy)dy ord (df ) O $ $ χ Z 1 = ψ(y(x + ))dy ord (df ) O $ $ χ 1 − 2 = (Nd) 1−$−ord$(dfχ)+d−1 . We get the local zeta function of fˆ Z ˆ − 1 1−s −1 × ζ(f, χˆ) = (Nd) 2 |x| χ0 (x)d x −$−ord$(dfχ)+d−1 1 Z − ord$(dfχ)(1−s) −1 × = (Nd) 2 (Np) χ0 (x)d x −ord$(dfχ) −$ (1+fχ)

−1 ord$(dfχ) Since χ0 (−$ (1 + y)) = χ0(−1) for any y ∈ fχ, we get Z − 1 1−s × ζ(f,ˆ χˆ) = χ0(−1)(Nd) 2 N(dfχ) d x . 1+fχ It thus follows that s s ζ(f, χ0| · | ) s− 1 ρ(χ0| · | ) = = N(dfχ) 2 ρ0(χ0). ˆ −1 1−s ζ(f, χ0 | · | ) 10 YICHAO TIAN

Remark 1.13. The number ρ0(χ0) in (3) is a generalization of (normalized) Gauss sum. By the same method as the classical case, we can show that |ρ0(χ0)| = 1. In general, it’s an interesting and diﬃcult problem to ﬁnd the exact argument of ρ0(χ0).

2. Global Theory

Let F be a number field, OF be its ring of integers. Let Σ be the set of all places of F , and Σf ⊂ Σ (resp. Σ∞ ⊂ Σ) be the subset of non-archimedean (resp. archimedean) places. For v ∈ Σ, we denote by Fv the completion of F at v. Let dxv be the self-dual Haar measure on Fv defined in 1.4. If v is finite, we denote by Ov the ring of integers of Fv, by pv the maximal ideal of Ov, and we fix a uniformizer $v ∈ pv. Let AF be the adèle Q ring of F , i.e. the subring of v∈Σ Fv consisting of elements x = (xv)v with xv ∈ Ov for almost all v, and AF,f be the ring of finite adèles.We choose the Haar measure on AF as Q dx = v dxv. It induces a quotient Haar measure on AF /F . Lemma 2.1. Under the notation above, we have R dx = 1. AF /F Proof. By Chinese reminders theorem, we have = F + Q O × Q F . We get AF v∈Σf v v∈Σ∞ v thus an isomorphism Y Y AF /F ' ( Ov × Fv)/OF . v∈Σf v∈Σ∞ Hence we have Z Y Z Z Y dx = dxv × dxv Q F /F Ov ( Fv)/OF A v∈Σf v∈Σ∞ v∈Σ∞ Y − 1 1/2 = (Ndv) 2 |∆F | ,

v∈Σf where dv denotes the diﬀerent of Fv and ∆F is the discriminant of F . If d denotes the diﬀerent of F/Q, then the lemma follows easily from the product formula: Y |∆F | = Nd = Ndv.

v∈Σf

For v ∈ Σ, let ψv be the additive character of the local field Fv defined in (1.1.1). Q It’s easy to check that ψ = v∈Σ ψv is trivial on additive group F , therefore it defines a character of the quotient AF /F . We call it the basic character of AF /F (or AF ). For any × ξ ∈ AF , let ψξ : AF → C be the character given by x 7→ ψ(xξ).

Proposition 2.2. The map Ψ: ξ 7→ ψξ defines an isomorphism between AF and its topological dual AbF . Moreover ψξ is a character of AF /F if and only if ξ ∈ F , i.e. ξ 7→ ψξ gives rise to an isomorphism of topological groups F ' A\F /F . 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 AF onto its image. Conversely, 0 × 0 0 let ψ : AF → C be a continuous character. The restriction ψv = ψ |Fv to the v-th local NOTES ON TATE’S THESIS 11 component defines a continuous character of Fv. By Proposition 1.2, there exists ξv ∈ Fv 0 0 such that ψv = ψv(ξv · ). Since ψ is continuous, there exists an open neighborhood Q Q 0 × v∈S Uv × v∈ /S Ov of 0 such that its image under ψ lies in B(1, 1/2) ⊂ C . As B(1, 1/2) 1 contains no non-trivial subgroups of S , we see that for any v∈ / S, we have ξv ∈ Ov. This 0 shows that ξ = (ξv)v∈Σ ∈ AF , and ψ = ψξ. This shows that Ψ : AF → AbF is a bijective continuous homomorphism of topological groups. To conclude that Ψ is an isomorphism, we need to show that if ξn ∈ AF is a sequence such that ψξn → 1 in AbF , we have ξn → 0 in AF as n → +∞. Actually, for any compact subset Uv ⊂ Fv with Uv = Ov for almost all v and any > 0, we have |ψ − 1|Q < for n sufficiently large. By Proposition ξn v Uv 1.2, for any finite subset S ⊂ Σ containing Σ∞, we can take (Uv)v∈S sufficiently large and Uv = Ov for v∈ / S such that |ξn|v < for v ∈ S and ξn ∈ Ov for v∈ / S. This means that ξn → 0 in AF . For the second part, let Γ ⊂ AF be the subgroup such that Ψ(Γ) ⊂ AbF consists of all characters trivial on F . It’s clear that F ⊂ Γ since ψ is trivial on F . To show that Γ = F , 1 1 Q we consider first the case F = Q. Let γ ∈ Γ. Since AQ = Q + (− 2 , 2 ] × p Zp, we can write γ = b + c, where b ∈ Q, c∞ ∈ (−1/2, 1/2] and cp ∈ Zp for all primes p. Then we have −2πic∞ 1 = ψγ(1) = ψ(γ) = ψ(b + c) = ψ(c) = e .

Hence we have c∞ = 0. Moreover, for any prime p and any integer n ≥ 0, we deduce from c 1 1 2πiλ( p ) 1 = ψ ( ) = ψ( (b + c)) = e pn γ pn pn n that cp ∈ p Zp, i.e. we have cp = 0. This shows γ = b, and hence Γ = Q. In the general case, we note that the basic character of AF is the composition of that on AQ with the trace map TrF/Q : AF → AQ. The following lemma will conclude the proof.

Lemma 2.3. Let x = (xv)v∈Σ ∈ AF such that TrF/Q(xy) ∈ Q ⊂ AQ for all y ∈ F . Then we have x ∈ F . ∗ Proof. Let (ei)1≤i≤d be a basis of F/Q, and (ei )1≤i≤d be the dual basis with respect to the perfect pairing F × F → Q given by (x, y) 7→ TrF/Q(xy). For any place p ≤ ∞ of Q, we have a canonical isomorphism of Qp-algebras Y F ⊗ Qp ' Fv. v|p Q Pd We put xp = (xv)v|p ∈ v|p Fv. Then we can write xp = i=1 ap,iei with ap,i ∈ Qp. As ∗ TrF/Q(xei ) ∈ Q ⊂ AQ for any i, we deduce that ap,i ∈ Q and it’s independent of p. This shows that x ∈ F .

Let S(AF ) be the space of Schwartz functions on AF , i.e. the space of ﬁnite linear Q combinations of functions on AF of the form f = v fv, where fv ∈ S(Fv) and fv = 1Ov for almost all v. For any f ∈ S(AF ), we deﬁne the Fourier transform of f to be Z (2.3.1) fˆ(ξ) = f(x)ψ(xξ)dx. AF 12 YICHAO TIAN

ˆ Proposition 2.4. (a) The Fourier transform f 7→ f preserves the space S(AF ), and ˆ fˆ(x) = f(−x). ˆ ˆ (b) If f = ⊗vfv with fv ∈ S(Fv) and fv = 1Ov for almost all v. Then f = ⊗vfv, where fˆv is the local Fourier transform (1.4.1) of fv. P (c) For any f ∈ S(A), the inﬁnite sum x∈F |f(x)| converges, and we have the Poisson formulae X X (2.4.1) f(x) = fˆ(ξ). x∈F ξ∈F 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 = ⊗vfv with fv ∈ S(Fv) and fv = 1O for almost all v. Then there exists an open compact v Q Q subgroup U ⊂ Af such that Supp(f) ⊂ U × v∈Σ Fv. Put OU = F ∩ (U × v∈Σ Fv). ∞ P ∞ This is a lattice in F . Each individual term in the summation x∈F f(x) is non-zero only ∞ ∞ if x ∈ OU . Write f = f f∞, where f = ⊗v∈Σf fv and f∞ = ⊗v∈Σ∞ fv. Then there exists a constant C > 0 such that |f ∞(x)| < C for all x ∈ U. Hence, we have X X X |f(x)| = |f(x)| < C |f∞(x)|.

x∈F x∈OU x∈OU 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 P g(x) = y∈F f(x + y), which converges for any x ∈ AF by the first part of (c). As g(x) is invariant under translation of F , we regard g(x) as a function on AF /F . Its Fourier transform of g(x) is Z gˆ(ξ) = g(x)ψ(xξ)dx (for ξ ∈ F ) AF /F Z = f(x)ψ(xξ)dx = fˆ(ξ). AF By the Fourier inverse formulae (a), we have X g(x) = gˆ(ξ)ψ(−xξ). ξ∈F The formulae (2.4.1) follows by setting x = 0. × Q × 2.5. Let IF = AF be the multiplicative group of idèlesof F , i.e. the subgroup of v∈Σ Fv × 1 consisting of elements x = (xv)v with xv ∈ Ov for almost all v, and IF be the subgroup of × 1 × IF of idèleswith norm 1. The diagonal embedding F ,→ IF identifies F with a discrete 1 1 subgroup of I for the induced restricted product topology on IF . A fundamental theorem 1 × in the theory of idèlessays that the quotient IF /F is compact [We74, IV §4 Thm.6]. We × Q × × consider the Haar measure d x = v d xv on IF , where d xv is the local Haar measure × 1 on Fv considered in 1.9. We use the same notation for the induced Haar measures on IF 1 × and IF /F . NOTES ON TATE’S THESIS 13

Proposition 2.6. Under the notation above, we have Z 2r1 (2π)r2 hR Vol( 1 /F ×) = d×x = , IF 1/2 1 × |∆ | w IF /F F where r1 (resp. r2) is the number of real places (resp. complex places) of F , h is the class number of F , ∆F is the discriminant, R is the regulator, and w denotes the number of roots of unity in F .

1 × 1 × Proof. Note ﬁrst that Vol(IF /F ) is ﬁnite, since IF /F is compact. For each x = (x ) ∈ , we denote by Div(x) = Q pordv(xv) be the fractional ideal associated v v∈Σ IF v∈Σf v with x. Then Div induces a short exact sequence

Y × × r1 × r2 × 0 → ( Ov × (R ) × (C ) ) × F → IF → ClF → 0, v∈Σf × r × r where ClF denotes the class group of F . Let Ω be the subgroup of (R ) 1 × (C ) 2 with Qr1 Qr2 product of absolute values i=1 |xi| × i=1 |zi|C = 1. The the exact sequence above induces a similar exact sequence

Y × × 1 0 → ( Ov × Ω) × F → IF → ClF → 0. v∈Σf Therefore, one gets Z Z d×x× = h d×x. 1 × Q × Q × × IF /F ( v Ov ×Ω)/( v Ov ×Ω)∩F Q × × Let UF denote the group of units of F . We have ( v Ov × Ω) ∩ F = UF , and hence Z Z Z 1 Z Y × × Y − 2 × = ( dxv ) × d x = Ndv d x. Q × × Q × × ( Ov ×Ω)/F ∩( Ov ×Ω) Ov Ω/UF Ω/UF v v v∈Σf v∈Σf

− 1 − 1 In view of the product formula Q Nd 2 = |∆ | 2 , to complete the proof, it suﬃces v∈Σf F to prove that Z 2r1 (2π)r2 R (2.6.1) d×x = . Ω/UF w Consider the map × r × r r +r Log : (R ) 1 × (C ) 2 → R 1 2 2 ((xi)1≤i≤r1 , (zj)1≤j≤r2 ) 7→ ((log |xi|)1≤i≤r1 , (log |zj| )1≤j≤r2 ). 1 × r +r Let S be the unit circle subgroup of C , and V be the subspace of R 1 2 deﬁned by the Pr1 Pr2 linear equation i=1 xi + j=1 yj = 0. Then the map Log induces a short exact sequence of abelian groups Log 0 → {±1}r1 × (S1)r2 → Ω −−→ V → 0. 14 YICHAO TIAN

r 1 r If µF denotes the group of roots of unity in F , we have ({±1} 1 × (S ) 2 ) ∩ UF = µF . Therefore, one obtains Z Z Z d×x = d×x × d×x. r 1 r Ω/UF {±1} 1 ×(S ) 2 /µF V/Log(UF ) × r By the deﬁnition of the Haar measure on IF , the induced measure on Ω ⊂ (R ) 1 × × r2 × × dx (C ) is determined as follows. On each copy of R , the measure is given by d x = |x| , × where dx is the usual Lebesgue measure on R; on each copy of C , the measure is given by dx ∧ dy d(r2) d×z = 2 = ∧ dθ, |z|2 r2 iθ where z = x + iy = re . Therefore, the Haar measure on IF induces the usual Lebesgue Qr2 1 r2 1 measure on Log(Ω) = V , and the measure j=1 dθj on (S ) . It follows that

Z 2r1 (2π)r2 d×x = . r 1 r {±1} 1 ×(S ) 2 /µF w By the deﬁnition of the regulator, we have R = R dx. Now the formula (2.6.1) V/Log(UF ) follows immediately. This ﬁnished the proof.

2.7. A Hecke character (or Grössencharacter) of F is a continuous homomorphism χ : × × IF /F → 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 F . We equip X with a structure of Riemann surface such that for each fixed character χ, the map s 7→ χ| · |s is a local isomorphism of C into X. × 1 × × × × Now we choose a splitting IF /F = IF /F × R+ of the norm map | · | : IF /F → R+. For every Hecke character χ, we put χ0 = χ| 1 × and denote still by χ0 its extension IF /F × × 1 × to IF /F by requiring χ0 is trivial on the chosen complement R+ of IF /F . Note that 1 × s χ0 is necessarily unitary since IF /F is compact and χ/χ0 is unramified, i.e. χ = χ0| · | with s ∈ C. We put σ(χ) = <(s), which is independent of the choice of the splitting. For v ∈ Σ, we put χ = χ| × . The local component χ is unramified for almost all v. v Fv v

Deﬁnition 2.8. Let f ∈ S(AF ) and χ be a Hecke character of F . We deﬁne the zeta function of f at χ to be Z ζ(f, χ) = f(x)χ(x)d×x. IF

Lemma 2.9. Let f ∈ S(AF ) and χ ∈ X. Then the zeta function ζ(f, χ) converges absolutely for σ(χ) > 1.

1 1 × R × Note that the ﬁniteness of Vol( /F ) implies that d x is ﬁnite, and hence Log(UF ) ⊂ V I V/Log(UF ) is a lattice. This actually gives another proof of Dirichlet’s theorem that UF has rank r1 + r2 − 1. NOTES ON TATE’S THESIS 15

s Proof. We may assume f = ⊗v∈Σfv with fv = 1Ov for almost all v ∈ Σf , and χ = χ0| · | 1 × 1 where χ0 : IF /F → S unitary. By definition, we have an Euler product Y ζ(f, χ) = ζ(fv, χv), v∈Σ where ζ(fv, χv) is the local zeta function defined in 1.10. We have seen in the proof of Theorem 1.12(3) that 1 1 s − 2 ζ(1Ov , | · |v) = (Ndv) −s . 1 − Npv Thus there exists a finite subset S of places such that

s Y s Y 1 |ζ(f, χ0| · | )| ≤ |ζ(fv, χ0,v| · |v)| −σ . 1 − Npv v∈S v∈ /S s Since each ζ(fv, χ0,v| · |v) converges for <(s) > 0, we are reduced to showing that the Q 1 product −σ converges absolutely for σ > 1. If F = Q, this is a well-known v∈ /S 1−Npv theorem of Euler. In the general case, we have 1 1 1 Y Y Y Y [F :Q] −σ ≤ −σ ≤ ( −σ ) . 1 − Npv 1 − Npv 1 − p v∈ /S p v|p p

Theorem 2.10 (Tate). Let f ∈ S(AF ). 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(f,ˆ χˆ), where fˆ is the Fourier transform of f (2.3.1), and χˆ = | · |χ−1. Moreover, ζ(f, χ) is holomorphic on the complex manifold X except for two simple poles at χ = 1 and χ = | · |, 1 × ˆ 1 × with residues −f(0)Vol(IF /F ) at χ = 1 and f(0)Vol(IF /F ) at χ = | · |, where r1 r2 1 × 2 (2π) hR Vol(IF /F ) = p . w |∆F |

≥1 ≤1 1 Proof. Let IF (resp. IF ) be the subset of IF with norm ≥ 1 (resp. ≤ 1). Since IF = ≥1 ≤1 IF ∩ IF has Haar measure 0 in IF , we have Z Z Z ζ(f, χ) = f(x)χ(x)d×x = f(x)χ(x)d×x + f(x)χ(x)d×x. ≥1 ≤1 IF IF IF R × Note that f is well-behaved when |x| → ∞, the first integral ≥1 f(x)χ(x)d x converges IF absolutely for all χ ∈ X, thus defines a holomorphic function on the whole complex manifold X. For the second integral, we have Z Z X f(x)χ(x)d×x = ( f(ξx))χ(x)d×x ≤1 ≤1/F × IF IF ξ∈F × 16 YICHAO TIAN by the triviality of χ on F ×. It’s easy to check that the Fourier transform of f(x · ) is −1 ˆ · |x| f( x ). It follows from the Poisson formulae (2.4.1) that X X 1 ξ 1 f(xξ) = fˆ( ) + fˆ(0) − f(0). |x| x |x| ξ∈F × ξ∈F × Therefore, we get Z Z X 1 ξ Z 1 f(x)χ(x)d×x = ( fˆ( ))χ(x)d×x+ ( fˆ(0)−f(0))χ(x)d×x. ≤1/F × ≤1/F × |x| x ≤1/F × |x| IF IF ξ∈F × IF 1 Making the change of variable y = x , the first term on the right hand side above becomes Z X ξ Z X ( fˆ( ))χ(x)d× = ( fˆ(yξ))ˆχ(y)d×y ≤1/F × x ≥1/F × IF ξ∈F × IF ξ∈F × Z = fˆ(y)ˆχ(y)d×y. ≥1 IF × 1 × × s We choose a splitting IF /F = IF /F × R+ as in 2.7, and write that χ = χ0| · | , with 1 × × χ0 : IF /F → C is unitary and s ∈ C. We have Z ˆ Z Z 1 ˆ f(0) × × f(0) s−1 ( − f(0))χ(x)d x = ( χ0(x)d x)( ( − f(0))t dt. ≤1 × |x| 1 × t IF /F IF /F t=0 We have Z ( × 1 × 0 if χ0 is non-trivial; χ0(x)d x = Vol(IF /F )δχ0,1 = 1 × 1 × Vol( /F ) if χ is trivial. IF /F IF 0 For the second term, we have Z 1 fˆ(0) fˆ(0) f(0) ( − f(0))ts−1dt = − . t=0 t s − 1 s Combining all the computations above, we get Z Z ˆ × ˆ × 1 × f(0) f(0) ζ(f, χ) = f(x)χ(x)d x + f(x)ˆχ(x)d x + Vol(IF /F )( − )δχ0,1. ≥1 ≥1 s − 1 s IF IF Now it’s clear that the right hand side of the equation above is invariant with f replaced by fˆ and χ replaced byχ ˆ. Thus (2.10.1) follows immediately. The moreover part follows from the fact that the first two integrals above define holomorphic functions on X. 2.11. 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 Y 1 X 1 ζF (s) = −s = s , 1 − Npv (Na) v∈Σ a⊂OF which converges absolutely for <(s) > 1. In the classical theory of Dedekind’s zeta function, we have NOTES ON TATE’S THESIS 17

Theorem 2.12. Let F be a number ﬁeld with r1 real places and r2 complex places. We put r1 r2 ZF (s) = G1(s) G2(s) ζF (s), s − 2 s 1−s where G1(s) = π Γ( 2 ), G2(s) = (2π) Γ(s). Then ZF (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 1 −s ZF (s) = |∆F | 2 ZF (1 − s). p 1 × 1 × Its residues at s = 0 and s = 1 are respectively − |∆F |Vol(IF /F ) and Vol(IF /F ). s Proof. We apply Tate’s theorem 2.10 to χ = | · | , and f = ⊗fv with  −πx2 e v if v is real;  −πx x¯ fv = e v v if v is complex;  1Ov if v is non-archimedean. By the local computations in 1.12, we have  s − 2 s π Γ( 2 ) if v is real; s  1−s ζ(fv, | · | ) = 2π Γ(s) if v is complex; − 1 1  2 (Ndv) 1−Np−s if v is non-archimedean. Therefore, we get 1 Y s r2s − ζ(f, χ) = ζ(fv, | · | ) = 2 |∆F | 2 ZF (s). v∈Σ

On the other hand, we have fˆ = ⊗vfˆv with fˆv = fv if v is real, fˆv(z) = 2fv(2z) if v is − 1 complex, and fˆ = (Nd ) 2 1 −1 if v is non-archimedean. The local zeta functions are v v dv  1−s − 2 1−s π Γ( 2 ) if v is real; 1−s  s s ζ(fˆv, | · | ) = 2 (2π) Γ(1 − s) if v is complex;  −s 1 (Ndv) 1−Nps−1 if v is non-archimedean. Hence, we obtain

ˆ s Y ˆ s r2s −s ζ(f, | · | ) = ζ(fv, | · | ) = 2 |∆F | ZF (1 − s). v∈Σ s ˆ 1−s The functional equation of ZF (s) follows immediately from ζ(f, | · | ) = ζ(f, | · | ). The s resides of ZF (s) follows from the residues of ζ(f, | · | ) and the fact that f(0) = 1 and 1 ˆ r2 − f(0) = 2 |∆F | 2 . References [Ta50] Tate, John T. (1950), Fourier analysis in number ﬁelds, and Hecke’s zeta-functions, Algebraic Number Theory (Proc. Instructional Conf., Brighton, 1965), Thompson, Washington, D.C., pp. 305347. [We74] A. Weil, Basic Number Theory, Third Edition, Springer-Verlag, (1974).