ESI The Erwin Schr�odinger International Boltzmanngasse 9Institute for Mathematical Physics A-1090 Wien, AustriaTrace Formula in Noncommutative Geometry andthe Zeros of the Riemann Zeta FunctionAlain Connes

Vienna, Preprint ESI 620 (1998) March 8, 1999Supported by Federal Ministry of Science and Transport, AustriaAvailable via http://www.esi.ac.at

Trace Formula in Noncommutative Geometry andthe Zeros of the Riemann Zeta FunctionAlain CONNESAbstract. We give a spectral interpretation of the critical zeros of the Rie-mann zeta function as an absorption spectrum, while eventual noncriticalzeros appear as resonances. We give a geometric interpretation of the explicitformulas of number theory as a trace formula on the noncommutative spaceof Adele classes. This reduces the Riemann hypothesis to the validity of thetrace formula and eliminates the parameter � of our previous approach.Table of contentsIntroduction.I Quantum chaos and the hypothetical Riemann ow.II Algebraic Geometry and global �elds of non zero characteristic.III Spectral interpretation of critical zeros.IV The distribution trace formula for ows on manifolds.V The action (�; x)! �x of K� on a local �eld K.VI The global case, and the formal trace computation.VII Proof of the trace formula in the S-local case.VIII The trace formula in the global case, and elimination of �.Appendix I, Proof of theorem 1.Appendix II, Explicit formulas.Appendix III, Distribution trace formulas.1

IntroductionWe shall give in this paper a spectral interpretation of the zeros of the Rie-mann zeta function and a geometric framework in which one can transposethe ideas of algebraic geometry involving the action of the Frobenius and theLefchetz formula. The spectral interpretation of the zeros of zeta will be asan absorption spectrum, i.e., as missing spectral lines. All zeros will play arole in the spectral side of the trace formula, but while the critical zeros willappear perse, the noncritical ones will appear as resonances and enter in thetrace formula through their harmonic potential with respect to the criticalline. Thus the spectral side is entirely canonical, and by proving positivity ofthe Weil distribution, we shall show that its equality with the geometric side,i.e., the global trace formula, is equivalent to the Riemann Hypothesis for allL-functions with Gr�ossencharakter.We shall model our discussion on the Selberg trace formula, but it di�ersfrom the latter in several important respects. We shall �rst explain in partic-ular why a crucial negative sign in the analysis of the statistical uctuationsof the zeros of zeta indicates that the spectral interpretation should be as anabsorption spectrum, or equivalently should be of a cohomological nature. Asit turns out, the geometric framework involves an innocent looking space, thespace X of Adele classes, where two adeles which belong to the same orbitof the action of GL1(k) (k a global �eld), are considered equivalent. Thegroup Ck = GL1(A)=GL1(k) of Idele classes (which is the class �eld theorycounterpart of the Galois group) acts by multiplication on X.Our �rst preliminary result (Theorem 1 of Section III) gives a spectralinterpretation of the critical zeros of zeta and L functions on a global �eldk from the action of the Idele class group on a space of square integrablefunctions on the space X = A=k� of Adele classes. Corollary 2 gives the cor-responding computation of the spectral trace. This result is only preliminarybecause it requires the use of an unnatural parameter � which plays the roleof a Sobolev exponent and allows us to see the absorption spectrum as a pointspectrum.Our second preliminary result is a formal computation (Section VI) of thecharacter of the representation of the Idele class group on the above L2 space.This formal computation gives the Weil distribution which is the essentialingredient of the Riemann-Weil explicit formula. At this point (which wasthe situation in [Co]), the main problems are to give a rigorous meaning tothe formal trace computation and to eliminate the unwanted parameter �.2

These two problems will be solved in the present paper. We �rst provea trace formula (Theorem 3 of Section V) for the action of the multiplicativegroup K� of a local �eld K on the Hilbert space L2(K), and (Theorem 4 ofSection VII) a trace formula for the action of the multiplicative group CS ofIdele classes associated to a �nite set S of places of a global �eld k, on theHilbert space of square integrable functions L2(XS), where XS is the quotientof Qv2S kv by the action of the group O�S of S-units of k. In both cases weobtain exactly the terms of the Weil explicit formulas which belong to the�nite set of places. This result is quite important since the space XS is highlynontrivial as soon as the cardinality of S is larger or equal to 3. Indeed thisquotient space is nontype I in the sense of Noncommutative Geometry, and itis reassuring that the trace formula continues to hold there.We check in detail (Theorem 6 of Appendix II) that the rewriting of theWeil explicit formulas which is predicted by the global trace formula is correct.Finally, we eliminate in Section VIII using ideas that are common both tothe Selberg trace formula and to the standard explanation of the absorptionlines in physics, the unpleasant parameter � which appeared as a label of thefunction spaces of Section III. We write the global trace formula as an analogueof the Selberg trace formula. The validity of the trace formula for any �niteset of places follows from Theorem 4 of Section VII, but in the global caseis left open and shown (Theorem 5 of Section VIII) to be equivalent to thevalidity of the Riemann Hypothesis for all L functions with Gr�ossencharakter.This equivalence, together with the plausibility of a direct proof of the traceformula along the lines of Theorem 4 (Section VII) constitute the main resultof this paper. The elimination of the parameter � is the main improvementof the present paper with respect to [Co].It is an old idea, due to Polya and Hilbert, that in order to understand thelocation of the zeros of the Riemann zeta function, one should �nd a Hilbertspace H and an operator D in H whose spectrum is given by the nontrivialzeros of the zeta function. The hope then is that suitable selfadjointnessproperties of D (of i �D � 12� more precisely) or positivity properties of � =D(1 � D) will be easier to handle than the original conjecture. The mainreasons why this idea should be taken seriously are �rst the work of A. Selberg([Se]) in which a suitable Laplacian � is related in the above way to ananalogue of the zeta function, and secondly the theoretical ([M][B][KS]) andexperimental evidence ([O][BG]) on the uctuations of the spacing betweenconsecutive zeros of zeta. The number of zeros of zeta whose imaginary part3

is less than E > 0,(1) N(E) = # of zeros � ; 0 < Im � < Ehas an asymptotic expression ([R]) given by(2) N(E) = E2� �log� E2��� 1�+ 78 + o(1) +Nosc(E)where the oscillatory part of this step function is(3) Nosc(E) = 1� Im log � �12 + iE�assuming that E is not the imaginary part of a zero and taking for the loga-rithm the branch which is 0 at +1.One shows (cf. [Pat]) that Nosc(E) is O(logE). In the decomposition (2)the two terms hN(E)i = N(E) �Nosc(E) and Nosc(E) play an independentrole. The �rst one hN(E)i which gives the average density of zeros just comesfrom Stirling's formula and is perfectly controlled. The second Nosc(E) isa manifestation of the randomness of the actual location of the zeros, andto eliminate the role of the density, one returns to the situation of uniformdensity by the transformation(4) xj = hN(Ej )i (Ej the jth imaginary part of zero of zeta) :Thus the spacing between two consecutive xj is now 1 on average and theonly information that remains is in the statistical uctuation. As it turns out([M][O]) these uctuations are the same as the uctuations of the eigenvaluesof a random hermitian matrix of very large size.H. Montgomery [M] proved (assuming RH) a weakening of the followingconjecture (with �; � > 0),Cardf(i; j) ; i; j 2 1; : : : ;M ; xi � xj 2 [�; �]g�M Z �� 1��sin(�u)�u �2! du(5)This law (5) is precisely the same as the correlation between eigenvalues ofhermitian matrices of the gaussian unitary ensemble ([M]). Moreover, numeri-cal tests due to A. Odlyzko ([O][BG]) have con�rmed with great precision the4

behaviour (5) as well as the analogous behaviour for more than two zeros. In[KS], N. Katz and P. Sarnak proved an analogue of the Montgomery-Odlyzkolaw for zeta and L-functions of function �elds over curves.It is thus an excellent motivation to try and �nd a natural pair (H;D)where naturality should mean for instance that one should not even have tode�ne the zeta function, let alone its analytic continuation, in order to obtainthe pair (in order for instance to avoid the joke of de�ning H as the `2 spacebuilt on the zeros of zeta).I Quantum chaos and the hypothetical Riemann owLet us �rst describe following [B] the direct atempt to construct the Polya-Hilbert space from quantization of a classical dynamical system. The originalmotivation for the theory of randommatrices comes from quantummechanics.In this theory the quantization of the classical dynamical system given bythe phase space X and hamiltonian h gives rise to a Hilbert space H and aselfadjoint operator H whose spectrum is the essential physical observable ofthe system. For complicated systems the only useful information about thisspectrum is that, while the average part of the counting function,(1) N(E) = # eigenvalues of H in [0; E]is computed by a semiclassical approximation mainly as a volume in phasespace, the oscillatory part,(2) Nosc(E) = N(E) � hN(E)iis the same as for a random matrix, governed by the statistic dictated by thesymmetries of the system.In the absence of a magnetic �eld, i.e., for a classical hamiltonian of theform,(3) h = 12m p2 + V (q)where V is a real-valued potential on con�guration space, there is a naturalsymmetry of classical phase space, called time reversal symmetry,(4) T (p; q) = (�p; q)5

which preserves h, and entails that the correct ensemble on the randommatri-ces is not the above GUE but rather the gaussian orthogonal ensemble: GOE.Thus the oscillatory part Nosc(E) behaves in the same way as for a randomreal symmetric matrix.Of course H is just a speci�c operator in H and, in order that it behavegenerically, it is necessary (cf. [B]) that the classical hamiltonian system(X;h) be chaotic with isolated periodic orbits whose instability exponents(i.e., the logarithm of the eigenvalues of the Poincar�e return map acting onthe transverse space to the orbits) are di�erent from 0.One can then ([B]) write down an asymptotic semiclassical approximationto the oscillatory function Nosc(E)(5) Nosc(E) = 1� ImZ 10 Trace(H � (E + i�))�1 id�using the stationary phase approximation of the corresponding functional in-tegral. For a system whose con�guration space is 2-dimensional, this gives([B] (15)),(6) Nosc(E) ' 1�X p 1Xm=1 1m 12sh�m�p2 � sin(Spm(E))where the p are the primitive periodic orbits, the label m corresponds tothe number of traversals of this orbit, while the corresponding instabilityexponents are ��p. The phase Spm(E) is up to a constant equal to mE T# where T# is the period of the primitive orbit p.The formula (6) gives very precious information ([B]) on the hypothetical\Riemann ow" whose quantization should produce the Polya-Hilbert space.The point is that the Euler product formula for the zeta function yields (cf.[B]) a similar asymptotic formula for Nosc(E) (3),(7) Nosc(E) ' �1� Xp 1Xm=1 1m 1pm=2 sin (mE log p) :Comparing (6) and (7) gives the following information,(A) The periodic primitive orbits should be labelled by the prime numbersp = 2; 3; 5; 7; : : :, their periods should be the log p and their instabilityexponents �p = � log p. 6

Moreover, since each orbit is only counted once, the Riemann ow should notpossess the symmetry T of (4) whose e�ect would be to duplicate the countof orbits. This last point excludes in particular the geodesic ows since theyhave the time reversal symmetry T . Thus we get(B) The Riemann ow cannot satisfy time reversal symmetry.However there are two important mismatches (cf. [B]) between the two for-mulas (6) and (7). The �rst one is the overall minus sign in front of formula(7), the second one is that though 2sh �m�p2 � � pm=2 when m ! 1, we donot have an equality for �nite values of m.These are two fundamental di�culties, and in order to overcome them weshall use the well known strategy of extending the problem to the case ofarbitrary global �elds. By specialising to the function �eld case, we shall thenobtain additional precious information.II Algebraic Geometry and global �elds of nonzero characteristicThe basic properties of the Riemann zeta function extend to zeta functionsassociated to an arbitrary global �eld, and it is unliquely that one can settlethe problem of the spectral interpretation of the zeros, let alone �nd theRiemann ow, for the particular case of the global �eld Q of rational numberswithout at the same time settling these problems for all global �elds. Theconceptual de�nition of such �elds k, is the following:A �eld k is a global �eld i� it is discrete and cocompact in a (nondiscrete)locally compact semisimple abelian ring A.As it turns out, A then depends functorially on k and is called the Adelering of k, often denoted by kA. Thus though the �eld k itself has no interestingtopology, there is a canonical and highly nontrivial topological ring which iscanonically associated to k. When the characteristic p of a global �eld k is> 0, the �eld k is the function �eld of a nonsingular algebraic curve � de�nedover a �nite �eld Fq included in k as its maximal �nite sub�eld, called the�eld of constants. One can then apply the ideas of algebraic geometry, �rstdeveloped over C , to the geometry of the curve � and obtain a geometricinterpretation of the basic properties of the zeta function of k; the dictionarycontains in particular the following lines7

Spectral interpretation of Eigenvalues of actionthe zeros of Frobenius on `-adic cohomologyFunctional equation Riemann Roch theorem(Poincar�e duality)Explicit formulas of Lefchetz formulanumber theory for the FrobeniusRiemann hypothesis Castelnuovo positivity(1) Since Fq is not algebraically closed, the points of � de�ned over Fq do notsu�ce and one needs to consider ��, the points of � on the algebraic closure�Fq of Fq , which is obtained by adjoining to Fq the roots of unity of orderprime to q. This set of points is a countable union of periodic orbits underthe action of the Frobenius automorphism; these orbits are parametrized bythe set of places of k and their periods are indeed given by the analogues ofthe log p of (A). Being a countable set it does not qualify as an analogue ofthe Riemann ow and it only aquires an interesting structure from algebraicgeometry. The minus sign which was problematic in the above discussionadmits here a beautiful resolution since the analogue of the Polya-Hilbertspace is given, if one replaces C by Q` the �eld of `-adic numbers ` 6= p, bythe cohomology group(2) H1et(��;Q`)which appears with an overall minus sign in the Lefchetz formula(3) X(�1)j Trace'�=Hj = X'(x)=x 1:For the general case this suggests(C) The Polya-Hilbert space H should appear from its negative H.In other words, the spectral interpretation of the zeros of the Riemann zetafunction should be as an absorption spectrum rather than as an emissionspectrum, to borrow the language of spectroscopy.The next thing that one learns from this excursion in characteristic p > 0is that in that case one is not dealing with a ow but rather with a single8

transformation. In fact taking advantage of abelian covers of � and of thefundamental isomorphismof class �eld theory, one �nds that the natural groupthat should replace R for the general Riemann ow is the Idele class group:(D) Ck = GL1(A)=k� :We can thus collect the information (A) (B) (C) (D) that we have obtainedso far and look for the Riemann ow as an action of Ck on an hypotheticalspace X.III Spectral interpretation of critical zerosThere is a third approach to the problem of the zeros of the Riemann zetafunction, due to G. P�olya [P] andM. Kac [K] and pursued further in [J] [BC]. Itis based on statistical mechanics and the construction of a quantum statisticalsystem whose partition function is the Riemann zeta function. Such a systemwas naturally constructed in [BC] and it does indicate using the �rst lineof the dictionary of Noncommutative Geometry (namely the correspondencebetween quotient spaces and noncommutative algebras) what the space Xshould be in general:(1) X = A=k�namely the quotient of the space A of adeles, A = kA by the action of themultiplicative group k�,(2) a 2 A ; q 2 k� ! aq 2 A :This space X already appears in a very implicit manner in the work of Tateand Iwasawa on the functional equation. It is a noncommutative space in that,even at the level of measure theory, it is a tricky quotient space. For instanceat the measure theory level, the corresponding von Neumann algebra,(3) R01 = L1(A)>/ k�where A is endowed with its Haar measure as an additive group, is the hy-per�nite factor of type II1.The idele class group Ck acts on X by(4) (j; a)! ja 8 j 2 Ck ; a 2 X9

and it was exactly necessary to divide A by k� so that (4) makes good sense.We shall come back later to the analogy between the action of Ck on R01 andthe action of the Galois group of the maximal abelian extension of k.What we shall do now is to construct the Hilbert space L2� of functionson X with growth indexed by � > 1. Since X is a quotient space we shall �rstlearn in the usual manifold case how to obtain the Hilbert space L2(M) ofsquare integrable functions on a manifoldM by working only on the universalcover fM with the action of � = �1(M). Every function f 2 C1c (fM ) givesrise to a function ef on M by(5) ef (x) = X�(ex)=x f(ex)and all g 2 C1(M) appear in this way. Moreover, one can write the Hilbertspace inner product RM ef1(x) ef2(x) dx, in terms of f1 and f2 alone. Thusk efk2 = R ���P 2� f( x)���2 dx where the integral is performed on a fundamentaldomain for � acting on fM . This formula de�nes a pre-Hilbert space norm onC1c (fM ) and L2(M) is just the completion of C1c (fM ) for that norm. Notethat any function of the form f �f has vanishing norm and hence disappearsin the process of completion. In our case of X = A=k�, we thus need to de�nethe analoguous norm on the Bruhat-Schwartz space S(A) of functions on A (cfAppendix I for the general de�nition of the Bruhat-Schwartz space). Since 0is �xed by the action of k�, the expressionP 2k� f( x) does not make sensefor x = 0 unless we require that f(0) = 0. Moreover, when jxj ! 0, the abovesums approximate, as Riemann sums, the product of jxj�1 by R f dx for theadditive Haar measure; thus we also require R f dx = 0. We can now de�nethe Hilbert space L2�(X)0 as the completion of the codimension 2 subspace(6) S(A)0 = ff 2 S(A) ; f(0) = 0 ; Z f dx = 0gfor the norm k k� given by(7) kfk2� = Z ���Xq2k� f(qx)���2 (1 + log2 jxj)�=2 jxj d�xwhere the integral is performed on A�=k� and d�x is the multiplicative Haarmeasure on A�=k�. The ugly term (1 + log2 jxj)�=2 is there to control thegrowth of the functions on the noncompact quotient. We shall see how toremove it later in Section VII. Note that jqxj = jxj for any q 2 k�.10

The key point is that we use the measure jxj d�x instead of the additive Haarmeasure dx. Of course for a local �eld K, one has dx = jxj d�x, but this failsin the above global situation. Instead one has,(8) dx = lim"!0 " jxj1+" d�x :One has a natural representation of Ck on L2�(X)0 given by(9) (U(j) f) (x) = f(j�1 x) 8x 2 A ; j 2 Ckand the result is independent of the choice of a lift of j in Jk = GL1(A)because the functions f � fq are in the kernel of the norm. The conditions(6) which de�ne S(A)0 are invariant under the action of Ck and give thefollowing action of Ck on the 2-dimensional supplement of S(A)0 � S(A);this supplement is C � C (1) where C is the trivial Ck module (correspondingto f(0)) while the Tate twist C (1) is the module(10) (j; �)! jjj�coming from the equality(11) Z f(j�1 x) dx = jjjZ f(x) dx :In order to analyze the representation (9) of Ck on L2�(X)0, we shall relate itto the left regular representation of the group Ck on the Hilbert space L2�(Ck)obtained from the following Hilbert space square norm on functions,(12) k�k2� = ZCk j�(g)j2 (1 + log2 jgj)�=2 d�g:Here we have normalized the Haar measure of the multiplicative group Ck,with module(13) j j : Ck ! R�+in such a way that (cf. [W3])(14) Zjgj2[1;�] d�g � log � when �! +1 :The left regular representation V of Ck on L2�(Ck) is(15) (V (a) �) (g) = �(a�1 g) 8 g; a 2 Ck :11

Note that because of the weight (1 + log2 jxj)�=2, this representation is notunitary but it satis�es the growth estimate(16) kV (g)k = 0 (log jgj)�=2 when jgj ! 1which follows from the inequality (valid for u; v 2 R)(17) �(u+ v) � 2�=2 �(u) �(v) ; �(u) = (1 + u2)�=2 :We letE be the linear isometry fromL2�(X)0 into L2�(Ck) given by the equality,(18) E(f) (g) = jgj1=2 Xq2k� f(qg) 8 g 2 Ck :By comparing (7) with (12) we see that E is an isometry and the factor jgj1=2is dictated by comparing the measures jgj d�g of (7) with d�g of (12).One has E(U(a) f) (g) = jgj1=2 Pk� (U(a) f) (qg) = jgj1=2 Pk� f(a�1qg) = jaj1=2 ja�1 gj1=2 Pk� f(q a�1 g) = jaj1=2 (V (a)E(f)) (g). Thus,(19) E U(a) = jaj1=2 V (a)E :This equivariance shows that the range of E in L2�(Ck) is a closed invariantsubspace for the representation V .The following theorem and its corollary show that the cokernel H =L2�(Ck)= Im(E) of the isometry E plays the role of the Polya-Hilbert space.Since ImE is invariant under the representation V , we let W be the corre-sponding representation of Ck on H.The abelian locally compact group Ck is (noncanonically) isomorphic toK �N where(20) K = fg 2 Ck ; jgj = 1g ; N = range j j � R�+ :For number �elds one has N = R�+, while for �elds of nonzero characteristicN ' Zis the subgroup qZ � R�+ (where q = p` is the cardinality of the �eldof constants).We choose (noncanonically) an isomorphism(21) Ck ' K �N :By construction the representationW satis�es (using (16)),(22) kW (g)k = 0(log jgj)�=212

and its restriction to K is unitary. Thus H splits as a canonical direct sum ofpairwise orthogonal subspaces,(23) H = ��2bK H� ; H� = f� ; W (g) � = � (g) � ; 8 g 2 Kgwhere � runs through the Pontrjagin dual group of K, which is the discreteabelian group bK of characters of K. Using the noncanonical isomorphism(21), i.e., the corresponding inclusion N � Ck, one can now restrict the rep-resentation W to any of the sectors H�. When char(k) > 0, then N ' Zand the condition (22) shows that the action of N on H� is given by a sin-gle operator with unitary spectrum. (One uses the spectral radius formulajSpecwj = Lim kwnk1=n.) When Char(k) = 0, we are dealing with an actionof R�+ ' R on H� and the condition (22) shows that this representation is gen-erated by a closed unbounded operator D� with purely imaginary spectrum.The resolvent R� = (D� � �)�1 is given, for Re� > 0, by the equality(24) R� = Z 10 W�(es) e��s dsand for Re� < 0 by,(25) R� = Z 10 W�(e�s) e�s dswhile the operator D� is de�ned by(26) D� � =lim"!0 1" (W�(e") � 1) � :Theorem 1. Let � 2 bK, � > 1, H� and D� be as above. Then D� hasdiscrete spectrum, SpD� � iR is the set of imaginary parts of zeros of theL function with Gr�ossencharakter e� which have real part equal to 12 ; � 2SpD, L �e�; 12 + �� = 0 and � 2 iR, where e� is the unique extension of � toCk which is equal to 1 on N . Moreover the multiplicity of � in SpD is equalto the largest integer n < 1+�2 , n � multiplicity of 12 + � as a zero of L.Theorem 1 has a similar formulation when the characteristic of k isnonzero. The following corollary is valid for global �elds k of arbitrary char-acteristic. 13

Corollary 2. For any Schwartz function h 2 S(Ck) the operator W (h) =R W (g)h(g) d� g in H is of trace class, and its trace is given byTraceW (h) = XL�e�; 12+��=0�2iR=N? bh(e�; �)where the multiplicity is counted as in Theorem 1 and where the Fourier trans-form bh of h is de�ned by,bh(e�; �) = ZCk h(u) e�(u) juj� d� u :Note that we did not have to de�ne the L functions, let alone theiranalytic continuation, before stating the theorem, which shows that the pair(27) (H�;D�)certainly quali�es as a Polya-Hilbert space.The case of the Riemann zeta function corresponds to the trivial character� = 1 for the global �eld k = Q of rational numbers.In general the zeros of the L functions can have multiplicity but oneexpects that for a �xed Gr�ossencharakter � this multiplicity is bounded, sothat for a large enough value of � the spectral multiplicity of D will be theright one. When the characteristic of k is > 0 this is certainly true.If we modify the choice of noncanonical isomorphism (21) this modi�esthe operator D by(28) D0 = D � i s ;where s 2 R is determined by the equality(29) e�0(g) = e�(g) jgji s 8 g 2 Ck :The coherence of the statement of the theorem is insured by the equality(30) L(e�0; z) = L(e�; z + i s) 8 z 2 C :When the zeros of L have multiplicity and � is large enough, the oper-ator D is not semisimple and has a nontrivial Jordan form (cf. Appendix I).14

This is compatible with the almost unitary condition (22) but not with skewsymmetry for D.The proof of Theorem 1, explained in Appendix I, is based on the dis-tribution theoretic interpretation by A. Weil [W2] of the idea of Tate andIwasawa on the functional equation. Our construction should be comparedwith [Bg] and [Z].As we expected from (C), the Polya-Hilbert space H appears as a cok-ernel. Since we obtain the Hilbert space L2�(X)0 by imposing two linearconditions on S(A),(31) 0! S(A)0 ! S(A) L! C � C (1) ! 0we shall de�ne L2�(X) so that it �ts in an exact sequence of Ck-modules(32) 0! L2�(X)0 ! L2�(X)! C � C (1) ! 0 :We can then use the exact sequence of Ck-modules(33) 0! L2�(X)0 ! L2�(Ck)!H ! 0together with Corollary 2 to compute in a formal manner what the characterof the module L2�(X) should be. Using (32) and (33) we obtain,(34) \Trace" (U(h)) = bh(0) + bh(1) � XL(�;�)=0Re�=12 bh(�; �) +1h(1)where bh(�; �) is de�ned by Corollary 2 and(35) U(h) = ZCk U(g)h(g) d� gwhile the test function h is in a suitable function space. Note that the traceon the left hand side of (34) only makes sense after a suitable regularizationsince the left regular representation of Ck is not of trace class. This situationis similar to the one encountered by Atiyah and Bott ([AB]) in their proof ofthe Lefchetz formula. We shall �rst learn how to compute in a formal mannerthe above trace from the �xed points of the action of Ck on X. In SectionVII, we shall show how to regularize the trace and completely eliminate theparameter �. 15

IV The distribution trace formula for ows on manifoldsIn order to understand how the left hand side of III(34) should be computed,we shall �rst give an account of the proof of the usual Lefchetz formula byAtiyah-Bott ([AB]) and describe the computation of the distribution theoretictrace for ows on manifolds, which is a variation on the theme of [AB] and isdue to Guillemin-Sternberg [GS]. We refer to Appendix III for a more detailedcoordinate independent treatment following [GS].Let us start with a di�eomorphism ' of a smooth compact manifold Mand assume that the graph of ' is transverse to the diagonal in M � M .One can then easily de�ne and compute the distribution theoretic trace of theoperator U : C1(M) ! C1(M),(1) (U�)(x) = �('(x)) :Indeed let k(x; y) be the Schwartz distribution on M �M such that(2) (U�)(x) = Z k(x; y) �(y) dy :The distributional trace of U is simply(3) \Trace" (U) = Z k(x; x) dx ;Near the diagonal and in local coordinates one gets(4) k(x; y) = �(y � '(x))where � is the Dirac distribution.Since, by hypothesis, the �xed points of ' are isolated, one can computethe trace (3) as a �nite sum Px;'(x)=x and get the contribution of each �xedpoint x 2M , '(x) = x, as(5) 1j1� '0(x)jwhere '0(x) is the Jacobian of ' and jAj = jdetAj. One just uses the invert-ibility of id� '0(x) to change variables in the integral,(6) Z �(y � '(y)) dy :16

One thus gets (cf. [AB]),(7) \Trace" (U) = Xx;'(x)=x 1j1� '0(x)j :This computation immediately extends to the action of ' on sections ofan equivariant vector bundle E such as the bundle ^kT � whose sections,C1(M;E) are the smooth forms of degree k. The alternate sum of the cor-responding distribution theoretic traces is the ordinary trace of the action of' on the de Rham cohomology, thus yielding the usual Lefchetz formula,(8) X(�1)j Trace'�=Hj = X'(x)=x sign det(1� '0(x)) :Let us refer to the appendix for more pedantic notations which show that thedistribution theoretic trace is coordinate independent.We shall now write down the analogue of formula (7) in the case of a owFt = exp(tv) of di�eomorphisms of M , where v 2 C1(M;T ) is a vector �eldon M . We get a one parameter group of operators acting on C1(M),(9) (Ut �)(x) = �(Ft(x)) 8 � 2 C1(M) ; x 2M ; t 2 R ;and we need the formula for,(10) �(h) = \Trace" �Z h(t)Ut dt� ; h 2 C1c (R) ; h(0) = 0 :The condition h(0) = 0 is required because we cannot expect that the identitymap F0 be transverse to the diagonal. In order to de�ne � as a distributionevaluated on the test function h, we let f be the following map:(11) f : X =M �R! Y =M ; f(x; t) = Ft(x) :The graph of f is the submanifold Z of X � Y ,(12) Z = f(x; t; y) ; y = Ft(x)g :One lets ' be the diagonal map,(13) '(x; t) = (x; t; x) ; ' :M �R! X � Yand one assumes the transversality '\j Z outside M � (0).17

Let � be the distribution,(14) � = '�(�(y � Ft(x)) dy) ;and let q be the second projection,(15) q(x; t) = t 2 R :Then by de�nition � is the pushforward q�(� ) of the distribution � . Onechecks (cf. Appendix III) that q�(� ) is a generalized function.Exactly as in the case of a single transformation, the contributions to(10) will come from the �xed points of Ft. The latter will come either froma zero of the vector �eld v, (i.e., x 2 M such that vx = 0) or from a periodicorbit of the ow and we call T# the length of such a periodic orbit. Underthe above transversality hypothesis, the formula for (10) is (cf. [GS], [G] andthe Appendix III), \Trace" �Z h(t)Ut dt�(16) = Xx;vx=0 Z h(t)j1� (Ft)�j dt+X XT T# 1j1� (FT=)�j h(T )where in the second sum is a periodic orbit with length T# , and T variesin ZT# while (FT=)� is the Poincar�e return map, i.e., the restriction of thetangent map to the transversal of the orbit.One can rewrite (16) in a better way as,(17) \Trace" �Z h(t)Ut dt� =X ZI h(u)j1� (Fu)�j d�u ;where the zeros x 2M , vx = 0, are considered also as periodic orbits , whileI � R is the isotopy subgroup of any x 2 , and d�u is the unique Haarmeasure in I such that the covolume of I is equal to 1, i.e., such that forthe unique Haar measure d� of total mass 1 on R=I and any f 2 C1c (R),(18) ZR f(t) dt = ZR=I �ZI f(u+ s) d�u� d�(s) :Also we still write (Fu)� for the restriction of the tangent map to Fu tothe transverse space of the orbits. 18

To understand what (Ft)� looks like at a zero of v we can replace v(x)for x near x0 by its tangent map. For simplicity we take the one dimensionalcase, with v(x) = x @@x , acting on R=M .One has Ft(x) = et x. Since Ft is linear the tangent map (Ft)� is(19) (Ft)� = etand (12) becomes(20) \Trace" �Z h(t)Ut dt� = Z h(t)j1� etj dt :Thus, for this ow, the distribution trace formula is(21) \Trace" (U(h)) = Z h(u)j1� uj d�uwhere we used the multiplicative notation so that R�+ acts on R by multiplica-tion, while U(h) = R U(v)h(v) d�v and d�v is the Haar measure of the groupR�+. One can treat in a similar way the action, by multiplication, of the groupof nonzero complex numbers on the manifold C .We shall now investigate the more general case of an arbitrary local �eld.V The action (�; x)! �x of K� on a local �eld KWe let K be a local �eld and consider the map,(1) f : K �K� ! K ; f(x; �) = �xtogether with the diagonal map,(2) ' : K �K� ! K �K� �K ; '(x; �) = (x; �; x)as in IV (11) and (12) above.When K is Archimedean we are in the framework of manifolds and wecan associate to f a �-section with support Z = Graph (f),(3) �Z = �(y � �x) dy :19

Using the projection q(x; �) = � from K � K� to K�, we then consider asabove the generalized function on K� given by(4) q�('� �Z) :The formal computation of this generalized function of � isZ �(x � �x) dx = Z �((1 � �)x) dx = Z �(y) d((1 � �)�1 y)= j1� �j�1 Z �(y) dy = j1� �j�1 :We want to justify it by computing the convolution of the Fourier transformsof �(x� y) and �(y��x) since this is the correct way of de�ning the productof two distributions in this local context. Let us �rst compute the Fouriertransform of �(ax+ by) where (a; b) 2 K2(6= 0). The pairing between K2 andits dual K2 is given by(5) h(x; y); (�; �)i = �(x � + y �) 2 U(1) ;where � is a �xed nontrivial character of the additive group K.Let (c; d) 2 K2 be such that ad�bc = 1 and consider the linear invertibletransformation of K2,(6) L � xy � = � a bc d � � xy � :The Fourier transform of ' � L is given by(7) (' � L)^ = jdetLj�1 b' � (L�1)t :Here one has detL = 1 and (L�1)t is(8) (L�1)t = � d �c�b a � :One �rst computes the Fourier transform of �(x), the additive Haar measuredx is normalized so as to be selfdual, and in one variable, �(x) and 1 areFourier transforms of each other. Thus(9) (� 1)^ = 1 � :20

Using (7) one gets that the Fourier transform of �(ax + by) is �(�b � + a �).Thus we have to compute the convolution of the two generalized functions�(� + �) and �(� + ��). NowZ f(�; �) �(� + �) d� d� = Z f(�;��) d�and Z f(�; �) �(� + ��) d� d� = Z f(���; �) d� :Thus we are dealing with two measures carried respectively by two distinctlines. Their convolution evaluated on f 2 C1c (K2) is R f(�+�) d�(�) d�(�) =R R f((�;��) + (���; �)) d� d� = R R f(� � ��;�� + �) d� d� = �R R f(�0 ; �0)d�0 d�0��jJ j�1 where J is the determinant of the matrix � 1 ���1 1 � = L,so that � �0�0 � = J � �� �. One has J = 1 � � and thus the convolution of thegeneralized functions �(� + �) and �(� + ��) gives as expected the constantfunction(10) j1� �j�1 1 :Correspondingly, the product of the distribution �(x� y) and �(y��x) givesj1� �j�1 �0 so that(11) Z �(x � y) �(y � �x) dx dy = j1� �j�1 :In this local case the Fourier transform alone was su�cient to make sense ofthe relevant product of distributions. In fact this would continue to makesense if we replace �(y � �x) by R h(��1) �(y � �x) d� � where h(1) = 0 .We shall now treat in detail the more delicate general case where h(1) isarbitrary.We shall prove a precise general result (Theorem 3) which handles thelack of transversality when h(1) 6= 0. We deal directly with the followingoperator in L2(K),(12) U(h) = Z h(�)U(�) d�� ;where the scaling operator U(�) is de�ned by(13) (U(�) �)(x) = �(��1 x) 8x 2 K21

and where the multiplicative Haar measure d�� is normalized by(14) Zj�j2[1;�] d�� � log � when �!1 :To understand the \trace" of U(h) we shall proceed as in the Selberg traceformula ([Se]) and use a cuto�. For this we use the orthogonal projection P�onto the subspace(15) P� = f� 2 L2(K) ; �(x) = 0 8x ; jxj > �g :Thus, P� is the multiplication operator by the function ��, where ��(x) =1 if jxj � �, and �(x) = 0 for jxj > �. This gives an infrared cuto�, and to getan ultraviolet cuto� we use bP� = FP�F�1 where F is the Fourier transform(which depends upon the basic character �). We let(16) R� = bP� P� :The main result of this section is thenTheorem 3. Let K be a local �eld with basic character �. Let h 2 S(K�) havecompact support. Then R� U(h) is a trace class operator and when �! 1,one has Trace (R� U(h)) = 2h(1) log0 �+ Z 0 h(u�1)j1� uj d�u+ o(1)where 2 log0 � = R�2K�; j�j2[��1;�] d��, and the principal value R 0 is uniquelydetermined by the pairing with the unique distribution on K which agrees withduj1�uj for u 6= 1 and whose Fourier transform vanishes at 1.Proof. We normalize as above the additive Haar measure to be the selfdualone on K. Let the constant � > 0 be determined by the equality(17) Z1�j�j�� d�j�j � � log � when �!1 :so that d�� = ��1 d�j�j . Let L be the unique distribution, extension of ��1 duj1�ujwhose Fourier transform vanishes at 1, bL(1) = 0. One then has by de�nition(18) Z 0 h(u�1)j1� uj d�u = �L; h(u�1)juj � ;22

where h(u�1)juj = 0 for u�1 outside the support of h.Let T = U(h). We can write the Schwartz kernel of T as(19) k(x; y) = Z h(��1) �(y � �x) d�� :Given any such kernel k we introduce its symbol,(20) �(x; �) = Z k(x; x + u)�(u�) duas its partial Fourier transform. The Schwartz kernel rt�(x; y) of the transposeRt� is given by(21) rt�(x; y) = ��(x) (c��) (x � y) :Thus, the symbol �� of Rt� is simply(22) ��(x; �) = ��(x) ��(�) :The operator R� is of trace class and one has(23) Trace (R� T ) = Z k(x; y) rt�(x; y) dx dy :Using the Parseval formula we thus get(24) Trace (R� T ) = Zjxj��;j�j�� �(x; �) dx d� :Now the symbol � of T is given by(25) �(x; �) = Z h(��1)�Z �(x + u� �x)�(u�) du� d�� :One has(26) Z �(x + u� �x)�(u�) du = �((� � 1)x�) ;thus (25) gives(27) �(x; �) = ��1 ZK g(�)�(�x�) d�23

where(28) g(�) = h((�+ 1)�1) j� + 1j�1 :Since h is smooth with compact support on K�, the function g belongs toC1c (K).Thus �(x; �) = ��1 bg(x�) and(29) Trace (R� T ) = ��1 Zjxj��;j�j�� bg(x�) dx d� :With u = x� one has dx d� = du dxjxj and, for juj � �2,(30) ��1 Z juj� �jxj�� dxjxj = 2 log0 �� log juj(using the precise de�nition of log0 � to handle the boundary terms). Thuswe can rewrite (29) as(31) Trace (R� T ) = Zjuj��2 bg(u) (2 log0 �� log juj) du :Since g 2 C1c (K), one has(32) Zjuj��2 jbg(u)j du = O(��N ) 8Nand similarly for jbg(u) log jujj. Thus(33) Trace (R� T ) = 2 g(0) log0 �� Z bg(u) log juj du + o(1):Now for any local �eld K and basic character �, if we take for the Haarmeasure da the selfdual one, the Fourier transform of the distribution '(u) =� log juj is given outside 0 by(34) b'(a) = ��1 1jaj ;with � determined by (17). To see this one lets P be the distribution on Kgiven by(35) P (f) = lim"!0"2Mod(K) Zjxj�" f(x) d�x+ f(0) log "! :24

One has P (fa) = P (f)�log jaj f(0) which is enough to show that the functionbP (x) is equal to � log jxj+ cst, and b' di�ers from P by a multiple of �0.Thus the Parseval formula gives, with the convention of Theorem 3,(36) �Z bg(u) log juj du = 1� Z 0 g(a) dajaj :Replacing a by �� 1 and applying (28) gives the desired result.We shall show in Appendix II that the privileged principal value, whichdepends upon the basic character �, is the same as in Weil's explicit formulas.VI The global case and the formal trace computationWe shall now consider the action of Ck on X and write down the analogue ofIV (17) for the distribution trace formula.Both X and Ck are de�ned as quotients and we let(1) � : A! X ; c : GL1(A)! Ckbe the corresponding quotient maps.As above we consider the graph Z of the action(2) f : X � Ck ! X ; f(x; �) = �xand the diagonal map(3) ' : X �Ck ! X � Ck �X '(x; �) = (x; �; x) :We �rst investigate the �xed points, '�1(Z), i.e., the pairs (x; �) 2 X�Cksuch that �x = x. Let x = �(~x) and � = c(j). Then the equality �x = xmeans that �(j~x) = �(~x). Thus there exists q 2 k� such that with ~j = qj,one has(4) ~j~x = ~x :Recall now that A is the restricted direct product A =�res kv of the local �eldskv obtained by completion of k with respect to the place v. The equality (4)means that ~jv~xv = ~xv; thus, if ~xv 6= 0, for all v it follows that ~jv = 1 8v and25

~j = 1. This shows that the projection of '�1(Z) \Cknf1g on X is the unionof the hyperplanes(5) [Hv ; Hv = �( ~Hv) ; ~Hv = fx ; xv = 0g :Each ~Hv is closed in A and is invariant under multiplication by elements ofk�. Thus each Hv is a closed subset of X and one checks that it is the closureof the orbit under Ck of any of its generic points(6) x ; xu = 0 () u = v :For any such point x, the isotropy group Ix is the image in Ck of the multi-plicative group k�v,(7) Ix = k�vby the map � 2 k�v ! (1; : : : ; 1; �; 1; : : :). This map already occurs in class�eld theory (cf [W1]) to relate local Galois theory to the global one.Both groups k�v and Ck are commensurable to R�+ by the module homo-morphism, which is proper with cocompact range,(8) G j j�! R�+ :Since the restriction to k�v of the module of Ck is the module of k�v, it followsthat(9) Ix is a cocompact subgroup of Ck :This allows us to normalize the respective Haar measures in such a way thatthe covolume of Ix is 1. This is in fact insured by the canonical normalizationof the Haar measures of modulated groups ([W3 ]),(10) Zjgj2[1;�] d�g � log � when �! +1 :It is important to note that though Ix is cocompact in Ck, the orbit of x isnot closed and one needs to close it, the result being Hv. We shall learn howto justify this point later in Section VII, in the similar situation of the actionof CS on XS . We can now in view of the results of the two preceding sections,write down the contribution of each Hv to the distributional trace.26

Since ~Hv is a hyperplane, we can identify the transverse space Nx to Hvat x with the quotient(11) Nx = A= ~Hv = kv ;namely the additive group of the local �eld kv. Given j 2 Ix one has ju =1 8u 6= v, and jv = � 2 k�v . The action of j on A is linear and �xes x; thusthe action on the transverse space Nx is given by(12) (�; a)! �a 8a 2 kv:We can thus proceed with some faith and write down the contribution of Hvto the distributional trace in the form(13) Zk�v h(�)j1� �j d��where h is a test function on Ck which vanishes at 1. We now have to takecare of a discrepancy in notation with the third section (formula 9), where weused the symbol U(j) for the operation(14) �U(j)f�(x) = f(j�1x) ;whereas we use j in the above discussion. This amounts to replacing the testfunction h(u) by h(u�1), and we thus obtain as a formal analogue of III(17)the following expression for the distributional trace(15) \Trace" (U(h)) =Xv Zk�v h(u�1)j1� uj d�u :Now the right-hand side of (15) is, when restricted to the hyperplane h(1) = 0,the distribution obtained by Andr�e Weil [W3] as the synthesis of the explicitformulas of number theory for all L-functions with Gr�ossencharakter. In par-ticular we can rewrite it as(16) h(0) + h(1)� XL(�;�)=0 h(�; �) +1 h(1)where this time the restriction Re(�) = 12 has been eliminated.Thus, equating (34) of Section III and (16) for h(1) = 0 would yield thedesired information on the zeros. Of course, this do requires �rst eliminating27

the role of �, and (as in [AB]) to prove that the distributional trace coincideswith the ordinary operator theoretic trace on the cokernel of E. This isachieved for the usual setup of the Lefchetz �xed point theorem by the use offamilies.A very important property of the right hand side of (15) (and of IV (17)in general) is that if the test function h; h(1) = 0 is positive,(17) h(u) � 0 8 u 2 Ckthen the right-hand side is positive. This indicated from the very start thatin order to obtain the Polya-Hilbert space from the Riemann ow, it is notquantization that should be involved but simply the passage to the L2 space,X ! L2(X). Indeed the positivity of IV (17) is typical of permutation ma-trices rather than of quantization. This distinction plays a crucial role in theabove discussion of the trace formula, in particular the expected trace formulais not a semi-classical formula but a Lefchetz formula in the spirit of [AB].The above discussion is not a rigorous justi�cation of this formula. The�rst obvious obstacle is that the distributional trace is only formal and togive it a rigorous meaning tied up to Hilbert space operators, one needs asin Section V, to perform a cuto�. The second di�culty comes from the pres-ence of the parameter � as a label for the Hilbert space, while � does notappear in the trace formula. As we shall see in the next two sections, thecuto� will completely eliminate the role of �, and we shall nevertheless show(by proving positivity of the Weil distribution) that the validity of the (�independent) trace formula is equivalent to the Riemann Hypothesis for allGr�ossencharakters of k.VII Proof of the trace formula in the S-local caseIn the formal trace computation of Section VI, we skipped over the di�cultiesinherent in the tricky structure of the space X. In order to understand howto handle trace formulas on such spaces, we shall consider the slightly simplersituation which arises when one only considers a �nite set S of places of k. Assoon as the cardinality of S is larger than 3, the corresponding space XS doshares most of the tricky features of the space X. In particular it is no longerof type I in the sense of Noncommutative Geometry.We shall nevertheless prove a precise general result (Theorem 4) whichshows that the above handling of periodic orbits and of their contribution to28

the trace is the correct one. It will in particular show why the orbit of the�xed point 0, or of elements x 2 A, such that xv vanishes for at least twoplaces do not contribute to the trace formula.At the same time, we shall handle as in Section V, the lack of transver-sality when h(1) 6= 0.Let us �rst describe the reduced framework for the trace formula. Welet k be a global �eld and S a �nite set of places of k containing all in�niteplaces. The group O�S of S-units is de�ned as the subgroup of k�,(1) O�S = fq 2 k�; jqvj = 1; v =2 Sg :It is cocompact in J1S where(2) JS = Yv2S k�vand,(3) J1S = fj 2 JS; jjj = 1g:Thus the quotient group CS = JS=O�S plays the same role as Ck, and acts onthe quotient XS of AS =Qv2S kv by O�S .To keep in mind a simple example, one can take k = Q, while S consistsof the three places 2, 3, and1. One checks in this example that the topologyof XS is not of type I since for instance the group O�S = f�2n3m; n;m 2 Zgacts ergodically on f0g �R� AS .We normalize the multiplicative Haar measure d�� of CS by(4) Zj�j2[1;�] d�� � log � when �!1 ;and normalize the multiplicative Haar measure d�� of JS so that it agreeswith the above on a fundamental domain D for the action of O�S on JS .There is no di�culty in de�ning the Hilbert space L2(XS) of squareintegrable functions on XS . We proceed as in Section III (without the �),and complete (and separate) the Schwartz space S(AS ) for the pre-Hilbertstructure given by(5) kfk2 = Z ��� Xq2O�S f(qx)���2 jxj d�x29

where the integral is performed on CS or equivalently on a fundamental do-main D for the action of O�S on JS . To show that (5) makes sense, one provesthat for f 2 S(AS), the function E0(f)(x) =Pq2O�S f(qx) is bounded aboveby a power of Logjxj when jxj tends to zero. To see this when f is the char-acteristic function of fx 2 AS ; jxvj � 1 ;8v 2 Sg, one uses the cocompactnessof O�S in J1S , to replace the sum by an integral. The latter is then comparableto(6) Zui�0;Pui=�LogjxjYdui;where the index i varies in S. The general case follows.The scaling operator U(�) is de�ned by(7) (U(�) �)(x) = �(��1 x) 8x 2 ASand the same formula, with x 2 XS , de�nes its action on L2(XS). Given asmooth compactly supported function h on CS , U(h) = R h(g)U(g)dg makessense as an operator acting on L2(XS ).We shall now show that the Fourier transform F on S(AS) does extendto a unitary operator on the Hilbert space L2(XS ).Lemma 1. a) For any fi 2 S(AS ) the series PO�S hf1; U(q) f2iA of innerproducts in L2(AS ) converges geometrically on the abelian �nitely generatedgroup O�S . Moreover its sum is equal to the inner product of f1 and f2 in theHilbert space L2(XS).b) Let � = Q�v be a basic character of the additive group AS and Fthe corresponding Fourier transformation. The map f ! F (f), f 2 S(AS)extends uniquely to a unitary operator in the Hilbert space L2(XS ).Proof. The map L : O�S ! RS, given by L(u)v = log juvj, has a �nite kerneland its range is a lattice in the hyperplaneH = f(yv);PS yv = 0g. On H onehas SupSyv � 1=2nP jyvj, where n = card(S). Thus one has the inequality(8) SupSjqvj � exp(d(q; 1) 8q 2 O�Sfor a suitable word metric d on O�S .Let Kn = fx 2 AS ; jxvj � n; 8v 2 Sg and kn be the characteristicfunction of Kn. Let (�n) be a sequence of rapid decay such that(9) jfi(x)j �X�n kn(x) 8x 2 AS :30

One has for a suitable constant c,(10) j kn; U(q�1) kn� j � c nm(SupS jqvj)�1where m = Card(S).Using (9) we thus see that hf1; U(q) f2iA decays exponentially on O�S .Applying Fubini's theorem yields the equality,(11) Z ��� Xq2O�S f(qx)���2 jxj d�x = XO�S hf; U(q) fiA :This proves a). To prove b), one just uses (11) and the equalities hFf; FfiA =hf; fiA and F (U(q) f) = U(q�1)F (f).Now exactly as above for the case of local �elds (Theorem V.3), we need touse a cuto�. For this we use the orthogonal projection P� onto the subspace,(12) P� = f� 2 L2(XS) ; �(x) = 0 8x ; jxj > �g :Thus, P� is the multiplication operator by the function ��, where ��(x) = 1if jxj � �, and �(x) = 0 for jxj > �. This gives an infrared cuto� and to getan ultraviolet cuto� we use bP� = FP�F�1 where F is the Fourier transform(Lemma 1) which depends upon the choice of the basic character � = Q�v.We let(13) R� = bP� P� :The main result of this section is thenTheorem 4. Let AS be as above, with basic character � = Q�v. Let h 2S(CS) have compact support. Then when �!1, one hasTrace (R� U(h)) = 2h(1) log0 �+Xv2S Z 0k�v h(u�1)j1� uj d�u+ o(1)where 2 log0 � = R�2CS; j�j2[��1;�] d��, each k�v is embedded in CS by the mapu! (1; 1; :::; u; :::; 1) and the principal value R 0 is uniquely determined by thepairing with the unique distribution on kv which agrees with duj1�uj for u 6= 1and whose Fourier transform relative to �v vanishes at 1.31

Proof. We normalize as above the additive Haar measure dx to be the selfdualone on the abelian group AS . Let the constant � > 0 be determined by theequality, (where the fundamental domain D is as above),Z�2D;1�j�j�� d�j�j � � log� when �!1 :so that d�� = ��1 d�j�j .We let f be a smooth compactly supported function on JS such that(14) Xq2O�S f(qg) = h(g) 8 g 2 CS :The existence of such an f follows from the discreteness of O�S in JS . We thenhave the equality U(f) = U(h), where(15) U(f) = Z f(�)U(�) d�� :To compute the trace of U(h) acting on functions on the quotient space XS ,we shall proceed as in the Selberg trace formula ([Se]). Thus for an operatorT , acting on functions on AS , which commutes with the action of O�S and isrepresented by an integral kernel,(16) T (�) = Z k(x; y)�(y) dy;and the trace of its action on L2(XS) is given by(17) Tr(T ) = Xq2O�S ZD k(x; qx)dx ;where D is as above a fundamental domain for the action of O�S on the subsetJS of AS , whose complement is negligible. Let T = U(f). We can write theSchwartz kernel of T as(18) k(x; y) = Z f(��1) �(y � �x) d�� :By construction one has(19) k(qx; qy) = k(x; y) q 2 O�S :32

For any q 2 O�S, we shall evaluate the integral,(20) Iq = Zx2D k(qx; y)rt�(x; y)dydxwhere the Schwartz kernel rt�(x; y) for the transpose Rt� is given by(21) rt�(x; y) = ��(x) (c��) (x � y) :To evaluate the above integral, we let y = x + a and perform a Fouriertransform in a. For the Fourier transform in a of rt�(x; x + a), one gets,(22) ��(x; �) = ��(x) ��(�) :For the Fourier transform in a of k(qx; x + a), one gets(23) �(x; �) = Z f(��1)�Z �(x + a � �qx)�(a�) da� d�� :One has(24) Z �(x + a� �qx)�(a�) da = �((�q � 1)x�) ;thus (23) gives(25) �(x; �) = ��1 ZAS gq(u)�(ux�) duwhere(26) gq(u) = f(q(u + 1)�1) ju+ 1j�1 :Since f is smooth with compact support on A�S , the function gq belongs toC1c (AS).Thus �(x; �) = ��1 bgq(x�) and, using the Parseval formula we get(27) Iq = Zx2D; jxj��;j�j�� �(x; �) dx d� :This gives(28) Iq = ��1 Zx2D; jxj��;j�j�� bgq(x�) dx d� :33

With u = x� one has dx d� = du dxjxj and, for juj � �2,(29) ��1 Zx2D; juj� �jxj�� dxjxj = 2 log0 �� log juj(using the precise de�nition of log0 � to handle the boundary terms). Thuswe can rewrite (28) as(30) Trace (R� T ) = Xq2O�S Zjuj��2 bgq(u) (2 log0 �� log juj) du :Now log juj =Pv2S log juvj, and we shall �rst prove that(31) Xq2O�S Z bgq(u) du = h(1) ;while for any v 2 S,(32) Xq2O�S Z bgq(u) (� log juvj) du = Z 0k�v h(u�1)j1� uj d�u :In fact all the sums in q will have only �nitely many nonzero terms. It willthen remain to control the error term, namely to show that,(33) Xq2O�S Z bgq(u) (log juj � 2 log0 �)+ du = 0(��N )for any N , where we used the notation x+ = 0 if x � 0 and x+ = x if x > 0.Now recall that gq(u) = f(q(u + 1)�1) ju+ 1j�1 ;so that R bgq(u) du = gq(0) = f(q). Since f has compact support in A�S ,the intersection of O�S with the support of f is �nite and by (14) we get theequality (31).To prove (32), we consider the natural projection prv from Ql2S k�l toQl6=v k�l . The image prv(O�S) is still a discrete subgroup of Ql6=v k�l , (since k�vis cocompact in CS); thus there are only �nitely many q 2 O�S such that k�vmeets the support of fq, where fq(a) = f(qa) for all a.For each q 2 O�S one has, as in Section V,(34) Z bgq(u) (� log juvj) du = Z 0k�v fq(u�1)j1� uj d�u ;34

and from what we have just seen, this vanishes except for �nitely many q0s,so that by (14) we get the equality (32). Let us prove (33). Let "�(u) =(log juj � 2 log0 �)+, and let(35) �q(�) = Z bgq(u) "�(u) dube the error term. We shall proveLemma 2. For any � the series PO�S j�q(�)j converges geometrically on theabelian �nitely generated group O�S . Moreover its sum �(�) is O(��N ) forany N .Proof. Let (cf. (8)), d be a suitable word metric on O�S such that(36) SupSjqvj � exp(d(q; 1)) 8q 2 O�S :Let � 2 S(AS ) be de�ned by �(x) = f(x�1)jx�1j for all x 2 A�S andextended by 0 elsewhere. One has gq(x) = �(q�1(1+x)) for all x 2 AS , so thatbgq(u) = R gq(x)�(ux) dx = �(�u) b�(q u). Now, �q(�) = R bgq(u) "�(u) du =R b�(q u)�(�u) "�(u) du = R b�(y)�(�q�1 y) "�(y) dy, since "�(q u) = "�(u)for all u.Thus we get, using the symbol F� for the inverse Fourier transform of �,the equality,(37) �q(�) = F ("�b�)(q�1):Let � 2]0; 1=2[ and consider the norm(38) k�k = Supx2AS jF (�)(x)SupS jxvj�j:In order to estimate (38), we �x a smooth function on R, equal to 1 in aneighborhood of 0 and with support in [�1; 1], and introduce the convolutionoperators(39) (C�;v � �)(x) = Zkv (j"j)(�(x + ")� �(x)) d"j"j1+� ;and the norms,(40) k�k(1;�;v) = kC�;v � �k1 ;35

where k k1 is the L1 norm.The Fourier transform on kv of the distribution C�;v behaves like jxvj�for jxvj ! 1 . Thus, using the equality F (C�;v � �) = F (C�;v)F (�), andthe control of the sup norm of F (g) by the L1 norm of g, we get an inequalityof the form(41) Supx2AS jF (�)(x)SupS jxvj�j � c�XS k�k(1;�;v) :Let us now show that for any � 2 S(AS ), and � < 1=2, one has(42) k"� �k(1;�;v) = O(��N ) ;for any N .One has j("�(x + ")�(x + ") � "�(x)�(x)) � "�(x)(�(x + ") � �(x))j �j("�(x + ") � "�(x))jj�(x + ")j. Moreover using the inequality(43) ja+ � b+j � ja� bj ;we see that j("�(x + ") � "�(x))j � j log jxv + "j � log jxvjj, for " 2 kv. Letthen(44) c0� = Zkv log j1 + yj dyjyj1+� :It is �nite for all places v 2 S provided � < 1=2, and one has(45) Zkv (j"j)(j log jx + "j � log jxjj) d"j"j1+� � c0�jxj�� :Thus one obtains the inequality(46) jC�;v � "� � � "� (C�;v � �)j(x) � c0�jxvj�� Sup"2kv;j"j�1 j�(x+ ")j :Since the function jxvj�� is locally integrable, for � < 1, one has for � 2S(AS ), and any N ,(47) ZX� jxvj�� Sup"2kv;j"j�1 j�(x+ ")jdx = O(��N ) ;where X� = f y + "; jyj � �; " 2 kv; j"j � 1 g.36

Moreover one has for any N ,(48) k"� (C�;v � �)k1 = O(��N ):Thus, using (46), we obtain the inequality (42).Taking � = b� and using (41), we thus get numbers ��, such that �� =O(��N ) for all N and(49) jF ("�b�)SupSjxvj�jj � �� 8x 2 AS 8�:Taking x = q 2 O�S , and using (36) and (37), we thus get(50) j�q(�)j � ��exp(�d(q; 1)) 8q 2 O�S ;which is the desired inequality.VIII The trace formula in the global case, and elimination of �The main di�culty created by the parameter � in Theorem 1 is that theformal trace computation of Section VI is independent of �, and thus cannotgive in general the expected value of the trace of Theorem 1, since in the lattereach critical zero � is counted with a multiplicity equal to the largest integern < 1+�2 , n � multiplicity of � as a zero of L. In particular for L functionswith multiple zeros, the �-dependence of the spectral side is nontrivial. It isalso clear that the function space L2�(X) arti�cially eliminates the noncriticalzeros by the introduction of the �.As we shall see, all these problems are eliminated by the cuto�. The latterwill be performed directly on the Hilbert space L2(X) so that the only valueof � that we shall use is � = 0. All zeros will play a role in the spectral side ofthe trace formula, but while the critical zeros will appear perse, the noncriticalones will appear as resonances and enter in the trace formula through theirharmonic potential with respect to the critical line. Thus the spectral side isentirely canonical and independent of �, and by proving positivity of the Weildistribution, we shall show that its equality with the geometric side, i.e., theglobal analogue of Theorem 4, is equivalent to the Riemann Hypothesis forall L-functions with Gr�ossencharakter.The abelian group A of adeles of k is its own Pontrjagin dual by meansof the pairing(1) ha; bi = �(ab)37

where � : A! U(1) is a nontrivial character which vanishes on k � A. Notethat such a character is not canonical , but that any two such characters �and �0 are related by k�(2) �0(a) = �(qa) 8a 2 A :It follows that the corresponding Fourier transformations on A are related by(3) f 0 = fq :This is yet another reason why it is natural to mod out by functions of theform f � fq, i.e., to consider the quotient space X.We �x the additive character � as above, � = Q�v and let d be adi�erental idele(4) �(x) = �0(dx) 8x 2 A ;where �0 =Q�0;v is the product of the local normalized additive characters(cf [W1]). We let S0 be the �nite set of places where �v is rami�ed.We shall �rst concentrate on the case of positive characteristic, i.e., offunction �elds, both because it is technically simpler and also because it allowsus to keep track of the geometric signi�cance of the construction (cf. SectionII). In order to understand how to perform in the global case, the cuto�R� = bP� P� of Section VII, we shall �rst analyze the relative position of thepair of projections bP�, P� when �!1. Thus, we let S � S0 be a �nite setof places of k, large enough so that mod(CS) = mod(Ck) = qZ and that forany fundamental domainD for the action of O�S on JS, the product D�QR�vis a fundamental domain for the action of k� on Jk.Both bP� and P� commute with the decomposition of L2(XS) as the directsum of the subspaces, indexed by characters �0 of CS;1,(5) L2�0 = f� 2 L2(XS) ; �(a�1x) = �0(a) �(x); 8x 2 XS ; a 2 CS;1gwhich corresponds to the projections P�0 = R �0(a)U(a) d1 a, where d1 a isthe Haar measure of total mass 1 on CS;1.Lemma 1. Let �0 be a character of CS;1. Then for � large enough bP� andP� commute on the Hilbert space L2�0.38

Proof. Let US be the image in CS of the open subgroupQR�v. It is a subgroupof �nite index l in CS;1. Let us �x a character � of US and consider the �nitedirect sum of the Hilbert spaces L2�0 where �0 varies among the characters ofCS;1 whose restriction to US is equal to �,(6) L2(XS)� = f� 2 L2(XS) ; �(a�1x) = �(a) �(x); 8x 2 XS ; a 2 USg :The corresponding orthogonal projection is U(h�), where h� 2 S(CS) is suchthat(7) Supp(h�) = US h�(x) = ��(x) 8x 2 US ;and the constant � = l= log(q) corresponds to our standard normalization ofthe Haar measure on CS . Let as in Section VII, f 2 S(JS) with supportQR�v be such that U(f) = U(h) and let � 2 S(AS ) be de�ned by �(x) =f(x�1)jx�1j for all x 2 A�S and extended by 0 elsewhere.Since � is locally constant, its Fourier transform has compact supportand the equality (37) of Section VII shows that for � large enough, one hasthe equality(8) Trace ( bP� P� U(h�)) = 2h�(1) log0 �+Xv2S Z 0k�v h�(u�1)j1� uj d�u :With � = qN , one has 2 log0 � = (2N + 1) log(q) so that(9) 2h�(1) log0 � = (2N + 1)l :The character � of QR�v is a product, � = Q�v and if one uses the standardadditive character �0 to take the principal value one has, (cf. [W1] AppendixIV),(10) Z 0R�v �v(u)j1� uj d�u = �fv log(qv)where fv is the order of rami�cation of �v. We thus get(11) Z 0k�v h�(u�1)j1� uj d�u = �fv deg(v) l + l log(jdvj)log(q)where qv = qdeg(v), and since we use the additive character �v, we had totake into account the shift log(jdvj)h�(1) in the principal value.39

Now one has jdj = Q jdvj = q2�2g, where g is the genus of the curve.Thus we get(12) Trace ( bP� P� U(h�)) = (2N + 1)l � f l + (2� 2g) lwhere f = PS fv deg(v) is the order of rami�cation of �, i.e., the degree ofits conductor.Let B� = Im(P�) \ Im( bP�) be the intersection of the ranges of theprojections P� and bP�, and let B�� be its intersection with L2(XS )�. Weshall exhibit for each character � of US a vector �� 2 L2(XS )� such that(13) U(g)(��) 2 B� 8g 2 CS ; jgj � �; jg�1j � q2�2g�f �;while the vectors U(g)(��) are linearly independent for g 2 DS , where DS isthe quotient of CS by the open subgroup US .With � = qN as above, the number of elements g of DS such thatjgj � �; jg�1j � q2�2g�f � is precisely equal to (2N+1)l�f l+(2�2g) l, whichallows us to conclude that the projections bP� andP� commute in L2(XS )�and that the subspace B�� is the linear span of the U(g)(��).Let us now construct the vectors �� 2 L2(XS )�. With the notations of[W1] Proposition VII.13, we let(14) �� = YS �vbe the standard function associated to � = Q�v so that for unrami�ed v, �vis the characteristic function of Rv, while for rami�ed v it vanishes outsideR�v and agrees with �v on R�v. By construction the support of �� is containedin R = QRv. Thus one has U(g)(��) 2 Im(P�) if jgj � �. Similarlyby [W1] Proposition VII.13, we get that U(g)(��) 2 Im( bP�) as soon asjg�1j � q2�2g�f �. This shows that �� satis�es (13) and it remains to showthat the vectors U(g)(��) are linearly independent for g 2 DS .Let us start with a nontrivial relation of the form(15) kX�gU(g)(��)k = 0where the norm is taken in L2(XS), (cf. VII. 5). Let then �� = QS �v 1Rwhere R = Qv=2S Rv. Let us assume �rst that � 6= 1. Then �� gives anelement of L2�(X)0 which is cyclic for the representation U of Ck in the direct40

sum of the subspaces L2�;�0(X)0 where �0 varies among the characters of Ck;1whose restriction to U is equal to �.Now (15) implies that in L2�(X)0 one has P�gU(g)(��) = 0. By thecyclicity of ��, one then gets P�gU(g) = 0 on any L2�;�0(X)0 which gives acontradiction (cf. Appendix 1, Lemma 3).The proof for � = 1 is similar but requires more care since 1R =2 S0(A).We can thus rewrite Theorem 4 in the case of positive characteristic asCorollary 2. Let Q� be the orthogonal projection on the subspace of L2(XS)spanned by the f 2 S(AS) which vanish as well as their Fourier transform forjxj > �. Let h 2 S(CS) have compact support. Then when �!1, one hasTrace (Q� U(h)) = 2h(1) log0 �+Xv2S Z 0k�v h(u�1)j1� uj d�u+ o(1)where 2 log0 � = R�2CS; j�j2[��1;�] d��, and the other notations are as in The-orem VII.4.In fact the proof of Lemma 1 shows that the subspaces B� stabilize veryquickly, so that the natural map � ! � 1R from L2(XS ) to L2(X 0S) forS � S0 maps BS� onto BS0� .We thus get from Corollary 2 an S-independent global formulation of thecuto� and of the trace formula. We let L2(X) be the Hilbert space L2�(X) ofSection III for the trivial value � = 0 which of course eliminates the unpleasantterm from the inner product, and we let Q� be the orthogonal projection onthe subspace B� of L2(X) spanned by the f 2 S(A) which vanish as well astheir Fourier transform for jxj > �. As we mentionned earlier, the proof ofLemma 1 shows that for S and � large enough (and �xed character �), thenatural map � ! � 1R from L2(XS )� to L2(X)� maps BS� onto B�.It is thus natural to expect that the following global analogue of the traceformula of Corollary 2 actually holds, i.e., that when �!1, one has(16) Trace (Q� U(h)) = 2h(1) log0 �+Xv Z 0k�v h(u�1)j1� uj d�u+ o(1)where 2 log0 � = R�2Ck; j�j2[��1;�] d��, and the other notations are as in The-orem VII.4.We can prove directly that (16) holds when h is supported by Ck;1 butare not able to prove (16) directly for arbitrary h (even though the right hand41

side of the formula only contains �nitely many nonzero terms since h 2 S(Ck)has compact support). What we shall show however is that the trace formula(16) implies the positivity of the Weil distribution, and hence the validity ofRH for k. Remember that we are still in positive characteristic where RH isactually a theorem of A.Weil. It will thus be important to check the actualequivalence between the validity of RH and the formula (16). This is achievedbyTheorem 5. Let k be a global �eld of positive characteritic and let Q� be theorthogonal projection on the subspace of L2(X) spanned by the f 2 S(A) suchthat f(x) and bf (x) vanish for jxj > � . Let h 2 S(Ck) have compact support.Then the following conditions are equivalent:a) When �!1, one hasTrace (Q� U(h)) = 2h(1) log0 �+Xv Z 0k�v h(u�1)j1� uj d�u+ o(1) ;b) All L functions with Gr�ossencharakter on k satisfy the Riemann Hy-pothesis.Proof. To prove that a) implies b), we shall prove (assuming a)) the positivityof the Weil distribution (cf. Appendix 2),(17) � = log jd�1j �1 +D �Xv Dv :First, by Theorem III.1 applied for � = 0, the map E,(18) E(f) (g) = jgj1=2 Xq2k� f(qg) 8 g 2 Ck ;de�nes a surjective isometry from L2(X)0 to L2(Ck) such that(19) E U(a) = jaj1=2 V (a)E ;where the left regular representation V of Ck on L2(Ck) is given by(20) (V (a) �) (g) = �(a�1 g) 8 g; a 2 Ck :Let S� be the subspace of L2(Ck) given by(21) S� = f� 2 L2(Ck) ; �(g) = 0; 8g ; jgj =2 [��1; �]g :42

We shall denote by the same letter the corresponding orthogonal projection.Let B�;0 be the subspace of L2(X)0 spanned by the f 2 S(A)0 such thatf(x) and bf(x) vanish for jxj > � and let Q�;0 be the corresponding orthogonalprojection. Let f 2 S(A)0 be such that f(x) and bf (x) vanish for jxj > �.Then E(f) (g) vanishes for jgj > �, and the equality (Appendix 1),(22) E(f)(g) = E( bf)�1g� f 2 S(A)0;shows that E(f) (g) vanishes for jgj < ��1.This shows that E(B�;0) � S�, so that if we let Q0�;0 = EQ�;0E�1, weget the inequality(23) Q0�;0 � S�and for any �, the following distribution on Ck is of positive type,(24) ��(f) = Trace ((S� � Q0�;0)V (f));i.e., one has,(25) ��(f � f�) � 0;where f�(g) = f (g�1) for all g 2 Ck.Let then f(g) = jgj�1=2 h(g�1), so that by (19) one has E U(h) = V ( ~f )Ewhere ~f (g) = f(g�1) for all g 2 Ck. By Lemma 3 of Appendix 2 one has(26) Xv Dv(f) � log jd�1j = Xv Z 0k�v h(u�1)j1� uj d�u:One has Trace (S� V (f)) = 2f(1) log0 �; thus using a) we see that the limitof �� when � ! 1 is the Weil distribution � (cf.(17)). The term D inthe latter comes from the nuance between the subspaces B� and B�;0. Thisshows using (24) that the distribution � is of positive type so that b) holds(cf. [W3]).Let us now show that b) implies a). We shall compute from the zeros ofL-functions and independently of any hypothesis the limit of the distributions�� when �!1.We choose (non canonically) an isomorphism(27) Ck ' Ck;1 �N :43

where N = range j j � R�+, N 'Zis the subgroup qZ � R�+ .For � 2 C we let d��(z) be the harmonic measure of � with respect tothe line iR� C . It is a probability measure on the line iR and coincides withthe Dirac mass at � when � is on the line.The implication b))a) follows immediately from the explicit formulas(Appendix 2) and the following lemma.Lemma 3. The limit of the distributions �� when �!1 is given by�1(f) = XL�e�; 12+��=0�2B=N? N(e�; 12 + �) Zz2i R bf (e�; z)d��(z) ;where B is the open strip B = f� 2 C ;Re(�) 2]�12 ; 12 [g, N(e�; 12 + �) is themultiplicity of the zero, d��(z) is the harmonic measure of � with respect tothe line iR� C , and the Fourier transform bf of f is de�ned bybf (e�; �) = ZCk f(u) e�(u) juj� d� u :Proof. Let � = qN . The proof of Lemma 1 gives the lower bound (2N +1) �f + (2� 2g) for the dimension of B�;� in terms of the order of rami�cation fof the character � of Ck;1, where we assume �rst that � 6= 1. We have seenmoreover that E(B�;�) � S�;� while the dimension of S�;� is 2N + 1.Now by Lemma 3 of Appendix 1, every element � 2 E(B�;�) satis�es theconditions(28) Z �(x)�(x) jxj� d�x = 0 8� 2 B=N?; L��; 12 + �� = 0 :This gives 2g � 2 + f linearly independent conditions (for N large enough),using [W1] Theorem VII.6, and shows that they actually characterize thesubspace E(B�;�) of S�;�.This reduces the proof of the lemma to the following simple computa-tion: One lets F be a �nite subset (possibly with multiplicity) of C � andEN the subspace of SN = f� 2 l2(Z); �(n) = 0 8n; jnj > Ng de�ned bythe conditions P �(n) zn = 0 8z 2 F . One then has to compute the limitwhen N ! 1 of Trace ((SN � EN)V (f)) where V is the regular represen-tation of Z(so that V (f) = P fk V k where V is the shift, V (�)n = �n�1).44

One then checks that the unit vectors �z 2 SN ; z 2 F; �z(n) = zn (jz2N+1j �jz�(2N+1)j)� 12 (jzj�jz�1j) 12 8n 2 [�N; N ], are asymptotically orthogonal andspan (SN �EN) (when F has multiplicity one has to be more careful). Theconclusion follows from(29) LimN!1hV (f)�z ; �zi = Zjuj=1Pz(u) bf (u) du;where Pz(u) is the Poisson kernel, and bf the Fourier transform of f .One should compare this lemma with Corollary 2 of Theorem III.1. Inthe latter only the critical zeros were coming into play and with a multiplicitycontrolled by �. In the above lemma, all zeros do appear and with their fullmultiplicity, but while the critical zeros appear perse, the noncritical onesplay the role of resonances as in the Fermi theory.Let us now explain how the above results extend to number �elds k. We�rst need to analyze, as above, the relative position of the projections P� andbP�. Let us �rst remind the reader of the well known geometry of pairs ofprojectors. Recall that a pair of orthogonal projections Pi in Hilbert spaceis the same as a unitary representation of the dihedral group � =Z=2 �Z=2.To the generators Ui of � correspond the operators 2Pi � 1. The group �is the semidirect product of the subgroup generated by U = U1 U2 by thegroup Z=2, acting by U 7! U�1. Its irreducible unitary representations areparametrized by an angle � 2 [0; �2 ], the corresponding orthogonal projectionsPi being associated to the one dimensional subspaces y = 0 and y = x tg(�)in the Euclidean x; y plane. In particular these representations are at mosttwo dimensional. A general unitary representation is characterized by theoperator � whose value is the above angle � in the irreducible case. It isuniquely de�ned by 0 � � � �2 and the equality(30) Sin(�) = jP1 � P2j;and commutes with Pi.The �rst obvious di�culty is that when v is an Archimedean place thereexists no nonzero function on kv which vanishes as well as its Fourier trans-form for jxj > �. This would be a di�cult obstacle were it not for the work ofLandau, Pollak and Slepian ([LPS]) in the early sixties, motivated by prob-lems of electrical engineering, which allows us to overcome it by showing thatthough the projections P� and bP� do not commute exactly even for large �,45

their angle is su�ciently well behaved so that the subspace B� makes goodsense.For simplicity we shall take k = Q, so that the only in�nite place is real.Let P� be the orthogonal projection onto the subspace(31) P� = f� 2 L2(R) ; �(x) = 0; 8x ; jxj > �gand bP� = FP�F�1, where F is the Fourier transform associated to the basiccharacter �(x) = e�2�ix. What the above authors have done is to analyze therelative position of the projections P�, bP� for �!1 in order to account forthe obvious existence of signals (a recorded music piece for instance) whichfor all practical purposes have �nite support both in the time variable andthe dual frequency variable.The key observation of ([LPS]) is that the following second order di�er-ential operator on R actually commutes with the projections P�, bP�,(32) H� (x) = �@((�2 � x2) @) (x) + (2��x)2 (x);where @ is ordinary di�erentiation in one variable. To be more precise theabove equality de�nes a symmetric operator with natural domain the Schwartzspace. One can show that this operator has both de�ciency indices equal to 4and admits a unique selfadjoint extension which commutes with the dihedralgroup � associated to the projections P�, bP�. We let H� be this selfadjointoperator. It commutes with Fourier transform F .If one restricts to [��;�], the operator H� has discrete simple spectrum,and was studied long before the work of [LPS]. It appears from the factor-ization of the Helmoltz equation � + k2 = 0 in one of the few separablecoordinate systems in Euclidean 3-space, called the prolate spheroidal coor-dinates. Its eigenvalues �n(�); n � 0 are simple, positive. The correspondingeigenfunctions n are called the prolate spheroidal wave functions and sinceP� bP� P� commutes with H�, they are the eigenfunctions of P� bP� P�. Alot is known about them; in particular one can take them to be real valued,they are even for n even and odd for n odd and they have exactly n zerosin the interval [��;�]. When � ! 1 the function n converges to theWeber-Hermite function of order n (cf. [Si]).The key result of [LPS] is that the eigenvalues �n of the operator P� bP� P�are simple and if we let �0 > �1 > ::: > �n > ::: be their list in decreasingorder, one has(33) P� bP� P� n = �n n46

They behave qualitatively in the following manner. They stay very close to1, �n ' 1 until n falls in an interval of size ' log(�) around the value 4�2.Their behaviour in this interval is governed by the relation ([Sl])(34) �n = (1 + e��)�1where � is the solution of smallest absolute value of the equation(35) (n+ 1=2)� = 4��2 + � Log(8��2)� � (Log(j�=2j) � 1)Beyond this interval the �n tend very rapidly to zero. Of course this givesthe eigenvalues of �. One can show that for n in the above interval andh 2 C1c (R�+) one has(36) kU(h) nk = O(��a)for some a > 0. This allows to de�ne the analogue of the subspace B� ofLemma 1, as the linear span of the n; n � 4�2, the exact value of theupper bound being irrelevant by (36). It also gives the justi�cation of thesemiclassical counting of the number of quantum mechanical states which arelocalized in the interval [��;�] as well as their Fourier transform as the areaof the corresponding square in phase space.We now know what is the subspace B� for the single place 1, and toobtain it for an arbitrary set of places (containing the in�nite one), we justuse the same rule as in the case of function �elds, i.e., we consider the map,(37) 7! 1R;which su�ces when we deal with the Riemann zeta function. Note also that inthat case we restrict ourselves to even functions on R. This gives the analogueof Lemma 1, Theorem 5, and Lemma 3.To end this section we shall come back to our original motivation ofSection I and show how the formula for the number of zeros(38) N(E) � (E=2�)(log(E=2�)� 1) + 7=8 + o(1) +Nosc(E)appears from our spectral interpretation.Let us �rst do a semiclassical computation for the number of quantummechanical states in one degree of freedom which ful�ll the conditions(39) jqj � �; jpj � �; jHj � E ;47

where H = qp is the Hamiltonian which generates the group of scaling trans-formations(40) (U(�)�)(x) = �(��1x) � 2 R�+; x 2 R; � 2 L2(R) ;as in our general framework.To comply with our analysis of Section III, we have to restrict ourselvesto even functions so that we exclude the region pq � 0 of the semiclassical(p; q) plane.Now the region given by the above condition is equal to D = D+ [ (�D+)where(41) D+ = f(p; q) 2 R+�R+; p � �; q � �; pq � Eg :Let us compute the area of D+ for the canonical symplectic form(42) ! = 12� dp ^ dq :By construction D+ is the union of a rectangle with sides E=�, � with thesubgraph, from q = E=� to q = �, of the hyperbola pq = E. Thus,(43) ZD+ ! = 12� E=���+ 12� Z �E=� E dqq = E2� + 2E2� log �� E2� logE :Now the above computation corresponds to the standard normalization of theFourier transform with basic character of R given by(44) �(x) = exp(ix) :But we need to comply with the natural normalization at the in�nite place,(45) �0(x) = exp(�2�ix) :We thus need to perform the transformation(46) P = p=2� ; Q = q :The symplectic form is now dP ^ dQ and the domain(47) D0 = f(P;Q); jQj � �; jP j � �; jPQj � E=2�g:48

The computation is similar and yields the result(48) ZD0+ ! = 2E2� log�� E2� �log E2� � 1� :In this formula we thus see the overall term hN(E)i which appears witha minus sign which shows that the number of quantum mechanical statescorresponding to D0 is less than 4E2� log � by the �rst approximation to thenumber of zeros of zeta whose imaginary part is less than E in absolutevalue (one just multiplies by 2 the equality (43) since D0 = D0+ [ (�D0+).Now 12� (2E)(2 log �) is the number of quantum states in the Hilbert spaceL2(R�+; d�x) which are localized in R�+ between ��1 and � and are localizedin the dual group R (for the pairing h�; ti = �it) between �E and E. Thuswe see clearly that the �rst approximation to N(E) appears as the lack ofsurjectivity of the map which associates to quantum states � belonging to D0the function on R�+,(49) E(�)(x) = jxj1=2Xn2Z �(nx)where we assume the additional conditions �(0) = R �(x)dx = 0.A �ner analysis, which is just what the trace formula is doing, would yieldthe additional terms 7=8 + o(1) + Nosc(E). The above discussion yields anexplicit construction of a large matrix whose spectrum approaches the zerosof zeta as �!1.It is quite remarkable that the eigenvalues of the angle operator � dis-cussed above also play a key role in the theory of random unitary matrices.To be more speci�c, let E(n; s) be the large N limit of the probability thatthere are exactly n eigenvalues of a random Hermitian N �N matrix in theinterval [� �p2N t; �p2N t], t = s=2. Clearly Pn E(n; s) = 1. Let Pt be as abovethe operator of multiplication by 1[�t;t] { characteristic function of the interval[�t; t] in the Hilbert space L2(R). In general (cf [Me]), E(n; s) is (�1)n timesthe n-th coe�cient of the Taylor expansion at z = 1 of �s(z) = 1Q1 (1�z�j (s)),where �j(s) are the eigenvalues of the operator cP�Pt. (Here we denote bycP� = FP�F�1, and F denotes the Fourier transform, F�(u) = R eixu�(x)dx.Note also that the eigenvalues of cPaPb only depend upon the product ab sothat the relation with the eigenvalues of � should be clear.)49

General remarks.a) There is a close analogy between the construction of the Hilbert spaceL2(X) in Section III, and the construction of the physical Hilbert space ([S]Theorem 2.1) in constructive quantum �eld theory, in the case of gauge theo-ries. In both cases the action of the invariance group (the group k� = GL1(k)in our case, the gauge group in the case of gauge theories) is wiped out bythe very de�nition of the inner product. Compare the comment after III (9)with ([S]) top of page 17.b) For global �elds of zero characteristic, the Idele class group has a nontrivial connected component of the identity and this connected group has sofar received no interpretation from Galois theory (cf [W4]). The occurrenceof type III factors in [BC] indicates that the classi�cation of hyper�nite typeIII factors [C] should be viewed as a re�nement of local class �eld theory forArchimedean places, and provide the missing interpretation of the connectedcomponent of the identity in the Idele class group. In particular hyper�nitetype III factors are classi�ed by closed (virtual) subgroups of R�+ (cf [C]) andthey all appear as \unrami�ed" extensions of the hyper�nite factor of typeIII1.c) Our construction of the Polya-Hilbert space bears some resemblanceto [Z] and in fact one should clarify their relation, as well as the relation ofthe space X of Adele classes with the hypothetical arithmetic site of Deninger[D]. Note that the division of A by k� eliminates the linear structure of A andthat it transforms drastically the formulas for dimensions of function spaces,replacing products by sums (cf. Theorem 4 of Section VII). It should be clearto the reader that the action of the Idele class group on the space of Adeleclasses is the analogue (through the usual dictionnary of class �eld theory) ofthe action of the Frobenius on the curve. (To be more speci�c one needs todivide �rst the space X of Adele classes by the action of the maximal compactsubgroup of the Idele class group).d) There is a super�cial resemblance between the way the function N(E)appears in the last computation of section VIII and the discussion in [BK],directly inspired from [Co]. It is amusing to note that the computation of [BK]is actually coincidental, the two rectangles are eliminated for no reason, whichchanges appropriately the sign of the term in E. What [BK] had not takeninto account is that the spectral interpretation of [Co] is as an absorptionspectrum rather than an emission spectrum.d) There is an even more super�cial resemblance with the work of D.50

Goldfeld [G]; in the latter the Weil distribution is used to de�ne a corre-sponding inner product on a function space on the Idele class group. Thepositivity of the inner product is of course equivalent to the positivity of theWeil distribution (and by the result of A. Weil to RH) but this does not giveany clue as to how to prove this positivity, nor does it give any explanation(except for a nice observation at the Archimedean place) for what the Weildistribution is, since it is introduced by hands in the formula for the innerproduct.e) The above framework extends naturally from the case of GL(1) tothe case of GL(n) where the Adele class space is replaced by the quotient ofMn(A) by the action on the right of GLn(k). Preliminary work of C. Souleshows that the analogue of Theorem III. 1 remains valid; the next step is towork out the analogue of the Lefchetz trace formula in this context.f) I have been told by P. Sarnak and E. Bombieri that Paul Cohen hasconsidered the space of Adele classes in connection with RH, but never gotany detail of his unpublished ideas.All the results of the present paper have been announced in the September98 conference on the Riemann hypothesis held in the Schrodinger Institute inVienna and have been published as a preprint of the Schrodinger Institute.We are grateful to the American Institute of Mathematics for its sponsoringof the conference.Appendix I, Proof of Theorem 1In this appendix we give the proof of Theorem 1. Let us �rst recall as apreliminary the results of Tate an Iwasawa as interpreted in [W 2]L functions and homogeneous distributions on A.In general for a non archimedean local �eld K we use the notations R forthe maximal compact subring, P for the maximal ideal of R, � for a generatorof the ideal P (i.e., P = �R).Let k be a global �eld and A the ring of Adeles of k. It is the restrictedproduct of the local �elds kv indexed by the set of places v of k, with respectto the maximal compact subrings Rv. Similarly, the Bruhat-Schwartz spaceS(A) is the restricted tensor product of the local Bruhat-Schwartz spacesS(kv), with respect to the vectors 1Rv .L functions on k are associated to Gr�ossencharakters, i.e., to characters51

of the Idele class group(1) Ck = Jk=k� :Let X be a character of the idele class group, we consider X as a character ofJk which is 1 on k�. As such it can be written as a product(2) X (j) = �Xv(jv) j = (jv) 2 Jk :By considering the restriction of X to the compact subgroup(3) G0 = �R�v � 1 � Jk ;it follows that for all �nite v but a �nite number, one has(4) Xv=R�v = 1 :One says that X is unrami�ed at v when this holds.Then Xv(x) only depends upon the module jxj, since(5) k�v=R�v = mod(kv) :Thus Xv is determined by(6) Xv(�v)which does not depend upon the choice of �v (modR�v).Let X be a quasi-character of Ck, of the form(7) X (x) = X0(x) jxjswhere s 2 C and X0 is a character of Ck. The real part � of s is uniquelydetermined by(8) jX (x)j = jxj� :Let P be the �nite set of �nite places where X0 is rami�ed. The L functionL(X0; s) is de�ned for � = Re(s) > 1 as(9) L(X0; s) =0B@ Yv finitev=2P (1 �X0;v(�v) q�sv )�11CA = 0B@ Yv finitev=2P (1�Xv(�v))�11CA52

where(10) j�vj = q�1v :Let us now recall from [W 2] how L(X0; s) appears as a normalization factorfor homogeneous distributions on A.First let K be a local �eld and X a quasi-character of K�,(11) X (x) = X0(x) jxjs ; X0 : K� ! U(1) :A distribution D on K is homogeneous of weight X i� one has(12) hfa;Di = X (a)�1 hf;Difor all test functions f and all a in K�, where by de�nition(13) fa(x) = f(ax) :When � = Re(s) > 0, there exists up to normalization only one homoge-neous distribution of weight X on K,(cf [W 2]). It is given by the absolutelyconvergent integral(14) ZK� f(x)X (x)d�x = �X (f) :In particular, let K be non archimedean. Then, for any compactly sup-ported locally constant function f on K, one has(15) f(x) � f(��1x) = 0 8x; jxj � � :Thus, for any s 2 C the integral(16) ZK� (f(x) � f(��1x)) jxjs d�x = �0s(f)with the multiplicative Haar measure d�x normalized by(17) h1R� ; d�xi = 1 :de�nes a distribution on K with the properties(18) h1R;�0si = 153

(19) hfa;�0si = jaj�s hf;�0siand(20) �0s = (1� q�s)�s;where j�j = q�1. (Let us check (18): : :(20). With f = 1R one has f(��1x) = 1i� ��1x 2 R, i.e., x 2 �R = P . Thus �0s(1R) = RR� d�x = 1. Let uscheck (20). One has R f(��1x) jxjs d�x = R f(y) j�js jyjs d�y = j�js�s(f).But j�j < 1, j�j = 1q . Note also that for s = 1 and f = 1R, one getsRR� dx = �1� 1q�RR dx.)Let then X be a quasi-character of Ck and write as above(21) X = �Xv ; X (x) = X0(x) jxjswhere s 2 C and X0 is a character. Let P be the �nite set of �nite placeswhere it is rami�ed. For any �nite place v =2 P , let �0v(s) be the uniquehomogeneous distribution of weight Xv normalized by(22) h�0v(s); 1Rv i = 1 :For any v 2 P or any in�nite place, let, for � = Re(s) > 0, �0v be given by(14) which is homogeneous of weight Xv but unnormalized. Then the in�nitetensor product(23) �0s = ��0v(s)makes sense as a continuous linear form on S(A) and is homogeneous of weightX . This solution is not equal to 0 since �0v 6= 0 for any v 2 P and any in�niteplace as well. It is �nite by construction of the space S(A) of test functionsas an in�nite tensor product(24) S(A) = (S(kv); 1Rv ) :Lemma 1. (cf [W 2]) For � = Re(s) > 1, the following integral convergesabsolutely Z f(x)X0(x) jxjs d�x = �s(f) 8 f 2 S(A)54

and �s(f) = L(X0; s)�0s(f).Proof. To get the absolute convergence, one can assume that f = 1R andX0 = 1. Then one has to control an in�nite product of local terms, givenlocally for the Haar measure d�x on k�v such that RR�v d�x = 1, by(25) ZR\k�v jxjs d�x (s real)which is 1 + q�sv + q�2sv + : : : = (1� q�sv )�1. Thus the convergence for � > 1is the same as for the zeta function.To prove the second equality, one only needs to consider the in�nite tensorproduct for �nite places v =2 P . Then by (20) one has �0v = (1 � q��vv )�vwhere(26) q��vv = Xv(�) = X0;v(�) q�svwith j�j = q�1v .Thus one gets �s =0B@ Yv finitev=2P (1�X0;v(�) q�sv )�11CA �0s = L(X0; s)�0s.By construction �0s makes sense whenever � > 0 and is a holomorphicfunction of s (for �xed f). Let us review brie y (cf. [W2]) how to extend thede�nition of �s.We let as above k be a global �eld, we �x a nontrivial additive character� of A, trivial on k,(27) �(x + y) = �(x)�(y) 2 U(1) ; �(q) = 1 8 q 2 k :We then identify the dual of the locally compact additive group A with Aitself by the pairing(28) hx; yi = �(xy) :One shows (cf. [W 1]) that the lattice k � A, i.e., the discrete and cocompactadditive subgroup k, is its own dual(29) hx; qi = 1 8 q 2 k , x 2 k :55

Since A is the restricted product of the local �elds kv, one can write � as anin�nite product(30) � = ��vwhere for almost all v, one has �v = 1 on Rv. Let us recall the de�nition ofthe space S(A)0(31) S(A)0 = ff 2 S(A) ; f(0) = 0 ; Z f dx = 0g :Lemma 2. Let f 2 S(A)0. Then the seriesE(f) (g) = jgj1=2 Xq2k� f(qg) 8 g 2 Ckconverges absolutely and one has8n ; 9 c; jE(f)(g)j � c e�nj log jgjj 8 g 2 Ckand E( bf ) (g) = E(f) (g�1).Proof. Let us �rst recall the formal de�nition ([Br]) of the Bruhat-Schwartzspace S(G) for an arbitrary locally compact abelian group G. One considersall pairs of subgroups G1; G2 of G such that G1 is generated by a compactneighborhood of 0 in G, while G2 is a compact subgroup of G1 such thatthe quotient group is elementary, i.e., is of the form RaTbZcF for F a �nitegroup. By de�nition the Bruhat-Schwartz space S(G) is the inductive limitof the Schwartz spaces S(G1=G2) where the latter have the usual de�nition interms of rapid decay of all derivatives. Since G1 is open in G, any element ofS(G1=G2) extended by 0 outside G1 de�nes a continuous function on G. Byconstruction S(G) is the union of the subspaces S(G1=G2) and it is endowedwith the inductive limit topology.Let bG be the Pontrjagin dual of G. Then the Fourier transform, whichdepends upon the normalization of the Haar measure on G, gives an isomor-phism of S(G) with S( bG).Let � be a lattice in the locally compact abelian group G. Then anyfunction f 2 S(G) is admissible for the pair G;� in the sense of [W 1], andthe Poisson summation formula (cf [W 1]) is the equality(32) Covol (�)X 2� f( ) = X�2�? bf (�) ;56

where �? is the dual of the lattice �, and(33) bf(�) = Z f(a)�(a) da :Both sides of (32) depend upon the normalization of the Haar measure on G.In our case we let A be as above the additive group of Adeles on k. Wenormalize the additive Haar measure dx on A by(34) Covol(k) = 1 :We then take � = xk, for some x 2 A�1. One has(35) Covol (xk) = jxj :The dual �? of the lattice xk, for x invertible in A, is the lattice �? = x�1k.Thus the Poisson formula (32) reads, for any f 2 S(A),(36) jxjXq2k f(xq) =Xq2k bf (x�1q) :Which we can rewrite as(37) jxjXk� f(xq) =Xk� bf (x�1q) + �� = �jxj f(0) + R f(y) dy.We can then rewrite (37) as the equality, valid for all f 2 S(A)0(38) E(f)(x) = E( bf)� 1x� f 2 S(A)0:It remains to control the growth of E(f)(x) on Ck, but by (38), it is enoughto understand what happens for jxj large.We only treat the case of number �elds, the general case is similar. LetA = Af � A1 be the decomposition of the ring of Adeles corresponding to�nite and in�nite places. Thus A1 =YS1kv where S1 is the set of in�niteplaces.Any element of S(A) is a �nite linear combination of test functions ofthe form(39) f = f0 f157

where f0 2 S(Af ) ; f1 2 S(A1) (cf. [W 5] 39); thus it is enough to controlthe growth of E(f)(x) for such f and jxj large.Let Jk;1 = fx 2 Jk; jxj = 1g be the group of Ideles of module one. SinceJk;1=k� is compact (cf. [W 1]), we shall �x a compact subset K1 of Jk;1 whoseimage in Jk;1=k� is this compact group.Let � be the diagonal embedding(40) � 2 R�+ ��! (�; : : : ; �) 2YS1 k�vwhich yields an isomorphism(41) Jk = Jk;1 � Im � :One has f0 2 S(Af ), hence (cf. [W 5]), f0 2 Cc(Af ) and we let K0 =Support f0. Since K0 is compact, one can �nd a �nite subset P of the set of�nite places, and C <1 such that(42) y 2 K = (Kf )�1K0 ) jyvj � 1 8 v =2 P; jyvj � C 8 v ;where Kf is the projection of K1 on Af .We let be the compact open subgroup of Af determined by(43) javj � 1 8 v =2 P; javj � C 8 v :By constructionE(f)(x) only depends upon the class of x in Jk=k�. Thus,to control the behaviour of E(f)(x) for jxj ! 1, we can take x = (xf ; x1) 2K1 and consider E(f)(�x) for � 2 R�+ ,� ! 1. Now let q = (qf ; q1) 2 k.Then,(44) f(q �x) = f0(qf xf ) f1(q1 �x1)and this vanishes unless qf xf 2 K0, i.e., unless qf 2 K. But then by (42)one has qf 2 . Let � be the lattice inYS1kv determined by(45) � = fq1; q 2 k; qf 2 ; g :The size of E(f)(�x) is thus controlled (up to the square root of j�xj ) by(46) C Xn2�� jf1(�x1 n)j ;58

where x1 varies in the projection K1 of K1 on YS1k�v .Since f1 2 S(A1), this shows that E(f)(x) decays faster than any powerof jxj for jxj ! 1.We have shown that E(f) has rapid decay in terms of jxj, for jxj ! 1.Using (38) and the stability of S(A)0 under Fourier, we see that it also hasexponential decay in terms of j log jxjj when j log jxjj ! 1.We then getLemma 3. (cf. [W 2]) For � = Re(s) > 0, and any character X0 of Ck, onehas Z E(f)(x)X0(x) jxjs�1=2 d�x = cL(X0; s)�0s(f) 8 f 2 S(A)0where the nonzero constant c depends upon the normalization of the Haarmeasure d�x on Ck.Proof. For � = Re(s) > 1, the equality follows from Lemma 1, but since bothsides are analytic in s it holds in general.As in Lemma 1, we shall continue to use the notation �s(f) for � =Re(s) > 0.Approximate units in the Sobolev spaces L2�(Ck)We �rst consider, for � > 1, the Hilbert space L2�(R) of functions �(u),u 2 R with square norm given by(1) ZR j�(u)j2 (1 + u2)�=2 du :We let �(u) = (1 + u2)�=2. It is comparable to (1 + juj)� and in particular,(2) �(u+ a)�(u) � c �(a) 8u 2 R ; a 2 Rwith c = 2�=2.We then let V (v) be the translation operator(3) (V (v)�)(u) = �(u� v) 8u; v 2 R :59

One has RR j�(u� v)j2 �(u) du = RR j�(u)j2 �(u+ v) du so that by (2) it is lessthan c RR j�(u)j2 �(u) �(v) du = c �(v) k�k2,(4) kV (v)k � (c �(v))1=2 :This shows that V (f) = R f(v)V (v) dv makes sense as soon as(5) Z jf(v)j �(v)1=2 dv <1 :This holds for all f 2 S(R).Lemma 4. There exists an approximate unit fn 2 S(R), such that bfn hascompact support, kV (fn)k � C 8n, andV (fn)! 1 strongly in L2�(R) :Proof. Let f be a function, f 2 S(R), whose Fourier transform bf has compactsupport, and such that R f dx = 1 (i.e., bf (0) = 1). Let then(6) fn(v) = nf(nv) n = 1; 2; : : :One has R jfn(v)j �(v)1=2 dv = R jf(u)j � �un�1=2 du � R jf(u)j �(u)1=2 du.Thus kV (fn)k is uniformly bounded.We can assume that bf is equal to 1 on [�1; 1]. Then bfn is equal to 1 on[�n; n] and V (fn)� = � for any � with Supp b� � [�n; n] . By uniformity onegets that V (fn)! 1 strongly.Let us now identify the dual (L2�)� of the Hilbert space L2� with L2�� bymeans of the pairing(7) h�; �i0 = ZR �(u) �(u) du :Since L2� is a Hilbert space, it is its own dual using the pairing(8) h�; �1i = ZR �(u) �1(u)(1 + u2)�=2 du :If we let �(u) = �1(u)(1 + u2)�=2, thenZ j�1(u)j2 (1 + u2)�=2 du = Z j�(u)j2 (1 + u2)��=2 du60

which is the natural norm square for L2��.Given a quasicompact group such as Ck with module(9) j j : Ck ! R�+ ;we let d�g be the Haar measure on Ck normalized by(10) Zjgj2[1;�] d�g � log � �!1and we let L2�(Ck) be de�ned by the norm(11) ZCk j�(g)j2 (1 + log jgj2)�=2 d�g :It is, when the module of k is R�+, a direct sum of spaces (1), labelled by thecharacters X0 of the compact group(12) Ck;1 = Kermod :The pairing between L2�(Ck) and L2��(Ck) is given by(13) h�; �i = Z �(g) �(g) d�g :The natural representation V of Ck by translations is given by(14) (V (a)�)(g) = �(a�1g) 8 g; a 2 Ck :It is not unitary but by (4) one has(15) kV (g)k = 0 j log jgjj�=2 ; j log jgjj ! 1 :Finally, one has, using Lemma 4 and the decomposition Ck = Ck;1 �N .Lemma 5. There exists an approximate unit fn 2 S(Ck), such that bfn hascompact support, kV (fn)k � C 8n, andV (fn)! 1 strongly in L2�(Ck) :61

Proof of Theorem III 1We �rst consider the subspace of codimension 2 of S(A) given by(1) f(0) = 0 ; Z f dx = 0 :On this subspace S(A)0 we put the inner product(2) ZCk jE(f)(x)j2 (1 + log jxj2)�=2 d�x :We let U be the representation of Ck on S(A) given by(3) (U(a)�)(x) = �(a�1x) 8 a 2 Ck ; x 2 A :We let L2�(X)0 be the separated completion of S(A)0 for the inner productgiven by (2). The linear map E : S(A)0 ! L2�(Ck) satis�es(4) kE(f)k2� = kfk2�by construction. Thus it extends to an isometry, still noted E,(5) E : L2�(X)0 ,! L2�(Ck) :One hasE(U(a)f)(g) = jgj1=2Xk� (U(a)f)(qg) = jgj1=2Xk� f(a�1qg)= jgj1=2Xk� f(qa�1g) = jaj1=2 ja�1gj1=2Xk� f(qa�1g)= jaj1=2(V (a)E(f))(g)(6) E U(a) = jaj1=2 V (a)E :The equality (6) shows that the natural representationU of Ck on L2�(X)0corresponds by the isometry E to the restriction of jaj1=2 V (a) to the invariantsubspace given by the range of E.In order to understand ImE we consider its orthogonal in the dual spaceL2��(Ck). The compact subgroup(7) Ck;1 = fg 2 Ck ; jgj = 1g62

acts by the representation V which is unitary when restricted to Ck;1. Thusone can decompose L2�(Ck) and its dual L2��(Ck) , in the direct sum of thesubspaces,(8) L2�;X0 = f� 2 L2�(Ck) ; �(a�1g) = X0(a) �(g) 8 g 2 Ck ; a 2 Ck;1gand(9) L2��;X0 = f� 2 L2��(Ck) ; �(a g) = X0(a) �(g) 8 g 2 Ck ; a 2 Ck;1gwhich corresponds to the projections PX0 = R X0(a)V (a) d1 a for L2� andP tX0 = R X0(a)V (a)t d1 a for the dual space L2��.In (9) we used the formula(10) (V (g)t �)(x) = �(gx)which follows from the de�nition of the transpose, hV (g)�; �i = h�; V (g)t�iusing Z �(g�1x) �(x) d�x = Z �(y) �(gy) d�y :In these formulas one only uses the character X0 as a character of thecompact subgroup Ck;1 of Ck. One now chooses, noncanonically, an extensioneX0 of X0 as a character of Ck(11) eX0(g) = X0(g) 8g 2 Ck;1 :This choice is not unique but any two such extensions di�er by a characterwhich is principal, i.e., of the form: g! jgjis0, s0 2 R.Let us �x a factorization Ck = Ck;1 �R�+, and �x eX0 as being equal to 1on R�+.We then write any element of L2��;X0(Ck) in the form(12) g 2 Ck ! �(g) = eX0(g) (jgj)where(13) Z j (jgj)2 (1 + (log jgj)2)��=2 d�g <1 :This vector is in the orthogonal of ImE i�(14) Z E(f)(x) eX0(x) (jxj) d�x = 0 8 f 2 S(A)0 :63

We �rst proceed formally and write (jxj) = R b (t) jxjit dt so that the lefthand side of (14) becomes(15) Z Z E(f)(x) eX0(x) jxjit b (t) d�xdt = Z �1=2+it(f) b (t) dt(using the notations of Lemmas 1 and 3).Let us justify this formal manipulation; since we deal with the orthogonalof an invariant subspace, we can assume that(16) V t(h) � = �;for some h such that bh has compact support. Indeed we can use Lemma 5 toonly consider vectors which belong to the range ofV t(h) = Z h(g)V (g)t d�g ; bh with compact support:Then, using (16), the Fourier transform of the tempered distribution on R�+ has compact support in R. Thus, since E(f)(x) has rapid decay,the equality between (14) and (15) follows from the de�nition of the Fouriertransform of the tempered distribution on R�+.Let us now describe suitable test functions f 2 S(A)0 in order to test thedistribution(17) Z � 12+it b (t) dt :We treat the case of characteristic zero; the general case is similar. Forthe �nite places we take(18) f0 = v=2P 1Rv fX0where fX0 is the tensor product over rami�ed places of the functions equal to0 outside R�v and to X 0;v on R�v. It follows then by the de�nition of �0s that(19) h�0s; f0 fi = Z f(x)X0;1(x) jxjs d�xfor any f 2 S(A1). Moreover if the set P of �nite rami�ed places is notempty, one has(20) f0(0) = 0 ; ZAf f0(x) dx = 064

so that f0 f 2 S(A)0 8 f 2 S(A1).Now let ` be the number of in�nite places of k and consider the map� : (R�+)` ! R�+ given by(21) �(�1; : : : ; �`) = �1 : : : �` :As soon as ` > 1 this map is not proper. Given a smooth function withcompact support, b 2 C1c (R�+), we need to �nd a 2 C1c ((R�+)`) such that thedirect image of the measure a(x) d�x is b(y) d�y where d�x = � d�xi is theproduct of the multiplicative Haar measures.Equivalently, one is dealing with a �nite dimensional vector space E anda linear form L : E ! R. One is given b 2 C1c (R) and asked to lift it. One canwrite E = R�E1 and the lift can be taken as a = b b1 where b1 2 C1c (E1),R b1 dx = 1.Thus we can in (19) take a function f of the form(22) f(x) = g(x)X 0;1 (x) ;where the function g 2 C1c (A1) only depends upon (jxjv), v 2 S1 and issmooth with compact support, disjoint from the closed set(x 2 Yv2S1kv ; 9 v ; xv = 0) :Thus, to any function b 2 C1c (R�+) we can assign a test function f = fbsuch that for any s (Re s > 0)(23) h�0s; f0 fbi = ZR�+ b(x) jxjs d�x :By Lemma 3, we get�Z � 12+it b (t) dt ; f0 fb� = �Z L (X0; 12 + it)�012+it b (t) dt ; f0 fb�= Z Z L (X0; 12 + it) b (t) b(x) jxj 12+it d�xdt :Thus, from (14) and (15) we conclude, using arbitrary test functions b that theFourier transform of the distribution L (X0; 1=2 + it) b (t) actually vanishes,(24) L (X0; 12 + it) b (t) = 0 :65

To justify the above equality, we need to control the growth of the Lfunction in the variable t. One has(25) jL(12 + it)j = 0 (jtjN ) :In particular, since L �12 + it� is an analytic function of t, we see that it is amultiplier of the algebra S(R) of Schwartz functions in the variable t. Thusthe product L ( 12+it) b (t) is still a tempered distribution, and so is its Fouriertransform. To say that the latter vanishes when tested on arbitrary functionswhich are smooth with compact support implies that it vanishes.The above argument uses the hypothesis X0=Ck;1 6= 1.In the case X0=Ck;1 = 1 we need to impose to the test function f used in(22) the condition R f dx = 0 which means(26) Z b(x) jxj d�x = 0 :But the space of functions b(x) jxj1=2 2 C1c (R�+) such that (26) holds is stilldense in the Schwartz space S(R�+).To understand the equation (24), let us consider an equation for distri-butions �(t) of the form(27) '(t)�(t) = 0 ;where we �rst work with distributions � on S1 and we assume that ' 2C1(S1) has �nitely many zeros xi 2 Z('), of �nite order ni. Let J be theideal of C1(S1) generated by '. One has 2 J , order of at xi is � ni.Thus the distributions �xi , �0xi ; : : : ; �(ni�1)xi form a basis of the space ofsolutions of (27).Now b (t) is, for � orthogonal to Im(E) and satisfying (16), a distribu-tion with compact support, and L �X0; 12 + it� b (t) = 0. Thus by the aboveargument we get that b is a �nite linear combination of the distributions(28) �(k)t ; L�X0; 12 + it� = 0 ; k < order of the zero; k < � � 12 :The condition k < order of the zero is necessary and su�cient to get thevanishing on the range of E. The condition k < ��12 is necessary and su�cientto ensure that belongs to L2��, i.e., that(29) Z (log jxj)2k (1 + j log jxjj2)��=2 d�x <166

which is 2k + � < �1, i.e., k < ��12 .Conversely, let s be a zero of L(X0; s) and k > 0 its order. By Lemma 3and the �niteness and analyticity of �0s (for Re s > 0) we get(30) � @@s�a �s(f) = 0 8 f 2 S(A)0 ; a = 0; 1; : : : ; k � 1 :We can di�erentiate the equality of Lemma 3 and get(31) � @@s�a �s(f) = ZCk E(f)(x)X0(x) jxjs�1=2 (log jxj)a d�x :Thus � belongs to the orthogonal of Im(E) and satis�es (16) i� it is a �nitelinear combination of functions of the form(32) �t;a(x) = X0(x) jxjit (log jxj)a;where(33) L�X0; 12 + it� = 0 ; a < order of the zero; a < � � 12 :The restriction to the subgroupR�+ of Ck of the transposed ofW is thus givenin the above basis by(34) W (�)t �t;a = aXb=0 Cba �it (Log(�))b �t;a�b:The multiplication operator by a function with bounded derivatives is abounded operator in any Sobolev space one checks directly using the density inthe orthogonal of Im(E) of vectors satisfying (16), that is, if L �X0; 12 + is� 6=0, then is does not belong to the spectrum of DtX0 .This determines the spectrum of the operator DtX0 and hence of its trans-pose DX0 as indicated in Theorem 1 and ends the proof of Theorem 1.Let us now prove the corollary. Let us �x h0 2 S(Ck) such that bh0 hascompact support contained in fX0g �R and bh0(X0; s) = 1 for s small.Let then hs be given by hs(g) = h0(g) jgjis. The Fourier transform bhs isthen the translate of bh0, and one can choose h0 such that(35) Xn2Z bhn(X0 ; u) = 1 u 2 R :67

When jsj ! 1, the dimension of the range of W t(hs) is of the order ofLogjsj as is the number of zeros of the L function in the translates of a �xedinterval (cf. [W 3]).Let h 2 S(Ck). One has W t(h) =Pn2Z W t(h � hn).It follows then from the polynomial growth of the norm of W t(g) thatthe operator(36) Z h(g)W (g)t d�gis of trace class for any h 2 S(Ck).Moreover using the triangular form given by (34) we get its trace, andhence the trace of its transpose W (h) as(37) TraceW (h) = XL(X ; 12+�)=0�2iR bh(X ; �)where the multiplicity is counted as in Theorem 1 and where the Fouriertransform bh of h is de�ned by(38) bh(X ; �) = ZCk h(u) eX (u) juj� d� u :Appendix II. Explicit formulasWe check in detail that the rewriting of the Weil explicit formulas whichis predicted by the global trace formula of Theorem VII 4 is correct. Ourcomputation is straightforward but can be better understood conceptuallyusing [H]. Let us �rst recall the Weil explicit formulas ([W3]). One lets k be aglobal �eld. One identi�es the quotient Ck=Ck;1 with the range of the module(1) N = fjgj ; g 2 Ckg � R�+ :One endows N with its normalized Haar measure d�x. Given a function F onN such that, for some b > 12 ,(2) jF (�)j = 0(�b) � ! 0 ; jF (�)j = 0(��b) ; � !1 ;one lets,(3) �(s) = ZN F (�) �1=2�s d�� :68

Given a Gr�ossencharakter X , i.e., a character of Ck and any � in the strip0 < Re(�) < 1, one lets N(X ; �) be the order of L(X ; s) at s = �. One lets(4) S(X ; F ) =X� N(X ; �)�(�)where the sum takes place over �'s in the above open strip. One then de�nesa distribution � on Ck by(5) � = log jd�1j �1 +D �Xv Dv ;where �1 is the Dirac mass at 1 2 Ck, where d is a di�erental idele of k sothat jdj�1 is up to sign the discriminant of k when char (k) = 0, and is q2g�2when k is a function �eld over a curve of genus g with coe�cients in the �nite�eld Fq .The distribution D is given by(6) D(f) = ZCk f(w) (jwj1=2 + jwj�1=2) d�w ;where the Haar measure d�w is normalized (cf. IIb). The distributions Dvare parametrized by the places v of k and are obtained as follows. For each vone considers the natural proper homomorphism(7) k�v ! Ck ; x! class of (1; : : : ; x; 1 : : :)of the multiplicative group of the local �eld kv in the idele class group Ck.One then has(8) Dv(f) = Pfw Zk�v f(u)j1� uj juj1=2 d�u ;where the Haar measure d�u is normalized (cf. IIb), and where the Weilprincipal value Pfw of the integral is obtained as follows, for a local �eldK = kv,(9) Pfw Zk�v 1R�v 1j1� uj d�u = 0 ;if the local �eld kv is non Archimedean, and otherwise(10) Pfw Zk�v '(u) d�u = PF0 ZR�+ (�) d�� ;69

where (�) = Rjuj=� '(u) d�u is obtained by integrating ' over the �bers,while(11) PF0 Z (�)d�� = 2 log(2�) c + limt!1�Z (1� f2t0 ) (�) d�� � 2c log t� ;where one assumes that � c f�11 is integrable on R�+, andf0(�) = inf(�1=2; ��1=2) 8 � 2 R�+; f1 = f�10 � f0 :The Weil explicit formula isTheorem 1. ([W]) With the above notations one has S(X ; F ) = �(F (jwj)X (w)).We shall now elaborate on this formula and in particular compare theprincipal values Pfw with those of Theorem V.3.Let us make the following change of variables,(12) jgj�1=2 h(g�1) = F (jgj)X0(g) ;and rewrite the above equality in terms of h.By (3) one has(13) ��12 + is� = ZCk F (jgj) jgj�is d�g :Thus, in terms of h,(14) Z h(g)X1(g) jgj1=2+is d�g = Z F (jg�1j)X0(g�1)X1(g) jgjis d�g ;which is equal to 0 if X1=Ck;1 6= X0=Ck;1 and for X1 = X0,(15) Z h(g)X0(g) jgj1=2+is d�g = ��12 + is� :Thus, with our notations we see that(16) Suppbh � X0 �R ; bh(X0; �) = �(�) :Thus we can write,(17) S(X0; F ) = XL(X ;�)=0;X2bCk;10<Re �<1 bh(X ; �)70

using a �xed decomposition Ck = Ck;1 �N .Let us now evaluate each term in (5). The �rst gives (log jd�1j)h(1). Onehas, using (6) and (12),hD;F (jgj)X0(g)i = ZCk jgj�1=2 h(g�1) (jgj1=2 + jgj�1=2) d�g= ZCk h(u) (1 + juj) d�u = bh(0) + bh(1) ;where for the trivial character of Ck;1 one uses the notation(18) bh(z) = bh(1; z) 8 z 2 C :Thus the �rst two terms of (5) give(19) (log jd�1j)h(1) + bh(0) + bh(1) :Let then v be a place of k. One has by (8) and (12),hDv ; F (jgj)X0(g)i = Pfw Zk�v h(u�1)j1� uj d�u :We can thus write the contribution of the last term of (5) as(20) �Xv Pfw Zk�v h(u�1)j1� uj d�u :Thus the equality of Weil can be rewritten as(21) bh(0) + bh(1) � XL(X ;�)=0;X2bCk;10<Re �<1 bh(X ; �) = (log jdj)h(1)+Xv Pfw Zk�v h(u�1)j1� uj d�u ;which now holds for �nite linear combinations of functions h of the form (12).This is enough to conclude when h(1) = 0.Let us now compare the Weil principal values, with those dictated byTheorem V.3. We �rst work with a local �eld K and compare (9), (10) withour prescription. Let �rst K be non Archimedean. Let � be a character of Ksuch that(22) �=R = 1 ; �=��1R 6= 1 :71

Then, for the Fourier transform given by(23) (Ff)(x) = Z f(y)�(y) dy ;with dy the selfdual Haar measure, one has(24) F (1R) = 1R :Lemma 2. With the above choice of � one hasZ 0 h(u�1)j1� uj d�u = Pfw Z h(u�1)j1� uj d�uwith the notations of Theorem 3.Proof. By construction the two sides can only di�er by a multiple of h(1).Let us recall from Theorem 3 that the left hand side is given by(25) �L; h(u�1)juj � ;where L is the unique extension of ��1 duj1�uj whose Fourier transform vanishesat 1, bL(1) = 0. Thus from (9) we just need to check that (25) vanishes forh = 1R� , i.e., that(26) hL; 1R� i = 0 :Equivalently, if we let Y = fy 2 K ; jy � 1j = 1g we just need to show, usingParseval, that(27) hlog juj ; b1Y i = 0 :One has b1Y (x) = RY �(xy) dy = �(x)b1R� (x), and 1R� = 1R � 1P , b1R� =1R � j�j 1��1R, thus, with q�1 = j�j,(28) b1Y (x) = �(x)�1R � 1q 1��1R� (x) :We now need to compute R log jxjb1Y (x) dx = A +B,(29) A = �1q Z��1R� �(x) (log q) dx ; B = �1� 1q�ZR log jxj dx :72

Let us show that A+B = 0. One has RR dx = 1, andA = � ZR� �(��1y) (log q) dy = � log q�ZR �(��1y) dy � ZP dy�= 1q log q ; since ZR �(��1y) dy = 0 as �=��1R 6= 1 :To compute B, note that R�nR� dy = q�n �1� 1q� so thatB = �1� 1q�2 1Xn=0(�n log q) q�n = �q�1 log q :and A +B = 0.Let us now treat the case of Archimedean �elds. We take K = R �rst,and we normalize the Fourier transform as(30) (Ff)(x) = Z f(y) e�2�ixy dyso that the Haar measure dx is selfdual.With the notations of (10) one has(31) Pfw ZR� f30 (juj) juj1=2j1� uj d�u = log � + where is Euler's constant, = ��0(1). Indeed integrating over the �bersgives f40 � (1 � f40 )�1, and one getsPF0 ZR�+ f40 � (1� f40 )�1 d�u = log(2�) + limt!1 ZR�+ (1 � f2t0 ) f40 (1 � f40 )�1 d�u� log t!!= log 2� + � log 2 :Now let '(u) = � log juj. It is a tempered distribution on R and one has(32) h'; e��u2 i = 12 log � + 2 + log 2 ;as one obtains from @@s R juj�s e��u2 du = @@s �� s�12 � �1�s2 �� evaluated ats = 0, using �0( 12 )�(12 ) = � � 2 log 2. 73

Thus by the Parseval formula one has(33) hb'; e��x2i = 12 log � + 2 + log 2 ;which gives, for any test function f ,(34) hb'; fi = lim"!0 Zjxj�" f(x) d�x + (log ") f(0)! + � f(0)where � = log(2�) + . In order to get (34), one uses the equality(35)lim"!0 Zjxj�" f(x) d�x + (log ") f(0)! = lim"!0�Z f(x) jxj" d�x� 1" f(0)� ;which holds since both sides vanish for f(x) = 1 if jxj � 1, f(x) = 0 otherwise.Thus from (34) one gets(36)Z 0R f(u) 1j1� uj d�u = � f(1) + lim"!0 Zj1�uj�" f(u)j1� uj d�u+ (log ") f(1)! :Taking f(u) = juj1=2 f30 (juj), the right hand side of (36) gives � � log 2 =log � + ; thus we conclude using (31) that for any test function f ,(37) Z 0R f(u) 1j1� uj d�u = Pfw ZR f(u) 1j1� uj d�u :Let us �nally consider the case K = C . We choose the basic character � as(38) �(z) = exp 2�i(z + z) ;the selfdual Haar measure is dz dz = jdz ^ dzj, and the function f(z) =exp�2�jzj2 is selfdual.The normalized multiplicative Haar measure is(39) d�z = jdz ^ dzj2�jzj2 :Let us compute the Fourier transform of the distribution(40) '(z) = � log jzjC = �2 log jzj :74

One has(41) h'; exp�2�jzj2i = log 2� + ;as is seen using @@" �R e�2�jzj2 jzj�2" jdz ^ dzj� = @@" ((2�)" �(1� ")).Thus hb'; exp�2�juj2i = log 2� + and one gets(42) hb'; fi = lim"!0 ZjujC�" f(u) d�u+ log " f(0)! + �0 f(0)where �0 = 2(log 2� + ).To see this one uses the analogue of (35) for K = C to compute the righthand side of (42) for f(z) = exp�2�jzj2.Thus, for any test function f , one has(43)Z 0C f(u) 1j1� ujC d�u = �0 f(1) + lim"!0 Zj1�ujC�" f(u)j1� ujC d�u+ (log ")f(1)! :Let us compare it with Pfw. When one integrates over the �bers of C � j jC�!R�+ the function j1� zj�1C , one gets(44) 12� Z 2�0 1j1� ei�zj2 d� = 11� jzj2 if jzj < 1 ; and 1jzj2 � 1 if jzj > 1 :Thus for any test function f on R�+ one has, by (10),(45) Pfw Z f(jujC ) 1j1� ujC d�u = PF0 Z f(�) 1j1� �j d��with the notations of (11). With f2(�) = � 12 f0(�) we thus get, using (11),(46) Pfw Z f2(jujC ) 1j1� ujC d�u = PF0 Z f0 f�11 d�� = 2(log 2� + ) :We shall now show that(47) lim"!0 Zj1�ujC�" f2(jujC )j1� ujC d�u+ log "! = 0 :It will then follow that, using (43),(48) Z 0C f(u) 1j1� ujC d�u = Pfw Z f(u) 1j1� ujC d�u :75

To prove (47) it is enough to investigate the integral(49) Zjzj�1;j1�zj�"((1� z)(1 � z))�1 jdz ^ dzj = j(") ;and show that j(") = � log "+ o(1) for "! 0. A similar statement then holdsfor Zjzj�1;j1�z�1j�"((1 � z)(1 � z))�1 jdz ^ dzj:One has j(") = RD jdZ ^ dZj, where Z = log(1� z) and the domain D iscontained in the rectangle(50) fZ = (x + iy) ; log " � x � log 2 ; ��2 � y � �=2g = R" ;and bounded by the curve x = log(2 cos y) which comes from the equation ofthe circle jzj = 1 in polar coordinates centered at z = 1. One thus gets(51) j(") = 4Z log 2log " Arc cos(ex=2) dx ;when " ! 0 one has j(") � 2� log(1="), which is the area of the followingrectangle (in the measure jdz ^ dzj),(52) fZ = (x + iy) ; log " � x � 0 ; ��=2 � y � �=2g :One has jR"j�2� log 2 = 2� log(1="). When "! 0 the area of R"nD convergesto(53) 4Z log 2�1 Arc sin(ex=2) dx = �4Z �=20 log(sinu) du = 2� log 2 ;so that j(") = 2� log(1=") + o(1) when "! 0.Thus we can assert that with the above choice of basic characters forlocal �elds one has, for any test function f ,(54) Z 0K f(u) 1j1� uj d�u = Pfw Z f(u) 1j1� uj d�u :Lemma 3. Let K be a local �eld, �0 a normalized character as above and �,�(x) = �0(�x) an arbitrary character of K. Let R 0 be de�ned as in TheoremV.3 relative to �. Then, for any test function f ,Z 0K f(u) 1j1� uj d�u = log j�j f(1) + Pfw Z f(u) 1j1� uj d�u :76

Proof. The new selfdual Haar measure is(55) da = j�j1=2 d0 awith d0 a selfdual for �0. Similarly, the new Fourier transform is given bybf(x) = Z �(xy) f(y) dy = Z �0(�xy) f(y) j�j1=2 d0 y :Thus(56) bf(x) = j�j1=2 bf0(�x) :Let then '(u) = � log juj. Its Fourier transform as a distribution is given by(57) hb'; fi = Z (� log juj) bf(u) du :One has Z (� log juj) bf (u) du = Z (� log juj) bf0(�u) j�j d0 u= Z (� log jvj) bf0(v) d0 v + Z log j�j bf0(v) d0 v := Z (� log jvj) bf0(v) d0 v + log j�j f(0) :Thus the lemma follows from (54).Let us now pass to the global case. Recall that if �, � 6= 1, is a characterof A such that �=k = 1, there exists a di�erental idele d = (dv) such that, (cf.[W1])(58) �v(x) = �0;v(dv x)where � = ��v and each local character �0;v is normalized as above.We can thus rewrite the Weil formula (Theorem 1) asTheorem 6. Let k be a global �eld, � a nontrivial character of A=k and� = ��v its local factors.Let h 2 S(Ck) have compact support. Thenbh(0) + bh(1)� XL(X ;�)=00<Re�<1 bh(X ; �) =Xv Z 0k�v h(u�1)j1� uj d�u77

where the normalization of R 0 is given by �v as in Theorem V.3, and bh(X ; z) =R h(u)X (u) jujz d�u.Proof. This follows from formula (21), Lemma 3 and the equality log jdj =Pvlog jdvj.Normalization of Haar measure on modulated groupWe let G be a locally compact abelian group with a proper morphism(1) g! jgj ; G! R�+whose range is cocompact in R�+.There exists a unique Haar measure d�g on G such that(2) Zjgj2[1;�] d�g � log� when �! +1 :Let G0 = Kermod = fg 2 G ; jgj = 1g. It is a compact group by hypothesis,and one can identify G=G0 with the range N of the module. Let us determinethe measure d�n on N � R�+ such that (2) holds for(3) Z f d�g = Z �Z f(ng0) dg0� d�nwhere the Haar measure dg0 is normalized by(4) ZG0 dg0 = 1 :We let �� be the function on G given by(5) ��(g) = 0 if jgj =2 [1;�] ; ��(g) = 1log� if g 2 [1;�] :The normalization (2) means that R �� d�g! 1 when �!1.Let �rst N = R�+. Then the unique measure satisfying (2) is(6) d�� = d�� :Let then N = �Z for some � > 1. Let us consider the measure(7) Z f d�g = �X f(�n) :78

We take f = ��. Then the right hand side is � Nlog � where N is the numberof �n 2 [1;�], i.e., � log �log� . This shows that (2) holds i�(8) � = log� :Let us show more generally that if H � G is a compact subgroup of G and ifboth d�g and d�h are normalized by (2), one has(9) Z �Z f(hy) d�h� d0 y = Z f d�gwhere d0 y is the Haar measure of integral 1 on G=H,(10) ZG=H d0 y = 1 :The left hand side of (9) de�nes a Haar measure on G and we just need toshow that it satis�es (2).One has k��(� y) � ��k1 ! 0 when �!1, and(11) Z ��(hy) d�h! 1 when�!1uniformly on compact sets of y 2 G. Thus(12) Z �Z ��(hy) d�h� d0 y ! 1 when �!1 :Appendix III. Distribution trace formulasIn this appendix we recall for the convenience of the reader the coordinate freetreatment of distributions of [GS] and give the details of the transversalityconditions.Given a vector space E over R, dimE = n, a density is a map, � 2 jE�j,(1) � : ^nE ! Csuch that �(�v) = j�j �(v) 8� 2 R; 8 v 2 ^nE. Given a linear mapT : E ! F we let jT �j : jF �j ! jE�j be the corresponding linear map; itdepends contravariantly on T . 79

A smooth compactly supported density � 2 C1c (M; jT �M j) on an arbitrarymanifold M has a canonical integral(2) Z � 2 C :One de�nes the generalized sections of a vector bundle L on M as the dualspace of C1c (M;L� jT �M j)(3) C�1(M;L) = dual of C1c (M;L� jT �M j)where L� is the dual bundle. One has a natural inclusion(4) C1(M;L) � C�1(M;L)given by the pairing(5) � 2 C1(M;L) ; s 2 C1c (M;L� jT �M j)! Z hs; �iwhere hs; �i is viewed as a density, hs; �i 2 C1c (M; jT �M j).One has a similar notion of generalized section with compact support.Given a smooth map ' : X ! Y , then if ' is proper, it gives a (contravari-antly) associated map(6) '� : C1c (Y;L)! C1c (X;'�(L)) ; ('� �)(x) = �('(x))where '�(L) is the pullback of the vector bundle L. Thus, given a lin-ear form on C1c (X;'�(L)) one has a (covariantly) associated linear formon C1c (Y;L). In particular with L trivial, we see that generalized densities� 2 C�1(X; jT �Xj) pushforward,(7) '�(�) 2 C�1(Y; jT �Y j)with h'�(�); �i = h�; '� �i 8 � 2 C1c (X).This gives the natural functoriality of generalized sections, they pushfor-ward under proper maps. However under suitable transversality conditionswhich are automatic for submersions, generalized sections also pull back. Forinstance, if ' is a �bration and � 2 C1c (X; jT �Xj) is a density then one canintegrate � along the �bers, the obtained density on Y , '�(�) is given as in(7) by(8) h'�(�); fi = h�; '� fi 8 f 2 C1(Y ):80

The point is that the result is not only a generalized section but a smoothsection '�(�) 2 C1c (Y; jT �Y j).It follows that if f 2 C�1(Y ) is a generalized function, then one obtainsa generalized function '�(f) on X by(9) h'�(f); �i = hf; '�(�)i 8 � 2 C1c (X; jT �Xj) :In general, the pullback '�(f) of a generalized function f , continues to makesense provided the following transversality condition holds,(10) d('�(l)) 6= 0 8l 2WF (f) :where WF (f) is the wave front set of f ([GS]).Next, let us recall the construction ([GS]) of the generalized section of avector bundle L on a manifold X associated to a submanifold Z � X and asymbol(11) � 2 C1(Z;L jNZ j) ;where NZ is the normal bundle of Z. The construction is the same as thatof the current of integration on a cycle. Given � 2 C1c (X;L� jT �Xj), theproduct � �=Z is a density on Z, since it is a section of jT �Z j = jT �X j jNZ j.One can thus integrate it over Z. When Z = X, one has NZ = f0g and jNZ jhas a canonical section, so that the current associated to � is just given by(5). Now let ' : X ! Y with Z a submanifold of Y and � as in (11). Let usassume that ' is transverse to Z, so that for each x 2 X with y = '(x) 2 Zone has(12) '�(Tx) + T'(x)(Z) = Ty Y :Let(13) �x = fX 2 Tx ; '�(X) 2 Ty(Z)g :Then '� gives a canonical isomorphism(14) '� : Tx(X)=�x ' Ty(Y )=Ty(Z) = Ny(Z) :81

And '�1(Z) is a submanifold of X of the same codimension as Z with anatural isomorphism of normal bundles(15) '� : N'�1(Z) ' '�NZ :In particular, given a (generalized) �-section of a bundle L with support Zand symbol � 2 C1(Z;LjNZ j), one has a corresponding symbol on '�1(Z)given by(16) '� �(x) = j('�)�1j�('(x)) 2 ('� L)x jNxjusing the inverse of the isomorphism (15), which requires the transversalitycondition.For any �-section associated to Z; �, the wave front set is contained inthe conormal bundle of the submanifold Z which shows that if ' is transverseto Z the pull back '� �Z;� of the distribution on Y associated to Z; � makessense, it is equal to �'�1(Z);'�(�).Let us now recall the formulation ([GS]) of the Schwartz kernel theorem.One considers a continuous linear map(17) T : C1c (Y )! C�1(X) :The statement is that one can write it as(18) (T �) (x) = Z k(x; y) �(y) dywhere k(x; y) dy is a generalized section(19) k 2 C�1(X � Y ; pr�Y (jT �Y j)) :Let f : X ! Y be a smooth map, and T = f� the operator(20) (T �) (x) = � (f(x)) 8 � 2 C1c (Y ) :The corresponding k is the �-section associated to the submanifold of X � Ygiven by(21) Graph(f) = f(x; f(x)) ; x 2 Xg = Zand its symbol, � 2 C1(Z;pr�Y (jT �Y j) jNZj) is obtained as follows.82

Given � 2 T �x (X), � 2 T �y (Y ), one has (�; �) 2 N�Z i� it is orthogonal to(v; f� v) for any v 2 Tx(X), i.e., hv; �i + hf� v; �i = 0 so that(22) � = �f t� � :Thus one has a canonical isomorphism j : T �y (Y ) ' N�Z , � j! (�f t� �; �). Thetransposed (j�1)t is given by (j�1)t(Y ) = class of (0; Y ) in NZ , 8Y 2 Ty(Y ).One has,(23) � = jj�1j 2 C1(Z;pr�Y (jT �Y j) jNZ j) :We denote the corresponding �-distribution by(24) k(x; y) dy = �(y � f(x)) dy :One then checks the formula(25) Z �(y � f(x)) �(y) dy = �(f(x)) 8 � 2 C1c (Y ) :Let us now consider a manifold M with a ow Ft(26) Ft(x) = exp(t v)x v 2 C1(M;TM )and the corresponding map f ,(27) f :M �R!M ; f(x; t) = Ft(x) :We apply the above discussion with X =M �R, Y =M . The graph of f isthe submanifold Z of X � Y(28) Z = f(x; t; y) ; y = Ft(x)g :One lets ' be the diagonal map(29) '(x; t) = (x; t; x) ; ' :M �R! X � Yand the �rst issue is the transversality '\j Z.We thus need to consider (12) for each (x; t) such that '(x; t) 2 Z, i.e.,such that x = Ft(x). One looks at the image by '� of the tangent spaceTxM�R toM �R at (x; t). One lets @t be the natural vector �eld on R. The83

image of (X;�@t) is (X;�@t ;X) for X 2 TxM;� 2 R. Dividing the tangentspace of M �R�M by the image of '�, one gets an isomorphism(30) (X;�@t ; Y )! Y �Xwith TxM . The tangent space to Z is f(X 0; � @t; (Ft)�X 0 + �vFt(x)); X 0 2TxM;� 2 Rg. Thus the transversality condition means that every element ofTxM is of the form(31) (Ft)�X �X + �vx X 2 TxM ; � 2 R :One has(32) (Ft)� �vx = �vxso that (Ft)� de�nes a quotient map, the Poincar�e return map(33) P : Tx=R vx! Tx=R vx = Nxand the transversality condition (31) means exactly that(34) 1� P is invertible:Let us make this hypothesis and compute the symbol � of the distribution(35) � = '�(�(y � Ft(x)) dy) :First, as above, let W = '�1(Z) = f(x; t) ; Ft(x) = xg. The codimension of'�1(Z) in M �R is the same as the codimension of Z in M � R�M , so itis dimM which shows that '�1(Z) is 1-dimensional. If (x; t) 2 '�1(Z), then(Fs(x); t) 2 '�1(Z). Thus, if we assume that v does not vanish at x, the map(36) (x; t) q! tis locally constant on the connected component of '�1(Z) containing (x; t).This allows us to identify the transverse space toW = '�1(Z) as the product(37) NWx;t ' Nx �R ;where to (X;�@t) 2 Tx;t(M � R), we associate the pair ( eX;�) given by theclass of X in Nx = Tx=R vx and � 2 R.84

The symbol � of the distribution (35) is a smooth section of jNW j tensoredby the pull back '�(L) where L = pr�Y jT �M j, and one has(38) '�(L) ' jp� T �M jwhere(39) p(x; t) = x 8 (x; t) 2M �R :To compute � one needs the isomorphism(40) NW(x;t) '�! T'(x;t)(M �R�M)=T'(x;t)(Z) = NZ :The map '� : NWx;t ! NZ is given by(41) '�(X;�@t) = (1� (Ft)�)X � � v X 2 Nx ; � 2 Rand the symbol � is just(42) � = j'�1� j 2 jp� T �M j jNW j :Let us now consider the second projection(43) q(x; t) = t 2 R ;and compute the pushforward q�(� ) of the distribution � . By constructionq�(� ) is a generalized function. We �rst look at the contribution of a periodicorbit. The corresponding part of '�1(Z) is of the form(44) '�1(Z) = V � � �M �R ;where � is a discrete cocompact subgroup of R, while V �M is a one dimen-sional compact submanifold of M .To compute q�(� ), we let h(t) jdtj be a 1-density on R and pull it back by qas the section on M �R of the bundle q� jT �j,(45) �(x; t) = h(t) jdtj :We now need to compute R'�1(Z) � �. We can look at the contribution ofeach component V � fTg, T 2 �. 85

One gets ([GS]),(46) T# 1j1� PT j h(T ) ;where T# is the length of the primitive orbit or equivalently the covolume of� in R for the Haar measure jdtj. We can thus write the contributions of theperiodic orbits as(47) X p X� Covol(�) 1j1� PT j h(T ) ;where the test function h vanishes at 0.The next case to consider is when the vector �eld vx has an isolated 0,vx0 = 0. In this case, the transversality condition (31) becomes(48) 1� (Ft)� invertible (at x0) :One has Ft(x0) = x0 for all t 2 R and now the relevant component of '�1(Z)is fx0g � R. The transverse space NW is identi�ed with Tx and the map'� : NW ' NZ is given by(49) '� = 1� (Ft)� :Thus the symbol � is the scalar function j1�(Ft)�j�1. The generalized sectionq� '�(�(y � Ft(x)) dy) is the function, t ! j1 � (Ft)�j�1. We can thus writethe contribution of the zeros of the ow as ([GS])(50) Xzeros Z h(t)j1� (Ft)�j dtwhere h is a test function vanishing at 0.We can thus collect the contributions 47 and 50 as(51) X ZI h(u)j1� (Fu)�j d�u ;where h is as above, I is the isotropy group of the periodic orbit , the Haarmeasure d�u on I is normalized so that the covolume of I is equal to one,and we still write (Fu)� for its restriction to the transverse space of .86

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