Tham khảo tài liệu 'analytic number theory a tribute to gauss and dirichlet episode 2 part 3', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 172 PHILIPPE MICHEL AND AKSHAY VENKATESH class of elliptic curves over C via z G H C Z zZ then HeK is identified with the set of elliptic curves with CM by OK. If f is a Maass form and X a character of ClK one has associated a twisted L-function L s f X x and it is known from the work of Waldspurger and Zhang Zha01 Zha04 that 2 L f 0 X 1 2 -2 x x f x 2. In other words the values L 1 f 0 x are the squares of the Fourier coefficients of the function x f x on the finite abelian group ClK. The Fourier transform being an isomorphism in order to show that there exists at least one X G ClK such that L 1 2 f 0 x is nonvanishing it will suffice to show that f x 0 for at least one x G ClK. There are two natural ways to approach this for D large enough 1 Probabilistically show this is true for a random x. It is known by a theorem of Duke that the points x x G ClK become equidistributed as D to . the Riemannian measure on Y thus f x is nonvanishing for a random x G ClK. 2 Deterministically show this is true for a special x. The class group ClK has a distinguished element namely the identity e G ClK and the corresponding point e looks very special it lives very high in the cusp. Therefore f e 0 for obvious reasons look at the Fourier expansion Thus we have given two fundamentally different proofs of the fact that there exists X such that L 22 f 0 x 0 Soft as they appear these simple ideas are rather powerful. The main body of the paper is devoted to quantifying these ideas further . pushing them to give that many twists are nonvanishing. Remark . The first idea is the standard one in analytic number theory to prove that a family of quantities is nonvanishing compute their average. It is an emerging philosophy that many averages in analytic number theory are connected to equidistribution questions and thus often to ergodic theory. Of course we note that in the above approach one does not really need to know that x x G ClK become equidistributed as D to it suffices to