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Analytic Continuation, Functional Equation: Examples 1. Dirichlet L-Functions L(S, Χ) for Even Dirichlet Characters

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Analytic Continuation, Functional Equation: Examples 1. Dirichlet L-Functions L(S, Χ) for Even Dirichlet Characters (October 24, 2018) Analytic continuation, functional equation: examples Paul Garrett [email protected] http:=/www.math.umn.edu/egarrett/ 1. L(s; χ) for even Dirichlet characters 2. L(s; χ) for odd Dirichlet characters 3. Dedekind zeta function ζo(s) for Gaussian integers o 4. Grossencharacter L-functions L-functions for Gaussian integers We try to imitate the argument used by Riemann for proving the analytic continuation of ζ(s) and its functional equation −s=2 s −(1−s)=2 1−s π Γ( 2 ) ζ(s) = π Γ( 2 ) ζ(1 − s) from the integral representation Z 1 −s=2 s s=2 θ(iy) − 1 dy π Γ( 2 ) ζ(s) = y 0 2 y in terms of the basic theta series [1] X 2 θ(iy) = e−πn y n2Z whose functional equation 1 i θ(iy) = p · θ y y is proven by Poisson summation. The discussion of L-functions L(s; χ) for Dirichlet characters (over Q) bifurcates into two families, depending upon the parity of χ, that is, whether χ(−1) = +1 or χ(−1) = −1. Analogous discussion of the zeta function of the Gaussian integers extends this discussion in a new direction. Treatment of the grossencharacter L-functions [2] for the Gaussian integers have no counterpart among Dirichlet L-functions for Z. L-functions of ideal class group characters deserve parallel treatment, but would require more background, concerning rings of algebraic integers that fail to be principal ideal domains, and other background. This background is important, but is not the immediate point, so we defer this part of the discussion. These examples are simple, but the book-keeping quickly becomes fragile. This should motivate receptiveness to the more abstract, but very clear, Iwasawa-Tate viewpoint, which we consider soon. 1. Dirichlet L-functions L(s; χ) for even Dirichlet characters Let χ be a non-trivial Dirichlet character mod N > 1, with L-function X χ(n) L(s; χ) = ns n≥1 [1] The reason that the argument is given as iy rather than y is that iy can be replaced by any z in the complex upper half-plane H. This aspect is important later. [2] The grossencharacters and their L-functions are also called Hecke characters and L-functions, as Hecke first studied them, about 1920. 1 Paul Garrett: Analytic continuation, functional equation: examples (October 24, 2018) By imitation of the corresponding discussion for ζ(s), we expect to define a theta series θχ with a functional equation provable via Poisson summation, to exhibit an integral representation of L(s; χ) in terms of θχ, and to use this integral representation to prove the analytic continuation and see the functional equation in the symmetry of the rewritten integral representation. We will find that the most obvious imitative approach succeeds only for even χ. Further, to have a symmetrical form of the functional equation, we must be able to determine the absolute value of the Gauss sum X hχ, i = χ(b) (b) (with (b) = e2πib=N ) b mod N This requires that χ be primitive mod N, that is, its conductor must be exactly N, not a proper divisor. That is, the non-zero parts of χ must not be well-defined modulo N 0 for any proper divisor of N. [1.1] The theta series, parity of χ The obvious imitation of the theta series for ζ suggests defining X −πn2y θχ(iy) = χ(n) e n2Z Non-trivial χ is extended by 0 to integers having a common factor with N, so χ(0) = 0. By design, 8 0 (for χ odd) θχ(iy) X χ(n) + χ(−n) 2 < = e−πn y = 2 2 : P −πn2y n≥1 n≥1 χ(n) e (for χ even) That is, this version of θχ cannot be correct for odd χ, since it vanishes identically. [1.2] The integral representation The integral representation is easy: for Re(s) > 1, Z 1 Z 1 s θχ(iy) dy X s −πn2y dy y 2 = χ(n) y 2 e 2 y y 0 n≥1 0 Z 1 X χ(n) s −y dy −s=2 s = y 2 e = π Γ( ) L(s; χ) πs=2 ns y 2 n≥1 0 by replacing y by y/πn2. This is the desired integral representation. [1.2.1] Remark: The measure dy=y is dilation-invariant, and also invariant under any change of variables y ! y=C for C > 0. Since these are the changes of variables the computation will require, it is vastly better to keep the y in the denominator of the measure, rather than absorb it into the ys=2. [1.3] Functional equation of θχ We anticipate that θχ has a functional equation similar to that of P −πn2y θ(iy) = n e , proven by application of the Poisson summation formula X X f(n) = fb(n) (for nice functions f) n2Z n2Z The minor but meaningful obstacle is that in this classical guise θχ is not simply a sum of a nice function over a lattice, so Poisson summation is not immediately applicable. However, since χ(n) depends only upon n mod N, we can break the sum expressing θχ into a finite sum of sums-over-lattices, and then apply Poisson summation. Specifically, X −πn2y X X −π(`N+b)2y θχ(iy) = χ(n) e = χ(b) e n b mod N `2Z 2 Paul Garrett: Analytic continuation, functional equation: examples (October 24, 2018) Thus, Poisson summation is to be applied to the function 2 f(x) = e−π(xN+b) y 2 Since this f differs from the Gaussian e−πx merely by dilations and translation, and the Fourier transform behaves cogently with respect to such, something reasonable will come out, by changing variables: Z −π(xN+b)2y −π(xN+b)2y −2πiξx e b(ξ) = e e dx R 1 Z 2 = e−π(x+b) y e−2πiξx=N dx (replacing x by x=N) N R 2πiξb=N e Z 2 = e−πx y e−2πiξx=N dx (replacing x by x − b) N R 2πiξb=N e Z 2 p p = p e−πx e−2πiξx=N y dx (replacing x by x= y) N y R 2πiξb=N e 2 2 = p e−πξ =N y (taking Fourier transform) N y Thus, by Poisson summation, 2πi`b=N X X 2 X X e 2 2 θ (iy) = χ(b) e−π(`N+b) y = χ(b) p e−π` =N y χ N y b mod N `2Z b mod N `2Z 1 X 2 2 X = p e−π` =N y χ(b)e2πi`b=N N y `2Z b mod N The obvious thing would be to replace b by b`−1 mod N, but the sum over ` does not obviously exclude ` with gcd(`; N) > 1. This is the first point where the primitivity of χ mod N enters, to make the inner sum vanish for gcd(`; N) > 1: [1.3.1] Claim: X χ(b) e2πi`b=N = 0 (for primitive χ mod N, for ` not invertible mod N) b mod N Proof: Let gcd(`; N) = N=m > 1. Then, replacing b by b · (1 + xm for any x mod N=m gives X X X χ(b) e2πi`b=N = = χ(b(1 + xm)) e2πi`β(1+xm)=N b mod N b mod N=m β mod m X X = χ(1 + xm) χ(b) e2πi`β=N b mod N=m β mod m This holds for all x. The primitivity of χ is exactly that x ! χ(1 + xm) cannot be trivial. Thus, the sum is 0, as claimed. === −1 Returning to the functional equation for θχ, now we can make the change of variables: replace b by b` mod N: 1 X 2 2 X θ (iy) = p χ−1(`) e−π` =N y χ(b)e2πib=N χ N y `2Z b mod N 3 Paul Garrett: Analytic continuation, functional equation: examples (October 24, 2018) The inner sum has become a Gauss sum hχ, i not depending on `, X hχ, i = χ(b) (b) (with (b) = e−2πib=N ) b mod N so we have the functional equation: hχ, i i θ (iy) = p · θ −1 (for even χ primitive mod N) χ N y χ N 2y [1.3.2] Remark: The most striking thing is that the functional equation relates θχ to θχ−1 , rather than to θχ itself. This is typical of a larger reality. [1.3.3] Remark: Note that the flip in the argument y is y ! 1=N 2y, not the earlier y ! 1=y. The fixed- point of this map is not 1, but 1=N. This influences the break-up of the integral in proof of the analytic continuation and functional equation, next. [1.4] Analytic continuation and functional equation Starting from the integral representation, Z 1 Z 1 Z 1=N s θχ(iy) dy s θχ(iy) dy s θχ(iy) dy −s=2 s 2 2 2 π Γ( 2 ) L(s; χ) = y = y + y 0 2 y 1=N 2 y 0 2 y As for ζ(s), the integral from 1=N to 1 is nicely convergent for all s 2 C, and gives an entire function. The goal is to use the functional equation to convert the integral from 0 to 1=N to the other sort, and then symmetrize the resulting expression as much as possible. That is, replace y by 1=N 2y in the problemmatical integral from 0 to 1=N, obtaining 1 1 Z s i dy Z s i dy 1 2 2 1 −s − 2 2 (1=N y) θχ 2 = 2 N y θχ 2 1=N N y y 1=N N y y Replacing χ by χ−1 in the functional equation above gives −1 hχ ; i i θ −1 (iy) = p · θ χ N y χ N 2y Rearranging, p i N y θ = θ −1 (iy) χ N 2y hχ−1; i χ Substituting this gives 1−s Z 1 N 1−s θχ−1 (iy) dy 2 −1 y hχ ; i 1=N 2 y That is, Z 1 1−s Z 1 s θχ(iy) dy N 1−s θχ−1 (iy) dy −s=2 s 2 2 π Γ( 2 ) L(s; χ) = y + −1 y 1=N 2 y hχ ; i 1=N 2 y This gives the analytic continuation, but is not quite symmetrized.
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