By Dorian Goldfeld, Joseph Hundley

This graduate-level textbook offers an uncomplicated exposition of the speculation of automorphic representations and L-functions for the overall linear staff in an adelic environment. The authors maintain definitions to a minimal and repeat them whilst reintroduced in order that the publication is available from any access aspect, and without previous wisdom of illustration thought. in addition they comprise concrete examples of either worldwide and native representations of GL(n), and current their linked L-functions. the idea is constructed from first rules for GL(1), then conscientiously prolonged to GL(2) with whole certain proofs of key theorems. a number of of the proofs are right here awarded for the 1st time, together with Jacquet's uncomplicated and chic facts of the tensor product theorem. eventually, the better rank scenario of GL(n) is given an in depth therapy. Containing quite a few routines, this publication will inspire scholars and researchers to start operating during this fertile box of analysis.

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**Extra resources for Automorphic Representations and L-Functions for the General Linear Group: Volume 1**

**Sample text**

3, and that for almost all p, the function f p is identically equal to 1 on Z p . Then f (x) d x = AQ R f ∞ (x∞ ) d x∞ · lim N →∞ p

In particular, |χ (a)|C = 1, ∀a ∈ (Z/qZ)× . It is standard practice in analytic number theory (see [Davenport, 2000]) to lift a Dirichlet character χ to Z by defining a new function χ1 : Z → C which satisfies χ1 (a) = 0, (∀a ∈ Z with (a, q) = 1) ; χ1 (a + mq) = χ (a), χ1 (ab) = χ1 (a)χ1 (b), (∀a, m ∈ Z with (a, q) = 1) ; (∀a, b ∈ Z) . We shall follow the standard practice of denoting the character χ1 by the sym× bol χ . Remarkably, it is also possible to lift χ to the idele group A× Q . Since AQ is a purely multiplicative group, the value of the lifted character can never be 0.

Step 4: It is enough to prove that f α. 11) α∈Q To see this fix x0 ∈ AQ and define a new function g(x) = f (x + x0 ) for x ∈ AQ . 11), we have gα . f (x0 ) = g(0) = α∈Q But gα = f (x + x0 )e(−αx) d x = e(αx0 ) Q\AQ f (x)e(−αx) d x = e(αx0 ) f α , Q\AQ from which it follows that f (x0 ) = f α e(αx0 ). 11) for functions f which satisfy the condition f (0) = 0. If this is not the case, consider the new function f (x) − f (0) which vanishes at 0. 11) is easy to prove for constant functions. 12) n∈Z where f satisfies f (0) = 0.