QUADRATIC BASE CHANGE and the ANALYTIC CONTINUATION of the ASAI L-FUNCTION: a NEW TRACE FORMULA APPROACH 1. Introduction Underst

QUADRATIC BASE CHANGE AND THE ANALYTIC CONTINUATION OF THE ASAI L-FUNCTION: A NEW TRACE FORMULA APPROACH

P. EDWARD HERMAN

Abstract. Using Langlands’s Beyond Endoscopy idea and analytic number theory techniques, we study the Asai L-function associated to a real quadratic ﬁeld K/Q. If the Asai L-function associated to an automorphic form over K has a pole, then the form is a base change from Q. We prove this and further prove the analytic continuation of the L-function. This is one of the ﬁrst examples of using a trace formula to get such information. A hope of Langlands is that general L-functions can be studied via this method.

1. Introduction Understanding the analytic continuation of L-functions is a central object of interest in number theory. If the L-function is associated to an automorphic form, then the analytic properties of the function are key in understanding the form itself. This is clearly demon- strated with the√ Asai L-function in this paper. We will now define it. Let K = Q( D) be a real quadratic extension of Q, with D a prime. We assume for computational clarity in this paper that K has class number one. Let Π be an automorphic form with level one and trivial nebentypus over the field K. The details we need of a automorphic form we save to next section, and for more elaborate details we refer to [BMP1], [BMP2], and [V]. If it is more comfortable, one can think of a Hilbert modular form instead of an automorphic form over K. The form has associated to it Fourier coefficients {cμ(Π)} parametrized by integral ideals μ. The standard L-function associated with the form is c (Π) L(s, Π) = μ . N(μ)s μ It is well known L(s, Π) has analytic continuation to the complex plane and satisfies a functional equation if Π is a cuspidal automorphic form. The Asai L-function for Π is a sort of subseries of L(s, Π). We define the series as ∞ c (Π) L(s, Π,Asai)=ζ(2s) n . ns n=1 The L-function is named after Asai, who in [A] showed the function has analytic continuation, uptoapoleats =1, an Euler product, and a functional equation. Further, if it has a pole then the associated form is a base change or a lift from a automorphic form over Q. Base change is usually denoted Π = BCK/Q(π), where π is an automorphic form over Q. In terms of L-functions, which is easier for the author to think about, the Asai L-function having a pole is equivalent to the decomposition of the standard L-function as product of 2 degree 2

This research was supported by an NSF Mathematical Sciences Research Institutes Post-Doctoral Fellow- ship and by the American Institute of Mathematics. 1 2P.EDWARDHERMAN

L-functions over Q. Namely,

L(s, Π) = L(s, π)L(s, π ⊗ χD), √ where χD is the character associated to the field K = Q( D). A key point is that even if Π is a base change its standard L-function is entire, just not primitive. The Asai L-function, on the other hand, ”detects” or sieves those forms for us! To study these forms and their associated Asai L-function we use the Kuznetsov trace formula. We use a technique originally formulated in [L], and modified in [S], to average over the spectrum of cuspidal automorphic forms over the quadratic field K, along with an averaging over the Fourier coefficients. In other words, we study (ignoring test functions and convergence issues of the Π-sum) 1 N(μ) (1.1) g( ) c (Π), X X μ μ Π ∈ ∞ R+ where g C0 ( ). One can think of this sum as an average of L-functions. The sum μ is over integral ideals of K. Using the trace formula on (1.1) would lead to studying the analytic properties of the standard L-function of Π. Presumably, this should follow without too much difficulty from imitating the procedure in [S] for a standard L- function of forms over Q. There the result is 1 g(n/X) c (π)=O(X−A) X n n π ∞ cn(π) for any integer A>0. This is equivalent to L(s, π)= n=1 ns being entire. One should be able to get the same bound for (1.1). The Beyond Endoscopic approach to the Asai L-function would be to average (1.1) not over integral ideals but rational integers. Before stating the main theorem, we ask what do we expect from such a calculation? Our calculation should, for large X, behave as 1 n (1.2) g( ) c (Π) ≈ Res L(s, Π,Asai)+O(X−δ),δ >0. X X n s=1 n Π Π As we said, if the Asai L-function has a pole at s =1, then Π is a base change from a form over Q, specifically a form π of level D with nebentypus χD, see [A]. Lets label these forms by B(D, χD). So our heuristic in (1.2) becomes 2 −δ (1.3) Ress=1L(s, Π,Asai) ≈ A(π)L(1,sym (π)) + O(X ),δ >0,

Π π∈B(D,χD) where A(π) is a certain constant associated to π. This heuristic is still not quite accurate as (1.3) says every π ∈ B(D, χD) has a cuspidal base change. This is not true as Maass’ cuspidal theta forms or dihedral forms constructed from Hecke characters over a quadratic ﬁeld have base changes that are Eisenstein series. Say a theta form θω is constructed from a Hecke character ω, then the associated Asai L-function decomposes as 2 2 L(s, θω,Asai)=L(s, χD) L(s, θω). Thus our almost complete heuristic for the calculation is, 2 −δ (1.4) Ress=1L(s, Π,Asai) ≈ A(π)L(1,sym (π)) + O(X ),δ >0.

Π π∈B(D,χD) π= θω QUADRATIC BASE CHANGE AND THE ANALYTIC CONTINUATION OF THE ASAI L-FUNCTION 3

The heuristic in some sense is lacking as it only tells us information at or near s =1. From the deﬁnition of the Asai L-function, we only get analyticity (ignoring whether there is a pole or not at s = 1) of the L-function up to the zeroes of ζ(2s). To remedy this we do an extra averaging in m ∈ Z, which should, and does, remove the poles created by the zeroes of the ζ(2s). Final heuristic!

1 m2n (1.5) g( ) c (Π) ≈ A(π)L(1,sym2(π)) + O(X−M ), X X n m n Π π∈B(D,χD) π= θω for any positive integer M>0. Our main result is: × ∈ ∞ R+ 2 ∈ ∞ R+ ∞ Theorem 1.1. Let V = V1 V2 C0 ( ) , and g C0 ( ), such that 0 g(x)dx =1. h(V,y) is a Bessel transform of V with index y deﬁned in the next section. For any positive integer M ≥ 0, and quadratic integer l, (l, D)=1, (1) {Cuspidal contribution}

1 g(m2n/X) h(V,ν )c (Π)c (Π) = X Π n l m n Π= 1 1 −M 2π(1+ ) h(V,tφ)a ll (φt,D)a1(φt,D)+ h(V,kφ)a ll (φk)a1(φk) +O(X ). D r2 r2 r∈N φt,D= θωμ φk,D r|l (2) {CSC:=Continuous spectrum contribution} kπ Let μk = , where 0 is the fundamental unit of K, and τit(n)= log0 a it | x |iμk ab=n χD(a)( b ) ,ψμk (y):= N(q)=y ωμk (q), where ωμk (x)= x , then 1 2 g(m n/X)CSCn,l = X m n ll 1 4πh(V1,μk)h(V2,μk)ψμk ( r2 )ψμk (1) 2π(1 + ) 2 + D L(1,χD)L(1,ω ) r∈N k∈Z μk r|l ∞ ll 1 h(V1,t)h(V2,t)τit( r2 )τit(1) −M 2 + O(X ). 4π −∞ |L(1 − 2it, χD)| r∈N r|l (3) {Coefficient comparison} If Π=BCK/Q(φ), then cl(Π) = r∈N a ll (φ). r|l r2 2. Details of Paper There have been several papers using the trace formula to prove quadratic base change. These include [La], [Sa], and [Y]. The first two references [La], [Sa] proved base change by comparing a trace formula over the ground field (in our case Q) with a certain ”twisted” trace formula over the quadratic field. The comparison is made through an equality of the associated test functions used for each trace formula. The last reference [Y] is the most similar 4P.EDWARDHERMAN to our approach as it uses the relative trace formula. The Kuznetsov trace formula, which we use, is a special case of the relative trace formula, see [KL] for a derivation. However, there is still a comparison of trace formulas involved with [Y]. Naively, the beyond endoscopy idea is to start with one trace formula and extra averaging over spectral data( fourier coefficients, whittaker functions,..) and by ”brute force” compute the answer. The same test function is used from start to end. To do such a calculation, one relies heavily upon analytic number theory techniques, not used in previous trace formula comparison papers. With these techniques, one gets a very good feel for the geometric side of the trace formula, and how one literally ”builds” the spectral sum of the forms over Q.

It seems a nice gift that the analytic continuation of the Asai L-function comes with the comparison. In the sense of getting analytic information on L-functions using a trace formula, the author is reminded of the beautiful paper of Jacquet and Zagier [JZ], which gets analytic continuation of the symmetric square L-function associated to a Maass form.

To the specifics of the paper, in section 3 we introduce the Kuznetsov trace formula stated in [BMP1]. They stated it for a general real number field, we only use it for a quadratic extension of Q. In section 4, we prove a crucial bijection between solutions of two different x sets of equations. This bijection gets rid of the difficult to handle e( c ) when one ”opens” the Kloosterman sum on the geometric side of the trace formula. With the bijection, in section 5, we implement it into the trace formula. Then, the hard analytic number theoretic computations are done in section√ 6. The problem is that all of our computations are done over the quadratic field Q( D). In section 7 we use a formula of Zagier, [Z] which in its simplest form is, for D|n,

r + r 1 e( )=√ SD(1, 1,n). n D r∈OK/(n) ≡ rr 1(n) x+x Here S (1, 1,n)= ∗ χ (a)e( ). This is our ”bridge” to the trace formula over Q. D a(n) D n √ In section 8, we deal with the continuous spectrum over Q( D). In section 9 we realize that the computation done in section 6 really is the geometric side of the trace formula over Q. This requires an important theorem on the convolution of Bessel√ transforms in [H]. In order to get an equality or comparison of just cusp forms from Q( D)toQ, we compare Fourier coefficients for the continuous spectrum over the 2 different fields. Then the continuous spectrum can be removed from both sides. From there with certain choices of test functions we can get equalities of a fixed archimedean parameter.

Acknowledgements. I would like to thank my advisor Jonathan Rogawski for proposing the idea of the studying the Asai L-function. I would like to also acknowledge the very useful conversations with Brian Conrey and Akshay Venkatesh.

3. Preliminaries √ For simplicity we focus on the field K = Q( D), with D prime and assume it is of class number one. This is to simplify already intensive calculations. The ring of integers will√ be denoted OK. Here the discriminant DK = D, and the different is generated by δ = D. Likewise, define the absolute norm of ideal c as N(c), and the norm of an element z ∈OK as QUADRATIC BASE CHANGE AND THE ANALYTIC CONTINUATION OF THE ASAI L-FUNCTION 5

N(z). We denote the non-trivial automorphism in this ﬁeld by x → x. We use the standard x x notation for the exponential e(TrK/Q(x)) := exp(2πi( δ + δ )). A bit of terminology is needed before we can deﬁne the Kuznetsov trace formula. We closely follow, and leave much of the details, to [BMP1] . We consider the algebraic group G = RK/Q(SL2)overQ obtained by restriction of scalars applied to SL2 over K.Wehave ∼ 2 ∼ (3.1) G := GR = SL2(R) , GQ = {(x, x ):x ∈ SL2(K)}, 2 G contains K := SO2(R) as a maximal compact subgroup.

The image of SL2(O) ⊂ SL2(F ) corresponds to GZ. This is a discrete subgroup of GR with ﬁnite covolume. It is called the Hilbert modular group.WelabelitΓ. In this paper, we are concerned with functions on Γ\G. We restrict ourselves to even functions: f(−g)=f(g). By L2(Γ\G)+ we mean the Hilbert space of (classes of) even functions that are left invariant under Γ, and square integrable on Γ\G for the measure induced by 2 \ + the Haar measure. This Hilbert space contains the closed subspace Lc (Γ G) generated by 2 \ + 2 \ + integrals of Eisenstein series. The orthogonal complement Ld(Γ G) of Lc (Γ G) is the clo- sure of Π VΠ,whereVΠ runs through an orthogonal family of closed irreducible subspaces for the G-action in L2(Γ\G) by right translation.

3.1. Irreducible unitary representations. Each representation Π has the form Π = Π1 ⊗ Π2,withΠj an even unitary irreducible representation of SL2(R). Table 1. of [BMP1] lists the possible isomorphism classes for each Πj. For each Π we deﬁne a spectral parameter νΠ =(νΠ,1,νΠ,2), with νΠ,j as in the last column of the corresponding table. There are Casimir 2 operators Cj acting on each coordinate for 1 ≤ j ≤ 2. The eigenvalue λΠ ∈ R is given by 1 − 2 ∈ 1 λΠ,j = 4 νΠ,j. WenotethatifΠj lies in the complementary series, λΠ,j (0, 4 ) and if Πj is ± ∈ Z ≥ b − b ∈ Z isomorphic to a discrete series representation Db , b 2 , b 2, then λΠ,j = 2 (1 2 ) ≤0. The constant functions give rise to Π = 1 := ⊗j1. It occurs with multiplicity one. If VΠ does not consist of the constant functions, then Πj =1forall j.

3.2. Automorphic forms and Fourier coefficients. The elements of VΠ that transform on the right according to a character of the maximal compact subgroup K are square integrable automorphic forms. The r-th Fourier coefficients of the automorphic form Π at the cusp κ, is ar,κ(fΠ)whereκ is the defined as in [BMP1]. They are essentially independent of the actual choice of the automorphic forms in VΠ. For the purposes of the Kuznetsov trace formula we define 2 (−1)q(νj )/2(2π|r |)−1/2 (3.2) d (r, ν):= j , κ Γ 1 + ν + 1 q(ν )sign(r ) j=1 2 j 2 j j where q(νj)=0ifΠj is the principal or complementary series, and equals b if Πj is the discrete series with minimal weight b. Then the coefficients associated to the trace formula are defined as c (Π) := ar,∞(Π) . This r d∞(r,ν) is very similar to Equation (17) of [BMP1]. We do not indicate the cusp in the definition of dκ(r, ν) as we are in class number one, and so we have to only consider the cusp ∞. The Kuznetsov trace formula then is (3.3) h(V,νΠ)cμ(Π)cν(Π) + {CSCμ,ν} = Π= 1 6P.EDWARDHERMAN

∞ 1 √ = S(ν, μ, c)V (4π μν/c), N(c) c=0 ∈O × ∞ R+ 2 where μ, ν K.V = V1 V2 is just a test function in C0 ( ) . The tranforms associated to the archimedean parameter ν are Π k ∞ −1 ∈ Z i 0 Vi(x)Jνφ−1(x)x dx if νφ 2 ; (3.4) h(Vi,νΠ)= ∞ −1 ∈ R 0 Vi(x)B2νφ (x)x dx if νφ i . −1 Here, B2it(x) = (2 sin(πit)) (J−2it(x) − J2it(x)), where Jμ(x) is the standard J-Bessel function of index μ. The Kloosterman sum rx + sx (3.5) S(r, s, c):= e(TrK/Q( )), ∗ ∗ c x∈(OK/cOK) where xx ≡ 1(c). We do not describe the continuous spectrum here as it is quite lengthy, but elaborate on it greatly in Section 8. Other descriptions and applications of the Kuznetsov trace formula for number ﬁelds include [BMP1], [BMP2], [KL], and [V].

4. Number-theoretic lemmas We prove some number-theoretic lemmas that are crucial to our calculation.

Deﬁnition 4.1. Let X(c, n) denote the equivalence classes of pairs x with x ∈OK such that (x, c)=1,m∈ Z, and δcx + δcx = n − mN(δc) Here we say that x is equivalent to y if x ≡ y (mod c). Let X(c, n) be a set of representatives for the classes in X(c, n). Proposition 4.2. Let γ =(−1)eηm where η is the√ fundamental unit of the ﬁeld K and e ∈{0, 1},m∈ Z. Let x ∈ X(c, 0), then c = γa or γb D, a, b ∈ N. Further, ηmx ≡ ηmx(δa), √ if c = γa and ηmx ≡−η mx(Db), if c = γb D. Proof. It is clear that x ∈ X(c, 0) implies√ c ≡ 0(c ), likewise c ≡ 0(c), This implies by [FT](V.1.16) c = ηma ∈ Z or c = ηmb D, a, b ∈ Z. Without loss of generality take c = γa, then implies δa(η mx − ηmx) ≡ 0(Da2), or η mx ≡ ηmx(δa). This implies ηmx ≡ η mx(δa). √ The identical calculation for c = γb D, gives η mx = −ηmx. It is assumed, unless stated otherwise, n =0 .

Proposition 4.3. Let x ∈ X(c, n). Let x ∈ OK be an inverse of x modulo c. Then there exists an r such that rr ≡ 1(modn) and δcr + δc (4.1) x = , n The r is uniquely determined modulo √n by the equivalence class of x, and the map from D X(c, n) to the set r modulo n is injective. Proof. Set nx − δc r = . δc Note that r is an integer in the ﬁeld K because nx − δc =(δcx + δcx)x − δc = δc(xx − 1) + δcxx ≡ 0(modc) QUADRATIC BASE CHANGE AND THE ANALYTIC CONTINUATION OF THE ASAI L-FUNCTION 7

It is clear that r is determined by x.Ifwereplacex by y = x + μc, r is replaced by n s = r + μ√ D If x and y in X(c, n) are both associated to r,thenx = y. Therefore x ≡ y (mod c). Finally, nx − δc nx − δc n2xx − nxδc − nyδc nxx − xδc − xδc rr = =1+ =1+n δc δc N(δc) N(δc) But nxx =(δcx + δcx)xx = δcxxx + δcxxx so we have δcxxx + δcxxx − xδc − xδc rr =1+n N(δc) δc(xx − 1)x + δc(xx − 1)x =1+n N(δc) xq xq =1+n + ,q ∈ OK. δ δ The expression in brackets is an integer of the ﬁeld, so rr ≡ 1(modn). Deﬁnition 4.4. Let c be an integer in OK. Set d =(c, c ). Assume that d|n.LetY (c, , n) ∗ be the set of classes r ∈ (OK/n) such that ≡ n (a’) (δc/d)r +(δ c /d) 0(modd ) ≡ n | (b’) (δc/d)r +(δ c /d) 0(modk ) if k d and k

5. Taking Geometric side of Trace formula Using the Kuznetsov formula we get 1 2 (5.1) (L):= g(m n/X) h(V,νΠ)cn(Π)cl(Π) + {CSCn,l} = X m ∈Z 1 ,n Π= √ √ 1 1 4π nl 4π nl g(m2n/X) S(n, l, c)V ( )V ( ). X N(c) 1 c 2 c m,n∈Z c We now break up the Kloosterman sums and gather all the n-terms. We can do this because c and n sum are ﬁnite due to the support of g and V.

1 1 lx lx (5.2) e( + ) X N(c) δc δc m c x(c)∗ √ √ x x 4π nl 4π nl e(n( + ))g(m2n/X)V ( )V ( ) , δc δc 1 c 2 c n∈Z where x is the multiplicative inverse of x(c). Since the term in brackets is smooth, we can and do apply Poisson summation to the n-sum as well as a change of variables t → Xt :

1 xl xl (5.3) e( + ) N(c) δc δc m c x(c)∗ √ √ ∞ xδc + xcδ − N(cδ)m 4π lXt 4π lXt m2 e(Xt( ))g( t)V1( )V2( )dt . −∞ N(cδ) c c m We hope the confusion of using m-andm-sums is not to big, the use of one will go away shortly. QUADRATIC BASE CHANGE AND THE ANALYTIC CONTINUATION OF THE ASAI L-FUNCTION 9

As we have fixed l, let √ √ ∞ Xtn 4π Xtl 4π Xtl m2 Im(n, c, X):= e( )g( t)V1( )V2( )dt −∞ N(δc) c c Then (L) is equal to 1 xl x l e( + ) Im(xδ c + x cδ − N(cδ)m, c, X) N(c) c c m c x(c)∗ m∈Z Note that for fixed X,thesumoverc is finite. Let X(c, n) be the set of solutions (x, m)of the equation δcx + δcx − N(cδ)m = n ∗ where x range over a fixed set of representatives of (OK/c) and m ∈ Z. Then (L)isequal to 1 xl xl e( + )Im(n, c, X) N(c) δc δc m n∈Z c (x,m)∈X(c,n) Note there is a bijection between the set (x, m) ∈ X(c, n) and the set of equivalence classes x in X(c, , n) from Definition 4.1. Thus we may replace the sum over X(c, n)witha sum over X(c, n): 1 xl xl (L)= e( + )Im(n, c, X) N(c) δc δc m n∈Z c x∈X(c.,n) Finally, let 1 xl xl (5.4) A := e( + )Im(n, c, X) n,X N(c) δc δc m c x∈X(c,,n)

Then (5.5) (L)= An,X ∈Z n√ Let us fix the generator of the different as D. Lemma 5.1. For n defined above, D|n. √ √ Proof. Let c := α + β D, x := w + z D, then √ √ √ √ √ √ (5.6) n =(− D)(α − β D)(w + z D)+( D)(α + β D)(w − z D)+ Dm(α2 − Dβ2). This equals −2D(αz − βw)+Dm(α2 − 2β2). We know see that (5.5) equals, (5.7) An,X . D|n∈Z

Now for n =0 , we can use the bijection of Proposition 4.5 to rewrite An,X as a sum over r ∈ Y (c, , n): 10 P. EDWARD HERMAN

rl + rl 1 −1 lc lc A = e( ) e( ( + ))Im(n, c, X) n,X N m n (c) n c c r∈OK/(n) c rr≡1(n) r ∈ Y (c, , n) Deﬁnition 5.2. Let X(r) be the set c such that r ∈ Y (c, n).

1 xl xl Deﬁnition 5.3. Let Hn,m(x, x ):= xx e( −nx + −nx )Im(n, x, x ,X). The main result of the calculations can be broken down into the cases: n =0, and n =0 . In Section 6.1 we show for any integer M ≥ 0, 1 ∞ ∞ δ(l, l )(1 + D ) dxdy −M (5.8) A0,X = V1(x)V2(y) + O(X ). 2 0 0 xy In Section 6.2 for n =0 , and for all r ∈ (Z/n)∗, and for any integer M ≥ 0, we have − ∞ ∞ 1 1 lc l c 1 −M (5.9) e( ( + ))Im(n, c, X)=(1+ ) H (x, y)dxdy+O(X ). N(c) n c c D n m c∈X(r) 0 0

In Section 7 for (l, D)=1, we get the r-sum in An,X is replaced by a sum of Kloosterman sums twisted by the character χD attached to the quadratic ﬁeld K. Speciﬁcally this is rl + rl 1 ll Da e( )=√ rSD( , 1, ). n D r2 r r∈OK/(n) r∈N rr≡1(n) r|l r|a Combining sections 6.2 and 7 we get 1 ll Da ∞ ∞ A =(1+ ) rS ( , 1, ) H (x, y)dxdy + O(X−M ). n,X D D r2 r n r∈N 0 0 r|l r|a Lastly, sections 6.1, 6.2, and 7 then show (5.10) ∞ δ(l, l)(1 + 1 ) ∞ ∞ dxdy 1 1 ll Da (L)= D V (x)V (y) +(1+ ) rS ( , 1, )× 2 1 2 xy D Da D r2 r 0 0 a=1 r∈N r|l r|a ∞ ∞ −M HDa(x, y)dxdy + O(X ). 0 0

6. Computing Aq,X.

We separate the cases of A0,X , and Aq,X,q =0 . The former is computed ﬁrst.

j 6.1. Evaluating A0,X . Using Proposition 4.2, we get from inside (L)forc = η a, xl xl xηjl xη jl xηj(l − l) e( + )= e( + )= e( ), δc δc δa δa δa x∈X(c,0) x(a)∗ x(a)∗ QUADRATIC BASE CHANGE AND THE ANALYTIC CONTINUATION OF THE ASAI L-FUNCTION 11 √ and for c = Db, we get xl xl xηj(l − l) e( + )= e( ). δc δc Db x∈X(c,0) x(Dj)∗

Introducing these to A0,X gives,

(6.1) ∞ √ √ 1 ηjx(l − l) ∞ 4π Xlt 4π Xlt A := e g(m2t)V ( )V ( )dt+ 0,X N j 2 1 j 2 j m ∈Z (η )a ∗ δa −∞ η a η a j a=1 x(a) ∞ √ √ 1 1 ηjx(l − l) ∞ 4π Xlt 4π Xlt m2 √ √ j 2 e g( t)V1( )V2( )dt . D N(η )b Db −∞ ηjb D η jb D b=1 x(Db)∗ Before we compute, we introduce 1ifl = l ; δ(l, l )= 0ifl = l. Proposition 6.1. For any integer M ≥ 0, 1 ∞ ∞ δ(l, l )(1 + D ) dxdy −M (6.2) A0,X = V1(x)V2(y) + O(X ). 2 0 0 xy Proof. Let √ √ ∞ 1 2 4π lt 4π l t Hm(x, y):= g(m t)V1( )V2( )dt, xy 0 x y then notice Hm(x, y)=H1(mx, my).

For economy we use H(x, y)forH1(x, y). After a change of variables x → ηjx, ∞ 1 x(l − l) mηja mηja (6.3) A0,X = e H( √ , √ )+ X δa X X m j∈Z a=1 x(a)∗ ∞ √ √ x(l − l) mηjb D −ηjb D e H( √ , √ ) . Db X X b=1 x(Db)∗ It is suﬃcient to focus on the ﬁrst part of (6.3). ∞ ∞ ˆ s1−1 s2−1 Let H(s1,s2):= 0 0 H(x, y)x y dxdy, then by Mellin inversion, the sum over equals √ s1 √ s2 1 X X Hˆ (s ,s ) L(s + s )ds ds . 2πi 1 2 mηj mηj 1 2 1 2 m j∈Z (σ1,σ2) ∞ fa(l−l ) xy Here L(s):= a=1 a2s , where fn(y)= x(n)∗ e n . n φ(n) Using the fact fn(y)=μ (y,n) φ( n ) , and multiplicativity, (y,n) 1 1 1 1 | − (1 + 2s−1 ) (6.4) L(s)= (1 − ) (1 − )(1 + )= p (l l ) p . p2s p2s p2s−1 ζ(2s) p(l−l) p|(l−l) 12 P. EDWARD HERMAN

If l = l then ∞ φ(a) ζ(2s − 1) L(s)= = . a2s ζ(2s) a=1 ˆ −N Since H is smooth and compactly supported H(s1,s2) N (1 + |t1|)(1 + |t2|) , and we can clearly interchange the integral and m-sum. Notice also the m sum inside of the integral is ζ(s1 + s2). So we have after recollecting terms, √ s1 √ s2 1 X X (6.5) Hˆ (s .s ) L(s + s )ζ(s + s )ds ds . 2πiX 1 2 ηj ηj 1 2 1 2 1 2 j (σ1,σ2) Lets assume l = l, notice now how the poles of L(s) are at the poles of ζ(2s), however these are exactly removed by our extra summation over m ∈ Z! Now using contour integration σi →−M,M > 0,i=1, 2, we get √ √ s1 s2 X ˆ 1 ˆ X X (6.6) H(1)+ H(s1.s2) m m L(s1 +s2)ζ(s1 +s2)ds1ds2. 2 2πiX − − η η j∈Z ( M, M) To execute the sum over the units in the ﬁeld, we ﬁrst need a lemma. ∗ Lemma 6.2. Let a, b ∈R,a+b>0. There exists a constant k such that for any f :(R )2 → C | |≤ | |a | |−b satisfying f(y1,y2) j=1,2 min( yj , yj ), we have a −b |f(y)|≤k(1 + log |N(y)| ) min (y1y2) , (y1y2) . ∈O∗ Proof. See Lemma 8.1of[BM]. Again there are no convergence issues and we can clearly interchange the integral and j-sum. Using Lemma 6.2, (6.6) is

6X −δ Hˆ (1) + O (X ). π2 l,l Similar analysis can be done for l = l, here however we get no pole from the L-function (6.4) and the term here is O(X−M ). For the b-sum we only note here φ(Db) D φ(b) = . (Db)2s D2s b2s b=1 b=1 X ˆ −M The same argument in (6.1) gives 2D H(1) + O(X ). Therefore (6.1) equals 1 X X (1 + 1 ) ( Hˆ (1) + Hˆ (1)) + O(X−M )= D Hˆ (1) + O(X−M ). X 2 2D 2 Then we note √ √ ∞ ∞ ∞ ˆ 4π lt 4π l t dxdy H(1) = g(t)V1( )V2( )dt . 0 0 0 x y xy √ √ → lt → lt ∞ A change of variables x 4πx,y 4πy , and the use of 0 g(x)dx =1, gives the result. QUADRATIC BASE CHANGE AND THE ANALYTIC CONTINUATION OF THE ASAI L-FUNCTION 13

6.2. Evaluating An,X ,n =0 . Now fix r, then by Proposition 4.5, X(r) is the set of (c, ) such that, setting d =(c, c), we have c c n (1) d , d are both prime to dδ . δcr+δc ≡ n (2) d 0(modd ) δcr+δc ≡ n (3) d 0(modk )ifk is a proper divisor of d. Now for each divisor d of n,letX(r, d) be the set of pairs (c, )inX(r) such that (c, c)=d. We would like to prove that there is a constant R(n, d) such that (6.7) − ∞ ∞ 1 1 lc l c 6 1 −M e( ( + ))Im(n, c, X)=R(n, d) H (x, y)dxdy + O(X ), N(c) n c c π2 n n m c∈X(r,d) 0 0 and R(n, d)=1. d|n Let us describe X(r, d) explicitly. If c ∈ X(r, d), then δc δc n (6.8) = − r + λ d d d c c Lemma 6.3. Fix c such that d is prime to d . Let λ be a whole number and defined by (6.8). Then c ∈ X(r, d) if and only if (λ, d)=1. c n | c Proof. If ( d , d )=p>1, then by (6.8), p d a contradiction. And if p divides c /d and n/d, then (6.8) gives p|r(c/d). But (r, n) = 1, so this implies that p divides c/d – again a contradiction. If d = 1, the requirements are met. If d>1, we must also require that δcr + δc n (6.9) ≡ 0(modp ) n d for all p|d.But δcr + δc n = λ d d Therefore (6.9) holds if and only if λ ≡ 0(modp) for all p|d. In other words, λ must be relatively prime to d. Perhaps it is more convenient to replace the c with dc where 1 = (c, c). Then the left-hand side of (6.7) is equal to 1 −1 lc lc (6.10) e( ( + ))Im(n, dc, X). N(dc) n c c m c:(c,c)=1 − n cr c =λ dδ (λ,d)=1 To study this more we also we need a further lemma counting solutions of (6.8). Lemma 6.4. Let n ∈ Z and rr ≡ 1(n).If f|d, d = d, and d|n, then the set nf nf #{c : cr − c ≡ 0( ) and (c, c)=1} = . dδ d − nf If d = d , then the set is of size Dd. 14 P. EDWARD HERMAN √ √ Proof. We assume = 1, the other case is similar. Let c := a + b D and r := x + y D. Then the condition nf (6.11) cr ≡ c( ), dδ − ≡ nf | ≡ implies a(x 1) + Dby 0( d ). As D n, (a, n)=1, implies a =0orx 1(D). If a =0, then (c, c ) > 1, a contradiction. √ Now assume x ≡ 1(D), or x =1+Dl, l ∈ N. Then r =1+Dl + y D and rr ≡ 1(n) implies (6.12) y2 ≡ Dl2 +2l(h), where n = Dh. ≡ hf Likewise, (6.11) implies al + by 0( d ). Assume the pair (a, b)and(a, b ) satisfy (6.11) − ≡ Dhf ≡ Dhf ≡ Dhf for b = b , then one gets (a a )l 0( d ). If l 0( d ), then (6.12) implies y 0( d ), ≡ Dhf { ≡ nf } nf or r 1( d ). Thus counting the integral elements c : c c ( dδ ) is clearly d . Otherwise, ≡ Dhf nf a a ( d ), and again the size of the set is d . The same argument works if we assume two different pairs (a, b)and(a,b). 1 xl xl Definition 6.5. Let Hn,m(x, x ):= xx e( −nx + −nx )Im(n, x, X). Like the n = 0 case, Hn,m(x, x )=Hn,1(mx, mx ), by a change of variables. Again for economy we label Hn(x, x )=Hn,1(x, x ). Then (6.10) equals 1 dc dc Hn(m√ , m√ ). X X X m c:(c,c )=1 − n cr c =λ dδ (λ,d)=1 Proposition 6.6. For any integer M ≥ 0, √ ∞ ∞ 1 dc dc D −3M −M Hn(m√ , m√ )= Hn(x, y)dxdy + O(n X ). X X X n m c:(c,c )=1 0 0 − n cr c =λ dδ (λ,d)=1 1 Proof. Ignoring the X factor, we get dc dc (6.13) Hn(m√ , m√ ). X X m c:(c,c )=1 − n cr c =λ dδ (λ,d)=1 Lets fix an m for the moment. The condition (λ, d) = 1 is now removed with Mobius inversion.Letting E := cr − c, this gives for (6.13) dc dc μ(a)Hn(m√ , m√ ). X X c a|d (c,c )=1 E a| n dδ Rewritten this is dc dc μ(a) Hn(m√ , m√ ). X X a|d c (c,c)=1 ≡ na E 0( dδ ) QUADRATIC BASE CHANGE AND THE ANALYTIC CONTINUATION OF THE ASAI L-FUNCTION 15

≡ na The condition E 0( dδ ) is removed via Dirichlet characters. Note from Lemma 6.4, the na size of the group of multiplicative characters on this group is φ( d ). We get (6.13) equals ⎛ ⎞ 1 dc dc ⎝ ⎠ m√ m√ μ(a) na χ(cr)χ(c ) Hn( , ). φ( ) na X X m a|d c d χ( ) (c,c)=1 d Lastly, (c, c)=1isequivalentto(c, l) = 1 for all l ∈ Z. This condition can be removed again by Mobius inversion. This provides ⎛ ⎞ ∞ 1 dlc dlc ⎝ ⎠ m√ m√ (6.14) μ(a) μ(l) na χ(lcr)χ(lc ) Hn( , ). m | φ( d ) na X X a d nal=1 c χ( d ) (l, d )=1 na The main contribution comes from the principal character χ0( d ). We label the term with the principal character as T M (n, d) and the rest of the characters as T R(n, d). Therefore, (6.24) equals T M (n, d)+T R(n, d).

6.3. Analysis of T M (n, d). We focus now on T M (n, d). This term fully written out is ∞ μ(a) dlc dlc (6.15) μ(l) Hn(m√ , m√ ). φ( na ) X X m a|d d l=1 nac na (c, )=1 (l, d )=1 d Now we perform Poisson summation on the c-sum. The relatively prime condition we remove with Mobius inversion to get (6.16) √ ∞ ∞ dlc dlc 2 dlx dly Hn(m√ , m√ )= D (1−1/p) Hn(m√ , m√ )e(mx+m y)dxdy. c X X | na ∈O −∞ −∞ X X na p d m K (c, d )=1 √ √ → Xx → Xy Including this into (6.15) and a change of variables x mdl and y mdl , we get

(6.17) √ ∞ X D 1 μ(a) μ(l) ∞ ∞ X (1−1/p)2 H (x, y)e( (mx+my))dxdy. 2 m2 na 2 n d m | φ( d ) l | na ∈O −∞ −∞ ld a d nal=1 p d m K (l, d )=1

Now Hn(x, y) is compactly supported away from zero, and integration by parts Q-times for m =0,onegets ∞ ∞ Q ∞ ∞ Q X ld ∂ Hn(x, y) X Hn(x, y)e( (mx+m y))dxdy = Q e( (mx+m y))dxdy. −∞ −∞ ld mX −∞ −∞ ∂x ld Further for a ﬁxed m =0 , we then have from (6.17)

√ ∞ 1 1 D μ(a) μ(l) (6.18) B (n, d, X):= (1 − 1/p)2× m Q m2−Q 2−Q Q na 2−Q X m d m | φ( d ) l | na a d nal=1 p d (l, d )=1 16 P. EDWARD HERMAN ∞ ∞ ∂H (x, y) X n e( (mx + my))dxdy. −∞ −∞ ∂x ld Now remember that in Deﬁnition 6.5 of Hn(x, x )is √ √ ∞ Xtn 4π Xtl 4π Xtl m2 Im(n, x, X)= e( )g( t)V1( )V2( )dt. −∞ N(δx) x x 1 Again, integration by parts K-times in this integral gives the bound O( nK ), and thus we also have the same bound for Hn(x, x ). We recall that ∞ μ(l) 1 1 = . 2s − 1 l ζ(2s) | na (1 p2s ) l=1 p d (l, na )=1 d m 1 Notice now that then the product of the and l sum is p| na − 1 . d (1 p2−Q ) Using trivial bounds on the a|d-sum and p| na -sum, and taking K ≥ 3Q, we get d −3Q −Q (6.19) Bm(n, d, X)=O(n X ). m=0 ∈OK Nowwefocusonthem = 0 term. Again the m-sum and the l-sum cancel, so the term to compute is √ − 1 ∞ ∞ DX μ(a) 1 p 2 na ( 1 ) Hn(x, y)dxdy. d φ( ) 1+ −∞ −∞ a|d d p| na p d 1 − 1 1 XR(n,d) ∞ ∞ Using φ(n) p|n(1 p )= n , our expression becomes D −∞ −∞ Hn(x, y)dxdy, where 1 μ(a) 1 (6.20) R(n, d):= 1 . dn | a | na (1 + p ) a d p d 1 Lemma 6.7. d|n R(n, d)= n . Proof. R(n, d) is multiplicative and so it remains to evaluate it at a prime power n = pl where d = pi,i =0, ..., l. It is suﬃcient to do this computation for i =0, 1 ≤ i ≤ l − 1, and i = l. For i =0, the sum has only one term 1 R(pl, 1) = . l 1 p (1 + p ) For 1 ≤ i ≤ l − 1, 1 − 1 l i 1 p R(p ,p)= l+i 1 . p 1+ p For i = l, 1 1 1 R(pl,pl)= 1 − = . 2l 1 2l 1 p p(1 + p ) p (1 + p ) Now l−1 − 1 1 − 1 1 1 l l = p − 1= p p . pi 1 − 1 1 − 1 i=1 p p QUADRATIC BASE CHANGE AND THE ANALYTIC CONTINUATION OF THE ASAI L-FUNCTION 17

So l − 1 1 − 1 1 1 1 l 1 (6.21) R(pl,pi)= + p p p + pl(1 + 1 ) pl 1+ 1 1 − 1 p2l(1 + 1 ) i=0 p p p p 1 (6.22) = pl + pl−1 − 1+1 2l 1 p (1 + p ) pl + pl−1 1 (6.23) = = . 2l 1 l p (1 + p ) p

1 Incorporating back the X factor we are left with √ D ∞ ∞ Hn(x, y)dxdy. n −∞ −∞ ∈ ∞ R+ But remember Hn(x, y) is deﬁned in terms of the test functions V1,V2 C0 ( ), so inte- ∞ ∞ gration is limited to 0 0 Hn(x, y)dxdy. The same compuation can be done for d = d , to get √ D ∞ ∞ Hn(x, y)dxdy. Dn −∞ −∞

6.4. Analysis of T R(n, d). Here (6.24) ∞ μ(a) 1 dlc dlc R m√ m√ T (n, d):= na μ(l) na χ(lcr)χ(lc ) Hn( , ). m | φ( d ) na φ( d ) X X a d χ=χ0( d ) nal=1 c (l, d )=1 na Now ﬁx a character χ = χ0 and break the c-sum into arithmetic progressions modulo ( d ) to get the inner sum equals dlc dlc (6.25) χ(q)χ(q ) Hn(m√ , m√ ). m na ≡ na X X q( d ) c q( d ) | |2 na ∈O Note we used χ(l) =1. Now using Poisson summation on c = q + d m, m K, we get, after a few change of variables, (6.25) equaling ∞ ∞ X mq X χ(q)χ(q ) e(TrK Q( )) H (x, y)e( (mx + m y))dxdy. m 2 / na n ( nla) na ∈O d −∞ −∞ ld q( d ) m K Now similar to the poisson summation for T M (n, d), we separate the sum into cases m =0 and m =0 . Again, the interchange of sums is valid due to Hn(x, y) being smooth and compactly supported. Now for m =0, we execute the q-sum which is zero since it is a complete character sum. As for m =0 , the exact same analysis used for T M (n, d) can be reused. Therefore, for any Q ≥ 0, T R(n, d)=O(n−3QX−Q). This completes Proposition 6.6. 18 P. EDWARD HERMAN

7. From exponential sums over a quadratic field to Kloosterman sums In order to classify base change, one needs a bridge between a trace formula over our quadratic field K to a trace formula over Q. This formula was discovered in [Z]. √ D1 We need a few definitions first. Let D = D1D2, then define ψ(D2)=( ) D2. Next D2 define ψ(D ) n H (n, m)= 2 HD1 ( ,m), b bD1 D2 D2 D=D1D2 D2|n (b,D2)=1 where 1 c D1 Hc (n, m)= ( )SD1 (nD2,m,c). c D2 Here d nD2d + md SD1 (nD2,m,c)= ( )e( ). D1 c d(c)∗ Proposition 7.1. Let D be prime such that D ≡ 1(4), then writing n = Da, a ∈ N rl + rl √ ll e( )=a D rH ( , 1). n a/r r2 r∈OK/(n) r∈N rr≡1(n) r|l r|a Corollary 7.2. If (l, D)=1then Proposition 7.1 implies rl + rl 1 ll Da e( )=√ rSD( , 1, ). n D r2 r r∈OK/(n) r∈N rr≡1(n) r|l r|a

8. The continuous spectrum The continuous spectrum contribution is ∞ (8.1) CSCρ.ξ := h(V1,t+ μ)h(V2,t+ μ )Ω(ρ, 1/2+it, iμ)Ω(ξ,1/2+it, iμ)dt. −∞ μ iμ ∗ More speciﬁcally, μ is in the lattice TrK/Q(x)=0, such that | | =1, for all ∈O. In πk our case, μ = ,k ∈ Z where 0 is the fundamental unit of our ﬁeld K. Let ω(x):= 2log0 iπk x | | 2log0 ∈ Z x ,k , and l σl,ω(n):= ω(a)N(a) . a|n We prove in this section Proposition 8.1. 1 2 1 h(V1,μ)h(V2,μ) m 2 (8.2) g( n/X)CSCn,ξ = (1 + ) 2 σ0,ωμ (ξ)+ X m D L(1,χD)L(1,ωμ) n μ ∞ N it σ−2it,0(ξ) (ξ) −M (8.3) h(V1,t)h(V2,t) 2 4 2 dt + O(X ), −∞ cosh(πt) |Γ(1/2+it)| |L(1 − 2it, χD)| for any integer M ≥ 0. QUADRATIC BASE CHANGE AND THE ANALYTIC CONTINUATION OF THE ASAI L-FUNCTION 19

From [BMP2], (8.4) 2N(r)it |π|i(t+μ) Ω(r, 1/2+it, iμ)= × cosh(π(t + μ)) cosh(π(t + μ)) Γ(1/2+i(t + μ))Γ(1/2+i(t + μ)) ω2(c)S(r, 0,c) , N(c)1+2it (c) where S(r, 0,c)= x(c)∗ e(xr/c). Then, using Mobius inversion 2 −2 ω (c)S(r, 0,c) ω(r) σ 2 (r) (8.5) = 2it,ω . N(c)1+2it N(r)2itL(1+2it, ω2) (c) We collect the n and m terms together, noting ω(n)2 =1, from (8.1) to get σ 2 (n) 1 (8.6) g(m2n/X) 2it,ω = G(s)Xsζ(2s)L(s)ds. N(n)it 2πi m n (σ) Here ∞ σ 2 (n) L(s):= 2it,ω , ns+2it n=1 is the Asai L-function associated to the Eisenstein series over the quadratic ﬁeld K.Using p p the multiplicativity of the divisor function, (8.6) equals for p split, say, p = 1 2, − 1 (1 p2s ) 2 p 2 p . − ω ( 1) − ω ( 2) − 1 − 1 (1 ps )(1 ps )(1 ps+2it )(1 ps−2it ) For p inert, ω2(p)=1, so we get − 1 1 (1 2s ) = p . 1 1 1 χD(p) 1 1 (1 − s it )(1 − s− it ) − − − − p +2 p 2 (1 ps )(1 ps )(1 ps+2it )(1 ps−2it ) p2 p For p = 1 ramiﬁed, χ( 1)=1, and likewise, 1 − 1 (1 + s ) (1 2s ) p = p . 1 1 1 χD(p) 1 1 (1 − s it )(1 − s− it ) − − − − p +2 p 2 (1 ps )(1 ps )(1 ps+2it )(1 ps−2it ) We now have 2 cases: k =0and k =0. 8.1. k = 0. Its clear L(s)haspolesats =1± 2it, but as well the ζ(2s) cancels with the ζ(2s), from the Asai L-function. Thus, shifting the contour of the integral to −M,M ≥ 0 we pick up pole at s =1+2it with residue ζ(1+4it)L(1+2it, ω2) X1+2it . ζ(2(1 + 2it)) The analogous argument follows for the pole at s =1− 2it. Now using the contour shift (using the fact we chose G(1) = 1.) and the functional equation of the gamma function π Γ(1/2+it)Γ(1/2 − it)= , cosh πt 20 P. EDWARD HERMAN

(8.7) 1 ∞ 1 2 (1 + D ) g(m n/X) h(V1,t+μ)h(V2,t+μ )Ω(n, 1/2+it, iμ)Ω(ξ,1/2+it, iμ)dt = X L(1,χ ) −∞ n,m D ∞ h(V1,t+ μ)h(V2,t+ μ )× −∞ −2 N it ω(ξ) σ−2it,ω2 (ξ) (ξ) × cosh(π(t + μ)) cosh(π(t + μ))|Γ(1/2+i(t + μ))Γ(1/2+i(t + μ))|2|L(1+2it, ω2)|2 X2itζ(1+4it)L(1+2it, ω2)dt + O(X−M ).

1 We note the term (1+ D ) , is mentioned in [J], but buried as a constant in [BMP1]. L(1,χD) Here ζ(1+4it) has a pole at t =0, and to understand this we use the following lemma. Lemma 8.2. Let H be a diﬀerentiable function in L1(R), then ∞ ikx dx ikx dx →± ± →∞ PV −∞ H(x)e x := lim→0+ |x|≥ H(x)e x πiH(0) as k . Proof. See [V], Lemma 10. Applying this lemma for k = − log X, we obtain (8.7) equals 1 −2 (1 + )h(V ,μ)h(V ,μ) ω(r) σ 2 (ξ) (8.8) D 1 2 0,ω + O(X−M ). | |2 2 L(1,χD)cosh(μ)cosh(μ ) Γ(1/2+iμ)Γ(1/2+iμ ) L(1,ωμ) Here we also used μ = −μ, and that h(Vj,t) is real valued function. So for k =0 , we get

1 −2 −M h(V ,μ)h(V ,μ)ω(r) σ 2 (ξ)+O(X ). L(1,ω2) 1 2 0,ωμ 8.2. k = 0. Loosely, the same calculations go as the previous section but here we have ζ(s +2it)ζ(s − 2it)ζ(s)L(s.χ ) L(s)= D . ζ(2s) Again the ζ(2s) cancels with the m-sum, and the poles of L(s)areats =1, 1 ± 2it. However, the poles at s =1± 2it, are cancelled by 1 1 = | 2 |2 | |2 L(1+2it, ω0) ζ(1+2it)L(1+2it, χD) found in (8.5) after expanding Ω(r, 1/2+it, 0)Ω(ξ,1/2+it, 0). Now In much more detail, we get for k =0, (8.1) after a contour shift to −M,M ≥ 0,

(8.9) 1 ∞ it (1 + ) σ− (ξ)N(ξ) D dth(V ,t)h(V ,t) 2it,0 × XL(1,χ ) 1 2 cosh(πt)2|Γ(1/2+it)|4|ζ(1+2it)L(1+2it, χ )|2 D −∞ D 2 1+2it G(1)X|ζ(1+2it)| L(1,χD)+G(1+2it)X ζ(1+4it)ζ(1+2it)L(1+2it, χD) 1−2it + G(1 − 2it)X ζ(1 − 4it)ζ(1 − 2it)L(1 − 2it, χD) QUADRATIC BASE CHANGE AND THE ANALYTIC CONTINUATION OF THE ASAI L-FUNCTION 21

We deal with the terms containing X1±2it first. Take the term with X1+2it, the other term will be analogous. This term is after some simplifications, 1 ∞ N it 2it (1 + D ) σ−2it,0(ξ) (ξ) G(1+2it)X ζ(1+4it) (8.10) dth(V1,t)h(V2,t) 2 4 L(1,χD) −∞ cosh(πt) |Γ(1/2+it)| ζ(1 − 2it)L(1 − 2it, χD) ζ(1+4it) Now − is analytic, so ζ(1 2it) it σ−2it,χ(0)N(ξ) G(1+2it)ζ(1+4it) F (t)=h(V1,t)h(V2,t) 2 4 cosh(πt) |Γ(1/2+it)| ζ(1 − 2it)L(1 − 2it, χD)ζ(2(1 + 2it)) is analytic and ∞ F (t)X2itdt O(X−M ), −∞ for any M>0. Now, we address the term in (8.9) coming from the pole at s =1. Similar to (8.10), we simplify to get ∞ it 1 σ−2it,0(ξ)N(ξ) (8.11) (1 + ) h(V1,t)h(V2,t) 2 4 2 dt D −∞ cosh(πt) |Γ(1/2+it)| |L(1 − 2it, χD)| 9. Putting it all together Equation (5.10) gives 1 (9.1) (L)= g(m2n/X) h(V,ν )c (Π)c (Π) + CSC = X Π n l n,l m n Π= 1 ∞ ∞ 1 dxdy (1 + ) δ(l, l ) V1(x)V2(y) + D 0 0 xy ∞ 1 ll Da ∞ ∞ + rS ( , 1, ) H (x, y)dxdy + O(X−M ). Da D r2 r Da a=1 r∈N 0 0 r|l r|a We only mention here that we used Proposition 6.6 to get ∞ O((Da)−3M X−M )=O(X−M ), a=1 for M large. Our aim now is to show (L) is a sum of different geometric sides of Kuznetsov trace formulas. Each one then gives a spectral side from which we then gather information about poles of the Asai L-function and hence information about base change. We now follow section 9 of [H] closely. Define ∞ ∞ i x y 1 8π2i (9.2) V1 ∗ V2(z):= exp z ( + ) exp ( ) × −∞ −∞ 2 y x z xy 4π 4π dx dy V ( )W ( ) . x y x y Recalling equation (5.3) and Definition 5.3 and the dependence on l, we get √ ∞ ∞ 4π ll (9.3) HDa(x, y)dxdy = V1 ∗ V2( ). 0 0 Da 22 P. EDWARD HERMAN

Now Theorem 3.1 of [H] states ∈ ∞ R+ ∗ Theorem 9.1. For all V,W C0 ( ) , h(V W, t)=Cth(V,t)h(W, t), where Ct =2π for t an even integer, and Ct = π for t purely imaginary. Let

M(t)=h(V1,t)h(V2,t), then using Sears-Titchmarsh inversion formula for V1 ∗ V2(z), see [H], we get ∞ (9.4) V1 ∗ V2(z)=4π M(t)tanh(πt)B2it(z)tdt + (k − 1)Jk−1(z)M(k) , 0 2k>0,k∈N Further from [H] we need

Proposition 9.2. Let M(t)=h(V1,t)h(V2,t) then, ∞ dx ∞ (9.5) V1(x)V2(x) =4 M(t) tanh(πt)tdt +2 (k − 1)M(k). x −∞ 0 2k>0,k∈N The limit (L) can be then rewritten as 1 ∞ (9.6) (L)=2π(1 + ) δ(l, l) M(t) tanh(πt)tdt + (k − 1)M(k) + D −∞ ⎛ 2k>0,k∈N ⎞ ⎜ √ √ ⎟ ⎜∞ ∞ ⎟ ⎜ 1 ll Da 4π ll 4π ll ⎟ ⎜ rSD( , 1, ) M(t) tanh(πt)B2it( )tdt + (k − 1)Jk−1( )M(k) ⎟ . Da r2 r Da Da ⎝a=1 r∈N 0 2k>0,k∈N ⎠ r|l r|a Deﬁnition 9.3. Denote the Maass form of eigenvalue 1/4+t2 and level D and neben- 1+it −1/2 τD,it(n) typus χD by φt,D, and let η(l, 1/2+it):=2π cosh(πt) , where Γ(1/2+it)L(1+2it,χD) it nx+mx τit(n)= ab=n χD(a)(a/b) . Let SD(n, m, c)= x(c)∗ χD(x)e( c ). Then the Kuznetsov trace formula for these Maass forms and its associated continuous spectrum (See [Iw])is ∞ 1 (9.7) G(tφ)al(φt)al(φt)+ G(t)η(l, 1/2+it)η(l , 1/2+it)dt = 4π −∞ φt

S (l, l,c) √ = G + D G+(4π ll/c), 0 c D|c δ(l,l ) ∞ + ∞ where G0 := π −∞ G(t) tanh(πt)tdt, and G (x):=4 0 G(t) tanh(πt)B2it(x)tdt.

Deﬁnition 9.4. Let the holomorphic forms of weight k, level D, and nebentypus χD be denoted as φk,D. Petersson’s trace formula then states

∞ δ(l, l) S (l, l,c) √ (9.8) G(k )a (φ )a(φ )= (k − 1)G(k)+ D G(4π ll/c), φ l k l k π c φk 2k>0,k∈N D|c QUADRATIC BASE CHANGE AND THE ANALYTIC CONTINUATION OF THE ASAI L-FUNCTION 23 where (9.9) G(x)=4 (k − 1)G(k)Jk−1(x). k>0,k even Notice that (9.6) almost looks like the geometric sides of these two trace formulas. We ﬁrst need to simplify the Kloosterman sums in (9.6). Inversion of the r and s-sum gives

∞ √ 1 ll Da ∞ 4π ll (9.10) rS ( , 1, ) M(t) tanh(πt)B ( )tdt = Da D r2 r 2it Da a=1 r∈N 0 r|l r|a ∞ √ 1 ll ∞ 4π ll S ( , 1,Da) M(t) tanh(πt)B ( )tdt = Da D r2 2it Dra r∈N a=1 0 | r l ∞ ∞ ll 1 ll 4π r2 S ( , 1,Da) M(t) tanh(πt)B ( )tdt Da D r2 2it Da r∈N a=1 0 r|l We rewrite the holomorphic side of the trace formula as well in this fashion in (9.6). Using the spectral sides of each trace formula we get (L)equals,

(9.11) ∞ ∞ ∞ 4π ll 1 1 ll r2 2π(1+ ) δ(l, l ) M(t) tanh(πt)tdt+ SD( 2 , 1,Da) M(t) tanh(πt)B2it( )tdt D −∞ Da r Da r∈N a=1 0 | r l ∞ 4π ll 1 ll r2 + δ(l, l ) (k − 1)M(k)+ S ( , 1,Da) (k − 1)J − ( )M(k) = Da D r2 k 1 Da 2k>0,k∈N a=1 2k>0,k∈N 1 =2π(1 + ) M(tφ)a ll (φt,D)a1(φt,D)+ D r2 r∈N φt,D | r l 1 ∞ ll ll M(t)η( 2 , 1/2+it)η(1, 1/2+it)dt + M(kφ)a (φk)a1(φk) . 4π −∞ r r2 φk,D We showed in Section 8,

1 2 1 h(V1,μ)h(V2,μ) −2 (9.12) g(m n/X)CSC =2π(1 + ) ω(l) σ 2 (l)+ X n,l D L(1,χ )L(1,ω2) 0,ωμ m n μ D μ ∞ N 2it σ−2it,0(l) (l) −M h(V1,t)h(V2,t) 2 4 2 dt + O(X ), −∞ cosh(πt) |Γ(1/2+it)| |L(1 − 2it, χD)| for M>0. 24 P. EDWARD HERMAN √ 9.1. Comparing Coefficients from Q( D) to Q. In this section we compare the coefficients of the continuous spectrum from (9.12) to those of (9.11). With the comparison we then show the continuous spectrum from (9.12) equals (9.11). Further we show the μ-sum in (9.12) is associated with a sub sum ( ”Theta functions”) of the cuspidal spectrum in (9.11). Define ψμ(k):= N(q)=k ωμ(q).

Lemma 9.5. it ll N ∈N (1) (l) σ−2it,0(l)= r τit( r2 ), r|l −2 ll 2 ∈N (2) ωμ(l) σ0,ωμ (l)= r ψμ( r2 ). r|l Proof. By multiplicativity, we do this for a prime power. Notice N it m 2it m (a) N(p ) σ− (p )= . 2it,0 N(b) ab=pm m it Remember τit(p )= ab=pm χD(a)(a/b) , Assume p inert prime, then we claim ll m p2m N(a) it (9.13) τ ( )= τ ( )= . it r2 it p2j N(b) r∈N j=0 ab=pm r|l Let X = pit, then the LHS of (9.13) equals

2(m−j) 1 m 1 m 1+(−1)2(m−j)X2(2(m−j)+1) X2j (−1)kX2k = X2j X2m X2m 1+X2 j=0 k=0 j=0 1 1 − X2(m+1) X2(2m+1)(1 − X−2(m+1)) = + X2m(1 + X2) 1 − X2 (1 − X−2) 1 1 − X2(m+1) X2m+2(X2(m+1) − 1) = + X2m(1 + X2) 1 − X2 X2(1 − X−2) 1 − X4(m+1) = . X2m(1 − X4) The RHS of (9.13) equals N(a) it m 1 − X4(m+1) = X2(2k−m) = , N(b) X2m(1 − X4) ab=pm k=0 so we are done. m For p = p1p2, we show for l = p1 , ll m N(a) it (9.14) τ ( )= τ (pm)= . it 2 it N ∈N r m (b) r j=0 ab=p1 r|l Let X = pit, then its clear both sides of (9.14) equals m X−2m X2j. j=0 QUADRATIC BASE CHANGE AND THE ANALYTIC CONTINUATION OF THE ASAI L-FUNCTION 25 √ For p = D, the identity is as transparent as the split prime case. The identity −2 ll ω (l) σ 2 (l)= ψ ( ) μ 0,ωμ μ r r∈N r|l is proven in an identical fashion.

Now using Lemma 9.5 and the duplication formula for the Gamma function, (9.12) equals ll 1 4πh(V1,μ)h(V2,μ)ψμ( r2 )ψμ(1) (9.15) 2π(1 + ) 2 + D L(1,χD)L(1,ω ) r∈N μ μ r|l

∞ ll 1 h(V1,t)h(V2,t)τit( r2 )τit(1) −M 2 + O(X ). 4π −∞ |L(1 − 2it, χD)| r∈N r|l This completes Theorem 1.1 (2.). Now it can be seen that the continuous spectrum part of (9.11) equals the LHS of (9.12). Putting all this together we get (9.16) ll 1 1 4πh(V ,μ)h(V ,μ)ψ ( 2 )ψ (1) m2 1 2 μ r μ g( n/X) h(V,νΠ)cn(Π)cl(Π)+2π(1+ ) 2 + X D L(1,χD)L(1,ω ) m n Π= 1 r∈N μ μ r|l ∞ ll 1 h(V1,t)h(V2,t)τit( r2 )τit(1) 2 = 4π −∞ |L(1 − 2it, χD)| r∈N | r l 1 2π(1 + ) M(tφ)a ll (φt,D)a1(φt,D)+ D r2 r∈N φt,D | r l ∞ 1 ll −M ll M(t)η( 2 , 1/2+it)η(1, 1/2+it)dt + M(kφ)a (φk)a1(φk) + O(X ) 4π −∞ r r2 φk,D Subtracting the continuous spectrums from both sides we get an equality of the discrete spectrum plus the μ-sum. 9.2. Addressing the Maass constructed theta forms. The μ sum parametrizes the theta forms. Remember from Section 8, μ = kπ ,k =0 ∈ Z. Each k = 0 corresponds to a log 0 cusp form constructed from Hecke characters over a quadratic field. The Fourier coefficients of these forms are ψμ(w):= N(q)=w ωμ(q), with the corresponding form being ∞ √ θωμ (z)= ψμ(n) yK iπk (2πny)e(nx). 2log0 n=1 ∈ Further θωμ Γ0(D, χd). For a more explicit explanation of these forms see [Bu]. The cuspidal theta forms base change to Eisenstein series. So it is to be expected that if the Asai L-function is detecting the cuspidal automorphic forms over the quadratic field K that are 26 P. EDWARD HERMAN base changes, they could not come from theta forms over Q. These are also the forms which give the poles of the symmetric square L-function in [V]. 9.3. Cuspidal spectrum. Now with the μ-sum in context (9.16) equals 1 (9.17) g(m2n/X) h(V,ν )c (Π)c (Π) = X Π n l m n Π= 1 kπ ll kπ kπ 4πM( log )ψ ( r2 )ψ (1) 1 0 log 0 log 0 ll − 2π(1 + ) M(tφ)a (φt,D)a1(φt,D) 2 D r2 L(1,χD)L(1,ω ) r∈N φt,D k=0 μ r|l k∈Z 1 −M +2π(1 + ) M(kφ)a ll (φk)a1(φk)+O(X ), D r2 φk,D for any integer M ≥ 0. 4π The factor 2 is needed for the orthonormal basis of the discrete spectrum im- L(1,χD)L(1,ωμ) plicit in the Kuznetsov trace formula. Therefore, we find (9.16) equals (9.18) 1 2 1 g(m n/X) h(V,νΠ)cn(Π)cl(Π) = 2π(1+ ) M(tφ)a ll (φt,D)a1(φt,D)+ X D r2 m n Π= 1 r∈N φt,D= θωμ r|l −M M(kφ)a ll (φk)a1(φk) + O(X ). r2 φk,D This completes Theorem 1.1 (1.). Now using specific test functions as in section 11 of [H], we get corresponding form matching Π = BCK/Q(φ), and equality of coefficients cl(Π) = a ll (φ). r2 r∈N r|l This completes Theorem 1.1 (3.). References

[A] Tetsuya Asai. On certain Dirichlet series associated with Hilbert modular forms and Rankin’s method. Math. Ann. , 226:81-94,1977. [BMP1] R.W.Bruggeman & R.J.Miatello & M.I.Pacharoni. Estimates for Kloosterman sums for totally real number fields. J. Reine Angew. Math., 535:103-164,2001. [BMP2] R.W.Bruggeman & R.J.Miatello & M.I.Pacharoni. Density results for automorphic forms on Hilbert modular groups. Geom. Funct. Anal.,13:681-719,2003. [BM] R.W.Bruggeman & R.J.Miatello & M.I.Pacharoni. Sum formula for SL2 over a number field and Selberg type estimate for exceptional eigenvalues GAFA.,8:627-655,1998. [Bu] Daniel Bump.Automorphic forms and Representations. Cambridge Studies in Advanced Mathematics, 55. 1996. [FT] A.Fröhlich, M.Taylor.Algebraic Number Theory,Cambridge University Press, 1991. [H] P. Edward Herman. Beyond Endoscopy for the Rankin-Selberg L-function, arXiv:1003.0462v1 [math.NT], submitted [Iw] H. Iwaniec, Spectral methods of automorphic forms. Second edition. Graduate Studies in Mathematics, 53. American Mathematical Society, Providence, RI; Revista Matemtica Iberoamericana, Madrid, 2002. [IK] H. Iwaniec and E. Kowalski, Analytic Number Theory. American Mathematical Society Colloquium Publications, 53. American Mathematical Society, Providence, RI, 2004. QUADRATIC BASE CHANGE AND THE ANALYTIC CONTINUATION OF THE ASAI L-FUNCTION 27

[JZ] H. Jacquet & D.ZagierEisenstein series and the Selberg trace formula. II. Trans. Amer. Math. Soc. 300 (1987), no. 1, 1–48. [J] D.Joyner, On the Kuznetsov-Bruggeman Formula for a Hilbert Modular Surface Having One Cusp, Mathematische Zeitschrift, 203:59-104. [KL] Andrew Knightly & Charles Li. A relative trace formula proof of the Petersson Trace formulaActa Arithmetica 122, no.3, 297-313, 2006. [La] Robert P. Langlands.Base change for GL(2). Institute for Advanced Study, Princeton, NJ, No. 96, 1980. [L] Robert P. Langlands. Beyond endoscopy. In Contributions to automorphic forms, geometry, and number theory, pp. 611697. Johns Hopkins Univ. Press, Baltimore, MD, 2004. [Sa] H.Saito. Automorphic forms and algebraic extensions of number ﬁelds,Lectures in Math.,no.8 Kinoku- niya Book Store Co. Ltd., Tokyo, Japan,1975. [S] Peter Sarnak. Comments on Langland’s Lecture. http://www.math.princeton.edu/sarnak/SarnakLectureNotes- 1.pdf. [V] Akshay Venkatesh. Beyond endoscopy and special forms on GL(2). J. Reine Angew. Math., 577:2380, 2004. [Y] Yangbo Ye. Kloosterman integrals and base change for GL(2).J. reine angew. Math., 400:57-121,1989. [Z] Don Zagier. Modular forms associated to real quadratic ﬁelds. Invent. Math., 30:1-46,1975.

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