Trace Formulae for Sl(2,R)

TRACE FORMULAE FOR SL(2, R)

DORIAN GOLDFELD

Contents 1. Trace Formula for ﬁnite groups 3 2. Trace Formula for the inﬁnite additive group R and subgroup Z 5 3. The Selberg Trace Formula for SL(2, R) (Spectral Side) 7 4. The Selberg Trace Formula for SL(2, R) (Geometric Side) 24 4.1. The Identity Contribution C(Id) 24 4.2. The Contribution C(∞) 25 4.3. Contribution of the Hyperbolic Conjugacy Classes C(P ) 31 4.4. Contribution of the Elliptic Conjugacy Class C(R) 35 5. The Selberg Trace Formula for SL(2, R) 39 6. The Kuznetsov Trace Formula for SL(2, R) 42 7. The SL(2, R) Kuznetsov Trace Formula (Spectral Side) 43 8. The SL(2, R) Kuznetsov Trace Formula (Geometric Side) 45 9. The Kuznetsov Trace Formula for SL(2, R) 48 9.1. Kontorovich-Lebedev Transform 48 9.2. Explicit choice of the pT,R test function 49 9.3. Bounds for the pT,R test function 50 9.4. Bounding the Geometric Side of the KTF 54 9.5. Computation of the Main Term M in the KTF 56 9.6. Explicit KTF for GL(2, R) with Error Term 58 9.7. Bounding the Eisenstein term in the KTF 58 10. The Kuznetsov Trace Formula for GL(3, R) 60 10.1. Whittaker functions and Maass forms on h3 63 11. The SL(3, R) Kuznetsov Trace Formula (Spectral Side) 64 11.1. Eisenstein contribution 66 References 69 1 2 DORIAN GOLDFELD

1. Trace Formula for ﬁnite groups Let G be a ﬁnite group and let Γ be a subgroup of G. We consider the C-vector space of functions n o

V := φ : G → C φ(γx) = φ(x), ∀γ ∈ Γ, ∀x ∈ G . We can think of V as the C-vector space of automorphic forms on Γ\G.

Definition 1.1. (Inner product on V ) For φ1, φ2 ∈ V we define X φ1, φ2 := φ1(x)φ2(x). x∈Γ\G Definition 1.2. (Orthonormal basis for V ) Let Γz ∈ Γ\G and define 1Γz ∈ V by ( 1 if x ∈ Γz, 1 (x) := Γz 0 otherwise.

The functions 1Γz for distinct cosets Γz form an orthonormal basis for V with respect to the above inner product.

Definition 1.3. (The kernel function Kf (x, y)) Let G be a finite group and f : G → C. Let Γ be a subgroup of G. For x, y ∈ G we define the kernel function X −1 Kf (x, y) := f x γy . γ∈Γ

Definition 1.4. (The linear map Kf ) Let φ ∈ V . We define a linear map Kf : V → V where for φ ∈ V we have Kf (φ) ∈ V where X (Kf φ)(x) := Kf (x, y) · φ(y). y∈Γ\G Definition 1.5. (Trace of a linear map on a finite dimensional vector space) Let V be a complex vector space with basis v1, v2, . . . , vn. A linear map L : V → V satisfies n n X X L(vi) = ai,jvj, for all 1 ≤ i ≤ n and ai,j ∈ C . i=1 j=1 Then associated to L we may define the matrix AL := ai,j 1≤i,j≤n. The trace of L (denoted Tr(L)) is defined to be n X Tr(L) = ai,i, i=1 i.e., it is the trace of the matrix AL. TRACE FORMULAE FOR SL(2, R) 3 Remarks on eigenvalues of a matrix: It is well known that for a matrix A with complex coefficients ai,j, (1 ≤ i, j ≤ n) that • Tr(A) = sum of the eigenvalues of A; • Det(A) = product of the eigenvalues of A.

We now compute the action of the linear map Kf : V → V on the orthonormal basis given in deﬁnition 1.2. Let z ∈ Γ\G. We have

X Kf 1Γz (x) = Kf (x, y) · 1Γz(y) y∈Γ\G

= Kf (x, z) X = Kf (x1, z) · 1Γx1 (x).

x1∈Γ\G

It follows that the matrix AK associated to the linear map Kf is f given by Kf (x1, z) , and x1,z∈Γ\G

X Tr(Kf ) = Kf (z, z). z∈Γ\G

Proposition 1.6. (Trace formula for a finite group G) Let G be a finite group and let Γ be a subgroup of G. Let Cl[Γ] denote the set −1 of conjugacy classes σ γσ σ ∈ Γ with γ ∈ Γ. For γ ∈ Γ we also define

−1 Γγ := σ ∈ Γ σ γσ = γ , −1 Gγ := g ∈ G g γg = γ .

It follows that

X #(Gγ) X −1 Tr(Kf ) = f z γz . # (Γγ) γ∈Cl[Γ] z∈Gγ \G 4 DORIAN GOLDFELD

Proof. We have X Tr(Kf ) = Kf (z, z) z∈Γ\G X X = f z−1γz z∈Γ\G γ∈Γ X X X = f z−1 · σ−1γσ · z

z∈Γ\G γ∈Cl[Γ] σ∈Γγ \Γ X X X = f z−1 · σ−1γσ · z

γ∈Cl[Γ] z∈Γ\G σ∈Γγ \Γ X X = f z−1γz

γ∈Cl[Γ] z∈Γγ \G X X X = f (gz)−1 · γ · gz

γ∈Cl[Γ] z∈Gγ \G g∈Γγ \Gγ X X X = f z−1 · g−1γg · z

γ∈Cl[Γ] z∈Gγ \G g∈Γγ \Gγ X X X = f z−1γz

γ∈Cl[Γ] z∈Gγ \G g∈Γγ \Gγ

X #(Gγ) X = f z−1γz . # (Γγ) γ∈Cl[Γ] z∈Gγ \G The trace formula has two sides:

X #(Gγ) X −1 Tr(Kf ) = f z γz . | {z } # (Γγ) spectral side γ∈Cl[Γ] z∈Gγ \G | {z } geometric side where the geometric side consists of a sum over conjugacy classes.

2. Trace Formula for the inﬁnite additive group R and subgroup Z

Consider the additive group G = R and subgroup of rational integers Γ = Z. Following the recipe in §1, we deﬁne the C vector space of smooth functions n o

V := φ : R → C φ(x + n) = φ(x), ∀n ∈ Z, ∀x ∈ R . TRACE FORMULAE FOR SL(2, R) 5

Deﬁnition 2.1. (Inner product on V ) Let φ1, φ2 ∈ V. We deﬁne the inner product Z Z 1

φ2, φ2 := φ1(x) φ2(x) dx = φ1(x) φ2(x) dx. 0 Z\R 2πinx Definition 2.2. (Orthonormal basis for V ) Let en(x) := e . It is well known that every periodic function is a linear combination of en(x) with n ∈ Z. Furthermore Z 1 ( 2πi(m−n)x 1 if m = n, em, en = e dx = 0 0 otherwise. This establishes that the en is an othonormal basis of V . n∈Z Definition 2.3. (The kernel function Kf (x, y)) Let f : R → C be a Schwartz function, i.e., a smooth function all of whose derivatives have rapid decay. For x, y ∈ R we define the kernel function X Kf (x, y) := f (−x + n + y) . n∈Z

Deﬁnition 2.4. (The linear map Kf ) Let φ ∈ V . We deﬁne the linear map Kf : V → V by Z 1 Kf φ (x) := Kf (x, y) · φ(y) dy. 0 Proposition 2.5. (The en are eigenfunctions of Kf ) Let n ∈ Z. Then Kf en (x) = fb(−n) · en(x), R ∞ −2πixξ where fb is Fourier transform of f given by fb(ξ) := −∞ f(x)·e dx. Proof. Z 1 Kf en (x) = Kf (x, y) · en(y) dy 0 1 Z X = f (−x + n + y) · en(y) dy 0 n∈Z ∞ ∞ Z Z = f(−x + y) · en(y) dy = f(y) · en(x + y) dy −∞ −∞

= fb(−n) · en(x), since en(x + y) = en(x) · en(y). 6 DORIAN GOLDFELD

Proposition 2.6. (The trace formula for the additive group R) The trace formula for the group R and subgroup Γ = Z is given by X X fb(n) = f(n) . n∈Z n∈Z | {z } | {z } spectral side geometric side

Remark: The trace formula is the well known Poisson summation formula.

Proof. We have shown that en is an eigenfunction of the linear map Kf with the eigenvalue fb(n). The trace of Kf is given by the sum of the eigenvalues which is just P fb(n). To ﬁnd the geometric side we must n∈Z compute 1 Z 1 Z X X K(x, x) dx = f(n) dx = f(n). 0 0 n∈Z n∈Z Note that since the additive group Z is abelian there is only one conjugacy class, namely Z itself.

3. The Selberg Trace Formula for SL(2, R) (Spectral Side) Let G = SL(2, R) and Γ = SL(2, Z). We also require the maximal compact subgroup cos(θ) sin(θ) K = SO(2, R) = 0 ≤ θ < 2π . − sin(θ) cos(θ) In this case we deﬁne the vector space V as a space of smooth functions φ given by n o

V := φ : G → C φ(γgk) = φ(g), ∀γ ∈ Γ, ∀k ∈ K, ∀g ∈ G . By the Iwasawa decomposition it is known that every g ∈ G can be uniquely expressed in the form 1 − 1 y 2 xy 2 cos(θ) sin(θ) g = − 1 · 0 y 2 − sin(θ) cos(θ) for some 0 ≤ θ < 2π, x ∈ R, y > 0. The action of SL(2, R) on the upper half plane h := {x + iy | x ∈ R, y > 0} given by a b az + b z := , ∀ ( a b ) ∈ SL(2, ), ∀z ∈ h , c d cz + d c d R TRACE FORMULAE FOR SL(2, R) 7 establishes a one-to-one correspondence between G/K and h given by

1 − 1 1 − 1 y 2 xy 2 cos(θ) sin(θ) y 2 xy 2 − 1 · i = − 1 i = x + iy. 0 y 2 − sin(θ) cos(θ) 0 y 2

1 − 1 y 2 xy 2 Notation involving z: We shall use the notation z = − 1 ∈ 0 y 2 G/K and also z = x + iy ∈ h interchangeably. The usage will depend on the context of the discussion.

Deﬁnition 3.1. (Petersson inner product on V ) For φ1, φ2 ∈ V we deﬁne the inner product Z × φ1, φ2 := φ1(g) φ2(g) d g

Γ\G/K where d×g denotes the left invariant measure on the coset space Γ\G/K. 1 1 y 2 xy− 2 If z = 1 ∈ G/K then the Petersson inner product can be 0 y− 2 written in a simple explicit manner as ZZ dxdy φ , φ := φ (z) φ (z) . 1 2 1 2 y2 z∈Γ\h We may then define L2 (Γ\G/K) as the Hilbert space completion of V. This space had been studied by Maass and elements of L2 (Γ\G/K) which are eigenfunctions of the Laplacian with rapid decay at the cusp are termed Maass forms for SL(2, Z). 2 2 Definition 3.2. (The space Lcusp of cusp forms) We let Lcusp denote the Hilbert space of Maass forms for SL(2, Z). It can be shown that each Maass form φ vanishes at the cusp, i.e., lim φ(x + iy) = 0. y→∞ It was not known before Selberg’s work on the trace formula whether infinitely many such Maass forms existed or not. In addition to the Maass forms there are are also Eisenstein series which are eigenfunctions of the Laplacian but are not in L2. 1 m Definition 3.3. Let Γ := . Let s ∈ with Re(s) > 1. ∞ 0 1 C m∈Z We define the Eisenstein series 1 X E(z, s) := Im(γz)s. 2 γ∈Γ∞\Γ 8 DORIAN GOLDFELD

Theorem 3.4. (Fourier expansion of Eisenstein series) The Eisen- stein series E(z, s) has the Fourier expansion

Deﬁnition 3.6. (The kernel function Kf ) Let f : K\G/K → C. For z, z0 ∈ SL(2, R) we deﬁne the kernel function

0 X −1 0 Kf (z, z ) := f z γz , γ∈Γ provided the sum converges absolutely.

0 Proposition 3.7. (Properties of Kf ) For all z, z ∈ SL(2, R) the Kernel function Kf (given in the above deﬁnition) satisﬁes the following properties:

0 0 0 0 • Kf (zk, z k ) = Kf (z, z ), (∀ k, k ∈ K), 0 0 0 • Kf (γz, γ z) = Kf (z, z ), (∀ γ, γ ∈ Γ).

Proof. Exercise for the reader.

To facilitate the computation of the trace of the linear operator Kf Selberg chose a class of especially nice test functions f which we now describe.

Deﬁnition 3.8. (The function τ) We deﬁne the function τ : G → C by

2 2 2 2 a b τ(g) := a + b + c + d − 2, for g = ( c d ) ∈ G . TRACE FORMULAE FOR SL(2, R) 9

1 − 1 y 2 xy 2 Proposition 3.9. (Properties of τ) Let z = − 1 , and let 0 y 2 1 1 ! 0 2 0 0− 2 0 y x y z = − 1 ∈ G/K. Then we have 0 y0 2 (i) τ(k) = 0, (∀ k ∈ K), (ii) τ(kz) = τ(zk) = τ(z), (∀ k ∈ K), (x − x0)2 + (y − y0)2 |z − z0|2 (iii) τz−1z0 = = , yy0 yy0 |σz − σz0|2 |z − z0|2 (iv) = for all σ ∈ SL(2, ). Im(σz) · Im(σz0) yy0 R Proof. Exercise for the reader.

Deﬁnition 3.10. (Selberg’s kernel function) Let f : R+ → C be −1− 0 a smooth function satisfying f(t) |t + 2| . Then for g, g ∈ SL(2, R), we deﬁne 0 X −1 0 Kf (g, g ) = f τ g γg . γ∈Γ Theorem 3.11. (Growth of Selberg’s kernel function at the 1 1 1 1 ! − 0 2 0 0− 2 y 2 xy 2 0 y x y cusps ) Let z = − 1 , z = − 1 ∈ G/K. Then, 0 y 2 0 y0 2 for every > 0 and y, y0 > 1, we have 0 X −1 0 − 0 1+ • |Kf (z, z )| ≤ f τ z γz y (y ) γ∈Γ

X −1 0 0 − • f τ z γz (yy ) . γ ∈ Γ−Γ∞ Proof. We compute (x − x0)2 + (y − y0)2 f τz−1z0 = f yy0 (x − x0)2 y y0 −1− + + yy0 y0 y (x − x0)2 y −1− + yy0 y0 10 DORIAN GOLDFELD

0 Next, for α ∈ Γ, we adopt the notation that zα = αz = xα + iyα. It follows that 2 ! 0 X X |z − zδα| Kf (z, z ) = f y yδα α∈Γ∞\Γ δ∈Γ∞ 2 ! X X |z − zα − m| = f y yα α∈Γ∞\Γ m∈Z 2 2 X X (x − xα − m) + (y − yα) = f y yα α∈Γ∞\Γ m∈Z 2 2 −1− X X (x − xα − m) y + y yα yyα α∈Γ∞\Γ m∈Z −1− 1+ X 1+ X 2 2 y yα (x − xα − m) + y α∈Γ∞\Γ m∈Z Now, we can bound the latter sum above as follows. −1− Z ∞ X 2 2 −2−2 2 2 −1− (x − xα − m) + y y + (u + y ) du 0 m∈Z y−1−2. It immediately follows from corollary 3.5 and the above calculations that 0 − 0 − 0 1+ Kf (z, z ) y E(z , 1 + ) y (y ) . This proves the ﬁrst assertion. For the second assertion we calculate the partial sums over Γ∞\Γ X −1 0 − X 1+ 0 − f τ z γz y yγ (yy ) γ ∈ Γ−Γ∞ Γ∞γ 6= Γ∞ 1+ − since E(z, 1 + ) − y y by corollary 3.5.

Deﬁnition 3.12. (The integral operator Kf ) Let z ∈ h. Then for 2 2 2 φ ∈ L (Γ\h) we deﬁne the integral operator Kf : L (Γ\h) → L (Γ\h) by Z dx0dy0 K φ(z) = K (z, z0) φ(z0) . f f (y0)2 Γ\h

0 0 Remarks Since Kf is symmetric, i.e., Kf z, z = Kf z , z , we see 0 1+ 0 − that Kf also satisﬁes the bound |Kf (z, z )| y (y ) . It follows TRACE FORMULAE FOR SL(2, R) 11 that for every z ∈ h that 1   2 Z 0 0 Z 0 0 0 2 dx dy 0 2 dx dy Kf φ (z) ≤  Kf (z, z ) · |φ(z )|   (y0)2 (y0)2  Γ\h Γ\h 1   2 Z dx0dy0 Z dx0dy0  (y0)−2 · |φ(z0)|2  z,  (y0)2 (y0)2  Γ\h Γ\h = Oz, 1 .

This shows that the integral defining the integral operator Kf converges absolutely. Definition 3.13. (Hilbert-Schmidt Operator) Let X be a locally compact space with a positive Borel measure. Assume that L2(X) is a separable Hilbert space. Let K : X × X → C be L2 in each variable. We define the integral operator Z K φ(x) := K(x, y) φ(y) dy, φ ∈ L2(X) . X The integral operator K is said to be of Hilbert-Schmidt type if ZZ |K(x, y)|2 dxdy < ∞.

X×X Theorem 3.14. (Hilbert-Schmidt) A Hilbert-Schmidt operator as in deﬁnition 3.13 is a compact operator. If K is self adjoint then 2 the space L (X) has an orthonormal basis of eigenfunctions φ1, φ2,... where Kφi = λi·φi, (i = 1, 2, 3,...) and λi → 0 as i → ∞. If the sum of the eigenvalues converges absolutely, then we say K is of trace class and ∞ X Z the trace of the operator K is given by Tr(K) = λi = K(x, x) dx. i=1 X

Proof. See [Bum97]

Warning: The integral operator Kf given in deﬁnition 3.12 is not Hilbert-Schmidt since it can be shown that Z Z dxdy dx0dy0 |K (z, z0)|2 = ∞. f (y)2 (y0)2 Γ\h Γ\h

We shall now construct a modiﬁcation of the integral operator Kf . The modiﬁcation will be done in two steps. 12 DORIAN GOLDFELD

Definition 3.15. (First modification of Kf ) We define Z 1 0 0 0 Kef (z, z ) := Kf (z, z ) − Kf (z, z + t) dt. 0 Proposition 3.16. Let z ∈ h. Fix a fundamental domain for Γ\h given by n o

D = z ∈ h − 1/2 ≤ Re(z) ≤ 1/2, |z| ≥ 1. 2 If φ ∈ Lcusp then we have

Z 0 0 0 0 dx dy Kef φ (z) = Kef (z, z ) φ(z ) = Kf φ (z). (y0)2 D Proof. Z Z 1 0 0 Z Z 1 0 0 0 0 dx dy 0 0 dx dy Kf (z, z + t) dt φ(z ) 0 2 = Kf (z, z ) φ(z − t) dt 0 2 = 0. 0 (y ) 0 (y ) D D

Definition 3.17. (Second modification of Kf ) We define

X |z − z0 + n|2 K#(z, z0) := K (z, z0) − f . f f yy0 n∈Z Theorem 3.18. (Growth of the Modiﬁed Kernel at the Cusps) Let z = x + iy, z0 = x0 + iy0 ∈ h. Then for every > 0 and all y, y0 > 1 Z ∞ 0 0 − 0 Kef (z, z ) (yy ) + |f (r)| dr. 0 Proof. We calculate 1 Z 0 2 # 0 0 0 X |z − z + n| K (z, z ) − Kef (z, z ) = Kf (z, z + t) dt − f f yy0 0 n∈Z 1 1 " # Z Z X |z − z0 + n + t|2 X |z − z0 + n|2 = K#(z, z0 + t) dt + f − f dt f yy0 yy0 0 0 n∈Z n∈Z 1 ∞ Z Z |z − z0 + t|2 = K#(z, z0 + t) dt + f dt − [t]. f yy0 0 −∞ TRACE FORMULAE FOR SL(2, R) 13 The ﬁrst term above is bounded by (yy0)− by theorem 3.11. For the second term we apply integration by parts. ∞ Z (x − x0 + t)2 + (y − y0)2 f dt − [t] yy0 −∞ ∞ Z (x − x0 + t)2 + (y − y0)2 = − t − [t] · df yy0 −∞ ∞ ∞ Z 2 0 2 Z 2 0 2 t + (y − y ) 0 t + (y − y ) 2t df = f dt yy0 yy0 yy0 −∞ −∞ | {z2 0 2 } put r= t +(y−y ) yy0 ∞ Z |f 0(r)| dr.

0 ∞ We have proved that K#(z, z0) − K (z, z0) (yy0)− + R |f 0(r)| dr. f ef 0 # 0 0 − The theorem follows from theorem 3.11 which says |Kf (z, z )| (yy ) .

Theorem 3.19. (The Kernel function Kef is Hilbert-Schmidt and of Trace Class) The kernel function Kef deﬁnes an integral 2 2 operator Kef : Lcusp → Lcusp which is Hilbert-Schmidt i.e., it satisﬁes Z Z 2 0 0 0 dxdy dx dy Kef (z, z ) < ∞. (y)2 (y0)2 D D

2 Furthermore, when restricted to the space Lcusp we have Kf = Kef . For 2 2 f real valued the integral operator Kef : Lcusp → Lcusp is self-adjoint and of trace class with trace given by Z dxdy Tr Kef = Kef (z, z) . (y)2 D Proof. It follows from theorem 3.18 that Z Z 2 0 0 Z Z 0 0 0 dxdy dx dy 0 − dxdy dx dy Kef (z, z ) (yy ) +1 1. (y)2 (y0)2 (y)2 (y0)2 D D D D 14 DORIAN GOLDFELD

Proposition 3.20. (The integral operator Kf commutes with the Laplacian) For z = x + iy ∈ h let ∂2 ∂2 ∆ := −y2 + z ∂x2 ∂y2

2 denote the Laplacian on L (Γ\h). Then ∆zKf = Kf ∆z. Proof. First note (by a brute force veriﬁcation) that

0 0 ∆zKf (z, z ) = ∆z0 Kf (z, z ). We compute, using integration by parts (Green’s theorem), that Z 0 0 0 0 dx dy 0 ∆zKf φ (z) = ∆z Kf (z, z ) · φ(z ) 0 2 Γ\h (y ) Z 2 2 ∂ ∂ 0 0 0 0 = − 0 2 + 0 2 Kf (z, z ) · φ(z ) dx dy Γ\h ∂(x ) ∂(y ) Z 2 2 0 ∂ ∂ 0 0 0 = Kf (z, z ) · − 0 2 + 0 2 φ(z ) dx dy Γ\h ∂(x ) ∂(y ) Z 0 0 0 0 dx dy 0 = Kf (z, z ) · ∆z φ(z ) 0 2 Γ\h (y ) = Kf ∆z φ (z).

Let λ ∈ C and assume the eigenspace ker(∆ − λ) is one dimensional. Then since ∆ and Kf commute, it would follow that if ∆φ = λ · φ for some φ ∈ L2(Γ\h) and λ ∈ C, then for each f there would exist a function hf (λ) ∈ C such that

Kf φ = hf (λ) · φ.

We will now show the existence and uniqueness of hf (λ) using diﬀer- ential equations. Proposition 3.21. (Eigenfunctions of ∆ are eigenfunctions of 2 Kf ) Assume that ∆zφ(z) = λ · φ(z) for some φ ∈ L (Γ\h). Then for + −1− every f : R → C (satisfying f(t) (t + 2) ) there exists a unique function hf : C → C such that Kf φ (z) = hf (λ) · φ(z).

Proof. We follow [Hej76]. We want to prove (Kf φ)(z) = hf (λ) · φ(z) which is equivalent to: TRACE FORMULAE FOR SL(2, R) 15

(3.22) Z X |z − γz0|2 dx0dy0 Z |z − z0|2 dx0dy0 f φ(z0) = f φ(z0) y · Im(γz0) (y0)2 y · y0 (y0)2 γ∈Γ Γ\h h

= hf (λ) · φ(z). The Cayley transformation c : h → U := {u ∈ C | |u| ≤ 1} which maps the upper half plane to the unit disk (taking i to 0) is given by i 1 z0 − i c(z0) := z0 = , (∀ z0 ∈ h). i −1 z0 + i It is easy to see that under this transformation |dz0| 2|dc| = . Im(z0) 1 − |c|2 For fixed z = x+iy ∈ h we define the modified Cayley transformation − 1 − 1 0 i 1 y 2 −xy 2 0 w(z ) := 1 z i −1 0 y 2 2y = 1 − i(x − x0) + y + y0 which maps the upper half plane h to the unit disk U (taking z to 0). Furthermore |z0 − z|2 4 |w|2 = . yy0 1 − |w|2 Let us now make the change of coordinates z0 → w (z0) in the integral 3.22 and define φ(z0) := Ψ(w). It follows that Z 2 ! 4 |w| f Ψ w dA w = hf (λ) · Ψ(0) 1 − |w|2 U where dA w denotes the area differential on U. We must prove that hf (λ) depends solely on f and λ. If we convert to polar coordinates on U the above can be rewritten as 1 Z 2 Z 2π 4r iθ rdr f 2 Ψ re dθ 2 2 = hf (λ) · Ψ(0). 1 − r 0 (1 − r ) r=0 16 DORIAN GOLDFELD

This can be rewritten as 1 Z 4r2 rdr f Ψ∗(r) = h (λ) · Ψ∗(0) 1 − r2 (1 − r2)2 f r=0 where Z 2π Ψ∗(w) := Ψ w · eiθ dθ. 0 Note that Z 2π Z 2π Ψ reiθ dθ = Ψ w · eiθ dθ 0 0 for all w ∈ U satisfying |w| = r from which it follows that Ψ∗(w) is a radially symmetric function which depends only on r = |w| and, hence, ∂ ∗ satisfies ∂θ Ψ (w) = 0. Furthermore, Ψ∗ is also a radially symmetric eigenfunction of the Laplacian and, therefore, satisfies the differential equation d2Ψ∗ 1 dΨ∗ 4λ (3.23) + + Ψ∗ = 0. dr2 r dr (1 − r2)2 Let λ ∈ C be fixed. A possible solution (up to a constant factor) to (3.23) must be of the form ∗ c 2 Ψ (r) = r 1 + a1r + a2r + ··· . We calculate 1 Ψ∗ 0(r) = crc−2 + a (c + 1)rc−1 + ··· r 1

∗ 00 c−2 c−1 Ψ (r) = c(c − 1)r + a1(c + 1)cr + ··· Now substitue into the differential equation and we get 1 4λ 0 = Ψ∗ 00(r) + Ψ∗ 0(r) + Ψ∗(r) = c2rc−2 + a (c + 1)2rc−1 + ··· r (1 − r2)2 1 It follows that c2 = 0 and higher order terms are defined recursively from the differential equation. There will be a second solution but it will be of the form log r(1 + b1r + ··· ) which is singular at the origin. Since a regular solution Ψ∗ exists and is uniquely determined it follows that hf (λ) exists and is also uniquely determined. Remark: The above proof shows that a solution to the equations

∆φ = λφ, (Kf φ)(z) = hf (λ)φ(z) TRACE FORMULAE FOR SL(2, R) 17 can be made radially symmetric around z in the non euclidean sense. Then ∆φ = λφ becomes an ordinary diﬀerential equation with a 1- dimensional space of solutions regular near z. This yields the constant hf (λ) as in the case of a one-dimensional eigenspace ker(∆ − λ).

Definition 3.24. (The Abel Transform) Let f : R+ → C be a smooth function such that f and f 0 are integrable on R+. For w ≥ 0 we define the Abel transform ∞ ∞ Z Z f(t) Q(w) := f w + ξ2 dξ = √ dt, t − w −∞ w provided the integral converges absolutely. Proposition 3.25. (Inverse Abel Transform) Let Q(w) be the Abel transform of f. Then ∞ −1 Z f(t) = Q0(t + w2) dw. π −∞ Proof. We have ∞ ∞ ∞ −1 Z −1 Z Z Q0(t + w2) dw = f 0(t + w2 + ξ2) dw dξ π π −∞ −∞ −∞ 2π ∞ −1 Z Z = f 0(t + r2) rdrdθ π 0 0 Z ∞ = − f 0(t + u) du = f(t). 0 1 2 Definition 3.26. (Selberg Transform hf ) Assume ∆φ = 4 + r ·φ for some φ ∈ L2 (SL(2, Z)\h). Let f : R+ → C be a smooth function −1− satisfying f(t) (t + 2) for t ≥ 0. Then the Selberg transform is defined as the unique function h = hf : C → C for which we have Kf φ = hf (r) · φ as in proposition 3.21.

Proposition 3.27. (Evaluation of the Selberg transform) Let Q be the Abel transform of f as in deﬁnition 3.24. Then we have Z ∞ h(r) = Q eu − 2 + e−u eiru du. −∞ 18 DORIAN GOLDFELD

1 +ir Proof. Choose φ(z) = y 2 in proposition 3.21. Now 2 2 1 +ir 2 ∂ ∂ 1 +ir 1 2 1 +ir ∆y 2 = −y + y 2 = + r · y 2 . ∂x2 ∂y2 4 1 +ir 2 Although y 2 is not in L (SL(2, Z)\h), the proof of proposition 3.21 still goes through. It follows that ∞ ∞ Z Z 0 2 0 2 0 0 1 +ir (x − x ) + (y − y ) 0 1 +ir dx dy h(r) · y 2 = f (y ) 2 yy0 y02 0 −∞ Make the transformation x − x0 dx0 ξ = √ , dξ = √ . yy0 yy0 Then ∞ ∞ Z Z 0 2 0 1 +ir (y − y ) 2 0 1 +irp dy h(r) · y 2 = f + ξ (y ) 2 yy0 dξ yy0 y02 0 −∞ ∞ Z 0 2 0 (y − y ) 0 1 +irp dy = Q (y ) 2 yy0 yy0 y02 0 ∞ Z 0 0 1 y y 0 ir dy = y 2 Q − 2 + (y ) y0 y y0 0 ∞ 0 1 Z 1 dy 2 +ir 0 0 ir = y Q 0 − 2 + y (y ) 0 0 y y 0 u The result follows upon making the transformation y = e . Deﬁnition 3.28. (The Selberg transform Functions) Start with + −1− a smooth function f : R → C satisfying f(t) (t + 2) . Selberg deﬁnes the following functions: Z ∞ Q(w) = f(w + ξ2) dξ, (Abel transform of f), −∞ u −u g(u) = Q e − 2 + e , (u ∈ R), ∞ Z h(r) = g(u)eiru du, (Fourier transform of g)

−∞ ∞ −1 Z g(u) = h(r)eiru dr, (inverse Fourier transform of h). 2π −∞ TRACE FORMULAE FOR SL(2, R) 19 Explicit Example of the Selberg Transform

−1 − 3 w f(t) = (t + 2) 2 ,Q(w) = 1 + , 2 πr g(u) = sech(u), h(r) = π · sech . 2 2 Since sech(u) = eu+e−u we see that g, h have exponential decay.

Proposition 3.29. Growth of the Selberg Transform Functions Let > 0. Then

−1− − 1 − f(t) (t + 2) ,Q(w) w 2 , −( 1 −)|u| −( 1 −)| πr | g(u) e 2 , h(r) e 2 2 .

Proof. Exercise for the reader.

q 3 For z ∈ h let φ0(z) := π be the constant function of Petersson norm one, and let φ1, φ2,... denote an orthonormal basis of Maass forms 1 2 consisting of eigenfunctions of the Laplacian where ∆φi = 4 + ri φi for i = 1, 2 ...

Theorem 3.30. (Selberg Spectral Decomposition) Let φ0 be the constant function of Petersson norm one, and let φ1, φ2, φ3,... denote an orthonormal basis of Maass forms consisting of eigenfunctions of the Laplacian. Let F ∈ L2 (SL(2, Z)\h). Then for z ∈ h

∞ ∞ X 1 Z F (z) = F, φ φ (z) + F,E(∗, 1/2 + ir) E(z, 1/2 + ir) dr. i i 4π i=0 −∞

Proof. See [Gol06].

Theorem 3.31. (Spectral Decomposition of Kf ) The Selberg 2 2 kernel function Kf : L (SL(2, Z)\h) → L (SL(2, Z)\h) deﬁned in 3.12 has the following spectral decomposition.

0 Kf (z, z ) = ∞ ∞ X 1 Z = h(r )φ (z)φ (z0) + h(r)E(z, 1/2 + ir)E(z0, 1/2 + ir) dr i i i 4π i=0 −∞ for z, z0 ∈ h. 20 DORIAN GOLDFELD

It follows from theorem 3.31, that  ∞  Z 1 Z dxdy (3.32) K (z, z) − h(r) · |E(z, 1/2 + ir)|2 dr  f 4π  y2 SL(2,Z)\h −∞ ∞ ∞ Z X dxdy X = h(r )φ (z)φ (z) = h(r ). i i i y2 j i=0 j=0 SL(2,Z)\h Unfortunately, each individual integral Z dxdy K (z, z) f y2 SL(2,Z)\h and ∞ 1 Z Z dxdy h(r) · |E(z, 1/2 + ir)|2 dr 4π y2 SL(2,Z)\h −∞ does not converge! Nevertheless, it turns out that each of these integrals blows up in exactly the same way, so that when they are subtracted one is left with a ﬁnite value. We, therefore, modify (3.32) as follows. First we modify the kernel function Kf by subtracting the contribution from the parabolic conju- a b gacy classes of SL(2, Z). A matrix γ = ( c d ) ∈ SL(2, Z) is parabolic if |Tr(γ)| = |a + d| = 2. Associated to γ we have the conjugacy class −1 [γ] := {σγσ | σ ∈ SL(2, Z)} and the centralizer

Γγ := {δ ∈ SL(2, Z) | δγ = γδ}. We deﬁne X X |z − δz0|2 K∗(z, z0) := K (z, z0) − f f f (Imz)(Imδz0) γ∈Conjpar(Γ) δ∈[γ] |z − z0|2 = K (z, z0) − f f (Imz)(Imz0) X X |z − σγσ−1z0|2 − f , (Imz)(Imσγσ−1z0) γ∈Conjpar(Γ) σ∈Γγ \Γ γ6=±I where Conjpar(Γ) is a set of representatives of parabolic matrices in Γ = SL(2,Z) under conjugation. TRACE FORMULAE FOR SL(2, R) 21

1 0 Lemma 3.33. Let γ 6= ± ( 0 1 ) be a parabolic matrix. Then 1 m Γγ = {± ( 0 1 ) | m ∈ Z} = ±Γ∞. 1 m Furthermore, {± ( 0 1 ) | m ∈ Z} constitutes a set of representatives of parabolic matrices in SL(2, Z). Proof. Exercise for the reader. Let n 1 1 o DY := x + iy ∈ h − ≤ x ≤ , x2 + y2 ≥ 1, y ≤ Y 2 2 denote a fundamental domain for SL(2, Z)\h truncated at Y > 1. We rewrite the left side of (3.32) as

 ∞  Z 1 Z dxdy (3.34) K (z, z) − h(r) · |E(z, 1/2 + ir)|2 dr  f 4π  y2 SL(2,Z)\h −∞ Z Z ∗ dxdy dxdy = Kf (z, z) + f(0) + lim I(Y ), y2 y2 Y →∞ SL(2,Z)\h SL(2,Z)\h where Z X X |z − σγσ−1z|2 dx dy f −1 2 DY (Imz)(Imσγσ z) y γ∈Conjpar(Γ) σ∈Γγ \Γ γ6=±I X X Z |σ−1z − γσ−1z|2 dx dy = f −1 −1 2 DY (Imσ z)(Imγσ z) y γ∈Conjpar(Γ) σ∈Γγ \Γ γ6=±I X X Z |σ−1z − (σ−1z + m)|2 dx dy = f −1 2 2 DY (Imσ z) y m∈Z σ∈±Γ∞\Γ m6=0 1 Y Z Z X m2 dx dy = f .. y2 y2 m∈Z x=0 y=0 m6=0 Hence Y 1 ∞ Z Z X m2 dxdy Z Z dxdy I(Y ) := f − h(r)·|E(z, 1/2+ir)|2 dr . y2 y2 4πy2 m∈ 0 0 Z DY −∞ m6=0 22 DORIAN GOLDFELD

On combining ( 3.32 ) and (3.34) we obtain the following preliminary version of the Selberg Trace Formula.

Proposition 3.35. (Preliminary Trace Formula) Let φ0 be the constant function of Petersson norm one, and let φ1, φ2, φ3,... denote 1 2 an orthonormal basis of Maass forms satisfying ∆φj = 4 + rj φj. Let n 1 1 o DY := x + iy ∈ h − ≤ x ≤ , x2 + y2 ≥ 1, y ≤ Y 2 2 denote a fundamental domain for SL(2, Z)\h truncated at Y > 1. Then

∞ Z X ∗ dxdy πf(0) h(rj) = Kf (z, z) + + lim I(Y ), y2 3 Y →∞ j=0 SL(2, )\h | {z } Z Spectral Side where

Y 1 ∞ Z Z X m2 dxdy Z Z dxdy I(Y ) := f − h(r) · |E(z, 1/2 + ir)|2 dr . y2 y2 4πy2 m∈ 0 0 Z DY −∞ m6=0 TRACE FORMULAE FOR SL(2, R) 23

4. The Selberg Trace Formula for SL(2, R) (Geometric Side) a b A matrix ( c d ) ∈ SL(2, R) is of 3 types: • Parabolic with |a + d| = 2, • Hyperbolic with |a + d| > 2, • Elliptic with |a + d| < 2.

We can decompose SL(2, Z) as a union over distinct conjugacy classes [ SL(2, Z) = [α] conj. classes α where [α] denotes the conjugacy class given by

−1 [α] := σασ σ ∈ SL(2, Z) . p Let φ0 = 3/π and let φ1, φ2,... be an orthonormal basis of Maass 2 2 cusp forms for Lcusp(SL(2, Z)\h) each satisfying ∆φj = 1/4 + rj φj. Fix Selberg transform functions as in deﬁnition 3.28. We assume that h(r) = h(−r). The Selberg Trace Formula for SL(2, Z) is the identity

∞ X X X h(rj) = C(Id) + C(P ) + C(R) + C(∞), j=0 P hyperbolic R elliptic | {z } conj. classes conj. classes Spectral Side | {z } Geometric Side where C(Id) is the contribution of the Identity element, C(P ) is the contribution of a hyperbolic conjugacy class [P ], while C(R) is the contribution of an elliptic conjugacy class [R], and C(∞) is the contribution of the sum of non-identity parabolic conjugacy classes minus the contribution of the continuous spectrum. In the following sections of these notes we will explicitly evaluate each of these contributions based on the preliminary trace formula derived in proposition 3.35.

4.1. The Identity Contribution C(Id). The identity contribution πf(0) C(Id) is just the term 3 in proposition 3.35. We want to express C(Id) in terms of the test function h. The ﬁnal result is.

∞ 1 Z Proposition 4.1. C(Id) = r tanh(πr) h(r) dr. 12 −∞ 24 DORIAN GOLDFELD

Proof. ∞ ∞ π 1 Z Q0(w) 1 Z g0(u) √ C(Id) = f(0) = − dw = − u − u du 3 3 w 3 e 2 − e 2 0 0 ∞ ∞ 1 Z Z sin(ru) = rh(r) u − u du dr. 6π e 2 − e 2 −∞ 0 ∞ Z sin(ru) π tanh(πr) Now u − u du = . The result follows. e 2 − e 2 2 0

4.2. The Contribution C(∞). Following proposition 3.35, we deﬁne   Y ∞ Z 2 Z Z  X m dy 2 dxdy  C(∞) = lim  f − h(r)|E(z, 1/2 + ir)| dr  . Y →∞ y2 y2 4πy2  m∈  0 Z DY −∞ m6=0 Proposition 4.2. For s ∈ C, let √ πΓ(s − 1/2)ζ(2s − 1) M(s) := . Γ(s)ζ(2s) 1 Let h(r) be an even function, holomorphic in some strip |Im(r)| ≤ 2 + −5|r| for > 0 which satisﬁes h(r) = O e for r ∈ R. Then

∞ 1 Z Γ0 M 0 C(∞) = − h(r) (1 + ir) + Re (1/2 + ir) dr 2π Γ M −∞ 1 − M(1/2) + − g(0) log 2. 4

Proof. The result follows from the next two lemmas. Lemma 4.3. For Y → ∞ we have Y Z X m2 dy h(0) f = g(0) log (Y/2) + y2 y2 4 0 m∈Z m6=0 ∞ 1 Z Γ0 log Y − h(r) (1 + ir) dr + O . 2π Γ Y −∞ TRACE FORMULAE FOR SL(2, R) 25

|m| Proof. After transforming, y 7→ y , and using

x X 1 1 = log x + c + O , m 0 x m=1 along with X 1 = 0 m 0

∞  ∞  Z 1 Z |f(y2)| = 2 [log(yY ) + c ] f y2 dy + O  dy 0 Y y  1/Y 1/Y

Since |f| 1, and f(y) y−1− as y → ∞ we have

∞ Y ∞ 1 Z |f(y2)| 1 Z 1 1 Z y−2−2 log Y dy = dy + dy Y y Y y Y y Y 1/Y 1/Y Y Further, using the bound |f| 1, we see that

1/Y Z 1 log(yY ) + c · |f(y2)| dy . 0 Y 0 It follows that Y ∞ ! Z X m2 dy Z X 1 f = 2 f y2 dy y2 y2 m m∈ 1≤m≤yY 0 Z 1/Y m6=0 ∞ ∞ Z Z log Y = 2(log Y ) f y2 dy + 2 f y2 (log y + c ) dy + O 0 Y 0 0 ∞ 1 Z dy log Y = g(0) log Y + c + f(y) log y √ + O . 0 2 y Y 0 26 DORIAN GOLDFELD

Next, by the Selberg transform,

∞ ∞ ∞ Z dy 1 Z Z log y (log y)f(y)√ = − √ √ Q0(w) dw dy y π y w − y 0 0 y ∞ w 1 Z Z log y = − √ √ Q0(w) dy dw π y w − y w=0 y=0 ∞  1  Z Z 1 log(wy) 0 = −  √ √ dy  Q (w) dw π y 1 − y 0 0 | {z } This integral = π log w − 2 log 2. ∞ Z = − log w − 2 log 2 Q0(w)dw

0 ∞ Z = − (log w) Q0(w) dw − 2(log 2)g(0) | {z } 0 Let w = eu + e−u − 2 ∞ Z = − log eu + e−u − 2 g0(u) du − 2(log 2)g(0)

Observe that eu + e−u − 2 = eu1 − e−u2 so that

log eu + e−u − 2 = u + 2 log 1 − e−u .

It follows that

∞ ∞ ∞ Z dy Z Z (log y)f(y)√ = −2 log 1 − e−u g0(u) du − u g0(u) du − 2(log 2)g(0) y 0 0 0 ∞ Z h(0) = −2 log 1 − e−u g0(u) du + − 2(log 2)g(0). 2 0 Now ∞ 1 Z g0(u) = ir h(r)eiru dr. 2π −∞ TRACE FORMULAE FOR SL(2, R) 27 It follows that ∞ ∞  ∞  Z i Z Z −2 log 1 − e−u dg(u) = h(r)r eiru log 1 − e−u du dr. π   0 −∞ 0 Further, ∞ Z Γ0(1 − ir) −ir eiru log 1 − e−u du = − − c . Γ(1 − ir) 0 0 If we insert this into the above we obtain ∞ ∞ Z 1 Z Γ0(1 + ir) −2 log 1 − e−u dg(u) = −2c g(0) − h(r) dr 0 π Γ(1 + ir) 0 −∞ since h is even. Combining all the previous computations completes the proof.

√ 3 Deﬁnition 4.4. (Truncated Eisenstein series ) Let Y > 2 . We deﬁne the truncated Eisenstein series EY (z, s) for z ∈ D by ( E(z, s) for y < Y, EY (z, s) := E(z, s) − ys − M(s)y1−s for y ≥ Y.

1 Lemma 4.5. (Maass-Selberg relation) Let s = σ + ir with σ > 2 and r 6= 0. Then Z 2σ−1 2 1−2σ 2ir −2ir Y 2 dxdy Y − |M(s)| Y M(s)Y − M(s)Y E (z, s) = + y2 2σ − 1 2ir D 1 where R E(x + iy, s) dx = ys + M(s)y1−s,M(s)M(1 − s) = 1, and 0 √ πΓ(s − 1/2)ζ(2s − 1) M(s) = . Γ(s)ζ(2s)

Proof. See [Iwa95].

We would like to use the Maass-Selberg relation to evaluate the term ∞   1 Z Z dxdy h(r) |E(z, 1/2 + ir)|2 dr 4π  y2  −∞ DY 28 DORIAN GOLDFELD which occurs in C(∞). There is an issue, however, because the Maass- R Y 2 dxdy Selberg relation is for D E (z, s) y2 while our goal is to compute R 2 dxdy DY |E(z, 1/2+ir)| y2 . We get around this following a method which was introduced by Selberg (see the Gottingen Lecture notes [Sel89]) Deﬁne

E∗(z, s) := E(z, s) − ys − M(s)y1−s.

Then for Y → ∞ we need to estimate the diﬀerence

1 ∞ Z Z Z Z Y 2 2 dxdy ∗ 2 dxdy E (z, s) − |E(z, s)| = |E (z, s)| y2 y2 D DY x=0 y=Y

1 when s = u + ir and 2 ≤ u 1. Now, with this choice of s,

s√ 2π y X 1 ∗ s− 2 2πinx E (z, s) = σ1−2s(n)|n| Ks− 1 (2π|n|y)e Γ(s)ζ(2s) 2 n6=0 e2re−πy.

The constants 2, π in the bound e2re−πy are not optimal. This bound follows from the exponential decay of the K-Bessel function (here we use the fact that t + t−1 ≥ 2) given by

∞ √ Z √ y −π|n|y t+ 1 1 du ( t ) u− 2 +ir y · Ku− 1 +ir(2π|n|y) = e t 2 2 u 0 − 3 |n|π e 2 and the exponential decay of the Gamma function given by Stirling’s asymptotic formula |Γ(u + ir)|2 ∼ 2π · |r|2u−1 e−π|r| together with the prime number theorem which says |ζ(1 + 2ir)|−1 log |r|, for |r| 1. We immediately deduce that

Z Z 2 dxdy Y 2 4r −2πY |E(z, u + ir)| = E (z, u + ir) + O e e y2 DY D TRACE FORMULAE FOR SL(2, R) 29 It follows from the Maass-Selberg relation and the above computations that ∞ Z 1 Z dxdy h(r) · |E(z, 1/2 + ir)|2 dr 4π y2 DY −∞ ∞ ! 1 Z Y 2σ−1 − Y 1−2σ 1 − |M(σ + ir)|2 · Y 1−2σ = h(r) · lim + dr 1 4π σ→ 2 2σ − 1 2σ − 1 −∞ ∞  ∞  1 Z M( 1 − ir)Y 2ir − M( 1 + ir)Y −2ir Z + h(r) · 2 2 dr + O e−2πY h(r)e4r dr 4π 2ir   −∞ −∞ −2πY = I1(Y ) + I2(Y ) + O e since we have assumed that h(r) e−5|r|.

Evaluation of I1(Y ): First of all, by the Selberg transform (see deﬁnition 3.28), we have ∞ 1 Z Y 2σ−1 − Y 1−2σ h(r) · lim dr = g(0) log Y. 1 4π σ→ 2 2σ − 1 −∞ It follows that ∞ 1 Z 1 − |M(σ + ir)|2 · Y 1−2σ I1(Y ) = g(0) log Y + lim h(r) · dr. 1 σ→ 2 4π 2σ − 1 −∞ To evaluate the limit in the integral above above we consider the 1 Taylor expansion around σ = 2 given by 1 − M(σ + ir)M(σ − ir) = 1 − M (1/2 + ir) M (1/2 − ir) | {z } this term = 0 1 1 1 1 1 + M 0 + ir M − ir + M + ir M 0 − ir · σ − 2 2 2 2 2 n o + higher powers of σ − 1/2 . We obtain ∞ " # 1 Z M 0 1 + ir M 0 1 − ir I (Y ) = g(0) log Y + h(r) · 2 + 2 dr. 1 8π 1 1 M 2 + ir M 2 − ir −∞ 30 DORIAN GOLDFELD

Evaluation of I2(Y ):

For the evaluation of I2(Y ), it is necessary to interpret the integrals involved in the sense of Cauchy Principal Value, i.e., Z ∞ f(x) Z f(x) (4.6) dx := lim dx. −∞ x →0 |x|≥ x It is easy to see that if f is C1 on R and decays moderately at ±∞, then the limit (4.6) exists. Next, we compute ∞ 1 Z M( 1 − ir)Y 2ir − M( 1 + ir)Y −2ir I (Y ) = h(r) · 2 2 dr 2 4π 2ir −∞ ∞ 1 1 Z Re M 2 − ir sin 2r log Y = h(r) · dr 4π r −∞ h(0)M 1 = 2 + o(1) 4 ∞ 1 R sin r·T as Y → ∞. This follows since F (0) = lim π F (r) r dr which T →∞ −∞ is a well known result in the theory of Fourier integrals. We have h(0)M 1 I (Y ) = 2 + o(1). 2 4 4.3. Contribution of the Hyperbolic Conjugacy Classes C(P ).

a b A matrix ( c d ) ∈ SL(2, Z) is termed hyperbolic if one of the following three equivalent conditions is satisﬁed: (1) |a + d| > 2,

a b awi + b (2)( c d ) has 2 distinct real ﬁxed points w1, w2 with = wi, cwi + b (3) There exists unique η ∈ SL(2, R) and ρ ∈ R with |ρ| > 1 such that a b ρ 0 −1 ( c d ) = η 0 ρ−1 η .

ρ 0 −1 Deﬁnition 4.7. If P = η 0 ρ−1 η ∈ SL(2, Z) is hyperbolic with |ρ| > 1, we deﬁne N(P ) := ρ2 to be the norm of P. TRACE FORMULAE FOR SL(2, R) 31

Deﬁnition 4.8. (Centralizer ΓP ) Let P ∈ SL(2, Z) be hyperbolic. We deﬁne the centralizer of P (denoted ΓP ) by

ΓP := {δ ∈ SL(2, Z) | δP = P δ}.

Lemma 4.9. Let P ∈ SL(2, Z) be hyperbolic. Then the centralizer ΓP is the direct product of an inﬁnite cyclic group generated by some P0 ∈ SL(2, Z) and the group {±1} of order 2. Here P0 is itself hyperbolic ` with the same ﬁxed points as P. Hence ΓP = {±P0 | ` ∈ Z}. m Furthermore P = ±P0 for some m ≥ 1, and this uniquely deter- mines P0. The element P0 is called a primitive hyperbolic element.

Proof. Assume δ ∈ ΓP ⇐⇒ δP = P δ. We now show that this condition forces δ to have the same ﬁxed points w1, w2 as P. In fact, for i = 1, 2, we have

δ(P wi) = δ wi = P (δwi). This implies that one of the following two cases must hold:

case 1: δw1 = w1, δw2 = w2 or case 2: δw1 = w2, δw2 = w1 .

2 We want to show case 2 cannot happen. In either case δ wi = wi. Now δ must have at least one fixed point which cannot be w1 or w2. Further, fixed points of δ must be fixed points of δ2. It follows that δ2 must have at least 3 distinct fixed points which is not possible unless δ2 is ±I where I is the identity matrix. So, we may assume δ2 = ±I and δ 6= ±I since we are in case 2. It a b follows that δ = c −a , hence tr(δ) = 0. It remains to show that tr(δ) 6= 0. Let η ∈ SL(2, R) such that ρ 0 −1 ρ 0 −1 ρ 0 −1 P = η 0 ρ−1 η . Then δ · η 0 ρ−1 η = η 0 ρ−1 η · δ which implies that

−1 ρ 0 ρ 0 −1 η δη · 0 ρ−1 = 0 ρ−1 · η δη .

ρ 0 The only matrices in SL(2, R) that commute with 0 ρ−1 are diagonal u 0 matrices of the form 0 u−1 with u ∈ R and u 6= 0. This implies that −1 u 0 η δη = 0 u−1 for some u ∈ R×. This is a contradiction since the trace of η−1δη is equal to the trace of δ which is zero. We have thus proved that any matrix δ which commutes with a hyperbolic matrix P must have the ` same ﬁxed points w1, w2 as P. It follows that ΓP = {±P0 | ` ∈ Z} 32 DORIAN GOLDFELD

Proposition 4.10. (Computation of C(P )) Let P ∈ SL(2, Z) be a m ρ 0 −1 hyperbolic matrix where P = ±P0 (m ≥ 1)and P0 = η 0 ρ−1 η as in lemma 4.9. Then

log N(P C(P ) = 0 · g log N(P ). 1 − 1 N(P ) 2 − N(P ) 2

Proof. The proof of proposition 4.10 is based on the following lemma.

ρ 0 −1 Lemma 4.11. Let P0 = η 0 ρ−1 η be a primitive hyperbolic matrix −1 in SL(2, Z). Then a fundamental domain for (η ΓP0 η) \h is given by

2 Dρ := z = x + iy ∈ h x ∈ R, 1 ≤ y ≤ ρ .

` −1 n ρ 0 o Proof. First of all η Γ η = −1 ` ∈ . Let z ∈ h. Then P0 0 ρ Z ρ 0 2 S 2` 0 ρ−1 z = ρ z. Clearly ρ ·Dρ = h. `∈Z

We now go on to prove proposition 4.10.

2 ! Z X |z − σP mσ−1z| dxdy C(P ) = f 0 Im(z) · Im(σP mσ−1z) y2 σ∈Γ \SL(2, ) 0 SL(2,Z)\h P0 Z 2 ! X Z |σz − σP mz| dxdy = f 0 Im(σz) · Im(σP mz) y2 σ∈Γ \SL(2, ) 0 P0 Z σ· SL(2,Z)\h | {z } z 7→ σz Z M 2 ! z − P0 z dxdy = f m 2 Im(z) · Im (P0 z) y Γ \h P0 | {z } This follows from proposition 3.9 (iv) TRACE FORMULAE FOR SL(2, R) 33 Next, make the transformation z 7→ η−1z. It follows from lemma 4.11 that

To proceed further let x = yξ in the last integral above. We obtain

ρ2 ∞ ! Z Z (ρ2m − 1)2 dξdy C(P ) = 2 f · 1 + ξ2 ρ2m y 1 0 ∞ ! Z (N(P ) − 1)2 = 2 log N(P ) f · 1 + ξ2 dξ 0 N(P ) 0

Next, let

2 1 (N(P ) − 1) N(P ) 2 u = · 1 + ξ2 , ξ = u − 1 . N(P ) (N(P ) − 1)2

It follows that

1 N(P ) 2 C(P ) = log N(P ) Q N(P ) + N(P )−1 − 2 0 N(P ) − 1 log N(P ) = 0 g log N(P ). 1 − 1 N(P ) 2 − N(P ) 2

34 DORIAN GOLDFELD

4.4. Contribution of the Elliptic Conjugacy Class C(R).

a b A matrix ( c d ) ∈ SL(2, Z) is termed elliptic if one of the following conditions is satisﬁed: (1) |a + d| < 2, a b (2)( c d ) has 2 complex conjugate ﬁxed points, cos θ sin θ (3) There exists unique η ∈ SL(2, R) and k(θ) = − sin θ cos θ such that a b −1 ( c d ) = η k(θ) η .

Deﬁnition 4.12. (Centralizer ΓR) Let R ∈ SL(2, Z) be elliptic. We deﬁne the centralizer of R (denoted ΓR) by

ΓR := {δ ∈ SL(2, Z) | δR = Rδ}.

Lemma 4.13. Let R ∈ SL(2, Z) be elliptic. Then the centralizer ΓR is a ﬁnite cyclic group of order e > 1 generated by some R0 ∈ SL(2, Z) which is itself elliptic with the same ﬁxed points as R. Hence ` ` ΓR = ΓR0 = {±R0 | ` ∈ Z, 1 ≤ ` ≤ e} = {R0 | ` ∈ Z, 1 ≤ ` ≤ 2e}. m Furthermore R = R0 for some 1 ≤ m < e, and this uniquely de- termines R0. The element R0 is called a primitive elliptic element and −1 takes the form R0 = η k(π/e) η . Proof. Exercise for the reader. Proposition 4.14. (Computation of C(R)) Let R ∈ SL(2, Z) be an m elliptic matrix of order e where R = ±R0 (with 1 ≤ m < e) and cos (π/e) sin (π/e) R = η η−1 0 − sin (π/e) cos (π/e) as in lemma 4.9. Then

∞ − 2πmr 1 Z e e C(R) = · h(r) dr. mπ 1 + e−2πr 2e sin e −∞ Proof. We follow Kubota’s proof in [Kub73]. Our goal is to compute 2 ! Z X |z − σRmσ−1z| dxdy C(R) = f 0 . Im(z) · Im(σRmσ−1z) y2 σ∈Γ \SL(2, ) 0 SL(2,Z)\h R0 Z

cos(π/e) sin(π/e) 2π Now − sin(π/e) cos(π/e) acts as a rotation of angle e around i.A −1 2π fundamental domain for η ΓR0 η is a hyperbolic sector of angle e . TRACE FORMULAE FOR SL(2, R) 35 It follows (as in the proof of proposition 4.10) that

Z m −1 2 ! |z − ηR0 η z| dxdy (4.15) C(R) = f m −1 2 Im(z) · Im (ηP0 η z) y −1 η ΓR0 η h Z m −1 2 ! 1 |z − ηR0 η z| dxdy = f m −1 2 e Im(z) · Im (ηR0 η z) y h

−1 because it takes e images of η ΓR0 η h to cover h exactly. Now, let cos (mπ/e) sin (mπ/e) α −β ηRm η−1 = := . 0 − sin (mπ/e) cos (mπ/e) β α We see that  2  αz−β ∞ ∞ z − 1 Z Z βz+α dxdy C(R) = f     2 e 0 −∞ αz−β y Im(z) · Im βz+α ! 1 Z ∞ Z ∞ |βz2 + β|2 dxdy = f 2 2 e 0 −∞ y y Z ∞ Z ∞ 2 2 2 2 (x + 1) 2 2 dxdy = f β 2 + 2(x − 1) + y 2 e 0 0 y y mπ Next, make the change of variables (recall that β = − sin e < 0): (x2 + 1)2 t = β2 + 2(x2 − 1) + y2 , y2 x2 + 1 + y2 4β2 r t dt = 4β2 x dx = 4 + x dx, y2 y β2 dx y = . dt 4|β|xpt + 4β2 It is clear that t = t(x) is strictly increasing for 0 ≤ x ≤ ∞, so 2 −1 2 2 −1 2 the minimum√ value of t is β (y − y) . Now t = β (y − y) when −1 t y − y = ± β . There are 2 cases. √ √ √ 2 −1 t t+ t+4|β| (1) y − y = |β| =⇒ y = 2|β| := y2, √ √ √ 2 −1 t − t+ t+4β (2) y − y = − |β| =⇒ y = 2|β| := y1. 36 DORIAN GOLDFELD

Hence 0 ≤ t < ∞ and y1 < y < y2. Furthermore

1 r 2 1 t 2 x = y 4 + − y2 − 1 = (y − y )(y − y) . β2 1 2 It follows that

∞ ∞ 2 Z Z dx dy C(R) = f(t) dt e dt y2 0 β2(y−1−y)2 ∞ ∞ ! 2 Z Z y dy = f(t) dt e 4|β|xpt + 4β2 y 0 β2(y−1−y)2

∞  y2  1 Z f(t) Z dy = dt p 2  1 1  2e|β| t + 4β y(y − y1) 2 (y2 − y) 2 0 y1 | {z } This integral = π because y1y2 = 1 ∞ π Z f(t) = dt 2e|β| pt + 4β2 0 ∞  ∞  π Z 1 −1 Z Q0(w) =  √ dw dt. 2e|β| pt + 4β2 π w − t 0 t To proceed further we interchange integrals and then integrate by parts to obtain

∞ w 1 Z Z dt C(R) = − dQ(w) 2e|β| p(4β2 + t)(w − t) w=0 t=0 ∞  w  1 Z Z dt = Q(w) d   2e|β| p(4β2 + t)(w − t) 0 0 ∞ 1 Z 2|β| = Q(w) √ dw 2e|β| w (w + 4β2) 0

| {z 2 } u − u Let w = e 2 − e 2

Finally, we obtain TRACE FORMULAE FOR SL(2, R) 37

∞ u − u 1 Z e 2 + e 2 C(R) = g(u) du u − u 2 2e e 2 − e 2 + 4β2 −∞ ∞  ∞  u − u 1 Z e 2 + e 2 Z = h(r)eiru dr du u − u 2   4πe e 2 − e 2 + 4β2 0 −∞ ∞  ∞  u − u 1 Z Z e 2 + e 2 = h(r) eiru du dr  u − u 2  4πe e 2 − e 2 + 4β2 −∞ −∞ ∞ − 2πmr 1 Z e e = · h(r) dr 2e|β| 1 + e−2πr −∞ 38 DORIAN GOLDFELD

5. The Selberg Trace Formula for SL(2, R) We now state the complete version of the Selberg Trace Formula for SL(2, R).

Theorem 5.1. Let h : C → C (with Fourier transform g) satisfy (1) h(r) = h(−r) for all r ∈ R. 1 (2) h(r) is holomorphic in the strip Im(r) < 2 + for some > 0 . (3) h(r) = O ((1 + r2)−1−) in the above strip.

Let φ0 be the constant function of Petersson norm one, and let φ1, φ2,... denote an orthonormal basis of Maass cusp forms which satisfy 1 2 ∆φi = 4 + ri φi for i = 1, 2 ... The Selberg trace formula is the following identity

∞ ∞ X 1 Z h(r ) = r tanh(πr) h(r) dr i 12 i=0 −∞ ∞ e−1 − 2πmr X X 1 Z e e + · h(r) dr mπ 1 + e−2πr 2e sin e R0 m=1 −∞ |{z} primitive elliptic ∞ X X log N(P0 + m − m · g m log N(P0) N(P0) 2 − N(P0) 2 P0 m=1 |{z} primitive hyperbolic ∞ 1 Z Γ0 M 0 − h(r) (1 + ir) + Re (1/2 + ir) dr 2π Γ M −∞ 1 − M(1/2) + − g(0) log 2. 4

Here √ πΓ(s − 1/2)ζ(2s − 1) M(s) = . Γ(s)ζ(2s) Proof. This follows immediately from propositions 4.1, 4.2, 4.10, 4.14 under the assumption that h(r) e−5|r| is even and holomorphic in a 1 strip |Im(r)| ≤ 2 + for > 0. We now show that the range in which the trace formula holds can be extended by allowing the test function TRACE FORMULAE FOR SL(2, R) 39 h to satisfy h(r) = O (1 + r2)−1− . We deﬁne Γ0 M 0 ω(r) := (1 + ir) + Re (1/2 + ir) . Γ M and move the last integral on the right side of the Selberg trace formula in theorem 5.1 to the left side (spectral side) which then becomes ∞ X 1 Z ∞ h(r ) + h(r) ω(r) dr. j 2π j=1 −∞

2 − r We now choose h(r) = e R2 for R → ∞, which satisfies the condition −5|r| 1 h(r) e and is even and holomorphic in the strip |Im(r)| ≤ 2 + . It follows that R X 1 Z 1 + ω(r) dr = O R2 2π |rj |≤R −R as R → ∞. Quoting from Selberg’s Gottingen lectures ([Sel89], page 668) with some modifications in the choice of variables and citations to previous equations: This implies that all series and integrals occurring in theorem 5.1 converge absolutely if h(r) only satisfies the usual conditions mentioned above. Also for a class of functions which satisfies these conditions uniformly, convergence of series and integrals is seen to be uniform. Considering now for a fixed h(r) the class h(r)e−r2 with 0 ≤ ≤ 1, these constitute such a class, we have that for > 0 that h(r)e−r2 e−5|r| is satisfied and so proposition 4.2 is valid. Making → 0 we obtain theorem 5.1. There is one final point in the proof of theorem 5.1 that needs to be clarified. It is necessary to make sure that the infinite sum over primitive hyperbolic conjugacy classes converges absolutely. ρ 0 −1 Let P0 = η 0 ρ−1 η be a primitive hyperbolic matrix with trace −1 2 t = ρ + ρ and norm N(P0) = ρ . Now t > 2, and √ √ 2 t + t2 − 4 t + t2 − 4 ρ = ,N(P ) = . 2 0 2 40 DORIAN GOLDFELD

We will show in §6 that the number of distinct primitive hyperbolic conjugacy classes P0 with the same trace t is exactly the class number √ of the real quadratic ﬁeld Q t2 − 4 . Since it is known that this class 2 1 + 1+ number is bounded by O (t − 4) 2 = O (t ), it follows that ∞ X X log N(P0 m − m · g m log N(P0) N(P0) 2 − N(P0) 2 P0 m=1 ∞ ∞ X X log t t1+ · g(m log t) t2m + t−2m t=3 m=1 which converges absolutely because of the rapid decay of the function g(u) as u → ∞. TRACE FORMULAE FOR SL(2, R) 41

6. The Kuznetsov Trace Formula for SL(2, R) There are two fundamental ways to express a group Γ as a disjoint union of interesting subsets of Γ. The first way breaks Γ into conjugacy classes [ Γ = [γ], [γ] = {σγσ−1 | σ ∈ Γ}. γ This was the basis for the Selberg trace formula with Γ = SL(2, Z). The second basic way is with double cosets. Let H,K be subgroups of Γ acting by left and right multiplication, respectively. Then we have the decomposition [ Γ = HγK. γ∈H\Γ/K The double coset decomposition is the basis for the Kuznetsov trace formula. Definition 6.1. (PoincaréSeries for SL(2, Z)) Let p : R → C be a smooth function satisfying ( y1+ if 0 < y ≤ 1, p(y) y−C if 1 < y, for some > 0 and some fixed constant C > 1. Let 1 m B := m ∈ Z 0 1 denote the Borel subgroup of Γ = SL(2, Z). Then for z ∈ h, we define the Poincaréseries Pm(∗, p): h → C by X 2πim·Re(γz) Pm(z, p) = p m · Im(γz) e . γ∈B\Γ

Let Pm(z, p), Qn(z, q) be two Poincaréseries with m, n ≥ 1 and smooth functions p, q : R → C as in definition 6.1 . The Kuznetsov trace formula is obtained by computing the Petersson inner product D E Z dxdy P (∗, p), Q (∗, q) = P (z, p) Q (z, q) m n m n y2 Γ\h in two different ways.

• First way, Spectral Side: Take the spectral expansion of Pm(∗, p) and unravel Qn(∗, q) in the inner product. • Second way, Geometric Side: Compute the Fourier expansion of Pm(∗, p) by double coset decomposition and unravel Qn(∗, q). 42 DORIAN GOLDFELD

7. The SL(2, R) Kuznetsov Trace Formula (Spectral Side) 2 1+ We ﬁrst need to show that Pm(∗, p) ∈ L (Γ\h). Since p(y) y it easily follows from the growth properties of p that

X 1+ |Pm(z, p)| Im γz γ∈B\Γ X p(y) + Im(γz)1+ γ∈B\Γ γ6=I2 1.

1 0 2 where I2 = ( 0 1 ). We have thus proved that Pm(∗, p) ∈ L (Γ\h). By the Selberg spectral decomposition given in theorem 3.31 we see that

∞ X D E φj(z) P (z, p) = P (∗, p), φ m m j hφ , φ i j=0 j j ∞ 1 Z D E + P (∗, p),E(∗, 1/2 + ir) E(z, 1/2 + ir) dr. 4π m −∞

It follows that the inner product of two Poincar´eseries is given by

(7.1) ∞ D E X Pm(∗, p), φj φj, Qn(∗, q) P (∗, p), Q (∗, q) = m n hφ , φ i j=0 j j ∞ 1 Z D ED E + P (∗, p),E(∗, 1/2 + ir) E(∗, 1/2 + ir), Q (∗, q) dr. 4π m n −∞

To proceed further we require formulae for the inner product of the Poincar´eseries Pm with a Maass form or Eisenstein series. Let φ0 be the constant function of Petersson norm one, and let φ1, φ2,... denote an orthonormal basis of Maass cusp forms which sat- 1 2 isfy ∆φi = 4 + ri φi for i = 1, 2 ... Each Maass form φj with j > 0 has a Fourier expansion of the form

X √ 2πi`x φj(z) = Aj(`) y Kirj (2π|`|y)e , `6=0 TRACE FORMULAE FOR SL(2, R) 43 where ∞ Z 1 − y u+ 1 ir du K (y) = e 2 ( u )u ir 2 u 0 is the K-Bessel function.

Lemma 7.2. (Inner product of Pm with a Maass form φj)

0 if j = 0, D E  ∞ Pm(∗, p), φj = 1 R dy A (m)m 2 p(y)K (2πy) if j > 0.  j irj 3 0 y 2 Proof. We compute D E Z X dxdy P (∗, p), φ = pm · Im(γz) e2πim Re(γz) · φ (z) m j j y2 Γ\h γ∈B\Γ Z dxdy = pmy e2πimx φ (z) j y2 B\h ∞  1  Z Z dy = e2πimxφ (x + iy) dx · p(my)  j  y2 0 0 | {z } This integral = 0 if j = 0. ∞ Z √ dy = A (m) y K (2πmy) · p(my) j irj y2 0 | {z } Here we assume j 6= 0.

Every Maass form φj for SL(2, Z) is self dual so Aj(m) = Aj(m).

Next, we consider the inner product of Pm and E(∗, s). The Fourier expansion of the Eisenstein series is given in theorem 3.4. Let

s s− 1 2π σ (m)m 2 A(m, s) := 1−2s Γ(s)ζ(2s) denote the mth arithmetic Fourier coeﬃcient of E(z, s).

Lemma 7.3. (Inner product of Pm with an Eisenstein series)

∞ D E 1 Z dy 2 Pm(∗, p),E(∗, s) = A(m, s) m p(y) Kirj (2πy) 3 . y 2 0 44 DORIAN GOLDFELD

Proof. The proof is the same as the proof for a Maass form in lemma 7.2 .

Proposition 7.4. (Kuznetsov trace formula, Spectral Side) Let Pm(z, p), Qn(z, q) be Poincar´eseries. Deﬁne the Bessel transforms: Z ∞ Z ∞ # dy # dy p (ir) := p(y)Kir(2πy) 3 , q (ir) := q(y)Kir(2πy) 3 . 0 y 2 0 y 2 Then the spectral side of the Kuznetsov trace formula is

∞ # # D E √ X p (irj)q (irj) P (∗, p), Q (∗, q) = mn A (m)A (n) · m n j j hφ , φ i j=1 j j √ ∞ mn Z 1 1 + A m, + ir A n, + ir · p#(ir) q#(ir) dr. 4π 2 2 −∞

Proof. This follows immediately from (7.1) and lemmas 7.2, 7.3.

8. The SL(2, R) Kuznetsov Trace Formula (Geometric Side) To compute the geometric side of the Kuznetsov tace formula we ﬁrst need to rewrite the Poincar´eseries Pm(∗, p) as a sum over double cosets B\Γ/B. We require the following lemma. Lemma 8.1. (Double coset decomposition of Γ = SL(2, Z)) Let 1 m B = {( 0 1 ) | m ∈ Z} ⊂ Γ. Then   [ [ ∗ ∗ Γ = B ∪ B B .  c d  c>0 d (mod c) Proof. Let c ≥ 1. Then we have the matrix identity 1 m a ∗ 1 n a + cm ∗ = . 0 1 c d 0 1 c d + cn ∗ ∗ It follows that the couble coset B ( c d ) B is determined uniquely by c and d (mod c) and this double coset is independent of the top row of ∗ ∗ ( c d ).

It follows from lemma 8.1 that the Poincaréseries Pm(z, p) can be rewritten in the following form. ∞ 0 2πimx X X my 2πim·Re az+b P (z, p) = p my e + p e ( cz+d ), m |cz + d|2 c=1 d∈Z TRACE FORMULAE FOR SL(2, R) 45 where the prime on the sum denotes that the sum is over (d, c) = 1. a b Further a, b is chosen so that ( c d ) ∈ SL(2, Z). One checks that the Poincaréseries is independent of the choice of a, b. The geometric side of the Kuznetsov trace formula involves classical Kloosterman sums which we now define.

Deﬁnition 8.2. (Kloosterman Sum) Let m, n, c ∈ Z with c ≥ 1. The Kloosterman sum is deﬁned by

0 X 2πi· am+dn S(m, n; c) := e c 1≤a≤c where the prime on the sum denotes that (a, c) = 1. Further d is chosen so that ad ≡ 1 (mod c).

Proposition 8.3. Kuznetsov tace formula, Geometric Side) Let Pm(z, p), Qn(z, q) be Poincar´eseries. The geometric side of the Kuznetsov trace formula is

∞ D E Z dy P (∗, p), Q (∗, q) = δ p(my) q(ny) m n m,n y2 0 ∞ ∞ Z Z ∞ my X −2πix m + ny dxdy + p q(ny) S(m, n; c) e c2y(x2+1) . c2y (x2 + 1) y 0 −∞ c=1

Proof. The proof of proposition 8.3 requires the Fourier expansion of the Poincar´eseries given in the following lemma.

Lemma 8.4. (Fourier expansion of Pm(z, p)) We have 1 R −2πinx Pm(z, p)e dx = δm,n · p(ny) 0 ∞ ∞ Z X my −2πix m + ny +y · S(m, n; c) p e c2y(x2+1) dx. c2y (x2 + 1) c=1 −∞

 1 m = n, where δm,n = is Kronecker’s symbol. 0 m 6= n, 46 DORIAN GOLDFELD

Proof. We compute. 1 1 Z Z −2πinx 2πi(m−n)x Pm(z, p)e dx = p my e dx 0 0 1 " ∞ # Z 0 X X my 2πim·Re az+b −2πinx + p e ( cz+d ) e dx |cz + d|2 0 c=1 d∈Z

= δm,n · p(my) + Ip(z).

Now az + b a 1 = − . cz + d c c(cz + d)

Next let d = `c + r. We see that Ip(z) is equal to   1 ∞ ∞ c Z 2πim·Re a − 1 X X X my c c(cz+`c+r  −2πinx  p e e dx  |cz + `c + r|2  0 c=1 `=−∞ r=1 (r,c)=1

| {z r } Make the change of variables x → x − ` − c −`− r +1 ∞ ∞ c Z c X X X my 2πim·Re( a − 1 ) −2πin(x−`− r ) = p e c c2z e c dx |cz|2 c=1 `=−∞ r=1 −`− r (r,c)=1 c ∞ ∞ Z X my 2πim·Re( −1 )e−2πinx = S(m, n; c) p e c2z dx c2 (x2 + y2) c=1 −∞ | {z } Let x 7→ xy. ∞ ∞ Z X my −2πix m + ny = y · S(m, n; c) p e c2y(x2+1) dx c2y2 (x2 + 1) c=1 −∞ The proof of proposition 8.3 follows from the above lemma together with the fact that by unraveling Qn ∞  1  D E Z Z dy P (∗, p), Q (∗, q) = P (z, p)e−2πinx dx q(ny) . m n  m  y2 0 0 TRACE FORMULAE FOR SL(2, R) 47

9. The Kuznetsov Trace Formula for SL(2, R) We now state the complete version of the Kuznetsov Trace Formula for SL(2, R). Theorem 9.1. Let p, q : R → C be a smooth functions satisfying ( y1+ if 0 < y ≤ 1, p(y), q(y) y−C if 1 < y, for some > 0 and some C > 1. Deﬁne the Bessel transforms Z ∞ Z ∞ # dy # dy p (ir) := p(y)Kir(2πy) 3 , q (ir) := q(y)Kir(2πy) 3 . 0 y 2 0 y 2 P √ 2πi`x Let φj(z) = Aj(`) y Kirj (2π|`|y)e , (j = 1, 2,...) denote an `6=0 1 2 orthonormal basis of Maass cusp forms where ∆φi = 4 + ri φi for 1 s s− 2π σ1−2s(m)m 2 th i ≥ 1. Let A(m, s) := Γ(s)ζ(2s) denote the m arithmetic Fourier coeﬃcient of E(z, s). The Kuznetsov trace formula is the following identity:

∞ # # √ X p (irj) q (irj) mn A (m)A (n) · j j hφ , φ i j=1 j j √ ∞ mn Z + A (m, 1/2 + ir) A (n, 1/2 + ir) · p#(ir) q#(ir) dr 4π −∞

∞ ∞ ∞ ∞ Z dy X Z Z = δ p(my) q(ny) + S(m, n; c) m,n y2 0 c=1 0 −∞ | {z } Main Term m −2πix m + ny dxdy · p q(ny) · e c2y(x2+1) . c2y2 (x2 + 1) y

9.1. Kontorovich-Lebedev Transform. The Kuznetsov trace formula involves a test function p : R → C and the transform Z ∞ # dy p (ir) := p(y)Kir(2πy) 3 . 0 y 2 48 DORIAN GOLDFELD

This type of transform is the well known Kontorovich-Lebedev transform (see [Leb72]) which was originally developed for applications in physics. The inverse transform (for a proof see [GKS11]) is given by ∞ 1 Z √ dr (9.2) p(y) = p#(ir) y K (2πy) . π ir Γ(ir)Γ(−ir) −∞ We also note the Mellin transform pair given by Z ∞ 1 s+ir 1 s−ir √ s dy Γ 4 + 2 Γ 4 + 2 y Kir(2πy) y = , 1 +s 0 y 4π 2 σ+i∞ Z 1 s+ir 1 s−ir √ 1 Γ 4 + 2 Γ 4 + 2 −s y Kir(2πy) = y ds, 1 +s 2πi 4π 2 s=σ−i∞ 1 for any σ > − 2 . It follows form (9.2) and the above Mellin transform that

∞ +i∞ Z Z s+ir s−ir 1 # Γ 2 Γ 2 −s 1 −s (9.3) p(y) = p (ir) π y 2 ds dr. 8π2i Γ(ir)Γ(−ir) −∞ −i∞

9.2. Explicit choice of the pT,R test function. We now present an explicit choice of test function for the Kuznetsov trace formula for SL(2, R) that is useful for many applications and generalizes to higher rank groups. We let T → ∞+ be a large real variable and let R be a sufficiently large (to be determined later) fixed positive real number. For applications it is convenient to define the test function on the spectral side to be given by 2 2 2 + R + 2α 2 + R − 2α (9.4) p# (α) := eα /T · Γ Γ . T,R 4 4 The main reason for choosing this type of test function is that if we assume, for example, that p = q in the Kuznetsov trace formula given in theorem 9.1 then the contribution of the Maass forms on the spectral side will be 2 −2r2/T 2 2+R+irj ∞ # 2 ∞ e j Γ X |pT,R(irj)| X 4 A (m)A (n) = A (m)A (n) j j hφ , φ i j j hφ , φ i j=1 j j j=1 j j which will turn out to naturally count sums of weighted products of Fourier coefficients Aj(m)Aj(n) of Maass forms with Lapace eigenvalue TRACE FORMULAE FOR SL(2, R) 49

2 2 2 1 2 −2rj /T λj T . This is due to the fact that λj = 4 +rj so the function e 2 is essentially supported on λj T . The constant R is needed later to allow us to shift the contour in α of a certain integral without crossing 2+R±α any poles of the Gamma function Γ 4 .

9.3. Bounds for the pT,R test function.

R Proposition 9.5. Let > 0 and R > 4 with 1 ≤ a < 4 . Then for y > 0 and T → +∞ we have

1 +2a + R + 3 −a pT,R(y) y 2 T 2 2 .

Proof. We recall the identity (9.3) which states that pT,R(y) is equal to

(9.6) ∞  +i∞  Z Z # 1 1 s + ir s − ir −s 1 −s pT,R(ir) dr Γ Γ π y 2 ds · . 4π  2πi 2 2  Γ(ir) Γ(−ir) −∞ −i∞ | {z } = I(y, r)

We may shift the line of integration in the above integral I(y, r) to the s+ir line Re(s) = −2a where a > 0 and a∈ / Z. Since Γ 2 has simple poles at s = −2` − ir for ` ∈ Z and ` ≥ 0 it follows from Cauchy’s residue theorem that

−2a+i∞ Z 1 s + ir s − ir −s 1 −s I(y, r) = Γ Γ π y 2 ds 2πi 2 2 −2a−i∞ X s + ir s − ir −s 1 −s + Res + Res Γ Γ π y 2 s = −2`+ir s = −2`−ir 2 2 0≤`≤a

−2a+i∞ Z 1 s + ir s − ir −s 1 −s = Γ Γ π y 2 ds 2πi 2 2 −2a−i∞ X ` 2`−ir 1 +2`−ir 2`+ir 1 +2`+ir + (−1) 2π y 2 Γ(−` + ir) + 2π y 2 Γ(−` − ir) . 0≤`≤a

We plug this expression into (9.6) and make the change of variable α = ir. It follows that 50 DORIAN GOLDFELD

s+α s−α 1 −s −1 Z Z Γ Γ y 2 p (y) = 2 2 · p# (α) ds dα T,R 8π2 Γ(α)Γ(−α) πs T,R Re(α)=0 Re(s)=−2a

2`−α 1 +2`−α 1 Z X π y 2 Γ(−` + α) − (−1)` πi Γ(α)Γ(−α) 0≤`≤a Re(α)=0

2`+α 1 +2`+α ! π y 2 Γ(−` − α) + · p# (α) dα Γ(α)Γ(−α) T,R

2 1 −s α s+α s−α 2+R+2α 2+R−2α −1 Z Z y 2 e T 2 Γ Γ Γ Γ = · 2 2 4 4 dsdα 8π2 πs Γ(α)Γ(−α) {0} {−2a} 2`−α 1 +2`−α 2+R+2α 2+R−2α 1 Z X π y 2 Γ Γ − (−1)` 4 4 πi (α − 1)(α − 2) ··· (α − `) · Γ(−α) 0≤`≤a Re(α)=0 1 2+R+2α 2+R−2α ! 2`+α 2 +2`+α 2 π y Γ 4 Γ 4 α + · e T 2 dα (−α − 1)(−α − 2) ··· (−α − `) · Γ(α)

(1) (2) (3) = IT,R(y) + IT,R(y) + IT,R(y).

(1) We will now bound the ﬁrst of the three integrals IT,R(y). To do so we use Stirling’s bound (for σ, t ∈ R with σ ﬁxed and |t| → ∞) given by

√ σ− 1 − π |t| (9.7) |Γ(σ + it)| ∼ 2π |t| 2 · e 2 . | {z } | {z } polynomial exponential term term

σ− 1 Here |t| 2 is called the polynomial factor because it has polynomial − π |t| growth or decay in |t| as |t| → ∞ and e 2 is called the exponential term because it has exponential decay in |t| as |t| → ∞. TRACE FORMULAE FOR SL(2, R) 51 Thus, setting s = −2a + iξ and α = ir, we see that the ﬁrst term for pT,R(y) is bounded by

Z Z 2 2+R+2α 2+R−2α s+α s−α (1) α Γ( 4 )Γ( 4 )Γ 2 Γ 2 I (y) e T 2 T,R Γ(α)Γ(−α) Re(α)=0 Re(s)=−2a 1 −s · y 2 ds dα

T 1+ ∞ R 2 +1 1 Z Z 1 + |r| 2 +2a y 1+2a 1+2a 1 + |ξ + r| 2 1 + |ξ − r| 2 r=0 ξ=−∞ − π |ξ+τ|+|ξ−τ|−2|τ| · e 4 dξ dτ. We rewrite this as

T 1+ ∞ Z Z (1) 1 +2a π I (y) y 2 P· exp − E dξ dτ T,R 4 r=0 ξ=−∞ where R +1 1 + |r| 2 P := 1+2a 1+2a 1 + |ξ + r| 2 1 + |ξ − r| 2 is the polynomal term and E := |ξ + r| + |ξ − r| − 2|r| is the exponential term. It is easy to see that E ≥ 0 for all ξ, r. Deﬁnition 9.8. (Exponential zero set of E) The exponential zero set of E is the set of all (ξ, r) ∈ R2 such that E = 0. Lemma 9.9. The exponential zero set of E := |ξ + r| + |ξ − r| − 2|r| is given by the set of (ξ, r) satisfying −r ≤ ξ ≤ r. One observes that there is exponential decay in the region of inte- (1) gration of the integral IT,R(y) that is outside the exponential zero set. It follows that 1+ T r R +1 Z Z 2 (1) 1 1 + r 2 +2a IT,R(y) y 1+2a 1+2a dξ dr. 1 + ξ + r 2 1 − ξ + r 2 r=0 ξ=−r To complete the proof we require the following lemma. 52 DORIAN GOLDFELD

Lemma 9.10. Suppose that C,C0 ≥ 0 (with C,C0 6= 1) and A > 0. Then

A Z dx 1 0 0 − 2 min 1+C, 1+C ,C+C 1+C 1+C0 A . (1 + x) 2 (1 + A − x) 2 x=0

Proof. Exercise for the reader.

So using Lemma 9.10, we obtain

1+ T  r R +1  Z Z 2 (1) 1 1 + r 2 +2a IT,R(y) y  1+2a 1+2a dξ dr 1 + ξ + r 2 1 − ξ + r 2 r=0 ξ=−r | {z } Make the change of variable x = ξ + r. T 1+ 2r 1 Z R +1 Z dx 2 +2a 2 y 1 + r 1+2a 1+2a dr (1 + x) 2 (1 + 2r − x) 2 r=0 x=0 T 1+ 1 Z R 1 +2a 2 +1 − min 1+2a, 4a y 2 1 + r (2r) 2 dr

r=0 1 +2a 1+ R +1 − 1 −a y 2 T · T 2 · T 2 1 +2a + R + 3 −a y 2 T 2 2 as claimed. (2) (3) We now bound the remaining two integrals: IT,R(y), IT,R(y). We have

1 2+R+2α 2+R−2α Z 2 2`−α 2 +2`−α (2) X α π y Γ 4 Γ 4 I (y) = e T 2 · dα T,R πi (1 − α)(2 − α) ··· (` − α) · Γ(−α) 0≤`≤a Re(α)=0

R Since we have assumed 1 < a < 2 we may shift the line of integration to Re(α) = 2` − a in the above integral without crossing any poles. It then follows from Stirling’s asymptotic formula that for α = 2`−2a+ir TRACE FORMULAE FOR SL(2, R) 53 and T → ∞ we have the bound 1 2+R+2α 2+R−2α Z 2 2`−α 2 +2`−α (2) X α π y Γ 4 Γ 4 I (y) = e T 2 · dα T,R πi (1 − α)(2 − α) ··· (` − α) · Γ(−α) 0≤`≤a Re(α)=2`−2a T 1+ Z Γ 2+R+4`−2a+2ir Γ 2+R−4`+2a−2ir) 1 +a X 4 4 y 2 · dr (1 + |r|)` · · Γ(2a − 2` − ir) 0≤`≤a r=0 T 1+ Z R 1 +a X (1 + r) 2 y 2 · dr 2a−2`− 1 (1 + r)` · (1 + r) 2 0≤`≤a r=0 1 +a + R + 3 −a y 2 T 2 2

(1) (3) Thus we get the same bound as for IT,R(y). The estimation of IT,R(y) is similar and gives the same bound. This completes the proof of proposition 9.5. 9.4. Bounding the Geometric Side of the KTF.

We consider the Kuznetsov trace formula with test functions p = q = pT,R with pT,R given by (9.4). In this case, the geometric side is

(9.11) ∞ ∞ ∞ X Z Z m G = S(m, n; c) p p (ny) T,R c2y2 (x2 + 1) T,R c=1 0 −∞ m dxdy · exp −2πix + ny . c2y(x2 + 1) y

We will now use the pT,R bound in proposition 9.5:

1 +2a R + 3 −a (9.12) pT,R(y) y 2 T 2 2 , (for 1 ≤ a < R/4), together with the Kloosterman bound

1 + (9.13) S(m, n; c) c 2 . The idea is to break the y-integral in (9.11) as follows

∞ 1 ∞ Z Z Z = +

y=0 y=0 y=1 54 DORIAN GOLDFELD

m and then to use diﬀerent pT,R bounds for pT,R c2y2(x2+1) and pT,R(ny) which appear in (9.11). In particular, we will use

1 +2a R + 3 −a pT,R(y) y 2 T 2 2

m to estimate pT,R c2y2(x2+1) and we will use

1 +2b R + 3 −b pT,R(y) y 2 T 2 2 to estimate pT,R(ny). Accordingly, we ﬁrst consider the integral with 0 ≤ y ≤ 1. Then after applying the above Kloosterman bounds we have to estimate

1 ∞ ∞ " 1 +2a # Z Z 2 X 1 + m R + 3 −a c 2 T 2 2 c2y2 (x2 + 1) c=1 y=0 x=−∞ 1 +2b R + 3 −b dxdy · (ny) 2 T 2 2 . y

R Here we take a = 1 and b = 4 − 1, which guarantees that the c- sum and the x and y integrals converge. We then obtain the bound R+3− R 3R +3 O T 4 = O T 4 .

Next, we consider 1 ≤ y ≤ ∞. In this case, we must bound

∞ ∞ ∞ " 1 +2a # Z Z 2 X 1 + m R + 3 −a c 2 T 2 2 c2y2 (x2 + 1) c=1 y=1 x=−∞ 1 +2b R + 3 −b dxdy · (ny) 2 T 2 2 . y

R Here we take a = 4 −1 and b = 1, which guarantees that the c-sum and 3R +3 the x and y integrals converge. We again obtain the bound O T 4 .

We have now proved the following bound. TRACE FORMULAE FOR SL(2, R) 55 Proposition 9.14. (Bounding the geometric side of KTF) Let

∞ ∞ ∞ X Z Z m G = S(m, n; c) p p (ny) T,R c2y2 (x2 + 1) T,R c=1 0 −∞ m dxdy · exp −2πix + ny . c2y(x2 + 1) y donote the geometric side of the Kuznetsov trace formula. Then

3R +3 G = O T 4 , (T → ∞).

9.5. Computation of the Main Term M in the KTF. The main term M only contributes when m = n and is given by ∞ Z dy M = m |p (y)|2 T,R y2 0 To evaluate M we shall need the Plancherel formula which we now √ derive. For α ∈ C and y > 0, let Wα(y) := y Kα(2πy) denote the GL(2, R) Whittaker function. We have the Kuznetsov-Lebedev transform pair: ∞ Z dy p#(α) = p(y) · W (y) α y2 0 1 Z dα p(y) = p#(α) W (y) . πi α Γ(α)Γ(−α) Re(α)=0 As a consequence we can now derive the Plancherel formula for the GL(2) Whittaker transform Proposition 9.15. (Plancherel Formula for GL(2)) Let p : [0, ∞] → C be a smooth function satisfying ( y1+ if 0 < y ≤ 1, p(y) y−C if 1 < y. Then ∞ ∞ Z Z 2 dy 1 # 2 dr |p(y)| = p (ir) . y2 π |Γ(ir)|2 0 −∞ 56 DORIAN GOLDFELD

Proof.

∞ ∞   Z dy Z 1 Z dα dy |p(y)|2 = p(y) ·  p#(α) W (y)  y2  πi α Γ(α)Γ(−α) y2 0 0 Re(α)=0 ∞ 1 Z Z dy dα = p#(α) p(y) W (y) πi α y2 Γ(α)Γ(−α) Re(α)=0 0 Z 1 # 2 dα = p (α) . πi Γ(α)Γ(−α) Re(α)=0 We are now in a position to obtain an asymptotic formula for the main term in the KTF as T → +∞ when the test functions p, q are chosen to be the same and equal to the pT,R function given in (9.4). Applying the Plancherel formula of proposition 9.15, the main term M is given by ∞ Z dy M = m |p (y)|2 T,R y2 0 2 ∞ 2+R+2ir 2+R−2ir Γ 4 Γ 4 2m Z −2r2 = e T 2 · dr. π Γ(ir)Γ(−ir) 0 It follows from Stirling’s asymptotic formula |Γ(σ + ir)|2 = 2π · |r|2σ−1 e−π|r|, σ ﬁxed, |r| → +∞ that ∞ Z 2 2m −2r R+1 M ∼ e T 2 · (1 + |r|) dr π 0 ∞ 2m Z 2 ∼ T R+2 · e−2r · T −1 + |r|R+1 dr π 0 R+2 ∼ c0 · T for some constant c0 > 0. With more care one can prove R+2 R+1 R M ∼ c0T + c1T + c2T + ··· for ﬁxed real constants c1, c2, c3,... TRACE FORMULAE FOR SL(2, R) 57

9.6. Explicit KTF for GL(2, R) with Error Term. Theorem 9.16. Let R > 4 be ﬁxed and let 2 2 2 + R + 2α 2 + R − 2α p# (α) := eα /T · Γ Γ . T,R 4 4 P √ 2πi`x Let φj(z) = Aj(`) y Kirj (2π|`|y)e , (j = 1, 2,...) denote an `6=0 1 2 orthonormal basis of Maass cusp forms where ∆φi = 4 + ri φi for i ≥ 1. Let s s− 1 2π σ (m)m 2 A(m, s) := 1−2s Γ(s)ζ(2s) denote the mth arithmetic Fourier coeﬃcient of E(z, s). Then we have

∞ # 2 √ X p (irj) mn A (m)A (n) · j j hφ , φ i j=1 j j ∞ √ Z mn # 2 + A (m, 1/2 + ir) A (n, 1/2 + ir) · p (ir) dr 4π −∞

R+2 R+1 R 3R +3 = δm,n · c0T + c1T + c2T + ··· + O T 4 ,

where c0 > 0 and c1, c2,... are ﬁxed real constants.

9.7. Bounding the Eisenstein term in the KTF. The Eisenstein term is given by ∞ √ Z mn # 2 A (m, 1/2 + ir) A (n, 1/2 + ir) · p (ir) dr 4π −∞ ∞ 2 −2r 2+R+2ir 2+R−2ir 2 Z e T 2 Γ Γ 4 4 dr |Γ(1/2 + ir)ζ(1 + 2ir)|2 −∞ T 1+ Z (1 + |r|)R dr. |ζ(1 + 2ir)| 0 58 DORIAN GOLDFELD

Here we have used Stirling’s asymptotic formula for the Gamma funciton. To complete the estimation we require the prime number theorem which says that ζ(1 + 2ir) 6= 0 for any r ∈ R. A more quanti- tative version is given by ζ(1 + 2ir)−1 log r. We have thus shown that the Eisenstein term in the Kuznetsov trace formula is bounded by

∞ √ Z mn # 2 R+1+ A (m, 1/2 + ir) A (n, 1/2 + ir) · p (ir) dr T . 4π −∞ TRACE FORMULAE FOR SL(2, R) 59

10. The Kuznetsov Trace Formula for GL(3, R) 3 × Let h := GL(3, R)/(O(3, R) · R ) and let U3 be the subgroup of GL(3) consisting of upper triangular unipotent matrices. Every coset representative in h3 can be written as g = xy with the specific paramet- rization     1 x12 x13 y1y2 (10.1) x =  1 x23  , y =  y1  , 1 1 3 (where x12, x13, x23 ∈ R, y1, y2 > 0) to assign coordinates on h . Let D3 denote the invariant differential operators on h3, namely all polynomials (with complex coefficients) in the variables ∂ ∂ ∂ ∂ ∂ , , , , ∂x12 ∂x13 ∂x23 ∂y1 ∂y2 which are invariant under all GL(3, R) transformations. Definition 10.2. (Langlands parameters) We define the Langlands parameters for GL(3) to be complex numbers {α1, α2, α3} which satisfy α1 + α2 + α3 = 0. By abuse of notation we often refer to the Langlands parameters as a vector α = (α1, α2, α3). Remark. These parameters are used to classify automorphic representations at the archimedean place. An important role is played by the eigenfunctions I(∗, α).

Deﬁnition 10.3. (The eigenfunction I(∗, α)) Let α = (α1, α2, α3) denote Langlands parameters. We deﬁne a power function on xy ∈ h3 by

1−α3 1+α1 (10.4) I(xy, α) := y1 y2 . The function I(∗, α) is an eigenfunction of D3.

2 2 2 α1+α2+α3 Lemma 10.5. The Laplace eigenvalue of I(∗, α) is 1 − 2 . Proof. The Laplacian ∆ on h3 is the second order diﬀerential operator given by 2 2 2 2 2 ∂ 2 ∂ ∂ 2 2 2 ∂ ∆ = −y1 2 − y2 2 + y1y2 − y1 x1,2 + y2 2 ∂y1 ∂y2 ∂y1∂y2 ∂x1,3 2 2 2 2 ∂ 2 ∂ 2 ∂ − y1 2 − y2 2 − 2y1x1,2 . ∂x2,3 ∂x1,2 ∂x2,3∂x1,3 The lemma can then be easily proved with a brute force computation. 60 DORIAN GOLDFELD

In order to develop the Kunetsov trace formula on SL(3, R) we need to introduce the relevant Poincaréseries. Recall from definition 6.1 that the Poincaréseries for SL(2, R) is given by X pm · Im(γz) e2πim·Re(γz).

γ∈U2(Z)\SL(2,Z) We need analogues of the test function p and the additive character e2πim·Re(z) when we move to SL(3, R). 2 Definition 10.6. (Additive character) Let M = (m1, m2) ∈ Z . 3 We shall define an additive character ψM : h → C as follows. First 3 of all for xy ∈ h (with x, y as in (10.1)) we require ψM (xy) = ψM (x). This guarantees that ψM is a character of the non-abelian unitary group U3(R). 1 x1,2 x1,3 Further, for x = 0 1 x2,3 ∈ U3(R), we define 0 0 1 2πi m1 x1,2+m2 x2,3 ψM (x) := e ,

0 0 0 which satisfies ψM (x · x ) = ψM (x)ψM (x ) for all x, x ∈ U3(R). Definition 10.7. (Test function p) Let p : h3 → C be a smooth func- y1y2 0 0 tion satisfying p(xy) = p(y) for all x ∈ U3(R) and all y = 0 y1 0 0 0 1 with y1, y2 > 0. We further assume that p(y) satisfies

B1 B2 p(y) y1 y2 , ( 1 + b if 0 < yi ≤ 1 for Bi = for sufficiently large b > 1 and −1 − b if 1 < yi i = 1, 2. We may now define the Poincaréseries.

2 Definition 10.8. (PoincaréSeries) Let M = (m1, m2) ∈ Z with 3 m1m2 6= 0. Let p : h → C be a smooth test function as in definition m m 0 0 ∗ 1 2 3 10.7 and define M := 0 m1 0 . Then for g ∈ h , we define the 0 0 1 3 Poincaréseries PM (∗, p): h → C by

X ∗ PM (g, p) := p M · γg ψM (γg).

γ∈U3(Z)\SL(3,Z) Recall that a locally compact group G is termed Hausdorff provided every pair of distinct elements in G lie in disjoint open sets. The general linear group GL(n, R) is a locally compact Hausdorff topological group. TRACE FORMULAE FOR SL(2, R) 61 Theorem 10.9. Let G be a locally compact Hausdorff topological group, and let H be a compact closed subgroup of G. Let µ be a Haar measure R on G, and let ν be a Haar measure on H, normalized so that H dν(h) = 1. Then there exists a unique (up to scalar multiple) quotient measure µ˜ on G/H. Furthermore Z Z Z f(g) dµ(g) = f(gh) dν(h) dµ˜(gH), G G/H H for all integrable functions f : G → C. Proof. See Halmos ????? Definition 10.10. (Left invariant measure on h3 ) Let g = xy ∈ h3 with     1 x12 x13 y1y2 x =  1 x23  , y =  y1  1 1 The left invariant measure d×g on h3 is given explicitly by

× dy1dy2 d g = dxdy = dx1,2dx1,3dx2,3 2 . (y1y2) Here d×g satisﬁes d×(αg) = d×g for all α ∈ GL(3, R).

Proof. The group GL(3, R) is generated by U3(R), diagonal matrices in GL(3, R) and the The Weyl group of GL(3, R) is given by n 1 0 0 1 0 0 0 1 0 0 1 0 0 0 1 0 0 1 o W3 = 0 1 0 , 0 0 1 , 1 0 0 , 0 0 1 , 1 0 0 , 0 1 0 . 0 0 1 0 1 0 0 0 1 1 0 0 0 1 0 1 0 0 It is enough to show that d×g is invariant under these three subgroups. It is clear that d×(xg) = d×g if x is an upper triangular matrix with ones on the diagonal. This is because the measures dxi,j (with 1 ≤ i < j ≤ 3) are all invariant under translation. One also checks that d×(ag) = d×g for all diagonal matrices a ∈ GL(3, R). × It remains to check the invariance of d g under the Weyl group W3. Now, if w ∈ W3 and   a3 0 0 a =  0 a2 0  ∈ GL(3, R) 0 0 a1 is a diagonal matrix, then waw−1 is again a diagonal matrix whose diagonal entries are a permutation of {a1, a2, a3}. The Weyl group is generated by the two transpositions ω1, ω2 which interchange (trans- pose) a1 and ai+1 when a is conjugated by ωi. With a brute force × × calculation one checks that d (ωig) = d g for i = 1, 2. 62 DORIAN GOLDFELD

With this measure, one may show that the volume is Z 3ζ(3) d×g = . 2π SL(3,Z)\h3

For M = (m1, m2) and N = (n1, n2) let PM (g, p), QN (g, q) (with m1m2n1n2 6= 0) be two Poincar´eseries with smooth functions p, q : R → C as in deﬁnition 10.8. The Kuznetsov trace formula is obtained by computing the Petersson inner product Z D E × PM (∗, p), QN (∗, q) = Pm(g, p) Qn(g, q) d g

SL(3,Z)\h3 in two diﬀerent ways.

• First way, Spectral Side: Take the spectral expansion of PM (∗, p) and unravel QN (∗, q) in the inner product. • Second way, Geometric Side: Compute the Fourier expansion of PM (∗, p) by double coset decomposition and unravel QN (∗, q).

10.1. Whittaker functions and Maass forms on h3. We now deﬁne the canonically normalized Whittaker function on GL(3, R) which appears in the Fourier expansion of automorphic forms for SL(3, Z).

0 0 1 3 Deﬁnition 10.11. Let w3 := 0 1 0 and α = (α1, α2, α3) ∈ . Then 1 0 0 C for g ∈ GL(3, R), the function Γ 1+αj −αk Y 2 Z W ±1(g) := · I(w ug, α) ψ (u) du α 1+αj −αk 3 1,1,±1 1≤j 0 and Re(α2 − α3) > 0 extends to a holomorphic function on the set of all 3 α ∈ C α1 + α2 + α3 = 0 . Remark. The integral is Jacquet’s integral representation. The prod- ± uct of Gamma factors is included so that Wα is invariant under all permutations of α1, α2, α3. Moreover, even though Jacquet’s integral often vanishes identically as a function of α, this normalization never ± does. If g is a diagonal matrix in GL(n, R) then the value of Wα (g) is TRACE FORMULAE FOR SL(2, R) 63 independent of sign, so we drop the ±. We also drop the ± if the sign is +1. Jacquet’s Whittaker function is characterized (up to scalar multiples depending on α) by the following properties:

± ± 3 • δWα (g) = λδ(α) · Wα (g), for all δ ∈ D , g ∈ GL(3, R) , ± ± • Wα (ugzk) = ψ(u)Wα (g), for all u ∈ U3(R), g ∈ GL(3, R), ? z ∈ R , k ∈ O3(R) , N ± • (y1y2y3) Wα (diag y) = O(1), for any N > 0 and yi ≥ 1 , ± • Wα is entire in α, ± ± 0 • Wα = Wα0 where α is any permutation of α = (α1, α2, α3).

Deﬁnition 10.12. (Maass cusp form for SL(3, Z)) A Maass cusp form φ : SL(3, Z)\h3 → C for SL(3, Z) is a joint eigenfunction of D3 which satisﬁes a bound of the form

−N |φ(xy)| N (y1y2y3) in the range y1, y2, y3 ≥ 1, for any N > 0. If the eigenvalues of φ agree with those of I(∗, α) then we term α the Langlands parameters of φ. The Maass form is termed a Hecke eigenform if it is a simultaneous eigenfunction of all the Hecke operators. Theorem 10.13. (Fourier expansion of Maass forms) Let φ : h3 → C be a Maass cusp form for SL(3, Z) with Langlands parameters 3 α = (α1, α2, α3) ∈ C . Then for g ∈ GL(3, R) ∞ A (M) X X X φ sgn(m2) ∗ γ 0 φ(g) = Wα M g , m1|m2| 0 1 γ∈U2(Z)\SL2(Z) m1=1 m26=0

|m m | ∗ 1 2 where M = (m1, m2) and M = |m1| with m1m2 6= 0.. 1

11. The SL(3, R) Kuznetsov Trace Formula (Spectral Side) By the Selberg spectral decomposition for GL(3, R) ﬁrst proved by Langlands ??? we have

∞ X D E φj(z) n o P (g, p) = P (∗, p), φ + Eisenstein contribution . M M j hφ , φ i j=0 j j 64 DORIAN GOLDFELD

Lemma 11.1. (Inner product of PM with a Maass form φj) Let M = (m1, m2) with m1m2 6= 0. Then

0 if j = 0, D E  ∞ ∞ PM (∗, p), φj = Aj (M) R R ∗ ∗ dy1dy2 W (j) (M y) · p(M y) 3 if j > 0.  m1m2 α (y1y2) 0 0

Proof. Let Γ = SL(3, Z). We compute Z D E X ∗ × PM (∗, p), φj = p M · γg ψM (γg) · φj(g) d g

Γ\h3 γ∈U3(Z)\Γ Z ∗ × = p M y ψM (x) φj(g) d g

3 U3(Z)\h ∞ ∞  1 1 1  Z Z Z Z Z ∗ dy1dy2 =  ψM (x)φj(xy) dx · p(M y) 3 (y1y2) 0 0 0 0 0 | {z } This integral = 0 if j = 0. ∞ ∞ Z Z Aj(M) ∗ ∗ dy1dy2 = Wα(j) (M y) · p(M y) 3 m1m2 (y1y2) 0 0 | {z } Here we assume j 6= 0.

# (j) = m1m2 Aj(M) · p α

∞ D E X PM (∗, p), φj φj, QN (∗, q) P (∗, p), Q (∗, q) = M N hφ , φ i j=1 j j n o + Eisenstein Contribution

∞ # (j) X p α q# (α(j)) = m m n n A (M)A (N) 1 2 1 2 j j φ , φ j=1 j j n o + Eisenstein Contribution . TRACE FORMULAE FOR SL(2, R) 65 11.1. Eisenstein contribution.

There are two types of Eisenstein series for GL(3, R), corresponding to whether one takes a minimal or maximal parabolic subgroup. Let

 ∗ ∗ ∗   ∗ ∗ ∗  B =  0 ∗ ∗  and P2,1 =  ∗ ∗ ∗  0 0 ∗ 0 0 ∗ denote the Borel and a specific maximal parabolic, respectively. Any parabolic subgroup of GL(3) is conjugate to a standard parabolic, (i.e., one containing B), and the only other standard parabolics of G are P1,2 (defined analogously to P2,1, but instead with zero entries in the bottom two entries of the first column) and the full group G itself.

Deﬁnition 11.2. (Borel Eisenstein series) The Borel Eisenstein series for G = GL(3, R) is deﬁned for Re(α1 −α2) and Re(α2 −α3) > 1 by the absolutely convergent sum

X 3 EB(g, α) := I(γg, α), (g ∈ h ),

γ∈U3(Z)\SL(3,Z) and for general α by meromorphic continuation.

Deﬁnition 11.3. (Induced Maass form Φ associated to P2,1) Let

 ∗ ∗ ∗  1 0 ∗ ∗ ∗ 0 P2,1 =  ∗ ∗ ∗  = 0 1 ∗ · ∗ ∗ 0 0 0 ∗ 0 0 1 0 0 ∗ | {z } | {z } N P2,1 M P2,1

Let φ : GL(2, R) → C be a Maass cusp form left-invariant under SL(2, Z). Let K = O(3, R). The Maass form Φ is then deﬁned on GL(3, R) = P2,1(R)K by the formula

a b P2,1 P2,1 Φ(nmk) := φ ( c d ) , (n ∈ N , m ∈ M , k ∈ K),

P a b 0 where m ∈ M 2,1 has the form m = c d 0 . 0 0 ∗ 66 DORIAN GOLDFELD

Definition 11.4. (Langlands Eisenstein series twisted by Φ) Consider the parabolic subgroup P2,1 with induced Maass form Φ as in the above definition. Let s = (1/2 + s1, −1 − 2s1) with s1 ∈ C and let α = (s1 − 1/2, s1 + 1/2, −2s1). The Langlands Eisenstein series determined by this data is defined by X (11.5) EP2,1,Φ(g, α) := Φ(γg) · I(γg, α)

γ ∈ (P2,1∩Γ)\Γ as an absolutely convergent sum for Re(s1) suﬃciently large, and it extends to s1 ∈ C by meromorphic continuation. In particular, if E is any one of the two Eisenstein series deﬁned above, then for M = (m1, m2) as above, we have Z AE(M, α) ∗ E(ug, s)ψM (u) du = Wα(M g), |m1m2| U3(Z)\U3(R)

th where Wα is a Whittaker function and AE(M, α) is termed the M arithmetic Fourier coefficient of E. The first coefficient of E is defined to be AE((1, 1), α). Furthermore, we have

AE(M, α) = AE((1, 1), α) · λE(M, α).

th where λE(M, α) is the M Hecke eigenvalue of E and λE((1, 1), α) = 1. Theorem 11.6. 3 (1) Let α = (α1, α2, α3) ∈ C with α1 + α2 + α3 = 0. Then the

ﬁrst coeﬃcient AEB ((1, 1), α) of the GL(3, R) Borel Eisenstein series is equal (up to a nonzero constant) to

−1 ∗ ∗ ∗ (11.7) ζ 1 + α1 − α2 ζ 1 + α2 − α3 ζ 1 + α1 − α3 .

1 (2) Let s = 2 + s1, −1 − 2s1 , α = (s1 − 1/2, s1 + 1/2, −2s1), and let φ be a Maass form for SL(2, Z) with Petersson norm one. Then the ﬁrst coeﬃcient A ((1, 1), α) of the GL(3, ) Eisenstein series EP2,1,Φ R

EP2,1,Φ(g, α) induced from φ is equal to

−1 ∗ 1/2 ∗ (11.8) L (1, Ad φ) · L (φ, 1 + 3s1) times an explicit non zero constant which is independent of φ. We can now state the Eisenstein contribution to the Langlands spectral decomposition of the Poincar´eseries. TRACE FORMULAE FOR SL(2, R) 67

Proposition 11.9. Let φj, j = 1, 2,... denote an orthonormal basis of Maass cusp forms for SL(3, Z). Then the Eisenstein contribution in the spectral decomposition of PM (∗, p) is given by 1 Z Z D E P (∗, p),E ∗, α E g, α dα dα (4πi)2 M B B 1 2 1 1 Re(α1)= 2 Re(α2)= 2 ∞ 1 X Z D E + P (∗, p),E ∗, α E g, α dα. 2πi M j j j=0 1 Re(α)= 2 We can now state the spectral side of the Kuznetsov trace formula for GL(3, R). Proposition 11.10. (Kuznetsov trace formula, Spectral Side) Let PM (z, p), QN (z, q) be Poincar´eseries. Deﬁne the Bessel transforms:

∞ ∞ Z Z # dy1dy2 p (α) := Wα (y) · p(y) 3 (y1y2) 0 0 ∞ ∞ Z Z # dy1dy2 q (α) := Wα (y) · q(y) 3 . (y1y2) 0 0 Then the spectral side of the Kuznetsov trace formula is

∞ # (j) D E X p α q# (α(j)) P (∗, p), Q (∗, q) = m m n n A (M)A (N) M N 1 2 1 2 j j φ , φ j=1 j j

1 Z Z p#(α) q#(α) + dα dα 2 ∗ ∗ ∗ 1 2 (4πi) ζ 1 + α1 − α2 ζ 1 + α2 − α3 ζ 1 + α1 − α3 1 1 Re(α1)= 2 Re(α2)= 2

∞ Z # # 1 P p (α) q (α) + 2πi ∗ 1/2 ∗ dα. j=0 L (1, Ad φ) · L (φ, 1 + 3s1) 1 Re(α)= 2

Acknowledgments: 68 DORIAN GOLDFELD

I would like to thank Kevin Kwan and David Soudry for carefully reading these notes and spotting many typos and errors which are now corrected.

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