Spectral analysis for Γ\H. Erez Lapid

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1 Spectral analysis for Γ\H Erez Lapid Spectral decomposition, hyperbolic lattice point problem(march, 9) Recall Bessel s inequality (u, e j ) u, j H is a Hilbert space, and{e j } is the orthogonal system. In our case, we consider functions in L (Γ\H),{e j } are orthogonal eigenfunctions of. We take kk(u(z, w)) C c (R>), k h is the Selberg transform. We define K(z, w) k(z,γw). Fix w and take u(z K(z, w)). If ( +λ j )ϕ j, whereλ j s j ( s j ), s j + it j, then Also, γ Γ (u,ϕ j ) ( K(, w),ϕ j ) ( k(u(, w),ϕ j )H h(t j)ϕ j (w). (u, The Bessel s inequality implies h(t j ) ϕ j (w) + j E(w; + ir)dr) h(r)e(w; + ir) dr. B We localize k, such that kχ [,δ],δ. We have K K(z, w) dµ(z) A Γ\H h(r) E(w; + ir) dr K(, w) Γ\H. If u(z, w)<δ and u(z,γw)<δ, then we have Then, by polar coordinates (.) k(u(γ z, w))k(u(γγ, w))dµ(z) Γ\Hγ,γ k(u(z, w))k(u(z, γw))dµ(z). (.) γ H u(w,γw)<4δ(δ+) 8δ. {z: u(z,w)<δ} vol{z : u(z, w)<δ} δ. {r : u(w,γw) 8δ}dµ(z) Lemma.δ h(t) δ, if t δ.

2 Proof. We have And h(t) k(u(z, i))y s dµ(z) H B δ {z : u(z,i)<δ} y s dµ(z). if t δ. Thus, Hence, From Bessel s inequality, we get If we takeδ T u(z, i)<δ y < δ y s δ y, y s dµ(z) vol(b δ ) B δ vol(b δ) δ ( u j (w) )+ t j 8δ vol(b δ) h(t) vol(b δ ). 8δ and T, we get weak local Weyl law t j <T u j (z) + T One can take it uniformly in z, T + TImz for Imz. Remark. By integrating over z, one gets T 8δ E(w; + ir) dr ) δ. E(w; + ir) dr T. {t j < T} t. Corollary. If k is any point-pair invariant function, K is the automorphic kernel, Suppose that K(z, w) where H is monotone, then j K(z, w) h(t j )u j (z)u j (w)+ <s j + h(t) H(t), t R, h(r)e(z; + ir)e(w; + ir)dr. h(t j )u j (z)u j (w)+o ( (H(t)+)t)dt ). Proof. Use partial summation or integral.

3 Hyperbolic lattice point problem: We study where z, w are fixed. All alternatively, {γ Γ : d(γz, w)<r)} as R, {γ Γ : u(γz, w)< X)}: P(X). (Recall that, inr,γz, this is the Gauss circle problem, {(a, b) : a + b < R}πR+O( R)), {(a, b) : a b < R} τ(n) area of a truncated hyperbolic, n<r {(a, b) : a b n} τ(n), whereτ(n) is the number of divisors. Our goal is to get an error term for P(X) X, P(X)K(z, w) for kχ [,X]. The main term should be the exceptional eigenvalue < s j. For that, we bound for h. Using k, where Y is a parameter. Recall P(X) K(z, w) P(X+ Y), q(v) v k(u) u v du, g(r)q(e r + e r ), h(t) + For our k, for any f, by partial integration we get e irt g(r)dr. where F(t) t f (x)dx. For f (u)k(u)du Y X+Y X F(t)dt, f (x) { x v, x>,, otherwise, we have F(t) min(, t v)q(v) Y X+Y X F(t)dt Y [ min(, (X+ Y v) 3/ ) min(, (X v) 3/ ) ], g (r)q (e r + e r )(e r e r ), 3

4 and We have Trivially, we get q (v) [ ] min(, (X+ Y v) min(, (X v) Y q () Y. h(t) e irt g(r)dr e irt g (r)dr. t h(t) g (r) dr t t total variation of g t max g Y So of extreme of g 5. IfΓS L (Z), t R, u For h( i ) k(u)du(x+ Y). P(X) area(γ\h). Consequently, if Y X /3 O(X /3 ), {( a b c d ) t X Y t X Y. area(γ\h) X+ O( Y+ X Y ), } S L (Z) : a + b + c + d X 3X+ O(X /3 ). {( ) a b Γ c d } SL (Z) : a d(), b c () which is conjugate toγ (), r(n) { (a, b) : na + b }, we have r(n)r(n+)4x+ O(X /3 ). n X We consider functions onγ\h, wherehs L (R)/K. We haveγ\hγ\s L (R)/K and have the following relations : functions onγ\hfunctions onγ\s L (R)/K leftγ-invariant and right K-invariant functions on GS L (R) Look at C(Γ\G) {left Γ-invariant functions on G} G acts on Γ\H by right translation. Therefore G acts on C(Γ\G). Take g G, f C(Γ\G) [R(g) f ](x) f (xg) 4

5 R(g) : C(Γ\G) R : G Aut(C(Γ\G)) is homomorphism of groups, i. e. (R, C(Γ\G)) is a representation of G. Consider L (Γ\G, dg), dg is Haar measure. R : G U(L (Γ\G)),Uis the group of unitary operators. Main problem: Decompose R into irreducible representation. C(Γ\G) K {v C(Γ\G) : R(k) v v, k K}, L (Γ\H)(L (Γ\G) K. One can classify irreducible unitary representation of G. For any irreducible representation (π, v) of G ( i.e. no closed G-invariant subspaces of V, or equivalently, no non-scalar operators which commute with π(g), g G). is at most one-dimensional. In the case whereγis uniform (co-compact). V K {v V;π(k)vv, k K} L (Γ\G) π i, this is the direct sum of Irreducible representation. Therefore L (Γ\H)(L (Γ\G) K π K i. (m(π)<, where m(π) multiplicity ofπin L (Γ\G)). preserves the isotypic compact ofπ. Eigenfunction of irreducible representationπwhich occurs in L (Γ\G) s. t.π K. dim{φ : ( +λ)φ}m(π), λ π λ. whereλ 4 +s, and s R [ i, i ]. This is a parametrization of irreducible unitary representation, such thatπ K. Selberg s conjecture: λ (Γ(N)) 4, s i R, i>. Representation withλ (i. e. s R) have decaying matrix coefficients. If (π, V) is an irreducible 4 admissible representation, a matric coefficient ofπis a function of the form where v V, and g matrix coefficient (π(g)v, v ), G gk ( a a ) k. f (g) +ε dg<, ε>. What about other representations which occur in L (Γ\G) modular forms+others. 5

6 Discrete series representations: for any k,,... Denoteσ k is the matrix coefficient in L (G), we have L (G) k σ k π 4 +s ds, (c.f. L (R) R eix dx) Denote m(σ k ) dimension of modular forms of weight k+, we have weight limit of discrete series. ForΓS L (z), Hecke operators :{T n } n Z. Recall strong approximation: S L (A)S L (Q) S L (R) S L (Z p ) s ir p< S L (Q)\S L (A) S L (Z)\S L (R) p< S L (Q)\S L (A)/ K p S L (Z)\S L (R)/K Γ()\H p S L (Q)\S L (A)/K K p (N) Γ(N)\S L R/K Γ(N)\H p< where K p (N) is congruence subgroup of K p. This is to say S L (Q)\S L (A)/K lim Γ(N)\H C (S L (Q)\S L (A)) K lim L (Γ(N)\H) The new problem: decomposition of L (S L (Q)\S L (A)). First part: what is the continuous spectrum? Answer: explicit construction using Eisenstein series. What about discrete part? The analogue of Selberg conjecture in the principal becomes a bound on the Fourier transform of Maass forms. Ramannujan-Petersson conjecture : a p. The bound was proved by Delign. Modular forms: a p a p, trivial bound: p k Let λ 4 +δ a p p. (s +δ), K p So for the best knownδis e. g. a p p δ + p δ. P(X) cx s j + O(X 3 ). c j< 6

7 Kim-Shahidi : Forδ, 6 π P(X) area(γ\h) X+ O(X 3 ). Higher weight case: G(Q)\G(A), G is linear reductive group. What is the decomposition into irreducible representation?γ\g/k ring of invariant differential operators. what is the continuous spectrum?(langlands). what is the discrete part? There is an analogue of Ramanujan conjecture(by Arthur). Langlands functoriality suggests a relation between the automorphic spectrum of L (G i (Q)\G i (A)) for many pairs (G i, G j ) of reductive groups. 7