Nonvanishing of geodesic periods over compact hyperbolic manifolds
aa r X i v : . [ m a t h . N T ] M a y Nonvanishing of geodesic periods over compacthyperbolic manifolds
Feng Su
Abstract.
Let X be a compact hyperbolic manifold with dimension d > n -dimensional ( n >
2) geodesic cycle of X . Let X be a compact hyperbolic manifold with dimension d > dx , { φ j } be an orthonor-mal basis of L ( X, dx ) such that each φ j is a Laplace eigenfunction: ∆ φ j = λ j φ j where λ j ∈ R > and ∆stands for the Laplace operator of X determined by its hyperbolic metric. In the theory of automorphicforms, φ j ’s are called (normalized) “Maass forms” (after H. Maass). The order of φ j ’s are arranged suchthat λ j ’s are nondecreasing as j grows. Let Y be a compact geodesic cycle of X with dimension n > dy (see Sect. 2.1), ψ be a normalized Maass form on Y with Laplace eigenvalue λ . Define the geodesic period ( period for short) as P Y ( φ j , ψ ) := Z Y φ j ( y ) ψ ( y ) dy. Such period fits into the general notion of automorphic period which has been playing a central role inthe study of automorphic forms thanks to its close relations with automorphic representations and special L -values (see [ II ], [ Wa ], [ Zh ], etc.). With notations and restrictions as above, the main conclusion ofthis paper is Theorem 1.
For any fixed ψ , there are infinitely many j such that P Y ( φ j , ψ ) = 0 . In literature the nonvanishing of (infinitely many) geodesic periods can follow from the asymptotics ofthese periods. See [ He ], [ G ], [ Ze ], [ P ], [ Ts ] and [ MW ]. In particular, [ Ts ] dealt with a class of arithmetichyperbolic manifolds and the periods defined over the codimension-1 geodesic cycles; the case where X is an arbitrary compact Riemann surface was treated in [ MW ] which has inspired us to work on higherdimensional situation.In the language of automorphic representations, Theorem 1 says that, under certain restrictions onΓ, there are infinitely many G -distinguished spherical automorphic representations where G ⊂ G = SO (1 , d ) is a closed subgroup which descends to Y (see Sect. 2.1).The content of this paper is organized as follows. In the next section we shall make some preparationson the necessary knowledge on hyperbolic space, representation theory and trace formula. These stuffwill be used later. In Sect. 3 we shall insert a special test function into the trace formula and discuss thespectral side of the trace formula. The analysis of the geometric side will be given in Sect. 4 where wesplit this side into two parts: the main and error terms; in particular, the contributions from these twoterms will be estimated. Theorem 1 then follows from the comparison of the spectral and geometric sides.The necessary conditions that the test function should fulfill (so that we can apply the trace formula)will be checked in Sect. 5. By the uniformization theorem, any d -dimensional orientable hyperbolic manifold X with finite volumecan be realized as a locally symmetric space: X ∼ = Γ \ G/K where G is the Lorentz group SO (1 , d ),Γ ∼ = π ( X ) is a torsion-free lattice of G and K ∼ = SO ( d ) is a maximal compact subgroup of G . Withoutloss of generality we assume that K = { diag(1 , k ) | k ∈ SO ( d ) } . The quotient space G/K is isomorphicto the hyperboloid model H d = n x = ( x , . . . , x d ) ∈ R d +1 (cid:12)(cid:12)(cid:12) x − X i =0 x i = 1 , x > o . Let Y be a compact geodesic cycle of X with dimension n and hyperbolic measure dy . Then, up to finitecover, Y is isomorphic to Γ \ G /K where G = { diag( τ , τ ) ∈ G | τ ∈ O (1 , n ) , τ ∈ O ( d − n ) } ,K = { diag( ρ , ρ ) ∈ K | ρ ∈ O ( n ) , ρ ∈ O ( d − n ) } is the maximal compact subgroup of G , and Γ ∼ = π ( Y ) is a torsion-free uniform lattice of G .An automorphic representation π of G is called G -distinguished if there exists φ ∈ π such that R Γ \ G φ ( x ) dx = 0. The notion of “distinguished representation” is used in the setting of groups overadeles. Nevertheless, we adopt this notion since this paper applies to (uniform) real arithmetic lattices. Inthis prospect, Theorem 1 can be rephrased as follows: there are infinitely many (spherical) automorphicrepresentations which are G -distinguished, provided that Γ is uniform. Let Θ be the Cartan involution on G : Θ( g ) = (cid:0) g T (cid:1) − (transpose inverse). The Cartan involution θ onLie algebra level (i.e., θ ( X ) = − X T ) gives rise to the vector space decomposition of the Lie algebra g of G : g = p ⊕ k where k = { X ∈ g | θ ( X ) = X } is a sub-Lie algebra of g and p = { X ∈ g | θ ( X ) = − X } . Let a be a maximal abelian subspace of p . For each linear functional α on a , define g α = { X ∈ g | [ H, X ] = α ( H ) X for all H ∈ a } . The set of those nonzero α such that g α = 0 is a root system,denoted as ( g , a ). Let E ij = ( e ij ) be a ( d +1) × ( d +1) matrix whose entries satisfy: e lk = 1 for ( l, k ) = ( i, j ), e ik = 0 otherwise. We choose a = R E where E = E + E . Then the root system ( g , a ) consists of twoelements ± α where α (the positive root) is defined by ad( E ). Let E i = E , i + E , i + E i, − E i, (1 i d − E ) E i = E i for each i . Hence, g α = n := R E ⊕ · · · ⊕ R E d − . Define the groups A , N , A + to be A = exp( a ), N = exp( n ), A + = exp ( R + E ) ⊂ A where R + means the set of positivereal numbers. We have the Iwasawa decomposition G = N AK and
KAK -decomposition G = KA + K both of which are unique (note that we have restricted the middle component A ( g ) of g in the KAK -decomposition to lie in A + ).Denote a + x = exp( xE ) , n u = exp d − X i =1 u i E i ! for x ∈ R , u = ( u , · · · , u d − ) ∈ R d − . It is easy to verify that a + x = cosh x sinh x · · · x cosh x · · ·
00 0 1 0 · · · · · · and n u = | u | − | u | u u · · · u d − | u | − | u | u u · · · u d − u − u · · · u − u · · · u d − − u d − · · · where | u | = P d − i =1 u i .The Killing form B ( X, Y ) = Tr (cid:0) ad( X )ad( Y ) (cid:1) on g , when restricted to p , induces a G -invariantRiemannian metric on G/K with which we have the distance between the two points g · o , e · o on G/K :d G/K ( g · o, e · o ) = B (cid:0) log A ( g ) , log A ( g ) (cid:1) / =: k g k . This invariant Riemannian metric on
G/K induces the metric and measure (denoted as µ ′ ) on Γ \ G/K .Let dk be a Haar measure of K . Throughout the paper we always assume that vol( K ) = R K dk = 1.Any Haar measure of G projects to a Radon measure µ of Γ \ G and the latter, up to a positive scalar,projects to the measure µ ′ on Γ \ G/K such that the quotient integral formula holds: R Γ \ G f ( x ) µ ( x ) = R Γ \ G/K f ( xk ) µ ′ ( x ) dk for any f ∈ C c (Γ \ G ).The group G acts on L (Γ \ G, µ ) via the right regular translation R : R ( f )( x ) = f ( xg ) for f ∈ L (Γ \ G, µ ), x ∈ Γ \ G . The Casimir operator (cid:3) acts on the dense subset of smooth functions of L (Γ \ G, µ )as a symmetric operator, and it has a unique self-adjoint extension to L (Γ \ G, µ ); the similar conclusionholds for ∆ and L (Γ \ G/K, µ ′ ) (see [ Ch ]). Any element in L (Γ \ G/K, µ ′ ) can be viewed as an elementin L (Γ \ G, µ ) that is K -invariant under the action R . When restricted to L (Γ \ G ) K , the two operators (cid:3) and ∆ are identical with each other.Assume that Γ is uniform. Then R is decomposed into irreducible representations (see Theorem9.2.2 of [ DE ]): R ∼ = M π ∈ b G N Γ ( π ) π (1)where b G denotes the unitary dual of G , i.e., the set of equivalent classes of unitary irreducible represen-tations of G , N Γ ( π ) < ∞ denotes the multiplicity of π . Hence L (Γ \ G/K ) ∼ = M π ∈ b G K N Γ ( π ) V Kπ (2)where b G K means the subset of b G whose element π satisfies the condition V Kπ = { } . Such π ’s arecalled spherical or class one representations. Here we use V π to denote the representation space of π .Let M ⊂ K denote the centralizer of A in K . Then M = { diag(1 , , k ) | k ∈ SO ( d − } . As a is ofdimension one, we can identify a ∗ C with C via the map ι : a ∗ C → C , α d − α ( E ). Let ρ be the halfsum of positive roots of the root system ( g , a ), then ι ( ρ ) = d − . From now on we shall not distinguish a ∗ C and C . It is known that any nontrivial irreducible spherical representation of G is equivalent to I ( ν ) = Ind GMAN ( ⊗ e ν ⊗ ) for some ν ∈ ( − ρ, ρ ) ∪ i R , and I ( ν ) ∼ = I ( − ν ) for such ν . The trivialrepresentation is isomorphic to the Langlands quotient of I ( ρ ) modulo its unique subrepresentation. Let ν ∈ a ∗ C and ( σ, V σ ) be a representation of M . Recall thatInd GMAN ( σ ⊗ e ν ⊗ ) = (cid:26) h : G → V σ (cid:12)(cid:12)(cid:12) h ( mang ) = e ( ν + ρ ) log a σ ( m ) h ( g ) for man ∈ M AN, g ∈ G ; h | K ∈ L ( K, V σ ) (cid:27) endowed with the action R of G : R ( g ) h ( x ) = h ( xg ) . Let { φ j } be an orthonormal basis of L (Γ \ G/K, µ ′ ) such that each φ j is a Maass form with Laplaceeigenvalue λ j = ρ − ν j . Denote by ˜ φ i ∈ L (Γ \ G ) the natural lift of φ j such that ˜ φ i ( xk ) = φ j ( x · o ) forany x ∈ Γ \ G , k ∈ K . Under the action R of G , ˜ φ i generates an irreducible unitary subrepresentation V λ j ⊂ L (Γ \ G ) of G . We have V λ j ∼ = I ( ν j ).Let G = M A N K be the Langlands decomposition of the group G where M = M ∩ G , A = A ∩ G = A , N = N ∩ G . Let n be the Lie algebra of N . The half sum of positive roots of thesystem ( g , a ) is ρ = n − . Identifying p = a ⊕ n with the tangent space of G /K at e · o , any geodesicon G /K can be translated by certain g ∈ G (via the left multiplication) to be a new geodesic whichpasses through e · o and is written as { exp( tX ) | t ∈ R } with proper X ∈ p (direction of the geodesic). ByProposition 5.13 of [ Kn ]: p = [ k ∈ K Ad( k ) a , there exists k ∈ K such that Ad( k ) X ∈ a . Taking proper conjugation if necessary, we may assume thatthere exists a closed geodesic C on Γ \ G /K which can be written as C = { exp( tX ) · o | t ∈ [0 , } withsome X ∈ a , or equivalently, C = Γ \ A · o where Γ = Γ ∩ AM = Γ ∩ AM (the second identity holdssince Γ is torsion-free). By Iwasawa decomposition, the subgroup
N A ⊂ G is topologically isomorphic to H d ∼ = G/K . We realizethis isomorphisim as S : N × A → H d , ( n, a ) S ( na ) = na · ξ where ξ = (1 , , · · · , ∈ H d . For r >
0, define a r := a +log r = exp(log r E ). There is a one-to-onecorrespondence between R d − × R + and N A : T : R d − × R + → N A, ( u, r ) n u a r . Let x = ( u, r ), y = ( v, t ) ∈ R d − × R + . The hyperbolic distance between two points a = S ◦ T ( x ), b = S ◦ T ( y ) on H d is d H d ( a, b ) = arccosh + (cid:20) | u − v | + r + t rt (cid:21) . (3)Here we use + to denote the nonnegative branch of the double valued function arccosh. The relationbetween d H d ( · , · ) and d G/K ( · , · ) is given byd H d ( a, b ) = d G/K (cid:0) T ( x ) · o, T ( y ) · o (cid:1) = (cid:13)(cid:13) T ( y ) − · T ( x ) (cid:13)(cid:13) . (4)For this fact, see Proposition I.7.3 and I.7.5 of [ FJ ]. Let U be a subset of G , f be a continuous function on G . Define f U : G → R > , g sup x, y ∈ U (cid:12)(cid:12) f ( xgy ) (cid:12)(cid:12) . We say f is uniformly integrable if there exists some compact neighborhood U of the unity e such that f U lies in L ( G ). Denote by C unif ( G ) the set of all continuous uniformly integrable functions over G . Given f ∈ C unif ( G ), define ( R ( f ) φ )( x ) = Z G f ( g ) R ( g ) φ ( x ) dg for φ ∈ L (Γ \ G ). Then R ( f ) is an integral operator by Lemma 1. ( R ( f ) φ )( x ) = Z Γ \ G K f ( x, y ) φ ( y ) µ ( y ) , where K f ( x, y ) = P γ ∈ Γ f ( x − γy ) is continuous on Γ \ G × Γ \ G . For details about C unif ( G ) and the proof of this lemma, see Sect. 9.2 of [ DE ]. The assumption in thereference, that H is uniform, is necessary for the decomposition (1), but not for this lemma.Let f be a bi- K -invariant function in C unif ( G ). Then R ( f ) acts on V Kλ j ⊂ L (Γ \ G ) K with the integralkernel K f since R ( f ) φ is still K -invariant for any φ ∈ V Kλ j . The space I ( ν j ) K is one-dimensional: any K -fixed function in I ( ν j ) is determined by its values at the points in P = M AN thanks to the Langlandsdecomposition G = M AN K and the the transformation law in I ( ν j ). Consequently, V Kλ j = C φ andthere exists a scalar h f ( ν j ) such that R ( f ) φ = h f ( ν j ) φ . In view of the bi- K -invariance of f and thedefinition of K f , we may identify L (Γ \ G ) K with L (Γ \ G/K ), and regard K f ( x, y ) as a function overΓ \ G/K × Γ \ G/K . Then R ( f ) induces the action of f on L (Γ \ G/K ). Likewise, R ( f ) acts on I ( ν j ):( R ( f ) h )( x ) = Z G f ( g ) R ( g ) h ( x ) dg, h ∈ I ( ν j )with integral kernel K f as above. Furthermore, we have R ( f ) η = h f ( ν j ) η for any nontrivial element η in I ( ν j ) K .To compute h f ( ν j ), we use the model I ( ν j ) and the action R ( f ). Let η ν j ∈ I ( ν j ) K be a complex-valued function over G defined as η ν j ( mank ) = e ( ν j + ρ ) log a . Since η ν j (1) = 1, it follows that( R ( f ) η ν j )(1) = h f ( ν j ) η ν j (1) = h f ( ν j ) . By definition, ( R ( f ) η ν j )(1) = Z G f ( g ) η ν j ( g ) dg ( a ) = Z A Z N Z K f ( ank ) η ν j ( ank ) dadndk = Z N Z A f ( an ) e ( ν j + ρ ) log a dadn (5)In the step (a) we have used the integral formula of functions on G where the variable is written in the AN K -order (see Corollary 5.3 of [ H ]). Now we choose the Haar measures on A and N . Let a = e X , n = e Y for X ∈ a , Y ∈ n . Since A and N are abelian groups, da := dX , dn = dY are Haar measureson A , N respectively, where dX , dY are Lebesgue measures on the Euclidean spaces a , n . Such choice ofmeasures holds for any semisimple groups. The reason is that the group N is nilpotent, while Lebesguemeasures on its Lie algebras induce Haar measures of N (see Theorem 2.1 of [ CG ]). To be more precise,we have: R N f ( n ) dn = R n f (exp Y ) dY for any f ∈ L ( N, dn ). Now (5) reads h f ( ν j ) = Z n Z a f (cid:0) e X · e Y (cid:1) e ν j ( X )+ ρ ( X ) dXdY. (6)We call h f ( ν j ) the Harish-Chandra – Selberg transform of f . The above formulation on h f ( ν j ) is due toSelberg [ Se ]. One can also use Harish-Chandra’s theory on spherical functions to describe h f ( ν j ).From now on we shall use h f ( λ j ) instead of h f ( ν j ). This is reasonable: as ν j is decided up to ± λ j , and I ( ν j ) ∼ = I ( − ν j ) for ν j ∈ ( − ρ, ρ ) ∪ i R , so we have h f ( ν j ) = h f ( − ν j ). Assume that f ∈ C unif ( G ) is a bi- K -invariant function such that the series k f ( z, w ) := ∞ X j =0 h f ( λ j ) φ j ( z ) φ j ( w ) , z, w ∈ Γ \ G/K locally uniformly converge everywhere.
Proposition 1. K f being viewed as a function over Γ \ G/K × Γ \ G/K , we have: K f = k f .Proof. By Lemma 1, R ( f ) is an integral operator with continuous integral kernel K f . Meanwhile R ( f ) φ j = h f ( λ j ) φ j . Define T k : L (Γ \ G/K ) → L (Γ \ G/K ) , φ Z Γ \ G/K k f ( z, w ) φ ( w ) µ ′ ( w ) . Then, by definition T k is an integral operator such that T k ( φ j ) = h f ( λ j ) φ j as φ j ’s are orthonormal toeach other. Hence, T k and R ( f ) are identical to each other as operators and their integral kernels areequal to each other except on a possible subset of measure zero. The locally uniform convergence of k f implies that k f is a continuous function as all φ j ’s are analytic over Γ \ G/K . It follows that K f = k f .By Lemma 1 and Proposition 1 one has X γ ∈ Γ f ( z − γw ) = ∞ X j =0 h f ( λ j ) φ j ( z ) φ j ( w ) , z, w ∈ Γ \ G/K.
Let ψ be a normalized Maass form on Γ \ G /K with Laplace eigenvalues λ = ρ − ν where ν ∈ ( − ρ , ρ ) ∪ i R . Integrating both sides on Γ \ G /K × Γ \ G /K with respect to the measure ψ ( z ) ψ ( w ) dzdw gives the (relative) trace formula Z Γ \ G /K Z Γ \ G /K X γ ∈ Γ f ( z − γw ) ψ ( z ) ψ ( w ) dzdw = ∞ X j =0 h f ( λ j ) P Y ( φ j , ψ ) P Y ( φ j , ψ ) . (7)The left (right) hand side of (7) is called geometric (resp. spectral ) side. For this identity to hold, the testfunction f should satisfy: (1) f ∈ C unif ( G ); (2) f is bi- K -invariant; (3) k f is locally uniformly convergent.These conditions will be checked in Sect. 5 for a special f chosen in the next section. In this section we choose a test function and apply it to the spectral side of (7). The bi- K -invarianceof f indicates that f ( g ) depends exactly on the hyperbolic distance between g · o and e · o (on G/K ).Let Φ µ ( x ) be a smooth function over R > for any µ ∈ R + . Define the test function f ∈ C ∞ ( G ) as f ( g ) = Φ µ (cid:0) d G/K ( g · o, e · o ) (cid:1) . Like the case of G/K , the quotient G /K ∼ = H n can be parameterized by R n − × R + via the maps T , S (see Sect. 2.3). By (6) we have h f ( λ j ) = Z R d − Z R Φ µ (cid:16) d G/K (cid:0) a + x n u · o, e · o (cid:1) (cid:17) · e ( ν j + ρ ) x dxdu = Z R d − Z R + Φ µ (cid:16) d G/K ( a r n u · o, e · o ) (cid:17) · r ν j + ρ − drdu where we have made the variable exchange x → log r in the second step. By (3) and (4) we haved G/K ( a r n u · o, e · o ) = d G/K ( e · o, n − u a r − · o ) = arccosh + (cid:18) | ru | + 1 + r r (cid:19) , noting that e = n a . Originally one would like to insert the heat kernel (see [ GN ]), but then it isdifficult to deal with the geometric side. In this paper the test function is chosen to beΦ µ ( x ) = exp( − µ · cosh x ) . It follows that h f ( λ j ) = Z R d − Z R + exp (cid:20) − µ (cid:18) | u | + 12 r + r (cid:19)(cid:21) r ν j + ρ − drdu. (8)The following two integral formulas on K -Bessel functions are useful to us: ∞ Z x ν − exp (cid:16) − αx − βx (cid:17) dx = 2 (cid:18) αβ (cid:19) ν K ν (cid:16) p αβ (cid:17) , Re( α ) > , Re( β ) > . (9) ∞ Z (cid:0) x + b (cid:1) ∓ ν K ν (cid:16) a p x + b (cid:17) cos( cx ) dx = r π a ∓ ν b ∓ ν (cid:0) a + c (cid:1) ± ν − K ± ν − (cid:16) b p a + c (cid:17) (10)where Re( a ) >
0, Re( b ) > c is a real number. These are the formulas 3.471.9 and 6.726.4 of [ GR ]respectively.Let α = µ , β = | u | +12 µ , ν = ν j + ρ in the formula (9), then the integration along r in (8) gives h f ( λ j ) = Z R d − (cid:0) | u | + 1 (cid:1) − νj + ρ K ν j + ρ (cid:16) µ p | u | + 1 (cid:17) du Let x = u , b = u + · · · + u d − + 1, a = µ , c = 0, ν = ν j + ρ in the (first case of) formula (10), then theintegration along u in the above integral is equal to2 d ∞ Z · · · ∞ Z r π µ (cid:16)q u + · · · + u d − + 1 (cid:17) − ( ν j + ρ ) K ν j + ρ − (cid:16) µ q u + · · · + u d − + 1 (cid:17) du · · · du d − (11)Let x = u , b = u + · · · + u d − + 1, a = µ , c = 0, ν = ν j + ρ − in the formula (10), then the integrationalong u in the above integral gives(11) = 2 d (cid:18)r π µ (cid:19) ∞ Z · · · ∞ Z (cid:16)q u + · · · + u d − + 1 (cid:17) − ( ν j + ρ ) K ν j + ρ − (cid:16) µ q u + · · · + u d − + 1 (cid:17) du · · · du d − Repeating this process, i.e., doing integrations along u , u , . . . , u d − step by step in the above fashion,we finally get h f ( λ j ) = 2 d (cid:18)r π µ (cid:19) d − K ν j + ρ − d − ( µ ) = 2 d (cid:18)r π µ (cid:19) d − K ν j ( µ ) . Now the spectral side of (7) reads: ∞ X j =0 d (cid:18)r π µ (cid:19) d − K ν j ( µ ) (cid:12)(cid:12) P Y ( φ j , ψ ) (cid:12)(cid:12) . Under our choice of f the geometric side of (7) splits as follows. Z Γ \ G /K Z Γ \ G /K X γ ∈ Γ f ( z − γw ) ψ ( z ) ψ ( w ) dzdw = Z Γ \ G /K Z Γ \ G /K X γ ∈ Γ Φ µ (cid:0) d G/K ( γw, z ) (cid:1) ψ ( z ) ψ ( w ) dzdw + O Z Γ \ G /K Z Γ \ G /K X γ , γ ∈ Γ X ˜ γ ∈ Γ \ Γ / Γ r { ˜1 } Φ µ (cid:0) d G/K ( γγ w, γ z ) (cid:1) ψ ( z ) ψ ( w ) dzdw = Z G /K Z Γ \ G /K Φ µ (cid:0) d G/K ( w, z ) (cid:1) ψ ( z ) ψ ( w ) dzdw + O Z G /K Z G /K X ˜ γ ∈ Γ \ Γ / Γ r { ˜1 } Φ µ (cid:0) d G/K ( γw, z ) (cid:1) ψ ( z ) ψ ( w ) dzdw where ˜ γ denotes a nontrivial double coset in Γ \ Γ / Γ . To simplify the notations, we still use dz , dw todenote the measure of space G /K . This is reasonable since the measure of G /K descends to that ofthe quotient space Γ \ G /K . It is not clear whether any element in Γ r Γ can be written as γ γγ withunique γ , γ ∈ Γ and some fixed representative element γ of a double coset class ˜ γ = ˜1. So we have touse the expression O ( · · · ) in the above formula. DenoteΣ = Z G /K Z Γ \ G /K Φ µ (cid:0) d G/K ( w, z ) (cid:1) ψ ( z ) ψ ( w ) dzdw and Σ = Z G /K Z G /K X ˜ γ ∈ Γ \ Γ / Γ r { ˜1 } Φ µ (cid:0) d G/K (˜ γw, z ) (cid:1) ψ ( z ) ψ ( w ) dzdw. In the next two sections we shall show that Σ is the main term of the geometric side, while Σ is theerror term. Like the case of φ ∈ L (Γ \ G/K ), the Maass form ψ also gives rise to an irreducible unitary sphericalrepresentation V ′ λ ⊂ L (Γ \ G ). Assume that V ′ λ ∼ = I ′ ( ν ) := Ind G M AN ( ⊗ e ν ⊗ ). Here, to distinguishthe representation of G from that of G , we use V ′ λ and I ′ ( ν ), instead of V λ and I ( ν ). Normalize theHaar measure of K such that vol( K ) = 1. By the quotient integral formula, we can rewrite Σ asΣ = Z G Z Γ \ G Φ µ (cid:0) d G/K ( g · o, x · o ) (cid:1) ˜ ψ ( x ) ˜ ψ (¯ g ) dxdg where ˜ ψ , as before, stands for the natural lift of ψ on Γ \ G , ¯ g denotes the element Γ · g ∈ Γ \ G , dx the invariant Radon measure of Γ \ G , dg the Haar measure of G . In such formulation, Σ defines anonzero ( K × K )-invariant functional L aut ν over V ′ λ × V ′ λ where each K × K acts on V ′ λ × V ′ λ via R × R (the right regular translation): L aut ν : V ′ λ × V ′ λ → C , ( h , h ) Z G Z Γ \ G Φ µ (cid:0) d G/K ( g · o, x · o ) (cid:1) h ( x ) h (¯ g ) dxdg. This functional is C -linear for its first entry and conjugate C -linear for its second entry. For z ∈ Γ \ G , L aut ν is well-defined although d G/K ( w · o, z · o ) is not. The space of ( K × K )-invariant functionals over V ′ λ × V ′ λ is one-dimensional. The reason is as follows. In view of the equivalence between V ′ λ and I ′ ( ν ), itsuffices to show that the space of K -invariant functional over I ′ ( ν ) is one dimensional, which is clearlytrue by the definition of I ′ ( ν ). As a consequence, there exists a nonzero scalar a ν ∈ C such that L aut ν ( h , h ) = a ν · L mod ν ( f , f )for any nonzero ( K × K )-invariant ( C × C )-linear functional L mod ν on I ′ ( ν ) × I ′ ( ν ) where f i ∈ I ′ ( ν )corresponds to h i ∈ V ′ λ . Define L mod ν to be L mod ν ( f , f ) = Z G Z Γ \ G Φ µ (cid:0) d G/K ( g · o, x · o ) (cid:1) f ( x ) f ( g ) dzdg, f , f ∈ I ′ ( ν ) . Then Σ = a ν · L mod ν ( η ν , η ν ). See Sect. 2.4 for the definition of η ν . Now we compute L mod ν ( η ν , η ν ). For g · o = n u a s · o ∈ G /K where u ∈ R n − , s ∈ R + , equip G /K with the hyperbolic measure d ( g · o ) = dsdus n .Let P ( Y ) be the completion of a subset of R n − × R + that is isomorphic to Γ \ G /K . Then P ( Y ) iscompact. The following commutativity property will be used frequently in this paper: a r n u = n ur a r . (12)One can verify (12) be a direct computation, or see Proposition I.4.2 of [ FJ ]. Actually (12) results froma simple fact in Lie algebra: ad( E ) E i = E i . As η ν is K -invariant and d G/K ( w, z ) = d G /K ( w, z ) for w , z ∈ G /K (note that G /K ⊂ G/K is totally geodesic), we have: L mod ν ( η ν , η ν ) = Z G /K Z Γ \ G /K Φ µ (cid:0) d G /K ( w, z ) (cid:1) η ν ( z ) η ν ( w ) dzdw = Z R n − × R + Z P ( Y ) Φ µ (cid:0) d G /K ( n u a s · o, n v a r · o ) (cid:1) r ν + ρ s ¯ ν + ρ drdvr n dsdus n = Z P ( Y ) Z R n − × R + exp (cid:18) − µ · | u − v | + s + r sr (cid:19) s ¯ ν + ρ dsdus n · r ν + ρ drdvr n = Z P ( Y ) Z R n − × R + exp " − µ (cid:12)(cid:12) u − vr (cid:12)(cid:12) + 1 sr + sr ! s ¯ ν + ρ dsdus n · r ν + ρ drdvr n Let u ′ = u − vr , s ′ = sr , then du = r n − du ′ and L mod ν ( η ν , η ν ) = Z P ( Y ) Z R n − × R + exp " − µ | u ′ | + 1 s ′ + s ′ ! s ′ (¯ ν + ρ − n ) ds ′ du ′ · r ¯ ν + ν +2 ρ − n drdv. The rest of the computation is merely a copy of that for h f ( λ j ). Firstly, apply (9) to the integration over s ′ , then the right hand side of the above identity is equal to2 Z P ( Y ) Z R n − (cid:0) | u ′ | (cid:1) ¯ ν + ρ − n +12 K ¯ ν + ρ − n +1 (cid:16) µ p | u ′ | (cid:17) du ′ · r ν ) − drdv. Secondly, apply the second case of (10) to the integration over u ′ step by step, then the above integral isequal to 2 n (cid:18)r π µ (cid:19) n − K ¯ ν ( µ ) Z P ( Y ) r ν ) − drdv. The subset P ( Y ) is compact and r > v, r ) ∈ P ( Y ). Hence, the integral I ν := Z P ( Y ) r ν ) − drdv converges and does not vanish. Denote b ν = 2 n a ν I ν = 0. Up to now we have shownΣ = b ν · (cid:18)r π µ (cid:19) n − K ¯ ν ( µ ) . (13) In this section we give a bound for Σ . The main conclusion isΣ ≪ µ − ( n +2) / e − µ . (14)It turns out that the error term Σ is more difficult to be treated than Σ . Like what we have done forΣ , we use the uniqueness of ( K × K )-invariant ( C × C )-linear functionals to reduce the computationof Σ to that of special integrals. Define ℓ aut ν : V ′ λ × V ′ λ → C , ( h , h ) Z G Z G X ˜ γ ∈ Γ \ Γ / Γ r { ˜1 } Φ µ (cid:0) d G/K ( γg · o, g · o ) (cid:1) h (¯ g ) h (¯ g ) dg dg g i denotes the element Γ · g i ∈ Γ \ G . It is clear that ℓ aut ν is a nonzero ( K × K )-invariant( C × C )-linear functional on V ′ λ × V ′ λ and Σ = ℓ aut ν (cid:0) ˜ ψ, ˜ ψ (cid:1) . As before, the space of such functionals on isone-dimensional. Thus, for a given nonzero ( K × K )-invariant ( C × C )-linear functional on I ′ ( ν ) × I ′ ( ν )there exists a scalar d ν ∈ C such that ℓ aut ν = d ν · ℓ mod ν . Note that d ν depends only on ν . Define ℓ mod ν : I ′ ( ν ) × I ′ ( ν ) → C , ( f , f ) Z G Z G X ˜ γ ∈ Γ \ Γ / Γ r { ˜1 } Φ µ (cid:0) d G/K ( g · o, g · o ) (cid:1) f ( g ) f ( g ) dg dg . Then ℓ mod ν is ( K × K )-invariant ( C × C )-linear on I ′ ( ν ) × I ′ ( ν ) and Σ = d ν · ℓ mod ν ( η ν , η ν ) where η ν isas before. Let w = n u a r · o , z = n v a t · o ∈ G /K where u , v ∈ R n − and r , t ∈ R + . We have ℓ mod ν ( η ν , η ν ) = Z R n − × R + Z R n − × R + X ˜ γ ∈ Γ \ Γ / Γ r { ˜1 } Φ µ (cid:0) d G/K ( γw, z ) (cid:1) t ν + ρ r ¯ ν + ρ dtdvt n drdur n . Write γ = a ( γ ) n ( γ ) k ( γ ) = a r n w u ! ∈ AN K where w = ( w , · · · , w ,d − ) ∈ R d − and u =( u ij ) ∈ SO d . Assume that γw = n v a s · o where v = ( v , · · · , v ,d − ). Denote J γ = Z R n − × R + Z R n − × R + Φ µ (cid:0) d G/K ( γw, z ) (cid:1) t ν + ρ r ¯ ν + ρ dtdvt n drdur n = Z R n − × R + Z R n − × R + exp (cid:18) − µ · | v − v | + s + t s t (cid:19) t ν + ρ − n r ¯ ν + ρ − n dtdvdrdu We treat the integration along t by using (9) and get J γ = 2 Z R n − × R + Z R n − (cid:0) | v − v | + s (cid:1) ν + ρ − n +12 K ν + ρ − n +1 µ s(cid:12)(cid:12)(cid:12)(cid:12) v − v s (cid:12)(cid:12)(cid:12)(cid:12) + 1 r ¯ ν + ρ − n dvdrdu = 2 Z R n − × R + Z R n − (cid:12)(cid:12)(cid:12)(cid:12) v − v s (cid:12)(cid:12)(cid:12)(cid:12) + 1 ! ν + ρ − n +12 K ν + ρ − n +1 µ s(cid:12)(cid:12)(cid:12)(cid:12) v − v s (cid:12)(cid:12)(cid:12)(cid:12) + 1 d (cid:18) vs (cid:19) s ν + ρ r ¯ ν + ρ − n drdu The first ( n −
1) entries in v s can be absorbed in to vs when we do the integration along vs , leavingthe last ( d − n ) components of v s . Denote | x | > n = x n + · · · + x d − for x = ( x , · · · , x d − ) ∈ R d − . Let v ′ = vs , then J γ = 2 Z R n − × R + Z R n − | v ′ | + (cid:12)(cid:12)(cid:12)(cid:12) v s (cid:12)(cid:12)(cid:12)(cid:12) > n + 1 ! ν + ρ − n +12 K ν + ρ − n +1 µ s | v ′ | + (cid:12)(cid:12)(cid:12)(cid:12) v s (cid:12)(cid:12)(cid:12)(cid:12) > n + 1 ! dv ′ s ν + ρ r ¯ ν + ρ − n drdu Applying the second case of (10) to dv ′ step by step (cid:0) just as what we have done for h f ( λ j ) and Σ (cid:1) gives J γ = 2 n (cid:18)r π µ (cid:19) n − Z R n − × R + (cid:12)(cid:12)(cid:12)(cid:12) v s (cid:12)(cid:12)(cid:12)(cid:12) > n + 1 ! ν K ν µ s(cid:12)(cid:12)(cid:12)(cid:12) v s (cid:12)(cid:12)(cid:12)(cid:12) > n + 1 ! s ν + ρ r ¯ ν + ρ − n drdu. (cid:0) one should distinguish w in below from the w that has appeared earlier (as apoint on Γ \ G /K or G /K ) (cid:1) : n w a s k = s + s − + s − | w | , · · · s − s − + s − | w | , · · · w · s − , · · · w · s − , · · · ... ... w d − · s − , · · · where w = ( w , · · · , w d − ) ∈ R d − and k ( γ ) n u a r = (cid:16) | u | (cid:17) r + r − − | u | r − r − , · · · (cid:18) u | u | + n P i =2 u i u i − (cid:19) r + r − + (cid:20) u (cid:16) − | u | (cid:17) − n P i =2 u i u i − (cid:21) r − r − , · · · (cid:18) u | u | + n P i =2 u i u i − (cid:19) r + r − + (cid:20) u (cid:16) − | u | (cid:17) − n P i =2 u i u i − (cid:21) r − r − , · · · ... ... (cid:18) u d | u | + n P i =2 u d i u i − (cid:19) r + r − + (cid:20) u d (cid:16) − | u | (cid:17) − n P i =2 u d i u i − (cid:21) r − r − , · · · . Let k ( γ ) n u a r = n w a s k for some k ∈ K . Then we have s + s − s − | w | = (cid:18) | u | (cid:19) r + r − − | u | r − r − s − s − s − | w | = u | u | n X i =2 u i u i − ! r + r − " u (cid:18) − | u | (cid:19) − n X i =2 u i u i − r − r − w i s − = u i +1 , | u | n X j =2 u i +1 , j u j − r + r − u i +1 , (cid:18) − | u | (cid:19) − n X j =2 u i +1 , j u j − r − r − s − = 1 − u r + (cid:18) u β (cid:19) r − (18)where β = (1 − u ) | u | − n P i =2 u i u i − . Let α i = u i +1 , | u | + n P j =2 u i +1 , j u j − , then (17) reads w i s − = u i +1 , r + (cid:16) α i − u i +1 , (cid:17) r − , i d − . (19)By the assumption that γw = n v a s · o , we have v = ( w + w ) r , s = r s . For any 1 i d −
1, denote m i = w i − u u i +1 , , n i = w i (cid:18) u β (cid:19) + (cid:16) α i − u i +1 , (cid:17) . Then the computation with the above terms shows that f γ ( u, r ) := (cid:12)(cid:12)(cid:12)(cid:12) v s (cid:12)(cid:12)(cid:12)(cid:12) > n + 1 = (cid:12)(cid:12)(cid:12)(cid:12) w + ws (cid:12)(cid:12)(cid:12)(cid:12) > n + 1 = M ( γ ) r + N u ( γ ) r − + Q u ( γ )2where M ( γ ) = d − X i = n m i , N u ( γ ) = d − X i = n n i , Q u ( γ ) = 1 + 2 d − X i = n m i n i . (20)Now we have: J γ = 2 n (cid:18)r π µ (cid:19) n − Z R n − × R + (cid:18)q f γ ( u, r ) (cid:19) ν K ν (cid:18) µ q f γ ( u, r ) (cid:19) s ν + ρ r ¯ ν + ρ − n drdu. Write f γ ( u, r ) = (cid:18)p M ( γ ) r − √ N u ( γ ) r (cid:19) + 2 p M ( γ ) N u ( γ ) + Q u ( γ ) and define δ u ( γ ) := 2 p M ( γ ) N u ( γ ) + Q u ( γ ) . The parameter u in the subscript of N , Q and δ indicates that these numbers depend on u as well as γ . Note that M depends only on γ . For simplicity, we do not write γ , u explicitly in the notations m i , n i . The number δ u ( γ ) has remarkable geometric meaning which can be interpreted from the followinginequalitiescosh (cid:0) d G/K ( γw, z ) (cid:1) = | v − v | + s s t + t s > s | v − v | + s s · s = s(cid:12)(cid:12)(cid:12)(cid:12) v − v s (cid:12)(cid:12)(cid:12)(cid:12) + 1 > s(cid:12)(cid:12)(cid:12)(cid:12) v s (cid:12)(cid:12)(cid:12)(cid:12) > n + 1 . The “=” at the first inequality can be achieved as t ranges among all positive numbers. The last stepfollows from the fact that v ranges among vectors in R n − (so the “=” can be achieved). To be moreprecise, the two “=” are simultaneously achieved at v = Pr n − ( v ) , t = q s + | v | > n where Pr n − means the projection map R d − → R n − , v = ( v , · · · , v d − ) ( v , · · · , v n − ). Since r ranges over all positive numbers, we have (cid:12)(cid:12)(cid:12) v s (cid:12)(cid:12)(cid:12) > n + 1 = f γ ( u, r ) > δ u ( γ ) where “=” can be obtained when p M ( γ ) r − √ N u ( γ ) r = 0, i.e., r = q N u ( γ ) M ( γ ) if M ( γ ) = 0, or r = ∞ if M ( γ ) = 0. So δ u ( γ ) measures theminimal hyperbolic distance between the two submanifolds γn u A · o and G /K :inf a ∈ A · o, z ∈ G /K d G/K ( γn u a · o, z ) = arccosh + (cid:16)p δ u ( γ ) (cid:17) . (21)By this formula we know that δ u ( · ) is well-defined over Γ \ Γ (but not on Γ / Γ ). It is clear from the abovediscussion that the number f γ ( u, r ) also has remarkable geometric meaning: it measures the (hyperbolic)distance between the point γn u a r · o and the submanifold G /K . More precisely,inf z ∈ G /K d G/K ( γn u a r · o, z ) = arccosh + (cid:18)q f γ ( u, r ) (cid:19) . The rest of this section is devoted to estimating Σ . The crucial ingredient in our argument is that,for each class ˜ γ ∈ Γ \ Γ / Γ r { ˜1 } we shall carefully choose a representative element (which satisfies someuniversal properties) and deduce the estimate to a lattice counting problem. We need some technicalconclusions whose proof will be postponed to Sect. 4.3. The first result to be used is Proposition 2.
Let n > . For any fixed u ∈ R n − , we can find a representative element γ in each class ˜ γ ∈ Γ \ Γ / Γ r { ˜1 } such that the following hold • M ( γ ) N u ( γ ) > c > where c is independent of u and the representative elements γ ’s. • δ u ( γ ) achieves its minimal value at some u γ ∈ F ⊂ R n − where F is a fixed compact domainindependent of γ ’s. • M ( γ ) > c > where c is independent of γ ’s. In what follows we shall select the representative elements that satisfy the three properties in this propo-sition. Let x = p M ( γ ) r − √ N u ( γ ) r , then r = x + p x +4 √ M ( γ ) N u ( γ )2 √ M ( γ ) (since M ( γ ) = 0 by Proposition 2)and J γ = 2 n (cid:18)r π µ (cid:19) n − Z R n − × R F γ ( u, x ) dxdu where F γ ( u, x ) = s ν + ρ x + q x + 4 p M ( γ ) N u ( γ )2 p M ( γ ) ν − ρ (cid:16)p x + δ u ( γ ) (cid:17) ν K ν (cid:16) µ p x + δ u ( γ ) (cid:17)q x + 4 p M ( γ ) N u ( γ ) . (22)Recall that s = r s = r − u r + (cid:0) u + β (cid:1) r − . The term u + β is nonnegative since by Cauchy inequality we have1 + u β = ( (1 − u ) | u | − n X i =2 u i u i − + (1 + u ) ) (cid:14) > q (1 − u ) · | u | − n X i =2 u i u i − > vuut n X i =2 u i · n X i =2 u i − − n X i =2 u i u i − > u i = 0 ( n + 1 i d ), u i − = t · u i (2 i n ) for some constant t , and(1 − u ) | u | = 1 + u . These conditions lead to at most one solution of u γ (up to ±
1) for any given γ .Since we are doing integration along u , the possible solution u γ can be neglected. Thus, s r p (1 − u )(1 + u ) + 2(1 − u ) β . In Sect. 4.3 we shall show
Proposition 3. sup γ ∈ Γ r Γ (cid:12)(cid:12) u ( γ ) (cid:12)(cid:12) < . This proposition indicates that (1 − u )(1 + u )+ 2(1 − u ) β is a polynomial of degree 2 (with respect toeach variable u i ). Hence, the denominator of s (i.e., s − ) grows (at least) polynomially with degree 1 andpositive minimum value (cid:0) as − u r + (cid:0) u + β (cid:1) r − is strictly positive if we neglect u γ (cid:1) . Multiplyingproper γ ∈ Γ to the left side of γ if necessary, we assume that r lies in a compact interval in R + . Case 1.
First we assume that (cid:12)(cid:12) Γ \ Γ / Γ (cid:12)(cid:12) < ∞ . The case (cid:12)(cid:12) Γ \ Γ / Γ (cid:12)(cid:12) = ∞ will be treated later. When µ is large, µ p x + δ u ( γ ) is also very large since δ u ( γ ) >
1. By the well-known asymptotic of K -Besselfunction: K z ( x ) ∼ r π x e − x , as x → ∞ (23)the function X ν K ν ( µX ) decreases with respect to X when µ is large. Hence (cid:12)(cid:12)(cid:12)(cid:16)p x + δ u ( γ ) (cid:17) ν K ν (cid:16) µ p x + δ u ( γ ) (cid:17)(cid:12)(cid:12)(cid:12) (cid:16)p x + 1 (cid:17) Re ν K ν (cid:16) µ p x + 1 (cid:17) µ → ∞ . Since K ν (cid:0) µ √ x + 1 (cid:1) ∼ K Re ν (cid:0) µ √ x + 1 (cid:1) , we have (cid:12)(cid:12)(cid:12)(cid:12)Z R (cid:16)p x + δ u ( γ ) (cid:17) ν K ν (cid:16) µ p x + δ u ( γ ) (cid:17) dx (cid:12)(cid:12)(cid:12)(cid:12) ≪ Z R (cid:16)p x + 1 (cid:17) Re ν K Re ν (cid:16) µ p x + 1 (cid:17) dx. Let y = x + 1 in right hand side of the above integral and apply the formula 6.592.12 of [ GR ] Z ∞ x − b ( x − c − K z (cid:0) a √ x (cid:1) dx = 2 c Γ( c ) a − c K b − c ( a ) , Re( a ) > , Re( c ) > , we get Z ∞ (cid:16)p x + δ u ( γ ) (cid:17) Re ν K Re ν (cid:16) µ p x + δ u ( γ ) (cid:17) dx = Γ(1 / √ µ − / K − Re ν − / ( µ ) . (24)By Cauchy inequality, we have p M ( γ ) N u ( γ ) > d − X i = n m i n i . The term n i expands as follows n i = | u | w i (1 − u ) + u i +1 , n X j =2 ( u i +1 , j − w i u j ) u j − + w i u − u i +1 , . (25)Note that m i = w i (1 − u )+ u i +1 , . Therefore, d − X i = n m i n i = | u | d − X i = n m i | {z } = M ( γ ) + d − X i = n n X j =2 m i ( u i +1 , j − w i u j ) u j − + d − X i = n m i (cid:18) w i u − u i +1 , (cid:19) . (26)This means that p M ( γ ) N u ( γ ) grows polynomially with degree 2 (with respect to each variable u i ) witha positive minimum value for all γ / ∈ Γ (by Proposition 2).When x >
0, one has (cid:18) x + q x + 4 p M ( γ ) N u ( γ ) (cid:19) Re ν − ρ (cid:16) p M ( γ ) N u ( γ ) (cid:17) Re ν − ρ noting that Re ν − ρ
0. As a consequence, (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x + q x + 4 p M ( γ ) N u ( γ )2 p M ( γ ) ν − ρ s ν + ρ q x + 4 p M ( γ ) N u ( γ ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≪ (cid:16) p M ( γ ) N u ( γ ) (cid:17) − ( ρ − Re ν +1) (cid:0) s − (cid:1) − ( ρ +Re ν ) . We have known that both p M ( γ ) N u ( γ ) and s − grow (at least) polynomially with degree 1 (with respectto each variable u i ) with positive minimum value. Since ρ = n − > n >
2, the following integral Z R n − (cid:16) p M ( γ ) N u ( γ ) (cid:17) − ( ρ − Re ν +1) ( s − ) − ( ρ +Re ν ) du converges as the function inside is positive and has polynomial degree (at most) − ( ρ − Re ν + 1) − ( ρ + Re ν ) = − ρ − < − . Note that the term (cid:16) p M ( γ ) (cid:17) ρ − ν in F γ ( u, x ) does not give essential contribution to the upper boundsince (cid:12)(cid:12) Γ \ Γ / Γ (cid:12)(cid:12) < ∞ . In view of (24), we get Z R n − Z ∞ F γ ( u, x ) dxdu ≪ µ − / K − Re ν − / ( µ ) ≪ µ − e − µ . (27)5When x − ω with any ω ∈ (0 , (cid:18) x + q x + 4 p M ( γ ) N u ( γ ) (cid:19) Re ν − ρ = q x + 4 p M ( γ ) N u ( γ ) − x p M ( γ ) N u ( γ ) ρ − Re ν = ( − x ) ρ − Re ν q √ M ( γ ) N u ( γ ) x + 14 p M ( γ ) N u ( γ ) ρ − Re ν = ( − x ) ρ − Re ν s M ( γ ) N u ( γ ) + 14 x p M ( γ ) N u ( γ ) + 14 p M ( γ ) N u ( γ ) ! ρ − Re ν ( ♯ ) ( − x ) ρ − Re ν s M ( γ ) N u ( γ ) + s x p M ( γ ) N u ( γ ) + 14 p M ( γ ) N u ( γ ) ! ρ − Re ν = ( − x ) ρ − Re ν (cid:16) p M ( γ ) N u ( γ ) (cid:17) − ( ρ − Re ν ) − x + 1 p M ( γ ) N u ( γ ) ! ρ − Re ν ( − x ) ρ − Re ν (cid:16) p M ( γ ) N u ( γ ) (cid:17) − ( ρ − Re ν ) (cid:18) − x + 1 √ c (cid:19) ρ − Re ν ≪ ( − x ) ρ − Re ν (cid:16) p M ( γ ) N u ( γ ) (cid:17) − ( ρ − Re ν ) (cid:18) ω (cid:19) ρ − Re ν provided that ω is small enough. At the step ( ♯ ) we have used the inequality √ a + b √ a + √ b . Thus, Z R n − Z − ω −∞ F γ ( u, x ) dxdu ≪ Z R n − Z − ω −∞ ( − x ) ρ − Re ν (cid:16) p M ( γ ) N u ( γ ) (cid:17) − ( ρ − Re ν ) ω ρ − Re ν × (cid:16) p M ( γ ) N u ( γ ) (cid:17) − ( s − ) − ( ρ +Re ν ) (cid:16)p x + δ u ( γ ) (cid:17) ν K ν (cid:16) µ p x + δ u ( γ ) (cid:17) dxdu As before, the integral Z R n − (cid:16) p M ( γ ) N u ( γ ) (cid:17) − ( ρ − Re ν +1) ( s − ) − ( ρ +Re ν ) du converges. Since K ν (cid:0) µ √ x + 1 (cid:1) ∼ q π µ √ x +1 e − µ √ x +1 as µ → ∞ , the ratio of K ν (cid:0) µ √ x + 1 (cid:1) at x = 0and µ − / is asymptotically equal to 1 p µ − · e − µ e − µ √ µ − +1 which converges to e / as µ → ∞ . Let ω = µ − / , then Z − µ − / −∞ ( − x ) ρ − Re ν (cid:16)p x + 1 (cid:17) ν K ν (cid:16) µ p x + 1 (cid:17) dx ≍ Z −∞ ( − x ) ρ − Re ν (cid:16)p x + 1 (cid:17) ν K ν (cid:16) µ p x + 1 (cid:17) dx, as µ → ∞ . Hence, we have Z R n − Z −∞ F γ ( u, x ) dxdu ≍ Z R n − Z − µ − / −∞ F γ ( u, x ) dxdu ≪ µ ( ρ − Re ν ) / Z − µ − / −∞ ( − x ) ρ − Re ν (cid:16)p x + δ u ( γ ) (cid:17) Re ν K Re ν (cid:16) µ p x + δ u ( γ ) (cid:17) dx< µ ( ρ − Re ν ) / Z −∞ ( − x ) ρ − Re ν (cid:16)p x + 1 (cid:17) Re ν K Re ν (cid:16) µ p x + 1 (cid:17) dx ( ∗ ) = µ ( ρ − Re ν ) / · ( ρ − Re ν − / µ − ( ρ − Re ν +1) / Γ (cid:18) ρ − Re ν + 12 (cid:19) K − Re ν − ( ρ − Re ν +1) / (cid:0) µ (cid:1) ≪ µ − e − µ (28)where the step ( ∗ ) is computed in the same way with (24). By (27) and (28), Z R n − Z R F γ ( u, x ) dxdu ≪ µ − e − µ which proves (14). Case 2.
Now we deal with the case (cid:12)(cid:12) Γ \ Γ / Γ (cid:12)(cid:12) = ∞ . By (22) and Proposition 2 we have (cid:12)(cid:12) J γ (cid:12)(cid:12) ≪ µ − ( n − / Z R n − Z R s ρ +Re ν (cid:16) p M ( γ ) (cid:17) ρ − Re ν (cid:18) x + q x + 4 p M ( γ ) N u ( γ ) (cid:19) Re ν − ρ × K ν (cid:16) µ p x + δ u ( γ ) (cid:17) dxdu. (29)By (23) and the inequality √ a + b > √ (cid:16) √ a + √ b (cid:17) , K ν (cid:16) µ p x + δ u ( γ ) (cid:17) ∼ s π µ p x + δ u ( γ ) e − µ √ x + δ u ( γ ) ≪ µ − / e − √ µ | x | e − √ µ √ δ u ( γ ) . (30)Combining (29) and (30) yields (cid:12)(cid:12) J γ (cid:12)(cid:12) ≪ µ − n/ Z R n − Z R s ρ +Re ν e − √ µ | x | (cid:18) x + q x + 4 p M ( γ ) N u ( γ ) (cid:19) Re ν − ρ (cid:16) p M ( γ ) (cid:17) ρ − Re ν e − √ µ √ δ u ( γ ) dxdu. The integral Z R s ρ +Re ν e − √ µ | x | (cid:18) x + q x + 4 p M ( γ ) N u ( γ ) (cid:19) Re ν − ρ dx converges and is uniformly upper bounded by a constant since the function inside has exponential decayand µ is large (note that s is a positive rational function of x ).As δ u ( γ ) grows polynomially with respect to the variables p M ( γ ) u i , for any ε ∈ (0 ,
1) we have X ˜ γ ∈ Γ \ Γ / Γ r { ˜1 } Z R n − (cid:16) p M ( γ ) (cid:17) ρ − Re ν e − √ µ √ δ u ( γ ) du = X ˜ γ ∈ Γ \ Γ / Γ r { ˜1 } Z R n − (cid:16) p M ( γ ) (cid:17) ρ − Re ν e − √ µ √ ε δ u ( γ )+(1 − ε ) δ u ( γ ) du X ˜ γ ∈ Γ \ Γ / Γ r { ˜1 } Z R n − (cid:16) p M ( γ ) (cid:17) ρ − Re ν e − µε √ δ u ( γ ) · e − µ √ − ε √ δ u ( γ ) du ≪ Z R n − (cid:16)p M ( γ ) (cid:17) − ρ − Re ν e − µε √ δ u ( γ ) d (cid:16)p M ( γ ) u (cid:17) × X ˜ γ ∈ Γ \ Γ / Γ r { ˜1 } e − µ √ − ε √ δ u ∗ ( γ ) u ∗ ∈ F . At the second step we have used the inequality √ a + b > √ ( √ a + √ b ). At thelast step we have used Proposition 2. Note that (cid:16)p M ( γ ) (cid:17) − ρ − Re ν is uniformly upper bounded since M ( γ ) > c >
0. Thus, the integral in the last step converges. It follows that J γ ≪ µ − n/ e − µ √ − ε √ δ u ∗ ( γ ) (31)In Sect. 4.3 we shall show Proposition 4.
For any fixed u ∈ R n − and sequence Λ u ⊂ Γ where each element in Λ u representsexactly one double class in Γ \ Γ / Γ , there exist a positive number c such that π Λ u ( x ) := { γ ∈ Λ u | δ u ( γ ) x } c · x ( d − n ) / , as x → ∞ . We may arrange the order of the elements in Λ u and get Λ u = { γ j } such that δ u ( γ j ) is nondecreasing as j grows. Then we have Proposition 5. δ u ( γ j ) ≫ j d − n ) / β for any fixed β > and u .Proof. It suffices to show that δ (cid:16) ˜ γ [ j ( d − n ) / β ] (cid:17) ≫ j , where [ x ] means taking the maximal integer thatdoes not exceed x . Assume that there exists a sequence { j i } ∞ i =1 such that j i increases as i grows and δ (cid:18) ˜ γ [ j ( d − n ) / βi ] (cid:19) j i → i → ∞ . Then all ˜ γ [ j ( d − n ) / βi ] are contained in the set { ˜ γ | δ (˜ γ ) j i } . This implies that π ( j i ) > (cid:2) j ( d − n ) / β i (cid:3) since δ ( γ j ) is nondecreasing as j grows, contradicting Proposition 4.Let u = u ∗ , then by Proposition 5 there exists j > δ u ∗ ( γ j ) > − ε j /d for any j > j .Denote J j = J γ j . By (31) we have ∞ X j = j J j ≪ µ − n/ ∞ X j = j e − µ j /d < µ − n/ Z ∞ e − µ x /d dx = dµ − n/ Z ∞ e − µ y y d − dy where we made the variable exchange x /d → y in the last step. Integration by parts shows that theintegral on the right hand side is upper bounded by µ − e − µ . Hence, ∞ X j = j J j ≪ µ − ( n +2) / e − µ . As for those J j where 1 j j , we apply the argument that has been done for the case (cid:12)(cid:12) Γ \ Γ / Γ (cid:12)(cid:12) < ∞ and get j P j =1 J j ≪ µ − ( n +1) / e − µ , thereby we have shown (14).Putting the data on geometric and spectral sides together, we get ∞ X j =0 d (cid:18)r π µ (cid:19) d − K ν j ( µ ) (cid:12)(cid:12) P Y ( φ j , ψ ) (cid:12)(cid:12) = b ν · (cid:18)r π µ (cid:19) n − K ¯ ν ( µ ) + O (cid:16) e − µ µ − ( n +1) / (cid:17) In view of the asymptotic (23) of K -Bessel function, multiplying 2 − d (cid:18)q µπ (cid:19) n e µ on both sides of thisformula and taking the limitation µ → ∞ yieldslim µ →∞ e µ (cid:18)r π µ (cid:19) d − n − ∞ X j =0 K ν j ( µ ) (cid:12)(cid:12) P Y ( φ j , ψ ) (cid:12)(cid:12) = c ν . (32)where c ν = 2 − d b ν . The K-Bessel function K ν i ( µ ) is positive for ν j ∈ i R , so the right hand side of (32)is nonnegative. Since b ν = 0 (see Sect. 4.1), the scalar c ν is positive. Proof of Theorem 1.
Assume that there are only finitely many φ j such that P Y ( φ j , ψ ) = 0. By theaysmptotic formula (23), the left hand side of (32) is equal to zero, a contradiction as c ν > In this section we prove the propositions that have been used in the previous section. The commutativityrelations listed below, as well as (12), will be used frequently:diag(1 , , k ) · n u = n uk T · diag(1 , , k ) , u ∈ R d − , k ∈ SO ( d − . (33)diag(1 , − , k ′ ) · a r = a r − · diag(1 , − , k ′ ) , k ′ ∈ O ( d − , det( k ′ ) = − . (34)These properties are easy to verify. Under our assumption on Γ (see Sect. 2.2), there exists γ = a ℓ k ∈ AM such that Γ = h γ i . It is clear that ℓ = 1. We might as well assume that ℓ > Lemma 2. u ( γ ) = ± for any γ ∈ Γ r Γ .Proof. Assume that u ( γ ) = 1, i.e., γ ∈ ( AN M ∩ Γ) r Γ . Write γ = a r n w k ( γ ) ∈ AN M where k ( γ ) = diag(1 , , k ) with k ∈ SO ( d − γ = a r diag(1 , , k ), γ = a r diag(1 , , k ) ∈ Γ (where k , k ∈ SO ( d − γ yields γ γγ = a r r r n w k T r − · diag(1 , , k kk ) . Here we have use the commutativity relations (12) and (33). With proper r , r , we may assume that r r r lies in (1 , ℓ ], and (cid:12)(cid:12) w k T r − (cid:12)(cid:12) = | w | r − is small enough. This means that γ γγ is close to AM ∩ Γ = Γ . The discreteness of Γ implies that γ γγ lies in Γ ⊂ Γ , whence γ ∈ Γ . Thiscontradicts the assumption. The case that u ( γ ) = − Proof of Proposition 3.
If there exists a sequence { γ i } ⊂ Γ r Γ such that u ( γ i ) →
1, then γ i is closeto AN M ∩ Γ. By Lemma Lemma 2 and the discreteness of Γ, γ i lies in AN M ∩ Γ = Γ for large enough i . However, we have assumed that γ / ∈ Γ . The case u ( γ i ) → − Lemma 3.
The image of Γ \ Γ under the map Γ \ Γ → Γ \ G is discrete and has no accumulation pointin Γ \ G with respect to the topology defined by the invariant Riemann metric of Γ \ G .Proof. We show the first part of the lemma since the second part follows in the same way. Assume thatthe sequence { Γ · γ i } ⊂ Γ \ G converges to Γ · γ where γ i , γ ∈ Γ, and Γ · γ i = Γ · γ j for i = j . Then thereexist a sequence { η i } ⊂ Γ and a compact neighborhood W i of e such that γ − η i γ i ∈ W i ∩ Γ. By passingto a subsequence if necessary, we may assume that (cid:8) γ − η i γ i (cid:9) converges. The compact neighborhood W i can be close to e arbitrarily for i large, which means that W i ∩ Γ = { e } for i large enough. Hence, γ − η i γ i = e for i large enough. As a consequence, Γ · γ i = Γ · γ j for i , j large enough, a contradiction. Proof of Proposition 4.
First we assume that M ( γ ) N u ( γ ) = 0. For given γ = a r n w k ( γ ) ∈ Λ u and u ∈ R n − there exist (unique) a r ∈ A and γ = a r · diag(1 , , k , k ) ∈ Γ with r ∈ (1 , ℓ ] and k ∈ O ( n − k ∈ O ( d − n ), such that γn u a rr · o = a t n v · o where | v | > n +1 = δ u ( γ ). Here, rr is the uniquepositive solution of the equation p M ( γ ) x − √ N u ( γ ) x = 0 (see Sect. 4.2 for the meaning of the notations).Thus, modulo (cid:8) ℓ n (cid:12)(cid:12) n ∈ Z (cid:9) (multiplicatively), r and r (then, a r and γ ) are unique. Clearly we have: γn u a rr · o = γγ a r n ( rr ) − uk · o (here we have used (12), (33)). For v = ( v , v , · · · , v d − ) ∈ R d − , denote v 0) and v > n = (0 , · · · , , v n , · · · , v d − ) . Let G ∗ denote a fundamental domain of Γ \ G in G which contains Γ · e . Then there exists (unique) γ ∈ Γ such that γ a t n v · o = γ a t n v For any sequence of pairs n(cid:0) γ ∗ i ( u i ) , γ ∗ i ( w i ) (cid:1) (cid:12)(cid:12)(cid:12) γ i , γ i ∈ Γ r Γ , ˜ γ i = ˜ γ i , γ ∗ i ( u i ) ∈ Ω u i ( ∞ ) , γ ∗ i ( w i ) ∈ Ω w i ( ∞ ) , u i , w i ∈ R n − o ∞ i =1 ,γ ∗ i ( u i ) and γ ∗ i ( w i ) can not be close enough (as i → ∞ ) with respect to the topology of G .Proof. Let γ ∗ i ( u i ) = γ ′ i γ i γ ′′ i a r i n u i ∈ Ω u i ( ∞ ) and γ ∗ i ( w i ) = γ ′ i γ i γ ′′ i a r i n u i ∈ Ω w i ( ∞ ) for some γ ′ i , γ ′ i ∈ Γ , γ ′′ i , γ ′′ i ∈ Γ , r i , r i ∈ (1 , ℓ ] and u i , u i ∈ R n − . Assume that γ ∗ i ( u i ) and γ ∗ i ( w i ) are closeenough as i → ∞ , then ( γ ∗ i ( u i )) − γ ∗ i ( w i ) → 1, that is, n − u i a r − i ( γ ′ i γ i γ ′′ i ) − ( γ ′ i γ i γ ′′ i ) a r i n u i ∈ U i where U i is a compact neighborhood of e that can be small enough for large i . It follows that η i :=( γ ′ i γ i γ ′′ i ) − ( γ ′ i γ i γ ′′ i ) lies in V i := a r i n u i U i n − u i a r − i , a compact neighborhood of a r i n ( u i − u i ) a r − i ∈ AN which is contained in AN V where V is a fixed compact neighborhood of e (note that V i is smallenough for large i ). Since Γ \ G is compact, the image of V i in Γ \ G is also compact. This implies that,passing to a subsequence if necessary we may assume that Γ · η i converges in Γ \ G . By Lemma 3, thesequence { Γ · η i } becomes stable for large enough i . Hence, ˜ γ i = ˜ γ j for large i and j , a contradiction.By Lemma 4, to count π u ( x ) it suffices to count the representative elements γ ∗ i ( u ) that lie in Ω u ( x ).Lemma 5 tells us that these γ ∗ i ( u ) are discrete and have no accumulation point with respect to thetopology of G . The topology of G , when restricted to Ω u ( x ) is equivalent to the Euclidean topology of R d − n since the components G ∗ and K of Ω u ( x ) are compact. Thus, π u ( x ) is (upper) bounded by thevolume of Ω u ( x ) which is of order x ( d − n ) / . This proves Proposition 4. Corollary 1. Assume that | Γ \ Γ / Γ | = ∞ , then for any fixed u ∈ R n − the unique accumulation pointof { δ u ( γ i ) | ˜ γ i = ˜ γ j } is ∞ . Lemma 6. If M ( γ ) N u ( γ ) = 0 , then Q u ( γ ) = δ u ( γ ) = 1 .Proof. This is clear in view of (20). Lemma 7. For any γ / ∈ Γ and u ∈ R n − , M ( γ ) and N u ( γ ) can not be zero simultaneously. Proof. Assume that M ( γ ) = N u ( γ ) = 0 for some γ / ∈ Γ and u ∈ R n − . As before, let γn u a r = a r s n w ws k (see Sect. 4.2). Then (cid:12)(cid:12) w + ws (cid:12)(cid:12) > n + 1 = M ( γ ) r + N u ( γ ) r + Q u ( γ ) ≡ r > γn u A · o ⊂ G /K . As n u A · o ⊂ G /K , we have γ ∈ Γ , a contradiction. Proof of Proposition 2. As before (see Sect. 4.2), let w = n u a r · o , z = n v a t · o ∈ G /K , γ = a r n w k ( γ ).Write k ( γ ) n u a r · o = n w a s · o . Then γn u a r · o = n v a s · o where v = ( w + w ) r , s = r s .First we show that, for each class ˜ γ = ˜1 there exists representative element γ such that M ( γ ) N u ( γ ) = 0for any u ∈ R n − . Recall that u + β = 0 has at most one solution u ( γ ) of u up to ± 1. Multiplying γ ∈ Γ to the left side of the geodesic C u := n u A · o ⊂ G /K and denote the point P ∈ γ C u as n u ′ a r ′ · o .Clearly, we can choose proper γ such that u ′ = u ( γ ) for any P ∈ γ C u . Replacing γ with γγ if necessary,we assume that u + β = 0. Note that the condition n > G /K should be large enough to allow inside the geodesic γ C u (and γ C u in below).If n = 1, such property does not hold. As r → ∞ , we have: s = r − u r + (cid:0) u + β (cid:1) r − → w i = u i +1 , r + (cid:0) α i − u i +1 , (cid:1) r − − u r + (cid:0) u + β (cid:1) r − → u i +1 , − u =: ˜ w i . Denote ˜ w = ( ˜ w , · · · , ˜ w d − ). Then ˜ w does not depend on u (as r → ∞ ). It follows that ξ γ :=lim r →∞ γn u a r · o = lim r → + n ( w + ˜ w ) r a r · o is a fixed point lying in the boundary X of G/K where X := { lim r → + n w a r · o | w ∈ R d − } . Note that ξ γ is independent of u . As r → 0, we have s → w i → α i − ui +1 , u + β =: ˚ w i . Denote ˚ w = (˚ w , · · · , ˚ w d − ) and ξ γ ( u ) = lim r → + n ( w +˚ w ) a r · o ∈ X . Thegeodesic on G/K is either a straight line or a semicircle, each of which is vertical to X (in the R d − × R + model). Our argument in above shows that γC u is a semicircle with two end points ξ γ , ξ γ ( u ) ∈ X .Multiply γ ∈ Γ on the left side of the geodesic C u and assume that γ C u is a semicircle on G /K with two end points P = lim r → + n u a r · o , P = lim r → + n u a r · o where u , u ∈ R n − . We can choose γ such that the resulting u , u are large enough, then u ( γ ) = v ′ for any n v ′ a t ′ · o ∈ γ C u . This impliesthat u + β = 0 for γγ ( u is the same with before). Define Y := { lim r → + n w a r · o | w ∈ R n − } , theboundary of G /K . Then Y is a proper subset of X . • If M ( γ ) = 0, then N u ( γ ) = 0 by Lemma 7. It follows from Lemma 6 that f γ ( u, r ) = N u ( γ ) /r +1 → δ u ( γ ). The (minimal) distance between the geodesic γC u and the submanifold G /K , whichis 0, is obtained at the two points n v a s · o ∈ γC u and n v a t · o ∈ G /K where v = ( w + ˜ w ) r , s = 0 + , v = Pr n − ( v ), t = q s + | v | > n . Hence, ξ γ lies in Y , i.e., the last ( d − n ) components of( w + ˜ w ) vanish. We claim that γP , γ P / ∈ Y . The reason is simple. Assume that γP ∈ Y , thenthe geodesic γn u A · o must lie in G /K since the two end points γP and lim r →∞ n u a r · o both liein Y . This implies that γ ∈ Γ , whereas we have assumed that γ / ∈ Γ . The case γP is shown inthe fashion. As a consequence, the geodesic γC u either intersects G /K or is away from G /K ,i.e., δ u ( γγ ) > 1. The first case indicates that the (minimal) distance between γγ C u and thesubmanifold G /K is achieved at a finite r , hence M ( γ ′ ) N u ( γ ′ ) = 0 (otherwise the distance willbe achieved at r = 0 or ∞ ). By Lemma 6, the second case directly indicates M ( γγ ) N u ( γγ ) = 0since δ u ( γγ ) > γ with γγ , we are done. • If N u ( γ ) = 0 for some u ∈ R n − , then M ( γ ) = 0 by Lemma 7, and f γ ( u, r ) → δ u ( γ ) as r → + .Due to the same reason for the case M ( γ ) = 0, ξ γ ( u ) lies in Y , i.e., the last ( d − n ) components of( w + ˚ w ) vanish. Meanwhile, ξ γ does not lie in Y , otherwise we have γC u ⊂ G /K which impliesthat γ ∈ Γ . As γ varies, u and u can be arbitrarily large. If there exist u , u such that γP , γP do not lie in Y , then the geodesic γγ C u meets or stays away from G /K . Both cases indicatesthat M ( γγ ) N u ( γγ ) = 0 (see the argument for the case M ( γ ) = 0). Let γ ′ = γγ , then we are1done. If γP or γP lies in X for any u , u , then for any i > n we have ( w + ˚ w ) i = 0 as r → w depends on u or u (not u ). This is equivalent to α i − u i +1 , u + β = − w i , ∀ i > n, for any u or u (which correspond to γ ). Assume that the above identity holds for infinitelymany u which can be large enough. Then we get − w i = u i +1 , − u ( i > n ), the proportion of thecoefficients of | u | of α i − u i +1 , and u + β . The proportion of the constant terms of α i − u i +1 , and u + β should also be equal to − w i = u i +1 , − u . This yields u i +1 , u = − u i +1 , − u from which weget u i +1 , = 0 ( i > n ). It follows that w i = 0 ( i > n ) and ξ γ lies in Y as ( w + ˜ w ) i = 0 for i > n ,but we have shown that ξ γ does not lie in Y .Now we show the first two properties of Proposition 2. Let γ = a r n w k ( γ ) and write k ( γ ) γ = n w a s k ∈ N AK where γ = a ℓ p k p ∈ AM ( p ∈ Z ). Then γγ = a r s n ( w + w ) /s k for some k ∈ K . Applying(18) and (19) to u = 0, we get the i -th component of w + ws : (cid:18) w + ws (cid:19) i = w i (1 − u ) + u i +1 , ℓ p + w i (1 + u ) − u i +1 , ℓ − p . Although M ( γ ) is nonzero under our assumption, it is possible that certain m i ( i > n ) might be zero.If m i = w i (1 − u )+ u i +1 , = 0 and w i (1+ u ) − u i +1 , = 0, then we can choose proper p (i.e., γ ) suchthat (cid:12)(cid:12)(cid:0) w + ws (cid:1) i (cid:12)(cid:12) is large enough. Here we require that p < | p | is large. Replacing γ with γγ ,we might as well assume that | w i | is large enough. If w i (1 − u )+ u i +1 , = w i (1+ u ) − u i +1 , = 0, then w i = u i +1 , = 0. It follows from (25) that n i = P nj =2 ( u i +1 ,j − w i u j ) u j − . Thus, | n i | achievesits minimal value at u = 0 ∈ F . There is nothing to prove in this case. Next we discuss the casewhere m i = 0 and w i = 0. It follows that w i (1+ u ) − u i +1 , = 0 (otherwise, w i = 0). The aboveformula shows that (cid:12)(cid:12)(cid:0) w + ws (cid:1) i (cid:12)(cid:12) can be large enough provided that p < | p | is large. Replacing γ with γγ we assume that | w i | is large enough. The case m i = 0 and w i = 0 is equivalent to the case w i (1 − u )+ u i +1 , = w i (1+ u ) − u i +1 , = 0 which has been discussed in above. In all, except the trivialcases (when w i = u i +1 , = 0) we can and will assume that | w i | is large enough. In view of Proposition3, this implies that both | m i | and M ( γ ) are large enough (again, note that M ( γ ) = 0, i.e., not all m i arezero). By (25) we can write n i as n P j =2 n ′ i,j − where n ′ i,j − = m i u j − + ( u i +1 ,j − w i u j ) u j − + w i (1 + u ) − u i +1 , n − . Each n ′ i,j − is a degree 2 polynomial with leading coefficient being m i . Consequently, the u at which N u ( γ ) achieves its minimal value must lie nearby u = ( u i ) where u i = − u i +1 ,j − w i u j m i . If | w i | issufficiently large, then u i is close to u j − u which lies in a bounded interval in R (independent of γ inview of Proposition 3). Likewise, based on (26) we can show that the u at which 2 P d − i = n m i n i achievesits minimal value also lies in a bounded domain in R n − . We omit the details.Finally, we show the third property of Proposition 2. Assume that there exist a sequence { γ i ∈ Γ r Γ | e γ i = e γ j for i = j } and u i ∈ R n − such that M ( γ i ) N u i ( γ i ) → i → ∞ . It follows from (20)that δ u i ( γ i ) → i → ∞ . Thus, δ u i ∗ ( γ ) → u i ∗ denote the point u at which δ u ( γ ) achievesits minimal value. Since u i ∗ lies in a fixed compact subset F ⊂ R n − , by passing to a subsequence ifnecessary, we assume that { u i ∗ } converges to u ∈ F . Then δ u ( γ i ) → i → ∞ . This contradictsCorollary 1. f and k f For the trace formula to be valid, f and k f should fulfill some conditions (see the end of Sect. 2.4). In thissection we examine these conditions. Remember that f ( g ) = Φ µ (cid:0) d G/K ( g · o, e · o ) (cid:1) for g ∈ G . Hence f is2bi- K -invariant and f = f K (see Sect. 2.4 for the definition of f U ). Next we show f ∈ C unif ( G ). Let U ⊂ G be a small enough compact neighborhood of e which is symmetric, i.e., U − = U (such neighborhoodexists: take U = V ∩ V − where V is a small neighborhood of e ). For any h , h ∈ U , we have (cid:12)(cid:12) d G/K ( h gh · o, e · o ) − d G/K ( g · o, e · o ) (cid:12)(cid:12) = (cid:12)(cid:12) d G/K ( gh · o, h − · o ) − d G/K ( g · o, e · o ) (cid:12)(cid:12) U )where diam( U ) means the diameter of U (modulo K ), i.e., diam( U ) = min u , u ∈ U d G/K ( u · o, u · o ). By thedefinition of f , one has f U ( g ) f ( g ) e µ diam( U ) . Hence, f U is integrable if f is so. The latter can be shown by the following computation Z G f ( g ) dg = Z N Z A Φ µ (cid:0) d G/K ( na · o, e · o ) (cid:1) dadn = 2 d (cid:18)r π µ (cid:19) d − K d − ( µ ) < ∞ where the second step is a copy of the computation for h f ( λ j ), dropping the term η ν j thereof (see Sect. 3).As a result, f ∈ C unif ( G ). For the rest of this section we check the locally uniform convergence of k f .The L ∞ -norm of any Laplace eigenfunction φ j on X satisfies the classical H¨ormander’s bound (see [ Ho ]):sup x ∈ X | φ j ( x ) | C λ ( d − / j k φ j k L ( X ) where λ j is the Laplace eigenvalue of φ j , C is uniform for all j . In our context, φ j ’s are orthonormal basisof L ( X ), so we have: sup x ∈ X | φ j ( x ) | C λ ( d − / j . For the convergence of k f , it suffices to consider those φ j ’s with large eigenvalues, i.e., ν j ∈ i R . Thus, we write ν j = i r j with r j ∈ R + . Then λ j = (cid:0) d − (cid:1) + r j .Substituting v = ir j , x = 1 into the following formula (see 8.432.5 of [ GR ]) K ν ( xz ) = Γ (cid:0) ν + (cid:1) (2 z ) ν x ν Γ (cid:0) (cid:1) Z ∞ cos xt dt ( t + z ) ν + , Re (cid:18) ν + 12 (cid:19) > , x > , | arg z | < π , we get K ir j ( z ) = Γ (1 / ir j )Γ(1 / 2) (2 z ) ir j Z ∞ cos t dt ( t + z ) / ir j , z > . The integration by parts shows that Z ∞ cos t dt ( t + z ) / ir j = (1 + 2 ir j ) Z ∞ t sin t dt ( t + z ) / ir j . The integral on the right hand side of the above equality absolutely converges for z ∈ R r { } . Thus, K ir j ( z ) is uniformly upper bounded by (cid:12)(cid:12) Γ (cid:0) + ir j (cid:1)(cid:12)(cid:12) r j for all z ∈ R r { } . By the Stirling formula onGamma function (where a , b ∈ R ): | Γ( a + ib ) | = √ π | b | a − / e −| b | π/ (cid:20) O (cid:18) | b | (cid:19)(cid:21) , as | b | → ∞ , we get a bound: K ir j ( µ ) = O (cid:0) r j e − π r j (cid:1) . Combining this bound with H¨ormander’s bound, we have: K ir j ( µ ) φ j ( z ) φ j ( w ) = O (cid:18) r j e − π r j h r ( d − / j i (cid:19) = O (cid:0) r dj e − π r j (cid:1) as j → ∞ . The spectrum { λ j } of the Laplacian is discrete with ∞ as the unique accumulation point and eacheigenvalue λ j occurs with finite multiplicity, so is { r j ∈ R + } . Let N ( x ) be the counting function ofLaplace eigenvalues with multiplicities over X : N ( x ) := X φ j : r j x . N ( x ) for large x (see [ MP ]). N ( x ) = vol( X )(4 π ) d/ Γ (cid:0) d + 1 (cid:1) x d + o (cid:0) x d (cid:1) , as x → ∞ . Let A j = p λ j − r j , then A j = O (cid:0) r − j (cid:1) as j → ∞ . With the bound on K ir j ( µ ) φ j ( z ) φ j ( w ) obtained inabove and the formula h f ( λ j ) = 2 d (cid:16)q π µ (cid:17) d − K ν j ( µ ), we have: k f ≪ X j r dj e − π r j < X j λ d j e − π ( √ λ j − A j ) ≍ X j λ d j e − π √ λ j = Z ∞ x d e − π √ x dN ( x ) . Here dN ( x ) means the measure on R + with mass 1 at x = r j (multiplicities counted), otherwise 0.Integration by parts shows that Z ∞ x d e − π √ x dN ( x ) = x d e − π √ x N ( x ) (cid:12)(cid:12)(cid:12) ∞ − Z ∞ e − π √ x (cid:18) d x d − − π x d − (cid:19) N ( x ) dx. Applying Weyl’s law on N ( x ) to the right hand side of this identity, we see that the integral on the lefthand side converges. Thus, we have shown the absolute and locally uniform convergence of k f . Acknowledgement. The author would like to thank Professor A. Deitmar for introducing him this topicas well as many helpful discussions. References [ CG] L. Corwin, F. P. Greenleaf, Representations of nilpotent Lie groups and their applications, Volume1 , Cambridge University Press, 2004[ Ch] P. R. Chernoff, Essential self-adjointness of powers of generators of hyperbolic equations , J. Func-tional Analysis (1973)[ DE] A. Deitmar, S. Echterhoff, Principles of harmonic analysis , Springer-Verlag, 2008[ FJ] J. 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Partial Diff.Equations (1992), No. 1-2[ Zh] W. Zhang, Fourier transform and the global Gan-Gross-Prasad conjecture for unitary groups ,( ) 2014, no. 3 Author’s address : Depart. of Mathematics, Bar-Ilan Univ., Ramat-Gan 52900, Israel Author’s email ::