46
GEODESIC RESTRICTIONS OF ARITHMETIC EIGENFUNCTIONS SIMON MARSHALL Abstract Let X be an arithmetic hyperbolic surface arising from a quaternion division alge- bra over Q. Let be a Hecke–Maass form on X , and let ` be a geodesic segment. We obtain a power saving over the local bound of Burq, Gérard, and Tzvetkov for the L 2 -norm of restricted to `, by extending the technique of arithmetic amplifica- tion developed by Iwaniec and Sarnak. We also improve the local bounds for various Fourier coefficients of along `. 1. Introduction If X is a compact Riemannian manifold and is a Laplace eigenfunction on X satisfying ĩ C 2 D 0, it is an interesting problem to study the extent to which can concentrate on small subsets of X . Two well-studied formulations of this problem are to normalize by k k 2 D 1, and either bound k k p for 2 p 1 or bound the L p -norms of restricted to some submanifold. We shall be interested in both of these problems in the case where X is two-dimensional and the submanifold we restrict to is a geodesic segment `. The basic upper bound for k k p in this case was proven by Sogge [30] (see also Avakumovi´ c[1] and Levitan [21] when p D1) and is k k p ı.p/ ; (1) where ı.p/ is given by ı.p/ D ´ 1=2 2=p; p 6; 1=4 1=2p; 2 p 6: The standard bound for k j ` k p is due to Burq, Gérard, and Tzvetkov [11] and is k j ` k p ı 0 .p/ ; (2) DUKE MATHEMATICAL JOURNAL Vol. 165, No. 3, © 2016 DOI 10.1215/00127094-3166736 Received 16 July 2013. Revision received 11 March 2015. First published online 10 December 2015. 2010 Mathematics Subject Classification. Primary 35P20; Secondary 11F25, 11F41. 463

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GEODESIC RESTRICTIONS OF ARITHMETICEIGENFUNCTIONS

SIMON MARSHALL

AbstractLet X be an arithmetic hyperbolic surface arising from a quaternion division alge-bra over Q. Let be a Hecke–Maass form on X , and let ` be a geodesic segment.We obtain a power saving over the local bound of Burq, Gérard, and Tzvetkov forthe L2-norm of restricted to `, by extending the technique of arithmetic amplifica-tion developed by Iwaniec and Sarnak. We also improve the local bounds for variousFourier coefficients of along `.

1. IntroductionIf X is a compact Riemannian manifold and is a Laplace eigenfunction on Xsatisfying � C�2 D 0, it is an interesting problem to study the extent to which can concentrate on small subsets ofX . Two well-studied formulations of this problemare to normalize by k k2 D 1, and either bound k kp for 2 � p �1 or boundthe Lp-norms of restricted to some submanifold. We shall be interested in bothof these problems in the case where X is two-dimensional and the submanifold werestrict to is a geodesic segment `. The basic upper bound for k kp in this case wasproven by Sogge [30] (see also Avakumovic [1] and Levitan [21] when p D1) andis

k kp� �ı.p/; (1)

where ı.p/ is given by

ı.p/D

´1=2� 2=p; p � 6;

1=4� 1=2p; 2� p � 6:

The standard bound for k j`kp is due to Burq, Gérard, and Tzvetkov [11] and is

k j`kp� �ı0.p/; (2)

DUKE MATHEMATICAL JOURNALVol. 165, No. 3, © 2016 DOI 10.1215/00127094-3166736Received 16 July 2013. Revision received 11 March 2015.First published online 10 December 2015.2010 Mathematics Subject Classification. Primary 35P20; Secondary 11F25, 11F41.

463

464 SIMON MARSHALL

where ı0.p/ is given by

ı0.p/D

´1=2� 1=p; p � 4;

1=4; 2� p � 4:

Both of these bounds are sharp when X is the round 2-sphere, but can be strength-ened under extra geometric assumptions on X such as negative curvature (see, e.g.,[12], [13], [31]–[33]). It should be noted that all such improvements in the negativelycurved case are by at most a power of log�.

We now let X be a compact congruence arithmetic hyperbolic surface arisingfrom a quaternion division algebra over Q. We let be a Hecke–Maass form on X ,which we shall always assume to be L2-normalized. In this context, it is more naturalto let � 2C be the spectral parameter of , which satisfies � C .1=4C �2/ D 0.As we are considering large-eigenvalue asymptotics, we will also assume that � 2R,in which case the bounds (1) and (2) remain unchanged. For theseX , Iwaniec and Sar-nak [20] have shown that the bound k k1� �1=2 given by (1) may be strengthenedby a power to k k1�� �

5=12C� . Their approach, known as arithmetic amplification,is to construct a projection operator onto by using the Hecke operators as well asthe wave group. It has been adapted by other authors to study the pointwise norms ofarithmetic eigenfunctions in various aspects (see, e.g., [6]–[9], [19], [34] as well asthe alternative approach taken in [4]). In this paper we apply amplification to a newkind of semiclassical problem, namely, improving the exponent in the bound (2) fork j`k2. Our main result is as follows.

THEOREM 1.1Let be a Hecke–Maass form on X with spectral parameter �. For any geodesicsegment ` of unit length we have

k j`k2�� �3=14C�; (3)

where the implied constant is independent of `.

We may combine Theorem 1.1 with the main theorem of [2] to obtain an improve-ment over the local bound k k4� �1=8.

COROLLARY 1.2We have k k4�� �

1=8�1=56C� .

This is much weaker than the bound k k4 �� �� announced by Sarnak and

Watson [28, Theorem 3] in the case of X D SL.2;Z/nH, although their result maybe conditional on the Ramanujan conjecture. See also [5] for results in the case of

GEODESIC RESTRICTIONS OF ARITHMETIC EIGENFUNCTIONS 465

holomorphic eigenforms. The results of [10] give an equivalence (up to factors of ��)between a sublocal bound for k k4 and one for k j`k2 that is uniform in `, andso the bound of Sarnak and Watson implies Theorem 1.1 with an exponent of 1=8.However, we feel that our method is of interest as it does not rely on special valueidentities or summation formulas, and we hope to apply it to restriction problems onother groups by combining it with the techniques of [23]; see, for instance, [22] forthe case of restricting an SL3 Maass form to a maximal flat subspace.

1.1. Fourier coefficients along geodesicsThe methods we use to prove Theorem 1.1 also allow us to prove bounds for Fouriercoefficients of along `. We shall describe the bounds that one expects on a gen-eral Riemannian surface before stating our improvements in the arithmetic case. Asbefore, let Y be a compact Riemannian surface, let be a Laplace eigenfunctionon Y , and let `� Y be a unit geodesic segment. We let ` W Œ0; 1�!X be an arc lengthparameterization of `, and let b 2 C10 .R/ be a function with supp.b/ � Œ0; 1�. Fort 2R, define

h ;beitxi D

Z 1�1

b.x/e�itx �`.x/

�dx:

The bound for h ;beitxi that one expects depends on both � and t . In the “classicalrange” when jt j � �, it is

h ;beitxi� �1=2�1C �� jt j

��1=2: (4)

This is proved in [14, Theorem 1.1] and [36, Section 2] when t D 0. Moreover, thearguments used there should imply (4) in full. When Y is hyperbolic, (4) follows fromLemma 3.8; see also [27] for the case when Y is hyperbolic, ` is closed, and t D 0. IfY is assumed to be negatively curved and t D 0, (4) was improved to h ;bi D o.1/in [14].

Now assume that is a Hecke–Maass form on an arithmetic surfaceX as before.We have the following improvement over (4).

THEOREM 1.3Let 1=2 > ı > 0, and let Iı D Œ�1C ı;�ı�[ Œı; 1� ı�.(a) We have h ;bi�� �

�1=12C� .(b) If t=� 2 Iı , then we have h ;beitxi��;ı �

�1=18C� .(c) Define ˇDmin j�˙ t j. If ˇ � �2=3, then we have h ;beitxi�� �

5=24C�.1C

ˇ/1=24.All of these bounds are uniform in t and `.

466 SIMON MARSHALL

RemarkThe bound ˇ � �2=3 in Theorem 1.3 could be replaced with �1�ı for any ı > 0;however, when ˇ � �1=7C� the bound h ;beitxi �� �

5=24C�.1C ˇ/1=24 is weakerthan the local bound of Lemma 3.8.

As in [20], Theorems 1.1 and 1.3 can both be strengthened under the assump-tion that the Hecke eigenvalues of are not small. In the case of Theorem 1.1 andTheorem 1.3(c), this assumption allows us to employ an amplifier of sufficient lengththat it becomes profitable to estimate the Hecke recurrence by using spectral methods,rather than the standard diophantine ones. We state the result we obtain in this way asa separate theorem. Let �.n/ be the automorphically normalized Hecke eigenvaluesof , and assume that they satisfy the boundsX

N�p�2N

ˇ�.p/

ˇ�� N

1�� (5)

for all N � 2 and ˇ�.p/

ˇ� 2p� (6)

for some � < 1=2 and p prime. Note that (6) is known with � D 7=64 (see [3]). Wethen prove the following.

THEOREM 1.4Suppose that the normalized Hecke eigenvalues �.n/ satisfy (5) and (6). We have

k j`k2�� �1=.8�8�/C�:

If ˇDmin jt ˙ �j and ˇ � �2=3, then we have

h ;beitxi�� ��=2C�.1C ˇ/1=4��=2: (7)

Both bounds are uniform in t and `.

In particular, Theorem 1.4 gives h ;beitxi �� �� when jt � �j � �� under the

assumption that � D 0. We note that (7) becomes weaker than the local bound ofProposition 3.8 when ˇ � �1=2C� .

1.2. Relations with L-values and subconvexityWhen ` is a closed geodesic, the integrals h ;eitxi are the Fourier coefficients of along `, and they may be expressed in terms of the L-values L.1=2; ˝ ��/by a formula of Waldspurger [35]. Here, � is a Grossencharacter of a real quadratic

GEODESIC RESTRICTIONS OF ARITHMETIC EIGENFUNCTIONS 467

field that corresponds to the chosen complex exponential on `, and �� is the associ-ated theta series on GL2. When jt j � .1� ı/�, the bound h ;eitxi � 1 correspondsto the convex bound for L.1=2; ˝ ��/, as one sees from explicit forms of Wald-spurger’s formula such as [24]. As a result, Theorem 1.3 gives a subconvex boundfor L.1=2; ˝ ��/ in this range. In the other direction, Michel and Venkatesh haveproven the strong bound h ;eitxi � ��ı uniformly for t 2 R, by combining Wald-spurger’s formula with their general solution of subconvexity for GL2 (see [25, Sec-tion 1.4], in particular equation (1.6)).

In the case whenX D SL.2;Z/nH and ` is the infinite vertical geodesic from i toi1, or a compact segment thereof, Ghosh, Reznikov, and Sarnak [18, Theorems 1.1,1.3] have proved the essentially optimal bound 1�k j`k2�� �

� by usingL-values.They prove a similar bound for restrictions to a closed horocycle.

1.3. Outline of the proofWe first give the rough idea behind the amplification method. The bound k j`k2��1=4 is proved by taking an approximate spectral projector T onto eigenfunctionswith frequency near �, defining R to be the operator of restriction to `, and boundingthe mapping norm RT W L2.X/! L2.`/. We obtain our improvement by letting T

be a Hecke operator that approximately projects onto functions with the same systemof Hecke eigenvalues as and bounding the norm of RT T . The asymptotic orthog-onality of systems of Hecke eigenvalues means that we expect RT T to have smallnorm on Maass forms other than . Correspondingly, we may prove a bound for thenorm of RT T by using geometric methods. Combining this with the fact that is aneigenfunction of T T with large eigenvalue gives the result.

We now describe our method in more detail. The basic amplification inequalitywe use is Proposition 3.1. We prove this in Section 3.1 by taking an amplified pretraceformula, but instead of evaluating at a point on the diagonal as in [20], we integrate itagainst a test function on `� `. The resulting identity may be thought of as a relativetrace formula. We then drop all terms on the spectral side except the one in which weare interested.

We must bound the geometric side of this inequality, which requires solving adiophantine problem and an analytic problem. The diophantine problem is estimatingthe number of times a Hecke operator maps ` back close to itself, in the sense that�` must be contained in a small tubular neighborhood of `. We do this in Section 3.2by adapting the proof of the corresponding lemma [20, Lemma 1.3], and in Section 4we prove a stronger bound by using spectral methods, conditional on (5) and (6). Theanalytic problem is to estimate an oscillatory integral over �` � `. This is the mainnew ingredient and occupies Sections 5–7 of the paper.

468 SIMON MARSHALL

By combining these two ingredients, we prove Theorem 1.3(a),(b) in Section 3.3,Theorem 1.1 and Theorem 1.3(c) in Section 3.4, and Theorem 1.4 in Section 4.

2. NotationThroughout the paper, the notation A�B will mean that there is a positive constantC such that jAj � CB , and AB will mean that there are positive constants C1 andC2 such that C1B �A� C2B .

2.1. Quaternion algebrasLetAD .a;b

Q/ be a quaternion division algebra over Q, where we assume that a; b 2 Z

are square-free with a > 0. We choose a basis 1;!;�;!� for A over Q that satisfies!2 D a, �2 D b, and !�C�! D 0. We denote the norm and trace by N.˛/D ˛˛and tr.˛/D ˛C ˛. We let R be a maximal order in A (or more generally an Eichlerorder, see [15]), and for m� 1 let

R0.m/D®˛ 2R

ˇN.˛/Dm

¯:

R0.1/ is the group of elements of norm 1; it acts on R0.m/ by multiplication on theleft, and R0.1/nR0.m/ is known to be finite (see [15]). Fix an embedding W A!M2.F /, the .2� 2)-matrices with entries in F DQ.

pa/ by

.˛/D

b�

!;

where

˛D x0C x1! C .x2C x3!/�D C ��:

For m > 0, we let R.m/ � PSL.2;R/ be the image of .R0.m// under projection.We define the lattice � DR.1/, which is cocompact as we assumed A to be a divisionalgebra and let X D �nH.

2.2. Hecke operatorsWe define the Hecke operators Tn WL2.X/!L2.X/, n� 1, by

Tnf .z/D1pn

X˛2R.1/nR.n/

f�.˛/z

�:

There is a positive integer q (depending on R) such that for .n; q/D 1, Tn has thefollowing properties (see [15]):

Tn D T�n ; that is, Tn is self-adjoint;

TmTn DX

d j.m;n/

Tmn=d2 :

GEODESIC RESTRICTIONS OF ARITHMETIC EIGENFUNCTIONS 469

2.3. Lie groups and algebrasWe let G D PSL.2;R/. We let K , A, and N be the standard subgroups of G, withparameterizations

k.�/D

�cos�=2 sin�=2� sin�=2 cos�=2

�; a.y/D

�ey 0

0 1

�; n.x/D

�1 x

0 1

�:

In particular, k.�/ represents an anticlockwise rotation by � about the point i . Wedenote the Lie algebra of G by g and equip g with the norm

k k W

�X1 X2

X3 �X1

�7!

qX21 CX

22 CX

23 : (8)

This norm defines a left-invariant metric on G, which we denote by d . We denote theLie algebras of K , A, and N by k, a, and n and write the Iwasawa decomposition as

gD n.g/ exp�A.g/

�k.g/D exp

�N.g/

�exp

�A.g/

�k.g/: (9)

We define

H D

�1=2 0

0 �1=2

�2 a; Xn D

�0 1

0 0

�2 n; XkD

�0 1=2

�1=2 0

�: (10)

We identify a'R under the mapH 7! 1 and consider A.g/ as a function A WG!R

under this identification, and likewise for n and N.g/. We let 's denote the standardspherical function with spectral parameter s on H or G, depending on the context.

We let dg be the Haar measure on G that is the extension of the usual hyperbolicvolume by the measure of mass 1 on K .

2.4. Maass formsWe let 2 L2.X/ be a Hecke–Maass form that is an eigenfunction of � and theoperators Tn with .n; q/D 1. We let �.n/ be the Hecke eigenvalues of and � be itsspectral parameter, so that

Tn D �.n/ ;

� C .1=4C �2/ D 0:

We assume that k k2 D 1 with respect to the hyperbolic volume on X . Note thatbecause � and Tn with .n; q/ D 1 are self-adjoint, we may assume that is real-valued.

3. AmplificationThis section contains all the arithmetic methods used in the proof of Theorems 1.1and 1.3. The amplification inequality that we derive from the pretrace formula is given

470 SIMON MARSHALL

in Proposition 3.1. The estimation of Hecke returns is carried out in Section 3.2. Theo-rems 1.3 and 1.1 are proven in Section 3.3 and Section 3.4, respectively, by combiningthese ingredients with bounds for oscillatory integrals proven in Sections 5–7.

We choose g0 2G and let our geodesic segment ` be ¹g0a.x/ W 0� x � 1º. Wefix a function b 2 C10 .0; 1/, which we may assume to be real-valued. If 2L1loc.R/,we shall study the integral

h ;bi D

Z 1�1

b.x/ �g0a.x/

�dx:

3.1. An amplification inequalityWe fix a real-valued function h 2 C1.R/ of Paley–Wiener type that is nonnegativeand satisfies h.0/D 1. Define h0

�by h0

�.s/D h.s � �/C h.�s � �/, and let k0

�be

the K-biinvariant function on H with Harish-Chandra transform h� (see [17] or [29]for definitions). The Paley–Wiener theorem of Gangolli [16] implies that k0

�is of

compact support that may be chosen arbitrarily small. Define k� D k0� � k0�

, whichhas Harish-Chandra transform h� D .h

0�/2. If 2L1loc.R/ and g 2G, we define

I.�;;g/D

“ 1

�1

b.x1/b.x2/k��a.�x1/ga.x2/

�dx1 dx2:

We assume that the supports of b and k� are small enough that I.�;;g/D 0 unlessd.g; e/ � 1. Let N � 1 be an integer, and let ¹˛nºNnD1 be complex numbers, both tobe chosen later. Define the Hecke operator

T DXn�N

˛nTn:

Our main amplification inequality is the following.

PROPOSITION 3.1We haveˇ

hT ;biˇ2�

Xm;n�N

j˛m˛njX

d j.m;n/

dpmn

X�2R.mn=d2/

ˇI.�;;g�10 �g0/

ˇ:

ProofConsider the kernel function

K.x;y/DX�2�

k�.x�1�y/

on �nG. Let ¹ iº be an orthonormal basis for L2.X/ consisting of Hecke–Maassforms with 2 ¹ iº. Let i be the spectral parameter of i . Then K.x;y/ has thespectral expansion

GEODESIC RESTRICTIONS OF ARITHMETIC EIGENFUNCTIONS 471

K.x;y/DXi

h�. i / i .x/ i .y/:

If we apply Tn to K.x;y/ in the x variable and equate the geometric and spectralexpansions, we obtain

1pn

X�2R.n/

k�.x�1�y/D

Xi

h�. i /Tn i .x/ i .y/:

Applying T T � therefore givesXm;n�N

˛m˛nX

d j.m;n/

dpmn

X�2R.mn=d2/

k�.x�1�y/D

Xi

h�. i /T i .x/T i .y/:

If we integrate this identity against b � b on g0A� g0A, we obtainXm;n�N

˛m˛nX

d j.m;n/

dpmn

X�2R.mn=d2/

I.�;;g�10 �g0/

DXi

h�. i /ˇhT i ; bi

ˇ2:

We have h0�. i / 2R; hence h�. i /� 0, for all i . We may therefore drop all terms on

the right-hand side except , which completes the proof.

3.2. Estimation of Hecke returnsWe now estimate how many times the Hecke operators map the geodesic g0A closeto itself, which amounts to giving a bound for the counting function

M.g;n; �/Dˇ®� 2R.n/

ˇd.g�1�g; e/� 1;d.g�1�g;A/� �

¯ˇ:

Our bound for M.g;n; �/ is Lemma 3.3 below. We first need a diophantine lemma.

LEMMA 3.2Let A;B > 0, y �A, and B � ı � 0. Then we haveˇ®

r; s 2 Z W jr2 � ys2 � nj � ın; r2C s2 �Bn¯ˇ� .n=ı/�.n

pıC 1/

for all � > 0, where the implied constant depends only on A, B , and �.

ProofLet Q � 1. We can find relatively prime integers p, q with 1� q �Q such thatˇp

q� y

ˇ�

1

qQ:

472 SIMON MARSHALL

The condition jr2 � ys2 � nj � ın then implies that

ˇr2 � .p=q/s2 � n

ˇ� ınC

s2

qQ;

jqr2 � ps2 � qnj � ınqCBn

Q:

Choose QD ı�1=2CA�1, which implies that p � 1. We then have

jqr2 � ps2 � qnj � ın.ı�1=2CA�1/CBnı1=2

� nı1=2.1CB CpB=A/:

If we let C D 1CB CpB=A, the number of r and s satisfying jqr2 � ps2 � qnj �

Cnı1=2 and r2C s2 �Bn is equal toXjm�qnj�Cnı1=2

ˇ¹qr2 � ps2 Dm;r2C s2 �Bnº

ˇ:

Let O denote the ring of integers in the field Q.ppq/, and let j j1 and j j2

denote its two archimedean valuations. Every solution of qr2 � ps2 Dm determinesan element z D qr C s

ppq 2O with Nz Dmq. We therefore haveˇ

¹qr2 � ps2 Dm;r2C s2 �Bnºˇ

�ˇ®z 2O WNz Dmq; jzj1C jzj2 � 2

pBn.qC

ppq/

¯ˇ:

The set on the right-hand side maps to the set of ideals in O with norm mp, and thenumber of such ideals is bounded by �.mq/�� .mq/

�� .n=ı/� . The fibers of thismap are orbits under multiplication by units, and the condition

jzj1C jzj2 � 2pBn.qC

ppq/� nı�1=2

and the fact that the regulators of real quadratic fields are bounded away from 0implies that the fibers have size�� .n=ı/

� . Summing this bound over m completesthe proof.

LEMMA 3.3If � � 1, we have the bound

M.g;n; �/�� .n=�/�.np� C 1/

uniformly in g.

GEODESIC RESTRICTIONS OF ARITHMETIC EIGENFUNCTIONS 473

ProofWe follow the proof of the corresponding lemma (see [20, Lemma 1.3]). We first givea parameterization of the group gAg�1 of translations along the geodesic gA. LetŒ˛;ˇ; �� be the quadratic form associated to gA, so that

ˇ2 � 4˛� D 1 (11)

and the roots of ˛z2C ˇzC � are the endpoints of gA. Note that the case when oneendpoint is at i1 corresponds to ˛D 0. We may parameterize gAg�1 as

gAg�1 D

²�t � ˇu �2�u

2˛u t C ˇu

� ˇt2 � u2 D 1; t > 0

³: (12)

As � is cocompact, we may assume that g lies in a fixed compact set �. This impliesthat ˛;ˇ; �� 1. If d.g�1�g; e/� 1, we have

d.g�1�g;A/� �! �D y CO.�/ with y 2 gAg�1: (13)

Because � 2R.n/, � is the image of an element z 2R0.n/. We have

.z/D

�x0 � x1

pa x2C x3

pa

bx2 � bx3pa x0C x1

pa

�(14)

with det..z//D n and Exi 2 Z for some E 2 Z depending on the order R. Com-bining (12), (13), and (14) gives

x0pnD t CO.�/; (15)

x1pa

pnD ˇuCO.�/; (16)

x2pnD�

�� �

˛

b

uCO.�/;

x3pa

pnD�

�� C

˛

b

uCO.�/:

Equation (11) implies that one of jˇj and j� ˙ ˛=bj must be � 1=4jbj. We shallassume that ˇ � 1=4jbj, as the other cases are similar. Because d.g�1�g; e/� 1, theentries of .z/=

pn must be bounded in terms of �, which gives xi �

pn for all i .

Combining this with equations (15) and (16) gives

1D t2 � u2 Dx20n�x21a

nˇ2CO.�/;

474 SIMON MARSHALL

or

x20 �x21a

ˇ2� n� �n:

Combining this with xi �pn, Lemma 3.2 implies that the number of choices for x0

and x1 is�� .n=�/�.np� C 1/. If we fix x0 and x1, then x2 and x3 must satisfy

�bx22 C abx23 D n� x

20 C ax

21� n

and xi �pn. By working with the field Q.

pa/ as in the proof of Lemma 3.2, we

see that the number of choices for x2 and x3 is�� n� . This completes the proof.

3.3. Bounds for geodesic periodsWe now prove Theorem 1.3(a),(b). As and b are real-valued, we may assume thatt � 0. We begin with a specialization of Proposition 3.1.

PROPOSITION 3.4If t=� 2 Œı; 1� ı�, we haveˇ

hT ;beitxiˇ2�ı;� N

����Xn�N

j˛nj2CN��1=3

�Xn�N

j˛nj2

:

We also haveˇhT ;bi

ˇ2�� N

����Xn�N

j˛nj2CN��1=2

�Xn�N

j˛nj2

:

ProofIf we apply Proposition 3.1 with .x/D eitx and denote I.�; eitx; g/ by I.�; t; g/,we haveˇhT ;beitxi

ˇ2�

Xm;n�N

j˛m˛njX

d j.m;n/

dpmn

X�2R.mn=d2/

ˇI.�; t; g�10 �g0/

ˇ: (17)

We shall estimate the right-hand side of (17) by combining Lemma 3.3 with the fol-lowing bounds for I.�; t; g/, proven in Section 7. We let ı be as in Theorem 1.3.

PROPOSITION 3.5If t=� 2 Œı; 1� ı�, we have

I.�; t; g/�ı

´.1C �d.g;A//�1=2; d.g;A/� ��1=3;

��1=3; d.g;A/� ��1=3:

GEODESIC RESTRICTIONS OF ARITHMETIC EIGENFUNCTIONS 475

We also have

I.�; 0; g/��1C �d.g;A/

��1=2:

We shall only consider the case t=� 2 Œı; 1 � ı�, as the case t D 0 is similar.The dependence of implied constants on the arbitrarily small � > 0 will be omit-ted for the remainder of the proof. If we assume that d.g�10 �g0; e/ � 1, then wehave d.g�10 �g0;A/ � 1, and we cover Œ0; 1� with the intervals I0 D Œ0; ��1�, Ik DŒek�1��1; ek��1� for 1 � k � 2

3log�, and I1 D Œe�1��1=3; 1�. When d.g�10 �g0;

A/ 2 I0 we apply the bounds

I.�; t; g�10 �g0/�ı 1;

M.g0; n;��1/ � n���.n��1=2C 1/

from Proposition 3.5 and Lemma 3.3 to obtainXm;n�N

Xd j.n;m/

d j˛m˛njpmn

X�2R.mn=d2/;

d.g�10�g0;A/2I0

ˇI.�; t; g�10 �g0/

ˇ

�ı N���

Xm;n�N

j˛m˛njX

d j.n;m/

dpmn

���1=2

mn

d2C 1

�ı N���

Xm;n�N

j˛m˛njX

d j.n;m/

�pmnd

��1=2Cdpmn

:

When d.g�10 �g0;A/ 2 Ik we have

I.�; t; g�10 �g0/�ı e�k=2;

M.g0; n; ek��1/ � n���.n��1=2ek=2C 1/;

which givesXm;n�N

Xd j.n;m/

d j˛m˛njpmn

X�2R.mn=d2/;

d.g�10�g0;A/2Ik

ˇI.�; t; g�10 �g0/

ˇ

�ı N���

Xm;n�N

j˛m˛njX

d j.n;m/

dpmn

���1=2

mn

d2C e�k=2

�ı N���

Xm;n�N

j˛m˛njX

d j.n;m/

�pmnd

��1=2Cdpmn

e�k=2:

476 SIMON MARSHALL

When d.g�10 �g0;A/ 2 I1 we have

I.�; t; g�10 �g0/�ı ��1=3;

M.g0; n; 1/ � n1C�;

so that Xm;n�N

Xd j.n;m/

d j˛m˛njpmn

X�2R.mn=d2/;

d.g�10�g0;A/2I1

ˇI.�; t; g�10 �g0/

ˇ

�ı N���

Xm;n�N

j˛m˛njX

d j.n;m/

pmn

d��1=3:

Combining these, and noting that we are summing over� log� values of k, weobtainˇhT ;beitxi

ˇ2�ı N

���X

m;n�N

j˛m˛njX

d j.n;m/

�pmnd

��1=3Cdpmn

: (18)

As in [20, p. 310], we haveXm;n�N

Xd j.n;m/

pmn

dj˛n˛mj �N 1C�

�Xn�N

j˛nj2; (19)

and Xm;n�N

Xd j.m;n/

dpmnj˛n˛mj �N �

Xn�N

j˛nj2: (20)

Combining (18) with (19) and (20) completes the proof.

To deduce Theorem 1.3(b) from Proposition 3.4, choose ¹˛nº to be the amplifierused in [20]. It follows as on [20, p. 311] thatˇ

h ;beitxiˇ2�ı;� N

���.N�1=2CN��1=3/;

and choosing N D �2=9 completes the proof. Case (a) follows by using the sameamplifier with N D �1=3.

3.4. Bounds for L2-normsTo prove Theorem 1.1, it suffices to bound the L2-norm of b.x/ .g0a.x// 2L2.R/for b 2 C10 .R/ real-valued with supp.b/ � Œ0; 1�, provided the bound is uniformin g0. If f 2 C10 .R/, define its Fourier transform bf by

GEODESIC RESTRICTIONS OF ARITHMETIC EIGENFUNCTIONS 477

bf ./D Z 1�1

f .x/e�i�x dx;

and extend this to an operator on L2.R/. Let ˇ be a parameter satisfying 1� ˇ � �.Define HC

ˇ;H�

ˇ�L2.R/ to be the spaces of functions whose Fourier support lies in

Œ˙��ˇ;˙�Cˇ�, and defineHˇ DHCˇCH�

ˇ. Let…ˇ be the orthogonal projection

onto Hˇ , and likewise for …˙ˇ

and H˙ˇ

.We shall bound …ˇb and .1�…ˇ /b separately, by applying Proposition 3.1

with 2Hˇ and 2H?ˇ

. As and b are real-valued it suffices to bound …Cˇb .

We shall need the following bounds for I.�;;g/.

PROPOSITION 3.6If 2H?

ˇsatisfies kk2 D 1 and e 2G is the identity, we have

I.�;; e/� �1=2ˇ�1=2:

PROPOSITION 3.7Suppose that ˇ � �2=3 and that 2HC

ˇsatisfies kk2 D 1. We haveˇ

I.�;;g/ˇ�� �

1=2C� (21)

for all g, while if �0 > 0 and d.g;A/� ��1=2C�0ˇ1=2 we haveˇI.�;;g/

ˇ��0;A �

�A: (22)

We shall prove Proposition 3.6 in Section 5, and Proposition 3.7 in Section 6.Using these, we may prove the following bounds for .1�…ˇ /b and …C

ˇb . The-

orem 1.1 follows from combining these with ˇ D �1=7, and Lemma 3.9 implies The-orem 1.3(c).

LEMMA 3.8We have k.1�…ˇ /b k2� �1=4ˇ�1=4.

ProofLet 2 H?

ˇwith kk2 D 1. We pass to a finite-index sublattice � 0 � � such that

� 0nH has injectivity radius � 10 and apply Proposition 3.1 with T the identity oper-ator. Then only � D e contributes to the geometric side, and we obtain jhb ;ij2 �jI.�;; e/j. Proposition 3.6 then gives k.1�…ˇ /b k2� �1=4ˇ�1=4.

LEMMA 3.9If ˇ � �2=3, we have k…C

ˇb k2�� �

5=24C�ˇ1=24 uniformly in ˇ and `.

478 SIMON MARSHALL

ProofWe again haveˇhT ;bi

ˇ2�

Xm;n�N

j˛m˛njX

d j.m;n/

dpmn

X�2R.mn=d2/

ˇI.�;;g�10 �g0/

ˇ: (23)

Choose �0 > 0. Proposition 3.7 implies that we only need to consider the terms in(23) with d.g�10 �g0; e/� 1 and d.g�10 �g0;A/� �

�1=2C�0ˇ1=2. Lemma 3.3 gives

M.g;n;��1=2C�0ˇ1=2/�� n���0C�.n��1=4ˇ1=4C 1/;

and so we haveXm;n�N

Xd j.n;m/

d j˛m˛njpmn

X�2R.mn=d2/

ˇI.�;;g�10 �g0/

ˇ�� N

���X

m;n�N

j˛m˛njX

d j.n;m/

dpmn

�1=2M.g;n;��1=2C�0ˇ1=2/

COA;�0.��A/

�� N���0C�

Xm;n�N

j˛m˛njX

d j.n;m/

�pmnd

�1=4ˇ1=4Cdpmn

�1=2

COA;�0.��A/: (24)

Combining (24) with (19) and (20) and choosing �0 small givesˇhT ;bi

ˇ2�� N

�����1=2

Xn�N

j˛nj2CN�1=4ˇ1=4

�Xn�N

j˛nj2COA.�

�A/;

and Lemma 3.9 now follows as in Section 3.3 by choosing N D �1=6ˇ�1=6.

4. Spectral estimation of Hecke returnsWe now prove Theorem 1.4 by improving the amplifier used in Lemma 3.9. Our newingredient is a spectral method for estimating M.g;n; �/, which works by estimatingthe sum of M.g;n; �/ over a range of n rather than one n at a time. The bound weobtain is Proposition 4.2, which allows us to prove the following result.

PROPOSITION 4.1Assume that satisfies (5) and (6), and that 1� ˇ � �2=3. We have the bound

k…Cˇb k2�� �

�=2C�ˇ1=4��=2:

GEODESIC RESTRICTIONS OF ARITHMETIC EIGENFUNCTIONS 479

Theorem 1.4 follows by choosing ˇ D �.1�2�/=.2�2�/ and combining this withLemma 3.8.

ProofWe maintain the notations of Section 3. Let �0 > 0, and letN D �1=2C�0ˇ�1=2. DefineT1 to be the operator

T1 DX

N=2<p<N

�.p/Tp:

Proposition 3.1 givesˇhT1 ;bi

ˇ2�

XN=2<p<N

X�2�

�.p/2ˇI.�;;g�10 �g0/

ˇC

XN=2<p1;p2<N

j�.p1/�.p2/jpp1p2

X�2R.p1p2/

ˇI.�;;g�10 �g0/

ˇ: (25)

We apply (21) and our assumption that j�.p/j � 2p� , which givesˇhT1 ;bi

ˇ2�� N

1C2��1=2C�

CN 2�X

N=2<p1;p2<N

1pp1p2

X�2R.p1p2/

ˇI.�;;g�10 �g0/

ˇ:

Enlarging the sum to one over all N 2=4 < n < N 2 with .n; q/D 1, where q is theinteger defined in Section 2.2, givesˇ

hT1 ;biˇ2�� N

1C2��1=2C�

CN 2�Xn�N2

.n;q/D1

1pn

X�2R.n/

ˇI.�;;g�10 �g0/

ˇ: (26)

By Proposition 3.7, we only need to consider the terms in the second sum withd.g�10 �g0; e/� 1 and d.g�10 �g0;A/� �

�1=2C�0ˇ1=2, which givesXn�N2

.n;q/D1

1pn

X�2R.n/

ˇI.�;;g�10 �g0/

ˇ

��

Xn�N2

.n;q/D1

�1=2C�pnM.g0; n;�

�1=2C�0ˇ1=2/COA;�0.��A/:

480 SIMON MARSHALL

The assumption that N D �1=2C�0ˇ�1=2 implies that we may apply Proposition 4.2with � D ��1=2C�0ˇ1=2 and M DN 2, which givesX

n�N2

.n;q/D1

�1=2C�pnM.g0; n;�

�1=2C�0ˇ1=2/��0;� N3��1=2C2�0C�ˇ:

Substituting this into (26) givesˇhT1 ;bi

ˇ2��0;� N

1C2��1=2C� CN 3C2���1=2C2�0C�ˇ:

If we estimate the action of T1 on by using our assumption (5) and substituteN �1=2C�0ˇ�1=2, and choose �0 sufficiently small, we obtainˇ

h ;biˇ�� �

�=2C�ˇ1=4��=2;

as required.

PROPOSITION 4.2If M , ı > 0, and 1 > � > 0 satisfy M � ��2�ı , we haveX

M=2<m<M

.m;q/D1

1pmM.g0;m; �/�q;ı �

2M 3=2;

where q is the integer defined in Section 2.2.

ProofLet b 2 C10 .g/ be a real nonnegative function that is supported in the ball of radius2 about the origin with respect to the norm k k defined in (8), and equal to 1 on theball of radius 1. Let C1 > 0 be a constant to be chosen later. Define b 2 C10 .g/ byb.X/D b.�

�1C1X/, and let eb 2 C10 .G/ be the pushforward of b under exp. LetA> 0, and define f 2 C10 .G/ by

f .g/D

Z A

�A

��2eb�a.�x/g�10 g�dx

so that f is an L1-normalized bump function around the geodesic segment ¹g0a.x/ W�A � x � Aº at scale �. We choose C1 small enough and A large enough that theconditions d.g�10 gg0; e/ � 1 and d.g�10 gg0;A/ � � imply that hf;gf i � 1, wherethe implied constant is independent of � and g0. If we define f 2L2.�nG/ by

f .g/DX�2�

f .�g/;

then kf k2 1 in L2.�nG/.

GEODESIC RESTRICTIONS OF ARITHMETIC EIGENFUNCTIONS 481

Choose h 2 C10 .0;1/ to be real, nonnegative, and satisfy h.x/D 1 for 1=2 �x � 1. If we define

S DX

.m;q/D1

h.m=M/Tm;

then we haveXM=2<m<M

.m;q/D1

1pmM.g0;m; �/�

X.m;q/D1

h.m=M/pm

X�2R.m/

hf; �f i D hf ;Sf i;

and we may estimate the right-hand side spectrally. Expand f with respect to adecomposition of L2.�nG/ into automorphic representations as

f DXi

˛i i ;

where i is an L2-normalized vector in an automorphic representation with eigen-value i under the Casimir operator C . We have

kC nf k2�n ��2n;

and if n is even and T > 0, this impliesXji j�T

ni j˛i j2 � hC nf ;f i;

Xji j�T

j˛i j2�n �

�2nT �n:

If we apply this with T D ��2�ı=2, we obtain

f Dhf ; 1i

Vol.X/C

X0

ji j��2�ı=2

˛i i C r;

where r 2L2.�nG/ satisfies krk2�A;ı �A. Here, †0 denotes the sum over the non-

trivial representations. Substituting this into hf ;Sf i gives

hf ;Sf i Dhf ; 1i2

Vol.X/2X

.m;q/D1

h.m=M/�.m/pm

CX

ji j��2�ı

j˛i j2X

.m;q/D1

h.m=M/�i .m/COA;ı.�A/

Xm�M

�.m/pm;

482 SIMON MARSHALL

where �i .m/ are the Hecke eigenvalues of i . The result now follows fromLemma 4.3 below, and the bounds hf ; 1i� � andX

n�N

�.n/pnN 3=2:

Note that our assumptions that M � ��2�ı and j i j � ��2�ı=2 guarantee that thehypothesis of the lemma is satisfied.

LEMMA 4.3If ı > 0 and M � j i j1Cı , we haveX

.m;q/D1

h.m=M/�i .m/�A;ı M�A;

where the implied constant is uniform in i .

ProofWe shall drop the subscript i , and assume that is a vector in a principal seriesrepresentation as the discrete series case is similar. We first consider the case q D 1.

Let r be the spectral parameter of , so that D 1=4C r2. By applying the func-tional equation and Stirling’s formula, we see that the L-function L.s; / satisfies theestimate

L.�AC i t; /�A;� .t2C r2C 1/AC1=2C� (27)

for A sufficiently large. If we let bh.s/ be the Mellin transform of h, which is entireand decays rapidly in vertical strips, we obtainX

m

h.m=M/�.m/D

Z.2/

L.s; /bh.s/M s ds:

If we shift the line of integration to � D�A and apply (27) and the rapid decay of bh,we have X

m

h.m=M/�.m/�A;� M�A.1C r2/AC1=2C�

�A;� M�A AC1=2C�

�A;� M�AM .1�ı/.AC1=2C�/

�B;ı M�B ;

as required. In the case when q > 1, we apply the same argument to the incompleteL-function obtained by removing the local factors at primes dividing q from L.s; /.

GEODESIC RESTRICTIONS OF ARITHMETIC EIGENFUNCTIONS 483

RemarkThe method we have used of estimating Hecke recurrences spectrally is unlikely towork in other situations. It requires us to choose an amplifier that makes the sumsof eigenvalues in Proposition 4.2 longer than the relevant analytic conductors, and inother cases (such as higher rank or when using the operators Tp2 on GL2 to give anunconditional theorem) this gives the amplifier so much mass that the “off-diagonal”term is worse than the trivial bound. The method also depends on the rapid decayof I.�;;g/ as d.g;A/ grows, and fails to improve the L1-bound of [20] under theassumption (5) because the rate of decay in the corresponding lemma [20, Lemma 1.1]is slower.

5. Bounding I.�;; e/We now prove Proposition 3.6. If we define p�.x/D k�.a.x//, we have I.�;; e/Dhb;p� � bi. Define

Iˇ D Œ��� ˇ=2;��C ˇ=2�[ Œ�� ˇ=2;�C ˇ=2�;

and write b D 1 C 2, where the Fourier transform of 2 is supported on Iˇ andthe transform of 1 is supported on R n Iˇ . Because b was a fixed smooth function,we have k2k2�A ˇ

�A. We have

hb;p� � bi D h1; p� � 1i C h2; p� � 2i

� supt…Iˇ

ˇcp�.t/ˇCOA.ˇ�A/kcp�k1:By [23, Lemma 2.7] (see also [11, Lemma 4.1]) we have

p�.x/� ��1C �jxj

��1=2; (28)

and this implies that kcp�k1� �1=2. It therefore suffices to prove the following esti-mate.

LEMMA 5.1We have jcp�.t/j � �1=2ˇ�1=2 for t … Iˇ .

ProofLet b1 2 C10 .R/ be a cutoff function that is equal to 1 on Œ�1; 1� and zero outsideŒ�2; 2�. We have

cp�.t/D Z 1�1

b1.x/k��a.x/

�e�itx dx:

We shall estimate this by inverting the Harish-Chandra transform and applying asymp-totics for 's from [23]. The first is [23, Theorem 1.3], which implies that

484 SIMON MARSHALL

b1.x/'s.x/��1C jsxj

��1=2: (29)

By [23, Theorem 1.5], there are functions f˙ 2 C1..0; 3/�R�0/ such that� @@x

nf˙.x; s/�n x

�n.sx/�1=2 (30)

and

's�a.x/

�D fC.x; s/e

isx C f�.x; s/e�isx COA

�.sx/�A

�(31)

for x 2 .0; 3/. The bound (29) givesZ 1�1

b1.x/'s�a.x/

�e�itx dx�

�1C jsj

��1=2for all s and t .

After inverting the Harish-Chandra transform and applying the rapid decay of h,it suffices to prove the boundZ 1

�1

b1.x/'s�a.x/

�e�itx dx� ��1=2ˇ�1=2

uniformly for t … Iˇ and js � �j � ˇ=4. We decompose the integral asZ 1�1

b1.x/b1.ˇx/'s.x/e�itx dx

C

1XnD1

Z 1�1

b1.x/�b1.2

�nˇx/� b1.2�nC1ˇx/

�'s.x/e

�itx dx:

The bound (29) implies that the first integral is� ��1=2ˇ�1=2. We shall estimate theintegrals in the sum by applying (31). The error term in (31) makes a contribution of

.�ˇ�12n/�Aˇ�12n� ��2ˇ2�n � ��1=2ˇ�1=22�n

to the nth term in the sum, which may be ignored. We shall only consider the integralover x > 0 and the main term involving fC, as the other terms are similar. We mustthen estimateZ 1

0

b1.x/�b1.2

�nˇx/� b1.2�nC1ˇx/

�fC.x; s/e

i.s�t/x dx:

If we replace x with 2nˇ�1x, this becomesZ 10

2nˇ�1b1.2nˇ�1x/

�b1.x/� b1.2x/

�fC.2

nˇ�1x; s/ei.s�t/2nˇ�1x dx: (32)

GEODESIC RESTRICTIONS OF ARITHMETIC EIGENFUNCTIONS 485

The conditions t … Iˇ and js � �j � ˇ=4 imply that j.s � t /2nˇ�1j � 2n�2, and (30)implies that the kth derivative of the amplitude factor in (32) is�k �

�1=22n=2ˇ�1=2.Integrating by parts once givesZ 1

0

2nˇ�1b1.2nˇ�1x/

�b1.x/� b1.2x/

�fC.2

nˇ�1x; s/ei.s�t/2nˇ�1x dx

� ��1=22�n=2ˇ�1=2;

and summing over n completes the proof.

6. Oscillatory integrals when t �In this section, we prove Proposition 3.7 by building up the integral I.�;;g/ inseveral steps. We begin with three calculations that we shall use repeatedly in thissection and in Section 7.

LEMMA 6.1Fix g 2 G, and define � W R=2�Z! R=2�Z by k.�/g 2 NAk.�.�//. Then � is adiffeomorphism.

ProofBy using the Cartan decomposition, we may reduce to the case where g D a.y/.Taking inverses gives a.�y/k.��/ 2 k.��.�//AN , and applying both sides to thepoint at infinity gives

e�y cot.�=2/D cot��.�/=2

�: (33)

This proves that � is a bijection, and a diffeomorphism everywhere except at � D 0.Rewriting the equation as ey tan.�=2/D tan.�.�/=2/ proves it at � D 0 also.

LEMMA 6.2Let y; z 2G, let � 2 R=2�Z, and let � be determined by k.�/y�1 2NAk.�/. Thenwe have

A�k.�/y�1z

�DA

�k.�/z

��A

�k.�/y

�:

ProofLet k.�/y�1 D nak.�/. We have

A�k.�/y�1z

�D A

�nak.�/z

�D A.na/CA

�k.�/z

486 SIMON MARSHALL

D A�k.�/z

��A

�n�1a�1k.�/

�D A

�k.�/z

��A

�k.�/y

�;

as required.

LEMMA 6.3Let g 2G have Iwasawa decomposition gD nak.�/. Then

@

@tA�ga.t/

�ˇtD0D cos�:

ProofLet H be as in (10), and let H� 2 g� be dual to H . We have

A�ga.t/

�D A.a/CA

�k.�/ exp.tH/k.��/

�D A.a/CA

�exp

�t Ad

�k.�/

�H��;

and therefore

@

@tA�ga.t/

�ˇtD0DH�

�Ad�k.�/

�H�D cos�:

6.1. A uniformization resultWe shall need the following uniformization lemma for the function A.

LEMMA 6.4Let D > 0. There exists ı; � > 0 depending on D, and a real analytic function W.�ı; ı/� .�D;D/2!R such that

@

@yA�k.�/n.x/a.y/

�D 1� �2.�; x; y/;

and ˇ.�; x; y/

ˇ� �; (34)ˇ @n

@yn.�; x; y/

ˇ�D;n 1;

for .�; x; y/ 2 .�ı; ı/� .�D;D/2.

ProofDefine the function ˛.�; x; y/ WR=2�Z� .�2D;2D/2!R=2�Z by requiring that

k.�/n.x/a.y/ 2NAk�˛.�; x; y/

�:

GEODESIC RESTRICTIONS OF ARITHMETIC EIGENFUNCTIONS 487

The analyticity of the Iwasawa decomposition implies that ˛ is analytic as a functionof .�; x; y/. Lemma 6.3 implies that

1�@

@yA�k.�/n.x/a.y/

�D 1� cos˛

D 2 sin2.˛=2/:

We choose ı such that sin.˛=2/ vanishes on .�2ı; 2ı/ � .�2D;2D/2 if and only if� D 0. Lemma 6.1 implies that @˛=@� never vanishes on ¹0º � .�2D;2D/2, and sobecause ˛ was analytic we see that there is a real analytic function 0 on .�2ı; 2ı/�.�2D;2D/2 such that sin.˛=2/D �0. Defining D 220 and restricting the domainto .�ı; ı/� .�D;D/2 gives the result.

6.2. Constituent integrals of I.�;;g/We now estimate two one-dimensional integrals that appear in I.�;;g/.

PROPOSITION 6.5Let C;D > 0 and �0 > 0 be constants, and let b 2 C10 .R/ be a function supported inŒ0; 1�. If x;y 2 Œ�D;D� and

j� j � Cs�1=2C�0ˇ1=2 and jt � sj � ˇ (35)

for some s;ˇ � 1 satisfying ˇ � 3s2=3, thenZ 1�1

b.z/ exp�i tz � isA

�k.�/n.x/a.y C z/

��dz� s�A: (36)

The implied constant depends only on A, C , D, �0, and the size of the first n deriva-tives of b, where n depends on �0 and A.

ProofBy applying Lemma 6.4, we see that there is some ı > 0 and a nonvanishing realanalytic function on .�ı; ı/� .�D � 2;DC 2/2 such that

@

@zA�k.�/n.x/a.y C z/

�D 1� �2.�; x; y C z/

when � 2 .�ı; ı/, x;y 2 Œ�D;D�, and z 2 Œ0; 1�. If Z.�;x;y/ is an antiderivative of with respect to y that is smooth as a function of .�; x; y/, we may integrate this toobtain

A�k.�/n.x/a.y/

�D y � �2Z.�;x;y/C c.x; �/

488 SIMON MARSHALL

for some function c.x; �/. If � 2 .�ı; ı/, we may use this to rewrite the integral (36)as Z 1

�1

b.z/ exp�i tz � isA

�k.�/n.x/a.y C z/

��dz

D e�is.yCc.�;x//Z 1�1

b.z/ exp�i.t � s/zC is�2Z.�;x;y C z/

�dz

D e�is.yCc.�;x//Z 1�1

b.z/ exp�is�2‰.z/

�dz; (37)

where we define ‰.z/DZ.�;x;y C z/C s�1��2.s � t /z.Our assumption (35) implies thatˇ

s�1��2.s � t /ˇ� s�1��2ˇ � C�2s�2�0 ;

so that

‰ D Z.�;x;y C z/CO.s�2�0/z and(38)

@‰

@zD .�; x; y C z/CO.s�2�0/:

It follows from (34) and (38) that for s sufficiently large, j@‰=@zj > �=2 for all� 2 .�ı; ı/, x;y 2 Œ�D;D� and z 2 Œ0; 1�. In addition, all other derivatives of ‰ arebounded from above. As (35) implies that s�2 � C 2s2�0ˇ � C 2s2�0 , the bound (36)follows by integration by parts in (37).

In the case where � … .�ı; ı/, Lemma 6.3 implies that .@=@z/A.k.�/n.x/a.yCz//� 1� c1 for some c1 > 0 depending only on ı, which gives

@

@z.i ts�1z �A

�k.�/n.x/a.y C z/

�� 1:

The result now follows by integration by parts.

The second one-dimensional integral that we shall estimate is as follows.

PROPOSITION 6.6Let C;D > 0 and �0 > 0 be constants, and let b 2 C10 .R/ be a function supported inŒ0; 1�. If x;y 2 Œ�D;D� and

jxj � Cs�1=2C�0ˇ1=2 and jt � sj � ˇ (39)

for some s;ˇ � 1 satisfying ˇ � 3s2=3, thenZ 1�1

b.z/ei tz's�n.x/a.y C z/

�dz� s�A: (40)

GEODESIC RESTRICTIONS OF ARITHMETIC EIGENFUNCTIONS 489

The implied constant depends only on A, C , D, �0, and the size of the first n deriva-tives of b, where n depends on �0 and A.

ProofIf we apply the functional equation 's D '�s , and substitute the formula for '�s asan integral of plane waves into the left-hand side of (40), it becomesZ 1

�1

Z 2�

0

b.z/ exp�i tzC .1=2� is/A

�k.�/n.x/a.y C z/

��d� dz:

Let f .x/ 2 C10 .R/ be a function with supp.f / � Œ�2; 2� and f .x/D 1 on Œ�1; 1�.Let C1 be a positive constant to be chosen later. Define b1 by b1.x/ D f .C�11 �s1=2��0ˇ�1=2x/ and set b2 D 1� b1, so that 1D b1.�/C b2.�/ is a smooth partitionof unity on R=2�Z with

supp.b1/ � Œ�2C1s�1=2C�0ˇ1=2; 2C1s

�1=2C�0ˇ1=2�;

supp.b2/ � R=2�Z n Œ�C1s�1=2C�0ˇ1=2;C1s

�1=2C�0ˇ1=2�:

Proposition 6.5 implies thatZ 1�1

Z 2�

0

b2.�/b.z/ exp�i tzC .1=2� is/A

�k.�/n.x/a.y C z/

��d� dz�A s

�A;

so that it suffices to estimateZ 1�1

Z 2�

0

b1.�/b.z/ exp�i tzC .1=2� is/A

�k.�/n.x/a.y C z/

��d� dz:

We shall do this by estimating the integralsZ 2�

0

b1.�/ exp��isA

�k.�/n.x/a.y/

��d� (41)

in � , where now D � jxj � Cs�1=2C�0ˇ1=2 and y 2 Œ�D;DC 1�.IfX 2 g, we letX� be the vector field on H whose value at p is @

@texp.tX/pjtD0.

It may be shown that these vector fields satisfy ŒX�; Y ��D�ŒX;Y ��, where the firstLie bracket is on H and the second is in g. We recall the vectors Xn and Xk definedin (10). It may be easily seen that the subset of H where the function X�

kA vanishes

is exactly A, and the following lemma implies that it vanishes to first-order there.

LEMMA 6.7We have X�nX

�kA.a.y//D ey for all y.

490 SIMON MARSHALL

ProofWe have

X�nX�k ADX

�k X�nAC ŒX

�n ;X

�k �A:

It may be seen that the first term vanishes, and we have

ŒX�n ;X�k �D�ŒXn;Xk�

� DH�;

which implies the lemma.

Lemma 6.7 implies that there exist �; ı > 0 depending on D such that if jxj< �and y 2 Œ�D � 1;DC 2�, then we have jX�

kA.n.x/a.y//j � ıjxj. Define

B D®n.x/a.y/

ˇjxj � �=2;y 2 Œ�D;DC 1�

¯;

B 0 D®n.x/a.y/

ˇjxj< �;y 2 Œ�D � 1;DC 2�

¯:

Let p D n.x/a.y/, where x and y are as in (41), and assume that p 2 B . We recallthe function N.p/ defined in (9), and assume without loss of generality that 1=8 > �0so that s�1=2C�0ˇ1=2 decays as s grows. If s is sufficiently large and C1 sufficientlysmall, depending on C , D, and � , and j� j � 2C1s�1=2C�0ˇ1=2, we have k.�/p 2B 0

and jN.k.�/p/j �C;D s�1=2C�0ˇ1=2. It follows thatˇ @

@� 0A�k.� 0/p

�ˇ� 0D�

ˇDˇX�k A

�k.�/p

�ˇ� ı

ˇN�k.�/p

�ˇ�C;D s

�1=2C�0ˇ1=2

when j� j � 2C1s�1=2C�0ˇ1=2. The proposition now follows from Lemma 6.8.If p …B , then there is � > 0 depending on D such that jX�

kA.k.�/p/j � � when

j� j � 2C1s�1=2C�0ˇ1=2 and s is sufficiently large (again using our assumption �0 <

1=8). The proposition again follows by integration by parts.

LEMMA 6.8If b 2 C10 .R/ is a cutoff function at scale 1� ı > 0, and 2 C1.R/ satisfies

0.x/� ı .n/.x/�n 1

for x 2 supp.b/ and n > 1, thenZb.x/ei t�.x/ dx�A ı.tı

2/�A:

ProofWe have Z

b.x/ei t�.x/ dx D ı

Zb.ıx/ei.tı

2/ı�2�.ıx/ dx:

GEODESIC RESTRICTIONS OF ARITHMETIC EIGENFUNCTIONS 491

The function b.ıx/ is now a cutoff function at scale 1, and ı�2.ıx/ satisfies

d

dxı�2.ıx/� 1

� ddx

nı�2.ıx/�n ı

n�2 � 1

for x 2 supp.b.ı// and n > 1. The lemma now follows by integration by parts.

6.3. Proof of Proposition 3.7If b0 2 C10 .0; 1/, t1; t2 2R, and 2L1loc.R/, we define

J.s; t1; t2; g/D

“ 1

�1

b0.x1/b0.x2/e�i.t1x1�t2x2/'s

�a.�x1/ga.x2/

�dx1 dx2;

J.s; ; g/D

“ 1

�1

b0.x1/b0.x2/'s�a.�x1/ga.x2/

�dx1 dx2:

We now combine Propositions 6.5 and 6.6 to bound J.s; t1; t2; g/, which will implyProposition 3.7 after integrating in the various spectral parameters.

PROPOSITION 6.9Let �0 > 0 be given. If g 2G satisfies

d.g; e/� 1 and d.g;A/� s�1=2C�0ˇ1=2

for some s;ˇ � 1 satisfying ˇ � 3s2=3, and t1; t2 2 Œs � ˇ; sC ˇ�, then

J.s; t1; t2; g/�A;�0 s�A: (42)

ProofDefine �.�; x1/ by k.�/a.�x1/ 2NAk.�.�; x1//. If we apply the functional equation's D '�s , write '�s as an integral over K , and applying Lemmas 6.1 and 6.2, weobtain

's�a.�x1/ga.x2/

�D

Zexp

�.1=2� is/

�A�k.�/ga.x2/

��A

�k.�/a.x1/

���d�d�

d�: (43)

Substituting this into the definition of J.s; t1; t2; g/ gives“ 1

�1

Z 2�

0

b0.x1/b0.x2/e�i.t1x1�t2x2/

� exp�.1=2� is/

�A�k.�/ga.x2/

��A

�k.�/a.x1/

���d�d�

d� dx1 dx2: (44)

492 SIMON MARSHALL

Let g D k.� 0/n.x0/a.y0/. The condition d.g; e/ � 1 implies that x0 and y0 arebounded. We then have k.�/ga.x2/ D k.� C � 0/n.x0/a.x2 C y0/. Choose a con-stant C > 0. If � … Œ�Cs�1=2C�0ˇ1=2;C s�1=2C�0ˇ1=2�, then integrating (44) in x1and applying Proposition 6.5 shows that the integral of (44) over x1 and x2 with thisvalue of � is �C;A;�0 s

�A, and if � C � 0 … Œ�Cs�1=2C�0ˇ1=2;C s�1=2C�0ˇ1=2� weobtain the same conclusion by integrating in x2 . Combining these, we see that (44)will be �C;A;�0 s

�A unless j� 0j � 2Cs�1=2C�0ˇ1=2, and we assume that this is thecase.

If C is chosen sufficiently small, the condition j� 0j � 2Cs�1=2C�0ˇ1=2 and ourassumption that d.g;A/� s�1=2C�0ˇ1=2 imply that jx0j � C1s�1=2C�0ˇ1=2 for someabsolute C1 > 0. It follows that if C is sufficiently small, we have N.a.�x1g// �C2s�1=2C�0ˇ1=2 for some absolute C2 and all 0� x1 � 1. The result now follows by

applying Proposition 6.6 to the integral in x2 defining J.s; t1; t2; g/ for each fixed x1.

Proof of Proposition 3.7To prove the bound (21), observe that equation (28) implies thatZ 1

�1

ˇk��ga.x/

�ˇ2dx� � log�

uniformly for g 2 G. It follows that b.x1/b.x2/k�.a.�x1/ga.x2// has norm ��

�1=2C� as an element of L2.R2/, and the result follows by Cauchy–Schwarz.We now prove (22). If we invert the Fourier transform of and note that kbk1 �

kbk2.2ˇ/1=2 D .2�/1=2.2ˇ/1=2, we may apply Proposition 6.9 with 2ˇ in place of ˇto obtain

J.s;;g/�A;�0 s�A (45)

for js � �j � ˇ. Inverting the Harish-Chandra transform of k� gives

I.�;;g/D1

2�

Z 10

J.s;;g/h�.s/s tanh.�s/ds: (46)

We may assume without loss of generality that ˇ > �� for some � > 0. We estimatethe integral of (46) over js � �j > ˇ by using the rapid decay of h and the integralover js � �j � ˇ by using (45), which completes the proof.

7. Oscillatory integrals when t < �We now prove Proposition 3.5. We will use the geometry of H to a greater extent thanin previous sections, and in particular we will distinguish between elements of G and

GEODESIC RESTRICTIONS OF ARITHMETIC EIGENFUNCTIONS 493

their images in H. We assume that all geodesics in H carry an orientation. When werefer to the unit tangent vector to a geodesic at a point, we shall always mean in thedirection of its orientation. If `1 and `2 are two intersecting geodesics, we shall denoteby ∠.`1; `2/ the angle between their unit tangent vectors at the point of intersectionmeasured in the counterclockwise direction from `1 to `2. We let ` be the geodesiccorresponding to A, and let ` W x 7! a.x/K be the standard parameterization. We give` the upward-pointing orientation, which we transfer to g` for any g.

As in the proof of Proposition 3.7, it suffices to bound J.s; t; g/ WD J.s; t; t; g/.After substituting the expression (43) for 's.a.�x1/ga.x2// into the integral definingJ.s; t; g/, we obtain an oscillatory integral in the variables � , x1, and x2 with phasee�is� , where

.x1; x2; �; g; �/D �.x1 � x2/�A�k.�/a.x1/

�CA

�k.�/ga.x2/

�and � D t=s � 0. We first assume that � 2 Œı; 1 � ı� for some 1=2 > ı > 0; the case� D 0 is simpler and treated at the end of Section 7.5. Define ˛ 2 Œ0;�=2� to bethe solution to cos˛ D �, which is bounded away from 0 and �=2. We shall studythe critical points of in Sections 7.1–7.4, before deriving a bound for I.s; t; g/from our results in Section 7.5. We shall write .x1; x2; �/ when g and � are notvarying.

7.1. The critical points of

LEMMA 7.1The phase function has a critical point at .x1; x2; �; g; �/ exactly when k.�/`.x1/and k.�/g`.x2/ lie on the same vertical geodesic v, which we give the upward-pointing orientation, and we have ∠.v; k.�/`/;∠.v; k.�/g`/ 2 ¹˙˛º.

ProofSuppose that .x01; x

02; �0/ is a critical point of . Define the functions x.�/, y.�/, and

ˇ.�/ by

k.�/a.x01/D n�x.�/

�a�y.�/

�k�ˇ.�/

�;

and let

n0 D n�x.� 0/

�; a0 D a

�y.� 0/

�; and ˇ0 D ˇ.� 0/:

It may be seen that v WD n0` is the upward-pointing geodesic through k.� 0/`.x01/ andthat ˇ0 D∠.v; k.� 0/`/. Lemma 6.3 then implies that ˇ0 D˙˛. The calculation in thecase of @=@x2 is identical.

494 SIMON MARSHALL

Lemma 6.2 gives

�A�k.�/a.x01/

�CA

�k.�/ga.x02/

�DA

�k�ˇ.�/

�a.x01/

�1ga.x02/�;

so that

@

@�.x01; x

02; �0/D

@

@�A�k�ˇ.�/

�a.x01/

�1ga.x02/�ˇ�D� 0

D@ˇ

@�.� 0/

@

@ˇA�k.ˇ/a.x01/

�1ga.x02/�ˇˇDˇ 0

: (47)

Because @ˇ=@� does not vanish by Lemma 6.1, and

@

@�A�k.�/g

�ˇ�D0D 0

if and only if g 2AK , we have @=@� D 0 if and only if k.ˇ0/a.x01/�1ga.x02/ 2AK;

that is, k.ˇ0/a.x01/�1g`.x02/ lies on `. Because k.ˇ0/a.x1/�1 D a0�1n0�1k.� 0/, this

is equivalent to the condition that k.� 0/ga.x02/ 2 n0AK , or that k.� 0/g`.x02/ lies on

the vertical geodesic v.We finish with an observation that will be useful in calculating the Hessian of .

We have k.ˇ0/a.x01/�1ga.x02/ 2 a.h/K for some h 2 R, and it may be seen that

k.� 0/a.x01/ 2 n0a0K and k.� 0/ga.x02/ 2 n

0a0a.h/K , so that h is the signed distancefrom k.� 0/`.x01/ to k.� 0/g`.x02/ along v.

Given a pair of geodesics `1 and `2, we say that a geodesic j is a criticalgeodesic for .`1; `2/ if j meets `1 and `2 at angles of˙˛. We may therefore rephraseLemma 7.1 as saying that .x1; x2; �; g; �/ is a critical point of exactly when .`; g`/has a critical geodesic j , `.x1/ and g`.x2/ both lie on j , and k.�/j is vertical. As inLemma 7.1, we define the aperture of a critical point to be the signed distance from`.x1/ to g`.x2/ on the geodesic j .

We shall now calculate the Hessian of at its critical points. Let .x01; x02; �0/ be a

critical point of , and define functions ˇi .�/ by

k.�/a.x01/ 2NAk�ˇ1.�/

�; k.�/ga.x02/ 2NAk

�ˇ2.�/

�: (48)

We let ˇ0i D ˇi .�0/. It follows from Lemma 7.1 that ˇ0i 2 ¹˙˛º. Let h be the aperture

of the critical point, so that

k.ˇ01/a.x01/�1ga.x02/ 2

�eh 0

0 1

�K:

We define � D @ˇ1@�.� 0/, which is nonzero by Lemma 6.1. The Hessian of at .x01; x

02;

� 0/ is given by the following proposition.

GEODESIC RESTRICTIONS OF ARITHMETIC EIGENFUNCTIONS 495

PROPOSITION 7.2The Hessian of at .x01; x

02; �0/ with respect to the coordinates .x1; x2; �/ is

D D

0@12

sin2 ˛ 0 � sinˇ010 �1

2sin2 ˛ ��eh sinˇ02

� sinˇ01 ��eh sinˇ02 �2.1� e2h/=2

1A :The determinant of D is

jDj D3

8�2 sin4 ˛.1� e2h/;

which is nonzero unless hD 0; that is, the points `.x01/ and g`.x02/ coincide in H.

ProofIt is clear that @2=@x1 @x2 is identically 0. To calculate @2=@x21 , define � W R!R=2�Z by the condition that k.� 0/a.x01 C t / 2NAk.�.t//. Our assumption that weare at a critical point implies that �.0/D ˇ01 D˙˛. Lemma 6.3 gives

@

@t.x01C t; x

02; �0/D �� cos�.t/

and

@2

@x21.x01; x

02; �0/D sinˇ01

@�

@t.0/: (49)

We have

k.� 0/a.x01C t / 2 NAk��.t/

�;

NAk.ˇ01/a.t/D NAk��.t/

�;

k.ˇ01/a.t/ 2 NAk��.t/

�:

Equation (33) then gives tan.�.t/=2/D et tan.ˇ01=2/, so that

@�

@tsec2

��.t/=2

�D et tan.ˇ01=2/;

@�

@t.0/D cos2.ˇ01=2/ tan.ˇ01=2/

D1

2sinˇ01:

Substituting this into (49) gives

@2

@x21.x01; x

02; �0/D

1

2sin2 ˇ01 D

1

2sin2 ˛:

The calculation of @2=@x22 is identical, with the exception of a change in sign.

496 SIMON MARSHALL

To calculate @2=@� @x1, we again have

@

@x1.x01; x

02; �/D �� cosˇ1.�/

and

@2

@� @x1.x01; x

02; �0/D sinˇ01

@ˇ1

@�.� 0/D � sinˇ01:

We likewise have

@2

@� @x2.x01; x

02; �0/D� sinˇ02

@ˇ2

@�.� 0/;

and we shall express @ˇ2@�.� 0/ in terms of � and h. We recall that

k.ˇ01/a.x01/�1ga.x02/D a.h/k.�0/

for some �0, and so

k.�/a.x01/k.�ˇ01/a.h/k.�0/D k.�/ga.x

02/:

Substituting both parts of (48) into this gives

NAk�ˇ1.�/

�k.�ˇ01/a.h/k.�0/D NAk

�ˇ2.�/

�;

k�ˇ1.�/� ˇ

01

�a.h/ 2 NAk

�ˇ2.�/� �0

�:

By setting � D � 0 we see that �0 D ˇ01. Equation (33) then gives

eh tan��ˇ1.�/� ˇ

01

�=2�D tan

��ˇ2.�/� ˇ

02

�=2�;

and differentiating both sides with respect to � and evaluating at � D � 0 gives

@ˇ2

@�.� 0/D �eh:

It follows that

@2

@� @x2.x01; x

02; �0/D��eh sinˇ02:

To calculate @2=@�2, we have as in (47) that

@

@�.x01; x

02; �/D

@ˇ1

@�

@

@ˇA�k.ˇ/a.x01/

�1ga.x02/�ˇˇDˇ1.�/

:

GEODESIC RESTRICTIONS OF ARITHMETIC EIGENFUNCTIONS 497

Because

@

@ˇA�k.ˇ/a.x01/

�1ga.x02/�ˇˇDˇ 0

1

D 0;

we have

@2

@�2.x01; x

02; �0/D �2

@2

@ˇ2A�k.ˇ/a.x01/

�1ga.x02/�ˇˇDˇ 0

1

D �2@2

@ˇ2A�k.ˇ � ˇ01/a.h/

�ˇˇDˇ 0

1

:

It is a standard calculation that

@2

@ˇ2A�k.ˇ/a.h/

�ˇˇD0D .1� e2h/=2;

and this completes the proof.

7.2. The function Define P DR=2�Z�G � Œı; 1� ı�, and define S �P to be the set where one of thegeodesics k.�/` and k.�/g` is vertical. Note that S is closed and contains at most 4values of � for each fixed .g; �/. We may define functions

1; 2 WP n S!R

by requiring that k.�/`.1.�; g; �// is the unique point on k.�/` at which the tangentvector to the geodesic makes an angle of ˛ with the upward-pointing vector, andlikewise for 2.�; g; �/ and k.�/g`. As 1 does not depend on g, we will omit thisargument of the function. We have

k.�/a�1.�; �/

�2NAk.�1˛/; k.�/ga

�2.�; g; �/

�2NAk.�2˛/ (50)

for �i 2 ¹˙1º, and so equation (33) gives

e�1.�; / tan.�=2/D tan.�1˛=2/; e�2.�;g; / tan.�=2/D tan.�2˛=2/:

Moreover, it may be seen that �1 D 1 if and only if the geodesic k.�/` runs fromright to left in the upper half-plane model of H, which is equivalent to � 2 .0;�/, andlikewise for �2.

It follows from Lemma 6.3 that 1.�; �/ and 2.�; g; �/may also be characterizedas the unique functions such that

@

@x1

�1.�; �/; x2; �

�D@

@x2

�x1; 2.�; g; �/; �

�D 0: (51)

498 SIMON MARSHALL

We define

WP n S ! R;

.�; g; �/ D �1.�; �/; 2.�; g; �/; �; g; �

�:

LEMMA 7.3.� 0; g0; �0/ is a critical point of exactly when .1.� 0/; 2.� 0/; � 0; g0; �0/ is a criticalpoint of . If .� 0; g0; �0/ is a critical point of , let � and h be the values associatedto the corresponding critical point of . We then have

@2

@�2.� 0; g0; �0/D�

3

2�2.1� e2h/:

ProofWe shall fix g and � and omit them from the arguments of and . LetD be the Hes-sian of calculated in Proposition 7.2. If we apply the chain rule to and substitute� D � 0, we obtain

@2

@�2.� 0/D

�@1@�.� 0/;

@2

@�.� 0/; 1

D�@1@�.� 0/;

@2

@�.� 0/; 1

t:

To calculate @�1@�.� 0/ and @�2

@�.� 0/, we differentiate (51) with respect to � and set � D � 0

to obtain

@2

@� @x1

�1.�

0/; 2.�0/; � 0

�C@1

@�.� 0/

@2

@x21

�1.�

0/; 2.�0/; � 0

�D 0: (52)

Substituting the second partial derivatives of calculated in Proposition 7.2 gives

@1

@�.� 0/D

�2�

sinˇ01;

and likewise

@2

@�.� 0/D

�2�eh

sinˇ02:

The lemma follows on substituting these into (52).

It follows that the set of .g; �/ for which the function .�;g; �/ has a degeneratecritical point is exactly those .g; �/ for which either ` D g` as oriented geodesics,so that g 2 A, or ∠.`; g`/ D ˙2˛. Note that these two cases are distinct, as ˛ 2.0;�=2/. In the first case the function .�;g; �/ vanishes identically. In the secondcase, .�;g; �/ has only a single degenerate critical point, as no oriented geodesic can

GEODESIC RESTRICTIONS OF ARITHMETIC EIGENFUNCTIONS 499

cross ` and g` making an angle of ˛ with both except at their point of intersection.We will show in Section 7.4 that this degeneracy is cubic. To determine the location ofthis critical point, the condition that ∠.`; g`/D˙2˛ implies that g 2 a.y/k.˙2˛/Afor some y 2 R, so that ` \ g` D `.y/. The angle bisector of the two geodesics atthe point `.y/ is a.y/k.˙˛/`, and the critical point of .�;g; �/ is the � such thatthe positive endpoint of k.�/a.y/k.˙˛/` is i1. This is equivalent to the conditionk.�/a.y/k.˙˛/ 2NA, and (33) then gives cot�=2D ey cot˛=2.

We define

D1 D®.�; g; �/ 2P n S

ˇg 2A

¯;

D˙2 D®.�; g; �/ 2P n S

ˇg 2 a.y/k.˙2˛/A; cot�=2D ey cot˛=2

¯to be the three sets on which has a degenerate critical point. We also define P D

G � Œı; 1� ı� and define

D1 D A� Œı; 1� ı�;

2 D®.g; �/ 2P

ˇg 2Ak.˙2˛/A

¯to be the projections of D1 and D˙2 to P .

7.3. The degenerate set D1

As .�;g; �/D .�;ga; �/ for a 2 A, we may study the degeneracy of near D1

by differentiating .�; exp.X/; �/ at X D 0 as in the following proposition.

PROPOSITION 7.4If X D

�0 X1X2 0

�2 g, then

@

@t ��; exp.tX/; �

�ˇtD0D � sin˛.e��2.�;e; /X1C e

�2.�;e; /X2/; (53)

where � is 1 if � 2 .0;�/ and �1 otherwise. In particular, @ =@t.�; exp.tX/; �/jtD0has no degenerate critical points unless X D 0.

ProofWe have

@

@t ��; exp.tX/; �

�ˇtD0D@

@t�1.�; �/; 2

��; exp.tX/; �

�; �; exp.tX/; �

�ˇtD0

D@

@x2

�1.�; �/; 2.�; e; �/; �; e; �

� @@t2��; exp.tX/; �

�ˇtD0

C@

@t�1.�; �/; 2.�; e; �/; �; exp.tX/; �

�ˇtD0:

500 SIMON MARSHALL

The first term vanishes by (51), so we are left with

@

@t ��; exp.tX/; �

�ˇtD0D@

@t�1.�; �/; 2.�; e; �/; �; exp.tX/; �

�ˇtD0

D@

@tA�k.�/ exp.tX/a

�2.�; e; �/

��ˇtD0:

We shall abbreviate 2.�; e; �/ to 2.�/ for the remainder of the proof. Write thefirst-order approximation to the Iwasawa decomposition of k.�/ exp.tX/a.2.�// as

k.�/ exp.tX/a�2.�/

�D n exp

�tXN CO.t

2/�a exp

�tXACO.t

2/�k exp

�tXK CO.t

2/�;

where XN 2 n, XA 2 a, and XK 2 k. As in equation (50), we have k D k.˛/ if � 2.0;�/ and k D k.�˛/ if � 2 .��;0/. We first assume that � 2 .0;�/. Rearrangingand equating first-order terms gives

X D Ad�a�2.�/

�k.˛/�1a�1

�XN CAd

�a�2.�/

�k.˛/�1

�XA

CAd�a�2.�/

��XK ;

Ad�k.˛/a

�2.�/

��1�X D Ad.a�1/XN CXACAd

�k.˛/

�XK :

As Ad.a�1/XN and Ad.k.˛//XK lie in a? � g, we see that XA is the orthogonalprojection of Ad.k.˛/a.2.�//�1/X to a. A calculation gives

XA D sin˛.e��2.�/X1C e�2.�/X2/

�1=2 0

0 �1=2

�;

so that

@

@tA�k.�/ exp.tX/a

�2.�/

��ˇtD0D sin˛.e��2.�/X1C e

�2.�/X2/:

This proves (53) when � 2 .0;�/, and the other case is identical.We now prove that @ =@t.�; exp.tX/; �/jtD0 has no degenerate critical points if

X ¤ 0 and � 2 .0;�/. We define f .x/D sin˛.X1e�x CX2ex/, so that

@

@tA�k.�/ exp.tX/a

�2.�/

��ˇtD0D f

�2.�/

�:

Differentiating equation (33) gives

@2

@�D�

1

2e��2.�/ tan.˛=2/ csc2.�=2/;

so that @2=@� is always nonzero. Suppose that X ¤ 0 and that � is a degeneratecritical point of @ =@t.�; exp.tX/; �/jtD0. We then have

GEODESIC RESTRICTIONS OF ARITHMETIC EIGENFUNCTIONS 501

0D@2

@� @tA�k.�/ exp.tX/a

�2.�/

��ˇtD0

D f 0�2.�/

�@2@�

D f 0�2.�/

�:

Differentiating again with respect to � gives

0D@3

@2� @tA�k.�/ exp.tX/a

�2.�/

��ˇtD0

D f 00�2.�/

��@2@�

2D f 00

�2.�/

�;

but this is a contradiction as it may be easily checked that f has no degenerate criticalpoints unless X D 0. The case of � 2 .��;0/ is identical.

Define P DR=2�Z�a?� Œı; 1�ı�, and define S D ¹.�;X;�/ 2 P j .�; exp.X/;�/ 2 Sº. S is again closed and contains at most 4 values of � for each fixed .X;�/.

LEMMA 7.5There is an open neighborhood 0 2 U � a? such that for all X 2 U , all � 2 Œı; 1� ı�,and all b 2 C10 .P n S/, we haveZ

b.�;X;�/eis .�;exp.X/; / d���1C skXk

��1=2;

where kXk is as in (8). The implied constant depends only on ı, kbk1, and k@b=@�k1.

ProofDefine the map X WR�R=2�Z! a? by

X.r; �/D

�0 r sin�

r cos� 0

�:

We define

eP DR=2�Z�R�R=2�Z� Œı; 1� ı�

and

eS D ®.�; r; �; �/� eP ˇ ��;X.r; �/; �

�2 S

¯:

502 SIMON MARSHALL

We define eb.�; r; �; �/ 2 C10 .eP n eS/ and e .�; r; �; �/ 2 C1.eP n eS/ to be the pull-backs of b and under X . We know that e vanishes when r D 0, and as e is smooth(in fact, real analytic) we have that e =r is again a smooth function. Proposition 7.4implies that e =r has no degenerate critical points when r D 0, and so there is some� > 0 such that it also has no degenerate critical points on the set supp.eb/\.R=2�Z�Œ��; ���R=2�Z� Œı; 1� ı�/. If we define U DX..��; �/�R=2�Z/, the result nowfollows from the van der Corput lemma (see, e.g., [26, Proposition 2, Section 1.2,Chapter VIII] and the corollary there).

COROLLARY 7.6If .a0; �0/ 2 D1, there is an open neighborhood .a0; �0/ 2 U � P such that for allb 2 C10 .P n S/ and all .g; �/ 2 U , we haveZ 2�

0

b.�; g; �/eis .�;g; / d���1C sd.g;A/

��1=2:

The implied constant depends only on ı, .a0; �0/, kbk1, and k@b=@�k1.

ProofLet UX � a? be as in Lemma 7.5. If g D exp.X/a.x/ for X 2 UX and jxj � B , wehave d.g;A/ kXk, where the implied constants depend only on B . As .�;ga;�/D .�;g; �/ for a 2A, the result follows from Lemma 7.5.

7.4. The degenerate set D˙2The next proposition proves that has a cubic degeneracy on D˙2 .

PROPOSITION 7.7If .� 0; g0; �0/ 2D˙2 , we have @3 =@�3.� 0; g0; �0/¤ 0.

ProofSuppose that g0 D a.y/k.2˛/a2. Let � > 0, and define g D a.y/k.2˛ C �/a2. If �is chosen sufficiently small, the pair .`; g`/ will have exactly two critical geodesics`1 and `2 as shown in Figure 1. The triangles AB1C1 and AB2C2 both have angulardefect, and hence area, �. Our assumption that ˛ was bounded away from 0 and �=2then implies that AB1 D AB2 �1=2 and B1C1 D B2C2 �1=2, where the impliedconstants depend only on ı. The critical points �i corresponding to `i are the solu-tions to

cot�1=2D�eyCAB1 cot˛=2; cot�2=2D�e

y�AB1 cot˛=2:

It follows that 0 > �1 >�˛ > �2 >�� and also that �1� �2 �1=2. The apertures hiof the critical points �i are given by h1 D �B1C1 ��1=2 and h2 D B2C2 �1=2,

GEODESIC RESTRICTIONS OF ARITHMETIC EIGENFUNCTIONS 503

Figure 1. Two degenerating critical geodesics.

so that Lemma 7.3 gives

@2

@�2.�1; g; �

0/��1=2;@2

@�2.�2; g; �

0/ �1=2:

It follows that there is �0 2 Œ�2; �1� at which

@3

@�3.�0; g; �

0/D

@2

@�2.�2; g; �

0/� @2

@�2.�1; g; �

0/

�2 � �1�1;

and shrinking � to 0 gives the result. The case g0 2Ak.�2˛/A is identical.

COROLLARY 7.8If .g0; �0/ 2D

˙

2 , there is an open neighborhood .g0; �0/ 2 U � P such that for allb 2 C10 .P n S/ and all .g; �/ 2 U , we haveZ 2�

0

b.�; g; �/eis .�;g; / d�� s�1=3:

The implied constant depends only on ı, .g0; �0/, kbk1, and k@b=@�k1.

ProofBy Proposition 7.7, there exist � > 0 and neighborhoods U� of � 0 and U of .g0; �0/,all depending on ı and .g0; �0/, such that .U� �U /\ S D ; and j@3 =@�3j � � > 0on U� � U . As .�;g0; �0/ only has a critical point at � 0, by shrinking U we may

504 SIMON MARSHALL

also assume that has no critical points on .R=2�Z nU� /�U n S . The result thenfollows from the van der Corput lemma.

7.5. Bounds for J.s; t; g/We shall use the results of the previous sections to prove the following proposition,which implies Proposition 3.5 in the case t=� 2 Œı; 1� ı� after inverting the Harish-Chandra transform.

PROPOSITION 7.9If g 2G satisfies d.g; e/� 1 and t=s 2 Œı; 1� ı�, we have

J.s; t; g/�ı

´s�1.1C sd.g;A//�1=2 when d.g;A/� s�1=3;

s�4=3 when d.g;A/� s�1=3:

ProofIf we substitute the expression (43) into the definition of J.s; t; g/, we obtainZ 2�

0

“ 1

�1

b0.x1/b0.x2/e�i t.x1�x2/

� exp�.1=2� is/

�A�k.�/ga.x2/

��A

�k.�/a.x1/

���d�d�

dx1 dx2 d�:

We let b 2 C10 .G/ be a function that is equal to 1 when d.g; e/� 1 and introduce afactor of b.g/ into the integral. When d.g; e/� 1 we then have

J.s; t; g/D

Z 2�

0

“ 1

�1

e�is�.x1;x2;�;g; /c.x1; x2; �; g; �/dx1 dx2 d�;

where c 2 C10 .R2 �P / is the combination of all of the amplitude factors. The fol-

lowing lemma eliminates the variables x1 and x2.

LEMMA 7.10There is a function c1 2 C10 .P n S/ such that for all .g; �/ 2G � Œı; 1� ı� we haveZ 2�

0

“ 1

�1

e�is�.x1;x2;�;g; /c.x1; x2; �; g; �/dx1 dx2 d�

D s�1Z 2�

0

e�is .�;g; /c1.�; g; �/d� CO.s�2/;

where the implied constant depends on ı and on .g; �/ in a locally uniform manner.

GEODESIC RESTRICTIONS OF ARITHMETIC EIGENFUNCTIONS 505

ProofWe shall apply stationary phase in the xi variables. For fixed .�; g; �/, the function.x1; x2/ has one critical point at .1.�; �/; 2.�; g; �// if .�; g; �/ … S , and noneotherwise. Moreover, it may be shown in the same way as the proof of Proposition 7.2that the Hessian at this critical point is

D D

�12

sin2 ˛ 0

0 �12

sin2 ˛

�;

so that the critical point is uniformly nondegenerate.Define

P0 D®.�; g; �/ 2P n S j

�1.�; �/; 2.�; g; �/; �; g; �

�2 supp.c/

¯;

so that P0 is compact and P0 \ S D;. If we define c1 2 C10 .P n S/ by

c1.�; g; �/D2�

sin2 ˛c�1.�; �/; 2.�; g; �/; �; g; �

�;

then we have supp.c1/�P0, and stationary phase gives“ 1

�1

e�is�.x1;x2;�/c.x1; x2; �; g; �/dx1 dx2

D e�is .�;g; /s�1c1.�; g; �/COı.s�2/ (54)

locally uniformly on P n S . We also have“ 1

�1

e�is�.x1;x2;�/c.x1; x2; �; g; �/dx1 dx2�A;ı s�A

locally uniformly on P nP0. Therefore, if we extend c1 to a function in C1.P / bymaking it 0 on S , then (54) holds locally uniformly on P and the lemma follows.

We now apply Corollaries 7.6 and 7.8. Let B D ¹g 2 G W d.g; e/ � 1º. Corol-lary 7.6 implies that there is an open neighborhood U1 of D1 \ .B � Œı; 1� ı�/ in P

such that

J.s; t; g/�ı s�1�1C sd.g;A/

��1=2when .g; �/ 2 U1 \ .B � Œı; 1 � ı�/, and Corollary 7.8 implies that there is a neigh-

borhood U2 of D˙

2 \ .B � Œı; 1 � ı�/ such that J.s; t; g/�ı s�4=3 when .g; �/ 2

U2 \ .B � Œı; 1 � ı�/. As has no degenerate critical points outside D1 [ D˙

2 ,we also have J.s; t; g/�ı s

�3=2 when .g; �/ 2 .B � Œı; 1� ı�/ n .U1 [ U2/. As thebound in Proposition 7.9 is the maximum of these three bounds, this completes theproof.

506 SIMON MARSHALL

It remains to discuss the case when t D 0, so that ˛ D �=2. The proof proceedsas before, until the analysis of the degenerate critical points of . These degeneraciesnow occur when

g 2A[

�0 1

�1 0

�A;

and the function vanishes identically at these points. These degeneracies may betreated in exactly the same way as D1 in Section 7.3, which gives the bound

J.s; 0; g/� s�1�1C sd.g;A/

��1=2when d.g; e/� 1. Inverting the Harish-Chandra transform completes the proof.

Acknowledgments. We would like to thank Xiaoqing Li, Peter Sarnak, ChristopherSogge, Nicolas Templier, Akshay Venkatesh, and Steve Zelditch for many helpfuldiscussions. We would also like to thank the referees for many helpful comments.

The author’s work was supported by National Science Foundation grant DMS-1501230.

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Department of Mathematics, University of Wisconsin–Madison, Madison, Wisconsin, USA;

[email protected]