EFFECTIVE EQUIDISTRIBUTION OF TRANSLATES OF LARGE SUBMANIFOLDS IN SEMISIMPLE HOMOGENEOUS SPACES 6 1 0 ADRIA´N UBIS 2 n a Abstract. Let G = SL2(R)d and Γ = Γd0 with Γ0 a lattice in J SL2(R). Let S be any “curved” submanifold of small codimen- 9 sion of a maximal horospherical subgroup of G relative to an R- 2 diagonalizableelementainthediagonalofG. ThenforS compact ] our result can be described by saying that anvolS converges in an S effective way to the volume measure of G/Γ when n , with D → ∞ vol the volume measure on S. S . h t a m [ 1. Introduction and results 1 Let G be a connected Lie group without compact factors and Γ a v lattice in G. For any a Ad-semisimple element of G, we can consider 0 2 (see [KSS02]) the expanding horospherical subgroup relative to a 0 8 U+ = g G : lim a−ngan = e . 0 { ∈ n→∞ } . 1 An element u G is in U+ whenever d (a−nu,a−n) 0 as n , 0 ∈ G → → ∞ 6 where dG is a fixed right G-invariant distance on G. This says that 1 the action u a−1u contracts regions of U+, so the opposite action : 7→ v u au expands regions of U+. Xi 7→Then, one would expect anVx0 to be quite large inside G/Γ for r any x0 G/Γ and V open set in U+. In fact, if U+ is maximal the a ∈ followingmuch strongerequidistribution result isknown[Vee77, Sha96] (see [KSS02, Theorem 3.7.8]): for any probability measure λ on U+ which is absolutely continuous with respect to a Haar measure on U+ we have anλ∗ µ where λ∗ is the image of λ onto G/Γ under the G → map g gx for a fixed x G/Γ, and µ is the probability Haar 0 0 G 7→ ∈ measure on G/Γ, namely (1.1) lim f(anux )dλ(u) = f(r)dµ (r) 0 G n→∞ZU+ ZG/Γ for any f C (G/Γ). In particular this implies anVx = G/Γ. c n 0 ∈ ∪ Actually the result in [Sha96] is much more general than (1.1). 1 2 ADRIA´NUBIS In [Gor07, Sha10] N. Shah raised the question of trying to gener- alize this result to some singular measures λ on U+, in particular to find conditions on a submanifold S of U+ which make the probability volume measure λ = λ supported on S satisfy (1.1). S In[Sha09b]thisquestionwassolvedforthecaseG = SO(d,1),where itwasshownthat(1.1)holdswhenever γ : [0,1] U+ isareal-analytic → curve with γ([0,1])x not contained in any proper subsphere of G/Γ. 0 This was extended to Cm curves in [Sha09a]. In this work we intend to study the case G = SL (R)d, Γ = Γd 2 0 with Γ a lattice in SL (R), a an element in the diagonal of G and S 0 2 a submanifold of U+ of small codimension not contained in an affine subspace of U+. Our methods will be Fourier-analytic and will give effective rates of decay; on the other hand, in order to prove (1.1) we will need to impose a curvature condition on S. I do not know whether the ideas from [Sha09b] could be applied to this case. In G = SL (R)d every semisimple element a in the diagonal of G 2 that generates a maximal horospherical subgroup is conjugate to √y 0 (1.2) a = (a(y),a(y),...,a(y)) a(y) = y (cid:18) 0 1/√y(cid:19) with 0 < y, so it is enough to study the horospherical subgroup corre- sponding to that element with 0 < y < 1, which is 1 0 (1.3) U+ = u : t Rd u = (u(t ),...,u(t )) u(t) = . { t ∈ } t 1 d (cid:18)t 1(cid:19) We then have that U+ and Rd are isomorphic Lie groups, so we can think of S as a submanifold of Rd. Now we are going to impose on S the following curvature condition, which is an strenghtening of the fact that S is not contained in any proper affine subspace of Rd. Definition 1.1 (Totally curved submanifold). Let S be a submanifold of Rd of codimension n d/2. For any p S, we shall say that S is ≤ ∈ totally curved at p if the second fundamental form (see [KN96]) II : T S T S (T S)⊥ p p p p × → satisfies II (V T S) = (T S)⊥ for every V subspace of T S of dimen- p p p p × sion n. We shall say that S is totally curved if the set of points at which S is curved is dense in S. Intuitively this condition is saying that the manifold is curved in every direction of Rd. In the case of S being an hypersurface (namely n = 1), S is totally curved at p precisely when it does not have zero curvatureatthatpoint. IngeneralweshallshowthatforS nottobeto- tally curved its coordinates must satisfy a certain differential equation, EFFECTIVE EQUIDISTRIBUTION 3 so a generic submanifold of Rd will be totally curved. That equation is just R(p) = 0, where R is the complex resultant of the polynomials s , 0 j n 1 defined in Propositon 4.1. j ≤ ≤ − An example of the exceptional submanifolds that we want to avoid with our curvature condition is S = γ S , with γ a curve in R2 and 1 S a submanifold of Rd−2; in order to p×rove (1.1) for λ = λ we would 1 S need to prove it for λ in the case d = 2, and then we would lose the γ small codimension condition for S. This indicates that our condition on S is a natural one for our context. As our results are quantitative, we need to control the smoothness of f to measure the decay in (1.1); for that purpose we shall use Sobolev norms Definition 1.2 (Sobolev norms). Any X in the Lie algebra g of G acts on C∞(G/Γ) by Xf(g) = df(etXg) . Thus, by fixing a basis B of dt |t=0 g we can define the L∞ Sobolev norms as f = Df k kSj k kL∞(G/Γ,µG) X deg(D)≤j for any j 0, where D runs over all the monomials in B of degree at ≥ most j. With those definitions, we can already state our main result. Theorem 1.3 (Main result). Let Γ be a lattice in SL (R). There 0 2 exists a constant γ > 0 such that for any S real-analytic totally curved submanifold of U+ of codimension n < γd, with U+ as in (1.3), and any λ probability measure with C∞ Radon-Nikodym derivative w.r.t. S c the volume measure on S we have f(a ux )dλ (u) f dµ yc f |Z y 0 S −Z G| ≪λS,x0 k kSd U+ G/Γ for some c > 0 and any f C∞(G/Γ), x G/Γ = SL (R)d/Γd and ∈ 0 ∈ 2 0 a as in (1.2) , where c just depends on d and S. y Remark. In the Theorem, the Sd norm can be changed by the S1 norm at the expense of diminishing the constant c. As pointed out in [Sha09b], this result can be seen as a hyperbolic equivalent of the equidistribution in Rd/Zd of dilations of a real-analytic submanifold of Rd not contained in any proper affine subspace. This result implies (1.1) for λ with a decay rate of ycn. The idea of S theproof is to see it as anequidistribution problem inRd, as in[Jon93], and to apply Fourier analysis there. Due to the conditions on S, it will be possible to show that λ almost always behaves in a way similar to S b 4 ADRIA´NUBIS the Fourier transform of the sphere, and then one can take advantage of it by applying the Weyl-Van der Corput Method together with the exponential mixing of one-parameter homogeneous flows of G on G/Γ. Actually, the approach described in the previous paragraph works just for functions f for which there is a quick decay for the mixing. But for G = SL (R)d this will always be the case except for functions 2 coming from SL (R)k with k small, and for those the theorem can be 2 directly proven. In principle, our approach could be applied to any G/Γ for which one has enough decay for the mixing, or if one can handle in another way the exceptional functions for which such a decay fails to exist. Our Fourier-analytic arguments are local, in the sense that we use Stationary Phase to evaluate the Fourier Transform of the submani- fold at each point. Perhaps it would be possible to extend our result to submanifolds withweaker curvatureconditions byusing globalFourier- analytic methods. On the other hand, curvature could be substituted by another kind of condition; for instance, in the case of S being an affinesubspace oflargedimension, assuggestedin[Ven05, Remark3.1], Fourier Analysis could be directly used to prove that our equidistribu- tion result is true provided its primitive dimension is also large (see Definition 4.2). Our methods cannot by themselves work for submani- folds of small dimension; the only advance on that difficult area seems to be [EMV09]. Throughout the paper we will use the notation e(t) = e2πit for t R, ∈ f(ξ) = f(x)e( ξx)dx the Fourier Transform of f in Rd with ξx Rd − the scalaRr product, and f g meaning f C g for some positive b ≪ | | ≤ | | constant C > 0. 2. Mixing and consequences Wearegoingtodeducetheexponential mixing forhomogeneous one- parameterflows ofGonG/ΓfromthecaseSL (R)onSL (R)/Γ . The 2 2 0 action hΓ ghΓ of an element g of G on G/Γ induces an action of 7→ G on C∞(G/Γ) described by g f(hΓ) = f(ghΓ). On the other hand, · we can consider the inner product in L2(G/Γ) with the Haar measure. With those definitions, the following is a direct consequence of (9.6) in [Ven10] (for a more general result see [KSS02]) Lemma 2.1 (Exponential mixing for SL (R)). For any g G and 2 f ,f C∞(SL (R)/Γ ), with f dµ = 0 we have ∈ 1 2 ∈ 2 0 2 SL2(R) R g f ,f g −r f f 1 2 1 S1 2 S1 h · i ≪ k k k k k k EFFECTIVE EQUIDISTRIBUTION 5 where g is the matrix norm of g and r is any number in (0,1/2] for k k which all nonzero eigenvalues of the hyperbolic Laplacian on /Γ are 0 H bounded from below by r r2. − Now, we are going to use the previous result in order to prove results similar to Lemmas 3.1 and 9.4 in [Ven10]. Their proofs will also be like the ones there. Lemma 2.2 (Equidistribution of translates of horocycles). For any Cx0∞∈(RS)Lw2(eRh)a/vΓe0, f ∈ C∞(SL2(R)/Γ0) with R f dµSL2(R) = 0 and ψ ∈ c f(a(y)u(t)x )ψ(t)dt yr/3 ψ f , 0 S1 S1 Z ≪ k k k k R where the implicit constant depends on x and the length of the smallest 0 interval containing the support of ψ. Remark. Here the Sobolev norms for ψ and f are the ones in R and SL (R)/Γ respectively. 2 0 Proof. The idea of the proof is that a(y) expands the u(t) direction and contracts the rest, so a ball flowed by a(y) transforms essentially into a segment in U+. Let ρ C∞(R) with ρ = 1. For δ (0,1) let us consider the ∈ c R ∈ measure R ν (f) = f(u(t )ta(et2)u(t )x )ρ(t )ρ (t )ψ(t )dt dt dt δ 1 3 0 1 δ 2 3 1 2 3 ZR3 with ρ (t) = δ−1ρ(t/δ). We can write δ ν (a(y) f) = f(a(y)u(t )ta(et2)u(t )x )ρ(t )ρ (t )ψ(t )dt dt dt δ 1 3 0 1 δ 2 3 1 2 3 · Z R3 and since a(y)u(t )t = u(yt )ta(y) we get that 1 1 ν (a(y) f) = f(u(yt )ta(et2)a(y)u(t )x )ρ(t )ρ (t )ψ(t )dt dt dt . δ 1 3 0 1 δ 2 3 1 2 3 · Z R3 Now, by the mean value theorem (see [Ven10], before Lemma 2.2) f(x ) f(x ) f d(x ,x ), with d a fixed right SL (R)-invariant 2 1 S1 2 1 2 d| istanc−e on SL| ≪(Rk)/kΓ , and then due to the identity ρ = 1 we have 2 0 R R ν (a(y) f) = I +O(δ+y) f ψ δ S1 S1 · k k k k with I the integral in the statement of the lemma. On the other hand, anyg SL (R)outsideasetofmeasurezerocanbeuniquelywrittenas 2 ∈ g = u(t )ta(et2)u(t ),andthenwecanwriteν (f) = f(g)H (g)dg 1 3 δ SL2(R) δ for some function H C∞(SL (R)) depending onR x , and covering δ ∈ c 2 0 6 ADRIA´NUBIS SL (R) by translations of a fundamental domain of SL (R)/Γ we 2 2 0 have ν (f) = f,h , δ δ h i for some h C∞(SL (R)/Γ ) with h δ−2 ψ . Here the δ ∈ c 2 0 k δkS1 ≪ k kS1 implicit constant depends both on x and on the length of the smallest 0 interval containing the support of ψ. Thus by Lemma 2.1 we have ν (a(y) f) = a(y) f,h yr f δ−2 ψ δ δ S1 S1 · h · i ≪ k k k k so by choosing δ = yr/3 we get the result. (cid:3) Lemma 2.3 (Uncorrelationoftranslatesofhorocyclesandcharacters). For f C∞(SL (R)/Γ ) with ψ C∞(R) we have ∈ 2 0 ∈ c r2 f(a(y)u(t)x0)ψ(t)e(ct)dt y6+12r ψ S1 f S1. Z ≪ k k k k R where the implicit constant depends on x and the length of the smallest 0 interval containing the support of ψ. Proof. It is enough to prove the result for ψ supported inside ( 1,1). Let ϕ(t) = f(a(y)u(t)x )ψ(t). For any v R we can write the in−tegral 0 ∈ in the statement of the lemma as I = ϕ(t)e(ct)dt = ϕ(t+v)e(ct)e(cv)dt Z Z R R so for y < δ < 1 and ρ C∞( 1,1) with ρ = 1 we have ∈ c − R I = ρ(s) ϕ(t+δs)e(ct)e(cδs)dtds ZR ZR and then 2 I ϕ(t+δs)ρ(s)e(cδs)ds dt. | | ≤ Z−2|ZR | By Cauchy’s inequality I 2 4 ϕ(t+δs)ρ(s)e(cδs)ds 2dt | | ≤ ZR|ZR | and by expanding the square and interchanging the integrals 1 1 I 2 4 ϕ(t+δs )ϕ(t+δs )dt ds ds , 1 2 1 2 | | ≤ Z−1Z−1|ZR | 2 4 ϕ(t+δs)ϕ(t)dt ds. ≤ Z |Z | −2 R EFFECTIVE EQUIDISTRIBUTION 7 But ϕ(t+w)ϕ(t)dt = f(a(y)u(w)u(t)x )f(a(y)u(t)x )ψ(t+w)ψ(t)dt 0 0 Z Z R R and since a(y)u(w) = u(w/y)a(y) we have ϕ(t+w)ϕ(t)dt = f (a(y)u(t)x )ψ (t)dt w 0 w Z Z R R with f (gΓ) = f(u(w/y)gΓ)f(gΓ) and ψ (t) = ψ(t+w)ψ(t). We can w w write f (gΓ) = f∗(gΓ)+ f (gΓ)dµ (g) = f∗(gΓ)+ u(w/y) f,f w w Z w SL2(R) w h · i with f∗ dµ = 0. Now, by applying Lemmas 2.2 and 2.1, and w SL2(R) usingRthat f u(w) f f w f 2 and ψ k wkS1 ≪ k y · kS1k kS1 ≪ yk kS1 k wkS1 ≪ ψ 2 [Ven10, Lemma 2.2] we have k kS1 ϕ(t+w)ϕ(t)dt [yr/3(w/y)+(y/w)r] ψ 2 f 2 . Z ≪ k kS1k kS1 R Using this bound for s > (y/δ)r/(1+r) and the trivial bound otherwise | | we have 2 Z |Z ϕ(t+δs)ϕ(t)dt|ds ≪ [(y/δ)1+rr +yr/3(δ/y)]kψk2S1kfk2S1 −2 R so by picking δ such that both summands are equal we get the bound (cid:3) in the statement. In order to transfer the exponential mixing from SL (R) to G in a 2 simple way, we are going to deal just with functions in G/Γ that can be written as products of functions in SL (R)/Γ . 2 0 Definition 2.4 (Factorizable function). We say that a function f : G/Γ C is factorizable if it can be written as → f((g ,...,g )Γ) = f (g Γ)...f (g Γ) 1 d 1 1 d d with f : SL (R)/Γ C. j 2 0 → Lemma 2.5 (Equidistribution of translates of horospheres). Let f ∈ C∞(G/Γ) be a factorizable function with k components having vanish- c ing integral. Then, we have f(a u x )ψ(t)dt (yr/3)k ψ f y t 0 S4d Sk Z ≪ k k k k Rd with the implicit constant depending on x and the volume of the small- 0 est ball containing the support of ψ. 8 ADRIA´NUBIS Proof. It is enough to prove it with the support of ψ contained in the unit ball. By choosing a fixed ρ C∞(( 2,2)) with ρ = 1 in ( 1,1) ∗ ∈ c − ∗ − and by setting ρ(t) equal to ρ (t )...ρ (t ) we can write ∗ 1 ∗ d (2.1) ψ(t) = ρ(t)ψ(t) = ψ(ξ)ρ (t)dξ ρ (t) = ρ(t)e(ξt) ξ ξ ZRd b By applying Lemma 2.2 on each factor with vanishing integral we have f(a u x )ρ (t)dt (yr/3)k(1+ ξ )k f y t 0 ξ Sk ZRd ≪ | | k k so the lemma follows from the bound ψ(ξ) ψ (1+ ξ )−j. (cid:3) j Sj ≪ k k | | Lemma 2.6 (Uncorrelation of translabtes of horospheres and charac- ters). Let 0 < y < 1 and f C∞(G/Γ) be a factorizable function. ∈ c Then for every c = (c ,...,c ) Rd with c y−r2/(24+48r) we have 1 d ∈ | | ≪ f(ayutx0)ψ(t)e(ct)dt c˜−2d ψ S4d max[y12+r224rl f Sl] ZRd ≪ k k l≤d k k with c˜the geometric mean of (1+ c ,...,1+ c ), with the implicit con- 1 d | | | | stant depending on x and the measure of the smallest ball containing 0 the support of ψ. Proof. We begin as in the proof of Lemma 2.5, by writing ψ in terms of ρ , using (2.1). Since f is factorizable we have ξ f(a u x )ρ (t)e(ct)dt = f (a(y)u(t)x )ρ (t )e(c t )dt , y t 0 ξ j 0,j ξ,j j j j j ZRd YZR j≤d so writing f as a constant plus a function having vanishing integral j and applying Lemma 2.3 we have r2 f(ayutx0)ρξ(t)e(ct)dt ( ρξ,j(cj) fj S0+y6+12r(1+ ξ ) fj S1). Z ≪ | |k k | | k k Rd Y j≤d b Now we use the bound ρ (c ) (1+ ξ 2)/(1+ c 2) and expand the ξ,j j j | | ≪ | | | | product; in the resulting sum, each term with l factors of the shape yr2/(6+12r)(1+ ξ ) f bis bounded by j S1 | | k k c˜−2d(1+ ξ )2d−l[y6+r122r(1+ c 2)]l f Sl, | | | | k k where we have used that the product of d l factors (1 + c )−2 is j − | | bounded by (1+ c )2l/ c˜2d. We finish by using the bound for c and | | | | | | ψ(ξ) ψ (1+ ξ )−j. (cid:3) j Sj ≪ k k | | b EFFECTIVE EQUIDISTRIBUTION 9 3. From G/Γ to Rd Our aim in this section is to transform the problem of equidistribu- tion of translates of S in G/Γ to a related problem in Rd. For that we shall use the mixing results from the previous section; but first we need to show that for equidistribution it is enough to handle factoriz- able functions. This is proven in the next two lemmas. Lemma 3.1 (Reduction to bounded support). Let 0 < δ < 1 and ν be a Borel probability measure on G/Γ. If (3.1) f dν f dµ δ f G Sd |Z −Z | ≫ k k G/Γ G/Γ then the same kind of inequality is true changing f by either a fac- torizable function or a function supported on B = Bd for some R 0,R R δ−1, with B the subset of elements gΓ in SL (R)/Γ satis- 0,R 0 2 0 ≪ fying gΓ 2 R, with gΓ = min gγ , the Frobenius k 0k ≤ k 0k γ0∈Γ0k 0k k · k matrix norm. Proof. Pick ψ C∞(SL (R)) non-negative with ψ(h)dh = 1, ∈ c 2 SL2(R) dh a Haar measure on SL (R), and consider the cRonvolution 2 ψ (gΓ ) = ψ(h)1 (h−1gΓ )dh. 0,R 0 ZSL2(R) B0,R 0 By the multiplicativity of one can show that the supports of ψ 0,R k·k and1 ψ arecontainedinB andBc respectively, withc ,c > − 0,R 0,c1R 0,c2R 1 2 0 two absolute constants. On the other hand, since ψ (gΓ ) = ψ(gh−1)1 (hΓ )dh 0,R 0 ZSL2(R) B0,R 0 we deduce that kψ0,RkSd ≪ 1. Then, since µSL2(R)(B0c,R) ≪ R−1 [Iwa02], we can write f = fψ + fǫ ̺ with i 2d 1, ψ (gΓ) = R i i i ≤ − R j≤dψ0,R(gjΓ0), ǫi = ±1 and ̺iPnon-negative factorizable functions Qwith ̺ 1 and supported on a set of µ measure O(R−1). By i Sd G k k ≪ submultiplicativity of the Sd norm and linearity we can change f by either fψ or f̺ (for some i) in (3.1). In the former case we are done R i because fψ is supported on B ; in the latter, if Rδ is larger than an R R absolute constant, taking into account the measure of the support of ̺ i and its non-negativity we can change f̺ by the factorizable function i f ̺ in (3.1). (cid:3) Sd i k k 10 ADRIA´NUBIS Lemma 3.2 (Reduction to factorizable functions). If we have (3.1) then there exists a factorizable f∗ such that (3.2) f∗dν f∗dµ δ4d+1 f∗ . G Sd |Z −Z | ≫ k k G/Γ G/Γ Proof. By Lemma 3.1 we can assume that the f in (3.1) is supported in B for some R δ−1. Now, for every 0 < β < 1 consider the R function η C∞(≪SL (R)) defined as η (h) = c φ(d (h,I)/β) with φ C∞(10,,2β)∈, d ca fixe2d right SL (R) in0v,βariant Rβiema0nnian distance on∈SLc(R) and0c a constant such2 that η (h)dh = 1. Thus 2 β SL2(R) 0,β c β−3 and η satisfies η β−3R−j. Let us define the factor- β 0,β 0,β Sj ≍ k k ≪ izable function η C∞(G) defined as η (g) = η (g ). We have β ∈ c β j≤d 0,β j η β−3d−j and it is supported on a ballQof radius O(β) around β Sj k k ≪ the identity in G, with the metric d induced from d . For any g G, 0 let us consider the function ηg(h) = η (gh). By invariance of the H∈aar β β measure we have ηg(h)dh = 1. G β Consider the mRap I : C (G) C (Γ G) defined as I(w)(Γh) = c c → \ w(γh) and the left G invariant measure µ∗ on Γ G defined by γ∈Γ \ P I(w)dµ∗ = w(h)dh (see [Rag72]). Applying this to ηg we have Γ\G G β R R (3.3) I(ηg)(Γh)dµ∗(Γh) = 1. Z β Γ\G One can see that I(ηg)(Γh) is invariant under the change g gγ, β 7→ γ Γ, and then we can write I(ηg)(Γh) = ρ (gΓ). Then, using (3.3) ∈ β β,Γh into (3.1) and Fubini we have (3.4) fρ dν fρ dµ dµ∗(Γh) δ f . β,Γh β,Γh G Sd Z |Z −Z | ≫ k k Γ\G G/Γ G/Γ Now, if gΓ is in the support of ρ we have d(gγh,I) β for some β,Γh ≪ γ Γ which by the mean value theorem (see [Ven10], before Lemma ∈ 2.2) f(gΓ) f(h−1Γ) f β. Hence, if β/δ is smaller than an S1 | − | ≪ k k absolute positive constant, from (3.3), (3.4) and H¨older inequality we deduce that ρ dν ρ dµ δ β,Γh β,Γh G |Z −Z | ≫ G/Γ G/Γ for some Γh with h−1Γ 2 δ−1, taking into account the support of f k k ≪ andthe factthat ifgΓis inthesupport ofρ we have d(g,h−1γ−1) β,Γh ≪ 1 for some γ−1 Γ. One can check that ρ is factorizable, so that β,Γh ∈ (3.2) will follow if we show that ρ η β−3d−j with β,Γh Sj β Sj k k ≪ k k ≪ β−1 δ−1. But this in turn follows from showing that in the sum ≪ defining ρ there is just one non-vanishing term. Suppose there β,Γh