Unbounded Harmonic Functions on homogeneous manifolds of negative curvature

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1 . Unbounded Harmonic Functions on homogeneous manifolds of negative curvature Richard Penney Purdue University February 23,

2 F is harmonic for a PDE L iff LF = 0 Theorem (Furstenberg, 1963). A bounded function F on a Riemannian symmetric space X is harmonic for the Laplace- Beltrami operator if and only if F is the Poisson integral of an L function over the Furstenberg boundary B. Symmetric space: X = G/K K maximal compact s.g. G = Semi-simple Lie group 2

3 Examples G = Sl(2,R) acts on H + (upper-half plane) via linear fractional transformation. The subgroup fixing i is the subgroup K of orthogonal elements in G. G/K = H +. Furstenberg boundary is R { }. Similarly, the unit disk is SU(1,1) modulo the subgroup fixing 0. Furstenberg boundary is { z = 1}. Iwasawa Decomposition: G = NAK N nilpotent A (R + ) r r = rank Example: G = Sl(n,R) N = upper-triangular, 1 on diagonal A = diagonal, positive entries K = SO(n,R) 3

4 Non-compact picture X = G/K = NAK/K = NA S N is a dense subset of the Furstenberg boundary which we approach as a 0 in (R + ) r. Laplace-Beltrami operator: = r A 2 i A A i,a 0 A A = L. alg. A, X i N n Xi 2 + X 0 1 N = L. alg. N (Left invariant vector fields) 4

5 Corollary 1. There is a positive function P on S N (the Poisson kernel function) such that F is a bounded -harmonic function on S if and only if F(z) = N P(f) for a unique f L (N). Generalizations: f(n)p(z, n) dn Theorem (Guivarc h (1977) and Raugi (1977)). A version of Corollary 1 holds for any Lie group G and any bounded µharmonic function on G where µ is a probability measure on G which is spread out and has finite first moment. In this case the integral is over N/N 1. 5

6 Theorem (Anderson, 1983). A version of Corollary 1 holds for bounded Laplace Beltrami harmonic functions on Riemannian manifolds of pinched negative curvature. Theorem (E. Damek, 1988). A version of Corollary 1 holds for bonded harmonic functions on any split solvable group S = NA for which (i) all of the roots are real (ii) the adjoint representation of A acts diagonally on S and (iii) there is a root functional λ of A such that < λ,a 0 > R +. In this case the integral is over N/N 1. 6

7 Theorem (Oshima, Sekiguchi). On a rank 1 Riemannian symmetric space (dima = 1), a -harmonic function F is the Poisson integral of a distribution iff it is of moderate growth i.e. F(x) Ce Kτ(x) τ(x) is the Riemannian distance of x to the base point. Higher rank: Poisson integral -harm., mod. growth -harm., mod. growth Poisson integral Our goal: Describe the harmonic functions of Metric growth : F(na) C( a k + a k )(1 + n ) m Moderate growth Metric growth 7

8 We assume S = NA is a Heintze group: S = NA ad A 1 A one-dimensional real part of all eigenvalues positive Theorem ( G. Heintze (1974)A homogeneous Riemannian manifold X has negative sectional curvature if and only if there is a Heintze group S that acts simply transitively on X. Our Case S = N s R + typical element (n, a) δ(a)n = (0, a)n(0, a 1 ) n N S δ(a) X i = a d i X i, X i basis of N, d i > 0 k L α = (a a ) 2 αa a + a 2d i X2 i + k c i a d i Xi 1 X i indep. of a 8 1

9 Example: S = R s R + = H + α = 1 L α = a 2 ( a 2 + x 2 ) Poisson kernel: P(z, n) = 1 π I ( 1 n z where n R and z = x + ia R R +. ) F(z) = R(z n ) F(x) = x n Not Poisson integrable for n N! 9

10 For n 0, let P(z, n) = 1 π I ( Q m (z, n) = 1 π I ( m 0 0 ) z k n k+1. ) z k n k+1 P m (z, n) = P(z, n) Q m (z, n) P m (z, n) is harmonic in z P m (z, n) C(1 + z ) a n (m+1) Q m (x, n) = 0 For x R f(x) 0 on a neighborhood of 0 in R P m (f)(z) = f(n)p m (z, n) dn. R 10

11 More generally: φ Cc (R) 0 φ(x) 1 φ(x) = 1, x [ 1,1] Set P mol m,φ(f) = P m ((1 φ)f) + P(φf) Lemma (Left to listener). If P mol and Pm mol,φ (f) are both defined then P mol m,φ(f)(z) P mol m,φ (f)(z) = Q(z) m,φ (f) where Q(z) is a harmonic polynomial on H + such that Q 0 on R. 11

12 Theorem. Suppose that F is a harmonic function on H + of metric growth. Then f(x) = lim F(x, a) a 0 + exists in the sense of tempered distributions on R. Furthermore, for any choice of φ and m for which P m,φ (f) is defined, there is a polynomial Q(x, a) such that Q(x, a) is harmonic Q(x, a) = aq 0 (x, a) Q 0 a polynomial in (x, a) F(x, a) = P m,φ (f)(x, a) + Q(x, a). Conversely, every expression of the form on the right above, where f is a tempered distribution, defines a harmonic function with metric growth whose boundary value is f. 12

13 Exactly the same theorem holds in Our Case. Let we have Ñ = span N {0} {d 1,..., d n }. Instead of Q m (z, n) = 1 π I ( m Q µ (x, a, n) = Ñ = { β<µ,β Ñ 0 ) z k n k+1 H β (x, a, n) n m i d i m i Z, m i 0} 1 where H β (x, a, n) is L α harmonic and H β (δ(t)x, ta, n) = t β H β (x, a, n) H β (x, a, δ(t)n) = t β d H β (x, a, n) d = d i 13

14 Just as before we set Now P µ (z, n) = P(z, n) Q µ (z, n) P µ (x, a, n) C(a a + a b )(1 + x ) τ n d α µ We define P mol µ,φ (f) just as before where now φ 1 on a neighborhood of e in N. We refer to the expansion P(x, a, n) β Ñ H β (x, a, n) as the expansion of P at n =. Definition A function Q on S is a polynomial if Q(x, a) = Q 0 (x, a d 1, a d 2,..., a d i ) where Q 0 is a polynomial on N R n. 14

15 Theorem. Suppose that in Our Case F is a harmonic function on S of metric growth. Then f(x) = lim F(x, a) a 0 + exists in the sense of tempered distributions on N. Furthermore, for any choice of φ and µ for which P µ,φ (f) is defined, there is a polynomial Q 0 (x, a) such that a α Q 0 (x, a) is harmonic F(x, a) = P µ,φ (f)(x, a) + a α Q 0 (x, a). Conversely, every expression of the form on the right above, where f is a tempered distribution on N and a α Q 0 (x, a) is harmonic with Q 0 a polynomial, defines a harmonic function with metric growth whose boundary value is f. 15

16 Corollary. Every Schwartz distribution f on N is the boundary value for a L harmonic function F of metric growth which is uniquely determined modulo harmonic functions of the from a α Q 0 (x, a) where Q 0 is a polynomial on S. Remark: Determining F from f modulo harmonic polynomials with null boundary value is the best we can do e.g. F(z) = I(z n ) has null boundary value. 16

17 For the sake of simplicity, we assume from this point on that α / Ñ. Definition We say that a function F on S is polynomial like relative to L if there are polynomial functions (in the sense just defined) p, q, and h on S such that F(x, a) = p(x, a) + a α q(x, a). Let P(N) be the set of polynomial functions on N. Theorem. There is an explicit bijective linear mapping between P(N) P(N) and the the set of harmonic polynomial like functions on S which maps 0 P(N) into the set of harmonic polynomials of the of the from a α Q 0 (x, a). In particular the set of such harmonic functions is infinite dimensional. 17

18 Proof Outline: 1. We prove our expansion of P(x, a, n) at n = using an asymptotic expansion of P(x, a, e) as a We use an asymptotic expansion to prove the existence of lim a 0 + F(x, a) in S(N). 3. Let F = F P mol µ,φ (f). We show that lim a 0 + F(x, a) = 0 in S (N). 4. We prove a Liouville Theorem that shows that F is a polynomial-like function which vanishes on N. 18

19 Liouville Theorem Classical Liouville theorem: A bounded harmonic function on C must be constant. Theorem. Suppose that F is L α harmonic on S and satisfies F(x, a) C(a b + a c )(1 + x ) k where b, c, k are all positive. Then F(x, a) = a α Q(x, a) for some polynomial Q(x, a). 19

20 Remark: If M is a complete, non-compact Riemannian maniold, with non-negative Ricci curvature, Colding and Minicozzi II proved that the space of harmonic functions with polynomial growth of order at most d is finite dimensional, proving a conjecture of S. Yau. Our results may be thought of as an extension of these results to a negatively curved case. 20

21 Asymptotic Expansions Assume that F is harmonic of metric growth i.e. F(na) C(a k +a k )(1 + n ) m Let φ C c (NA). Replace F with φ F. X I F(na) C I (a k + a k )(1 + n ) m I X I = X i Xi n n 21

22 Theorem. There are unique functions F β, G β in C (N) indexed by Ñ such that for all µ Ñ F(x, a) = β<µ a β F β (x) + a α β<µ a β G β (x) where + R Asy µ (x, a) R Asy µ (x, a) Ca µ (1 + x ) k µ for 0 < a 1. (If α logarithmic terms.) Ñ, there will also be 22

23 Proof L α = L A + a d L N d > 0 k k L N = a b i X2 i + c i a c i Xi b i, c i 0 1 L A = a a (a a α) 1 We right-invert L A. For each b 0, the operator Λ b F(a) = a b s 1 F(s) ds is a right inverse for a a on R +. We use Λ 0 and Λ 1. If F(a) Ca c, then Λ 1 F(a) C(1 + a c ) Λ 0 F(a) Ca c c > 0 (Λ 0 is defined only if c > 0.) 23

24 (C changes line to line c does not.) Right inverse for a a α: Λ α b = a α Λ b a α (Λ α 0 is defined only if c > α.) For b 0, c 0 Λ b,c = Λ α b Λ c is a right inverse for L A. Hence Λ b,c L A F(a) = F(a) + A + Ba α. Then, for F(x, a) harmonic L A F = a d L N F Λ 1,1 L A F = Λ 1,1 (a d L N )F F(x, a) = A(x) + B(x)a α Λ 1,1 (a d L N )F(x, a) 24

25 Hence Formally F = (I + N 1,1 )F = A(x) + B(x)a α N 1,1 = Λ 1,1 (a d L N ) ( 1) n N1,1 n (A(x) + a α B(x)). 0 This series typically does not not converge. However, let F n = n ( 1) k N1,1 k (A(x) + a α B(x)). 0 Then F n F where F = { ( a β F β (x) + a α+β G β (x) ) I < } β I Ñ Also F F n = ( 1) n+1 N n+1 1,1 F. 25

26 For C and M generic constants, F(x, a) C(a k + a k )(1 + x ) M Hence for 0 < a < 1 a d L N F(x, a) C(a k+d + a k+d )(1 + x ) M N 1,1 F(x, a) C(1 + a k+d )(1 + x ) M Thus, for n > k d F(x, a) F n (x, a) C(1 + a k+nd )(1 + x ) M C(1 + x ) M But This is as far as we can go with N 1,1. G(x, a) = F(x, a) F n (x, a) D(N 1,0 ) N 1,0 = Λ 1,0 (a d L N ) and LG 0 mod F 26

27 Hence Q = (I + N 1,0 )G 0 mod F. Let G m = m ( 1) k N1,0Q. k 0 Then G m F and G G m = ( 1) m+1 N m+1 1,0 G F (F n + G m ) = ( 1) m+1 N m+1 1,0 G It is easily seen that for m sufficiently large We let N m+1 1,0 G D(N 0,0 ). H = F (F n + G m ). Then LH 0 mod F. 27

28 Now Q = (I + N 0,0 )H 0 mod F. Let H l = l ( 1) k ln0,0q. k 0 Then H l F and H H l = ( 1) l+1 N l+1 1,0 G F (F n + G m + H l ) = ( 1) l+1 N l+1 0,0 H The existence of the asymptotic expansion follows from the fact that N p 0,0 H(x, a) Capd (1 + x ) M. 28

29 The Expansion of P at P(x, a, n) = P(n 1 x, a, e). Let P(x, a) = P(x, a, e). Known: For all multi-indecies I X I P(x, a) Ca α (a + x ) (d+α+ I ) where I = d j i j d = d j For each n N, P(x, a, n) is harmonic in (x, a). In articular P is harmonic. 29

30 Theorem. There is a a sequence P β C (N \ {0}) indexed by Ñ, where each P β is δ(a)-homogeneous of degree β d α, such that for all µ Ñ P(x, a) = a α β<µa β P β (x) + R µ (x, a) where for all multi-indecies I R µ (x, a) Ca α+µ x (d+α+µ) a (0,1]. Furthermore, the P β are unique. For each β, we expand P β (n 1 x) in a δ- homogeneous Taylor series in n about n = 0, substituting the result into the asymptotic series, we obtain a double expansion for P(x, a, n). Our theorem follows by grouping these terms by homogeneity. 30

31 Liouville Theorem Let χ L = χ (0,1] and χ R = χ [1, ). Lemma. and satisfies Assume that F is L-harmonic F(x, a) C(a α χ L (a) + a b χ R (a))(1 + x ) k where b < α 4. Then F 0. For the proof we show that X I F = 0 whenever I is sufficiently large. i.e. for all φ C c (S), < F, X I φ >= 0 We show that for I sufficiently large, there is a C function ψ I such that L α ψ I = X I φ 31

32 Formally L α = (L α ). Hence,with luck, it should follow that < F, X I φ > =< F, L α ψ I > =< L α F, ψ I >= 0. For this to work, we need ψ I and its derivatives to tend to 0 sufficiently fast at 0 and. To prove the existence of ψ I and to prove the decay of the derivatives, we use a Green s function for L α constructed by R. Urban, together with his estimates on the decay of the Green s kernel. The case where F can grow as a is considerably subtler and involves reducing to the case just described. 32