.

Unbounded Harmonic Functionson homogeneous manifolds

of negative curvature

Richard PenneyPurdue University

February 23, 2006

1

F is harmonic for a PDE L iff

LF = 0

Theorem (Furstenberg, 1963) . Abounded function F on a Riemannian sym-metric space X is harmonic for the Laplace-Beltrami operator ∆ if and only if F is thePoisson integral of an L∞ function over theFurstenberg boundary B.

Symmetric space:

X = G/K K maximal compact s.g.

G = Semi-simple Lie group

2

Examples G = Sl(2,R) acts on H+

(upper-half plane) via linear fractional trans-formation. The subgroup fixing i is the sub-group K of orthogonal elements in G. G/K =H+. Furstenberg boundary is R ∪ {∞}.Similarly, the unit disk is SU(1, 1) modulothe subgroup fixing 0. Furstenberg bound-ary 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)

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Non-compact picture

X = G/K

= NAK/K

= NA ≡ S

N is a dense subset of the Furstenbergboundary which we approach as a → 0 in(R+)r.

Laplace-Beltrami operator:

∆ =

r∑

1

A2i − A0 +n

∑

1

X2i + X0

Ai,A0 ∈ A Xi ∈ N

A = L. alg. A, N = L. alg. N

(Left invariant vector fields)

4

Corollary 1. There is a positive functionP on S × N (the Poisson kernel function)such that F is a bounded ∆-harmonic func-tion on S if and only if

F (z) =

∫

N

f(n)P (z, n) dn

≡ P (f)

for a unique f ∈ L∞(N).

Generalizations:

Theorem (Guivarc’h (1977) and Raugi(1977)) . A version of Corollary 1 holdsfor any Lie group G and any bounded “µ-harmonic function” on G where µ is a prob-ability measure on G which is “spread out”and has “finite first moment.” In this casethe integral is over N/N1.

5

Theorem (Anderson, 1983) . A versionof Corollary 1 holds for bounded Laplace Bel-trami harmonic functions on Riemannianmanifolds of “pinched negative curvature.”

Theorem (E. Damek, 1988) . A versionof Corollary 1 holds for bonded harmonicfunctions on any split solvable group S =NA for which (i) all of the roots are real(ii) the adjoint representation of A acts di-agonally on S and (iii) there is a root func-tional λ of A such that < λ,A0 >∈ R

+. Inthis case the integral is over N/N1.

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Theorem (Oshima, Sekiguchi) . On arank 1 Riemannian symmetric space(dimA = 1), a ∆-harmonic function F isthe Poisson integral of a distribution iff itis of moderate growth–i.e.

|F (x)| ≤ CeKτ(x)

τ(x) is the Riemannian distance of x to thebase point.

Higher rank:

Poisson integral ⇒∆-harm., mod. growth

∆-harm., mod. growth 6⇒ Poisson integral

Our goal: Describe the harmonic func-tions of “Metric growth”:

|F (na)| ≤ C(|a|k+|a|−k)(1 + |n|)m

Moderate growth ⇒ Metric growth

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We assume S = NA is a Heintze group:

S = NA A one-dimensional

ad A1 real part of all eigenvalues positive

Theorem ( G. Heintze (1974)A homoge-neous Riemannian manifold X has negativesectional curvature if and only if there is aHeintze group S that acts simply transitivelyon 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)∗Xi = adiXi, Xi basis of N , di > 0

Lα = (a∂a)2 − αa∂a +

k∑

1

a2diX̃2i

+k

∑

1

ciadiX̃i

X̃i indep. of a

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Example:

S = R ×s R+ = H+

α = 1

Lα = a2(

∂2a + ∂2x

)

Poisson kernel:

P (z, n) =1

πℑ

(

1

n − z

)

where n ∈ R and z = x + ia ∈ R × R+.

F (z) = ℜ(zn)

F (x) = xn

Not Poisson integrable for n ∈ N!

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P (z, n) =1

πℑ

(

∞∑

0

zk

nk+1

)

.

For n 6= 0, let

Qm(z, n) =1

πℑ

(

m∑

0

zk

nk+1

)

Pm(z, n) = P (z, n) − Qm(z, n)

Pm(z, n) is harmonic in z

|Pm(z, n)| ≤ C(1 + |z|)a|n|−(m+1)

Qm(x, n) = 0 x ∈ R

For

f(x) ≡ 0 on a neighborhood of 0 in R

Pm(f)(z) =

∫

R

f(n)Pm(z, n) dn.

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More generally:

φ ∈ C∞c (R)

0 ≤ φ(x) ≤ 1

φ(x) = 1, x ∈ [−1, 1]

Set

Pmolm,φ (f) = Pm((1 − φ)f) + P (φf)

Lemma (Left to listener) . If Pmolm,φ (f)

and Pmolm′,φ′(f) are both defined then

Pmolm,φ (f)(z) − Pmolm′,φ′(f)(z) = Q(z)

where Q(z) is a harmonic polynomial onH+ such that Q ≡ 0 on R.

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Theorem . Suppose that F is a harmonicfunction on H+ of metric growth. Then

f(x) = lima→0+

F (x, a)

exists in the sense of tempered distributionson R. Furthermore, for any choice of φ andm for which Pm,φ(f) is defined, there is apolynomial Q(x, a) such that

Q(x, a) is harmonic

Q(x, a) = aQ0(x, a) Q0 a polynomial in (x, a)

F (x, a) = Pm,φ(f)(x, a) + Q(x, a).

Conversely, every expression of the form onthe right above, where f is a tempered dis-tribution, defines a harmonic function withmetric growth whose boundary value is f .

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Exactly the same theorem holdsin Our Case.

Let

Ñ = span N∪{0}{d1, . . . , dn}.

Instead of

Qm(z, n) =1

πℑ

(

m∑

0

zk

nk+1

)

we have

Qµ(x, a, n) =∑

β

Just as before we set

Pµ(z, n) = P (z, n) − Qµ(z, n)

Now

|Pµ(x, a, n)| ≤ C(aa + ab)(1 + |x|)τ |n|−d−α−µ

We definePmolµ,φ (f)

just as before where now φ ≡ 1 on a neigh-borhood of e in N .

We refer to the expansion

P (x, a, n) ≈∑

β∈Ñ

Hβ(x, a, n)

as the “expansion of P at n = ∞.”Definition A function Q on S is a polyno-mial if

Q(x, a) = Q0(x, ad1 , ad2 , . . . , adi)

where Q0 is a polynomial on N × Rn.

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Theorem . Suppose that in Our CaseF is a harmonic function on S of metricgrowth. Then

f(x) = lima→0+

F (x, a)

exists in the sense of tempered distributionson N . Furthermore, for any choice of φand µ for which Pµ,φ(f) is defined, there isa polynomial Q0(x, a) such that

aαQ0(x, a) is harmonic

F (x, a) = Pµ,φ(f)(x, a) + aαQ0(x, a).

Conversely, every expression of the form onthe right above, where f is a tempered dis-tribution on N and aαQ0(x, a) is harmonicwith Q0 a polynomial, defines a harmonicfunction with metric growth whose bound-ary value is f .

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Corollary . Every Schwartz distributionf on N is the boundary value for a L har-monic function F of metric growth which isuniquely determined modulo harmonic func-tions of the from aαQ0(x, a) where Q0 is apolynomial on S.

Remark: Determining F from f moduloharmonic “polynomials” with null bound-ary value is the best we can do e.g. F (z) =ℑ(zn) has null boundary value.

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For the sake of simplicity, we as-sume from this point on that α /∈ Ñ .

Definition We say that a function F onS is “polynomial like” relative to L if thereare polynomial functions (in the sense justdefined) p, q, and h on S such that

F (x, a) = p(x, a) + aαq(x, a).

Let P(N) be the set of polynomial func-tions on N .

Theorem . There is an explicit bijectivelinear mapping between P(N) × P(N) andthe the set of harmonic polynomial like func-tions on S which maps 0 × P(N) into theset of harmonic polynomials of the of thefrom aαQ0(x, a). In particular the set ofsuch harmonic functions is infinite dimen-sional.

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Proof Outline:1 . We prove our “expansion of P (x, a, n)

at n = ∞” using an asymptotic expan-sion of P (x, a, e) as a → 0.

2 . We use an asymptotic expansion toprove the existence oflima→0+ F (x, a) in S(N)

′.3 . Let F̃ = F − Pmolµ,φ (f). We show that

lima→0+ F̃ (x, a) = 0 in S′(N).

4 . We prove a “Liouville Theorem” thatshows that F̃ is a polynomial-like func-tion which vanishes on N .

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Liouville Theorem

Classical Liouville theorem:A bounded harmonic function on C mustbe constant.

Theorem . Suppose that F is Lα harmonicon S and satisfies

|F (x, a)| ≤ C(ab + ac)(1 + |x|)k

where b, c, k are all positive. Then

F (x, a) = aαQ(x, a)

for some polynomial Q(x, a).

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Remark: If M is a complete, non-compactRiemannian maniold, with non-negative Riccicurvature, Colding and Minicozzi II provedthat the space of harmonic functions withpolynomial growth of order at most d isfinite dimensional, proving a conjecture ofS. Yau. Our results may be thought of asan extension of these results to a negativelycurved case.

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Asymptotic Expansions

Assume that F is harmonic of metricgrowth–i.e.

|F (na)| ≤ C(ak+a−k)(1 + |n|)m

Let φ ∈ C∞c (NA). Replace F with φ ∗ F .

|XIF (na)| ≤ CI(ak + a−k)(1 + |n|)mI

XI = Xi11 . . . Xinn

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Theorem . There are unique functionsFβ , Gβ in C

∞(N) indexed by Ñ such that

for all µ ∈ Ñ

F (x, a) =∑

β

Proof

Lα = LA + adLN d > 0

LN =k

∑

1

abiX̃2i +

k∑

1

ciaciX̃i bi, ci ≥ 0

LA = a∂a(a∂a − α)

We right-invert LA. For each b ≥ 0,the operator

ΛbF (a) =

∫ a

b

s−1F (s) ds

is a right inverse for a∂a on R+.

We use Λ0 and Λ1.If F (a) ≤ Cac, then

|Λ1F (a)| ≤ C(1 + ac)

|Λ0F (a)| ≤ Cac c > 0

(Λ0 is defined only if c > 0.)

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(C changes line to line–c does not.)Right inverse for a∂a − α:

Λαb = aαΛba

−α

(Λα0 is defined only if c > α.)

For b ≥ 0, c ≥ 0

Λb,c = Λαb Λc

is a right inverse for LA. Hence

Λb,cLAF (a) = F (a) + A + Baα.

Then, for F (x, a) harmonic

LAF = −adLNF

Λ1,1LAF = −Λ1,1(adLN )F

F (x, a) = A(x) + B(x)aα

− Λ1,1(adLN )F (x, a)

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Hence

(I + N1,1)F = A(x) + B(x)aα

N1,1 = Λ1,1(adLN )

Formally

F =∞∑

0

(−1)nNn1,1 (A(x) + aαB(x)) .

This series typically does not not converge.However, let

Fn =n

∑

0

(−1)kNk1,1 (A(x) + aαB(x)) .

Then Fn ∈ F where

F = {∑

β∈I⊂Ñ

(

aβFβ(x) + aα+βGβ(x)

)

| |I| < ∞}

Also

F − Fn = (−1)n+1Nn+11,1 F.

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For C and M generic constants,

|F (x, a)| ≤ C(ak + a−k)(1 + |x|)M

Hence for 0 < a < 1

|adLNF (x, a)| ≤ C(ak+d + a−k+d)(1 + |x|)M

|N1,1F (x, a)| ≤ C(1 + a−k+d)(1 + |x|)M

Thus, for n > kd

|F (x, a) − Fn(x, a)| ≤ C(1 + a−k+nd)(1 + |x|)M

≤ C(1 + |x|)M

This is as far as we can go with N1,1.But

G(x, a) = F (x, a) − Fn(x, a) ∈ D(N1,0)

N1,0 = Λ1,0(adLN )

andLG ≡ 0 mod F

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Hence

Q = (I + N1,0)G ≡ 0 mod F .

Let

Gm =

m∑

0

(−1)kNk1,0Q.

Then Gm ∈ F and

G − Gm = (−1)m+1Nm+11,0 G

F − (Fn + Gm) = (−1)m+1Nm+11,0 G

It is easily seen that for m sufficiently large

Nm+11,0 G ∈ D(N0,0).

We let

H = F − (Fn + Gm).

ThenLH ≡ 0 mod F .

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Now

Q̃ = (I + N0,0)H ≡ 0 mod F .

Let

H l =l

∑

0

(−1)klNk0,0Q.

Then H l ∈ F and

H − H l = (−1)l+1N l+11,0 G

F − (Fn + Gm + H l) = (−1)l+1N l+10,0 H

The existence of the asymptotic expansionfollows from the fact that

|Np0,0H(x, a)| ≤ Capd(1 + |x|)M .

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The Expansion of P at ∞

P (x, a, n) = P (n−1x, a, e).

LetP̃ (x, a) = P (x, a, e).

Known: For all multi-indecies I

|X̃I P̃ (x, a)| ≤ Caα(a + |x|)−(d+α+‖I‖)

where‖I‖ =

∑

djij

d =∑

dj

For each n ∈ N , P (x, a, n) is harmonicin (x, a). In articular P̃ is harmonic.

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Theorem . There is a a sequence Pβ ∈

C∞(N \ {0}) indexed by Ñ, where each Pβis δ(a)-homogeneous of degree −β − d − α,such that for all µ ∈ Ñ

P̃ (x, a) = aα∑

β

Liouville Theorem

Let χL = χ(0,1] and χR = χ[1,∞).

Lemma . Assume that F is L-harmonicand satisfies

|F (x, a)| ≤ C(aαχL(a) + abχR(a))(1 + |x|)

k

where b < −α − 4. Then F ≡ 0.For the proof we show that

X̃IF = 0

whenever |I| is sufficiently large. i.e. for allφ ∈ C∞c (S),

< F, X̃Iφ >= 0

We show that for |I| sufficiently large,there is a C∞ function ψI such that

L−αψI = X̃Iφ

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FormallyL−α = (Lα)∗.

Hence,with luck, it should follow that

< F, X̃Iφ > =< F,L−αψI >

=< LαF, ψI >= 0.

For this to work, we need ψI and its deriva-tives to tend to 0 sufficiently fast at 0 and∞.

To prove the existence of ψI and toprove the decay of the derivatives, we usea Green’s function for L−α constructed byR. Urban, together with his estimates onthe decay of the Green’s kernel.

The case where F can grow as a → ∞ isconsiderably subtler and involves reducingto the case just described.

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