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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)
3
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.
6
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
7
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
8
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!
9
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.
10
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.
11
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 .
12
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.
14
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 .
15
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.
16
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.
17
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 .
18
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).
19
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.
20
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
21
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.)
23
(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)
24
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.
25
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
26
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 .
27
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 .
28
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.
29
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
31
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.
32