BP in HnC
Convex Geometry
|x|
Connection with kxk−2
(1 − ( kxk )2 )−1
K
K
Solution of the Problem
The Busemann-Petty Problem in the
Complex Hyperbolic Space
Susanna Dann
University of Missouri-Columbia
Texas A&M University, July 19, 2012
BP in HnC
Convex Geometry
|x|
Connection with kxk−2
(1 − ( kxk )2 )−1
K
K
Solution of the Problem
The Busemann-Petty Problem in Rn
Let K and L be two origin-symmetric convex bodies in Rn .
Suppose that for every ξ ∈ Sn−1
Voln−1 (K ∩ ξ ⊥ ) ≤ Voln−1 (L ∩ ξ ⊥ ).
Does it follow that
Voln (K ) ≤ Voln (L)?
The answer is affirmative for n ≤ 4 and negative for n ≥ 5.
posed in 1956
solved in 90’
BP in HnC
Convex Geometry
|x|
Connection with kxk−2
(1 − ( kxk )2 )−1
K
K
Solution of the Problem
The Busemann-Petty Problem in other spaces
Yaskin : in real hyperbolic and spherical spaces
For the real spherical space the answer is the same as for Rn and
for the real hyperbolic space the answer is affirmative for n ≤ 2
and negative for n ≥ 3.
Koldobsky, König, Zymonopoulou : Cn
For the complex version of the Busemann-Petty problem the
answer is affirmative for the complex dimension n ≤ 3 and negative
for n ≥ 4.
BP in HnC
Convex Geometry
|x|
Connection with kxk−2
(1 − ( kxk )2 )−1
K
K
Solution of the Problem
The Ball Model of HnC
HnC can be identified with the open unit ball in Cn
Bn := {z ∈ Cn : (z, z) < 1}.
The volume element on HnC is
dµn = 8n
r 2n−1 drdσ
,
(1 − r 2 )n+1
where dσ is the volume element on the unit sphere S2n−1 .
For K ⊂ Bn denote by HVol2n (K ) the volume of K with respect to
dµn .
BP in HnC
Convex Geometry
|x|
Connection with kxk−2
(1 − ( kxk )2 )−1
K
K
Solution of the Problem
Hyperplanes in Cn
For ξ ∈ Cn , |ξ| = 1, denote by Hξ := {z ∈ Cn : (z, ξ) = 0} the
complex hyperplane through the origin perpendicular to ξ.
Identify Cn with R2n via the mapping
(ξ11 + iξ12 , . . . , ξn1 + iξn2 ) 7→ (ξ11 , ξ12 , . . . , ξn1 , ξn2 ) .
Under this mapping the hyperplane Hξ turns into a
(2n − 2)-dimensional subspace of R2n orthogonal to the vectors
ξ = (ξ11 , ξ12 , . . . , ξn1 , ξn2 ) and ξ⊥ := (−ξ12 , ξ11 , . . . , −ξn2 , ξn1 ) .
BP in HnC
Convex Geometry
|x|
Connection with kxk−2
(1 − ( kxk )2 )−1
K
K
Solution of the Problem
Origin Symmetric Convex Sets in Cn
K ⊂ Cn is origin symmetric if x ∈ K implies −x ∈ K .
Origin symmetric convex bodies in Cn are unit balls of norms on
Cn . We call them complex convex bodies.
Norms on Cn satisfy: kλzk = |λ|kzk for λ ∈ C.
A star body K in R2n is called Rθ -invariant, if for every θ ∈ [0, 2π]
and every ξ = (ξ11 , ξ12 , . . . , ξn1 , ξn2 ) ∈ R2n
kξkK = kRθ (ξ11 , ξ12 ), . . . , Rθ (ξn1 , ξn2 )kK ,
where Rθ stands for the counterclockwise rotation by an angle θ
around the origin in R2 .
Thus complex convex bodies in Cn are Rθ -invariant convex bodies
in R2n .
BP in HnC
Convex Geometry
|x|
Connection with kxk−2
(1 − ( kxk )2 )−1
K
K
Solution of the Problem
Formulation of the problem
The Busemann-Petty problem in HnC can be posed as follows.
Given two Rθ -invariant convex bodies K and L in R2n contained in
the unit ball such that
HVol2n−2 (K ∩ Hξ ) ≤ HVol2n−2 (L ∩ Hξ )
for all ξ ∈ S2n−1 , does it follow that
HVol2n (K ) ≤ HVol2n (L)?
BP in HnC
Convex Geometry
|x|
Connection with kxk−2
(1 − ( kxk )2 )−1
K
K
Solution of the Problem
Basic Definitions
A distribution f on Rn is even homogeneous of degree p ∈ R, if
D
x E
= |α|n+p hf , ϕi
f (x), ϕ
α
for every test function ϕ and every α ∈ R, α 6= 0.
The Fourier transform of an even homogeneous distribution of
degree p on Rn is an even homogeneous distribution of degree
−n − p.
A distribution
f is positive definite, if fˆ is a positive distribution,
D
E
ˆ
i.e. f , ϕ ≥ 0 for every non-negative test function ϕ.
A star body K is k-smooth, k ∈ N ∪ {0}, if k · kK belongs to the
class C k (Sn−1 ) of k times continuously differentiable functions on
the unit sphere.
If k · kK ∈ C k (Sn−1 ) for any k ∈ N, then a star body K is said to
be infinitely smooth.
BP in HnC
Convex Geometry
|x|
Connection with kxk−2
(1 − ( kxk )2 )−1
K
K
Solution of the Problem
Approximation Results
The radial function of K is ρK (x) = kxk−1
K .
One can approximate any convex body K in Rn in the radial metric
ρ(K , L) := max |ρK (x) − ρL (x)|
x∈Sn−1
by a sequence of infinitely smooth convex bodies with the same
symmetries as K .
BP in HnC
Convex Geometry
|x|
Connection with kxk−2
(1 − ( kxk )2 )−1
K
K
Solution of the Problem
Fourier Approach to Sections
For an infinitely smooth origin symmetric star body K in Rn and
0 < p < n, the Fourier transform of the distribution kxk−p
K is an
infinitely smooth function on Rn \ {0}, homogeneous of degree
−n + p.
Lemma
Let K and L be infinitely smooth origin symmetric star bodies in
Rn , and let 0 < p < n. Then
Z
Z
−p ∧
−n+p ∧
−n+p
n
(k·kK ) (θ)(k·kL
) (θ)dθ = (2π)
kθk−p
dθ .
K kθkL
Sn−1
Sn−1
BP in HnC
Convex Geometry
|x|
Connection with kxk−2
(1 − ( kxk )2 )−1
K
K
Solution of the Problem
Fourier Approach to Sections
Let 0 < k < n and let H be an (n − k)-dimensional subspace of Rn .
Fix an orthonormal basis e1 , . . . , ek in the orthogonal subspace H ⊥ .
For a star body K in Rn , define the (n − k)-dimensional parallel
section function AK ,H as a function on Rk such that for u ∈ Rk
AK ,H (u) = Voln−k (K ∩ {H + u1 e1 + · · · + uk ek }).
If K is infinitely smooth, the function AK ,H is infinitely
differentiable at the origin.
Lemma
Let K be an infinitely smooth origin symmetric star body in Rn
and 0 < k < n. Then for every (n − k)-dimensional subspace H of
Rn and for every m ∈ N ∪ {0}, m < (n − k)/2,
Z
(−1)m
∆m AK ,H (0) =
(kxk−n+2m+k
)∧ (ξ)dξ ,
K
(2π)k (n − 2m − k) Sn−1 ∩H ⊥
where ∆ denotes the Laplacian on Rk .
BP in HnC
Convex Geometry
|x|
Connection with kxk−2
(1 − ( kxk )2 )−1
K
K
Solution of the Problem
Fourier and Radon Transforms of Rθ -invariant Functions
Lemma
Suppose that K is an infinitely smooth Rθ -invariant star body in
R2n . Then for every 0 < p < 2n and ξ ∈ S2n−1 the Fourier
transform of the distribution kxk−p
K is a constant function on
S2n−1 ∩ Hξ⊥ .
Proof : Since kxkK is Rθ -invariant, so is the Fourier transform of
⊥
kxk−p
K . The two-dimensional space Hξ is spanned by two vectors ξ
and ξ⊥ .
Every vector in S2n−1 ∩ Hξ⊥ is the image of ξ under one of the
coordinate-wise rotations Rθ , so the Fourier transform of kxk−p
K is
a constant function on S2n−1 ∩ Hξ⊥ .
BP in HnC
Convex Geometry
|x|
Connection with kxk−2
(1 − ( kxk )2 )−1
K
K
Solution of the Problem
Fourier and Radon Transforms of Rθ -invariant Functions
Denote by Cθ (S2n−1 ) the space of Rθ -invariant continuous
functions on the unit sphere S2n−1 , i.e. continuous real-valued
functions f satisfying f (ξ) = f (Rθ ξ) for any ξ ∈ S2n−1 and any
θ ∈ [0, 2π].
The complex spherical Radon transform is an operator
Rc : Cθ (S2n−1 ) → Cθ (S2n−1 ) defined by
Z
Rc f (ξ) =
f (x)dx.
S2n−1 ∩Hξ
BP in HnC
Convex Geometry
|x|
Connection with kxk−2
(1 − ( kxk )2 )−1
K
K
Solution of the Problem
Fourier and Radon Transforms of Rθ -invariant Functions
Lemma
Let f ∈ Cθ (S2n−1 ). Extend f to a homogeneous function of degree
−2n + 2, f · r −2n+2 , then the Fourier transform of this extension is
a homogeneous function of degree −2 on R2n , whose restriction to
the unit sphere is continuous. Moreover, for every ξ ∈ S2n−1
Rc f (ξ) =
1
(f · r −2n+2 )∧ (ξ) .
2π
BP in HnC
|x|
Connection with kxk−2
(1 − ( kxk )2 )−1
K
Convex Geometry
K
Solution of the Problem
Volume of Sections
Let K be an Rθ -invariant star body contained in the unit ball of
R2n with n ≥ 2. Let ξ ∈ S2n−1 , we compute:
HVol2n−2 (K ∩ Hξ )
Z
n−1
=8
kxk−1
K
r 2n−3
drdx
(1 − r 2 )n
S2n−1 ∩Hξ 0
Z |x|
Z
kxkK
r 2n−3
n−1
−2n+2
=8
|x|
drdx
(1 − r 2 )n
S2n−1 ∩Hξ
0
!∧
Z |x|
2n−3
kxkK
r
8n−1
|x|−2n+2
dr
(ξ) .
=
2π
(1 − r 2 )n
0
Z
BP in HnC
Convex Geometry
|x|
Connection with kxk−2
(1 − ( kxk )2 )−1
K
K
Solution of the Problem
Counterexample for n ≥ 4
Theorem
There exist Rθ -invariant convex bodies K and L contained in the
unit ball of R2n with n ≥ 4 satisfying
HVol2n−2 (K ∩ Hξ ) ≤ HVol2n−2 (L ∩ Hξ )
for every ξ ∈ S2n−1 , but
HVol2n (K ) > HVol2n (L) .
Proof: Let K , L be Rθ -invariant convex bodies in R2n with n ≥ 4
that provide a counterexample to the complex BPP, i.e.
Vol2n−2 (K ∩ Hξ ) ≤ Vol2n−2 (L ∩ Hξ )
(1)
2n−1
for every ξ ∈ S
, but
Vol2n (K ) > Vol2n (L) .
(2)
Can assume that bodies K and L are infinitely smooth.
BP in HnC
|x|
Connection with kxk−2
(1 − ( kxk )2 )−1
K
Convex Geometry
K
Solution of the Problem
Counterexample for n ≥ 4
Since (2) is strict, we can dilate the body L by α > 1 to make (1)
strict as well. By continuity of ξ 7→ AK ,Hξ (0), there is an > 0 so
that
(1 + )Vol2n−2 (K ∩ Hξ ) ≤ Vol2n−2 (L ∩ Hξ )
for every ξ ∈ S2n−1 , but
Vol2n (K ) > (1 + )Vol2n (L) .
Can dilate K and L by α > 0.
Choose α so small that αK and αL lie in the ball of radius s that
satisfies the inequality
1≤
1
≤ 1 + .
(1 − s 2 )n+1
Then K and L provide a counterexample.
BP in HnC
|x|
Connection with kxk−2
(1 − ( kxk )2 )−1
K
Convex Geometry
K
Solution of the Problem
Positive Part I
Lemma
Let K be an Rθ -invariant convex body contained in the unit ball of
R2n such that
kxk−2
K 2
|x|
1− kxk
is a positive definite distribution on R2n .
K
And let L be an Rθ -invariant star body contained in the unit ball of
R2n so that
HVol2n−2 (K ∩ Hξ ) ≤ HVol2n−2 (L ∩ Hξ )
for every ξ ∈ S2n−1 . Then
HVol2n (K ) ≤ HVol2n (L) .
BP in HnC
|x|
Connection with kxk−2
(1 − ( kxk )2 )−1
K
Convex Geometry
Solution of the Problem
K
Positive Part II
Proof : Since the function
a2
1 − a2
b
r 2n−3
dr =
2 n
a (1 − r )
Z
r2
1−r 2
is increasing on (0, 1),
b
r 2n−1
a2 (1 − r 2 )
dr ≤
2 n+1 1 − a2
r2
a (1 − r )
Z
b
r 2n−1
dr
2 n+1
a (1 − r )
Z
for a, b ∈ (0, 1). True in case a ≤ b as well as in case b ≤ a.
Integrating both sides in the above inequality over the unit sphere
−1
S2n−1 with a = kxk−1
K and b = kxkL we obtain:
Z
S2n−1
kxk−2
K
1 − kxk−2
K
Z
kxkL−1
kxk−1
K
r 2n−3
drdx ≤
(1 − r 2 )n
Z
S2n−1
Z
kxk−1
L
kxk−1
K
r 2n−1
drdx .
(1 − r 2 )n+1
Show that the LHS is positive. Let dµ0 be the measure
corresponding to the Fourier transform of the positive definite
kxk−2
distribution |x|K 2 , then we obtain:
1−
kxkK
BP in HnC
Convex Geometry
|x|
Connection with kxk−2
(1 − ( kxk )2 )−1
K
K
Solution of the Problem
Positive Part III
(2π)2n
Z kxk−1
K
kxk−2
r 2n−3
K
drdx
−2
(1
− r 2 )n
0
S2n−1 1 − kxkK
Z
Z |x|
kxkK
kxk−2
r 2n−3
−2n+2
K
= (2π)2n
drdx
|x|
2
(1 − r 2 )n
|x|
S2n−1 1 −
0
kxkK
!∧
Z
Z |x|
kxkK
r 2n−3
−2n+2
=
|x|
(ξ) dµ0 (ξ)
dr
(1 − r 2 )n
S2n−1
0
Z
2π
= n−1
HVol2n−2 (K ∩ Hξ ) dµ0 (ξ)
8
2n−1
ZS
2π
≤ n−1
HVol2n−2 (L ∩ Hξ ) dµ0 (ξ)
8
S2n−1
!∧
Z |x|
Z
kxkL
r 2n−3
dr
(ξ) dµ0 (ξ)
=
|x|−2n+2
(1 − r 2 )n
0
S2n−1
Z
Z kxk−1
L
kxk−2
r 2n−3
K
= (2π)2n
drdx ,
−2
(1 − r 2 )n
S2n−1 1 − kxkK
0
Z
BP in HnC
Convex Geometry
|x|
Connection with kxk−2
(1 − ( kxk )2 )−1
K
K
Solution of the Problem
Positive Part IV
Thus the RHS is positive as well, which shows
Z
S2n−1
Z
0
kxk−1
K
r 2n−1
drdx ≤
(1 − r 2 )n+1
Z
S2n−1
Z
kxk−1
L
0
That is,
HVol2n (K ) ≤ HVol2n (L) .
r 2n−1
drdx .
(1 − r 2 )n+1
BP in HnC
Convex Geometry
|x|
Connection with kxk−2
(1 − ( kxk )2 )−1
K
Solution of the Problem
K
Negative Part I
Lemma
Suppose there is an infinitely smooth complex convex body K in
Bn with strictly positive curvature so that
kxk−2
K 2
|x|
1− kxk
is not a
K
positive definite distribution on R2n . Then one can perturb the
body K to construct another complex convex body L in Bn so that
for every ξ ∈ S2n−1
HVol2n−2 (L ∩ Hξ ) ≤ HVol2n−2 (K ∩ Hξ ) ,
but
HVol2n (L) > HVol2n (K ) .
BP in HnC
Convex Geometry
|x|
Connection with kxk−2
(1 − ( kxk )2 )−1
K
K
Solution of the Problem
Negative Part II
Proof : There
is an open subset Ω ⊂ S2n−1 on which
!
∧
kxk−2
K 2
|x|
1− kxk
is negative. Note: Ω is Rθ -invariant.
K
Choose a smooth non-negative Rθ -invariant function f on S2n−1
with supp(f) ⊂ Ω and extend f to an Rθ -invariant homogeneous
x
|x|−2 of degree −2 on R2n .
function f |x|
The Fourier transform of this extension is an Rθ -invariant infinitely
smooth function on R2n \ {0}, homogeneous of degree −2n + 2, i.e.
∧
y
x
−2
f
|x|
(y ) = g
|y |−2n+2
|x|
|y |
with g ∈ Cθ∞ (S2n−1 ).
BP in HnC
Convex Geometry
|x|
Connection with kxk−2
(1 − ( kxk )2 )−1
K
K
Solution of the Problem
Negative Part III
Define a body L in B n by
|x|−2n+2
Z
0
|x|
kxkL
r 2n−3
dr = |x|−2n+2
(1 − r 2 )n
Z
0
|x|
kxkK
r 2n−3
dr −g
(1 − r 2 )n
x
|x|
|x|−2n+2
for some > 0.
For small enough , the body L is convex.
K and L are the bodies we seek.
BP in HnC
Convex Geometry
|x|
Connection with kxk−2
(1 − ( kxk )2 )−1
K
K
Solution of the Problem
Parallel section function
For an infinitely smooth Rθ -invariant star body K ⊂ R2n and
m ∈ N ∪ {0}, m < n − 1,
Z
(−1)m
(kxk−2n+2m+2
)∧ (ν)dν
∆m AK ,Hξ (0) =
K
(2π)2 (2n − 2m − 2)
S2n−1 ∩Hξ⊥
=
(−1)m
(kxk−2n+2m+2
)∧ (ξ) .
K
2π(2n − 2m − 2)
In complex dimension two there is only one choice for m, namely
m = 0, and we get
1
∧
AK ,Hξ (0) =
(kxk−2
K ) (ξ) .
4π
In complex dimension three, for m = 1, we obtain
−1
∧
∆AK ,Hξ (0) =
(kxk−2
K ) (ξ) .
4π
BP in HnC
|x|
Connection with kxk−2
(1 − ( kxk )2 )−1
K
Convex Geometry
K
Solution of the Problem
n=2
Lemma
For any Rθ -invariant star body K in R4 contained in the unit ball,
the distribution
−2
kxkK
2
|x|
1− kxk
is positive definite.
K
Proof : Can assume K is smooth.
Define M by kxk−2
M =
kxk−2
K 2
|x|
1− kxk
.
K
M is smooth and Rθ -invariant, thus
AM,Hξ (0) =
1
∧
(kxk−2
M ) (ξ) ,
4π
−2
and consequently kxkM
is positive definite.
BP in HnC
Convex Geometry
|x|
Connection with kxk−2
(1 − ( kxk )2 )−1
K
K
Solution of the Problem
n=3
Lemma
There is a smooth Rθ -invariant convex body K in R6 contained in
kxk−2
the unit ball for which the distribution |x|K 2 is not positive
1−
kxkK
definite.
Proof : For ξ ∈ R6 , ξ = (ξ11 , ξ12 , ξ21 , ξ22 , ξ31 , ξ32 ), denote by
˜ ξ3 ).
ξ˜ = (ξ11 , ξ12 , ξ21 , ξ22 ) and by
ξ3 = (ξ31, ξ32 ), then ξ = (ξ,
q
Consider the map (r , θ) 7→
r2
,θ
1−r 2
.
This map takes the line x = 1a to the hyperbola
(a2 − 1)x 2 − y 2 = 1, provided that a2 − 1 > 0, and it takes the
ellipse x 2 + (1 + b 2 )y 2 = 1 to the line y = b1 .
Denote the equation of the elliptic arc by e and the equation of the
hyperbolic arc by h. Define K by
˜ ≤ 1 and |ξ3 | ≤ e(|ξ|)}
˜ .
K = {ξ ∈ R6 : |ξ|
a
BP in HnC
|x|
Connection with kxk−2
(1 − ( kxk )2 )−1
K
Convex Geometry
Solution of the Problem
K
n=3
For a > 1, K is contained in the unit ball.
Define a star body M by
kxk−2
M =
kxk−2
K
2 .
|x|
1 − kxkK
It can be described as
˜ ≤ h(|ξ3 |) with |ξ3 | ≤
M = {ξ ∈ R6 : |ξ|
1
}.
b
By construction both bodies are Rθ -invariant and we may assume
that they are smooth.
Let x = (x̃, x3 ) ∈ M with x3 6= (0, 0). Choose ξ ∈ S5 in the
direction of x3 . Fix an orthonormal basis {e1 , e2 } for Hξ⊥ . For
u ∈ R2 with |u| ≥ b1 , AM,Hξ (u) = 0, and otherwise
BP in HnC
|x|
Connection with kxk−2
(1 − ( kxk )2 )−1
K
Convex Geometry
K
Solution of the Problem
n=3
AM,Hξ (u) = Vol4 (M ∩ {Hξ + u1 e1 + u2 e2 })
Z
=
χ(kxkM )dx
{x∈R2n : (x,e1 )=u1 ,(x,e2 )=u2 }
h(|u|)
3
Z Z
=
r drdθ
S3
= |S3 |
=
0
h(|u|)4
4
π2
h(|u|)4 ,
2
Setting a = 2, we get AM,Hξ (u) =
∆AM,Hξ (u)
π2
2
1+|u|2
3
2
and consequently
2
= 4π9 (1 + 2|u|2 ). Since M is smooth we have
16π 3
∧
(kxk−2
)
(ξ)
=
−4π
∆A
(0)
=
−
.
M,Hξ
M
9