Connection with � x � − 2 � x � K ) 2 ) − 1 | x | BP in H n Convex Geometry (1 − ( Solution of the Problem C K The Busemann-Petty Problem in the Complex Hyperbolic Space Susanna Dann University of Missouri-Columbia Texas A&M University, July 19, 2012
Connection with � x � − 2 � x � K ) 2 ) − 1 | x | BP in H n Convex Geometry (1 − ( Solution of the Problem C K The Busemann-Petty Problem in R n Let K and L be two origin-symmetric convex bodies in R n . Suppose that for every ξ ∈ S n − 1 Vol n − 1 ( K ∩ ξ ⊥ ) ≤ Vol n − 1 ( L ∩ ξ ⊥ ) . Does it follow that Vol n ( K ) ≤ Vol n ( L )? The answer is affirmative for n ≤ 4 and negative for n ≥ 5. posed in 1956 solved in 90’
Connection with � x � − 2 � x � K ) 2 ) − 1 | x | BP in H n Convex Geometry (1 − ( Solution of the Problem C K 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 R n and for the real hyperbolic space the answer is affirmative for n ≤ 2 and negative for n ≥ 3. onig, Zymonopoulou : C n Koldobsky, K¨ For the complex version of the Busemann-Petty problem the answer is affirmative for the complex dimension n ≤ 3 and negative for n ≥ 4.
Connection with � x � − 2 � x � K ) 2 ) − 1 | x | BP in H n Convex Geometry (1 − ( Solution of the Problem C K The Ball Model of H n C H n C can be identified with the open unit ball in C n B n := { z ∈ C n : ( z , z ) < 1 } . The volume element on H n C is d µ n = 8 n r 2 n − 1 drd σ (1 − r 2 ) n +1 , where d σ is the volume element on the unit sphere S 2 n − 1 . For K ⊂ B n denote by HVol 2 n ( K ) the volume of K with respect to d µ n .
Connection with � x � − 2 � x � K ) 2 ) − 1 | x | BP in H n Convex Geometry (1 − ( Solution of the Problem C K Hyperplanes in C n For ξ ∈ C n , | ξ | = 1, denote by H ξ := { z ∈ C n : ( z , ξ ) = 0 } the complex hyperplane through the origin perpendicular to ξ . Identify C n with R 2 n via the mapping ( ξ 11 + i ξ 12 , . . . , ξ n 1 + i ξ n 2 ) �→ ( ξ 11 , ξ 12 , . . . , ξ n 1 , ξ n 2 ) . Under this mapping the hyperplane H ξ turns into a (2 n − 2)-dimensional subspace of R 2 n orthogonal to the vectors ξ = ( ξ 11 , ξ 12 , . . . , ξ n 1 , ξ n 2 ) and ξ ⊥ := ( − ξ 12 , ξ 11 , . . . , − ξ n 2 , ξ n 1 ) .
Connection with � x � − 2 � x � K ) 2 ) − 1 | x | BP in H n Convex Geometry (1 − ( Solution of the Problem C K Origin Symmetric Convex Sets in C n K ⊂ C n is origin symmetric if x ∈ K implies − x ∈ K . Origin symmetric convex bodies in C n are unit balls of norms on C n . We call them complex convex bodies. Norms on C n satisfy: � λ z � = | λ |� z � for λ ∈ C . A star body K in R 2 n is called R θ -invariant, if for every θ ∈ [0 , 2 π ] and every ξ = ( ξ 11 , ξ 12 , . . . , ξ n 1 , ξ n 2 ) ∈ R 2 n � ξ � K = � R θ ( ξ 11 , ξ 12 ) , . . . , R θ ( ξ n 1 , ξ n 2 ) � K , where R θ stands for the counterclockwise rotation by an angle θ around the origin in R 2 . Thus complex convex bodies in C n are R θ -invariant convex bodies in R 2 n .
Connection with � x � − 2 � x � K ) 2 ) − 1 | x | BP in H n Convex Geometry (1 − ( Solution of the Problem C K Formulation of the problem The Busemann-Petty problem in H n C can be posed as follows. Given two R θ -invariant convex bodies K and L in R 2 n contained in the unit ball such that HVol 2 n − 2 ( K ∩ H ξ ) ≤ HVol 2 n − 2 ( L ∩ H ξ ) for all ξ ∈ S 2 n − 1 , does it follow that HVol 2 n ( K ) ≤ HVol 2 n ( L )?
Connection with � x � − 2 � x � K ) 2 ) − 1 | x | BP in H n Convex Geometry (1 − ( Solution of the Problem C K Basic Definitions A distribution f on R n is even homogeneous of degree p ∈ R , if � x � �� = | α | n + p � f , ϕ � f ( x ) , ϕ α for every test function ϕ and every α ∈ R , α � = 0. The Fourier transform of an even homogeneous distribution of degree p on R n is an even homogeneous distribution of degree − n − p . A distribution f is positive definite, if ˆ f is a positive distribution, � � 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 belongs to the class C k ( S n − 1 ) of k times continuously differentiable functions on the unit sphere. If � · � K ∈ C k ( S n − 1 ) for any k ∈ N , then a star body K is said to be infinitely smooth.
Connection with � x � − 2 � x � K ) 2 ) − 1 | x | BP in H n Convex Geometry (1 − ( Solution of the Problem C K Approximation Results The radial function of K is ρ K ( x ) = � x � − 1 K . One can approximate any convex body K in R n in the radial metric ρ ( K , L ) := max x ∈ S n − 1 | ρ K ( x ) − ρ L ( x ) | by a sequence of infinitely smooth convex bodies with the same symmetries as K .
Connection with � x � − 2 � x � K ) 2 ) − 1 | x | BP in H n Convex Geometry (1 − ( Solution of the Problem C K Fourier Approach to Sections For an infinitely smooth origin symmetric star body K in R n and 0 < p < n , the Fourier transform of the distribution � x � − p is an K infinitely smooth function on R n \ { 0 } , homogeneous of degree − n + p . Lemma Let K and L be infinitely smooth origin symmetric star bodies in R n , and let 0 < p < n. Then � � S n − 1 ( �·� − p K ) ∧ ( θ )( �·� − n + p S n − 1 � θ � − p K � θ � − n + p ) ∧ ( θ ) d θ = (2 π ) n d θ . L L
Connection with � x � − 2 � x � K ) 2 ) − 1 | x | BP in H n Convex Geometry (1 − ( Solution of the Problem C K Fourier Approach to Sections Let 0 < k < n and let H be an ( n − k )-dimensional subspace of R n . Fix an orthonormal basis e 1 , . . . , e k in the orthogonal subspace H ⊥ . For a star body K in R n , define the ( n − k )-dimensional parallel section function A K , H as a function on R k such that for u ∈ R k A K , H ( u ) = Vol n − k ( K ∩ { H + u 1 e 1 + · · · + u k e k } ) . If K is infinitely smooth, the function A K , H is infinitely differentiable at the origin. Lemma Let K be an infinitely smooth origin symmetric star body in R n and 0 < k < n. Then for every ( n − k ) -dimensional subspace H of R n and for every m ∈ N ∪ { 0 } , m < ( n − k ) / 2 , ( − 1) m � ∆ m A K , H (0) = S n − 1 ∩ H ⊥ ( � x � − n +2 m + k ) ∧ ( ξ ) d ξ , K (2 π ) k ( n − 2 m − k ) where ∆ denotes the Laplacian on R k .
Connection with � x � − 2 � x � K ) 2 ) − 1 | x | BP in H n Convex Geometry (1 − ( Solution of the Problem C K Fourier and Radon Transforms of R θ -invariant Functions Lemma Suppose that K is an infinitely smooth R θ -invariant star body in R 2 n . Then for every 0 < p < 2 n and ξ ∈ S 2 n − 1 the Fourier transform of the distribution � x � − p is a constant function on K S 2 n − 1 ∩ H ⊥ ξ . Since � x � K is R θ -invariant, so is the Fourier transform of Proof : � x � − p K . The two-dimensional space H ⊥ ξ is spanned by two vectors ξ and ξ ⊥ . Every vector in S 2 n − 1 ∩ H ⊥ ξ is the image of ξ under one of the coordinate-wise rotations R θ , so the Fourier transform of � x � − p is K a constant function on S 2 n − 1 ∩ H ⊥ ξ . �
Connection with � x � − 2 � x � K ) 2 ) − 1 | x | BP in H n Convex Geometry (1 − ( Solution of the Problem C K Fourier and Radon Transforms of R θ -invariant Functions Denote by C θ ( S 2 n − 1 ) the space of R θ -invariant continuous functions on the unit sphere S 2 n − 1 , i.e. continuous real-valued functions f satisfying f ( ξ ) = f ( R θ ξ ) for any ξ ∈ S 2 n − 1 and any θ ∈ [0 , 2 π ]. The complex spherical Radon transform is an operator R c : C θ ( S 2 n − 1 ) → C θ ( S 2 n − 1 ) defined by � R c f ( ξ ) = f ( x ) dx . S 2 n − 1 ∩ H ξ
Connection with � x � − 2 � x � K ) 2 ) − 1 | x | BP in H n Convex Geometry (1 − ( Solution of the Problem C K Fourier and Radon Transforms of R θ -invariant Functions Lemma Let f ∈ C θ ( S 2 n − 1 ) . Extend f to a homogeneous function of degree − 2 n + 2 , f · r − 2 n +2 , then the Fourier transform of this extension is a homogeneous function of degree − 2 on R 2 n , whose restriction to the unit sphere is continuous. Moreover, for every ξ ∈ S 2 n − 1 R c f ( ξ ) = 1 2 π ( f · r − 2 n +2 ) ∧ ( ξ ) .
Connection with � x � − 2 � x � K ) 2 ) − 1 | x | BP in H n Convex Geometry (1 − ( Solution of the Problem C K Volume of Sections Let K be an R θ -invariant star body contained in the unit ball of R 2 n with n ≥ 2. Let ξ ∈ S 2 n − 1 , we compute: HVol 2 n − 2 ( K ∩ H ξ ) � � x � − 1 r 2 n − 3 � K = 8 n − 1 (1 − r 2 ) n drdx S 2 n − 1 ∩ H ξ 0 | x | r 2 n − 3 � � � x � K = 8 n − 1 | x | − 2 n +2 (1 − r 2 ) n drdx S 2 n − 1 ∩ H ξ 0 � ∧ | x | � = 8 n − 1 r 2 n − 3 � � x � K | x | − 2 n +2 ( ξ ) . (1 − r 2 ) n dr 2 π 0
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