On the maximal perimeter of sections of the cube Hermann Knig Kiel, - PowerPoint PPT Presentation

On the maximal perimeter of sections of the cube Hermann König Kiel, Germany Jena, September 2019 Hermann König (Kiel) Sections of the cube Jena, September 2019 1 / 23

Introduction Keith Ball showed that the hyperplane section of the unit n -cube B n ∞ 1 perpendicular to a max := 2 ( 1 , 1 , 0 , . . . , 0 ) has maximal √ ( n − 1 ) -dimensional volume among all hyperplane sections, i.e. for any a ∈ S n − 1 ⊂ R n √ vol n − 1 ( B n ∞ ∩ a ⊥ ) ≤ vol n − 1 ( B n ∞ ∩ a ⊥ max ) = 2 , where a ⊥ is the central hyperplane orthogonal to a . Oleszkiewicz and Pełczyński proved a complex analogue of this result, with the same hyperplane a ⊥ max . Hermann König (Kiel) Sections of the cube Jena, September 2019 2 / 23

Introduction Keith Ball showed that the hyperplane section of the unit n -cube B n ∞ 1 perpendicular to a max := 2 ( 1 , 1 , 0 , . . . , 0 ) has maximal √ ( n − 1 ) -dimensional volume among all hyperplane sections, i.e. for any a ∈ S n − 1 ⊂ R n √ vol n − 1 ( B n ∞ ∩ a ⊥ ) ≤ vol n − 1 ( B n ∞ ∩ a ⊥ max ) = 2 , where a ⊥ is the central hyperplane orthogonal to a . Oleszkiewicz and Pełczyński proved a complex analogue of this result, with the same hyperplane a ⊥ max . Pełczyński asked whether the same hyperplane section is also maximal for intersections with the boundary of the n -cube, i.e. whether for all a ∈ S n − 1 ⊂ R n √ vol n − 2 ( ∂ B n ∞ ∩ a ⊥ ) ≤ vol n − 2 ( ∂ B n ∞ ∩ a ⊥ max ) = 2 (( n − 2 ) 2 + 1 ) . He proved it for n = 3 when vol 1 ( ∂ B 3 ∞ ∩ a ⊥ ) is the perimeter of the quadrangle or hexagon of intersection. Hermann König (Kiel) Sections of the cube Jena, September 2019 2 / 23

Introduction Abbildung: Cubic sections Hermann König (Kiel) Sections of the cube Jena, September 2019 3 / 23

Introduction Notation. Let K ∈ { R , C } , α = 1 1 2 for K = R and α = √ π for K = C . Let || · || ∞ and | · | be the maximum and the Euclidean norm on K n and ∞ := { x ∈ K n | || x || ∞ ≤ α } B n be the n -cube of volume 1 in K n . For K = C , identify C k = R 2 k for volume calculations. For a ∈ K n with | a | = 1 and t ∈ K , the parallel section function A is defined by ∞ ∩ ( a ⊥ + α ta )) , A n − 1 ( a , t ) := vol l ( n − 1 ) ( B n with l = 1 if K = R and l = 2 if K = C . Put A n − 1 ( a ) = A n − 1 ( a , 0 ) . Hermann König (Kiel) Sections of the cube Jena, September 2019 4 / 23

Introduction Notation. Let K ∈ { R , C } , α = 1 1 2 for K = R and α = √ π for K = C . Let || · || ∞ and | · | be the maximum and the Euclidean norm on K n and ∞ := { x ∈ K n | || x || ∞ ≤ α } B n be the n -cube of volume 1 in K n . For K = C , identify C k = R 2 k for volume calculations. For a ∈ K n with | a | = 1 and t ∈ K , the parallel section function A is defined by ∞ ∩ ( a ⊥ + α ta )) , A n − 1 ( a , t ) := vol l ( n − 1 ) ( B n with l = 1 if K = R and l = 2 if K = C . Put A n − 1 ( a ) = A n − 1 ( a , 0 ) . By Ball and Oleszkiewicz-Pełczyński, we have for all a ∈ K n with | a | = 1 √ 2 ) l . A n − 1 ( a ) ≤ A n − 1 ( a max ) = ( Hermann König (Kiel) Sections of the cube Jena, September 2019 4 / 23

Introduction Notation. Let K ∈ { R , C } , α = 1 1 2 for K = R and α = √ π for K = C . Let || · || ∞ and | · | be the maximum and the Euclidean norm on K n and ∞ := { x ∈ K n | || x || ∞ ≤ α } B n be the n -cube of volume 1 in K n . For K = C , identify C k = R 2 k for volume calculations. For a ∈ K n with | a | = 1 and t ∈ K , the parallel section function A is defined by ∞ ∩ ( a ⊥ + α ta )) , A n − 1 ( a , t ) := vol l ( n − 1 ) ( B n with l = 1 if K = R and l = 2 if K = C . Put A n − 1 ( a ) = A n − 1 ( a , 0 ) . By Ball and Oleszkiewicz-Pełczyński, we have for all a ∈ K n with | a | = 1 √ 2 ) l . A n − 1 ( a ) ≤ A n − 1 ( a max ) = ( For a ∈ K n with | a | = 1, define the perimeter of the cubic section by a ⊥ as P n − 2 ( a ) := vol l ( n − 2 ) ( ∂ B n ∞ ∩ a ⊥ ) , l as before. Hermann König (Kiel) Sections of the cube Jena, September 2019 4 / 23

The main result The answer to Pełczyński’s problem for P n − 2 ( a ) := vol l ( n − 2 ) ( ∂ B n ∞ ∩ a ⊥ ) is affirmative. This is a joint result with A. Koldobsky: Theorem 1 2 ( 1 , 1 , 0 , · · · , 0 ) ∈ K n . Then for any a ∈ K n with 1 Let n ≥ 3 and a max := √ | a | = 1 we have P n − 2 ( a ) ≤ P n − 2 ( a max ) , (1) We have √ P n − 2 ( a max ) = 2 (( n − 2 ) 2 + 1 ) , K = R and P n − 2 ( a max ) = 2 π (( n − 2 ) 2 + 1 ) , K = C . Hermann König (Kiel) Sections of the cube Jena, September 2019 5 / 23

An application of the Busemann-Petty type As a consequence of Theorem 1 we find a counterexample to a surface area version of the Busemann-Petty for large dimensions (König, Koldobsky): Theorem 2 For each n ≥ 14 , there exist origin-symmetric convex bodies K , L in R n such that for all a ∈ S n − 1 vol n − 2 ( ∂ K ∩ a ⊥ ) ≤ vol n − 2 ( ∂ L ∩ a ⊥ ) but vol n − 1 ( ∂ K ) > vol n − 1 ( ∂ L ) . Example. Let K = B n ∞ be the unit cube in R n . Let L be the Euclidean ball of radius r in R n so that the perimeters of hyperplane sections of L are all equal to the maximal perimeter of sections of K . Hermann König (Kiel) Sections of the cube Jena, September 2019 6 / 23

An application of the Busemann-Petty type ∞ be the unit cube in R n . Let L be the Euclidean ball of radius r in R n so that Let K = B n √ vol n − 2 ( ∂ K ∩ a ⊥ ) ≤ vol n − 2 ( ∂ K ∩ a ⊥ max ) = 2 (( n − 2 ) 2 + 1 ) = vol n − 2 ( rS n − 2 ) = r n − 2 2 π ( n − 1 ) / 2 , Γ( n − 1 2 ) √ 1 2 + 1 )Γ( n − 1 r = [(( n − 2 ) 2 )] n − 2 . π ( n − 1 ) / ( 2 ( n − 2 )) Hermann König (Kiel) Sections of the cube Jena, September 2019 7 / 23

An application of the Busemann-Petty type ∞ be the unit cube in R n . Let L be the Euclidean ball of radius r in R n so that Let K = B n √ vol n − 2 ( ∂ K ∩ a ⊥ ) ≤ vol n − 2 ( ∂ K ∩ a ⊥ max ) = 2 (( n − 2 ) 2 + 1 ) = vol n − 2 ( rS n − 2 ) = r n − 2 2 π ( n − 1 ) / 2 , Γ( n − 1 2 ) √ 1 2 + 1 )Γ( n − 1 r = [(( n − 2 ) 2 )] n − 2 . π ( n − 1 ) / ( 2 ( n − 2 )) The opposite inequality for the surface areas of K and L happens when ∞ ) = 2 n > vol n − 1 ( rS n − 1 ) = r n − 1 2 π n / 2 vol n − 1 ( ∂ B n 2 ) , Γ( n √ n − 1 1 > π n / 2 2 + 1 )Γ( n − 1 [(( n − 2 ) 2 )] 1 n − 2 2 ) r n − 1 = =: BP ( n ) . n Γ( n n Γ( n π 1 / ( 2 ( n − 2 )) 2 ) Then BP is decreasing in n , with BP ( x 0 ) = 1 for x 0 ≃ 13 . 70, so BP ( n ) < 1 for all n ≥ 14. In the complex case, a similar counterexamples exists for all n ≥ 11. Hermann König (Kiel) Sections of the cube Jena, September 2019 7 / 23

Perimeter formulas and idea of the proof of Theorem 1 For a ∈ K n with | a | = 1 let a ⋆ denote the non-increasing rearrangement of the sequence ( | a k | ) n k = 1 . Then A n − 1 ( a , t ) = A n − 1 ( a ⋆ , | t | ) P n − 2 ( a ) = P n − 2 ( a ⋆ ) . , Thus assume that a = ( a k ) n k = 1 satisfies a 1 ≥ · · · ≥ a n ≥ 0, | a | = 1 and t ≥ 0. Hermann König (Kiel) Sections of the cube Jena, September 2019 8 / 23

Perimeter formulas and idea of the proof of Theorem 1 For a ∈ K n with | a | = 1 let a ⋆ denote the non-increasing rearrangement of the sequence ( | a k | ) n k = 1 . Then A n − 1 ( a , t ) = A n − 1 ( a ⋆ , | t | ) P n − 2 ( a ) = P n − 2 ( a ⋆ ) . , Thus assume that a = ( a k ) n k = 1 satisfies a 1 ≥ · · · ≥ a n ≥ 0, | a | = 1 and t ≥ 0. Then Proposition 1 ∞ n A n − 1 ( a , t ) = 2 � sin ( a k s ) � cos ( ts ) ds , K = R , (2) π a k s k = 1 0 ∞ n A n − 1 ( a , t ) = 1 � � j 1 ( a k s ) J 0 ( ts ) s ds , K = C , (3) 2 k = 1 0 where j 1 ( t ) = 2 J 1 ( t ) sin ( a k s ) and J ν denote the Bessel functions of index ν . If a k = 0 , t a k s and j 1 ( a k s ) have to be read as 1 in formulas (2) and (3) . Hermann König (Kiel) Sections of the cube Jena, September 2019 8 / 23

Perimeter formulas and idea of the proof of Theorem 1 For a ∈ K n with | a | = 1 let a ⋆ denote the non-increasing rearrangement of the sequence ( | a k | ) n k = 1 . Then A n − 1 ( a , t ) = A n − 1 ( a ⋆ , | t | ) P n − 2 ( a ) = P n − 2 ( a ⋆ ) . , Thus assume that a = ( a k ) n k = 1 satisfies a 1 ≥ · · · ≥ a n ≥ 0, | a | = 1 and t ≥ 0. Then Proposition 1 ∞ n A n − 1 ( a , t ) = 2 � sin ( a k s ) � cos ( ts ) ds , K = R , (2) π a k s k = 1 0 ∞ n A n − 1 ( a , t ) = 1 � � j 1 ( a k s ) J 0 ( ts ) s ds , K = C , (3) 2 k = 1 0 where j 1 ( t ) = 2 J 1 ( t ) sin ( a k s ) and J ν denote the Bessel functions of index ν . If a k = 0 , t a k s and j 1 ( a k s ) have to be read as 1 in formulas (2) and (3) . Formula (2) is due to Pólya 1913 and was used by Ball in his proof. Both formulas can be shown by taking the Fourier transform of A n − 1 ( a , · ) , using Fubini’s theorem and taking the inverse Fourier transform. The sin t and j 1 ( t ) functions occur as Fourier t transforms of the interval in R and the disc in C = R 2 , respectively. Hermann König (Kiel) Sections of the cube Jena, September 2019 8 / 23