Optimizing volume with prescribed diameter or minimum width B. Gonz - PowerPoint PPT Presentation

Optimizing volume with prescribed diameter or minimum width B. Gonz´ alez Merino* (joint with T. Jahn, A. Polyanskii, M. Schymura, and G. Wachsmuth) Berlin *Author partially funded by Fundaci´ on S´ eneca, proyect 19901/GERM/15, and by MINECO, project MTM2015-63699-P, Spain. Departament of Mathematical Analysis, University of Sevilla, Spain Einstein Workshop Discrete Geometry and Topology, FU Berlin, Germany, π -Day 2018.

Kakeya problem (1917): Which region D minimizes area for which a needle of length 1 can be rotated inside D by 2 π radians?

Kakeya problem (1917): Which region D minimizes area for which a needle of length 1 can be rotated inside D by 2 π radians? Deltoid? No!

Kakeya problem (1917): Which region D minimizes area for which a needle of length 1 can be rotated inside D by 2 π radians? Besicovitch (1919) Deltoid? No!

Kakeya problem (1917): Which region D minimizes area for which a needle of length 1 can be rotated inside D by 2 π radians? Besicovitch (1919) Arbitrary small area Deltoid? No!

Convex Kakeya problem: Which convex region K minimizes area for which the breadth in each direction is at least 1?

Convex Kakeya problem: Which convex region K minimizes area for which the breadth in each direction is at least 1?

Convex Kakeya problem: Which convex region K minimizes area for which the breadth in each direction is at least 1? w ( K ) = min u w ( K , u )

Convex Kakeya problem: Which convex region K minimizes area for which w ( K ) ≥ 1? w ( K ) = min u w ( K , u )

Theorem (P´ al, 1921) Let K be a planar convex set. Then A ( K ) 1 w ( K ) 2 ≥ √ . 3

Theorem (P´ al, 1921) Let K be a planar convex set. Then A ( K ) 1 w ( K ) 2 ≥ √ . 3 Equality holds iff K is an equilateral triangle.

P´ al’s problem Let K ∈ K n . Find K 0 ∈ K n and C n > 0 s.t. vol ( K ) w ( K ) n ≥ vol ( K 0 ) w ( K 0 ) n = C n .

P´ al’s problem Let K ∈ K n . Find K 0 ∈ K n and C n > 0 s.t. vol ( K ) w ( K ) n ≥ vol ( K 0 ) w ( K 0 ) n = C n . Observation: K 0 is w -minimal under inclusion.

P´ al’s problem Let K ∈ K n . Find K 0 ∈ K n and C n > 0 s.t. vol ( K ) w ( K ) n ≥ vol ( K 0 ) w ( K 0 ) n = C n . Definition (Reduced set): is w -minimal under inclusion. K

Lassak’s question (1990): Does it exist reduced polytopes in dimension n ≥ 3?

Lassak’s question (1990): Does it exist reduced polytopes in dimension n ≥ 3? Theorem (Martini, Swanepoel ’04, Averkov, Martini ’08) Let P ∈ K n be a polytope in n ≥ 3 . If . . .

Lassak’s question (1990): Does it exist reduced polytopes in dimension n ≥ 3? Theorem (Martini, Swanepoel ’04, Averkov, Martini ’08) Let P ∈ K n be a polytope in n ≥ 3 . If . . . 1 it is a simplex, then . . . . . . P is not reduced.

Lassak’s question (1990): Does it exist reduced polytopes in dimension n ≥ 3? Theorem (Martini, Swanepoel ’04, Averkov, Martini ’08) Let P ∈ K n be a polytope in n ≥ 3 . If . . . 1 it is a simplex, then . . . 2 it is a pyramid, then . . . . . . P is not reduced.

Lassak’s question (1990): Does it exist reduced polytopes in dimension n ≥ 3? Theorem (Martini, Swanepoel ’04, Averkov, Martini ’08) Let P ∈ K n be a polytope in n ≥ 3 . If . . . 1 it is a simplex, then . . . 2 it is a pyramid, then . . . 3 it has n + 2 facets or n + 2 vertices, then . . . . . . P is not reduced.

Theorem 1 (G.M., Jahn, Polyanskii, Wachsmuth ’17) 3 4 5 6 8 7 12 11 10 9 2 1 Figure: A reduced polytope with 12 vertices and 16 facets

Question (f-vector) If P ∈ K n is a reduced polytope, find best c ( n ) , C ( n ) > 0 s.t. f 0 ( P ) + c ( n ) ≤ f n − 1 ( P ) ≤ f 0 ( P ) + C ( n ) ( c ( n ) , C (3) − 4 ≥ 0).

Isodiametric inequality (Bieberbach 1915) Let K ∈ K n . Then D ( K ) n ≤ vol ( B n vol ( K ) 2 ) 2 ) n = 2 − n κ n . D ( B n Equality holds iff K = B n 2 .

Reverse isodiametric? If K = [ − a , a ] × [ − a − 1 , a − 1 ], for a > 0 arbitrarily large, then A ( K ) 4 D ( K ) 2 = � → 0 . a 2 + 1 � 4 a 2

Reverse isodiametric? If K = [ − a , a ] × [ − a − 1 , a − 1 ], for a > 0 arbitrarily large, then A ( K ) 4 D ( K ) 2 = � → 0 . a 2 + 1 � 4 a 2 Idea: Affine Geometry

Definition (Isodiametric position, Behrend ’37, G.M., Schymura ’18+) K ∈ K n is in isodiametric position if vol ( K ) vol ( A ( K )) D ( K ) n = sup D ( A ( K )) n . A ∈ GL ( n , R )

Definition (Isodiametric position, Behrend ’37, G.M., Schymura ’18+) K ∈ K n is in isodiametric position if vol ( K ) vol ( A ( K )) D ( K ) n = sup D ( A ( K )) n . A ∈ GL ( n , R ) Definition V ⊂ S n − 1 is the set of diameters of K , and fulfills v ∈ V iff ∃ x ∈ K s . t . x + D ( K )[0 , v ] ⊂ K .

Theorem 2 (G.M., Schymura ’18+) Let K ∈ K n be in IDP and V be its set of diameters. Then for every L i -dimensional subspace, i ∈ [ n − 1], �� � i ∃ v ∈ V s.t. ∢ ( L , v ) ≥ arccos . n

Theorem 2 (G.M., Schymura ’18+) Let K ∈ K n be in IDP and V be its set of diameters. Then for every L i -dimensional subspace, i ∈ [ n − 1], �� � i ∃ v ∈ V s.t. ∢ ( L , v ) ≥ arccos . n

Theorem 2 (G.M., Schymura ’18+) Let K ∈ K n be in IDP and V be its set of diameters. Then for every L i -dimensional subspace, i ∈ [ n − 1], �� � i ∃ v ∈ V s.t. ∢ ( L , v ) ≥ arccos . n Sharpness? Identifying WHEN K is in isodiametric position!

L¨ owner position K ⊆ B n owner position if B n 2 is in L¨ 2 is the ellipsoid of minimum volume containing K .

L¨ owner position K ⊆ B n owner position if B n 2 is in L¨ 2 is the ellipsoid of minimum volume containing K . Theorem (John 1948, Ball 1992) Let K ∈ K n be 0-symmetric s.t. K ⊆ B n 2 . The following are equivalent: K is in L¨ owner position. � n +1 There exist u i ∈ K ∩ S n − 1 , λ i ≥ 0, i ∈ [ m ], n ≤ m ≤ � , 2 s.t. m � λ i u i u T = I n . i i =1

Theorem 3 (G.M., Schymura ’18+) Let K ∈ K n . The following are equivalent:

Theorem 3 (G.M., Schymura ’18+) Let K ∈ K n . The following are equivalent: K is in isodiametric position.

Theorem 3 (G.M., Schymura ’18+) Let K ∈ K n . The following are equivalent: K is in isodiametric position. 1 D ( K ) ( K − K ) is in L¨ owner position.

Remark 1 (Dvoretzky-Rogers Factorization 1950) Let u i ∈ S n − 1 , λ i ≥ 0, s.t. I n = � m i =1 λ i u i u T i . Then there exist 1 ≤ i 1 < · · · < i n ≤ m s.t. �� � j ∢ ( L j , u i j +1 ) ≥ arccos where L j = span ( u i 1 , . . . , u i j ) . n

Remark 1 (Dvoretzky-Rogers Factorization 1950) Let u i ∈ S n − 1 , λ i ≥ 0, s.t. I n = � m i =1 λ i u i u T i . Then there exist 1 ≤ i 1 < · · · < i n ≤ m s.t. �� � j ∢ ( L j , u i j +1 ) ≥ arccos where L j = span ( u i 1 , . . . , u i j ) . n Remark 2 1 √ n ( ± 1 , . . . , ± 1), i ∈ [2 n ] and The vectors u i = L j = span ( e 1 , . . . , e j ) shows that Theorem 2 is best possible.

Theorem 4 (G.M., Schymura ’18+) � n +1 Let u i ∈ S n − 1 , λ i ≥ 0, i ∈ [ m ], with n ≤ m ≤ � , be such that 2 m � λ i u i u T I n = i . i =1 Then � � n � � m � 2 1 ≤ i < j ≤ m | u T 2 min i u j | ≤ 1 − . � m � n 2 � n +1 � The inequality is sharp for m = n , n + 1, and sometimes for . 2

Theorem 4 (G.M., Schymura ’18+) � n +1 Let u i ∈ S n − 1 , λ i ≥ 0, i ∈ [ m ], with n ≤ m ≤ � , be such that 2 m � λ i u i u T I n = i . i =1 Then � � n � � m � 2 1 ≤ i < j ≤ m | u T 2 min i u j | ≤ 1 − . � m � n 2 � n +1 � The inequality is sharp for m = n , n + 1, and sometimes for . 2  � � 1 arccos DR-F √ n  ∢ ( u i 1 , u i 2 ) ≥ � � 1 arccos GMS √ n +2 

Lemma 1 (Gen. Cauchy-Binet formula) � [ n ] Let A ∈ R n × m , B ∈ R m × n , and I , J ∈ � . Then i � det(( AB ) I , J ) = det( A I , P ) det( B P , J ) . P ∈ ( [ m ] i )

Lemma 2 � n +1 Let u i ∈ S n − 1 , λ i ≥ 0, i ∈ [ m ], n ≤ m ≤ � , be 2 s.t. I n = � m i =1 λ i u i u T i . Then for every i ∈ [ n ] � n � � λ J det(( U J ) T U J ) , = i J ∈ ( [ m ] i ) where λ J = � j ∈ J λ j and U J = ( u j : j ∈ J ).

Lemma 2 � n +1 Let u i ∈ S n − 1 , λ i ≥ 0, i ∈ [ m ], n ≤ m ≤ � , be 2 s.t. I n = � m i =1 λ i u i u T i . Then for every i ∈ [ n ] � n � � λ J det(( U J ) T U J ) , = i J ∈ ( [ m ] i ) where λ J = � j ∈ J λ j and U J = ( u j : j ∈ J ). For instance, n = � m i =1 λ i .

Lemma 3 Let m ∈ N , m ≥ 2, λ i ≥ 0, i ∈ [ m ], and c ≥ 0 be such that c = � m i =1 λ i . If � f ( λ 1 , . . . , λ m ) = λ i λ j , 1 ≤ i < j ≤ m then λ i ≥ 0 f ( λ 1 , . . . , λ m ) = f ( c / m , . . . , c / m ) = c 2 ( m − 1) max . 2 m

Theorem 4 (G.M., Schymura ’18+) � n +1 Let u i ∈ S n − 1 , λ i ≥ 0, i ∈ [ m ], with n ≤ m ≤ � , be such that 2 m � λ i u i u T I n = i . i =1 Then � � n � � m � 2 1 ≤ i < j ≤ m | u T 2 min i u j | ≤ 1 − . � m � n 2 � n +1 � The inequality is sharp for m = n , n + 1, and sometimes for . 2