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

Optimizing volume with prescribed diameter

• r minimum width
• B. Gonz´

alez Merino* (joint with T. Jahn, A. Polyanskii, M. Schymura, and G. Wachsmuth)

Berlin *Author partially funded by Fundaci´

• n 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 2π radians?

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

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

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

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) = minu w(K, u)

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

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

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

P´ al’s problem Let K ∈ Kn. Find K0 ∈ Kn and Cn > 0 s.t. vol(K) w(K)n ≥ vol(K0) w(K0)n = Cn.

P´ al’s problem Let K ∈ Kn. Find K0 ∈ Kn and Cn > 0 s.t. vol(K) w(K)n ≥ vol(K0) w(K0)n = Cn. Observation: K0 is w-minimal under inclusion.

P´ al’s problem Let K ∈ Kn. Find K0 ∈ Kn and Cn > 0 s.t. vol(K) w(K)n ≥ vol(K0) w(K0)n = Cn. Definition (Reduced set): K is w-minimal under inclusion.

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 ∈ Kn 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 ∈ Kn 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 ∈ Kn 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 ∈ Kn 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)

1 2 3 4 5 6 7 8 9 10 11 12

Figure: A reduced polytope with 12 vertices and 16 facets

Question (f-vector) If P ∈ Kn is a reduced polytope, find best c(n), C(n) > 0 s.t. f0(P) + c(n) ≤ fn−1(P) ≤ f0(P) + C(n) (c(n), C(3) − 4 ≥ 0).

Isodiametric inequality (Bieberbach 1915) Let K ∈ Kn. Then vol(K) D(K)n ≤ vol(Bn

2)

D(Bn

2)n = 2−nκn.

Equality holds iff K = Bn

2.

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

• a2 + 1

a2

→ 0.

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

• a2 + 1

a2

→ 0. Idea: Affine Geometry

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

A∈GL(n,R)

vol(A(K)) D(A(K))n .

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

A∈GL(n,R)

vol(A(K)) D(A(K))n . Definition V ⊂ Sn−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 ∈ Kn be in IDP and V be its set of diameters. Then for every L i-dimensional subspace, i ∈ [n − 1], ∃ v ∈ V s.t. ∢(L, v) ≥ arccos

• i

n

• .

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

• i

n

• .

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

• i

n

• .

Sharpness? Identifying WHEN K is in isodiametric position!

L¨

• wner position

K ⊆ Bn

2 is in L¨

• wner position if Bn

2 is the ellipsoid of minimum

volume containing K.

L¨

• wner position

K ⊆ Bn

2 is in L¨

• wner position if Bn

2 is the ellipsoid of minimum

volume containing K. Theorem (John 1948, Ball 1992) Let K ∈ Kn be 0-symmetric s.t. K ⊆ Bn

• 2. The following are

equivalent: K is in L¨

• wner position.

There exist ui ∈ K ∩ Sn−1, λi ≥ 0, i ∈ [m], n ≤ m ≤ n+1

2

• ,

s.t.

m

• i=1

λiuiuT

i

= In.

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

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

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

1 D(K)(K − K) is in L¨

• wner position.

Remark 1 (Dvoretzky-Rogers Factorization 1950) Let ui ∈ Sn−1, λi ≥ 0, s.t. In = m

i=1 λiuiuT i . Then there exist

1 ≤ i1 < · · · < in ≤ m s.t. ∢(Lj, uij+1) ≥ arccos

• j

n

• where

Lj = span(ui1, . . . , uij).

Remark 1 (Dvoretzky-Rogers Factorization 1950) Let ui ∈ Sn−1, λi ≥ 0, s.t. In = m

i=1 λiuiuT i . Then there exist

1 ≤ i1 < · · · < in ≤ m s.t. ∢(Lj, uij+1) ≥ arccos

• j

n

• where

Lj = span(ui1, . . . , uij). Remark 2 The vectors ui =

1 √n(±1, . . . , ±1), i ∈ [2n] and

Lj = span(e1, . . . , ej) shows that Theorem 2 is best possible.

Theorem 4 (G.M., Schymura ’18+) Let ui ∈ Sn−1, λi ≥ 0, i ∈ [m], with n ≤ m ≤ n+1

2

• , be such that

In =

m

• i=1

λiuiuT

i .

Then min

1≤i<j≤m |uT i uj| ≤

• 1 −

n

2

• m

2

• m

n 2 . The inequality is sharp for m = n, n + 1, and sometimes for n+1

2

• .

Theorem 4 (G.M., Schymura ’18+) Let ui ∈ Sn−1, λi ≥ 0, i ∈ [m], with n ≤ m ≤ n+1

2

• , be such that

In =

m

• i=1

λiuiuT

i .

Then min

1≤i<j≤m |uT i uj| ≤

• 1 −

n

2

• m

2

• m

n 2 . The inequality is sharp for m = n, n + 1, and sometimes for n+1

2

• .

∢(ui1, ui2) ≥    arccos

• 1

√n

• DR-F

arccos

• 1

√n+2

• GMS

Lemma 1 (Gen. Cauchy-Binet formula) Let A ∈ Rn×m, B ∈ Rm×n, and I, J ∈ [n]

i

• . Then

det((AB)I,J) =

• P∈([m]

i )

det(AI,P) det(BP,J).

Lemma 2 Let ui ∈ Sn−1, λi ≥ 0, i ∈ [m], n ≤ m ≤ n+1

2

• , be

s.t. In = m

i=1 λiuiuT i . Then for every i ∈ [n]

n i

• =
• J∈([m]

i )

λJ det((UJ)TUJ), where λJ =

j∈J λj and UJ = (uj : j ∈ J).

Lemma 2 Let ui ∈ Sn−1, λi ≥ 0, i ∈ [m], n ≤ m ≤ n+1

2

• , be

s.t. In = m

i=1 λiuiuT i . Then for every i ∈ [n]

n i

• =
• J∈([m]

i )

λJ det((UJ)TUJ), where λJ =

j∈J λj and UJ = (uj : j ∈ J).

For instance, n = m

i=1 λi.

Lemma 3 Let m ∈ N, m ≥ 2, λi ≥ 0, i ∈ [m], and c ≥ 0 be such that c = m

i=1 λi. If

f (λ1, . . . , λm) =

• 1≤i<j≤m

λiλj, then max

λi≥0 f (λ1, . . . , λm) = f (c/m, . . . , c/m) = c2(m − 1)

2m .

Theorem 4 (G.M., Schymura ’18+) Let ui ∈ Sn−1, λi ≥ 0, i ∈ [m], with n ≤ m ≤ n+1

2

• , be such that

In =

m

• i=1

λiuiuT

i .

Then min

1≤i<j≤m |uT i uj| ≤

• 1 −

n

2

• m

2

• m

n 2 . The inequality is sharp for m = n, n + 1, and sometimes for n+1

2

• .

Proof. n 2

• =
• 1≤i<j≤m

λiλj det

• 1

uT

i uj

uT

i uj

1

• [Lem. 2]

Proof. n 2

• =
• 1≤i<j≤m

λiλj det

• 1

uT

i uj

uT

i uj

1

• [Lem. 2]

=

• 1≤i<j≤m

λiλj(1 − (uT

i uj)2)

Proof. n 2

• =
• 1≤i<j≤m

λiλj det

• 1

uT

i uj

uT

i uj

1

• [Lem. 2]

=

• 1≤i<j≤m

λiλj(1 − (uT

i uj)2)

≤ max

λk=n, λk≥0

 

• 1≤i<j≤m

λiλj   · max

1≤i<j≤m(1 − (uT i uj)2)

Proof. n 2

• =
• 1≤i<j≤m

λiλj det

• 1

uT

i uj

uT

i uj

1

• [Lem. 2]

=

• 1≤i<j≤m

λiλj(1 − (uT

i uj)2)

≤ max

λk=n, λk≥0

 

• 1≤i<j≤m

λiλj   · max

1≤i<j≤m(1 − (uT i uj)2)

= n m 2 m 2 1 − min

1≤i<j≤m(uT i uj)2

• [Lem. 3]

Proof of equality. λ1 = · · · = λm = n m and |uT

i uj| =

m − n n(m − 1),

Proof of equality. λ1 = · · · = λm = n m and |uT

i uj| =

m − n n(m − 1), i.e., {u1, . . . , um} is set of equiangular lines with ∢(ui, uj) = arccos m − n n(m − 1)

• .

.

m = n : |uT

i uj| = 0

m = n : |uT

i uj| = 0

m = n + 1 : |uT

i uj| = 1 n

m = n : |uT

i uj| = 0

m = n + 1 : |uT

i uj| = 1 n

m = n+1

2

• : |uT

i uj| = 1 √n+2

m = n : |uT

i uj| = 0

m = n + 1 : |uT

i uj| = 1 n

m = n+1

2

• : |uT

i uj| = 1 √n+2

Lemmens-Seidel ’73: n = 4 Not sharp :(

m = n : |uT

i uj| = 0

m = n + 1 : |uT

i uj| = 1 n

m = n+1

2

• : |uT

i uj| = 1 √n+2

Lemmens-Seidel ’73: n = 4 Not sharp :( but for n = 7 and n = 23 is sharp too.

Corollary (Behrend 1937) Let K ∈ K2 be in isodiametric position. Then vol(K) D(K)2 ≥ √ 3 4 . Equality holds iff K is an equilateral triangle.

Proof. D(K) = 1, V = {ui}i∈[m], C := conv([xi, xi + ui] : i ∈ [m]) ⊆ K.

Proof. D(K) = 1, V = {ui}i∈[m], C := conv([xi, xi + ui] : i ∈ [m]) ⊆ K. Then vol(K) ≥ vol(C)

Proof. D(K) = 1, V = {ui}i∈[m], C := conv([xi, xi + ui] : i ∈ [m]) ⊆ K. Then vol(K) ≥ vol(C) ≥ vol(conv([x1, x1 + u1] ∪ [x2, x2 + u2])) [Thm. 4]

Proof. D(K) = 1, V = {ui}i∈[m], C := conv([xi, xi + ui] : i ∈ [m]) ⊆ K. Then vol(K) ≥ vol(C) ≥ vol(conv([x1, x1 + u1] ∪ [x2, x2 + u2])) [Thm. 4] ≥ vol(1 2conv({±u1, ±u2})) [Betke-Henk’93]

Proof. D(K) = 1, V = {ui}i∈[m], C := conv([xi, xi + ui] : i ∈ [m]) ⊆ K. Then vol(K) ≥ vol(C) ≥ vol(conv([x1, x1 + u1] ∪ [x2, x2 + u2])) [Thm. 4] ≥ vol(1 2conv({±u1, ±u2})) [Betke-Henk’93] = 1 2

• 1 − (uT

1 u2)2

Proof. D(K) = 1, V = {ui}i∈[m], C := conv([xi, xi + ui] : i ∈ [m]) ⊆ K. Then vol(K) ≥ vol(C) ≥ vol(conv([x1, x1 + u1] ∪ [x2, x2 + u2])) [Thm. 4] ≥ vol(1 2conv({±u1, ±u2})) [Betke-Henk’93] = 1 2

• 1 − (uT

1 u2)2

≥ 1 2

• 1 − 1/4 =

√ 3 4 [Thm. 4]

Bibliography

• F. Behrend, ¨

Uber einige Affinvarianten konvexer Bereiche,

• Math. Ann. 113 (1937), no. 1, 713–747.
• B. Gonz´

alez Merino, T. Jahn, A. Polyanskii, G. Wachsmuth, Hunting for reduced polytopes , Discrete Comput. Geom., 2018. P.W.H. Lemmens, J.J. Seidel, Equiangular lines, J. Algebra 24 (1973), 494–512.

• J. P´

al, Ein Minimal problem f¨ ur Ovale, Math. Ann. 83 (1921), 311-319.

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