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Extremal problems for convex lattice polytopes Imre Brny Rnyi Institute, Hungarian Academy of Sciences & Department of Mathematics, University College London A sample problem Jarnk proved in 1926 that if R 2 is a (closed)

  1. Notations: P = P d the set of primitive vectors in Z d K = K d the set of convex bodies in R d (convex compact sets with non-empty interior) P = P d set of convex lattice polytopes, for P ∈ P , f 0 ( P ) = number of vertices of P , f s ( P ) = number of s -dim faces of P

  2. Notations: P = P d the set of primitive vectors in Z d K = K d the set of convex bodies in R d (convex compact sets with non-empty interior) P = P d set of convex lattice polytopes, for P ∈ P , f 0 ( P ) = number of vertices of P , f s ( P ) = number of s -dim faces of P

  3. Notations: P = P d the set of primitive vectors in Z d K = K d the set of convex bodies in R d (convex compact sets with non-empty interior) P = P d set of convex lattice polytopes, for P ∈ P , f 0 ( P ) = number of vertices of P , f s ( P ) = number of s -dim faces of P

  4. Notations: P = P d the set of primitive vectors in Z d K = K d the set of convex bodies in R d (convex compact sets with non-empty interior) P = P d set of convex lattice polytopes, for P ∈ P , f 0 ( P ) = number of vertices of P , f s ( P ) = number of s -dim faces of P

  5. Notations: P = P d the set of primitive vectors in Z d K = K d the set of convex bodies in R d (convex compact sets with non-empty interior) P = P d set of convex lattice polytopes, for P ∈ P , f 0 ( P ) = number of vertices of P , f s ( P ) = number of s -dim faces of P

  6. THE PROBLEMS 1. Minimal volume. Determine or estimate V d ( n ) = min { Vol P : P ∈ P d and f 0 ( P ) = n } 2. Minimal surface area. Determine or estimate S d ( n ) = min { S ( P ) : P ∈ P d and f 0 ( P ) = n } just solved it for d = 2. 3. Minimal lattice width. Determine or estimate w d ( n ) = min { w ( P ) : P ∈ P d and f 0 ( P ) = n } where w ( P ) is the lattice width of P ∈ P d

  7. THE PROBLEMS 1. Minimal volume. Determine or estimate V d ( n ) = min { Vol P : P ∈ P d and f 0 ( P ) = n } 2. Minimal surface area. Determine or estimate S d ( n ) = min { S ( P ) : P ∈ P d and f 0 ( P ) = n } just solved it for d = 2. 3. Minimal lattice width. Determine or estimate w d ( n ) = min { w ( P ) : P ∈ P d and f 0 ( P ) = n } where w ( P ) is the lattice width of P ∈ P d

  8. THE PROBLEMS 1. Minimal volume. Determine or estimate V d ( n ) = min { Vol P : P ∈ P d and f 0 ( P ) = n } 2. Minimal surface area. Determine or estimate S d ( n ) = min { S ( P ) : P ∈ P d and f 0 ( P ) = n } just solved it for d = 2. 3. Minimal lattice width. Determine or estimate w d ( n ) = min { w ( P ) : P ∈ P d and f 0 ( P ) = n } where w ( P ) is the lattice width of P ∈ P d

  9. THE PROBLEMS 1. Minimal volume. Determine or estimate V d ( n ) = min { Vol P : P ∈ P d and f 0 ( P ) = n } 2. Minimal surface area. Determine or estimate S d ( n ) = min { S ( P ) : P ∈ P d and f 0 ( P ) = n } just solved it for d = 2. 3. Minimal lattice width. Determine or estimate w d ( n ) = min { w ( P ) : P ∈ P d and f 0 ( P ) = n } where w ( P ) is the lattice width of P ∈ P d

  10. THE PROBLEMS 1. Minimal volume. Determine or estimate V d ( n ) = min { Vol P : P ∈ P d and f 0 ( P ) = n } 2. Minimal surface area. Determine or estimate S d ( n ) = min { S ( P ) : P ∈ P d and f 0 ( P ) = n } just solved it for d = 2. 3. Minimal lattice width. Determine or estimate w d ( n ) = min { w ( P ) : P ∈ P d and f 0 ( P ) = n } where w ( P ) is the lattice width of P ∈ P d

  11. Definition K ∈ K d , z ∈ Z d and z � = 0, then w ( K , z ) = max { z · ( x − y ) : x , y ∈ K } . The lattice width of K is w ( K ) = min { w ( K , z ) : z ∈ Z d , z � = 0 } . How many parallel lattice hyperplanes meet K ? FACT. For P ∈ P d , w ( P ) + 1 = minimal number of parallel lattice lines meeting P .

  12. Definition K ∈ K d , z ∈ Z d and z � = 0, then w ( K , z ) = max { z · ( x − y ) : x , y ∈ K } . The lattice width of K is w ( K ) = min { w ( K , z ) : z ∈ Z d , z � = 0 } . How many parallel lattice hyperplanes meet K ? FACT. For P ∈ P d , w ( P ) + 1 = minimal number of parallel lattice lines meeting P .

  13. Definition K ∈ K d , z ∈ Z d and z � = 0, then w ( K , z ) = max { z · ( x − y ) : x , y ∈ K } . The lattice width of K is w ( K ) = min { w ( K , z ) : z ∈ Z d , z � = 0 } . How many parallel lattice hyperplanes meet K ? FACT. For P ∈ P d , w ( P ) + 1 = minimal number of parallel lattice lines meeting P .

  14. x z K y w ( K ) is invariant under lattice preserving affine transformations

  15. x z K y w ( K ) is invariant under lattice preserving affine transformations

  16. 4. Arnold’s question. How many convex lattice polytopes are there? P , Q ∈ P d are equivalent if P can be carried to Q by a lattice preserving affine transformation. Equivalent polytopes have the same volume. Arnold’s question. (1980) How many equivalence classes are there in R d , of volume ≤ V ? not an extremal question yet ..

  17. 4. Arnold’s question. How many convex lattice polytopes are there? P , Q ∈ P d are equivalent if P can be carried to Q by a lattice preserving affine transformation. Equivalent polytopes have the same volume. Arnold’s question. (1980) How many equivalence classes are there in R d , of volume ≤ V ? not an extremal question yet ..

  18. 4. Arnold’s question. How many convex lattice polytopes are there? P , Q ∈ P d are equivalent if P can be carried to Q by a lattice preserving affine transformation. Equivalent polytopes have the same volume. Arnold’s question. (1980) How many equivalence classes are there in R d , of volume ≤ V ? not an extremal question yet ..

  19. 4. Arnold’s question. How many convex lattice polytopes are there? P , Q ∈ P d are equivalent if P can be carried to Q by a lattice preserving affine transformation. Equivalent polytopes have the same volume. Arnold’s question. (1980) How many equivalence classes are there in R d , of volume ≤ V ? not an extremal question yet ..

  20. 4. Arnold’s question. How many convex lattice polytopes are there? P , Q ∈ P d are equivalent if P can be carried to Q by a lattice preserving affine transformation. Equivalent polytopes have the same volume. Arnold’s question. (1980) How many equivalence classes are there in R d , of volume ≤ V ? not an extremal question yet ..

  21. 5. Maximal polytopes. Assume K ∈ K d is “large”. Determine max { f 0 ( P ) : P ∈ P d , P ⊂ K } . equivalently, determine or estimate the maximal number of points in K ∩ Z d that are in convex position, i.e., none of them is in the convex hull of the others answers: order of magnitude, asymptotic, precise..

  22. 5. Maximal polytopes. Assume K ∈ K d is “large”. Determine max { f 0 ( P ) : P ∈ P d , P ⊂ K } . equivalently, determine or estimate the maximal number of points in K ∩ Z d that are in convex position, i.e., none of them is in the convex hull of the others answers: order of magnitude, asymptotic, precise..

  23. 5. Maximal polytopes. Assume K ∈ K d is “large”. Determine max { f 0 ( P ) : P ∈ P d , P ⊂ K } . equivalently, determine or estimate the maximal number of points in K ∩ Z d that are in convex position, i.e., none of them is in the convex hull of the others answers: order of magnitude, asymptotic, precise..

  24. 1. Minimal volume V d ( n ) Theorem (Andrews ’63) If P ∈ P d and Vol P > 0 , then d + 1 d − 1 ≤ c d Vol P . f 0 ( P ) or with better notation: d + 1 d − 1 ≪ Vol P . f 0 ( P ) Corollary d + 1 d − 1 ≪ V d ( n ) . n Several proofs, none easy: Andrews ’63, Arnold ’80 ( d = 2), Konyagin, Sevastyanov ’84 , ( d ≥ 2), W. Schmidt ’86, B.-Vershik ’92, B.-Larman ’98, Reisner-Schütt-Werner ’01, and more

  25. 1. Minimal volume V d ( n ) Theorem (Andrews ’63) If P ∈ P d and Vol P > 0 , then d + 1 d − 1 ≤ c d Vol P . f 0 ( P ) or with better notation: d + 1 d − 1 ≪ Vol P . f 0 ( P ) Corollary d + 1 d − 1 ≪ V d ( n ) . n Several proofs, none easy: Andrews ’63, Arnold ’80 ( d = 2), Konyagin, Sevastyanov ’84 , ( d ≥ 2), W. Schmidt ’86, B.-Vershik ’92, B.-Larman ’98, Reisner-Schütt-Werner ’01, and more

  26. 1. Minimal volume V d ( n ) Theorem (Andrews ’63) If P ∈ P d and Vol P > 0 , then d + 1 d − 1 ≤ c d Vol P . f 0 ( P ) or with better notation: d + 1 d − 1 ≪ Vol P . f 0 ( P ) Corollary d + 1 d − 1 ≪ V d ( n ) . n Several proofs, none easy: Andrews ’63, Arnold ’80 ( d = 2), Konyagin, Sevastyanov ’84 , ( d ≥ 2), W. Schmidt ’86, B.-Vershik ’92, B.-Larman ’98, Reisner-Schütt-Werner ’01, and more

  27. Definition A tower of P ∈ P d is F 0 ⊂ F 1 ⊂ .. ⊂ F d − 1 where F i is an i -dim face of P . T ( P ) = number of towers of P . Theorem If P ∈ P d and Vol P > 0 , then d + 1 d − 1 ≪ Vol P . T ( P ) implies the same bound for f i ( P ) . OPEN PROBLEM. For all polytopes P ∈ K d T ( P ) ≪ f 0 ( P ) + f 1 ( P ) + . . . f d − 1 ( P )????

  28. Definition A tower of P ∈ P d is F 0 ⊂ F 1 ⊂ .. ⊂ F d − 1 where F i is an i -dim face of P . T ( P ) = number of towers of P . Theorem If P ∈ P d and Vol P > 0 , then d + 1 d − 1 ≪ Vol P . T ( P ) implies the same bound for f i ( P ) . OPEN PROBLEM. For all polytopes P ∈ K d T ( P ) ≪ f 0 ( P ) + f 1 ( P ) + . . . f d − 1 ( P )????

  29. Definition A tower of P ∈ P d is F 0 ⊂ F 1 ⊂ .. ⊂ F d − 1 where F i is an i -dim face of P . T ( P ) = number of towers of P . Theorem If P ∈ P d and Vol P > 0 , then d + 1 d − 1 ≪ Vol P . T ( P ) implies the same bound for f i ( P ) . OPEN PROBLEM. For all polytopes P ∈ K d T ( P ) ≪ f 0 ( P ) + f 1 ( P ) + . . . f d − 1 ( P )????

  30. Definition A tower of P ∈ P d is F 0 ⊂ F 1 ⊂ .. ⊂ F d − 1 where F i is an i -dim face of P . T ( P ) = number of towers of P . Theorem If P ∈ P d and Vol P > 0 , then d + 1 d − 1 ≪ Vol P . T ( P ) implies the same bound for f i ( P ) . OPEN PROBLEM. For all polytopes P ∈ K d T ( P ) ≪ f 0 ( P ) + f 1 ( P ) + . . . f d − 1 ( P )????

  31. FACT. n ( d + 1 ) / ( d − 1 ) is best possible estimate Example 1. (Arnold 80’) G is the graph of the parabola y = x 2 , | x | ≤ t , and P = P t = conv ( G ∩ Z 2 ) . Then f 0 ( P ) = 2 t + 1 and Area P ≈ 2 3 t 3 . in d -dim, G = G t is given by x d = x 2 1 + · · · + x 2 d − 1 ≤ t , P t = conv ( G t ∩ Z d ) . f 0 ( P ) ≈ t d − 1 and Vol P ≈ t d + 1 .

  32. FACT. n ( d + 1 ) / ( d − 1 ) is best possible estimate Example 1. (Arnold 80’) G is the graph of the parabola y = x 2 , | x | ≤ t , and P = P t = conv ( G ∩ Z 2 ) . Then f 0 ( P ) = 2 t + 1 and Area P ≈ 2 3 t 3 . in d -dim, G = G t is given by x d = x 2 1 + · · · + x 2 d − 1 ≤ t , P t = conv ( G t ∩ Z d ) . f 0 ( P ) ≈ t d − 1 and Vol P ≈ t d + 1 .

  33. FACT. n ( d + 1 ) / ( d − 1 ) is best possible estimate Example 1. (Arnold 80’) G is the graph of the parabola y = x 2 , | x | ≤ t , and P = P t = conv ( G ∩ Z 2 ) . Then f 0 ( P ) = 2 t + 1 and Area P ≈ 2 3 t 3 . in d -dim, G = G t is given by x d = x 2 1 + · · · + x 2 d − 1 ≤ t , P t = conv ( G t ∩ Z d ) . f 0 ( P ) ≈ t d − 1 and Vol P ≈ t d + 1 .

  34. FACT. n ( d + 1 ) / ( d − 1 ) is best possible estimate Example 1. (Arnold 80’) G is the graph of the parabola y = x 2 , | x | ≤ t , and P = P t = conv ( G ∩ Z 2 ) . Then f 0 ( P ) = 2 t + 1 and Area P ≈ 2 3 t 3 . in d -dim, G = G t is given by x d = x 2 1 + · · · + x 2 d − 1 ≤ t , P t = conv ( G t ∩ Z d ) . f 0 ( P ) ≈ t d − 1 and Vol P ≈ t d + 1 .

  35. FACT. n ( d + 1 ) / ( d − 1 ) is best possible estimate Example 1. (Arnold 80’) G is the graph of the parabola y = x 2 , | x | ≤ t , and P = P t = conv ( G ∩ Z 2 ) . Then f 0 ( P ) = 2 t + 1 and Area P ≈ 2 3 t 3 . in d -dim, G = G t is given by x d = x 2 1 + · · · + x 2 d − 1 ≤ t , P t = conv ( G t ∩ Z d ) . f 0 ( P ) ≈ t d − 1 and Vol P ≈ t d + 1 .

  36. FACT. n ( d + 1 ) / ( d − 1 ) is best possible estimate Example 1. (Arnold 80’) G is the graph of the parabola y = x 2 , | x | ≤ t , and P = P t = conv ( G ∩ Z 2 ) . Then f 0 ( P ) = 2 t + 1 and Area P ≈ 2 3 t 3 . in d -dim, G = G t is given by x d = x 2 1 + · · · + x 2 d − 1 ≤ t , P t = conv ( G t ∩ Z d ) . f 0 ( P ) ≈ t d − 1 and Vol P ≈ t d + 1 .

  37. rB P

  38. Example 2. (B.-Balog ’92 (d=2), B.-Larman ’98, all d ) P r = conv ( rB d ∩ Z d ) the integer convex hull of rB d Vol P r ≈ r d implies via Andrews’s theorem f 0 ( P r ) ≪ ( Vol P r ) ( d − 1 ) / ( d + 1 ) ≈ r d ( d − 1 ) / ( d + 1 ) . needed: f 0 ( P r ) ≫ r d ( d − 1 ) / ( d + 1 ) .

  39. Example 2. (B.-Balog ’92 (d=2), B.-Larman ’98, all d ) P r = conv ( rB d ∩ Z d ) the integer convex hull of rB d Vol P r ≈ r d implies via Andrews’s theorem f 0 ( P r ) ≪ ( Vol P r ) ( d − 1 ) / ( d + 1 ) ≈ r d ( d − 1 ) / ( d + 1 ) . needed: f 0 ( P r ) ≫ r d ( d − 1 ) / ( d + 1 ) .

  40. Example 2. (B.-Balog ’92 (d=2), B.-Larman ’98, all d ) P r = conv ( rB d ∩ Z d ) the integer convex hull of rB d Vol P r ≈ r d implies via Andrews’s theorem f 0 ( P r ) ≪ ( Vol P r ) ( d − 1 ) / ( d + 1 ) ≈ r d ( d − 1 ) / ( d + 1 ) . needed: f 0 ( P r ) ≫ r d ( d − 1 ) / ( d + 1 ) .

  41. Example 2. (B.-Balog ’92 (d=2), B.-Larman ’98, all d ) P r = conv ( rB d ∩ Z d ) the integer convex hull of rB d Vol P r ≈ r d implies via Andrews’s theorem f 0 ( P r ) ≪ ( Vol P r ) ( d − 1 ) / ( d + 1 ) ≈ r d ( d − 1 ) / ( d + 1 ) . needed: f 0 ( P r ) ≫ r d ( d − 1 ) / ( d + 1 ) .

  42. Lemma Vol ( rB d \ P r ) ≪ r d ( d − 1 ) / ( d + 1 ) . The proof uses the Flatness Theorem, combined with a statement from approximation theory: Lemma If P ⊂ B d is a polytope with f 0 ( P ) ≤ n, then n − 2 / ( d − 1 ) ≪ Vol ( B d \ P ) . f 0 ( P r ) − 2 / ( d − 1 ) ≪ Vol ( rB d \ P r ) ≪ r d ( d − 1 ) / ( d + 1 ) ≪ r − 2 d / ( d + 1 ) . Vol rB d r d implies f 0 ( P r ) ≫ r d ( d − 1 ) / ( d + 1 ) .

  43. Lemma Vol ( rB d \ P r ) ≪ r d ( d − 1 ) / ( d + 1 ) . The proof uses the Flatness Theorem, combined with a statement from approximation theory: Lemma If P ⊂ B d is a polytope with f 0 ( P ) ≤ n, then n − 2 / ( d − 1 ) ≪ Vol ( B d \ P ) . f 0 ( P r ) − 2 / ( d − 1 ) ≪ Vol ( rB d \ P r ) ≪ r d ( d − 1 ) / ( d + 1 ) ≪ r − 2 d / ( d + 1 ) . Vol rB d r d implies f 0 ( P r ) ≫ r d ( d − 1 ) / ( d + 1 ) .

  44. Lemma Vol ( rB d \ P r ) ≪ r d ( d − 1 ) / ( d + 1 ) . The proof uses the Flatness Theorem, combined with a statement from approximation theory: Lemma If P ⊂ B d is a polytope with f 0 ( P ) ≤ n, then n − 2 / ( d − 1 ) ≪ Vol ( B d \ P ) . f 0 ( P r ) − 2 / ( d − 1 ) ≪ Vol ( rB d \ P r ) ≪ r d ( d − 1 ) / ( d + 1 ) ≪ r − 2 d / ( d + 1 ) . Vol rB d r d implies f 0 ( P r ) ≫ r d ( d − 1 ) / ( d + 1 ) .

  45. Lemma Vol ( rB d \ P r ) ≪ r d ( d − 1 ) / ( d + 1 ) . The proof uses the Flatness Theorem, combined with a statement from approximation theory: Lemma If P ⊂ B d is a polytope with f 0 ( P ) ≤ n, then n − 2 / ( d − 1 ) ≪ Vol ( B d \ P ) . f 0 ( P r ) − 2 / ( d − 1 ) ≪ Vol ( rB d \ P r ) ≪ r d ( d − 1 ) / ( d + 1 ) ≪ r − 2 d / ( d + 1 ) . Vol rB d r d implies f 0 ( P r ) ≫ r d ( d − 1 ) / ( d + 1 ) .

  46. Lemma Vol ( rB d \ P r ) ≪ r d ( d − 1 ) / ( d + 1 ) . The proof uses the Flatness Theorem, combined with a statement from approximation theory: Lemma If P ⊂ B d is a polytope with f 0 ( P ) ≤ n, then n − 2 / ( d − 1 ) ≪ Vol ( B d \ P ) . f 0 ( P r ) − 2 / ( d − 1 ) ≪ Vol ( rB d \ P r ) ≪ r d ( d − 1 ) / ( d + 1 ) ≪ r − 2 d / ( d + 1 ) . Vol rB d r d implies f 0 ( P r ) ≫ r d ( d − 1 ) / ( d + 1 ) .

  47. REMARK. Works for all K ∈ K d (instead of B d ) with smooth enough boundary. OPEN PROBLEM. Does lim n − d + 1 d − 1 V d ( n ) exist???? will come back when d = 2.

  48. REMARK. Works for all K ∈ K d (instead of B d ) with smooth enough boundary. OPEN PROBLEM. Does lim n − d + 1 d − 1 V d ( n ) exist???? will come back when d = 2.

  49. REMARK. Works for all K ∈ K d (instead of B d ) with smooth enough boundary. OPEN PROBLEM. Does lim n − d + 1 d − 1 V d ( n ) exist???? will come back when d = 2.

  50. 2. Minimal surface area S d ( n ) Isoperimetric inequality: For all K ∈ K d S ( K ) d S ( B d ) d ( Vol K ) d − 1 ≥ ( Vol B d ) d − 1 implies S ( P ) ≫ ( Vol P ) ( d − 1 ) / d ≫ f 0 ( P ) ( d + 1 ) / d Corollary n ( d + 1 ) / d ≪ S d ( n ) Example 2 shows that this is best possible OPEN PROBLEM. Does lim n − d + 1 d S d ( n ) exist???? d = 2 Jarník

  51. 2. Minimal surface area S d ( n ) Isoperimetric inequality: For all K ∈ K d S ( K ) d S ( B d ) d ( Vol K ) d − 1 ≥ ( Vol B d ) d − 1 implies S ( P ) ≫ ( Vol P ) ( d − 1 ) / d ≫ f 0 ( P ) ( d + 1 ) / d Corollary n ( d + 1 ) / d ≪ S d ( n ) Example 2 shows that this is best possible OPEN PROBLEM. Does lim n − d + 1 d S d ( n ) exist???? d = 2 Jarník

  52. 2. Minimal surface area S d ( n ) Isoperimetric inequality: For all K ∈ K d S ( K ) d S ( B d ) d ( Vol K ) d − 1 ≥ ( Vol B d ) d − 1 implies S ( P ) ≫ ( Vol P ) ( d − 1 ) / d ≫ f 0 ( P ) ( d + 1 ) / d Corollary n ( d + 1 ) / d ≪ S d ( n ) Example 2 shows that this is best possible OPEN PROBLEM. Does lim n − d + 1 d S d ( n ) exist???? d = 2 Jarník

  53. 2. Minimal surface area S d ( n ) Isoperimetric inequality: For all K ∈ K d S ( K ) d S ( B d ) d ( Vol K ) d − 1 ≥ ( Vol B d ) d − 1 implies S ( P ) ≫ ( Vol P ) ( d − 1 ) / d ≫ f 0 ( P ) ( d + 1 ) / d Corollary n ( d + 1 ) / d ≪ S d ( n ) Example 2 shows that this is best possible OPEN PROBLEM. Does lim n − d + 1 d S d ( n ) exist???? d = 2 Jarník

  54. 2. Minimal surface area S d ( n ) Isoperimetric inequality: For all K ∈ K d S ( K ) d S ( B d ) d ( Vol K ) d − 1 ≥ ( Vol B d ) d − 1 implies S ( P ) ≫ ( Vol P ) ( d − 1 ) / d ≫ f 0 ( P ) ( d + 1 ) / d Corollary n ( d + 1 ) / d ≪ S d ( n ) Example 2 shows that this is best possible OPEN PROBLEM. Does lim n − d + 1 d S d ( n ) exist???? d = 2 Jarník

  55. 2. Minimal surface area S d ( n ) Isoperimetric inequality: For all K ∈ K d S ( K ) d S ( B d ) d ( Vol K ) d − 1 ≥ ( Vol B d ) d − 1 implies S ( P ) ≫ ( Vol P ) ( d − 1 ) / d ≫ f 0 ( P ) ( d + 1 ) / d Corollary n ( d + 1 ) / d ≪ S d ( n ) Example 2 shows that this is best possible OPEN PROBLEM. Does lim n − d + 1 d S d ( n ) exist???? d = 2 Jarník

  56. 2. Minimal surface area S d ( n ) Isoperimetric inequality: For all K ∈ K d S ( K ) d S ( B d ) d ( Vol K ) d − 1 ≥ ( Vol B d ) d − 1 implies S ( P ) ≫ ( Vol P ) ( d − 1 ) / d ≫ f 0 ( P ) ( d + 1 ) / d Corollary n ( d + 1 ) / d ≪ S d ( n ) Example 2 shows that this is best possible OPEN PROBLEM. Does lim n − d + 1 d S d ( n ) exist???? d = 2 Jarník

  57. 3. Minimal lattice width w d ( n ) first d = 2. w ( P ) + 1 is the minimal number of consecutive lattice lines intersecting P . each such line contains at most two vertices of P = ⇒ f 0 ( P ) ≤ 2 ( w ( P ) + 1 ) FACT. w 2 ( n ) = ⌈ n 2 ⌉ − 1 FACT. w d ( n ) = 1 OPEN PROBLEM. Modify the question!!!

  58. 3. Minimal lattice width w d ( n ) first d = 2. w ( P ) + 1 is the minimal number of consecutive lattice lines intersecting P . each such line contains at most two vertices of P = ⇒ f 0 ( P ) ≤ 2 ( w ( P ) + 1 ) FACT. w 2 ( n ) = ⌈ n 2 ⌉ − 1 FACT. w d ( n ) = 1 OPEN PROBLEM. Modify the question!!!

  59. 3. Minimal lattice width w d ( n ) first d = 2. w ( P ) + 1 is the minimal number of consecutive lattice lines intersecting P . each such line contains at most two vertices of P = ⇒ f 0 ( P ) ≤ 2 ( w ( P ) + 1 ) FACT. w 2 ( n ) = ⌈ n 2 ⌉ − 1 FACT. w d ( n ) = 1 OPEN PROBLEM. Modify the question!!!

  60. 3. Minimal lattice width w d ( n ) first d = 2. w ( P ) + 1 is the minimal number of consecutive lattice lines intersecting P . each such line contains at most two vertices of P = ⇒ f 0 ( P ) ≤ 2 ( w ( P ) + 1 ) FACT. w 2 ( n ) = ⌈ n 2 ⌉ − 1 FACT. w d ( n ) = 1 OPEN PROBLEM. Modify the question!!!

  61. 3. Minimal lattice width w d ( n ) first d = 2. w ( P ) + 1 is the minimal number of consecutive lattice lines intersecting P . each such line contains at most two vertices of P = ⇒ f 0 ( P ) ≤ 2 ( w ( P ) + 1 ) FACT. w 2 ( n ) = ⌈ n 2 ⌉ − 1 FACT. w d ( n ) = 1 OPEN PROBLEM. Modify the question!!!

  62. 3. Minimal lattice width w d ( n ) first d = 2. w ( P ) + 1 is the minimal number of consecutive lattice lines intersecting P . each such line contains at most two vertices of P = ⇒ f 0 ( P ) ≤ 2 ( w ( P ) + 1 ) FACT. w 2 ( n ) = ⌈ n 2 ⌉ − 1 FACT. w d ( n ) = 1 OPEN PROBLEM. Modify the question!!!

  63. 4. Arnold’s question P , Q ∈ P d are equivalent if a lattice preserving affine transformation maps P to Q . FACT. P ∼ Q = ⇒ f 0 ( P ) = f 0 ( Q ) , w ( P ) = w ( Q ) , Vol P = Vol Q . N d ( V ) = number of equivalent classes of P ∈ P d with Vol P ≤ V N 2 ( A ) for d = 2 motivation

  64. 4. Arnold’s question P , Q ∈ P d are equivalent if a lattice preserving affine transformation maps P to Q . FACT. P ∼ Q = ⇒ f 0 ( P ) = f 0 ( Q ) , w ( P ) = w ( Q ) , Vol P = Vol Q . N d ( V ) = number of equivalent classes of P ∈ P d with Vol P ≤ V N 2 ( A ) for d = 2 motivation

  65. 4. Arnold’s question P , Q ∈ P d are equivalent if a lattice preserving affine transformation maps P to Q . FACT. P ∼ Q = ⇒ f 0 ( P ) = f 0 ( Q ) , w ( P ) = w ( Q ) , Vol P = Vol Q . N d ( V ) = number of equivalent classes of P ∈ P d with Vol P ≤ V N 2 ( A ) for d = 2 motivation

  66. 4. Arnold’s question P , Q ∈ P d are equivalent if a lattice preserving affine transformation maps P to Q . FACT. P ∼ Q = ⇒ f 0 ( P ) = f 0 ( Q ) , w ( P ) = w ( Q ) , Vol P = Vol Q . N d ( V ) = number of equivalent classes of P ∈ P d with Vol P ≤ V N 2 ( A ) for d = 2 motivation

  67. 4. Arnold’s question P , Q ∈ P d are equivalent if a lattice preserving affine transformation maps P to Q . FACT. P ∼ Q = ⇒ f 0 ( P ) = f 0 ( Q ) , w ( P ) = w ( Q ) , Vol P = Vol Q . N d ( V ) = number of equivalent classes of P ∈ P d with Vol P ≤ V N 2 ( A ) for d = 2 motivation

  68. 4. Arnold’s question P , Q ∈ P d are equivalent if a lattice preserving affine transformation maps P to Q . FACT. P ∼ Q = ⇒ f 0 ( P ) = f 0 ( Q ) , w ( P ) = w ( Q ) , Vol P = Vol Q . N d ( V ) = number of equivalent classes of P ∈ P d with Vol P ≤ V N 2 ( A ) for d = 2 motivation

  69. Theorem (Arnold 1980) A 1 / 3 ≪ log N 2 ( A ) ≪ A 1 / 3 log A. lower bound: let P be the polytope from Example 1 or 2. ⇒| W | ≈ A 1 / 3 . Its vertex set W = For each subset U ⊂ W , conv U ∈ P 2 . there are 2 | W | ≈ 2 A 1 / 3 such subpolygons. Most of them distinct. for the upper bound we need: Lemma (Square lemma) For every P ∈ P 2 there is Q ∼ P which is contained in the square [ 0 , 36 A ] 2 . So each equivalence class is represented in this square. Proof follows from Andrews theorem + Square lemma

  70. Theorem (Arnold 1980) A 1 / 3 ≪ log N 2 ( A ) ≪ A 1 / 3 log A. lower bound: let P be the polytope from Example 1 or 2. ⇒| W | ≈ A 1 / 3 . Its vertex set W = For each subset U ⊂ W , conv U ∈ P 2 . there are 2 | W | ≈ 2 A 1 / 3 such subpolygons. Most of them distinct. for the upper bound we need: Lemma (Square lemma) For every P ∈ P 2 there is Q ∼ P which is contained in the square [ 0 , 36 A ] 2 . So each equivalence class is represented in this square. Proof follows from Andrews theorem + Square lemma

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