The minimal volume of simplices containing a convex body

Transcription

1 The minimal volume of simplices containing a convex body Mariano Merzbacher (Joint work with Damián Pinasco and Daniel Galicer) Universidad de Buenos Aires and CONICET NonLinear Functional Analysis, Valencia 2017

2 The basics Notation A convex body K R n is a compact convex set with nonempty interior.

3 The basics Notation A convex body K R n is a compact convex set with nonempty interior.

4 The basics Notation A convex body K R n is a compact convex set with nonempty interior. The barycenter of K is bar(k) := 1 vol(k) K xdx.

5 The basics Notation A simplex in R n is always an n-simplex, the convex hull of n + 1 affinely independents points. 3-Simplex

6 The problem Given a convex body K,approximate it a by a simpler one, in this case a simplex S. K

7 The problem Given a convex body K,approximate it a by a simpler one, in this case a simplex S. K S

8 The problem Given a convex body K,approximate it a by a simpler one, in this case a simplex S. K S Approximating: enclosing it and having similar volume (Lebesgue measure).

9 Definition For a convex body K we define the outer simplex ratio of K as Sout(K) := min ( ) vol(s) 1/n, vol(k) the minimum is taken over all simplices S R n containing K.

10 Definition For a convex body K we define the outer simplex ratio of K as Sout(K) := min ( ) vol(s) 1/n, vol(k) the minimum is taken over all simplices S R n containing K. Problem How small can Sout(K) be?

11 Historical background This is the generalization of an old geometrical problem. n = 2 Wilhelm Gross at the early 20 s: every convex body K R 2 can be inscribed in a triangle of area 2vol(K).

12 Historical background This is the generalization of an old geometrical problem. n = 2 Wilhelm Gross at the early 20 s: every convex body K R 2 can be inscribed in a triangle of area 2vol(K). In this case the square is the worst posible fit.

13 n = 3 No exact bound has been given, Sout(K) 9 2 is conjectured. For a parallelepiped K a tethraedron of volume 9 2vol(K) can be found, but is not know if it is the best posible bound.

14 n = 3 No exact bound has been given, Sout(K) 9 2 is conjectured. For a parallelepiped K a tethraedron of volume 9 2vol(K) can be found, but is not know if it is the best posible bound. For n 4 no exact bound was even conjectured.

15 Some examples The Euclidean ball, B n 2

16 Some examples The Euclidean ball, B n 2 B n 2 S,the regular simplex, and is the smallest one.

17 Some examples The Euclidean ball, B n 2 B n 2 S,the regular simplex, and is the smallest one. vol(b n 2 ) = π n 2 Γ( n 2 +1) vol(s) = (n+1) n+1 2 n n 2 n!

18 Some examples The Euclidean ball, B n 2 B n 2 S,the regular simplex, and is the smallest one. vol(b n 2 ) = π n 2 Γ( n 2 +1) vol(s) = (n+1) n+1 2 n n 2 n! Applying Stirling s formula: ( ) 1 Sout(B2 n) = vol(s) n vol(b2 n) n 1 2

19 Some examples The unit cube, C = [0, 1] n.

20 Some examples The unit cube, C = [0, 1] n. C S = {0, ne 1,..., ne n }

21 Some examples The unit cube, C = [0, 1] n. C S = {0, ne 1,..., ne n } vol(c) = 1 vol(s) = nn n!

22 Some examples The unit cube, C = [0, 1] n. C S = {0, ne 1,..., ne n } vol(c) = 1 vol(s) = nn n! Applying Stirling s formula: Sout(C) ( ) 1 vol(s) n vol(c) 1

23 Some examples The unit cube, C = [0, 1] n. C S = {0, ne 1,..., ne n } vol(c) = 1 vol(s) = nn n! Applying Stirling s formula: Sout(C) ( ) 1 vol(s) n vol(c) For higher dimensions this is no longer the worst fit. 1

24 Known general bounds

25 Known general bounds Chakerian in 1973, Sout(K) n n 1 n n,

26 Known general bounds Chakerian in 1973, Sout(K) n n 1 n n, Giannopoulos-Hartzoulaki 2002/Pivovarov-Paoruis 2017, Sout(K) C log(n) n.

27 Known general bounds Chakerian in 1973, Sout(K) n n 1 n n, Giannopoulos-Hartzoulaki 2002/Pivovarov-Paoruis 2017, Sout(K) C log(n) n. Theorem (Galicer, Pinasco, M.) Sout(K) C n for some absolute constant C > 0.

28 Known general bounds Chakerian in 1973, Sout(K) n n 1 n n, Giannopoulos-Hartzoulaki 2002/Pivovarov-Paoruis 2017, Sout(K) C log(n) n. Theorem (Galicer, Pinasco, M.) Sout(K) C n for some absolute constant C > 0. Up to an absolute constant this bound cannot be lessened, as Sout(B n 2 ) n.

29 Polar bodies Definition Given a convex body K R n, with 0 int(k), its polar body is K = {y R n : x, y 1 x K}.

30 Polar bodies Definition Given a convex body K R n, with 0 int(k), its polar body is K = {y R n : x, y 1 x K}.

31 Polar bodies Definition Given a convex body K R n, with 0 int(k), its polar body is K = {y R n : x, y 1 x K}. Some facts If K is the unit ball of E = (R n, K ), K is the ball of E.

32 Polar bodies Definition Given a convex body K R n, with 0 int(k), its polar body is K = {y R n : x, y 1 x K}. Some facts If K is the unit ball of E = (R n, K ), K is the ball of E. The polar of a simplex is also a simplex.

33 Polar bodies Definition Given a convex body K R n, with 0 int(k), its polar body is K = {y R n : x, y 1 x K}. Some facts If K is the unit ball of E = (R n, K ), K is the ball of E. The polar of a simplex is also a simplex. If 0 / int(k), K is unbounded.

34 Polar bodies Definition Given a convex body K R n, with 0 int(k), its polar body is K = {y R n : x, y 1 x K}. Some facts If K is the unit ball of E = (R n, K ), K is the ball of E. The polar of a simplex is also a simplex. If 0 / int(k), K is unbounded. (K ) = K and K L L K.

35 Polarity and volume Theorem (Balschke-Santaló and Bourgain-Milman) If K has barycenter at the origin then (vol(k)vol(k )) 1 n vol(b n 2 ) 2 n 1 n

36 Polarity and volume Theorem (Balschke-Santaló and Bourgain-Milman) If K has barycenter at the origin then (vol(k)vol(k )) 1 n vol(b n 2 ) 2 n 1 n Observations The barycenter condition is central for this inequality to hold.

37 Polarity and volume Theorem (Balschke-Santaló and Bourgain-Milman) If K has barycenter at the origin then (vol(k)vol(k )) 1 n vol(b n 2 ) 2 n 1 n Observations The barycenter condition is central for this inequality to hold. In this case polarity reverses volume.

38 A dual problem Two equivalent problems

39 A dual problem Two equivalent problems Original problem Approximating a convex body by a simplex enclosing it of small volume.

40 A dual problem Two equivalent problems Original problem Approximating a convex body by a simplex enclosing it of small volume. Dual problem Approximating a convex body by a simplex inside it of large volume.

41 A dual problem Two equivalent problems Original problem Approximating a convex body by a simplex enclosing it of small volume and same barycenter. Dual problem Approximating a convex body by a simplex inside it of large volume and same barycenter.

42 Inner volume ratio S inn (K) := max S K ( ) 1 vol(s) n vol(k)

43 Inner volume ratio Theorem (dual version) S inn (K) := max S K ( ) 1 vol(s) n vol(k) S inn (K) C n for some absolute constant C > 0.

44 Goal Given K R n find a simplex S K satisfying: 1 vol(s) 1 n C vol(k) 1 n n 2 bar(s) = bar(k) with the same barycenter and large volume.

45 A random approach

46 A random approach Objective A simplex inside K with same barycenter and large volume.

47 A random approach Objective A simplex inside K with same barycenter and large volume. Strategy Define an adequate probabilty measure on K.

48 A random approach Objective A simplex inside K with same barycenter and large volume. Strategy Define an adequate probabilty measure on K. Consider the random simplex S := co{0, X 1... X n }, with X i distributed according to this measure.

49 A random approach Objective A simplex inside K with same barycenter and large volume. Strategy Define an adequate probabilty measure on K. Consider the random simplex S := co{0, X 1... X n }, with X i distributed according to this measure. Prove that with postive probability there is a simplex similar to the objective.

50 A random approach Objective A simplex inside K with same barycenter and large volume. Strategy Define an adequate probabilty measure on K. Consider the random simplex S := co{0, X 1... X n }, with X i distributed according to this measure. Prove that with postive probability there is a simplex similar to the objective. Adjust it.

51 The uniforme measure on a general body can be hard to handle.

52 The uniforme measure on a general body can be hard to handle. Affine invariance S inn (K) = S inn (TK) for every affin transformation T.

53 The uniforme measure on a general body can be hard to handle. Affine invariance S inn (K) = S inn (TK) for every affin transformation T. S inn (K) is a property of the affine class of K rather than a property of K.

54 The uniforme measure on a general body can be hard to handle. Affine invariance S inn (K) = S inn (TK) for every affin transformation T. S inn (K) is a property of the affine class of K rather than a property of K. Choose a representative with a controllable uniform measure.

55 Isotropic bodies Definition A convex body K is isotropic if:

56 Isotropic bodies Definition A convex body K is isotropic if: vol(k) = 1, bar(k) = 0, there is L K R such that K x ix j = L 2 K δ ij. ( K X X = L K Id )

57 Isotropic bodies Definition A convex body K is isotropic if: vol(k) = 1, bar(k) = 0, there is L K R such that K x ix j = L 2 K δ ij. ( K X X = L K Id )

58 Isotropic bodies Definition A convex body K is isotropic if: vol(k) = 1, bar(k) = 0, there is L K R such that K x ix j = L 2 K δ ij. ( K X X = L K Id ) Fact Every convex body K has a unique (up to orthogonal transformations) affine isotropic image.

59 Random simplex on isotropic bodies Proposition Let K be an isotropic convex body and X 1,... X n independent random variables distributed uniformly on K. S = co{0, X 1,..., X n }. With probabilty grater than 1 e n :

60 Random simplex on isotropic bodies Proposition Let K be an isotropic convex body and X 1,... X n independent random variables distributed uniformly on K. S = co{0, X 1,..., X n }. With probabilty grater than 1 e n : bar(s) = 1 n+1 n i=1 X i L K C 1

61 Random simplex on isotropic bodies Proposition Let K be an isotropic convex body and X 1,... X n independent random variables distributed uniformly on K. S = co{0, X 1,..., X n }. With probabilty grater than 1 e n : bar(s) = 1 n+1 n i=1 X i L K C 1 vol(s) = det X 1...X n n! C n 2 n n 2 Ln K

62 Random simplex on isotropic bodies Proposition Let K be an isotropic convex body and X 1,... X n independent random variables distributed uniformly on K. S = co{0, X 1,..., X n }. With probabilty grater than 1 e n : bar(s) = 1 n+1 n i=1 X i L K C 1 vol(s) = det X 1...X n n! C n 2 n n 2 Ln K S has large volume and barycenter close to the origin.

63 0 bar(s) S K

64 0 bar(s) S K

65 A non-probabilistic approach A convex body K is in John s position if B2 n ellipsoide inside it. is the maximal volume

66 A non-probabilistic approach A convex body K is in John s position if B2 n ellipsoide inside it. is the maximal volume Lemma (Dvoresky-Rogers) Let K R n be a convex body in John s position, there are y 1... y n Bd(B2 n ) Bd(K) such that ( ) 1 n i + 1 P span{y1...y i 1 } (y 2 i). n

67 A non-probabilistic approach A convex body K is in John s position if B2 n ellipsoide inside it. is the maximal volume Lemma (Dvoresky-Rogers) Let K R n be a convex body in John s position, there are y 1... y n Bd(B2 n ) Bd(K) such that ( ) 1 n i + 1 P span{y1...y i 1 } (y 2 i). n det y 1... y n = y 1 n i=2 P span{y 1...y i 1 } (y i) ( ) n n 1 2 n!

68 Case K symmetric

69 Case K symmetric Consider the points y 1,..., y n given by Dvoretsky-Rogers. 0 y 1 y 2

70 Case K symmetric y 2 0 y 1 y 1 y 2 Consider the points y 1,..., y n given by Dvoretsky-Rogers. K is symmetric, so y 1,..., y n Bd(K)

71 Case K symmetric y 2 0 y 1 y 1 y 2 Consider the points y 1,..., y n given by Dvoretsky-Rogers. K is symmetric, so y 1,..., y n Bd(K) P = n i { x, y i 1} P

72 Case K symmetric y 2 0 y 1 y 1 y 2 Consider the points y 1,..., y n given by Dvoretsky-Rogers. K is symmetric, so y 1,..., y n Bd(K) P = n i { x, y i 1} 2 vol(p) = n det y 1...y C n n P

73 Case K symmetric y 2 0 y 1 P y 1 y 2 Consider the points y 1,..., y n given by Dvoretsky-Rogers. K is symmetric, so y 1,..., y n Bd(K) P = n i { x, y i 1} 2 vol(p) = n det y 1...y C n n vol(p) vol(k) vol(p) vol(b n 2 ) C n n n 2

74 Case K symmetric S y 2 0 y 1 P y 1 y 2 Consider the points y 1,..., y n given by Dvoretsky-Rogers. K is symmetric, so y 1,..., y n Bd(K) P = n i { x, y i 1} 2 vol(p) = n det y 1...y C n n vol(p) vol(p) C n n n vol(k) vol(b2 n) 2 Finally take S so that ( vol(s) vol(p) ) 1 n C

75 Case K symmetric S y 2 0 y 1 P y 1 y 2 Consider the points y 1,..., y n given by Dvoretsky-Rogers. K is symmetric, so y 1,..., y n Bd(K) P = n i { x, y i 1} 2 vol(p) = n det y 1...y C n n vol(p) vol(p) C n n n vol(k) vol(b2 n) 2 Finally take S so that ( vol(s) vol(p) ) 1 n C

76 General case Rogers-Shephard Inequality Given a convex body K with barycenter at the origin, vol(k K) 4 n vol(k).

77 General case Rogers-Shephard Inequality Given a convex body K with barycenter at the origin, vol(k K) 4 n vol(k). A centered body can be approximated by outside by a symmetric body with essentialy the same volume. ( vol(k K) vol(k) ) 1 n 4.

78 Thank you!