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On the isotropy constant of projections of polytopes Julio Bernués (joint work with D. Alonso-Gutiérrez, J. Bastero and P. Wolff) University of Zaragoza College Station, July 20th 2009 Introduction Let K ⊂ Rd be a symmetric convex body. Its isotropy constant LK is ) ( Z 1 2 2 |x| dx : T ∈ GL(d) dLK := mı́n 2 |TK |1+ d TK Introduction Let K ⊂ Rd be a symmetric convex body. Its isotropy constant LK is ) ( Z 1 2 2 |x| dx : T ∈ GL(d) dLK := mı́n 2 |TK |1+ d TK LK is a linear invariant. Introduction Let K ⊂ Rd be a symmetric convex body. Its isotropy constant LK is ) ( Z 1 2 2 |x| dx : T ∈ GL(d) dLK := mı́n 2 |TK |1+ d TK LK is a linear invariant. By Hölder’s inequality, LK ≥ LB d ' 0,24 2 Introduction Let K ⊂ Rd be a symmetric convex body. Its isotropy constant LK is ) ( Z 1 2 2 |x| dx : T ∈ GL(d) dLK := mı́n 2 |TK |1+ d TK LK is a linear invariant. By Hölder’s inequality, LK ≥ LB d ' 0,24 2 Problem: Estimate LK from above. LK is known to be bounded by an absolute constant for many families of convex bodies. Introduction Let K ⊂ Rd be a symmetric convex body. Its isotropy constant LK is ) ( Z 1 2 2 |x| dx : T ∈ GL(d) dLK := mı́n 2 |TK |1+ d TK LK is a linear invariant. By Hölder’s inequality, LK ≥ LB d ' 0,24 2 Problem: Estimate LK from above. Conjecture: LK ≤ LB∞ d ' 0,29 LK is known to be bounded by an absolute constant for many families of convex bodies. Introduction For random convex bodies, (2008-2009) Introduction For random convex bodies, (2008-2009) Theorem (Klartag-Kozma) Let G0 , . . . , Gn be i.i.d. gaussian random vectors in Rd , n > d. For K = conv{±G0 , . . . , ±Gn } we have that with probability greater than 1 − Ce −cd LK ≤ C . (extra arguments needed for n ≤ cd) Theorem (Alonso-Gutierrez and Dafnis, Giannopoulos, Guedon) If n ≥ cd and {Pi }ni=0 are independent random points on S d−1 on an isotropic unconditional convex body in Rd and K = conv{±P1 , . . . , ±Pn } then with probability ≥ 1 − c1 e −c2 d LK ≤ C Back Introduction We will focus on the isotropy constant of projections of convex bodies. Introduction We will focus on the isotropy constant of projections of convex bodies. Notation: From now on, E is a d dimensional subspace of Rn and PE is the orthogonal projection onto E . Introduction We will focus on the isotropy constant of projections of convex bodies. Notation: From now on, E is a d dimensional subspace of Rn and PE is the orthogonal projection onto E . For any d and 1 < p ≤ ∞, (M. Junge, 1994, 1995; E. Milman, 2006) √ 1 1 LPE (Bpn ) ≤ c p 0 , p + p 0 = 1. LPE (B1n ) ≤ c log n. PE (B1n ) is a d-dimensional convex polytope of at most n vertices. An observation. Random projections of B1n Recall Gn,d := {E ⊆ Rn : E subspace d-dimensional}. There exists a unique probability measure, µn,d (Haar measure) on Gn,d invariant under orthogonal transformations. An observation. Random projections of B1n Recall Gn,d := {E ⊆ Rn : E subspace d-dimensional}. There exists a unique probability measure, µn,d (Haar measure) on Gn,d invariant under orthogonal transformations. If G is an n × d, G : Rd → Rn , random matrix with entries gij i.i.d. standard gaussian r.v.’s then by uniqueness for every borelian A ⊂ Gn,d , µn,d (A) = P{Im G ∈ A} An observation. Random projections of B1n If G is an n × d matrix and E = Im G ∈ Gn,d then (linear algebra) t G|E PE B1n = conv{±G1 , . . . , ±Gn } where Gi are the rows of G . An observation. Random projections of B1n If G is an n × d matrix and E = Im G ∈ Gn,d then (linear algebra) t G|E PE B1n = conv{±G1 , . . . , ±Gn } where Gi are the rows of G . Thus, Klartag-Kozma result reads µn,d {E ∈ Gn,d : LPE B1n ≤ C } > 1 − ce −c 0d An observation. Random projections of B1n If G is an n × d matrix and E = Im G ∈ Gn,d then (linear algebra) t G|E PE B1n = conv{±G1 , . . . , ±Gn } where Gi are the rows of G . Thus, Klartag-Kozma result reads µn,d {E ∈ Gn,d : LPE B1n ≤ C } > 1 − ce −c 0d For d as small as d < c log n Dvoretzky’s theorem says that PE B1n is typically like the euclidean ball B2d and so, µn,d {E ∈ Gn,d : LPE B1n ≤ C } > 1 − ce −c 0 log n An observation. Random projections of B1n In conclusion, there exists C , c, c 0 > 0 such that for all 1 ≤ d ≤ n, µn,d {E ∈ Gn,d : LPE B1n ≤ C } > 1 − ce −c 0 máx{d,log n} An observation. Random projections of B1n In conclusion, there exists C , c, c 0 > 0 such that for all 1 ≤ d ≤ n, µn,d {E ∈ Gn,d : LPE B1n ≤ C } > 1 − ce −c 0 máx{d,log n} On the other hand recall the following: Fact If LPE B1n ≤ C for every d-dimensional E and every 1 ≤ d ≤ n, then LK ≤ C for every symmetric convex body K ⊂ Rd in any dimension d. An observation. Random projections of B1n In conclusion, there exists C , c, c 0 > 0 such that for all 1 ≤ d ≤ n, µn,d {E ∈ Gn,d : LPE B1n ≤ C } > 1 − ce −c 0 máx{d,log n} On the other hand recall the following: Fact If LPE B1n ≤ C for every d-dimensional E and every 1 ≤ d ≤ n, then LK ≤ C for every symmetric convex body K ⊂ Rd in any dimension d. In this sense, “most” symmetric convex bodies K ⊆ Rd have isotropy constant bounded. Main result. Projections of B1n Theorem r L PE (B1n ) , LPE (Sn ) ≤ C where Sn is the n-dimensional regular simplex. n d Main result. Projections of B1n Theorem r L PE (B1n ) , LPE (Sn ) ≤ C n d where Sn is the n-dimensional regular simplex. m Corollary Let K ⊆ Rd be a convex polytope with n vertices. Then r n LK ≤ C d Proof: Starting point For any convex polytope K ⊂ Rn , dL2PE (K ) ≤ 1 1 |PE (K )|2/d |PE (K )| Z PE (K ) |x|2 λE (dx) Proof: Starting point For any convex polytope K ⊂ Rn , dL2PE (K ) ≤ 1 1 |PE (K )|2/d |PE (K )| 1 Estimate from above |PE (K )| Estimate from below |PE (K )| Z PE (K ) Z |x|2 λE (dx) PE (K ) |x|2 λE (dx) Integrating on PE (K ) K PE ↓ E Integrating on PE (K ) K PE ↓ E Main Lemma Let K ⊂ Rn be a convex polytope and E a d dimensional subspace of Rn . There exists a subset F̃ of d dimensional faces of K such that {PE (relint F ) F ∈ F̃} is a family of pair-wise disjoint sets, S {PE (F ); F ∈ F̃} = PE (K ), a.e. λE For each F ∈ F̃, PE |F : F → PE (F ) is an affine isomorphism. Integrating on PE (K ) Integrating on PE (K ) Corollary Let K ⊂ Rn be a convex polytope and E a d dimensional subspace of Rn . There exists a subset F̃ of d dimensional faces of K such that for any integrable function f : PE (K ) → R, Z XZ f (x) λE (dx) = f (x) λE (dx) = PE (K ) F ∈F̃ PE (F ) X |PE (F )| Z = f (PE y ) λaffF (dy ) |F | F F ∈F̃ In particular for f ≡ 1, |PE (K )| = X F ∈F̃ |PE (F )| Sketch of the proof 1 |PE (K )| XZ 1 |x| dx = |x|2 dx = |P (K )| E PE (K ) PE (F ) Z 2 F ∈F̃ Sketch of the proof 1 |PE (K )| XZ 1 |x| dx = |x|2 dx = |P (K )| E PE (K ) PE (F ) Z 2 F ∈F̃ X |PE (F )| Z X |PE (F )| 1 Z 1 2 |PE y | dy ≤ |y |2 dy |PE (K )| |F | |P (K )| |F | E F F F ∈F̃ F ∈F̃ Sketch of the proof 1 |PE (K )| XZ 1 |x| dx = |x|2 dx = |P (K )| E PE (K ) PE (F ) Z 2 F ∈F̃ X |PE (F )| Z X |PE (F )| 1 Z 1 2 |PE y | dy ≤ |y |2 dy |PE (K )| |F | |P (K )| |F | E F F F ∈F̃ 1 ≤ sup F ∈F̃ |F | F ∈F̃ Z F |y |2 dy since 1= X |PE (F )| |PE (K )| F ∈F̃ In the symmetric case, K = B1n , denoting ∆d the d-dimensional regular simplex in he1 , . . . , ed+1 i, 1 |PE (B1n )| Z |x|2 dx ≤ sup PE (B1n ) F ∈F̃ 1 |F | Z F |y |2 dy = 1 |∆d | Z ∆d |x|2 dx = 2 d +2 In the symmetric case, K = B1n , denoting ∆d the d-dimensional regular simplex in he1 , . . . , ed+1 i, 1 |PE (B1n )| Z |x|2 dx ≤ sup PE (B1n ) F ∈F̃ 1 |F | Z |y |2 dy = F 1 |∆d | Z ∆d |x|2 dx = 2 d +2 On the other hand, n−1/2 B2n ⊆ B1n =⇒ n−1/2 PE (B2n ) ⊆ PE (B1n ). Therefore c |PE (B1n )|1/d ≥ √ nd In the symmetric case, K = B1n , denoting ∆d the d-dimensional regular simplex in he1 , . . . , ed+1 i, 1 |PE (B1n )| Z |x|2 dx ≤ sup PE (B1n ) F ∈F̃ 1 |F | Z |y |2 dy = F 1 |∆d | Z ∆d |x|2 dx = 2 d +2 On the other hand, n−1/2 B2n ⊆ B1n =⇒ n−1/2 PE (B2n ) ⊆ PE (B1n ). Therefore c |PE (B1n )|1/d ≥ √ nd Combining all estimates we get L2PE (B n ) ≤ 1 1 1 1 d |PE (B1n )|2/d |PE (B1n )| Z PE (B1n ) |x|2 dx ≤ 1 nd 2 n ≤ c0 d c2 d + 2 d Application Our result, Theorem Let K ⊆ Rd be a convex polytope with n vertices. Then r n LK ≤ C d Application Our result, Theorem Let K ⊆ Rd be a convex polytope with n vertices. Then r n LK ≤ C d solves the difficulties in the papers above Rdm : Application Our result, Theorem Let K ⊆ Rd be a convex polytope with n vertices. Then r n LK ≤ C d solves the difficulties in the papers above Rdm : If the number of vertices is proportional to the dimension, n ≤ cd then LK ≤ C deterministically. The isotropy constant of projections of Bpn , 1 < p ≤ 2. For hyperplanes, Theorem. There exists C > 0 such that for every 1 < p ≤ 2 and any hyperplane H = θ⊥ , LPH (Bpn ) ≤ C The isotropy constant of projections of Bpn , 1 < p ≤ 2. For hyperplanes, Theorem. There exists C > 0 such that for every 1 < p ≤ 2 and any hyperplane H = θ⊥ , LPH (Bpn ) ≤ C Proof of the Theorem. (Ideas from Barthe-Naor). From the definition, Z 1 1 (n − 1)L2PH (Bpn ) ≤ · |x|2 dx 2 n )| |P (B n n H PH (Bp ) p |PH (Bp )| n−1 The isotropy constant of hyperplane projections of Bpn , 1 < p ≤ 2. The first step is Cauchy’s formula: For good f we have Z Z 1 n f (x)dx = |∂Bp | f (PH (y ))|hN(y ), θi|dσpn (y ) 2 PH (Bpn ) ∂Bpn where N(y ) is the unit normal vector to ∂Bpn and σpn the normalized area measure. We apply it to f = 1 and f (y ) = |y |2 . The isotropy constant of hyperplane projections of Bpn , 1 < p ≤ 2. The first step is Cauchy’s formula: For good f we have Z Z 1 n f (x)dx = |∂Bp | f (PH (y ))|hN(y ), θi|dσpn (y ) 2 PH (Bpn ) ∂Bpn where N(y ) is the unit normal vector to ∂Bpn and σpn the normalized area measure. We apply it to f = 1 and f (y ) = |y |2 . Question. In order to extend the proof to lower dimensional projections, E , we need a formula for Z f (x)dx PE (Bpn ) The proof Fact 1 Naor, Romik 2003. The relation between two measures on ∂Bpn : σpn (the normalized area measure) and µnp (the cone probability measure) is given by n|Bpn | dσpn (x) = |5(k · kp )(x)| dµnp |∂Bpn | a.e. x ∈ ∂Bpn The proof Fact 1 Naor, Romik 2003. The relation between two measures on ∂Bpn : σpn (the normalized area measure) and µnp (the cone probability measure) is given by n|Bpn | dσpn (x) = |5(k · kp )(x)| dµnp |∂Bpn | a.e. x ∈ ∂Bpn Fact 2 Schechtmann, Zinn 1990. There is a representation of the cone measure µnp . Let g1 , . . . , gn be i.i.d. copies of a r.v. with density P p cp e −|t| , t ∈ R. Define S = ( ni=1 |gi |p )1/p . Then the probability determined by (g1 /S, . . . , gn /S) ∈ ∂Bpn is µnp . Moreover, (g1 /S, . . . , gn /S) is independent of S. By Cauchy’s formula and good f , Z Z 1 n f (x)dx = |∂Bp | f (PH (y ))|hN(y ), θi|dσpn (y ) 2 n n PH (Bp ) ∂Bp (N(y ) is the unit normal vector to ∂Bpn ) Z n = |Bpn | f (PH (y ))|h5(k · kp )(y ), θi|dµnp (y ) 2 n ∂Bp Z n X n n p−1 = |Bp | f (PH (y )) |yi | sgn(yi )θi dµnp (y ) 2 n ∂Bp i=1 Use it with f = 1 (for the volume) and f (y ) = |y |2 (and also |PH (y )| ≤ |y |). Now use the representation of dµnp . yi → gi S. 1 |PH (Bpn )| Z |x|2 dx ≤ PH (Bpn ) E |gi |2 i=1 S 2 Pn P n i=1 P E ni=1 P Since S = ( ni=1 |gi |p )1/p , ES p−1 c ≤ 2/p p+1 ES n |gi |p−1 sgn(gi )θi S p−1 = (by independence) Pn n ES p−1 X E|gi |2 i=1 |gi |p−1 sgn(gi )θi P = ES p+1 E | ni=1 |gi |p−1 sgn(gi )θi | i=1 |gi |p−1 sgn(gi )θi S p−1 n n X X |εi gi |p−1 εi sgn(gi )θi |gi |p−1 sgn(gi )θi = Eε Eg E i=1 i=1 ≥ (by Khinchine’s inequality) !1/2 n X 2p−2 2 ≥ C Eg |gi | θi i=1 ≥ (by Jensen’s inequality) n X ≥C E |gi |p−1 θi2 = C E|g |p−1 ≥ C > 0 i=1 With the same argument, n X p−1 E|gi | |gi | sgn(gi )θi ≤ C 2 i=1 The isotropy constant of hyperplane projections of Bpn , 1 < p ≤ 2. Putting all estimates together, 1 |PH (Bpn )| 2 By the same argument, |PH (Bpn )| n−1 ≥ (n − 1)L2PH (Bpn ) ≤ 1 2 |PH (Bpn )| n−1 · Z |x|2 dx ≤ C PH (Bpn ) C n2/p n n2/p and so, 1 |PH (Bpn )| Z PH (Bpn ) |x|2 dx ≤ Cn The isotropy constant of hyperplane projections of Bpn , 1 < p ≤ 2. 1 Putting all estimates together, |PH (Bpn )| 2 By the same argument, |PH (Bpn )| n−1 ≥ (n − 1)L2PH (Bpn ) ≤ 1 2 |PH (Bpn )| n−1 Z |x|2 dx ≤ C PH (Bpn ) C n2/p n n2/p and so, 1 · |PH (Bpn )| Z |x|2 dx ≤ Cn PH (Bpn ) Question. In order to extend the proof to lower dimensional projections, E , we need a formula for Z f (x)dx PE (Bpn )