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 )