![High-dimensional Convex Bodies: A geometric and a probabilistic viewpoint
Kaifeng Chen, Faculty Advisor: Beatrice-Helen Vritsiou
Illinois Geometry Lab
Geometry Labs United (GLU) Conference, Aug 28-30, 2015
Some definitions
Convex Body: convex, compact set with non-empty
interior
Random vector X = (X1, . . . , Xn) distributed uniformly in
convex body K means P(X ∈ A) =
vol(K ∩ A)
vol(K)
1-dimensional marginals of X:
Can be thought of as projections of X onto 1-dim subspaces, that
is lines.
Are of the form X, θ where θ is a unit vector in Rn
and where
P( X, θ ∈ B) =
B
vol(K ∩ (tθ + θ⊥
))
vol(K)
dt.
Thus the distribution function P(| X, θ | ≤ t) of the marginals is
exactly the portion of volume of the body found between two
hyperplanes perpendicular to θ and at height t and −t from the
origin.
One innocent-looking, but surprisingly hard question
HYPERPLANE CONJECTURE: ∃ a constant c > 0 such
that, any convex body K of volume one (in any
dimension) has at least one hyperplane section
K ∩ (tθ + θ⊥
) with volume ≥ c.
Easier to state the conjecture (see next column) when
the body K is in isotropic position: this means that the
centre of mass of K is at the origin and that all
1-dimensional marginals have the same variance, i.e.
K
x, θ 2
dx =
K
x2
1 dx = L2
K .
WHY CAN WE DO THAT?: Every convex body K of
volume 1 can be made isotropic by a translation and a
volume-preserving linear transformation.
Hyperplane Conjecture: An equivalent formulation
In isotropic position every central hyperplane section (that is, every hyperplane section passing through
the origin) has roughly the same ((n − 1)-dimensional) volume, 1/LK . Thus the question can be restated
as: does ∃ a constant C such that, for every dimension n and for every isotropic convex body K in Rn
, LK ≤ C?
Known bounds: LK ≤ C 4
√
n log n (Bourgain, 1987), LK ≤ C 4
√
n (Klartag, 2006).
Some examples of gaussian-like behaviour
n-dimensional cube Qn = [−1/2, 1/2]n
A uniform random vector X on Qn has n independent and identically distributed coordinates, each with mean 0
and variance 1/12. Thus, by the Central Limit Theorem, if θ = ( 1√
n
, . . . , 1√
n
), for all t > 0
P( X, θ ≤ t) ∼
√
6
√
π
t
−∞
exp(−6s2
) ds.
In fact, by refinements of CLT, the above is true for “most” unit directions θ (with respect to the uniform
measure on the unit sphere).
Euclidean ball Dn of radius
√
n + 2
The random vector X ∼ Unif(Dn) has identically distributed coordinates of mean 0 and variance 1, but they
are not independent.
However, for every θ the marginal distribution function P( X, θ ≤ t) (which is the volume of the ball found
below a hyperplane ⊥ θ at height t, see 1st Figure) is the same, can be computed exactly, and is very close to
the standard gaussian distribution N(0, 1).
Here we have a Central Limit-type Theorem without independence because of convexity.
Central Limit Problem for Convex Bodies
Again more convenient to place the bodies K in isotropic position (this time so that all 1-dimensional marginals
are of mean 0 and have the same variance, = L2
K ).
QUESTION: We want a sequence n 0 such that, given an isotropic convex body K ⊂ Rn
,
sup
t>0
P( X, θ ≤ t) −
1
√
2πLK
t
−∞
e−s2
/2L2
K ds) < n
for most unit directions θ.
The problem was solved by Klartag in 2007 (estimates for n also by Fleury-Gu´edon-Paouris, Fleury,
Gu´edon-Milman). Best known estimate: n ∼
(log n)1/3
n1/6
(Gu´edon-Milman, 2011).
Problem equivalent to a sort of volume concentration, or else:
Concentration Hypothesis for the Euclidean norm X 2 of X
Does ∃ n 0 such that, for every isotropic convex body K ⊂ Rn
and X ∼ Unif(K),
P X 2 −
√
nLK ≥ n
√
nLK ≤ n?
Another form of the question (slightly stronger): Bound well
σ2
K = Var( X 2) E ( X 2 −
√
nLK )2
.
What we currently know
Large Deviation Estimates
Prob({x ∈ K : x 2 ≥ 2t
√
nLk}) ≤ exp(−t
√
n)
for every t ≥ 1 (Paouris, 2006).
Thus the part of the body below that is found outside the
outer light blue circle with radius 2
√
nLK has
exponentially small volume.
Small Ball Estimates
Prob({x ∈ K : x 2 ≤ 0.5
√
nLk}) ≤ exp −log
c0 √
n
for every ∈ (0, 1) (Paouris, 2012).
Thus the part of the body that is found inside the inner
light blue circle with radius 0.5
√
nLK also has
exponentially small volume.
What the Thin-Shell Conjecture says
The part of the body found outside the dark blue ring
also has negligible volume, but now the width of the ring
is much smaller than its average radius (which is
√
nLK ).
By the currently known estimates for σK , the width of the
dark blue ring can be chosen n1/3
.
If σ2
K ≤ C, then the width of the blue ring will be at most
(log n)2
LK (much, much smaller than
√
nLK !!!).
FUN BONUS FACT
If the thin-shell conjecture is true for all isotropic
convex bodies K ⊂ Rn
, then the hyperplane conjecture
is true for all isotropic convex bodies K ⊂ Rn
:
sup
K⊂Rn
isotropic
LK ≤ C sup
K⊂Rn
isotropic
σK
These posters are made with the support of University of Illinois at Urbana-Champaign Public Engagement Office](https://image.slidesharecdn.com/c8f36ea0-14fe-4395-9ff4-d3b05cd61fec-150907005429-lva1-app6892/85/Kaifeng_final-version1-1-320.jpg)
Kaifeng_final version1
- 1. High-dimensional Convex Bodies: A geometric and a probabilistic viewpoint Kaifeng Chen, Faculty Advisor: Beatrice-Helen Vritsiou Illinois Geometry Lab Geometry Labs United (GLU) Conference, Aug 28-30, 2015 Some definitions Convex Body: convex, compact set with non-empty interior Random vector X = (X1, . . . , Xn) distributed uniformly in convex body K means P(X ∈ A) = vol(K ∩ A) vol(K) 1-dimensional marginals of X: Can be thought of as projections of X onto 1-dim subspaces, that is lines. Are of the form X, θ where θ is a unit vector in Rn and where P( X, θ ∈ B) = B vol(K ∩ (tθ + θ⊥ )) vol(K) dt. Thus the distribution function P(| X, θ | ≤ t) of the marginals is exactly the portion of volume of the body found between two hyperplanes perpendicular to θ and at height t and −t from the origin. One innocent-looking, but surprisingly hard question HYPERPLANE CONJECTURE: ∃ a constant c > 0 such that, any convex body K of volume one (in any dimension) has at least one hyperplane section K ∩ (tθ + θ⊥ ) with volume ≥ c. Easier to state the conjecture (see next column) when the body K is in isotropic position: this means that the centre of mass of K is at the origin and that all 1-dimensional marginals have the same variance, i.e. K x, θ 2 dx = K x2 1 dx = L2 K . WHY CAN WE DO THAT?: Every convex body K of volume 1 can be made isotropic by a translation and a volume-preserving linear transformation. Hyperplane Conjecture: An equivalent formulation In isotropic position every central hyperplane section (that is, every hyperplane section passing through the origin) has roughly the same ((n − 1)-dimensional) volume, 1/LK . Thus the question can be restated as: does ∃ a constant C such that, for every dimension n and for every isotropic convex body K in Rn , LK ≤ C? Known bounds: LK ≤ C 4 √ n log n (Bourgain, 1987), LK ≤ C 4 √ n (Klartag, 2006). Some examples of gaussian-like behaviour n-dimensional cube Qn = [−1/2, 1/2]n A uniform random vector X on Qn has n independent and identically distributed coordinates, each with mean 0 and variance 1/12. Thus, by the Central Limit Theorem, if θ = ( 1√ n , . . . , 1√ n ), for all t > 0 P( X, θ ≤ t) ∼ √ 6 √ π t −∞ exp(−6s2 ) ds. In fact, by refinements of CLT, the above is true for “most” unit directions θ (with respect to the uniform measure on the unit sphere). Euclidean ball Dn of radius √ n + 2 The random vector X ∼ Unif(Dn) has identically distributed coordinates of mean 0 and variance 1, but they are not independent. However, for every θ the marginal distribution function P( X, θ ≤ t) (which is the volume of the ball found below a hyperplane ⊥ θ at height t, see 1st Figure) is the same, can be computed exactly, and is very close to the standard gaussian distribution N(0, 1). Here we have a Central Limit-type Theorem without independence because of convexity. Central Limit Problem for Convex Bodies Again more convenient to place the bodies K in isotropic position (this time so that all 1-dimensional marginals are of mean 0 and have the same variance, = L2 K ). QUESTION: We want a sequence n 0 such that, given an isotropic convex body K ⊂ Rn , sup t>0 P( X, θ ≤ t) − 1 √ 2πLK t −∞ e−s2 /2L2 K ds) < n for most unit directions θ. The problem was solved by Klartag in 2007 (estimates for n also by Fleury-Gu´edon-Paouris, Fleury, Gu´edon-Milman). Best known estimate: n ∼ (log n)1/3 n1/6 (Gu´edon-Milman, 2011). Problem equivalent to a sort of volume concentration, or else: Concentration Hypothesis for the Euclidean norm X 2 of X Does ∃ n 0 such that, for every isotropic convex body K ⊂ Rn and X ∼ Unif(K), P X 2 − √ nLK ≥ n √ nLK ≤ n? Another form of the question (slightly stronger): Bound well σ2 K = Var( X 2) E ( X 2 − √ nLK )2 . What we currently know Large Deviation Estimates Prob({x ∈ K : x 2 ≥ 2t √ nLk}) ≤ exp(−t √ n) for every t ≥ 1 (Paouris, 2006). Thus the part of the body below that is found outside the outer light blue circle with radius 2 √ nLK has exponentially small volume. Small Ball Estimates Prob({x ∈ K : x 2 ≤ 0.5 √ nLk}) ≤ exp −log c0 √ n for every ∈ (0, 1) (Paouris, 2012). Thus the part of the body that is found inside the inner light blue circle with radius 0.5 √ nLK also has exponentially small volume. What the Thin-Shell Conjecture says The part of the body found outside the dark blue ring also has negligible volume, but now the width of the ring is much smaller than its average radius (which is √ nLK ). By the currently known estimates for σK , the width of the dark blue ring can be chosen n1/3 . If σ2 K ≤ C, then the width of the blue ring will be at most (log n)2 LK (much, much smaller than √ nLK !!!). FUN BONUS FACT If the thin-shell conjecture is true for all isotropic convex bodies K ⊂ Rn , then the hyperplane conjecture is true for all isotropic convex bodies K ⊂ Rn : sup K⊂Rn isotropic LK ≤ C sup K⊂Rn isotropic σK These posters are made with the support of University of Illinois at Urbana-Champaign Public Engagement Office