Approximately gaussian marginals and the hyperplane conjecture

Introduction

Little is currently known about the uniform measure on a general high-dimensional convex body. Many aspects of the Euclidean ball or the unit cube are easy to analyze, yet it is difficult to answer even some of the simplest questions regarding arbitrary convex bodies, lacking symmetries and structure. For example,

Here, of course, $Vol_{k}$ stands for $k$-dimensional volume. A convex body is a bounded, open convex set. Question 1.1 is referred to as the “slicing problem” or the “hyperplane conjecture”, and was raised by Bourgain in relation to the maximal function in high dimensions. It was demonstrated by Ball that Question 1.1 and similar questions are most naturally formulated in the broader class of logarithmically concave densities.

where $f$ is the log-concave density of $\mu$. It is well-known (see, e.g., [17, Lemma 3.1]) that $L_{f}>c$, for some universal constant $c>0$. Define

for a universal constant $C>0$. Again, up to a universal constant, one may restrict attention in (2) to random vectors that are distributed uniformly in centrally-symmetric convex bodies. This essentially follows from the same technique as in the case of the parameter $L_{n}$ mentioned above.

The importance of the parameter $\sigma_{n}$ stems from the central limit theorem for convex bodies . This theorem asserts that most of the one-dimensional marginals of an isotropic, log-concave random vector are approximately gaussian. The Kolmogorov distance to the standard gaussian distribution of a typical marginal has roughly the order of magnitude of $\sigma_{n}/\sqrt{n}$. Therefore, the conjectured bound (3) actually concerns the quality of the gaussian approximation to the marginals of high-dimensional log-concave measures. Our main result reads as follows:

Inequality 1.1 may be sharpened, in view of Lemma 1.3, to the bound

for a universal constant $C>0$. This is explained in the proof of Inequality 1.1 in Section 3. Our argument involves a certain Riemannian structure, which is presented in Section 2.

We write $\nabla\varphi$ for the gradient of the function $\varphi$, and $\nabla^{2}\varphi$ for the hessian matrix. For $\theta\in S^{n-1}$ we write $\partial_{\theta}$ for differentiation in direction $\theta$, and $\partial_{\theta\theta}(\varphi)=\partial_{\theta}(\partial_{\theta}\varphi)$.

Acknowledgements. We would like to thank Daniel Dadush, Vitali Milman, Leonid Polterovich, Misha Sodin and Boris Tsirelson for interesting discussions related to this work, and to Shahar Mendelson for pointing out that there is a difference between extremal points and exposed points.

A Riemannian metric associated with a convex body

where $b(\mu_{\xi})$ is the barycenter of the probability measure $\mu_{\xi}$ and $Cov(\mu_{\xi})$ is the covariance matrix. We learned the following lemma from Gromov’s work . A proof is provided for the reader’s convenience.

Suppose $X=(U,g,\Psi,x_{0})$ and $Y=(V,h,\Phi,y_{0})$ are Riemannian packages. A map $\varphi:U\rightarrow V$ is an isomorphism of $X$ and $Y$ if the following conditions hold:

$\varphi$ is a Riemannian isometry between the Riemannian manifolds $(U,g)$ and $(V,h)$.

$\Phi(\varphi(x))=\Psi(x)$ for any $x\in U$.

When such an isomorphism exists we say that $X$ and $Y$ are isomorphic, and we write $X\cong Y$.

Let us describe an additional construction of the same Riemannian package associated with $\mu$, a construction which is dual to the one mentioned above. Consider the Legendre transform

Then $h_{\mu}$ is a Riemannian metric on $K$. Note the identity

A moment of reflection reveals that the definition (12) of the positive-definite bilinear form $g_{\xi}$ is equivalent to the definition (10) given above. Additionally, there exists a linear operator $A_{\xi}:V^{*}\rightarrow V^{*}$, which is self-adjoint and positive-definite with respect to the bilinear form $g_{0}$, that satisfies

Hence we may define $\Psi(\xi)=\log\det A_{\xi}$, which coincides with the definition (11) of $\Psi_{\mu}$ above. Therefore, $X_{\mu}=(V^{*},g,\Psi,0)$ is the Riemannian package associated with $\mu$. Back to the lemma, we see that $X_{\mu}$ is constructed from exactly the same data as $X_{\nu}$, hence they must be isomorphic. $\square$

Proof: The only difference from Lemma 2.3 is that the map $T$ is assumed to be affine, and not linear. It is clearly enough to deal with the case where $T$ is a translation, i.e.,

Suppose $X=(U,g,\Psi,\xi_{0})$ is an $n$-dimensional Riemannian package of log-concave type. Let $\xi_{1}\in U$. Denote

In order to prove the lemma, we need to demonstrate that

In order to see that $\varphi$ is indeed an isomorphism, note that (15) yields

Inequalities

This proves the inequality on the left in (5). Regarding the inequality on the right, we use the bound

which follows from Paouris theorem . Here $1_{|X|>C\sqrt{n}}$ is the random variable that equals one when $|X|>C\sqrt{n}$ and vanishes otherwise. Apply again the identity $|X|^{2}-n=(|X|-\sqrt{n})(|X|+\sqrt{n})$ to conclude that

Suppose $X=(U,g,\Psi,\xi_{0})$ is an $n$-dimensional Riemannian package of log-concave type. Then, for any $\xi\in U$,

Proof: Suppose first that $\xi=\xi_{0}$. We need to establish the bound

Since $\mu$ is isotropic, $\nabla^{2}\Lambda_{\mu}(0)=Cov(\mu)=Id$, where $Id$ is the identity matrix. Consequently, the desired bound (20) is equivalent to

A straightforward computation shows that $\partial_{\theta}\log\det\nabla^{2}\Lambda_{\mu}(\xi)$ equals the trace of the matrix $\left(\nabla^{2}\Lambda_{\mu}(\xi)\right)^{-1}\nabla^{2}\partial_{\theta}\Lambda_{\mu}(\xi)$. Since $\mu$ is isotropic,

where $d(\xi,\eta)$ is the Riemannian distance between $\xi$ and $\eta$, with respect to the Riemannian metric $g_{\mu}$. In particular, when the barycenter of $\mu$ lies at the origin,

Proof: The bound (21) is obvious when $\xi=\eta$. When $\xi\neq\eta$, we need to exhibit a path from $\eta$ to $\xi$ whose Riemannian length is at most the expression on the right in (21). Set $\theta=(\xi-\eta)/|\xi-\eta|$ and $R=|\xi-\eta|$. Consider the interval

This path connects $\eta$ and $\xi$, and its Riemannian length is

according to the Cauchy-Schwartz inequality. Clearly, $\int_{0}^{R}dt/(2R-t)=\log 2\leq 1$. Regarding the other integral, recall Taylor’s formula with integral remainder:

The inequality (21) is thus proven. Furthermore, $\Lambda(0)=0$, and when the barycenter of $\mu$ lies at the origin, also $\nabla\Lambda(0)=0$. Thus (22) follows from (21). $\square$

Proof: It suffices to prove the lemma under the additional assumption that $t$ is an integer. According to Lemma 22,

Note that the restriction of $\Lambda$ to the subspace $E$ is the logarithmic Laplace transform of $(Proj_{E})_{*}\mu$. It is proven in [17, Lemma 2.8] that

Since $K_{t}\subseteq B_{t}$, the bound (23) follows. $\square$

Since $\Psi_{\mu}(\xi)=\log\det\nabla^{2}\Lambda_{\mu}(\xi)$ and $\Psi_{\mu}(0)=0$, then

as $\Lambda_{\mu}$ is convex and hence $\det\nabla^{2}\Lambda_{\mu}(\xi)\geq 0$ for all $\xi$. Lemma 3.3 yields that

thanks to Lemma 3.4. In view of (1), the bound $L_{n}\leq C\underline{\sigma}_{n}$ is proven. The desired inequality (4) now follows from Lemma 1.3. $\square$

where $\sigma_{n-1}$ is the uniform Lebesgue probability measure on the sphere $S^{n-1}$, and $C>0$ is a universal constant.

is a homogenous, harmonic polynomial of degree three. In other words, the restriction $F|_{S^{n-1}}$ decomposes into spherical harmonics as

Since spherical harmonics of different degrees are orthogonal to each other,

admits tight concentration bounds. For instance,

References