Generating an equidistributed net on a unit n-sphere using random rotations

TL;DR

This paper introduces a randomized algorithm that efficiently generates an $ ext{epsilon}$-net on an n-sphere, enabling accurate approximation of integrals of Lipschitz functions with high probability.

Contribution

It presents a novel randomized method for constructing an $ ext{epsilon}$-net on the sphere using a specific number of random rotations and word combinations, ensuring equidistribution.

Findings

Algorithm succeeds with high probability

Produces an $ ext{epsilon}$-net with controlled size

Effective for approximating integrals of Lipschitz functions

Abstract

We develop a randomized algorithm (that succeeds with high probability) for generating an $\epsilon$-net in a sphere of dimension n. The basic scheme is to pick $O(n \ln(1/n) + \ln(1/\delta))$ random rotations and take all possible words of length $O(n \ln(1/\epsilon))$ in the same alphabet and act them on a fixed point. We show this set of points is equidistributed at a scale of $\epsilon$. Our main application is to approximate integration of Lipschitz functions over an n-sphere.

Equations

$$\mathbb{E}\left[\begin{array}{c}|{\lambda }_2^{\ast }(\chi (G,S))|\end{array}\right]<\delta .$$

$${\left(\begin{array}{c}\frac{2\ln 2}{ϵ}+o(1)\end{array}\right)}^2\log |D|$$

$$(2|S|{)}^{-1}\sum _{s\in S}(R(s)+R(s^{-1})),$$

$$z_{\mu }(f)(x)={\int }_{G}f(xg)d\mu (g)$$

$$k\ge 8\ln 2\left(\begin{array}{c}(n+4)+2\ln \left(\begin{array}{c}\frac{1}{\delta }\end{array}\right)+6n(1+a_n)\ln \left(\begin{array}{c}\frac{1}{ϵ}\end{array}\right)-\ln (n!)\end{array}\right),$$

$$l=\frac{n}{2}{\log }_2\left(\begin{array}{c}\frac{1}{rϵ}\end{array}\right)+(4+3a_n){\log }_2\left(\begin{array}{c}\frac{1}{ϵ}\end{array}\right).$$

$$W_1(\sigma ,\nu )\le ϵ,$$

$$\sigma (B)=\frac{\lambda (\hat{B})}{\lambda (D)},$$

$${\Omega }_n:=\lambda (S^n)=\frac{2{\pi }^{\frac{n+1}{2}}}{\Gamma \left(\begin{array}{c}\frac{n+1}{2}\end{array}\right)}.$$

$$-\Delta (g)={\Delta }_{{\mathbb{R}}^n}(\tilde{g}){\mid }_{S^n}$$

$$L^2(S^n)=\underset{k=0}{\overset{\infty }{⨁}}{H}_k(S^n)$$

$$h_k:=\left(\begin{array}{c}n+k\\ n\end{array}\right)-\left(\begin{array}{c}n+k-2\\ n\end{array}\right)$$

$$\sum _{a=0}^k\dim H_a(S^n)$$

$$=\left(\begin{array}{c}n+k\\ k\end{array}\right)+\left(\begin{array}{c}n+k-1\\ k-1\end{array}\right)$$

$$=\dim H_k(S^{n+1})$$

$$\underset{k\to \infty }{\lim }\frac{\sum _{k\le a\le k\sqrt[n]{2}}\dim H_a(S^n)}{k^n}$$

$$=2\underset{l\to \infty }{\lim }\frac{\sum _{0\le a\le l}\dim H_a}{l^n}-\underset{k\to \infty }{\lim }\frac{\sum _{0\le a\le k}\dim H_a}{k^n}$$

$$=\underset{k\to \infty }{\lim }\frac{\sum _{0\le a\le k}\dim H_a}{k^n}$$

$$=\underset{k\to \infty }{\lim }\frac{\dim H_k(S^{n+1})}{k^n}$$

$$=\underset{k\to \infty }{\lim }{k}^{-n}\left(\begin{array}{c}n+k\\ k\end{array}\right)+\underset{k\to \infty }{\lim }{k}^{-n}\left(\begin{array}{c}n+k-1\\ k-1\end{array}\right)$$

$$=\frac{2}{n!}$$

$${\left(\begin{array}{c}n+k\\ k\end{array}\right)}^2$$

$$\text{mbox}AM\ge \text{mbox}GM\Rightarrow$$

$$<\frac{2^n{k}^{2n}}{(n!{)}^2}$$

$$\underset{\lambda \to \infty }{\lim }\frac{\sum _{\lambda \le a\le \lambda \sqrt[n]{4}}\dim H_a}{{\lambda }^{\frac{n}{2}}}$$

$$\frac{\sum _{\lambda \le a\le \lambda \sqrt[n]{4}}\dim H_a}{{\lambda }^{\frac{n}{2}}}$$

$$\frac{\partial u}{\partial t}$$

$$\underset{t\to 0^{+}}{\lim }u(t,x)$$

$$H_t(x)$$

$$=\sum _{k=0}^{\infty }{e}^{-{\lambda }_{k}t}{h}_k{P}_{k,n}(x\cdot x_0)$$

$$P_{k,n}(t)=\sum _{j=0}^{\lfloor \frac{k}{2}\rfloor }{C}_{2j}{t}^{k-2j}(1-t^2{)}^j$$

$$C_0=1,C_{2j}=(-1{)}^j\frac{k(k-1)\cdots (k-2j+1)}{(2\cdot 4\cdots 2j)(n(n+2)\cdots (n+2j-2))}$$

$${\int }_{S^n}{H}_t(x{)}^{2}d\sigma (x)$$

$$=\sum _{k=0}^{\infty }{e}^{-2{\lambda }_{k}t}{h}_k$$

$$H(t,x,x)=\sum _{k=0}^{\infty }{e}^{-{\lambda }_{k}t}{h}_k$$

Full text

1. Introduction

In the present article, we develop a randomized algorithm (with high success probability) for the generation of an $\epsilon$-net in an unit sphere of dimension $n$. The basic scheme is to pick $k:={O}(n\ln(1/\epsilon)+\ln(1/\delta))$ random rotations and take all possible words on length $l:={O}(n\ln(1/\epsilon))$ in the same alphabet and act them on a fixed point. We show this set of points is equidistributed at a scale of $\epsilon$. The group $SU(2)$ can be identified with the three dimensional sphere $S^{3}$, thus we obtain a scheme for producing an $\epsilon$-net of $SU(2)$, a task which is relevant to quantum computing in the context of the Solovay-Kitaev algorithm (although we do not address the issue of generating a sequence of elementary gates to efficiently approximate any given gate, in a non-exhaustive fashion). Our main application is to integration of Lipschitz functions over an $n$-sphere. The net we produce, is $\epsilon$-close in Hausdorff distance to the $n$-sphere, and is also equidistributed in the following sense (with high probability): The uniform counting measure $\mu_{\operatorname{net}}$ over the net is close to the uniform measure on the $n$-sphere $\mu_{\operatorname{haar}}$ in $1$-Wasserstein distance. This implies that the integral of every $1$-Lipschitz function on $S^{n}$ with respect to $\mu_{\operatorname{net}}$ is $\epsilon$-close to its integral with respect to $\mu_{\operatorname{haar}}$.

In [2], Alon and Roichman proved that, given any $\delta>0$, there exists a $c(\delta)>0$ such that for any finite group $G$, and a random subset $S\subset G$ of order at least $c(\delta)\log|G|$, the induced Cayley graph $\chi(G,S)$ has small normalized second largest eigenvalue (in absolute value):

[TABLE]

Considering random walk on expander multigraphs, it follows that every element $g\in G$ is an $S$-word of length at most $\log|G|$. For an irreducible representation $\rho\in{\hat{G}}$, let $d_{\rho}$ be its dimension; let $R$ be the regular representation of $G$, and $D=\sum_{\rho\in{\hat{G}}}d_{\rho}$. In [13], Russel and Landau proved that (1.1) holds for all random subsets $S\subset G$ of order at least

[TABLE]

This was obtained via an application of tail bounds for operator-valued random variables, as in Ahlswede and Winter [1], building upon the following observation: the normalized adjacency matrix of $\chi(G,S)$ is the operator

[TABLE]

presented in terms of the standard basis of $\mathbb{C}[G]$.

Now let $G$ be a compact Lie group, and $\mu$ a left-invariant Borel probability measure on $G$. One considers the averaging operator $z_{\mu}:L^{2}(G)\rightarrow L^{2}(G)$, given by

[TABLE]

In [5], Bourgain and Gamburd established that if $G=SU(d)$ then $z_{\mu}$ has spectral radius $<1$ when $\mbox{supp}(\mu)$ is finite algebraic subset generating a nonabelian free subgroup of $G$. This has since been extended to all compact connected simple Lie groups by Benoist and de Saxcé in [3], where it was shown that $z_{\mu}$ has a spectral gap if and only if $\mu$ is almost diophantine. A corollary to the main result in [3] is the following: if $\mu$ is finitely-supported almost diophantine then the set of words in $\mbox{supp}(\mu)$ of fixed length approaches $G$ in Hausdorff distance.

In [15], a quantitative version of the spectral gap question was considered. It was shown that the Hausdorff distance between $G$, a compact connected Lie group, and the subset of fixed length words on a random essentially small finite alphabet $S\subset G$ decays exponentially in the length of the words, with high probability. This was done via an analysis of the heat kernel with respect to a suitable finite dimensional subspace of $L^{2}(G)$ and an application of tail bounds for operator-valued random variables. We note that the results of the present article are not implied by the results of [15], because the dimension of the Lie group $SO_{n}$ is $n(n+1)/2$, and so the bounds from [15] for the length of the words and the number of generators, that apply for general compact Lie groups would be quadratic in $n$ rather than linear in $n$. In the special case of the unitary group $U_{n}$, such a result with a quadratic dependence on dimension for the length of the words $n$ was previously obtained by Hastings and Harrow in Theorem 5 of [10], however in their result the number of generators is specified in a indirect manner, whose dependence on $n$ is not obvious. On the other hand, the bounds obtained in the present work are linear in $n$, both for the number of generators and the length of the words. In fact, for these parameters, the value of $(2k)^{l}$ is close to the volumetric lower bound of $(1/\epsilon)^{\Omega(n)}$ on the size of an $\epsilon-$net of $S^{n}$.

The two main results of this paper are stated below. The numberings correspond to their appearances in Sections $3$ and $4$ respectively. In the following statements, $C_{n}$ denotes a certain positive constant depending on $n$. For a finite set $S$, the $l$-fold product $S^{l}$ consists of all $S$-words of length $l$ inside the free group generated by $S$.

**Theorem 3.15:

Let $\epsilon\in(0,{\frac{1}{3n}})$ and $\delta\in(0,1)$; let $r=2\epsilon\sqrt{\ln{\frac{3C_{n}}{\epsilon^{2n-1}}}}$. Let $S\subset SO_{n+1}$ consist of $k$ iid random points, drawn from the Haar measure on $SO_{n+1}$, where**

[TABLE]

and $a_{n}:={\frac{2\log_{2}\log_{2}(5n)}{\log_{2}(5n)}}$. Let ${\hat{S}}:=S\sqcup S^{-1}$ be the (multi)set of all elements in $S$ and their inverses. Let $l={\frac{n}{2}}\log_{2}\begin{pmatrix}{\frac{1}{r\epsilon}}\end{pmatrix}+(4+3a_{n})n\log_{2}\begin{pmatrix}{\frac{1}{\epsilon}}\end{pmatrix}$; if $r$ is sufficiently small then the probability that $x_{0}{\hat{S}}^{l}\subseteq S^{n}$ is an $r$-net in $S^{n}$ is at least $1-\delta$. $\Box$

**Theorem 4.14:

For $n>1$, let $\sigma$ be the probability measure on $S^{n}$ corresponding to haar probability measure on the group of rotations $SO_{n+1}$. Let $\epsilon,\delta>0$ be sufficiently small and $r=2\epsilon\sqrt{\ln{\frac{3C_{n}}{\epsilon^{2n-1}}}}$. Let $S\subseteq SO_{n+1}$ be a random subset such that $|S|=k$ satisfies**

[TABLE]

where $a_{n}:={\frac{2\log_{2}\log_{2}(5n)}{\log_{2}(5n)}}$. Let ${\hat{S}}:=S\sqcup S^{-1}$ be the (multi)set of all elements in $S$ and their inverses. Let $x_{0}\in S^{n}$ and let $\nu$ be the probability measure on $S^{n}$, uniformly supported on ${\hat{S}}^{l}x_{0}$, where

[TABLE]

If $r$ is sufficiently small, then the following inequality holds with probability at least $1-\delta$:

[TABLE]

Here $W_{1}$ is used to denote the $1-$Wasserstein distance between two measures supported on $S^{n}$. $\Box$

For the remainder of this section, we assume given a real number model of computation, in which only standard algebraic operations are allowed on Gaussian random vectors, but bits are not manipulated. Thus, for $k$ and $l$ as described in Theorem 4.14, choose a set $S$ consisting of $k$ orthogonal matrices, each chosen independently from the Haar measure of $SO_{n+1}$. Consider all $(2k)^{l}$ words of length $l$ in these generators and their inverses. Apply the resulting matrices to the vector ${\bf e}_{n+1}=(0,0,\cdots,0,1)^{T}$. Then these $(2k)^{l}$ points form an equidistributed net, that can be used for integrating a $1-$Lipschitz to within an additive error of $\epsilon$. If we assume an oracle that outputs independent $n-$dimensional Gaussian random vectors when queried, then the whole process requires only $kn$ queries to this oracle. Note that the obvious procedure of producing an equidistributed net, would require $\epsilon^{-\Omega(n)}$ calls to the Gaussian oracle (which would then be normalized to lie on the sphere). The latter method uses exponentially more randomness than our procedure using random rotations.

Acknowledgments

SC would like to thank Sandeep Juneja and Jaikumar Radhakrishnan for helpful conversations. HN was partially supported by a Ramanujan fellowship.

2. Application of Spherical Harmonics

This section briefly reviews the basics of harmonic analysis on the unit $n$-sphere $S^{n}\subset\mathbb{R}^{n+1}$. The two lemmas in this section will be used in an essential manner in deriving the computations in this section.

Let $\sigma$ denote the standard euclidean surface probability measure on $S^{n}$. For any Borel set $B\subset S^{n}$, if ${\hat{B}}:=\{\alpha x:x\in B,\alpha\in[0,1]\}$ then

[TABLE]

where $\lambda$ is the standard Lebesgue measure on ${\mathbb{R}}^{n+1}$ and $D\subseteq{\mathbb{R}}^{n+1}$ is the unit disk centered at origin, so that $S^{n}=\partial D$. We recall that the Lebesgue measure of the unit $n$-sphere in ${\mathbb{R}}^{n}$ is

[TABLE]

Now, the fact that $\sigma$ is $SO_{n+1}$- invariant follows from usual rotation invariance of Lebesgue measure on ${\mathbb{R}}^{n+1}$. Hence (by unimodularity of the compact Lie group $SO_{n+1}$ of rotations of $S^{n}$), the measure $\sigma$ is the unique probability measure on $S^{n}$ induced by the haar measure on $SO_{n+1}$. For $n>1$, let $\Delta:=\Delta_{S^{n}}$ be the negative of the Laplace-Beltrami operator on $S^{n}$. Thus, given $g\in C^{2}(S^{n})$, one has

[TABLE]

where ${\tilde{g}}:{\mathbb{R}}^{n+1}-\{0\}\rightarrow{\mathbb{R}}$ is defined by ${\tilde{g}}(x)=g(|x|^{-1}x)$. It is well-known that the Hilbert-product space $L^{2}(S^{n})$ decomposes into a direct sum of the eigenspaces of $\Delta$, in the sense that the $L^{2}$-closure of the direct sum is $L^{2}(S^{n})$:

[TABLE]

Recall that $H_{k}(S^{n})$ is the space of degree-$k$ homogeneous harmonic polynomials in $n+1$ variables, restricted to $S^{n}$; the dimension of $H_{k}(S^{n})$ is

[TABLE]

and the corresponding eigenvalue is $\lambda_{k}:=k(n+k-1)$. Note that, for any $n,k>0$, one has

[TABLE]

One has

[TABLE]

Moreover, notice that, when $k>2n+1$, one has

[TABLE]

Equivalently, writing $H_{\lambda}$ for the eigenspace corresponding to eigenvalue $\lambda$, one gets

[TABLE]

and for all $\lambda>6n^{2}+3n$, the inequality $\lambda_{ak}<a^{2}\lambda_{k}$ implies

[TABLE]

where, the sum ranges over all eigenvalues in $[\lambda,\lambda\sqrt[n]{4}]$. Fix a point $x_{0}\in S^{n}$. Let $H_{t}(x)$ be the heat kernel on $S^{n}$, corresponding to Brownian motion started at $x_{0}$. That is, $H_{t}(x)$ is the fundamental solution to the problem

[TABLE]

where the convergence is taken to be in the $\mbox{weak}^{\ast}$ topology. A Brownian motion on $S^{n}$, started at $x_{0}\in S^{n}$ has infinitesimal generator $H_{\frac{t}{2}}(x)$. Fixing orthonormal basis $\phi_{k,1},\cdots,\phi_{k,h_{k}}$ of $H_{k}:=H_{\lambda_{k}}$ for each $k\geq 0$, one has

[TABLE]

where the last equality is due to addition theorem of spherical harmonics (see theorem 2.26 of [14]) that correspond to the usual addition formula for trigonometric functions when $n=1$; here $P_{k,n}(t)$ denotes the Legendre polynomial of degree $k$ and dimension $n+1$; in explicit terms, this polynomial is

[TABLE]

where the coefficients are given by

[TABLE]

One has

[TABLE]

where the last equation is a well-known properties of Legendre polynomials (see theorem 2.29 of [14]). We note that by theorem 2.29 of [14], one has

[TABLE]

for all $t>0$. It is known that $H_{t}(x)>0$ for all $t>0$.

For $M>0$, let $H_{t,M}(x)$ be defined by

[TABLE]

Lemma 2.7**.**

Suppose that $t\in(0,6^{-1})$ and for any $\eta>0$, let $M\geq 4^{\frac{k_{0}}{n}}$ where

[TABLE]

then the following inequality holds:

[TABLE]

Proof.

When $k>n\log_{2}(3n)$, one has $4^{\frac{k}{n}}>6n^{2}+3n$. For $k\geq 0$, write

[TABLE]

For $k_{0}>n\log_{2}(3n)$, one has

[TABLE]

Suppose that an integer $k>k_{0}$ satisfies

[TABLE]

Then

[TABLE]

Consider the inequality

[TABLE]

By monotone property of the logarithm function, the following inequality is equivalent to (2.9) above:

[TABLE]

For $x\in(2^{e},+\infty)$, the function

[TABLE]

satisfies $0<g(x)<1$, has global minima $g(2^{e})=1-e^{-1}\log_{2}e>0.46$, and is increasing. Since $t\in(0,6^{-1})$ and $n>1$, the condition $n/t>2^{e}$ is satisfied; because $k\geq n\log_{2}(3n)$, the following inequality implies (2.10):

[TABLE]

We claim that, for $n>1$ and $t\in(0,6^{-1})$, the following inequality holds:

[TABLE]

This is equivalent to

[TABLE]

Writing $x={\frac{n}{t}}$, this is equivalent to $\sqrt{x}\geq\log_{2}x$. The function $h(x)=\sqrt{x}-\log_{2}x$ has derivative $h^{\prime}(x)={\frac{1}{2\sqrt{x}}}-{\frac{1}{x\ln 2}}$, which is increasing for $x>{\frac{4}{(\ln 2)^{2}}}$, and $h(12)>0$. This proves the claim.

Thus, $k_{0}=n\log_{2}\begin{pmatrix}{\frac{n}{t}}\end{pmatrix}+2n\log_{2}\log_{2}\begin{pmatrix}{\frac{n}{t}}\end{pmatrix}$ implies

[TABLE]

∎

Remark 2.12*.*

If $\epsilon\in(0,{\frac{1}{3n}})$ and $t=\epsilon^{2}$, one has

[TABLE]

We write

[TABLE]

Then the lemma above implies

[TABLE]

for $M=4^{\frac{k_{0}}{n}}$ where $k_{0}>\max\{\log_{2}\begin{pmatrix}{\frac{1}{\eta}}\end{pmatrix},{\frac{3n}{2}}\begin{pmatrix}1+a_{n}\end{pmatrix}\log_{2}\begin{pmatrix}{\frac{1}{t}}\end{pmatrix}\}$.

Lemma 2.13**.**

Let $c=\min\{\ln\sqrt{2},n^{-1}\}$. For all $t\in(0,c)$ and all $M>0$, following inequality holds:

[TABLE]

Proof.

For any $M>0$, addition theorem implies

[TABLE]

The function $\phi(x)=(x+a)te^{-2xt}$, for $x\in(0,\infty)$, satisfies

[TABLE]

which shows that $\phi(x)\leq(2e)^{-1}e^{2at}$. This yields

[TABLE]

If $t\in(0,\ln\sqrt{2})$ then $1-e^{-2t}>t$, which implies that, for $t\in(0,c)$ where $c=\min\{\ln\sqrt{2},n^{-1}\}$, and any $M>0$, one has

[TABLE]

∎

3. Hausdorff distance

We recall the following definition:

Definition 3.1**.**

Given a subset ${\hat{S}}\subset S^{n}$, and $\epsilon\geq 0$, let ${\hat{S}}_{\epsilon}$ be the union of all $\epsilon$-neighbourhoods of points in ${\hat{S}}$; the Hausdorff distance $d_{H}({\hat{S}},S^{n})$ is defined to be

[TABLE]

Our analysis in this section will be based on an application of the following theorem, first appeared in [1].

Theorem 3.2** (Ahlswede-Winter).**

Let $V$ be a finite dimensional Hilbert space, with $\dim V=D$. Let $A_{1},\cdots,A_{k}$ be independent identically distributed random variables taking values in the cone of positive semidefinite operators on $V$, such that $\mathbb{E}[A_{i}]=A\geq\mu I$ for some $\mu\geq 0$, and $A_{i}\leq I$. Then, for all $\epsilon\in[0,0.5]$, the following holds:

[TABLE]

Let $S\subset SO_{n+1}$ be a non-empty subset, with $|S|=k$. For $M>9n^{2}$, let

[TABLE]

Recall (inequality 2.3) that $\dim E_{M}\leq{\frac{2(2M)^{\frac{n}{2}}}{n!}}$. Because $\Delta:=\Delta_{S^{n}}$ is $SO_{n+1}$-invariant, the subspace $E_{M}$ is invariant under the operators

[TABLE]

Due to rotation invariance of the surface probability measure $\sigma$, the operators $A_{s}:E_{M}\rightarrow E_{M}$ turns out to be self-adjoint. Positive semidefiniteness of $A_{s}$ follows from the identity

[TABLE]

Moreover, writing $\mu$ for the unique right-invariant Haar (probability) measure on $SO_{n+1}$, one has

[TABLE]

Writing $\tau:SO_{n+1}\rightarrow S^{n}$ for the map $s\mapsto xs$, one has

[TABLE]

where $\tau_{\ast}\mu(E)=\mu(\tau^{-1}(E))$. Since $\tau_{\ast}\mu$ is rotation-invariant measure on $S^{n}$, one has $\sigma=\tau_{\ast}\mu$. Because $f\in E_{M}\subset L^{2}_{0}(S^{n})$, one has

[TABLE]

Therefore, $\begin{pmatrix}\mathbb{E}_{s\sim\mu}[A_{s}]\end{pmatrix}f(x)={\frac{1}{2}}f(x)$, which makes

[TABLE]

Furthermore, the operators $I-A_{s}$ are positive semidefinite for all $s\in SO(n+1)$, because

[TABLE]

Theorem 3.6**.**

Let $S\subset SO_{n+1}$ be a set of order $|S|=k$, chosen independently, and uniformly at random from the Haar measure on $SO_{n+1}$ and let ${\hat{S}}:=S\sqcup S^{-1}$ be the (multi)set of all elements in $S$ and their inverses. Let $\eta>0$ satisfy

[TABLE]

where $a_{n}:={\frac{2\log_{2}\log_{2}(5n)}{\log_{2}(5n)}}$. Let $t\in(0,c)$ where $c=\min\{{\frac{1}{6}},{\frac{1}{n}}\}$. For

[TABLE]

and any integer $l>0$ satisfying $2^{l}\geq{\frac{t^{-\frac{n}{2}}}{\eta}}$, the following inequality holds:

[TABLE]

Proof.

Since $t\in(0,c)$ and $c=\min\{{\frac{1}{6}},{\frac{1}{n}}\}$, one has $\eta^{-\frac{2}{n}}\geq 9n^{2}$. Setting $\epsilon=0.5$ and $M=\eta^{-\frac{2}{n}}$ in (3.3) yields

[TABLE]

Therefore, for all $f\in E_{M}$, the following inequality holds:

[TABLE]

In particular, writing ${\tilde{H}}_{t,M}:=1_{S^{n}}-H_{t,M}$, one has

[TABLE]

Iterating this inequality $l>0$ times fetches

[TABLE]

Let ${\tilde{H}}_{t}=H_{t}-1_{S^{n}}$; then

[TABLE]

Hence, using lemma 2.7 and 2.13 in inequality (3.8), one derives

[TABLE]

Since $H_{t}(xs)=H_{t}(x_{0}s^{-1},x)$ and ${\hat{S}}$ is inverse-symmetric, this produces (3.7). ∎

The following theorem has appeared in [16]:

Theorem 3.9** (Nowak-Sjögren-Szarek).**

Let $H(t,x_{0},x)$ be the heat kernel on the sphere $S^{n}$, corresponding to Brownian motion initiated at $x_{0}\in S^{n}$. Let $n\geq 1$ and fix $T>0$. Let $\phi(x):=\arccos\langle x,x_{0}\rangle$ be the Riemannian distance, so that $\phi(x)\in[0,\pi]$. Then, for all $0<t\leq T$, the inequality

[TABLE]

holds for some constants $c,C>0$ depending only on $n$ and $T$.

Let $\epsilon,\eta\in(0,1)$ and let $x\in S^{n}$ be such that $\phi(x)>2\epsilon\sqrt{\ln{\frac{1}{\eta\epsilon^{2n-1}}}}$. Then, taking $T=1$ in the above upperbound, one has

[TABLE]

Notice that the constant $C_{S^{n}}$ is independent of the initial point $x_{0}\in S^{n}$. Letting $C_{n}:=1+C_{S^{n}}$, we have

[TABLE]

Lemma 3.11**.**

Let $\epsilon\in(0,c)$ where $c=(n+4)^{-1}$. If $r=2\epsilon\sqrt{\ln{\frac{3C_{n}}{\epsilon^{2n-1}}}}$ is sufficiently small, then the following inequality implies that $x_{0}{\hat{S}}^{l}\subseteq S^{n}$ is an $r$-net:

[TABLE]

Proof.

Let $0<\eta<(3C_{n})^{-1}$; then, for any $\epsilon\in(0,1)$, and all $x\in S^{n}$ satisfying the inequality

[TABLE]

it follows from (3), and positivity of the heat kernel, that $0<H_{t}(x_{0}s,x)<{\frac{1_{S^{n}}}{3}}$, and (hence)

[TABLE]

Write $B(x_{0}s,r)\subseteq S^{n}$ for the Riemannian disk of radius $r$, centered at $x_{0}s\in S^{n}$. Let $B_{n}\subseteq{\mathbb{R}}^{n}$ denote the unit euclidean ball; one has (see [8])

[TABLE]

Here “vol” denotes the standard Lebesgue volume. Now suppose, if possible, that (3.12) is satisfied, and yet, $x_{0}{\hat{S}}^{l}\subseteq S^{n}$ is not an $r$-net, so that there is $x\in S^{n}$ such that $d(x,x_{0}{\hat{S}}^{l})>r$. Writing

[TABLE]

we derive from (3.12) and (3.13)

[TABLE]

which produces

[TABLE]

Considering (3), this is impossible if $r>0$ is sufficiently small. ∎

Theorem 3.15**.**

Let $\epsilon\in(0,{\frac{1}{3n}})$ be small, and $r=2\epsilon\sqrt{\ln{\frac{3C_{n}}{\epsilon^{2n-1}}}}$. Let $S\subset SO_{n+1}$ consist of $k$ iid random points, drawn from the Haar measure on $SO_{n+1}$, where

[TABLE]

with $a_{n}:={\frac{2\log_{2}\log_{2}(5n)}{\log_{2}(5n)}}$. Let $l={\frac{n}{2}}\log_{2}\begin{pmatrix}{\frac{1}{r\epsilon}}\end{pmatrix}+(4+3a_{n})n\log_{2}\begin{pmatrix}{\frac{1}{\epsilon}}\end{pmatrix}$; if $r$ is sufficiently small then the probability that $x_{0}{\hat{S}}^{l}\subseteq S^{n}$ is an $r$-net in $S^{n}$ is at least $1-\delta$.

Proof.

One sees by the remark 2.12 (following lemma 2.7) that for any $\eta>0$, if

[TABLE]

and $M=4^{\frac{k_{0}}{n}}$, then the following inequality holds:

[TABLE]

Let $\eta:=\epsilon^{3n(1+a_{n})}$, so that for sufficiently large $l>0$ (to be determined à la theorem 3.2) the parameter $M=\eta^{-\frac{2}{n}}$ ensures

[TABLE]

Taking logarithm of (3.16), we find that it suffices to take

[TABLE]

Suppose that $l>0$ is large enough so that $2^{-l}t^{-\frac{n}{2}}\leq r^{\frac{n}{2}}$; for this to be true, we require $l\geq{\frac{n}{2}}\log_{2}\begin{pmatrix}{\frac{1}{r\epsilon^{2}}}\end{pmatrix}$. We enforce the inequality $2^{-l}t^{-\frac{n}{2}}\leq\epsilon^{3n(1+a_{n})}$ by requiring

[TABLE]

Threfore, if $l=(4+3a_{n})n\log_{2}\begin{pmatrix}{\frac{1}{\epsilon}}\end{pmatrix}+{\frac{n}{2}}\log_{2}\begin{pmatrix}{\frac{1}{r\epsilon}}\end{pmatrix}$, then $2^{-l}t^{-\frac{n}{2}}\leq\min\{r^{\frac{n}{2}},\eta\}$ holds. Since $\epsilon>0$ is small, one has $\alpha_{n}r^{\frac{n}{2}}>6\eta=6\epsilon^{3n(1+a_{n})}$. Thus, by theorem 3.6, the following inequality holds:

[TABLE]

The proof is complete by lemma 3.11.

∎

4. Equidistribution and Wasserstein Distance

Let $(Y,d)$ be a compact connected metric space. Let $C(Y)$ be the Banach space of continuous functions on $Y$, and ${\mathcal{M}}(Y)$ its dual — consisting of linear functionals on $C(Y)$ — equipped with $\mbox{weak}^{\ast}$ topology; recall that, by compactness of $Y$, every linear functional is bounded, and hence, continuous. Let ${\mathscr{M}}(Y)$ be the space of all finite Borel measures on $Y$. By Reisz-Markov theorem, there is a bijection ${\mathscr{M}}(Y)\cong{\mathcal{M}}(Y)$, defined by

[TABLE]

that is closed under addition and scalar multiplication. The space ${\mathscr{M}}(Y)$ inherits the sequential topology on ${\mathcal{M}}(Y)$ via this bijection. Thus, one says $\mu_{n}\Rightarrow\mu$ if and only if

[TABLE]

for every $f\in C(Y)$. Since the Lipschitz functions are dense in $C(Y)$, it suffices to consider only the 1-Lipschitz functions in the above limit.

For probability measures $\mu,\nu\in{\mathscr{M}}(Y)$, the Prokhorov distance $d_{P}(\mu,\nu)\geq 0$ is defined to be

[TABLE]

where $B_{\epsilon}:=\{y\in Y:\exists~{}b\in B,~{}d(b,y)<\epsilon\}$. This gives a metric on the convex subspace ${\mathscr{P}}(Y)\subset{\mathscr{M}}(Y)$ of probability measures on $Y$, and — by Prokhorov’s theorem — the induced metric topology on ${\mathscr{P}}(Y)$ is the subspace of the weak topology on ${\mathscr{M}}(Y)$; moreover, the space ${\mathscr{P}}(Y)$ is compact.

We recall that, in a metric space $(Y,d)$, the 1-Wasserstein distance between two regular Borel probability measures $\mu$ and $\nu$ on $X$ is defined to be

[TABLE]

where $\Pi(\mu,\nu)$ is the space of couplings of $\mu$ and $\nu$; that is, $\Pi(\mu,\nu)$ is the space of all regular Borel probability measures on $Y\times Y$ such that the following holds: $\lambda\in\Pi(\mu,\nu)$ if and only if for every Borel set $B\in{\mathscr{B}}(Y)$, one has

[TABLE]

Let $\operatorname{Lip}_{1}(Y)$ be the space of all 1-Lipschitz functions on $Y$; we recall that, for any $c>0$, one says $f\in\operatorname{Lip}_{c}(Y)$ if and only if $|f(x)-f(y)|\leq c\cdot d(x,y)$ for all $x,y\in Y$. The following duality theorem first appeared in [12].

Theorem 4.1** (Kantorovič - Rubinšteín).**

For any $\mu,\nu\in{\mathscr{P}}(Y)$, the following equality holds:

[TABLE]

Definition 4.3**.**

Let $\epsilon>0$. Let $\mu$ be a Borel probability measure on the metric space $(Y,d)$. A finite nonempty subset $U\subset Y$ is said to be strongly $(\mu,\epsilon)$-equidistributed if the following inequality holds:

[TABLE]

As mentioned before, for $U\subset Y$ to be strongly $(\mu,\epsilon)$-equidistributed, it suffices to have a constant $c:=c(\mu)$ such that

[TABLE]

Below we show strong $(\mu,\epsilon)$-equidistribution of a subset of $S^{n}$ of appropriate size and low degree of randomness.

The following lemma will be useful in course of proving the main theorem of this subsection.

Lemma 4.6** (Fourier convergence).**

Fix $y\in S^{n}$, and let $0<a<b$; then the Fourier-Laplace expansion of $H_{t}(y,x)$ converges uniformly to $H_{t}(y,x)$ in $[a,b]\times S^{n}$.

Proof.

Fix an orthonormal basis $\phi_{1,k},\cdots,\phi_{h_{k},k}$ for the eigenspace $H_{k}(S^{n})$. Then the Fourier-Laplace expansion of the heat kernel based at $y\in S^{n}$ is

[TABLE]

Write

[TABLE]

so that

[TABLE]

One has

[TABLE]

by rotation invariance of the sum (see Lemma 2.19 and 2.29, [14])

[TABLE]

Therefore,

[TABLE]

and since

[TABLE]

this forces

[TABLE]

The Weierstrass’ $M-$test implies uniform convergence of $H_{t}(x,y)$, to a continous function on $[a,b]\times S^{n}$; the claim follows by uniqueness of the continuous limit. ∎

Lemma 4.8**.**

Let $d(\cdot,\cdot)$ be the metric distance on $S^{n}$. Let $\sigma$ be the uniform surface probability measure on $S^{n}$. For all $t>0$, one has

[TABLE]

Proof.

Without loss of generality we may assume that $S^{n}$ is embedded in $\mathbb{R}^{n+1}$ as the unit sphere with center at $-{\bf e}_{n+1}=(0,\cdots,0,-1)$, and $y={\bf 0}$. We write $H_{t}(x):=H_{t}(0,x)$, and let $\sigma_{t}^{\ast}$ be the Borel measure whose Radon-Nikodym derivative is

[TABLE]

Let $\{X_{u}\mid u\in[0,t]\}$ be a standard Brownian motion on $S^{n}$ with infinitesimal generator $H_{\frac{t}{2}}(x)$. For each positive integer $m>0$, consider the equi-partition

[TABLE]

where $t_{i+1}-t_{i}=m^{-1}t$ for $i=0,1,\cdots,m-1$. Now define $\{X_{i}^{(m)}\}_{i=0}^{m}$ as follows:

[TABLE]

By linearity of expectation, for any integer $m>0$ one has

[TABLE]

Fix a realization of the Brownian motion $\{X_{u}\mid u\in[0,t]\}$. For integer $1\leq i\leq m$, we consider the tangent space $T_{X_{i-1}^{(m)}}(S^{n})$. Write

[TABLE]

In explicit terms, one has

[TABLE]

Orthogonality relations such as

[TABLE]

are immediate; moreover, one has

[TABLE]

where ${\bf n}=-{\bf e}_{n+1}-X_{i+1}^{(m)}$ is the unit normal to $T_{X_{i-1}^{(m)}(S^{n})}$ pointing inward. From the inequalities

[TABLE]

one has $\langle X_{i}^{(m)}-Y_{i}^{(m)},Z_{i-1}^{(m)}\rangle\leq 0$. Hence,

[TABLE]

Suppose $X_{i-1}^{(m)}=z$ and $X_{i}=z^{\prime}$ in $S^{n}$. Let $z^{\prime\prime}:=z^{\prime\prime}(z,z^{\prime})\in S^{n}$ be such that

[TABLE]

Since the function

[TABLE]

takes arbitrarily small positive values, such a point $z^{\prime\prime}\in S^{n}$ — that satisfies (4.10) — exists by continuity of $g(z^{\prime\prime})$ and compactness of $S^{n}$. Note that

[TABLE]

by independence of increments for Brownian motion on euclidean space. From

[TABLE]

we derive

[TABLE]

It thus suffices to prove

Lemma 4.11**.**

[TABLE]

Proof.

We will use the stereographic projection of $S^{n}\setminus\{0\}$ onto $\mathbb{R}^{n}$. It can be shown (see for example [7]) that the image $Y_{t}$ of a standard Brownian motion on $S^{n}$ via the stereographic projection onto $\mathbb{R}^{n}$, where $r=|Y_{t}|$, with $\Delta_{\mathbb{R}^{n}}$ being the Laplacian on $\mathbb{R}^{n}$, has an infinitesimal generator $(1/2)\Delta_{S^{n}}$ that satisfies

[TABLE]

Applying this to the function $f(x)=\|x\|^{2}$, we see that

[TABLE]

It follows that for any $i$,

[TABLE]

converges as $m\rightarrow\infty$ to $nt$, proving the lemma. ∎

∎

Corollary 4.12**.**

Let $n>1$ and for integers $k\geq 0$, let $P_{k,n}(t)$ be the Legendre polynomial of degree $k$ and dimension $n+1$. Let $h_{k}=\dim H_{k}(S^{n})$ and

[TABLE]

Then the following inequality holds for all $n\geq 4$:

[TABLE]

Proof.

We recall the Hecke-Funk formula: for any function $\chi:[-1,1]\rightarrow{\mathbb{R}}$, which satisfies the inequality

[TABLE]

and any eigenfunction $\phi\in{\mathcal{H}}_{k}(S^{n})$ and point $y\in S^{n}$ one has

[TABLE]

Consider the function $\chi(t)=\sqrt{2}(1-t)^{\frac{1}{2}}$, taking values in $[0,2]$; this satisfies the hypothesis in Hecke-Funk formula, and since $d(y,x)=\sqrt{2}(1-\langle y,x\rangle)^{\frac{1}{2}}$, one gets

[TABLE]

Consider the Fourier- Laplace expansion of the heat kernel, as in lemma 4.6 above. By uniform convergence (lemma 4.6) of the Fourier-Laplace expansion of the heat kernel, one has

[TABLE]

Note that

[TABLE]

for all $n\geq 4$. Hence, Lemma 4.8 together with Hölder inequality implies

[TABLE]

∎

Theorem 4.14**.**

For $n>1$, let $\mu$ be the probability measure on $S^{n}$ corresponding to $\sigma$. Let $\epsilon,\delta>0$ be sufficiently small and $r=2\epsilon\sqrt{\ln{\frac{3C_{n}}{\epsilon^{2n-1}}}}$. Let $S\subseteq SO(n+1)$ be a random subset such that $|S|=k$ satisfies the inequality in theorem 3.15, namely

[TABLE]

where $a_{n}:={\frac{2\log_{2}\log_{2}(5n)}{\log_{2}(5n)}}$. Let $x_{0}\in S^{n}$ and let $\nu$ be the uniform probability measure on $S^{n}$, supported on ${\hat{S}}^{l}x_{0}$, where ${\hat{S}}=S\cup S^{-1}$ as before and

[TABLE]

Then, with probability at least $1-\delta$, the following inequality holds:

[TABLE]

Proof.

Let $\operatorname{Lip}_{1,0}(S^{n})$ be the set of mean-zero $\operatorname{Lip}_{1}$-functions on $S^{n}$. By theorem 4.1, it suffices to show that

[TABLE]

For any such function $\phi\in\operatorname{Lip}_{1,0}(S^{n})$, if $\phi(x_{0})=||\phi||_{L^{\infty}}$ then

[TABLE]

For sufficiently small $t>0$, we let $\nu_{t}^{\ast}$ be the Borel probability measure on $S^{n}$ whose density is

[TABLE]

Then, for $t={\epsilon^{2}}$, one has

[TABLE]

with probability at least $1-\delta$.

For any function $\phi\in\operatorname{Lip}_{1,0}(S^{n})$, define ${\tilde{\phi}}_{t}:S^{n}\rightarrow{\mathbb{R}}$ to be

[TABLE]

From uniform convergence of the Fourier-Laplace expansion of heat-kernel, it follows that $\int_{S^{n}}H_{t}(y,x)~{}d\sigma(x)=1$; hence, putting $t=\epsilon^{2}$, one has

[TABLE]

Moreover,

[TABLE]

by lemma 4.8 and Hölder inequality applied to $d(y,x)=d(y,x)\cdot 1_{S^{n}}$. Therefore, for $t=\epsilon^{2}>0$ sufficiently small, equations (4.16), (4), and (4) yield

[TABLE]

∎

5. Conclusion

We proved two results about the finite time behavior of a random Markov Chain on the sphere $S^{n}$ whose transitions correspond to rotations chosen uniformly at random. The first result states that for $k=O(n\ln\frac{1}{\epsilon}+\frac{1}{\delta})$ random rotations and $\ell={O}(n\ln 1/\epsilon)$, if one takes the image of the north pole on the sphere under all possible words of length $\ell$ in the $k$ alphabets and their inverses, one obtains an $\epsilon-$net with high probability. For these parameters, the value of $(2k)^{\ell}$ is close to the volumetric lower bound of $(1/\epsilon)^{\Omega(n)}$ on the size of an $\epsilon-$net of $S^{n}$. Secondly, we show that this $\epsilon-$net is equidistributed with probability at least $1-\delta$ in the sense that the $1-$Wasserstein distance of the uniform measure on the net is within $\epsilon$ of the uniform measure on $S^{n}$.

These results can respectively be applied to approximately minimize a $1-$ Lipschitz function on the sphere (by evaluation on the $\epsilon-$net) and in to approximately integrate a $1-$Lipschitz function on the sphere. In both cases the approximation is within an additive $\epsilon$ of the true value.