Rigidity Hierarchy in Random Point Fields: Random Polynomials and Determinantal Processes

Abstract

In certain point processes, the configuration of points outside a bounded domain determines, with probability 1, certain statistical features of the points within the domain. This notion, called rigidity, was introduced in Ghosh and Peres (Duke Math J 166(10):1789–1858, 2017). In this paper, rigidity and the related notion of tolerance are examined systematically and point processes with rigidity of various degrees are introduced. Natural classes of point processes such as determinantal point processes, zero sets of Gaussian entire functions and perturbed lattices are examined from the point of view of rigidity, and general conditions are provided for them to exhibit specified nature of spatially rigid behaviour. In particular, we examine the rigidity of determinantal point processes in terms of their kernel, and demonstrate that a necessary condition for determinantal processes to exhibit rigidity is that their kernel must be a projection. We introduce a one parameter family of point processes which exhibit arbitrarily high levels of rigidity (depending on the choice of parameter value), answering a natural question on point processes with higher levels of rigidity (beyond the known examples of rigidity of local mass and center of mass). Our one parameter family is also related to a natural extension of the standard planar Gaussian analytic function process and their zero sets.

Appendix: Paradigms for Rigidity and Tolerance

In this section, we establish technical results pertaining to various notions related to rigidity and tolerance.

   We begin by recalling Theorem 6.1 from [GP], which has been demonstrated to be useful for our purposes in this paper. In fact, in what follows, we state and prove a slightly more general version of this theorem.

Theorem 7.1

(Theorem 6.1 in [GP]). Let $\Pi$ be a point process on second countable Hausdorff locally compact topological space $\Xi$, and let D be a measurable set in $\Xi$. Let $\mathcal {S}_D$ denote the space of locally finite point configurations on D. Let $\varphi$ be a compactly supported measurable function on $\Xi$. Suppose that for any $1>\varepsilon >0$, we have a compactly supported measurable function $\Phi ^{\varepsilon }$ such that $\Phi ^{\varepsilon }=\varphi$ on D, and $\mathrm {Var}\left( \sum _{z \in \Pi } \Phi ^{\varepsilon }(z) \right) < \varepsilon$. Then the function $T:\mathcal {S}_D \rightarrow {\mathbb {C}}$ given by $T(\Upsilon )=\sum _{z \in \Upsilon \cap D} \varphi (z)$ is rigid.

Proof

In the rest of this proof, for any locally finite point configuration $\Upsilon$ and a measurable function $\psi$ on $\Xi$, we will denote by $\int \psi \mathrm {d}[\Upsilon ]$ the sum $\sum _{z \in \Upsilon } \psi (z)$.

Consider the sequence of functions $\Phi ^{2^{-n}}, n \ge 1$. Note that ${\mathbb {E}}\left[ \int _{\Xi } \Phi ^{2^{-n}} \mathrm {d}[\Pi ] \right] = \int _{\Xi } \Phi ^{2^{-n}} \mathrm {d} \rho _1$ where $\rho _1$ is the first intensity measure of $\Pi$. It follows from Chebyshev’s inequality that

$$\begin{aligned} {\mathbb {P}}\left( \left| \int _{\Xi } \Phi ^{2^{-n}} \mathrm {d}[\Pi ] - {\mathbb {E}}\left[ \int _{\Xi } \Phi ^{2^{-n}} \mathrm {d}[\Pi ] \right] \right| > 2^{-n/4} \right) \le 2^{-n/2}. \end{aligned}$$

The Borel Cantelli lemma implies that with probability 1, as $n \rightarrow \infty$ we have

$$\begin{aligned} \left| \int _{\Xi } \Phi ^{2^{-n}} \mathrm {d}[\Pi ] - {\mathbb {E}}\left[ \int _{\Xi } \Phi ^{2^{-n}} \mathrm {d}[\Pi ] \right] \right| \rightarrow 0. \end{aligned}$$

But

$$\begin{aligned} \int _{\Xi } \Phi ^{2^{-n}} \mathrm {d}[\Pi ] = \int _{D} \Phi ^{2^{-n}} \mathrm {d}[\Pi ] + \int _{D^c} \Phi ^{2^{-n}} \mathrm {d}[\Pi ]. \end{aligned}$$

Thus we have, as $n \rightarrow \infty$

$$\begin{aligned} \left| \int _{D} \Phi ^{2^{-n}} \mathrm {d}[\Pi ] + \int _{D^c} \Phi ^{2^{-n}} \mathrm {d}[\Pi ] - \int _{\Xi } \Phi ^{2^{-n}} \mathrm {d} \rho _1 \right| \rightarrow 0. \end{aligned}$$

(19)

If we know $\Pi _{D^c}$, we can compute $\int _{D^c} \Phi ^{2^{-n}} \mathrm {d}[\Pi ]$ exactly, also $\rho _1$ is known explicitly as a functional of the point process $\Pi$. Hence, from the limit in (19), a.s. we can obtain $\int _{D} \Phi ^{2^{-n}} \mathrm {d}[\Pi ] = \int _{D} \varphi \mathrm {d}[\Pi ]$ as the limit

$$\begin{aligned} \lim _{n \rightarrow \infty } \left( \int _{\Xi } \Phi ^{2^{-n}} \mathrm {d} \rho _1 - \int _{D^c} \Phi ^{2^{-n}} \mathrm {d}[\Pi ] \right) . \end{aligned}$$

$\square$

We move on with a lemma that pertains to the nature of the domains for which rigidity of numbers fails to hold (c.f. 2.1).

Lemma 7.2

Let $\Pi$ be a point process on a space E such that the point count in a domain U, i.e. $N_U(\Pi )$, is not rigid. Then, for any domain $V \subset E$ such that $U \subseteq V$, the point count in V, i.e. $N_V(\Pi )$, is also not rigid.

Proof

Suppose $N_V(\Pi )$ is rigid. Write $N_U(\Pi )=N_V(\Pi )-N_{V\cap U^c}(\Pi )$. The second term is clearly in $\sigma (\Pi _{U^c})$, while the rigidity of $N_V(\Pi )$ implies that the first is in $\sigma (\Pi _{V^c})$ which is a subset of $\sigma (\Pi _{U^c})$. Therefore, $N_U(\Pi )\in \sigma (\Pi _U^c)$, showing that $N_U(\Pi )$ must also be rigid. $\quad \square$

   We now discuss the relationship between tolerance and strong tolerance. For that we state an equivalent condition for tolerance.

Lemma 7.3

Consider the following statements.

• (A) Let $(\Pi ,D)$ be tolerant subject to $\{f_{1},\ldots ,f_{m}\}$.

• (B) With positive probability, the conditional distribution of $\Pi _{D}$ given $\Pi _{D^{c}}$ is mutually absolutely continuous to the conditional distribution of $\Pi _{D}$ given $\sigma \{f_{1}(\Pi _{D}),\ldots ,f_{m}(\Pi _{D})\}$.

• (C) With probability one, the conditional distribution of $\Pi _{D}$ given $\Pi _{D^{c}}$ is mutually absolutely continuous to the conditional distribution of $\Pi _{D}$ given $\sigma \{f_{1}(\Pi _{D}),\ldots ,f_{m}(\Pi _{D})\}$.

Then (C) $\implies$ (B) $\implies$ (A).

   As (C) is the same as strong tolerance and (A) is the same as tolerance, in particular it follows that strong tolerance implies tolerance. The intermediate statement (B) is put in to show that there are other possibilities. Let us first give an example.

Example 7.1

Let $\Pi _{1}$ be the infinite Ginibre ensemble. Let $\Pi _{2}$ be a Poisson process with unit intensity on the plane. Let $\xi$ be a $\text{ Ber }(1/2)$ random variable. Assume that $\Pi _{1},\Pi _{2},\xi$ are independent. Set $\Pi =\Pi _{1}$ if $\xi =0$ and $\Pi =\Pi _{2}$ is $\xi =1$. Then $\Pi$ is a stationary point process.

   If D is a bounded set, then from $\Pi _{D^{c}}$ we can deduce $\xi$ and then, we may or may not be able to deduce the number of points in D depending on the value of $\xi$. Hence ${\mathcal {R}}_{D}$ is trivial. Further, if $\xi =0$, then the conditional distribution of $\Pi _{D}$ given $\Pi _{D^{c}}$ is Poisson process on D. If $\xi =1$, the conditional distribution of $\Pi _{D}$ is supported on collection of N points where N is a function of $\Pi _{D^{c}}$. Thus, $\Pi$ satisfies (A) and (B) but not (C).

   Playing the same game with Ginibre and GAF zeroes, we can get a stationary point process which is rigid for numbers, which is tolerant subject to numbers, but which is not strongly tolerant subject to numbers.

   One might wonder if the point process in the above example is somehow unnatural. For instance, it is not ergodic. Here is an example to show that such a difference can occur even under ergodicity, mixing and tail triviality.

Example 7.2

Let $\xi _{n}$, $n\in {\mathbb {Z}}$, be i.i.d. with ${\mathbb {P}}\{\xi _{n}=0\}={\mathbb {P}}\{\xi _{n}=1\}=\frac{1}{2}$. Define

$$\begin{aligned} X_{n}={\left\{ \begin{array}{ll} 1 &{} \text{ if } \xi _{n}=1 \text{ or } \xi _{n-1}=\xi _{n+1}=1, \\ 0 &{} \text{ otherwise }. \end{array}\right. } \end{aligned}$$

Then $(X_{n})_{n\in {\mathbb {Z}}}$ is stationary. Since $(X_{k})_{k\le n}$ and $(X_{k})_{k\ge n+3}$ are independent (a property known as 2-dependent), X is tail-trivial, ergodic and mixing in every sense.

   Condition on $(X_{k})_{k\not = 0}$. If $X_{1}=X_{-1}=1$, then we can easily deduce that $X_{0}=1$. However, for any of the other three possible values for $(X_{1},X_{-1})$, the conditional distribution of $X_{0}$ gives positive probability to both 0 and 1. It should also be noted that $(X_{1},X_{-1})$ takes all four values with positive probability.

   Regard X as a point process on ${\mathbb {Z}}$ and let $D=\{0\}$. Then ${\mathcal {R}}_{D}$ is trivial, (B) holds, but not (C).

It is also possible to satisfy (A) without satisfying (B).

Example 7.3

Let $\Omega =\{1,2\}\times \{1,2,3\}$ with uniform probability distribution. Let $X(i,j)=j$ and

$$\begin{aligned} Y(1,2)=Y(1,3)=1, \;\; Y(2,1)=Y(2,3)=2, \;\; Y(1,1)=Y(2,2)=3. \end{aligned}$$

Then, $\sigma \{X\}\cap \sigma \{Y\}$ is trivial. However, for any $\omega \in \Omega$, the conditional distribution of X given Y is supported on exactly two of the points 1, 2, 3.

   To make an example involving point processes, let (i, j) be sampled uniformly from $\Omega$. Given (i, j), let $\Pi$ be a Poisson process on ${\mathbb {R}}$ with intensity X(i, j) on $(-1,1)$ and intensity Y(i, j) on $(-1,1)^{c}$. Then the process $\Pi$ satisfies (A) but not (B).

Proof of Lemma 7.3

We only need to prove that (B) implies (A).

If (A) were not true, we could find $f:\mathcal {S}_{D}\mapsto {\mathbb {R}}$ such that (a) f is rigid for $\Pi$, (b) there is no random variable measurable with respect to $\sigma \{f_{1}(\Pi _{D}),\ldots ,f_{m}(\Pi _{D})\}$ that is equal to $f(\Pi _{D})$ a.s. In particular, $f(\Pi _{D})$ is not a constant and in fact, with positive probability, the conditional distribution of $f(\Pi _{D})$ given $\sigma \{f_{1}(\Pi _{D}),\ldots ,f_{m}(\Pi _{D})\}$ is not degenerate. Hence, with positive probability, the conditional distribution of $\Pi _{D}$ given $\sigma \{f_{1}(\Pi _{D}),\ldots ,f_{m}(\Pi _{D})\}$ gives probability less than one to any level set of f. But the rigidity of f implies that, almost surely, the conditional distribution of $\Pi _{D}$ given $\Pi _{D^{c}}$ is supported inside a level set of f. This contradicts (B). $\quad \square$