Does a dense rotation subgroup force a spectral gap for the rotation-averaging operator on $L^2(\mathrm{SO}(d))$?

Lindenstrauss and Varjú (Duke Math. J., 2016) proved a local limit theorem with exponentially small error for random walks by Euclidean isometries, and absolute continuity of associated self-similar measures, under the hypothesis that the Markov operator averaging the rotation part of the law, $T\varphi(\sigma)=\mathbb{E}[\varphi(\theta(X)^{-1}\sigma)]$ on $L^2(\mathrm{SO}(d))$, has a spectral gap. They recorded as an open problem whether this hypothesis can be weakened to the assumption that the support of the rotation part merely generates a dense subgroup of $\mathrm{SO}(d)$, $d\ge 3$: no counterexample is known, and the Bourgain–Gamburd method establishes the gap only under arithmetic (algebraicity or noncommutative Diophantine) assumptions. The problem asks to prove that denseness implies the spectral gap — in the general-law setting, or in the finitely supported setting where this amounts to removing the algebraicity hypothesis from the theorems of Bourgain–Gamburd and Benoist–de Saxcé — or to construct a probability measure whose rotation support generates a dense subgroup of $\mathrm{SO}(d)$ while $T$ fails to have a spectral gap. The question is open in both directions.

Background

Random walks driven by random isometries of Euclidean space go back to Arnold and Krylov: given i.i.d. random orientation-preserving isometries $X_1,X_2,\dots$ of $\mathbb{R}^d$ and a starting point $x_0$, one studies the random points $Y_l=X_l\cdots X_1(x_0)$. Central limit theorems describe $Y_l$ on the diffusive scale $\sqrt{l}$ (Tutubalin and others), and in dimension $2$ ratio limit theorems describe it on bounded scales; the harder question, addressed by Lindenstrauss and Varjú (2016), is a uniform local description on all intermediate scales, in particular on balls whose radii shrink exponentially in $l$.

The hypothesis under which they proved their local limit theorem is a spectral property of the rotation part. Write $\theta:\mathrm{Isom}(\mathbb{R}^d)\to\mathrm{SO}(d)$ for the canonical projection sending an isometry to its linear part, and for a random isometry $X$ define the Markov averaging operator $T\varphi(\sigma)=\mathbb{E}[\varphi(\theta(X)^{-1}\sigma)]$ on $L^2(\mathrm{SO}(d))$ (with respect to Haar measure, the translation-invariant probability measure on the group), and let $L^2_0$ be the subspace orthogonal to the constants. One says that $T$ has a spectral gap if $\|T^l\|_{L^2_0}<1$ for some integer $l$: informally, averaging the rotation part of the law over $l$ steps strictly contracts every zero-mean square-integrable function. Under this hypothesis, together with $d\ge 3$, a moment condition, and absence of an almost-sure fixed point, their Theorem 1.1 approximates the law of $Y_l$ by a Gaussian with error $O(e^{-cl})$ in a Sobolev norm, uniformly down to spatial scales $e^{-cl}$. The same hypothesis drives their companion result (Theorem 1.3) that stationary measures of random contracting similarities of $\mathbb{R}^d$ in dimensions $d\ge 3$ — higher-dimensional analogues of Bernoulli convolutions, the laws of random power series $\sum\pm\lambda^n$ — are absolutely continuous with $n$ times differentiable densities when all contraction ratios are sufficiently close to $1$.

The spectral gap is the nontrivial input, and its status is the subject of this problem. The Bourgain–Gamburd method (an $L^2$-flattening argument for the convolution powers of the measure, combined with a non-concentration estimate for the walk) shows that the gap holds under arithmetic assumptions: if the measure is uniform on a finite symmetric subset of a compact simple Lie group that generates a dense subgroup and consists of elements with algebraic entries (entries that are algebraic numbers, i.e. roots of rational polynomials) — or more generally satisfies a noncommutative Diophantine condition — then $T$ has a spectral gap (Bourgain–Gamburd 2008 for $\mathrm{SU}(2)$/SO(3), Bourgain–Gamburd 2012 for $\mathrm{SU}(d)$, Benoist–de Saxcé 2016 for arbitrary compact simple Lie groups). Benoist and de Saxcé also proved that the spectral gap is equivalent to a weak Diophantine property of the generating set, namely that the random walk approaches proper closed subgroups only at an exponentially small rate. All known proofs use arithmetic structure to establish this non-concentration; Lindenstrauss and Varjú write that the algebraicity condition "seems essential for the argument and its removal would probably require significant new ideas". In the other direction, they note that no example is known of a probability law whose rotation support generates a dense subgroup of $\mathrm{SO}(d)$ while $T$ lacks a spectral gap, and they raise the possibility that denseness alone might force the gap.

The question is genuinely non-abelian. For $d=2$ the rotation group $\mathrm{SO}(2)$ is a circle: a single rotation by an irrational multiple of $2\pi$ already generates a dense cyclic subgroup, while the averaging operator over a one-point (or two-point symmetric) rotation support never has a spectral gap, since its Fourier multipliers $\cos(n\alpha)$ have supremum $1$ — an elementary observation that explains why the problem concerns $d\ge 3$. For $d\ge 3$, the closure of any abelian subgroup of $\mathrm{SO}(d)$ lies in a maximal torus (a maximal connected abelian subgroup, of dimension strictly smaller than $\mathrm{SO}(d)$), so any subgroup generating a dense subgroup must be non-abelian, and no counterexample is known; a 2018 survey of expansion in simple groups (Breuillard–Lubotzky) highlights precisely this as the "worst case" of the theory.

Problem Statement

Let $d\ge 3$. Let $X$ be a random orientation-preserving isometry of $\mathbb{R}^d$ with an arbitrary probability law, let $\theta(X)\in\mathrm{SO}(d)$ denote its rotation (linear) part, and let $T\varphi(\sigma)=\mathbb{E}[\varphi(\theta(X)^{-1}\sigma)]$ be the associated Markov averaging operator on $L^2(\mathrm{SO}(d))$, with $L^2_0$ the closed subspace of functions of Haar mean zero. Recall that $T$ has a spectral gap if there is an integer $l\ge 1$ with $\|T^l\|_{L^2_0}<1$.

Question: if the subgroup generated by the support of the rotation part $\theta(X)$ is dense in $\mathrm{SO}(d)$, must $T$ have a spectral gap?

A positive answer would prove that denseness of the generated subgroup implies the spectral gap, for arbitrary probability laws; in the finitely supported case it says that for every finite symmetric subset $S\subset\mathrm{SO}(d)$ generating a dense subgroup, the averaging operator over $S$ has a spectral gap, which would remove the algebraicity/Diophantine hypothesis from the theorems of Bourgain–Gamburd and Benoist–de Saxcé.

A negative answer would construct, for some $d\ge 3$, a probability measure on $\mathrm{SO}(d)$ (or a law of isometries of $\mathbb{R}^d$ whose rotation part has this law) whose support generates a dense subgroup of $\mathrm{SO}(d)$, yet for which $T$ fails to have a spectral gap on $L^2(\mathrm{SO}(d))$.

The question is restricted to $d\ge 3$ because for $d=2$ the rotation group $\mathrm{SO}(2)$ is a circle and the implication fails by an elementary Fourier computation (a single irrational rotation generates a dense cyclic subgroup while the averaging operator over a symmetric one-generator support has Fourier multipliers $\cos(n\alpha)$ with supremum $1$); for $d\ge 3$ dense generation forces the subgroup to be non-abelian, since closures of abelian subgroups lie in proper maximal tori, and no counterexample is known. A positive answer would immediately extend the Lindenstrauss–Varjú local limit theorem with exponentially small error, and its self-similar-measure corollaries, from the spectral-gap hypothesis to the minimal, easily checked denseness hypothesis; a negative answer would exhibit the first known topological generating set of a compact simple-type Lie group without a spectral gap.

References

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