Fractal Kintchine

We are at the Wework in Pune, India. This was a working group activity organized by Anish Ghosh.

Fractal Kintchine

Kintchine theorem is a fundamental theorem that tells us about how well approximable is a typical point in the real numbers. The fractal Kintchine does this for Cantor sets.

A probability measure $\sigma$ on $\mathbb{R}$ is self-similar if it satisfies \[\begin{equation} \sigma = \sum \lambda_{i}(\phi_{i})_{*} \sigma, \end{equation}\] where $m \geq 1$, $\lambda_{i} > 0$ ( $\sum_{}^{} \lambda_{i} = 1$), and $\phi_{i}$ are invariant affine transformation of $\mathbb{R}$ transformation of $\mathbb{R}$ without common fixed point.

$s \in \mathbb{R}$ is $\psi$-approximable if \[\begin{equation} |q s - p | < \psi(q) \end{equation}\] has infinitely many solutions for $p \in \mathbb{Z}$ and $q \in \mathbb{N}$.

(Timothée Bénard, He, Zhang, 2024)

Let $\sigma$ be self-similar on $\mathbb{R}$ and let $\psi: \mathbb{N} \rightarrow \mathbb{R}$ be a non-increasing. Then \[\begin{equation} \sigma(s: \text{ $\psi$-approximable}) = \begin{cases} 0 & \text{ if } \sum_{q \geq 0}{\psi(q)} < \infty \\ 1 & \text{ if } \sum_{q \geq 0}^{} \psi(q) = \infty . \end{cases}. \end{equation}\]

Something tangential

Take $X = \mathrm{SL}_{2}(\mathbb{R} )/\mathrm{SL}_{2}(\mathbb{Z})$. Take $\mu$ to be some probability measure on $\mathrm{SL}_{2}(\mathbb{R})$. This can be used to define a random walk on $X$. Just take a point $x \in X$ and act with a $g \in \mathrm{SL}_{2}(\mathbb{R})$. Then we can $n$-step distribution which gives us $\mu ^{*n} * \delta_{x}$.

\[\begin{equation} \mu^{*n} * \delta_{x} = \text{ law of }g_{1},\dots,g_{n} \text{ as } g_{i} \sim \mu^{\otimes \mathbb{N}_{*}} \end{equation}\]

(BHZ)

Assume support of $\mu$ in $\left(\begin{smallmatrix} * & * \\ 0 & * \end{smallmatrix}\right)$, finite and not simultaneously diagonalizable, and \[\begin{equation} \int_{}^{} \log \| g_{1,1}\|\, \mathrm{d}\mu (g) > 0 \end{equation}\] (This means that the top-left coordiate is bigger than 1 on average).

Let $x \in X$ and $f \in C^{1}(X), n \geq 1$. Then \[\begin{equation} | \mu^{n} * \delta_{n}(f) - \mathrm{Haar}_{X}(f) | \leq A \|f\|_{C} \cdot \mathrm{inj}(x)^{-1} e^{- \Sigma n}, \end{equation}\] where $A, \Sigma > 0$ depend on $\sigma$ only.

Consider $\Gamma_{\mu} = \langle \mathrm{supp}\mu \rangle , W_{R}= \{x \in X \mid |\Gamma_{\mu} x| \leq R\}$.

Assume $\Gamma_{\mu}$ Zariski dense in $\mathrm{SL}_{2}(\mathbb{R})$. Then for any $\rho > 0$, any $x$, $n \geq 1$ \[\begin{equation} | \mu^{*n} * \delta_{n}(f) -\mathrm{Haar}_{X}(f) | \leq \rho \end{equation}\] provided \[\begin{equation} n \geq A ( |\log \rho| + |\log \|f\|_{\mathrm{Lip}}| + |\log \mathrm{inj}(x)| + |\log \mathrm{dist}(x,W_{\rho^{-A}})|) \end{equation}\]

Sketch of proof

Step 1: Positive dimension

$\exists A, \kappa > 0$ such that $\forall n \geq 1$, $\forall \rho \geq e^{-n}$, $\forall x,y \in X$ \[\begin{equation} \mu^{n} * \delta_{x} (B_{\rho} y ) \leq A \rho^{\kappa}. \end{equation}\]

Step 2: Dimensional bootstrap

Go from $\kappa$ to $\dim X - \varepsilon$ using more conditions by $\mu$. Uses harmonic analysis.

Step 3: Go from high dimension to equidistribution

We use spectral gap of $f \mapsto f * \mu$ on $L^{2}(X)$ where \[\begin{equation} f * \mu (x) = \int_{}^{} f(gx)\, \mathrm{d}\mu(g) \end{equation}\]

There is a connection to Kakeya conjecture which uses dimensional bootstrap. There is also a 3-step strategy used in some upcoming work of Lindenstraus and some more people.

More details of the three step procedure

Proof of Step 1

We will follow what happens in the case of an example. Define \[\begin{equation} a_{t} = \begin{bmatrix} t^{\frac{1}{2}} & 0 \\ 0 & t^{-\frac{1}{2}} \end{bmatrix}, \mu_{s} = \begin{bmatrix} 1 & s \\ 0 & 1 \end{bmatrix}. \end{equation}\] Take $\mu := \tfrac{1}{2} \delta_{a_{3}} + \tfrac{1}{2} \delta_{a_{3} u_{ 2/3}}$. Observe that \[\begin{equation} \mu^{*n} = a_{3^{n}} \mu(\sigma^{(n)}) , \text{ where } \sigma^{(n)} \underbrace{=}_{\text{ law }} \sum_{i=1}^{n} \tfrac{2 \varepsilon_{i}}{3}, \end{equation}\] where $\varepsilon_{i} \in \mathrm{Bool}$. So it is an approximation of the Cantor set.

To prove step 1: Let $A, \kappa, n,\rho$ parameters. Assume \[\begin{equation} \mu^{n} * \delta_{x}(B_{\rho} y) \geq \rho^{\kappa} \text{ for some } x,y \in X. \end{equation}\] Assume $n \geq |\log \rho|$, take $c > 0$ (small problem to be decided later). Write $m = \lfloor c |\log \rho | \rfloor$. Note $\exists x'$ such that \[\begin{equation} \mu^{m} * \delta_{x'}(B_{\rho} y ) \geq \rho^{ \kappa}. \end{equation}\]

Then \[\begin{equation} (\mu^{m})^{\otimes 2} \left( (g_{1},g_{2}) \mid g_{1} x', g_{2} x' \in B_{\rho}y\right) \geq \rho^{2 \kappa} \end{equation}\] This can be seen as measuring $g_{1},g_{2}$ such that $g_{2}^{-1} g_{1} x' \in B_{\rho^{1-c}} x'$.

Note $g_{2}^{-1} g_{1} \underbrace{=}_{\text{law}} \left(\begin{smallmatrix} 1 & s^{m} \\ 0 & 1 \end{smallmatrix}\right)$, where $s^{m} := \sigma^{(m)} \ominus\sigma^{(m)}$. This is the law of $s_{1} - s_{2}$ where $s_{1},s_{2} \sim \sigma^{(m)}$.

Hence $\{s \in [-1,1] \mid u(s)x' \in B_{\rho^{1-c}}(x') \} \subseteq \cup_{\substack{\text{ finite } \\ \text{ card $O(1)$ }}} B_{\rho^{1-c}(x')}$.

Because $\rho$ is Frostman (means positive dimension), one has \[\begin{equation} s^{(m)} ( [\rho^{1- 2c}, \rho^{1-2c}] + s_{0}) \leq \rho^{\alpha}, \end{equation}\] where $\alpha$ is absolute. There is a contradiction here when $\kappa$ is small enough.

$\sigma$ is Frostman if $\sigma([-r,r] + s) \leq C r^{\alpha}$ for every $r,s$ and where $C, \alpha$ depend only on $\sigma$.

Step 3

Fact: $f \in C^{\infty}(X), P_{\mu} f(x) = \int_{}^{} f(gx) \, \mathrm{d}\mu g \in C^{\infty}(X)$. This $P_{\mu}$ is called the Markov operator. Then \[\begin{equation} \| P_{\mu}^{n} f \|_{L^{2}} \leq e^{- s n } \|f\|_{C'} , \end{equation}\] for $\int_{}^{} f = 0$.

To get from here to the fact, we write \[\begin{equation} \langle P_{\mu}^{n} f , P_{\mu}^{n} f \rangle = \int_{}^{} \langle f \circ g_{1}, f \circ g_{2} \rangle \, \mathrm{d}\mu^{n} (g_{1}) \, \mathrm{d}\mu^{n}(g_{2}). \end{equation}\]

Consider $\nu$ a probability measure on $X$ such that for some parameters $\varepsilon, \rho > 0$ \[\begin{equation} \sup_{y \in X} \nu( B_{\rho} y) \leq \rho^{3 - \varepsilon}. \end{equation}\] This $3$ is the dimension of $X$. Let $f \in C^{\infty}(X)$, $\mathrm{Haar}_{X}(f) = 0$. We want to choose $n = n(\rho)$ such that $|\mu^{n} * \nu(f)| \leq \rho^{*}$.

We compute \[\begin{equation} | \mu^{n} * \nu(f) | = | \nu(P^{n}_{\mu}f) | = | \nu_{\rho}(P_{\mu}^{n} f ) | + \underbrace{O(\rho \mathrm{Lip}(P_{\mu}^{n} f )) }_{O(\rho e^{O_{\mu}(n)}) \mathrm{Lip}(f)} \end{equation}\] Here the main term term before the big-$O$ is \[\begin{equation} \int_{}^{} (P_{\mu}^{n} f )(x) \varphi(x)\, \mathrm{d}\mathrm{Haar}_{X}(x) \leq \underbrace{\|\varphi\|_{L^{\infty}}}_{\rho^{- \varepsilon}} \underbrace{\| P_{\mu}^{n} f \|_{L^{2}}}_{e^{-sn} \text{ by spectral gap}} . \end{equation}\]

To conclude, take $\varepsilon$ to be very small. Then $n =$ small multiple of $|\log \rho|$. Then $| \mu^{n} * \nu| \leq \rho^{*}$

Step 2: Bootstrapping the dimension

We have $\mu^{k} * \delta_{x}$, we know that \[\begin{equation} \mu^{k} * \delta_{x}(B_{l}) \leq e^{\kappa}. \end{equation}\] We want to show that for some $n = n(\rho)$, $\varepsilon > 0$, \[\begin{equation} \mu^{n + k} * \delta_{x}( B_{\rho^{1/2}}) \leq \rho^{\frac{k + \varepsilon}{2}} \end{equation}\]

Bourgain’s projection theorem

Consider $A \subseteq \mathbb{R}^{2}$ to be any subset. Let $\pi_{\theta}:\mathbb{R}^{2} \xrightarrow[ ]{ } L_{\theta}$ be the projection map to a line at origin with angle $\theta$ (say from the $x$-axis).

Mastrand theorem

For Lebesgue-measure 1, with respect to any $\theta \in \mathbb{R}/ \mathbb{Z}$, one has $\dim \pi_{\theta} A = \min \{1, \dim A\}$.

Bourgain: Take $\sigma$ probability measure which is Frostman. That is $\sigma([-r,r]+s) \leq C r^{\kappa}$ Then for $\sigma$-almost every $\theta$, one has \[\begin{equation} \dim ( \pi_{{\theta}} A ) \geq \tfrac{1}{2} \dim A. \end{equation}\]

\[\begin{equation} A' \leq \pi_{\theta_{1}'} (A') \times \pi_{\theta_{2}'}(A') \end{equation}\]

So \[\begin{equation} \dim A' \leq \dim\pi_{\theta_{1}'} (A') + \dim \pi_{\theta_{2}'}(A') \end{equation}\]

Let $\eta \in (0,1/2),$ $A \subseteq B^{\mathbb{R}^{2}}$ with $\dim A \in (\eta,2-\eta)$. Then for $\sigma$-almost $\theta$, \[\begin{equation} \dim (\pi_{\theta} A) \geq \tfrac{1}{2} \dim(A) + \varepsilon_{0}. \end{equation}\]

So the dimension gain is upto $\varepsilon_{0}$. In fact, Bourgain’s theorem is established at “macroscopic” scales.

Bourgain 2nd version: Let $\rho \in (0,1)$. Let $\kappa, \eta \in(0,1)$, let $C>1$, Assume $\sigma$ satisfies \[\begin{equation} \forall r \geq \rho, \sigma([-r,r] + s) \leq C r^{\kappa}, \end{equation}\]

Let $\alpha \in [\eta, 2- \eta], A \subseteq B^{\mathbb{R}^{2}}_{1}$ such that: \[\begin{equation} \forall x \in \mathbb{R}^{2}, \forall r \geq \rho \tfrac{N_{\rho} ( A \cap B_{r}(x))}{N_{\rho}(A)} \leq C r^{\alpha}. \end{equation}\] Here $N_{\rho}$ is the smallest number of $\rho$-balls needed to cover $A$.

If $\rho, \varepsilon_{0}$ small enough depending on $C,\kappa,\eta$ then with $\sigma$-prob is $\geq 1 - \rho^{\Sigma_{0}}$ one has: \[\begin{equation} N_{\rho} ( \pi_{\theta}(A)) \geq \rho^{ - \alpha/2 - \varepsilon_{0}}. \end{equation}\]

We can replace $A$ by a probablity measure $\nu$ on $B_{1}^{\mathbb{R}^{2}}$ such that \[\begin{equation} \nu(B_{r}(x)) \leq C r^{\alpha} , \forall r \geq \rho, x \in X. \end{equation}\]

In other words, we have the assumption that \[\begin{equation} \nu(B_{r}(x)) \leq C \mathrm{vol}_{\mathbb{R}^{2}}(B_{r}(x))^{\alpha/2} , \forall r \geq \rho, x \in X. \end{equation}\]

The conclusion is then upto ignoring a small subset of $\nu$ \[\begin{equation} \underbrace{ \pi_{\theta} ( \nu) (B_{\rho}^{L_{\theta}}(x)) }_{\pi_{\theta}^{-1} B_{\rho}(x)}\leq \underbrace{\rho^{\alpha/2 + \varepsilon}}_{= \mathrm{vol}_{B_{1}^{\mathbb{R}^{2}}(\pi^{-1}_{\theta} B_{\rho}(x)) }}, \end{equation}\]

This is a gain in dimension from $\alpha/2$ to $\alpha/2 + \varepsilon_{0}$.

Back to step 3

Now back to $X$, $\mu$-radom walk on $X$. We take $\nu = \mu^{k} * \delta_{x}$. Note $\nu(B_{r}(u)) \leq C r^{\kappa}$ thanks to Step 1.

Let $n \geq 1$, \[\begin{equation} \mu^{n} * v (B_{\rho}(x)) = \int_{\mathrm{SL}_{2}(\mathbb{R})}^{} \underbrace{\nu(g^{-1} B_{\rho}(x))}_{\substack{\text{ looks like box } \\ \mathrm{Ad}(g^{-1})B_{\rho}^{\mathrm{sl}_{2}(\mathbb{R})} \text{ in exp chart}} } \, \mathrm{d}\mu^{n}(g) \end{equation}\]

Now, the thing to do is prove an abstract version of Bourgain’s theorem to random boxes of the form \[\begin{equation} \underbrace{R_{\theta}}_{\text{ rotation } \in \mathrm{SO}(3)}( [0,1] e_{1} + [0,\rho^{1/2}] e_{2} + [0,\rho] e_{3}). \end{equation}\] And then, we have to use this three dimensional version to get what we need for $\mathrm{sl}_{2}(\mathbb{R})$.

The abstract dimensional gain theorem we need in $\mathbb{R}^{3}$ is the following. Consider the cuboid in the picture below.

Consider $\nu$ probability measure on $B_{1}^{\mathbb{R}^{3}}$ with the following property: \[ \forall r \geq \rho, x \in \mathbb{R}^{3}, \nu(B_{r} x ) \leq \mathrm{vol}(B_{r}(x))^{\alpha} \text{ where } \alpha \in [\eta,1-\eta].\] Take $\sigma$ to be a probability measure on $\mathrm{SO}(3)$ such that \[\begin{equation} \forall P \subseteq \mathbb{R}^{3}, 2-\text{planes}, \forall L \subseteq \mathbb{R}^{3} \text{ lines } \end{equation}\] one has \[\begin{equation} \sigma\{\theta \in \mathrm{SO}(3) \mid d_{\angle}( R_{\theta}(\mathbb{R} e_{1}, P)) \leq r \} \leq C r^{\kappa}, \end{equation}\] and \[\begin{equation} \sigma\{\theta \in \mathrm{SO}(3) \mid d_{\angle}( R_{\theta}(\mathbb{R} e_{1} \oplus \mathbb{R} e_{2}, L)) \leq r \} \leq C r^{\kappa}. \end{equation}\]

Then, provided $\rho, \varepsilon$ are small enough depending on $(C,\kappa,\eta)$, for most $\theta$, all $x$ \[\begin{equation} \nu( R_{\theta}( \text{ box } + x)) \leq \mathrm{vol}( \text{ box })^{\alpha + \varepsilon}. \end{equation}\]